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let 's see if we can divide 30.24 divided by 0.42 . and try pausing the video and solving it on your own before i work through it . so there is a couple of ways you can think about it . we could just write it as 30.24 divided by 0.42 . but what do you do now ? well the important realization is , is when you 're doing a division problem like this , you will get the same answer as long as you multiply or divide both numbers by the same thing . and to understand that , rewrite this division as 30.42 over 0.42 . we could write it really as a fraction . and we know that when we have a fraction like this we 're not changing the value of the fraction if we multiple the numerator and the denominator by the same quantity . and so what could we multiply this denominator by to make it a whole number ? well we can multiply it by 10 and then another 10 . so we can multiply it by a 100 . so lets do that . if we multiply the denominator by 100 in order to not change the value of this , we also need to multiply the numerator by 100 . we are essentially multiplying by 100 over 100 , which is just 1 . so we 're not changing the value of this fraction . or , you could view this , this division problem . so this is going to be 30.42 times 100 . move the decimal two places to the right , gets you 3,042 . the decimal is now there if you care about it . and , 0.42 times 100 . once again move the decimal one , two places to the right , it is now 42 . so this is going to be the exact same thing as 3,042 divided by 42 . so once again we can move the decimal here , two to the right . and if we move that two to the right , then we can move this two to the right . or we need to move this two to the right . and so this is where , now the decimal place is . you could view this as 3,024 . let me clear that 3024 divided by 42 . let me clear that . and we know how to tackle that already , but lets do it step by step . how many times does 42 go into 3 ? well it does not go at all , so we can move on to 30 . how many times does 42 go into 30 ? well it does not go into 30 so we can move on to 302 . how many times does 42 go into 302 ? and like always this is a bit of an art when your dividing by a two-digit or a multi-digit number , i should say . so lets think about it a little bit . so this is roughly 40 . this is roughly 300 . so how many times does 40 go into 300 ? well how many times does 4 go into 30 ? well , it looks like it 's about seven times , so i 'm going to try out a 7 , see if it works out . 7 times 2 is 14 . 7 times 4 is 28 . plus 1 is 29 . and now i can subtract . do a little bit of re-grouping here . so lets see , if i regroup -- i take a 100 from the 300 . that becomes a 200 . then our zero tens , now i have 10 tens , but i 'm going to need one of those 10 tens , so that 's going to be 9 tens . and i 'm going to give it over here . so this is going to be a 12 . 12 minus 4 is 8 . 9 minus 9 is 0 . 2 minus 2 is 0 . so what i got left over is less than 42 , so i know that 7 is the right number . i want to go as many times as possible into 302 without going over . so now lets bring down the next digit . lets bring down this 4 over here . how many times does 42 go into 84 ? well that jumps out at you , or hopefully it jumps -- it goes two times . 2 times 2 is 4 . 2 times 4 is 8 . you subtract , and we have no remainder . so 3,042 divided by 42 is the same thing as 30.42 divided by 0.42 . and it 's going to be equal to 72 . actually , i did n't have to copy and paste that , i 'll just write this . this is equal to 72 . just like that .
and we know that when we have a fraction like this we 're not changing the value of the fraction if we multiple the numerator and the denominator by the same quantity . and so what could we multiply this denominator by to make it a whole number ? well we can multiply it by 10 and then another 10 . so we can multiply it by a 100 .
what happens if you have to multiply one number by 100 and you only need to multiply the second number by 10 ?
let 's see if we can divide 30.24 divided by 0.42 . and try pausing the video and solving it on your own before i work through it . so there is a couple of ways you can think about it . we could just write it as 30.24 divided by 0.42 . but what do you do now ? well the important realization is , is when you 're doing a division problem like this , you will get the same answer as long as you multiply or divide both numbers by the same thing . and to understand that , rewrite this division as 30.42 over 0.42 . we could write it really as a fraction . and we know that when we have a fraction like this we 're not changing the value of the fraction if we multiple the numerator and the denominator by the same quantity . and so what could we multiply this denominator by to make it a whole number ? well we can multiply it by 10 and then another 10 . so we can multiply it by a 100 . so lets do that . if we multiply the denominator by 100 in order to not change the value of this , we also need to multiply the numerator by 100 . we are essentially multiplying by 100 over 100 , which is just 1 . so we 're not changing the value of this fraction . or , you could view this , this division problem . so this is going to be 30.42 times 100 . move the decimal two places to the right , gets you 3,042 . the decimal is now there if you care about it . and , 0.42 times 100 . once again move the decimal one , two places to the right , it is now 42 . so this is going to be the exact same thing as 3,042 divided by 42 . so once again we can move the decimal here , two to the right . and if we move that two to the right , then we can move this two to the right . or we need to move this two to the right . and so this is where , now the decimal place is . you could view this as 3,024 . let me clear that 3024 divided by 42 . let me clear that . and we know how to tackle that already , but lets do it step by step . how many times does 42 go into 3 ? well it does not go at all , so we can move on to 30 . how many times does 42 go into 30 ? well it does not go into 30 so we can move on to 302 . how many times does 42 go into 302 ? and like always this is a bit of an art when your dividing by a two-digit or a multi-digit number , i should say . so lets think about it a little bit . so this is roughly 40 . this is roughly 300 . so how many times does 40 go into 300 ? well how many times does 4 go into 30 ? well , it looks like it 's about seven times , so i 'm going to try out a 7 , see if it works out . 7 times 2 is 14 . 7 times 4 is 28 . plus 1 is 29 . and now i can subtract . do a little bit of re-grouping here . so lets see , if i regroup -- i take a 100 from the 300 . that becomes a 200 . then our zero tens , now i have 10 tens , but i 'm going to need one of those 10 tens , so that 's going to be 9 tens . and i 'm going to give it over here . so this is going to be a 12 . 12 minus 4 is 8 . 9 minus 9 is 0 . 2 minus 2 is 0 . so what i got left over is less than 42 , so i know that 7 is the right number . i want to go as many times as possible into 302 without going over . so now lets bring down the next digit . lets bring down this 4 over here . how many times does 42 go into 84 ? well that jumps out at you , or hopefully it jumps -- it goes two times . 2 times 2 is 4 . 2 times 4 is 8 . you subtract , and we have no remainder . so 3,042 divided by 42 is the same thing as 30.42 divided by 0.42 . and it 's going to be equal to 72 . actually , i did n't have to copy and paste that , i 'll just write this . this is equal to 72 . just like that .
7 times 4 is 28 . plus 1 is 29 . and now i can subtract .
when u divide decimals is it the same thing as dividing over 1 ?
let 's see if we can divide 30.24 divided by 0.42 . and try pausing the video and solving it on your own before i work through it . so there is a couple of ways you can think about it . we could just write it as 30.24 divided by 0.42 . but what do you do now ? well the important realization is , is when you 're doing a division problem like this , you will get the same answer as long as you multiply or divide both numbers by the same thing . and to understand that , rewrite this division as 30.42 over 0.42 . we could write it really as a fraction . and we know that when we have a fraction like this we 're not changing the value of the fraction if we multiple the numerator and the denominator by the same quantity . and so what could we multiply this denominator by to make it a whole number ? well we can multiply it by 10 and then another 10 . so we can multiply it by a 100 . so lets do that . if we multiply the denominator by 100 in order to not change the value of this , we also need to multiply the numerator by 100 . we are essentially multiplying by 100 over 100 , which is just 1 . so we 're not changing the value of this fraction . or , you could view this , this division problem . so this is going to be 30.42 times 100 . move the decimal two places to the right , gets you 3,042 . the decimal is now there if you care about it . and , 0.42 times 100 . once again move the decimal one , two places to the right , it is now 42 . so this is going to be the exact same thing as 3,042 divided by 42 . so once again we can move the decimal here , two to the right . and if we move that two to the right , then we can move this two to the right . or we need to move this two to the right . and so this is where , now the decimal place is . you could view this as 3,024 . let me clear that 3024 divided by 42 . let me clear that . and we know how to tackle that already , but lets do it step by step . how many times does 42 go into 3 ? well it does not go at all , so we can move on to 30 . how many times does 42 go into 30 ? well it does not go into 30 so we can move on to 302 . how many times does 42 go into 302 ? and like always this is a bit of an art when your dividing by a two-digit or a multi-digit number , i should say . so lets think about it a little bit . so this is roughly 40 . this is roughly 300 . so how many times does 40 go into 300 ? well how many times does 4 go into 30 ? well , it looks like it 's about seven times , so i 'm going to try out a 7 , see if it works out . 7 times 2 is 14 . 7 times 4 is 28 . plus 1 is 29 . and now i can subtract . do a little bit of re-grouping here . so lets see , if i regroup -- i take a 100 from the 300 . that becomes a 200 . then our zero tens , now i have 10 tens , but i 'm going to need one of those 10 tens , so that 's going to be 9 tens . and i 'm going to give it over here . so this is going to be a 12 . 12 minus 4 is 8 . 9 minus 9 is 0 . 2 minus 2 is 0 . so what i got left over is less than 42 , so i know that 7 is the right number . i want to go as many times as possible into 302 without going over . so now lets bring down the next digit . lets bring down this 4 over here . how many times does 42 go into 84 ? well that jumps out at you , or hopefully it jumps -- it goes two times . 2 times 2 is 4 . 2 times 4 is 8 . you subtract , and we have no remainder . so 3,042 divided by 42 is the same thing as 30.42 divided by 0.42 . and it 's going to be equal to 72 . actually , i did n't have to copy and paste that , i 'll just write this . this is equal to 72 . just like that .
you subtract , and we have no remainder . so 3,042 divided by 42 is the same thing as 30.42 divided by 0.42 . and it 's going to be equal to 72 .
why did you put a 1 on top of the 42 and add it ?
let 's see if we can divide 30.24 divided by 0.42 . and try pausing the video and solving it on your own before i work through it . so there is a couple of ways you can think about it . we could just write it as 30.24 divided by 0.42 . but what do you do now ? well the important realization is , is when you 're doing a division problem like this , you will get the same answer as long as you multiply or divide both numbers by the same thing . and to understand that , rewrite this division as 30.42 over 0.42 . we could write it really as a fraction . and we know that when we have a fraction like this we 're not changing the value of the fraction if we multiple the numerator and the denominator by the same quantity . and so what could we multiply this denominator by to make it a whole number ? well we can multiply it by 10 and then another 10 . so we can multiply it by a 100 . so lets do that . if we multiply the denominator by 100 in order to not change the value of this , we also need to multiply the numerator by 100 . we are essentially multiplying by 100 over 100 , which is just 1 . so we 're not changing the value of this fraction . or , you could view this , this division problem . so this is going to be 30.42 times 100 . move the decimal two places to the right , gets you 3,042 . the decimal is now there if you care about it . and , 0.42 times 100 . once again move the decimal one , two places to the right , it is now 42 . so this is going to be the exact same thing as 3,042 divided by 42 . so once again we can move the decimal here , two to the right . and if we move that two to the right , then we can move this two to the right . or we need to move this two to the right . and so this is where , now the decimal place is . you could view this as 3,024 . let me clear that 3024 divided by 42 . let me clear that . and we know how to tackle that already , but lets do it step by step . how many times does 42 go into 3 ? well it does not go at all , so we can move on to 30 . how many times does 42 go into 30 ? well it does not go into 30 so we can move on to 302 . how many times does 42 go into 302 ? and like always this is a bit of an art when your dividing by a two-digit or a multi-digit number , i should say . so lets think about it a little bit . so this is roughly 40 . this is roughly 300 . so how many times does 40 go into 300 ? well how many times does 4 go into 30 ? well , it looks like it 's about seven times , so i 'm going to try out a 7 , see if it works out . 7 times 2 is 14 . 7 times 4 is 28 . plus 1 is 29 . and now i can subtract . do a little bit of re-grouping here . so lets see , if i regroup -- i take a 100 from the 300 . that becomes a 200 . then our zero tens , now i have 10 tens , but i 'm going to need one of those 10 tens , so that 's going to be 9 tens . and i 'm going to give it over here . so this is going to be a 12 . 12 minus 4 is 8 . 9 minus 9 is 0 . 2 minus 2 is 0 . so what i got left over is less than 42 , so i know that 7 is the right number . i want to go as many times as possible into 302 without going over . so now lets bring down the next digit . lets bring down this 4 over here . how many times does 42 go into 84 ? well that jumps out at you , or hopefully it jumps -- it goes two times . 2 times 2 is 4 . 2 times 4 is 8 . you subtract , and we have no remainder . so 3,042 divided by 42 is the same thing as 30.42 divided by 0.42 . and it 's going to be equal to 72 . actually , i did n't have to copy and paste that , i 'll just write this . this is equal to 72 . just like that .
so there is a couple of ways you can think about it . we could just write it as 30.24 divided by 0.42 . but what do you do now ?
did you mean to put 30.42 instead of 30.24 ?
let 's see if we can divide 30.24 divided by 0.42 . and try pausing the video and solving it on your own before i work through it . so there is a couple of ways you can think about it . we could just write it as 30.24 divided by 0.42 . but what do you do now ? well the important realization is , is when you 're doing a division problem like this , you will get the same answer as long as you multiply or divide both numbers by the same thing . and to understand that , rewrite this division as 30.42 over 0.42 . we could write it really as a fraction . and we know that when we have a fraction like this we 're not changing the value of the fraction if we multiple the numerator and the denominator by the same quantity . and so what could we multiply this denominator by to make it a whole number ? well we can multiply it by 10 and then another 10 . so we can multiply it by a 100 . so lets do that . if we multiply the denominator by 100 in order to not change the value of this , we also need to multiply the numerator by 100 . we are essentially multiplying by 100 over 100 , which is just 1 . so we 're not changing the value of this fraction . or , you could view this , this division problem . so this is going to be 30.42 times 100 . move the decimal two places to the right , gets you 3,042 . the decimal is now there if you care about it . and , 0.42 times 100 . once again move the decimal one , two places to the right , it is now 42 . so this is going to be the exact same thing as 3,042 divided by 42 . so once again we can move the decimal here , two to the right . and if we move that two to the right , then we can move this two to the right . or we need to move this two to the right . and so this is where , now the decimal place is . you could view this as 3,024 . let me clear that 3024 divided by 42 . let me clear that . and we know how to tackle that already , but lets do it step by step . how many times does 42 go into 3 ? well it does not go at all , so we can move on to 30 . how many times does 42 go into 30 ? well it does not go into 30 so we can move on to 302 . how many times does 42 go into 302 ? and like always this is a bit of an art when your dividing by a two-digit or a multi-digit number , i should say . so lets think about it a little bit . so this is roughly 40 . this is roughly 300 . so how many times does 40 go into 300 ? well how many times does 4 go into 30 ? well , it looks like it 's about seven times , so i 'm going to try out a 7 , see if it works out . 7 times 2 is 14 . 7 times 4 is 28 . plus 1 is 29 . and now i can subtract . do a little bit of re-grouping here . so lets see , if i regroup -- i take a 100 from the 300 . that becomes a 200 . then our zero tens , now i have 10 tens , but i 'm going to need one of those 10 tens , so that 's going to be 9 tens . and i 'm going to give it over here . so this is going to be a 12 . 12 minus 4 is 8 . 9 minus 9 is 0 . 2 minus 2 is 0 . so what i got left over is less than 42 , so i know that 7 is the right number . i want to go as many times as possible into 302 without going over . so now lets bring down the next digit . lets bring down this 4 over here . how many times does 42 go into 84 ? well that jumps out at you , or hopefully it jumps -- it goes two times . 2 times 2 is 4 . 2 times 4 is 8 . you subtract , and we have no remainder . so 3,042 divided by 42 is the same thing as 30.42 divided by 0.42 . and it 's going to be equal to 72 . actually , i did n't have to copy and paste that , i 'll just write this . this is equal to 72 . just like that .
and so this is where , now the decimal place is . you could view this as 3,024 . let me clear that 3024 divided by 42 .
doing 302 - 294 seemed extremely confusing for such a simple subtraction problem , what is that method called ?
let 's see if we can divide 30.24 divided by 0.42 . and try pausing the video and solving it on your own before i work through it . so there is a couple of ways you can think about it . we could just write it as 30.24 divided by 0.42 . but what do you do now ? well the important realization is , is when you 're doing a division problem like this , you will get the same answer as long as you multiply or divide both numbers by the same thing . and to understand that , rewrite this division as 30.42 over 0.42 . we could write it really as a fraction . and we know that when we have a fraction like this we 're not changing the value of the fraction if we multiple the numerator and the denominator by the same quantity . and so what could we multiply this denominator by to make it a whole number ? well we can multiply it by 10 and then another 10 . so we can multiply it by a 100 . so lets do that . if we multiply the denominator by 100 in order to not change the value of this , we also need to multiply the numerator by 100 . we are essentially multiplying by 100 over 100 , which is just 1 . so we 're not changing the value of this fraction . or , you could view this , this division problem . so this is going to be 30.42 times 100 . move the decimal two places to the right , gets you 3,042 . the decimal is now there if you care about it . and , 0.42 times 100 . once again move the decimal one , two places to the right , it is now 42 . so this is going to be the exact same thing as 3,042 divided by 42 . so once again we can move the decimal here , two to the right . and if we move that two to the right , then we can move this two to the right . or we need to move this two to the right . and so this is where , now the decimal place is . you could view this as 3,024 . let me clear that 3024 divided by 42 . let me clear that . and we know how to tackle that already , but lets do it step by step . how many times does 42 go into 3 ? well it does not go at all , so we can move on to 30 . how many times does 42 go into 30 ? well it does not go into 30 so we can move on to 302 . how many times does 42 go into 302 ? and like always this is a bit of an art when your dividing by a two-digit or a multi-digit number , i should say . so lets think about it a little bit . so this is roughly 40 . this is roughly 300 . so how many times does 40 go into 300 ? well how many times does 4 go into 30 ? well , it looks like it 's about seven times , so i 'm going to try out a 7 , see if it works out . 7 times 2 is 14 . 7 times 4 is 28 . plus 1 is 29 . and now i can subtract . do a little bit of re-grouping here . so lets see , if i regroup -- i take a 100 from the 300 . that becomes a 200 . then our zero tens , now i have 10 tens , but i 'm going to need one of those 10 tens , so that 's going to be 9 tens . and i 'm going to give it over here . so this is going to be a 12 . 12 minus 4 is 8 . 9 minus 9 is 0 . 2 minus 2 is 0 . so what i got left over is less than 42 , so i know that 7 is the right number . i want to go as many times as possible into 302 without going over . so now lets bring down the next digit . lets bring down this 4 over here . how many times does 42 go into 84 ? well that jumps out at you , or hopefully it jumps -- it goes two times . 2 times 2 is 4 . 2 times 4 is 8 . you subtract , and we have no remainder . so 3,042 divided by 42 is the same thing as 30.42 divided by 0.42 . and it 's going to be equal to 72 . actually , i did n't have to copy and paste that , i 'll just write this . this is equal to 72 . just like that .
let 's see if we can divide 30.24 divided by 0.42 . and try pausing the video and solving it on your own before i work through it .
why did sal switch the decimals ?
let 's see if we can divide 30.24 divided by 0.42 . and try pausing the video and solving it on your own before i work through it . so there is a couple of ways you can think about it . we could just write it as 30.24 divided by 0.42 . but what do you do now ? well the important realization is , is when you 're doing a division problem like this , you will get the same answer as long as you multiply or divide both numbers by the same thing . and to understand that , rewrite this division as 30.42 over 0.42 . we could write it really as a fraction . and we know that when we have a fraction like this we 're not changing the value of the fraction if we multiple the numerator and the denominator by the same quantity . and so what could we multiply this denominator by to make it a whole number ? well we can multiply it by 10 and then another 10 . so we can multiply it by a 100 . so lets do that . if we multiply the denominator by 100 in order to not change the value of this , we also need to multiply the numerator by 100 . we are essentially multiplying by 100 over 100 , which is just 1 . so we 're not changing the value of this fraction . or , you could view this , this division problem . so this is going to be 30.42 times 100 . move the decimal two places to the right , gets you 3,042 . the decimal is now there if you care about it . and , 0.42 times 100 . once again move the decimal one , two places to the right , it is now 42 . so this is going to be the exact same thing as 3,042 divided by 42 . so once again we can move the decimal here , two to the right . and if we move that two to the right , then we can move this two to the right . or we need to move this two to the right . and so this is where , now the decimal place is . you could view this as 3,024 . let me clear that 3024 divided by 42 . let me clear that . and we know how to tackle that already , but lets do it step by step . how many times does 42 go into 3 ? well it does not go at all , so we can move on to 30 . how many times does 42 go into 30 ? well it does not go into 30 so we can move on to 302 . how many times does 42 go into 302 ? and like always this is a bit of an art when your dividing by a two-digit or a multi-digit number , i should say . so lets think about it a little bit . so this is roughly 40 . this is roughly 300 . so how many times does 40 go into 300 ? well how many times does 4 go into 30 ? well , it looks like it 's about seven times , so i 'm going to try out a 7 , see if it works out . 7 times 2 is 14 . 7 times 4 is 28 . plus 1 is 29 . and now i can subtract . do a little bit of re-grouping here . so lets see , if i regroup -- i take a 100 from the 300 . that becomes a 200 . then our zero tens , now i have 10 tens , but i 'm going to need one of those 10 tens , so that 's going to be 9 tens . and i 'm going to give it over here . so this is going to be a 12 . 12 minus 4 is 8 . 9 minus 9 is 0 . 2 minus 2 is 0 . so what i got left over is less than 42 , so i know that 7 is the right number . i want to go as many times as possible into 302 without going over . so now lets bring down the next digit . lets bring down this 4 over here . how many times does 42 go into 84 ? well that jumps out at you , or hopefully it jumps -- it goes two times . 2 times 2 is 4 . 2 times 4 is 8 . you subtract , and we have no remainder . so 3,042 divided by 42 is the same thing as 30.42 divided by 0.42 . and it 's going to be equal to 72 . actually , i did n't have to copy and paste that , i 'll just write this . this is equal to 72 . just like that .
actually , i did n't have to copy and paste that , i 'll just write this . this is equal to 72 . just like that .
would n't you have to divide 72 by 10000 ?
let 's see if we can divide 30.24 divided by 0.42 . and try pausing the video and solving it on your own before i work through it . so there is a couple of ways you can think about it . we could just write it as 30.24 divided by 0.42 . but what do you do now ? well the important realization is , is when you 're doing a division problem like this , you will get the same answer as long as you multiply or divide both numbers by the same thing . and to understand that , rewrite this division as 30.42 over 0.42 . we could write it really as a fraction . and we know that when we have a fraction like this we 're not changing the value of the fraction if we multiple the numerator and the denominator by the same quantity . and so what could we multiply this denominator by to make it a whole number ? well we can multiply it by 10 and then another 10 . so we can multiply it by a 100 . so lets do that . if we multiply the denominator by 100 in order to not change the value of this , we also need to multiply the numerator by 100 . we are essentially multiplying by 100 over 100 , which is just 1 . so we 're not changing the value of this fraction . or , you could view this , this division problem . so this is going to be 30.42 times 100 . move the decimal two places to the right , gets you 3,042 . the decimal is now there if you care about it . and , 0.42 times 100 . once again move the decimal one , two places to the right , it is now 42 . so this is going to be the exact same thing as 3,042 divided by 42 . so once again we can move the decimal here , two to the right . and if we move that two to the right , then we can move this two to the right . or we need to move this two to the right . and so this is where , now the decimal place is . you could view this as 3,024 . let me clear that 3024 divided by 42 . let me clear that . and we know how to tackle that already , but lets do it step by step . how many times does 42 go into 3 ? well it does not go at all , so we can move on to 30 . how many times does 42 go into 30 ? well it does not go into 30 so we can move on to 302 . how many times does 42 go into 302 ? and like always this is a bit of an art when your dividing by a two-digit or a multi-digit number , i should say . so lets think about it a little bit . so this is roughly 40 . this is roughly 300 . so how many times does 40 go into 300 ? well how many times does 4 go into 30 ? well , it looks like it 's about seven times , so i 'm going to try out a 7 , see if it works out . 7 times 2 is 14 . 7 times 4 is 28 . plus 1 is 29 . and now i can subtract . do a little bit of re-grouping here . so lets see , if i regroup -- i take a 100 from the 300 . that becomes a 200 . then our zero tens , now i have 10 tens , but i 'm going to need one of those 10 tens , so that 's going to be 9 tens . and i 'm going to give it over here . so this is going to be a 12 . 12 minus 4 is 8 . 9 minus 9 is 0 . 2 minus 2 is 0 . so what i got left over is less than 42 , so i know that 7 is the right number . i want to go as many times as possible into 302 without going over . so now lets bring down the next digit . lets bring down this 4 over here . how many times does 42 go into 84 ? well that jumps out at you , or hopefully it jumps -- it goes two times . 2 times 2 is 4 . 2 times 4 is 8 . you subtract , and we have no remainder . so 3,042 divided by 42 is the same thing as 30.42 divided by 0.42 . and it 's going to be equal to 72 . actually , i did n't have to copy and paste that , i 'll just write this . this is equal to 72 . just like that .
how many times does 42 go into 302 ? and like always this is a bit of an art when your dividing by a two-digit or a multi-digit number , i should say . so lets think about it a little bit .
is there a simpler way of dividing decimals ?
let 's see if we can divide 30.24 divided by 0.42 . and try pausing the video and solving it on your own before i work through it . so there is a couple of ways you can think about it . we could just write it as 30.24 divided by 0.42 . but what do you do now ? well the important realization is , is when you 're doing a division problem like this , you will get the same answer as long as you multiply or divide both numbers by the same thing . and to understand that , rewrite this division as 30.42 over 0.42 . we could write it really as a fraction . and we know that when we have a fraction like this we 're not changing the value of the fraction if we multiple the numerator and the denominator by the same quantity . and so what could we multiply this denominator by to make it a whole number ? well we can multiply it by 10 and then another 10 . so we can multiply it by a 100 . so lets do that . if we multiply the denominator by 100 in order to not change the value of this , we also need to multiply the numerator by 100 . we are essentially multiplying by 100 over 100 , which is just 1 . so we 're not changing the value of this fraction . or , you could view this , this division problem . so this is going to be 30.42 times 100 . move the decimal two places to the right , gets you 3,042 . the decimal is now there if you care about it . and , 0.42 times 100 . once again move the decimal one , two places to the right , it is now 42 . so this is going to be the exact same thing as 3,042 divided by 42 . so once again we can move the decimal here , two to the right . and if we move that two to the right , then we can move this two to the right . or we need to move this two to the right . and so this is where , now the decimal place is . you could view this as 3,024 . let me clear that 3024 divided by 42 . let me clear that . and we know how to tackle that already , but lets do it step by step . how many times does 42 go into 3 ? well it does not go at all , so we can move on to 30 . how many times does 42 go into 30 ? well it does not go into 30 so we can move on to 302 . how many times does 42 go into 302 ? and like always this is a bit of an art when your dividing by a two-digit or a multi-digit number , i should say . so lets think about it a little bit . so this is roughly 40 . this is roughly 300 . so how many times does 40 go into 300 ? well how many times does 4 go into 30 ? well , it looks like it 's about seven times , so i 'm going to try out a 7 , see if it works out . 7 times 2 is 14 . 7 times 4 is 28 . plus 1 is 29 . and now i can subtract . do a little bit of re-grouping here . so lets see , if i regroup -- i take a 100 from the 300 . that becomes a 200 . then our zero tens , now i have 10 tens , but i 'm going to need one of those 10 tens , so that 's going to be 9 tens . and i 'm going to give it over here . so this is going to be a 12 . 12 minus 4 is 8 . 9 minus 9 is 0 . 2 minus 2 is 0 . so what i got left over is less than 42 , so i know that 7 is the right number . i want to go as many times as possible into 302 without going over . so now lets bring down the next digit . lets bring down this 4 over here . how many times does 42 go into 84 ? well that jumps out at you , or hopefully it jumps -- it goes two times . 2 times 2 is 4 . 2 times 4 is 8 . you subtract , and we have no remainder . so 3,042 divided by 42 is the same thing as 30.42 divided by 0.42 . and it 's going to be equal to 72 . actually , i did n't have to copy and paste that , i 'll just write this . this is equal to 72 . just like that .
so there is a couple of ways you can think about it . we could just write it as 30.24 divided by 0.42 . but what do you do now ?
6 sal switched 30.24 as 30.42 was that a mistake ?
let 's see if we can divide 30.24 divided by 0.42 . and try pausing the video and solving it on your own before i work through it . so there is a couple of ways you can think about it . we could just write it as 30.24 divided by 0.42 . but what do you do now ? well the important realization is , is when you 're doing a division problem like this , you will get the same answer as long as you multiply or divide both numbers by the same thing . and to understand that , rewrite this division as 30.42 over 0.42 . we could write it really as a fraction . and we know that when we have a fraction like this we 're not changing the value of the fraction if we multiple the numerator and the denominator by the same quantity . and so what could we multiply this denominator by to make it a whole number ? well we can multiply it by 10 and then another 10 . so we can multiply it by a 100 . so lets do that . if we multiply the denominator by 100 in order to not change the value of this , we also need to multiply the numerator by 100 . we are essentially multiplying by 100 over 100 , which is just 1 . so we 're not changing the value of this fraction . or , you could view this , this division problem . so this is going to be 30.42 times 100 . move the decimal two places to the right , gets you 3,042 . the decimal is now there if you care about it . and , 0.42 times 100 . once again move the decimal one , two places to the right , it is now 42 . so this is going to be the exact same thing as 3,042 divided by 42 . so once again we can move the decimal here , two to the right . and if we move that two to the right , then we can move this two to the right . or we need to move this two to the right . and so this is where , now the decimal place is . you could view this as 3,024 . let me clear that 3024 divided by 42 . let me clear that . and we know how to tackle that already , but lets do it step by step . how many times does 42 go into 3 ? well it does not go at all , so we can move on to 30 . how many times does 42 go into 30 ? well it does not go into 30 so we can move on to 302 . how many times does 42 go into 302 ? and like always this is a bit of an art when your dividing by a two-digit or a multi-digit number , i should say . so lets think about it a little bit . so this is roughly 40 . this is roughly 300 . so how many times does 40 go into 300 ? well how many times does 4 go into 30 ? well , it looks like it 's about seven times , so i 'm going to try out a 7 , see if it works out . 7 times 2 is 14 . 7 times 4 is 28 . plus 1 is 29 . and now i can subtract . do a little bit of re-grouping here . so lets see , if i regroup -- i take a 100 from the 300 . that becomes a 200 . then our zero tens , now i have 10 tens , but i 'm going to need one of those 10 tens , so that 's going to be 9 tens . and i 'm going to give it over here . so this is going to be a 12 . 12 minus 4 is 8 . 9 minus 9 is 0 . 2 minus 2 is 0 . so what i got left over is less than 42 , so i know that 7 is the right number . i want to go as many times as possible into 302 without going over . so now lets bring down the next digit . lets bring down this 4 over here . how many times does 42 go into 84 ? well that jumps out at you , or hopefully it jumps -- it goes two times . 2 times 2 is 4 . 2 times 4 is 8 . you subtract , and we have no remainder . so 3,042 divided by 42 is the same thing as 30.42 divided by 0.42 . and it 's going to be equal to 72 . actually , i did n't have to copy and paste that , i 'll just write this . this is equal to 72 . just like that .
well the important realization is , is when you 're doing a division problem like this , you will get the same answer as long as you multiply or divide both numbers by the same thing . and to understand that , rewrite this division as 30.42 over 0.42 . we could write it really as a fraction .
sal writes 30.42 instead of 30.24. ?
let 's see if we can divide 30.24 divided by 0.42 . and try pausing the video and solving it on your own before i work through it . so there is a couple of ways you can think about it . we could just write it as 30.24 divided by 0.42 . but what do you do now ? well the important realization is , is when you 're doing a division problem like this , you will get the same answer as long as you multiply or divide both numbers by the same thing . and to understand that , rewrite this division as 30.42 over 0.42 . we could write it really as a fraction . and we know that when we have a fraction like this we 're not changing the value of the fraction if we multiple the numerator and the denominator by the same quantity . and so what could we multiply this denominator by to make it a whole number ? well we can multiply it by 10 and then another 10 . so we can multiply it by a 100 . so lets do that . if we multiply the denominator by 100 in order to not change the value of this , we also need to multiply the numerator by 100 . we are essentially multiplying by 100 over 100 , which is just 1 . so we 're not changing the value of this fraction . or , you could view this , this division problem . so this is going to be 30.42 times 100 . move the decimal two places to the right , gets you 3,042 . the decimal is now there if you care about it . and , 0.42 times 100 . once again move the decimal one , two places to the right , it is now 42 . so this is going to be the exact same thing as 3,042 divided by 42 . so once again we can move the decimal here , two to the right . and if we move that two to the right , then we can move this two to the right . or we need to move this two to the right . and so this is where , now the decimal place is . you could view this as 3,024 . let me clear that 3024 divided by 42 . let me clear that . and we know how to tackle that already , but lets do it step by step . how many times does 42 go into 3 ? well it does not go at all , so we can move on to 30 . how many times does 42 go into 30 ? well it does not go into 30 so we can move on to 302 . how many times does 42 go into 302 ? and like always this is a bit of an art when your dividing by a two-digit or a multi-digit number , i should say . so lets think about it a little bit . so this is roughly 40 . this is roughly 300 . so how many times does 40 go into 300 ? well how many times does 4 go into 30 ? well , it looks like it 's about seven times , so i 'm going to try out a 7 , see if it works out . 7 times 2 is 14 . 7 times 4 is 28 . plus 1 is 29 . and now i can subtract . do a little bit of re-grouping here . so lets see , if i regroup -- i take a 100 from the 300 . that becomes a 200 . then our zero tens , now i have 10 tens , but i 'm going to need one of those 10 tens , so that 's going to be 9 tens . and i 'm going to give it over here . so this is going to be a 12 . 12 minus 4 is 8 . 9 minus 9 is 0 . 2 minus 2 is 0 . so what i got left over is less than 42 , so i know that 7 is the right number . i want to go as many times as possible into 302 without going over . so now lets bring down the next digit . lets bring down this 4 over here . how many times does 42 go into 84 ? well that jumps out at you , or hopefully it jumps -- it goes two times . 2 times 2 is 4 . 2 times 4 is 8 . you subtract , and we have no remainder . so 3,042 divided by 42 is the same thing as 30.42 divided by 0.42 . and it 's going to be equal to 72 . actually , i did n't have to copy and paste that , i 'll just write this . this is equal to 72 . just like that .
well it does not go into 30 so we can move on to 302 . how many times does 42 go into 302 ? and like always this is a bit of an art when your dividing by a two-digit or a multi-digit number , i should say .
the method i was taught : 3024/42=72 294 how often does 42 go into 302 ?
let 's see if we can divide 30.24 divided by 0.42 . and try pausing the video and solving it on your own before i work through it . so there is a couple of ways you can think about it . we could just write it as 30.24 divided by 0.42 . but what do you do now ? well the important realization is , is when you 're doing a division problem like this , you will get the same answer as long as you multiply or divide both numbers by the same thing . and to understand that , rewrite this division as 30.42 over 0.42 . we could write it really as a fraction . and we know that when we have a fraction like this we 're not changing the value of the fraction if we multiple the numerator and the denominator by the same quantity . and so what could we multiply this denominator by to make it a whole number ? well we can multiply it by 10 and then another 10 . so we can multiply it by a 100 . so lets do that . if we multiply the denominator by 100 in order to not change the value of this , we also need to multiply the numerator by 100 . we are essentially multiplying by 100 over 100 , which is just 1 . so we 're not changing the value of this fraction . or , you could view this , this division problem . so this is going to be 30.42 times 100 . move the decimal two places to the right , gets you 3,042 . the decimal is now there if you care about it . and , 0.42 times 100 . once again move the decimal one , two places to the right , it is now 42 . so this is going to be the exact same thing as 3,042 divided by 42 . so once again we can move the decimal here , two to the right . and if we move that two to the right , then we can move this two to the right . or we need to move this two to the right . and so this is where , now the decimal place is . you could view this as 3,024 . let me clear that 3024 divided by 42 . let me clear that . and we know how to tackle that already , but lets do it step by step . how many times does 42 go into 3 ? well it does not go at all , so we can move on to 30 . how many times does 42 go into 30 ? well it does not go into 30 so we can move on to 302 . how many times does 42 go into 302 ? and like always this is a bit of an art when your dividing by a two-digit or a multi-digit number , i should say . so lets think about it a little bit . so this is roughly 40 . this is roughly 300 . so how many times does 40 go into 300 ? well how many times does 4 go into 30 ? well , it looks like it 's about seven times , so i 'm going to try out a 7 , see if it works out . 7 times 2 is 14 . 7 times 4 is 28 . plus 1 is 29 . and now i can subtract . do a little bit of re-grouping here . so lets see , if i regroup -- i take a 100 from the 300 . that becomes a 200 . then our zero tens , now i have 10 tens , but i 'm going to need one of those 10 tens , so that 's going to be 9 tens . and i 'm going to give it over here . so this is going to be a 12 . 12 minus 4 is 8 . 9 minus 9 is 0 . 2 minus 2 is 0 . so what i got left over is less than 42 , so i know that 7 is the right number . i want to go as many times as possible into 302 without going over . so now lets bring down the next digit . lets bring down this 4 over here . how many times does 42 go into 84 ? well that jumps out at you , or hopefully it jumps -- it goes two times . 2 times 2 is 4 . 2 times 4 is 8 . you subtract , and we have no remainder . so 3,042 divided by 42 is the same thing as 30.42 divided by 0.42 . and it 's going to be equal to 72 . actually , i did n't have to copy and paste that , i 'll just write this . this is equal to 72 . just like that .
so there is a couple of ways you can think about it . we could just write it as 30.24 divided by 0.42 . but what do you do now ?
why did sal say 30.42 instead of 30.24 ?
let 's see if we can divide 30.24 divided by 0.42 . and try pausing the video and solving it on your own before i work through it . so there is a couple of ways you can think about it . we could just write it as 30.24 divided by 0.42 . but what do you do now ? well the important realization is , is when you 're doing a division problem like this , you will get the same answer as long as you multiply or divide both numbers by the same thing . and to understand that , rewrite this division as 30.42 over 0.42 . we could write it really as a fraction . and we know that when we have a fraction like this we 're not changing the value of the fraction if we multiple the numerator and the denominator by the same quantity . and so what could we multiply this denominator by to make it a whole number ? well we can multiply it by 10 and then another 10 . so we can multiply it by a 100 . so lets do that . if we multiply the denominator by 100 in order to not change the value of this , we also need to multiply the numerator by 100 . we are essentially multiplying by 100 over 100 , which is just 1 . so we 're not changing the value of this fraction . or , you could view this , this division problem . so this is going to be 30.42 times 100 . move the decimal two places to the right , gets you 3,042 . the decimal is now there if you care about it . and , 0.42 times 100 . once again move the decimal one , two places to the right , it is now 42 . so this is going to be the exact same thing as 3,042 divided by 42 . so once again we can move the decimal here , two to the right . and if we move that two to the right , then we can move this two to the right . or we need to move this two to the right . and so this is where , now the decimal place is . you could view this as 3,024 . let me clear that 3024 divided by 42 . let me clear that . and we know how to tackle that already , but lets do it step by step . how many times does 42 go into 3 ? well it does not go at all , so we can move on to 30 . how many times does 42 go into 30 ? well it does not go into 30 so we can move on to 302 . how many times does 42 go into 302 ? and like always this is a bit of an art when your dividing by a two-digit or a multi-digit number , i should say . so lets think about it a little bit . so this is roughly 40 . this is roughly 300 . so how many times does 40 go into 300 ? well how many times does 4 go into 30 ? well , it looks like it 's about seven times , so i 'm going to try out a 7 , see if it works out . 7 times 2 is 14 . 7 times 4 is 28 . plus 1 is 29 . and now i can subtract . do a little bit of re-grouping here . so lets see , if i regroup -- i take a 100 from the 300 . that becomes a 200 . then our zero tens , now i have 10 tens , but i 'm going to need one of those 10 tens , so that 's going to be 9 tens . and i 'm going to give it over here . so this is going to be a 12 . 12 minus 4 is 8 . 9 minus 9 is 0 . 2 minus 2 is 0 . so what i got left over is less than 42 , so i know that 7 is the right number . i want to go as many times as possible into 302 without going over . so now lets bring down the next digit . lets bring down this 4 over here . how many times does 42 go into 84 ? well that jumps out at you , or hopefully it jumps -- it goes two times . 2 times 2 is 4 . 2 times 4 is 8 . you subtract , and we have no remainder . so 3,042 divided by 42 is the same thing as 30.42 divided by 0.42 . and it 's going to be equal to 72 . actually , i did n't have to copy and paste that , i 'll just write this . this is equal to 72 . just like that .
so there is a couple of ways you can think about it . we could just write it as 30.24 divided by 0.42 . but what do you do now ?
did sal mean 30.24 instead of 30.42 ?
let 's see if we can divide 30.24 divided by 0.42 . and try pausing the video and solving it on your own before i work through it . so there is a couple of ways you can think about it . we could just write it as 30.24 divided by 0.42 . but what do you do now ? well the important realization is , is when you 're doing a division problem like this , you will get the same answer as long as you multiply or divide both numbers by the same thing . and to understand that , rewrite this division as 30.42 over 0.42 . we could write it really as a fraction . and we know that when we have a fraction like this we 're not changing the value of the fraction if we multiple the numerator and the denominator by the same quantity . and so what could we multiply this denominator by to make it a whole number ? well we can multiply it by 10 and then another 10 . so we can multiply it by a 100 . so lets do that . if we multiply the denominator by 100 in order to not change the value of this , we also need to multiply the numerator by 100 . we are essentially multiplying by 100 over 100 , which is just 1 . so we 're not changing the value of this fraction . or , you could view this , this division problem . so this is going to be 30.42 times 100 . move the decimal two places to the right , gets you 3,042 . the decimal is now there if you care about it . and , 0.42 times 100 . once again move the decimal one , two places to the right , it is now 42 . so this is going to be the exact same thing as 3,042 divided by 42 . so once again we can move the decimal here , two to the right . and if we move that two to the right , then we can move this two to the right . or we need to move this two to the right . and so this is where , now the decimal place is . you could view this as 3,024 . let me clear that 3024 divided by 42 . let me clear that . and we know how to tackle that already , but lets do it step by step . how many times does 42 go into 3 ? well it does not go at all , so we can move on to 30 . how many times does 42 go into 30 ? well it does not go into 30 so we can move on to 302 . how many times does 42 go into 302 ? and like always this is a bit of an art when your dividing by a two-digit or a multi-digit number , i should say . so lets think about it a little bit . so this is roughly 40 . this is roughly 300 . so how many times does 40 go into 300 ? well how many times does 4 go into 30 ? well , it looks like it 's about seven times , so i 'm going to try out a 7 , see if it works out . 7 times 2 is 14 . 7 times 4 is 28 . plus 1 is 29 . and now i can subtract . do a little bit of re-grouping here . so lets see , if i regroup -- i take a 100 from the 300 . that becomes a 200 . then our zero tens , now i have 10 tens , but i 'm going to need one of those 10 tens , so that 's going to be 9 tens . and i 'm going to give it over here . so this is going to be a 12 . 12 minus 4 is 8 . 9 minus 9 is 0 . 2 minus 2 is 0 . so what i got left over is less than 42 , so i know that 7 is the right number . i want to go as many times as possible into 302 without going over . so now lets bring down the next digit . lets bring down this 4 over here . how many times does 42 go into 84 ? well that jumps out at you , or hopefully it jumps -- it goes two times . 2 times 2 is 4 . 2 times 4 is 8 . you subtract , and we have no remainder . so 3,042 divided by 42 is the same thing as 30.42 divided by 0.42 . and it 's going to be equal to 72 . actually , i did n't have to copy and paste that , i 'll just write this . this is equal to 72 . just like that .
let 's see if we can divide 30.24 divided by 0.42 . and try pausing the video and solving it on your own before i work through it .
a ) how much did a pen cost ?
let 's see if we can divide 30.24 divided by 0.42 . and try pausing the video and solving it on your own before i work through it . so there is a couple of ways you can think about it . we could just write it as 30.24 divided by 0.42 . but what do you do now ? well the important realization is , is when you 're doing a division problem like this , you will get the same answer as long as you multiply or divide both numbers by the same thing . and to understand that , rewrite this division as 30.42 over 0.42 . we could write it really as a fraction . and we know that when we have a fraction like this we 're not changing the value of the fraction if we multiple the numerator and the denominator by the same quantity . and so what could we multiply this denominator by to make it a whole number ? well we can multiply it by 10 and then another 10 . so we can multiply it by a 100 . so lets do that . if we multiply the denominator by 100 in order to not change the value of this , we also need to multiply the numerator by 100 . we are essentially multiplying by 100 over 100 , which is just 1 . so we 're not changing the value of this fraction . or , you could view this , this division problem . so this is going to be 30.42 times 100 . move the decimal two places to the right , gets you 3,042 . the decimal is now there if you care about it . and , 0.42 times 100 . once again move the decimal one , two places to the right , it is now 42 . so this is going to be the exact same thing as 3,042 divided by 42 . so once again we can move the decimal here , two to the right . and if we move that two to the right , then we can move this two to the right . or we need to move this two to the right . and so this is where , now the decimal place is . you could view this as 3,024 . let me clear that 3024 divided by 42 . let me clear that . and we know how to tackle that already , but lets do it step by step . how many times does 42 go into 3 ? well it does not go at all , so we can move on to 30 . how many times does 42 go into 30 ? well it does not go into 30 so we can move on to 302 . how many times does 42 go into 302 ? and like always this is a bit of an art when your dividing by a two-digit or a multi-digit number , i should say . so lets think about it a little bit . so this is roughly 40 . this is roughly 300 . so how many times does 40 go into 300 ? well how many times does 4 go into 30 ? well , it looks like it 's about seven times , so i 'm going to try out a 7 , see if it works out . 7 times 2 is 14 . 7 times 4 is 28 . plus 1 is 29 . and now i can subtract . do a little bit of re-grouping here . so lets see , if i regroup -- i take a 100 from the 300 . that becomes a 200 . then our zero tens , now i have 10 tens , but i 'm going to need one of those 10 tens , so that 's going to be 9 tens . and i 'm going to give it over here . so this is going to be a 12 . 12 minus 4 is 8 . 9 minus 9 is 0 . 2 minus 2 is 0 . so what i got left over is less than 42 , so i know that 7 is the right number . i want to go as many times as possible into 302 without going over . so now lets bring down the next digit . lets bring down this 4 over here . how many times does 42 go into 84 ? well that jumps out at you , or hopefully it jumps -- it goes two times . 2 times 2 is 4 . 2 times 4 is 8 . you subtract , and we have no remainder . so 3,042 divided by 42 is the same thing as 30.42 divided by 0.42 . and it 's going to be equal to 72 . actually , i did n't have to copy and paste that , i 'll just write this . this is equal to 72 . just like that .
let 's see if we can divide 30.24 divided by 0.42 . and try pausing the video and solving it on your own before i work through it .
b ) how much did a pencil cost ?
let 's see if we can divide 30.24 divided by 0.42 . and try pausing the video and solving it on your own before i work through it . so there is a couple of ways you can think about it . we could just write it as 30.24 divided by 0.42 . but what do you do now ? well the important realization is , is when you 're doing a division problem like this , you will get the same answer as long as you multiply or divide both numbers by the same thing . and to understand that , rewrite this division as 30.42 over 0.42 . we could write it really as a fraction . and we know that when we have a fraction like this we 're not changing the value of the fraction if we multiple the numerator and the denominator by the same quantity . and so what could we multiply this denominator by to make it a whole number ? well we can multiply it by 10 and then another 10 . so we can multiply it by a 100 . so lets do that . if we multiply the denominator by 100 in order to not change the value of this , we also need to multiply the numerator by 100 . we are essentially multiplying by 100 over 100 , which is just 1 . so we 're not changing the value of this fraction . or , you could view this , this division problem . so this is going to be 30.42 times 100 . move the decimal two places to the right , gets you 3,042 . the decimal is now there if you care about it . and , 0.42 times 100 . once again move the decimal one , two places to the right , it is now 42 . so this is going to be the exact same thing as 3,042 divided by 42 . so once again we can move the decimal here , two to the right . and if we move that two to the right , then we can move this two to the right . or we need to move this two to the right . and so this is where , now the decimal place is . you could view this as 3,024 . let me clear that 3024 divided by 42 . let me clear that . and we know how to tackle that already , but lets do it step by step . how many times does 42 go into 3 ? well it does not go at all , so we can move on to 30 . how many times does 42 go into 30 ? well it does not go into 30 so we can move on to 302 . how many times does 42 go into 302 ? and like always this is a bit of an art when your dividing by a two-digit or a multi-digit number , i should say . so lets think about it a little bit . so this is roughly 40 . this is roughly 300 . so how many times does 40 go into 300 ? well how many times does 4 go into 30 ? well , it looks like it 's about seven times , so i 'm going to try out a 7 , see if it works out . 7 times 2 is 14 . 7 times 4 is 28 . plus 1 is 29 . and now i can subtract . do a little bit of re-grouping here . so lets see , if i regroup -- i take a 100 from the 300 . that becomes a 200 . then our zero tens , now i have 10 tens , but i 'm going to need one of those 10 tens , so that 's going to be 9 tens . and i 'm going to give it over here . so this is going to be a 12 . 12 minus 4 is 8 . 9 minus 9 is 0 . 2 minus 2 is 0 . so what i got left over is less than 42 , so i know that 7 is the right number . i want to go as many times as possible into 302 without going over . so now lets bring down the next digit . lets bring down this 4 over here . how many times does 42 go into 84 ? well that jumps out at you , or hopefully it jumps -- it goes two times . 2 times 2 is 4 . 2 times 4 is 8 . you subtract , and we have no remainder . so 3,042 divided by 42 is the same thing as 30.42 divided by 0.42 . and it 's going to be equal to 72 . actually , i did n't have to copy and paste that , i 'll just write this . this is equal to 72 . just like that .
the decimal is now there if you care about it . and , 0.42 times 100 . once again move the decimal one , two places to the right , it is now 42 .
how do you divide when you have say 0.677 divided by 0.98 how do you move the decimal when you multiply the numbers each by 100 but one number is not whole ?
let 's see if we can divide 30.24 divided by 0.42 . and try pausing the video and solving it on your own before i work through it . so there is a couple of ways you can think about it . we could just write it as 30.24 divided by 0.42 . but what do you do now ? well the important realization is , is when you 're doing a division problem like this , you will get the same answer as long as you multiply or divide both numbers by the same thing . and to understand that , rewrite this division as 30.42 over 0.42 . we could write it really as a fraction . and we know that when we have a fraction like this we 're not changing the value of the fraction if we multiple the numerator and the denominator by the same quantity . and so what could we multiply this denominator by to make it a whole number ? well we can multiply it by 10 and then another 10 . so we can multiply it by a 100 . so lets do that . if we multiply the denominator by 100 in order to not change the value of this , we also need to multiply the numerator by 100 . we are essentially multiplying by 100 over 100 , which is just 1 . so we 're not changing the value of this fraction . or , you could view this , this division problem . so this is going to be 30.42 times 100 . move the decimal two places to the right , gets you 3,042 . the decimal is now there if you care about it . and , 0.42 times 100 . once again move the decimal one , two places to the right , it is now 42 . so this is going to be the exact same thing as 3,042 divided by 42 . so once again we can move the decimal here , two to the right . and if we move that two to the right , then we can move this two to the right . or we need to move this two to the right . and so this is where , now the decimal place is . you could view this as 3,024 . let me clear that 3024 divided by 42 . let me clear that . and we know how to tackle that already , but lets do it step by step . how many times does 42 go into 3 ? well it does not go at all , so we can move on to 30 . how many times does 42 go into 30 ? well it does not go into 30 so we can move on to 302 . how many times does 42 go into 302 ? and like always this is a bit of an art when your dividing by a two-digit or a multi-digit number , i should say . so lets think about it a little bit . so this is roughly 40 . this is roughly 300 . so how many times does 40 go into 300 ? well how many times does 4 go into 30 ? well , it looks like it 's about seven times , so i 'm going to try out a 7 , see if it works out . 7 times 2 is 14 . 7 times 4 is 28 . plus 1 is 29 . and now i can subtract . do a little bit of re-grouping here . so lets see , if i regroup -- i take a 100 from the 300 . that becomes a 200 . then our zero tens , now i have 10 tens , but i 'm going to need one of those 10 tens , so that 's going to be 9 tens . and i 'm going to give it over here . so this is going to be a 12 . 12 minus 4 is 8 . 9 minus 9 is 0 . 2 minus 2 is 0 . so what i got left over is less than 42 , so i know that 7 is the right number . i want to go as many times as possible into 302 without going over . so now lets bring down the next digit . lets bring down this 4 over here . how many times does 42 go into 84 ? well that jumps out at you , or hopefully it jumps -- it goes two times . 2 times 2 is 4 . 2 times 4 is 8 . you subtract , and we have no remainder . so 3,042 divided by 42 is the same thing as 30.42 divided by 0.42 . and it 's going to be equal to 72 . actually , i did n't have to copy and paste that , i 'll just write this . this is equal to 72 . just like that .
so we 're not changing the value of this fraction . or , you could view this , this division problem . so this is going to be 30.42 times 100 .
are there other methods for decimal division ?
let 's see if we can divide 30.24 divided by 0.42 . and try pausing the video and solving it on your own before i work through it . so there is a couple of ways you can think about it . we could just write it as 30.24 divided by 0.42 . but what do you do now ? well the important realization is , is when you 're doing a division problem like this , you will get the same answer as long as you multiply or divide both numbers by the same thing . and to understand that , rewrite this division as 30.42 over 0.42 . we could write it really as a fraction . and we know that when we have a fraction like this we 're not changing the value of the fraction if we multiple the numerator and the denominator by the same quantity . and so what could we multiply this denominator by to make it a whole number ? well we can multiply it by 10 and then another 10 . so we can multiply it by a 100 . so lets do that . if we multiply the denominator by 100 in order to not change the value of this , we also need to multiply the numerator by 100 . we are essentially multiplying by 100 over 100 , which is just 1 . so we 're not changing the value of this fraction . or , you could view this , this division problem . so this is going to be 30.42 times 100 . move the decimal two places to the right , gets you 3,042 . the decimal is now there if you care about it . and , 0.42 times 100 . once again move the decimal one , two places to the right , it is now 42 . so this is going to be the exact same thing as 3,042 divided by 42 . so once again we can move the decimal here , two to the right . and if we move that two to the right , then we can move this two to the right . or we need to move this two to the right . and so this is where , now the decimal place is . you could view this as 3,024 . let me clear that 3024 divided by 42 . let me clear that . and we know how to tackle that already , but lets do it step by step . how many times does 42 go into 3 ? well it does not go at all , so we can move on to 30 . how many times does 42 go into 30 ? well it does not go into 30 so we can move on to 302 . how many times does 42 go into 302 ? and like always this is a bit of an art when your dividing by a two-digit or a multi-digit number , i should say . so lets think about it a little bit . so this is roughly 40 . this is roughly 300 . so how many times does 40 go into 300 ? well how many times does 4 go into 30 ? well , it looks like it 's about seven times , so i 'm going to try out a 7 , see if it works out . 7 times 2 is 14 . 7 times 4 is 28 . plus 1 is 29 . and now i can subtract . do a little bit of re-grouping here . so lets see , if i regroup -- i take a 100 from the 300 . that becomes a 200 . then our zero tens , now i have 10 tens , but i 'm going to need one of those 10 tens , so that 's going to be 9 tens . and i 'm going to give it over here . so this is going to be a 12 . 12 minus 4 is 8 . 9 minus 9 is 0 . 2 minus 2 is 0 . so what i got left over is less than 42 , so i know that 7 is the right number . i want to go as many times as possible into 302 without going over . so now lets bring down the next digit . lets bring down this 4 over here . how many times does 42 go into 84 ? well that jumps out at you , or hopefully it jumps -- it goes two times . 2 times 2 is 4 . 2 times 4 is 8 . you subtract , and we have no remainder . so 3,042 divided by 42 is the same thing as 30.42 divided by 0.42 . and it 's going to be equal to 72 . actually , i did n't have to copy and paste that , i 'll just write this . this is equal to 72 . just like that .
you could view this as 3,024 . let me clear that 3024 divided by 42 . let me clear that .
is 3024 divided by 42 the same as 1512 divided by 21 ?
let 's see if we can divide 30.24 divided by 0.42 . and try pausing the video and solving it on your own before i work through it . so there is a couple of ways you can think about it . we could just write it as 30.24 divided by 0.42 . but what do you do now ? well the important realization is , is when you 're doing a division problem like this , you will get the same answer as long as you multiply or divide both numbers by the same thing . and to understand that , rewrite this division as 30.42 over 0.42 . we could write it really as a fraction . and we know that when we have a fraction like this we 're not changing the value of the fraction if we multiple the numerator and the denominator by the same quantity . and so what could we multiply this denominator by to make it a whole number ? well we can multiply it by 10 and then another 10 . so we can multiply it by a 100 . so lets do that . if we multiply the denominator by 100 in order to not change the value of this , we also need to multiply the numerator by 100 . we are essentially multiplying by 100 over 100 , which is just 1 . so we 're not changing the value of this fraction . or , you could view this , this division problem . so this is going to be 30.42 times 100 . move the decimal two places to the right , gets you 3,042 . the decimal is now there if you care about it . and , 0.42 times 100 . once again move the decimal one , two places to the right , it is now 42 . so this is going to be the exact same thing as 3,042 divided by 42 . so once again we can move the decimal here , two to the right . and if we move that two to the right , then we can move this two to the right . or we need to move this two to the right . and so this is where , now the decimal place is . you could view this as 3,024 . let me clear that 3024 divided by 42 . let me clear that . and we know how to tackle that already , but lets do it step by step . how many times does 42 go into 3 ? well it does not go at all , so we can move on to 30 . how many times does 42 go into 30 ? well it does not go into 30 so we can move on to 302 . how many times does 42 go into 302 ? and like always this is a bit of an art when your dividing by a two-digit or a multi-digit number , i should say . so lets think about it a little bit . so this is roughly 40 . this is roughly 300 . so how many times does 40 go into 300 ? well how many times does 4 go into 30 ? well , it looks like it 's about seven times , so i 'm going to try out a 7 , see if it works out . 7 times 2 is 14 . 7 times 4 is 28 . plus 1 is 29 . and now i can subtract . do a little bit of re-grouping here . so lets see , if i regroup -- i take a 100 from the 300 . that becomes a 200 . then our zero tens , now i have 10 tens , but i 'm going to need one of those 10 tens , so that 's going to be 9 tens . and i 'm going to give it over here . so this is going to be a 12 . 12 minus 4 is 8 . 9 minus 9 is 0 . 2 minus 2 is 0 . so what i got left over is less than 42 , so i know that 7 is the right number . i want to go as many times as possible into 302 without going over . so now lets bring down the next digit . lets bring down this 4 over here . how many times does 42 go into 84 ? well that jumps out at you , or hopefully it jumps -- it goes two times . 2 times 2 is 4 . 2 times 4 is 8 . you subtract , and we have no remainder . so 3,042 divided by 42 is the same thing as 30.42 divided by 0.42 . and it 's going to be equal to 72 . actually , i did n't have to copy and paste that , i 'll just write this . this is equal to 72 . just like that .
so there is a couple of ways you can think about it . we could just write it as 30.24 divided by 0.42 . but what do you do now ?
why did you change the 30.24 to 30.42 ?
let 's see if we can divide 30.24 divided by 0.42 . and try pausing the video and solving it on your own before i work through it . so there is a couple of ways you can think about it . we could just write it as 30.24 divided by 0.42 . but what do you do now ? well the important realization is , is when you 're doing a division problem like this , you will get the same answer as long as you multiply or divide both numbers by the same thing . and to understand that , rewrite this division as 30.42 over 0.42 . we could write it really as a fraction . and we know that when we have a fraction like this we 're not changing the value of the fraction if we multiple the numerator and the denominator by the same quantity . and so what could we multiply this denominator by to make it a whole number ? well we can multiply it by 10 and then another 10 . so we can multiply it by a 100 . so lets do that . if we multiply the denominator by 100 in order to not change the value of this , we also need to multiply the numerator by 100 . we are essentially multiplying by 100 over 100 , which is just 1 . so we 're not changing the value of this fraction . or , you could view this , this division problem . so this is going to be 30.42 times 100 . move the decimal two places to the right , gets you 3,042 . the decimal is now there if you care about it . and , 0.42 times 100 . once again move the decimal one , two places to the right , it is now 42 . so this is going to be the exact same thing as 3,042 divided by 42 . so once again we can move the decimal here , two to the right . and if we move that two to the right , then we can move this two to the right . or we need to move this two to the right . and so this is where , now the decimal place is . you could view this as 3,024 . let me clear that 3024 divided by 42 . let me clear that . and we know how to tackle that already , but lets do it step by step . how many times does 42 go into 3 ? well it does not go at all , so we can move on to 30 . how many times does 42 go into 30 ? well it does not go into 30 so we can move on to 302 . how many times does 42 go into 302 ? and like always this is a bit of an art when your dividing by a two-digit or a multi-digit number , i should say . so lets think about it a little bit . so this is roughly 40 . this is roughly 300 . so how many times does 40 go into 300 ? well how many times does 4 go into 30 ? well , it looks like it 's about seven times , so i 'm going to try out a 7 , see if it works out . 7 times 2 is 14 . 7 times 4 is 28 . plus 1 is 29 . and now i can subtract . do a little bit of re-grouping here . so lets see , if i regroup -- i take a 100 from the 300 . that becomes a 200 . then our zero tens , now i have 10 tens , but i 'm going to need one of those 10 tens , so that 's going to be 9 tens . and i 'm going to give it over here . so this is going to be a 12 . 12 minus 4 is 8 . 9 minus 9 is 0 . 2 minus 2 is 0 . so what i got left over is less than 42 , so i know that 7 is the right number . i want to go as many times as possible into 302 without going over . so now lets bring down the next digit . lets bring down this 4 over here . how many times does 42 go into 84 ? well that jumps out at you , or hopefully it jumps -- it goes two times . 2 times 2 is 4 . 2 times 4 is 8 . you subtract , and we have no remainder . so 3,042 divided by 42 is the same thing as 30.42 divided by 0.42 . and it 's going to be equal to 72 . actually , i did n't have to copy and paste that , i 'll just write this . this is equal to 72 . just like that .
let 's see if we can divide 30.24 divided by 0.42 . and try pausing the video and solving it on your own before i work through it .
why do n't we do that when we divide decimals ?
let 's see if we can divide 30.24 divided by 0.42 . and try pausing the video and solving it on your own before i work through it . so there is a couple of ways you can think about it . we could just write it as 30.24 divided by 0.42 . but what do you do now ? well the important realization is , is when you 're doing a division problem like this , you will get the same answer as long as you multiply or divide both numbers by the same thing . and to understand that , rewrite this division as 30.42 over 0.42 . we could write it really as a fraction . and we know that when we have a fraction like this we 're not changing the value of the fraction if we multiple the numerator and the denominator by the same quantity . and so what could we multiply this denominator by to make it a whole number ? well we can multiply it by 10 and then another 10 . so we can multiply it by a 100 . so lets do that . if we multiply the denominator by 100 in order to not change the value of this , we also need to multiply the numerator by 100 . we are essentially multiplying by 100 over 100 , which is just 1 . so we 're not changing the value of this fraction . or , you could view this , this division problem . so this is going to be 30.42 times 100 . move the decimal two places to the right , gets you 3,042 . the decimal is now there if you care about it . and , 0.42 times 100 . once again move the decimal one , two places to the right , it is now 42 . so this is going to be the exact same thing as 3,042 divided by 42 . so once again we can move the decimal here , two to the right . and if we move that two to the right , then we can move this two to the right . or we need to move this two to the right . and so this is where , now the decimal place is . you could view this as 3,024 . let me clear that 3024 divided by 42 . let me clear that . and we know how to tackle that already , but lets do it step by step . how many times does 42 go into 3 ? well it does not go at all , so we can move on to 30 . how many times does 42 go into 30 ? well it does not go into 30 so we can move on to 302 . how many times does 42 go into 302 ? and like always this is a bit of an art when your dividing by a two-digit or a multi-digit number , i should say . so lets think about it a little bit . so this is roughly 40 . this is roughly 300 . so how many times does 40 go into 300 ? well how many times does 4 go into 30 ? well , it looks like it 's about seven times , so i 'm going to try out a 7 , see if it works out . 7 times 2 is 14 . 7 times 4 is 28 . plus 1 is 29 . and now i can subtract . do a little bit of re-grouping here . so lets see , if i regroup -- i take a 100 from the 300 . that becomes a 200 . then our zero tens , now i have 10 tens , but i 'm going to need one of those 10 tens , so that 's going to be 9 tens . and i 'm going to give it over here . so this is going to be a 12 . 12 minus 4 is 8 . 9 minus 9 is 0 . 2 minus 2 is 0 . so what i got left over is less than 42 , so i know that 7 is the right number . i want to go as many times as possible into 302 without going over . so now lets bring down the next digit . lets bring down this 4 over here . how many times does 42 go into 84 ? well that jumps out at you , or hopefully it jumps -- it goes two times . 2 times 2 is 4 . 2 times 4 is 8 . you subtract , and we have no remainder . so 3,042 divided by 42 is the same thing as 30.42 divided by 0.42 . and it 's going to be equal to 72 . actually , i did n't have to copy and paste that , i 'll just write this . this is equal to 72 . just like that .
so this is going to be 30.42 times 100 . move the decimal two places to the right , gets you 3,042 . the decimal is now there if you care about it . and , 0.42 times 100 .
why did the put decimal in before they knew it needed to be there ?
let 's see if we can divide 30.24 divided by 0.42 . and try pausing the video and solving it on your own before i work through it . so there is a couple of ways you can think about it . we could just write it as 30.24 divided by 0.42 . but what do you do now ? well the important realization is , is when you 're doing a division problem like this , you will get the same answer as long as you multiply or divide both numbers by the same thing . and to understand that , rewrite this division as 30.42 over 0.42 . we could write it really as a fraction . and we know that when we have a fraction like this we 're not changing the value of the fraction if we multiple the numerator and the denominator by the same quantity . and so what could we multiply this denominator by to make it a whole number ? well we can multiply it by 10 and then another 10 . so we can multiply it by a 100 . so lets do that . if we multiply the denominator by 100 in order to not change the value of this , we also need to multiply the numerator by 100 . we are essentially multiplying by 100 over 100 , which is just 1 . so we 're not changing the value of this fraction . or , you could view this , this division problem . so this is going to be 30.42 times 100 . move the decimal two places to the right , gets you 3,042 . the decimal is now there if you care about it . and , 0.42 times 100 . once again move the decimal one , two places to the right , it is now 42 . so this is going to be the exact same thing as 3,042 divided by 42 . so once again we can move the decimal here , two to the right . and if we move that two to the right , then we can move this two to the right . or we need to move this two to the right . and so this is where , now the decimal place is . you could view this as 3,024 . let me clear that 3024 divided by 42 . let me clear that . and we know how to tackle that already , but lets do it step by step . how many times does 42 go into 3 ? well it does not go at all , so we can move on to 30 . how many times does 42 go into 30 ? well it does not go into 30 so we can move on to 302 . how many times does 42 go into 302 ? and like always this is a bit of an art when your dividing by a two-digit or a multi-digit number , i should say . so lets think about it a little bit . so this is roughly 40 . this is roughly 300 . so how many times does 40 go into 300 ? well how many times does 4 go into 30 ? well , it looks like it 's about seven times , so i 'm going to try out a 7 , see if it works out . 7 times 2 is 14 . 7 times 4 is 28 . plus 1 is 29 . and now i can subtract . do a little bit of re-grouping here . so lets see , if i regroup -- i take a 100 from the 300 . that becomes a 200 . then our zero tens , now i have 10 tens , but i 'm going to need one of those 10 tens , so that 's going to be 9 tens . and i 'm going to give it over here . so this is going to be a 12 . 12 minus 4 is 8 . 9 minus 9 is 0 . 2 minus 2 is 0 . so what i got left over is less than 42 , so i know that 7 is the right number . i want to go as many times as possible into 302 without going over . so now lets bring down the next digit . lets bring down this 4 over here . how many times does 42 go into 84 ? well that jumps out at you , or hopefully it jumps -- it goes two times . 2 times 2 is 4 . 2 times 4 is 8 . you subtract , and we have no remainder . so 3,042 divided by 42 is the same thing as 30.42 divided by 0.42 . and it 's going to be equal to 72 . actually , i did n't have to copy and paste that , i 'll just write this . this is equal to 72 . just like that .
so how many times does 40 go into 300 ? well how many times does 4 go into 30 ? well , it looks like it 's about seven times , so i 'm going to try out a 7 , see if it works out .
when you divide decimals , can you make the dividend an integer as well as the divisor ?
there are four seasons each year . rafael gets a haircut two times each season . one at arlo 's barber shop , and the other at awesome snips . which choices represent the number of haircuts rafael will get this year ? and they 've even made a little diagram for us right over here . so they tell us there 's four seasons in the year , spring , summer , autumn , winter . and that each season rafael gets two haircuts . he gets one at arlo 's and one at awesome snips . so in spring this is the one that he gets at arlo 's , this is the one that he gets at awesome snips . and summer , this is the one he gets at arlos ' , this is the one he gets at awesome snips . autumn this is the he gets at arlo 's , this is the one at awesome snips . and winter , this is the one at arlo 's , this is the one at awesome snips . so they ask us , which choices represent the number of haircuts rafael will get this year ? well one way to this about it is : in spring he will get two haircuts then in summer he is going to get another 2 haircuts . in autumn he is going to get another two haircuts . and then in winter he is going to get another two haircuts . so he is going to get two in spring , plus two in summer , plus two in autumn , plus two in winter . which is going to be equal to eight haircuts . so which of these represent eight ? well let 's see , two plus two , that 's not eight . four plus four plus four plus four , that 's not eight , that 's 16 . two plus two is four . four plus four is eight , plus four is 12 plus four is 16 , that 's not eight . four plus four , that is equal to eight . that is the number of haircuts rafael is going to get this year . so we can check that one . two plus two plus two plus two , that 's exactly what we wrote here , and it is eight . so that is equal to the number of haircuts he is going to get this year . and then two plus four , well that 's not equal to eight , that 's six , so that 's not right . so these are the two that represent the number of haircuts rafael will get this year .
there are four seasons each year . rafael gets a haircut two times each season .
would you be able to use a quadratic equation for this sort of problem ?
there are four seasons each year . rafael gets a haircut two times each season . one at arlo 's barber shop , and the other at awesome snips . which choices represent the number of haircuts rafael will get this year ? and they 've even made a little diagram for us right over here . so they tell us there 's four seasons in the year , spring , summer , autumn , winter . and that each season rafael gets two haircuts . he gets one at arlo 's and one at awesome snips . so in spring this is the one that he gets at arlo 's , this is the one that he gets at awesome snips . and summer , this is the one he gets at arlos ' , this is the one he gets at awesome snips . autumn this is the he gets at arlo 's , this is the one at awesome snips . and winter , this is the one at arlo 's , this is the one at awesome snips . so they ask us , which choices represent the number of haircuts rafael will get this year ? well one way to this about it is : in spring he will get two haircuts then in summer he is going to get another 2 haircuts . in autumn he is going to get another two haircuts . and then in winter he is going to get another two haircuts . so he is going to get two in spring , plus two in summer , plus two in autumn , plus two in winter . which is going to be equal to eight haircuts . so which of these represent eight ? well let 's see , two plus two , that 's not eight . four plus four plus four plus four , that 's not eight , that 's 16 . two plus two is four . four plus four is eight , plus four is 12 plus four is 16 , that 's not eight . four plus four , that is equal to eight . that is the number of haircuts rafael is going to get this year . so we can check that one . two plus two plus two plus two , that 's exactly what we wrote here , and it is eight . so that is equal to the number of haircuts he is going to get this year . and then two plus four , well that 's not equal to eight , that 's six , so that 's not right . so these are the two that represent the number of haircuts rafael will get this year .
there are four seasons each year . rafael gets a haircut two times each season .
how are you supposed to know your seasons if you do n't know how to add ?
there are four seasons each year . rafael gets a haircut two times each season . one at arlo 's barber shop , and the other at awesome snips . which choices represent the number of haircuts rafael will get this year ? and they 've even made a little diagram for us right over here . so they tell us there 's four seasons in the year , spring , summer , autumn , winter . and that each season rafael gets two haircuts . he gets one at arlo 's and one at awesome snips . so in spring this is the one that he gets at arlo 's , this is the one that he gets at awesome snips . and summer , this is the one he gets at arlos ' , this is the one he gets at awesome snips . autumn this is the he gets at arlo 's , this is the one at awesome snips . and winter , this is the one at arlo 's , this is the one at awesome snips . so they ask us , which choices represent the number of haircuts rafael will get this year ? well one way to this about it is : in spring he will get two haircuts then in summer he is going to get another 2 haircuts . in autumn he is going to get another two haircuts . and then in winter he is going to get another two haircuts . so he is going to get two in spring , plus two in summer , plus two in autumn , plus two in winter . which is going to be equal to eight haircuts . so which of these represent eight ? well let 's see , two plus two , that 's not eight . four plus four plus four plus four , that 's not eight , that 's 16 . two plus two is four . four plus four is eight , plus four is 12 plus four is 16 , that 's not eight . four plus four , that is equal to eight . that is the number of haircuts rafael is going to get this year . so we can check that one . two plus two plus two plus two , that 's exactly what we wrote here , and it is eight . so that is equal to the number of haircuts he is going to get this year . and then two plus four , well that 's not equal to eight , that 's six , so that 's not right . so these are the two that represent the number of haircuts rafael will get this year .
which choices represent the number of haircuts rafael will get this year ? and they 've even made a little diagram for us right over here . so they tell us there 's four seasons in the year , spring , summer , autumn , winter .
repeated addition is multiplying , right ?
so we were talking in a previous video about hypertension and the damage that it causes on the body . and what i 'd like to do is pick up where we left off and finish that conversation . and we had kind of divided things up into receiving high blood pressures and generating high blood pressures . and receiving high pressure -- really , that means the vessels , right ? that 's where the pressure is going to be seen . and so the vessels can get damaged . and we talked about some of the different ways that that can happen . and in fact , there are the larger arteries and also the smaller arteries . but now , let 's focus our attention on generating high pressures . because this might be less intuitive , but actually making the pressure go so high is not an easy thing to do . and it 's actually a very challenging task for the heart to accomplish . and in trying to figure it out , sometimes the heart can run into troubles . so there was an equation , and that equation was delta p equals q times r. and i thought we had put some words to this equation . so the idea that delta p -- that 's basically saying the arterial pressure minus the venous pressure . so if you have blood and it 's exiting here , and that 's got a certain pressure on it . and as it enters again on the other side into the atrium , it 's got a pressure on the venous side . so we 're saying that the pressure leaving in the arteries minus the pressure returning in the veins is what 's going to help you take a certain flow of blood , q , past the resistance in the vessels . so that 's a way of thinking about it . and specifically -- let me actually draw out some of this resistance . you have , of course , resistance once it gets into the arteries . and so you 've got some , r , resistance there . then you 've got more resistance once you get into the arterioles . in fact , i 'm going to triple underline that one because there 's lots of resistance in the arterioles . and then you 've got resistance in the capillaries , and you 've got resistance in the veins . and then you 've got the blood returning into the venous side . so there 's resistance in the circulatory system . and there 's a flow of blood that you 're actually trying to move around , right ? so anything , anything at all , that increases the resistance or increases the flow is going to force the amount of pressure you have to put in on the arterial side to go up . and now if i draw it out , you 'll see even more clearly why this is a problem . so if you have , let 's say , the left ventricle , and i 'll leave a little space for the valves . and this is the chamber of the heart that 's doing the pumping now . and it 's pumping the blood out into the aorta . this is the aorta . and this is another valve . and this valve is called the aortic valve -- pretty easy to remember name , aortic valve . and it separates the aorta from the left ventricle -- lv for left ventricle . so the left ventricle is basically going to have to try to get blood through this door , through this valve . and the way to do that is to apply a force to that blood and force it through that door . now , the aortic valve has a certain area . and so if you remember , any time you have a force over an area -- i 'll write that over here -- anytime you have a force over an area , you have a pressure . and in this case , that is the arterial pressure . so when i talk about arterial pressure , i 'm actually talking about the force that the left ventricle generates on that aortic valve , that area of aortic valve . so now , think about it . if you have , let 's say , more resistance . let 's say there 's lots and lots of resistance in the vessels , so that all these numbers are slowly going up , up , up . well , that 's going to cause the overall resistance to go up , and now the left ventricle has to put more force . or you could say , what if you have larger flow ? you actually have a larger volume of blood you 're trying to move around . again , the aortic pressure is going to have to go up . so the way that the left ventricle accomplishes this or tries to accomplish this is by basically building more muscle . so this is one strategy . it 'll say , ok , well , if i need to generate more force , why not generate more muscle ? so this becomes very muscular in here . if you actually look at a heart that has had high pressures over the years , oftentimes you have what 's called left ventricular hypertrophy . so on the surface , this sounds great , right ? left ventricular hypertrophy sounds like you 're making the muscle of the heart stronger . and that sounds like a good thing . you know , that 's certainly why i go to the gym . i want my muscles to get bigger . so it 's confusing , then , to hear that that 's not a good thing , and here 's why . so now , imagine you have all this muscle , and you 're feeling really good . but then i tell you that , well , this muscle 's using more oxygen . so it 's using more oxygen . but you notice i did n't draw any new blood vessels yet . so there are no new blood vessels getting oxygen to this muscle . but the muscle is definitely using a lot of energy , and it needs a lot of oxygen . and so what happens is that you have areas where blood is basically able to get to those areas and other areas where blood is not able to get there . and where the blood is unable to get enough oxygen to that heart muscle , you could have a heart attack or a myocardial infarction . so left ventricular hypertrophy is a way of compensating for needing more force . but it comes with a little bit of a risk because now you might have a risk of an mi or myocardial infarction . the other possibility is that the left heart does n't compensate , and you have what 's called left heart failure . and what that basically means is that the left heart says , you know what , i ca n't generate the pressures that are being required to move all that blood around , and so i wo n't . i wo n't generate those pressures . i 'll do the best i can , and maybe i ca n't move , let 's say , six liters of blood around the body , but i can do 5 liters or 5 and 1/2 liters . so it 'll try to do what it can to move as much blood as possible , but it is essentially not compensating completely . and that 's why you sometimes see people with heart failure after having years and years of high blood pressure . their heart simply ca n't keep up . and so you experience heart failure .
because this might be less intuitive , but actually making the pressure go so high is not an easy thing to do . and it 's actually a very challenging task for the heart to accomplish . and in trying to figure it out , sometimes the heart can run into troubles .
what actually causes the heart muscle to grow thicker ?
so we were talking in a previous video about hypertension and the damage that it causes on the body . and what i 'd like to do is pick up where we left off and finish that conversation . and we had kind of divided things up into receiving high blood pressures and generating high blood pressures . and receiving high pressure -- really , that means the vessels , right ? that 's where the pressure is going to be seen . and so the vessels can get damaged . and we talked about some of the different ways that that can happen . and in fact , there are the larger arteries and also the smaller arteries . but now , let 's focus our attention on generating high pressures . because this might be less intuitive , but actually making the pressure go so high is not an easy thing to do . and it 's actually a very challenging task for the heart to accomplish . and in trying to figure it out , sometimes the heart can run into troubles . so there was an equation , and that equation was delta p equals q times r. and i thought we had put some words to this equation . so the idea that delta p -- that 's basically saying the arterial pressure minus the venous pressure . so if you have blood and it 's exiting here , and that 's got a certain pressure on it . and as it enters again on the other side into the atrium , it 's got a pressure on the venous side . so we 're saying that the pressure leaving in the arteries minus the pressure returning in the veins is what 's going to help you take a certain flow of blood , q , past the resistance in the vessels . so that 's a way of thinking about it . and specifically -- let me actually draw out some of this resistance . you have , of course , resistance once it gets into the arteries . and so you 've got some , r , resistance there . then you 've got more resistance once you get into the arterioles . in fact , i 'm going to triple underline that one because there 's lots of resistance in the arterioles . and then you 've got resistance in the capillaries , and you 've got resistance in the veins . and then you 've got the blood returning into the venous side . so there 's resistance in the circulatory system . and there 's a flow of blood that you 're actually trying to move around , right ? so anything , anything at all , that increases the resistance or increases the flow is going to force the amount of pressure you have to put in on the arterial side to go up . and now if i draw it out , you 'll see even more clearly why this is a problem . so if you have , let 's say , the left ventricle , and i 'll leave a little space for the valves . and this is the chamber of the heart that 's doing the pumping now . and it 's pumping the blood out into the aorta . this is the aorta . and this is another valve . and this valve is called the aortic valve -- pretty easy to remember name , aortic valve . and it separates the aorta from the left ventricle -- lv for left ventricle . so the left ventricle is basically going to have to try to get blood through this door , through this valve . and the way to do that is to apply a force to that blood and force it through that door . now , the aortic valve has a certain area . and so if you remember , any time you have a force over an area -- i 'll write that over here -- anytime you have a force over an area , you have a pressure . and in this case , that is the arterial pressure . so when i talk about arterial pressure , i 'm actually talking about the force that the left ventricle generates on that aortic valve , that area of aortic valve . so now , think about it . if you have , let 's say , more resistance . let 's say there 's lots and lots of resistance in the vessels , so that all these numbers are slowly going up , up , up . well , that 's going to cause the overall resistance to go up , and now the left ventricle has to put more force . or you could say , what if you have larger flow ? you actually have a larger volume of blood you 're trying to move around . again , the aortic pressure is going to have to go up . so the way that the left ventricle accomplishes this or tries to accomplish this is by basically building more muscle . so this is one strategy . it 'll say , ok , well , if i need to generate more force , why not generate more muscle ? so this becomes very muscular in here . if you actually look at a heart that has had high pressures over the years , oftentimes you have what 's called left ventricular hypertrophy . so on the surface , this sounds great , right ? left ventricular hypertrophy sounds like you 're making the muscle of the heart stronger . and that sounds like a good thing . you know , that 's certainly why i go to the gym . i want my muscles to get bigger . so it 's confusing , then , to hear that that 's not a good thing , and here 's why . so now , imagine you have all this muscle , and you 're feeling really good . but then i tell you that , well , this muscle 's using more oxygen . so it 's using more oxygen . but you notice i did n't draw any new blood vessels yet . so there are no new blood vessels getting oxygen to this muscle . but the muscle is definitely using a lot of energy , and it needs a lot of oxygen . and so what happens is that you have areas where blood is basically able to get to those areas and other areas where blood is not able to get there . and where the blood is unable to get enough oxygen to that heart muscle , you could have a heart attack or a myocardial infarction . so left ventricular hypertrophy is a way of compensating for needing more force . but it comes with a little bit of a risk because now you might have a risk of an mi or myocardial infarction . the other possibility is that the left heart does n't compensate , and you have what 's called left heart failure . and what that basically means is that the left heart says , you know what , i ca n't generate the pressures that are being required to move all that blood around , and so i wo n't . i wo n't generate those pressures . i 'll do the best i can , and maybe i ca n't move , let 's say , six liters of blood around the body , but i can do 5 liters or 5 and 1/2 liters . so it 'll try to do what it can to move as much blood as possible , but it is essentially not compensating completely . and that 's why you sometimes see people with heart failure after having years and years of high blood pressure . their heart simply ca n't keep up . and so you experience heart failure .
and what that basically means is that the left heart says , you know what , i ca n't generate the pressures that are being required to move all that blood around , and so i wo n't . i wo n't generate those pressures . i 'll do the best i can , and maybe i ca n't move , let 's say , six liters of blood around the body , but i can do 5 liters or 5 and 1/2 liters .
also what happens to the extra liter that is n't pumped ?
so we were talking in a previous video about hypertension and the damage that it causes on the body . and what i 'd like to do is pick up where we left off and finish that conversation . and we had kind of divided things up into receiving high blood pressures and generating high blood pressures . and receiving high pressure -- really , that means the vessels , right ? that 's where the pressure is going to be seen . and so the vessels can get damaged . and we talked about some of the different ways that that can happen . and in fact , there are the larger arteries and also the smaller arteries . but now , let 's focus our attention on generating high pressures . because this might be less intuitive , but actually making the pressure go so high is not an easy thing to do . and it 's actually a very challenging task for the heart to accomplish . and in trying to figure it out , sometimes the heart can run into troubles . so there was an equation , and that equation was delta p equals q times r. and i thought we had put some words to this equation . so the idea that delta p -- that 's basically saying the arterial pressure minus the venous pressure . so if you have blood and it 's exiting here , and that 's got a certain pressure on it . and as it enters again on the other side into the atrium , it 's got a pressure on the venous side . so we 're saying that the pressure leaving in the arteries minus the pressure returning in the veins is what 's going to help you take a certain flow of blood , q , past the resistance in the vessels . so that 's a way of thinking about it . and specifically -- let me actually draw out some of this resistance . you have , of course , resistance once it gets into the arteries . and so you 've got some , r , resistance there . then you 've got more resistance once you get into the arterioles . in fact , i 'm going to triple underline that one because there 's lots of resistance in the arterioles . and then you 've got resistance in the capillaries , and you 've got resistance in the veins . and then you 've got the blood returning into the venous side . so there 's resistance in the circulatory system . and there 's a flow of blood that you 're actually trying to move around , right ? so anything , anything at all , that increases the resistance or increases the flow is going to force the amount of pressure you have to put in on the arterial side to go up . and now if i draw it out , you 'll see even more clearly why this is a problem . so if you have , let 's say , the left ventricle , and i 'll leave a little space for the valves . and this is the chamber of the heart that 's doing the pumping now . and it 's pumping the blood out into the aorta . this is the aorta . and this is another valve . and this valve is called the aortic valve -- pretty easy to remember name , aortic valve . and it separates the aorta from the left ventricle -- lv for left ventricle . so the left ventricle is basically going to have to try to get blood through this door , through this valve . and the way to do that is to apply a force to that blood and force it through that door . now , the aortic valve has a certain area . and so if you remember , any time you have a force over an area -- i 'll write that over here -- anytime you have a force over an area , you have a pressure . and in this case , that is the arterial pressure . so when i talk about arterial pressure , i 'm actually talking about the force that the left ventricle generates on that aortic valve , that area of aortic valve . so now , think about it . if you have , let 's say , more resistance . let 's say there 's lots and lots of resistance in the vessels , so that all these numbers are slowly going up , up , up . well , that 's going to cause the overall resistance to go up , and now the left ventricle has to put more force . or you could say , what if you have larger flow ? you actually have a larger volume of blood you 're trying to move around . again , the aortic pressure is going to have to go up . so the way that the left ventricle accomplishes this or tries to accomplish this is by basically building more muscle . so this is one strategy . it 'll say , ok , well , if i need to generate more force , why not generate more muscle ? so this becomes very muscular in here . if you actually look at a heart that has had high pressures over the years , oftentimes you have what 's called left ventricular hypertrophy . so on the surface , this sounds great , right ? left ventricular hypertrophy sounds like you 're making the muscle of the heart stronger . and that sounds like a good thing . you know , that 's certainly why i go to the gym . i want my muscles to get bigger . so it 's confusing , then , to hear that that 's not a good thing , and here 's why . so now , imagine you have all this muscle , and you 're feeling really good . but then i tell you that , well , this muscle 's using more oxygen . so it 's using more oxygen . but you notice i did n't draw any new blood vessels yet . so there are no new blood vessels getting oxygen to this muscle . but the muscle is definitely using a lot of energy , and it needs a lot of oxygen . and so what happens is that you have areas where blood is basically able to get to those areas and other areas where blood is not able to get there . and where the blood is unable to get enough oxygen to that heart muscle , you could have a heart attack or a myocardial infarction . so left ventricular hypertrophy is a way of compensating for needing more force . but it comes with a little bit of a risk because now you might have a risk of an mi or myocardial infarction . the other possibility is that the left heart does n't compensate , and you have what 's called left heart failure . and what that basically means is that the left heart says , you know what , i ca n't generate the pressures that are being required to move all that blood around , and so i wo n't . i wo n't generate those pressures . i 'll do the best i can , and maybe i ca n't move , let 's say , six liters of blood around the body , but i can do 5 liters or 5 and 1/2 liters . so it 'll try to do what it can to move as much blood as possible , but it is essentially not compensating completely . and that 's why you sometimes see people with heart failure after having years and years of high blood pressure . their heart simply ca n't keep up . and so you experience heart failure .
but it comes with a little bit of a risk because now you might have a risk of an mi or myocardial infarction . the other possibility is that the left heart does n't compensate , and you have what 's called left heart failure . and what that basically means is that the left heart says , you know what , i ca n't generate the pressures that are being required to move all that blood around , and so i wo n't .
could somebody please explain the difference between left and right sided heart failure and what the body experiences with each of them , or is there a video explaining this already and i have n't found it yet ?
so we were talking in a previous video about hypertension and the damage that it causes on the body . and what i 'd like to do is pick up where we left off and finish that conversation . and we had kind of divided things up into receiving high blood pressures and generating high blood pressures . and receiving high pressure -- really , that means the vessels , right ? that 's where the pressure is going to be seen . and so the vessels can get damaged . and we talked about some of the different ways that that can happen . and in fact , there are the larger arteries and also the smaller arteries . but now , let 's focus our attention on generating high pressures . because this might be less intuitive , but actually making the pressure go so high is not an easy thing to do . and it 's actually a very challenging task for the heart to accomplish . and in trying to figure it out , sometimes the heart can run into troubles . so there was an equation , and that equation was delta p equals q times r. and i thought we had put some words to this equation . so the idea that delta p -- that 's basically saying the arterial pressure minus the venous pressure . so if you have blood and it 's exiting here , and that 's got a certain pressure on it . and as it enters again on the other side into the atrium , it 's got a pressure on the venous side . so we 're saying that the pressure leaving in the arteries minus the pressure returning in the veins is what 's going to help you take a certain flow of blood , q , past the resistance in the vessels . so that 's a way of thinking about it . and specifically -- let me actually draw out some of this resistance . you have , of course , resistance once it gets into the arteries . and so you 've got some , r , resistance there . then you 've got more resistance once you get into the arterioles . in fact , i 'm going to triple underline that one because there 's lots of resistance in the arterioles . and then you 've got resistance in the capillaries , and you 've got resistance in the veins . and then you 've got the blood returning into the venous side . so there 's resistance in the circulatory system . and there 's a flow of blood that you 're actually trying to move around , right ? so anything , anything at all , that increases the resistance or increases the flow is going to force the amount of pressure you have to put in on the arterial side to go up . and now if i draw it out , you 'll see even more clearly why this is a problem . so if you have , let 's say , the left ventricle , and i 'll leave a little space for the valves . and this is the chamber of the heart that 's doing the pumping now . and it 's pumping the blood out into the aorta . this is the aorta . and this is another valve . and this valve is called the aortic valve -- pretty easy to remember name , aortic valve . and it separates the aorta from the left ventricle -- lv for left ventricle . so the left ventricle is basically going to have to try to get blood through this door , through this valve . and the way to do that is to apply a force to that blood and force it through that door . now , the aortic valve has a certain area . and so if you remember , any time you have a force over an area -- i 'll write that over here -- anytime you have a force over an area , you have a pressure . and in this case , that is the arterial pressure . so when i talk about arterial pressure , i 'm actually talking about the force that the left ventricle generates on that aortic valve , that area of aortic valve . so now , think about it . if you have , let 's say , more resistance . let 's say there 's lots and lots of resistance in the vessels , so that all these numbers are slowly going up , up , up . well , that 's going to cause the overall resistance to go up , and now the left ventricle has to put more force . or you could say , what if you have larger flow ? you actually have a larger volume of blood you 're trying to move around . again , the aortic pressure is going to have to go up . so the way that the left ventricle accomplishes this or tries to accomplish this is by basically building more muscle . so this is one strategy . it 'll say , ok , well , if i need to generate more force , why not generate more muscle ? so this becomes very muscular in here . if you actually look at a heart that has had high pressures over the years , oftentimes you have what 's called left ventricular hypertrophy . so on the surface , this sounds great , right ? left ventricular hypertrophy sounds like you 're making the muscle of the heart stronger . and that sounds like a good thing . you know , that 's certainly why i go to the gym . i want my muscles to get bigger . so it 's confusing , then , to hear that that 's not a good thing , and here 's why . so now , imagine you have all this muscle , and you 're feeling really good . but then i tell you that , well , this muscle 's using more oxygen . so it 's using more oxygen . but you notice i did n't draw any new blood vessels yet . so there are no new blood vessels getting oxygen to this muscle . but the muscle is definitely using a lot of energy , and it needs a lot of oxygen . and so what happens is that you have areas where blood is basically able to get to those areas and other areas where blood is not able to get there . and where the blood is unable to get enough oxygen to that heart muscle , you could have a heart attack or a myocardial infarction . so left ventricular hypertrophy is a way of compensating for needing more force . but it comes with a little bit of a risk because now you might have a risk of an mi or myocardial infarction . the other possibility is that the left heart does n't compensate , and you have what 's called left heart failure . and what that basically means is that the left heart says , you know what , i ca n't generate the pressures that are being required to move all that blood around , and so i wo n't . i wo n't generate those pressures . i 'll do the best i can , and maybe i ca n't move , let 's say , six liters of blood around the body , but i can do 5 liters or 5 and 1/2 liters . so it 'll try to do what it can to move as much blood as possible , but it is essentially not compensating completely . and that 's why you sometimes see people with heart failure after having years and years of high blood pressure . their heart simply ca n't keep up . and so you experience heart failure .
and so what happens is that you have areas where blood is basically able to get to those areas and other areas where blood is not able to get there . and where the blood is unable to get enough oxygen to that heart muscle , you could have a heart attack or a myocardial infarction . so left ventricular hypertrophy is a way of compensating for needing more force .
could someone please specify the types of resistance in blood vessels that ( forces the heart to pump out blood at an increased pace ) or refer a video pertaining to that information ?
so we were talking in a previous video about hypertension and the damage that it causes on the body . and what i 'd like to do is pick up where we left off and finish that conversation . and we had kind of divided things up into receiving high blood pressures and generating high blood pressures . and receiving high pressure -- really , that means the vessels , right ? that 's where the pressure is going to be seen . and so the vessels can get damaged . and we talked about some of the different ways that that can happen . and in fact , there are the larger arteries and also the smaller arteries . but now , let 's focus our attention on generating high pressures . because this might be less intuitive , but actually making the pressure go so high is not an easy thing to do . and it 's actually a very challenging task for the heart to accomplish . and in trying to figure it out , sometimes the heart can run into troubles . so there was an equation , and that equation was delta p equals q times r. and i thought we had put some words to this equation . so the idea that delta p -- that 's basically saying the arterial pressure minus the venous pressure . so if you have blood and it 's exiting here , and that 's got a certain pressure on it . and as it enters again on the other side into the atrium , it 's got a pressure on the venous side . so we 're saying that the pressure leaving in the arteries minus the pressure returning in the veins is what 's going to help you take a certain flow of blood , q , past the resistance in the vessels . so that 's a way of thinking about it . and specifically -- let me actually draw out some of this resistance . you have , of course , resistance once it gets into the arteries . and so you 've got some , r , resistance there . then you 've got more resistance once you get into the arterioles . in fact , i 'm going to triple underline that one because there 's lots of resistance in the arterioles . and then you 've got resistance in the capillaries , and you 've got resistance in the veins . and then you 've got the blood returning into the venous side . so there 's resistance in the circulatory system . and there 's a flow of blood that you 're actually trying to move around , right ? so anything , anything at all , that increases the resistance or increases the flow is going to force the amount of pressure you have to put in on the arterial side to go up . and now if i draw it out , you 'll see even more clearly why this is a problem . so if you have , let 's say , the left ventricle , and i 'll leave a little space for the valves . and this is the chamber of the heart that 's doing the pumping now . and it 's pumping the blood out into the aorta . this is the aorta . and this is another valve . and this valve is called the aortic valve -- pretty easy to remember name , aortic valve . and it separates the aorta from the left ventricle -- lv for left ventricle . so the left ventricle is basically going to have to try to get blood through this door , through this valve . and the way to do that is to apply a force to that blood and force it through that door . now , the aortic valve has a certain area . and so if you remember , any time you have a force over an area -- i 'll write that over here -- anytime you have a force over an area , you have a pressure . and in this case , that is the arterial pressure . so when i talk about arterial pressure , i 'm actually talking about the force that the left ventricle generates on that aortic valve , that area of aortic valve . so now , think about it . if you have , let 's say , more resistance . let 's say there 's lots and lots of resistance in the vessels , so that all these numbers are slowly going up , up , up . well , that 's going to cause the overall resistance to go up , and now the left ventricle has to put more force . or you could say , what if you have larger flow ? you actually have a larger volume of blood you 're trying to move around . again , the aortic pressure is going to have to go up . so the way that the left ventricle accomplishes this or tries to accomplish this is by basically building more muscle . so this is one strategy . it 'll say , ok , well , if i need to generate more force , why not generate more muscle ? so this becomes very muscular in here . if you actually look at a heart that has had high pressures over the years , oftentimes you have what 's called left ventricular hypertrophy . so on the surface , this sounds great , right ? left ventricular hypertrophy sounds like you 're making the muscle of the heart stronger . and that sounds like a good thing . you know , that 's certainly why i go to the gym . i want my muscles to get bigger . so it 's confusing , then , to hear that that 's not a good thing , and here 's why . so now , imagine you have all this muscle , and you 're feeling really good . but then i tell you that , well , this muscle 's using more oxygen . so it 's using more oxygen . but you notice i did n't draw any new blood vessels yet . so there are no new blood vessels getting oxygen to this muscle . but the muscle is definitely using a lot of energy , and it needs a lot of oxygen . and so what happens is that you have areas where blood is basically able to get to those areas and other areas where blood is not able to get there . and where the blood is unable to get enough oxygen to that heart muscle , you could have a heart attack or a myocardial infarction . so left ventricular hypertrophy is a way of compensating for needing more force . but it comes with a little bit of a risk because now you might have a risk of an mi or myocardial infarction . the other possibility is that the left heart does n't compensate , and you have what 's called left heart failure . and what that basically means is that the left heart says , you know what , i ca n't generate the pressures that are being required to move all that blood around , and so i wo n't . i wo n't generate those pressures . i 'll do the best i can , and maybe i ca n't move , let 's say , six liters of blood around the body , but i can do 5 liters or 5 and 1/2 liters . so it 'll try to do what it can to move as much blood as possible , but it is essentially not compensating completely . and that 's why you sometimes see people with heart failure after having years and years of high blood pressure . their heart simply ca n't keep up . and so you experience heart failure .
and so if you remember , any time you have a force over an area -- i 'll write that over here -- anytime you have a force over an area , you have a pressure . and in this case , that is the arterial pressure . so when i talk about arterial pressure , i 'm actually talking about the force that the left ventricle generates on that aortic valve , that area of aortic valve .
what is mean arterial pressure ?
the goal of this video is to try to compare these two quantities . i have 0.17 and i have 0.2 , and i wan na figure out which of these two is larger or maybe they 're equal , and i encourage you to pause the video to try to figure that out . one way to do it is to try to visualize these . each of these large squares you can view as a whole . and this one on the left , i have it split into 100 smaller sections 'cause you notice we have 10 rows , and each row has 10 squares , so we have 100 squares here . so , each of these squares represents 1/100 of the whole . and so this number up here , 0.17 , we could view that as 17/100 . so let 's color in 17/100 . 17/100 is going to be , so that 's 1/100 , two , three , four , five , six , seven , eight , nine , 10/100 . and notice 10/100 , i filled in one out of the 10 rows , so this is the same thing as 1/10 . so that 's 10/100 , 11/100 , 12/100 , 13 , 14 , 15 , 16 , 17 . so , one way to visualize 0.17 , which is 17/100 , is the fraction of this whole that is filled in magenta . now what about 0.2 ? 0.2 is the same thing as 2/10 . so we could take our whole and divide it into 1/10 , divide it into 10 equal sections which we 've done here . notice each of these sections is equal to 10/100 . and that makes sense , 1/10 is equal to 10/100 . let 's fill in two of them now 'cause we 're dealing with 2/10 . so let 's fill in two of them . we have 1/10 and 2/10 . which of these is larger ? we see we 're filling in more of the whole when we 're doing 2/10 than when we 're doing 17/100 . and that makes sense because 2/10 is the same thing . 2/10 is the same thing . another way of me saying 2/10 is you could write it as 20/100 . 20/100 is greater than 17/100 . to get the 20/100 , you 'd have to fill in these three as well . notice when you fill out those three , you 're filling out the same fraction of your whole . so , which one 's larger ? 2/10 is . and how do we write that down ? when we write it in equality , we wan na open to the larger number . you want to open to that one . so we have 0.17 or 17/100 , is less than 2/10 . now , another way that we could 've tackled this , even without having to draw all of this , is we could 've just gone to the largest place value . so if you went to the one 's place , we have zero ones . and we have zero ones here , so that does n't help us . then you go to the next largest , so you go to the 1/10 place . here you have 1/10 , here you have 2/10 . so immediately , without even looking at what comes after this , this could be 0.17358 , it could keep going , but the bottom line is you have more 1/10 here than you have 1/10 here , which tells you that this one over here is going to be larger . so the general thing is , look at the largest place value . if one of the numbers has a larger digit in the largest place value , it 's gon na be larger . if they 're equal , go to the next larger place value . and you could keep going like that , but another way to think about it is just to visualize it , just like we did over here .
you want to open to that one . so we have 0.17 or 17/100 , is less than 2/10 . now , another way that we could 've tackled this , even without having to draw all of this , is we could 've just gone to the largest place value .
is n't 17 larger than 2 ?
the goal of this video is to try to compare these two quantities . i have 0.17 and i have 0.2 , and i wan na figure out which of these two is larger or maybe they 're equal , and i encourage you to pause the video to try to figure that out . one way to do it is to try to visualize these . each of these large squares you can view as a whole . and this one on the left , i have it split into 100 smaller sections 'cause you notice we have 10 rows , and each row has 10 squares , so we have 100 squares here . so , each of these squares represents 1/100 of the whole . and so this number up here , 0.17 , we could view that as 17/100 . so let 's color in 17/100 . 17/100 is going to be , so that 's 1/100 , two , three , four , five , six , seven , eight , nine , 10/100 . and notice 10/100 , i filled in one out of the 10 rows , so this is the same thing as 1/10 . so that 's 10/100 , 11/100 , 12/100 , 13 , 14 , 15 , 16 , 17 . so , one way to visualize 0.17 , which is 17/100 , is the fraction of this whole that is filled in magenta . now what about 0.2 ? 0.2 is the same thing as 2/10 . so we could take our whole and divide it into 1/10 , divide it into 10 equal sections which we 've done here . notice each of these sections is equal to 10/100 . and that makes sense , 1/10 is equal to 10/100 . let 's fill in two of them now 'cause we 're dealing with 2/10 . so let 's fill in two of them . we have 1/10 and 2/10 . which of these is larger ? we see we 're filling in more of the whole when we 're doing 2/10 than when we 're doing 17/100 . and that makes sense because 2/10 is the same thing . 2/10 is the same thing . another way of me saying 2/10 is you could write it as 20/100 . 20/100 is greater than 17/100 . to get the 20/100 , you 'd have to fill in these three as well . notice when you fill out those three , you 're filling out the same fraction of your whole . so , which one 's larger ? 2/10 is . and how do we write that down ? when we write it in equality , we wan na open to the larger number . you want to open to that one . so we have 0.17 or 17/100 , is less than 2/10 . now , another way that we could 've tackled this , even without having to draw all of this , is we could 've just gone to the largest place value . so if you went to the one 's place , we have zero ones . and we have zero ones here , so that does n't help us . then you go to the next largest , so you go to the 1/10 place . here you have 1/10 , here you have 2/10 . so immediately , without even looking at what comes after this , this could be 0.17358 , it could keep going , but the bottom line is you have more 1/10 here than you have 1/10 here , which tells you that this one over here is going to be larger . so the general thing is , look at the largest place value . if one of the numbers has a larger digit in the largest place value , it 's gon na be larger . if they 're equal , go to the next larger place value . and you could keep going like that , but another way to think about it is just to visualize it , just like we did over here .
so , one way to visualize 0.17 , which is 17/100 , is the fraction of this whole that is filled in magenta . now what about 0.2 ? 0.2 is the same thing as 2/10 .
how are the decimals in 0.009 and 0.09 related ?
the goal of this video is to try to compare these two quantities . i have 0.17 and i have 0.2 , and i wan na figure out which of these two is larger or maybe they 're equal , and i encourage you to pause the video to try to figure that out . one way to do it is to try to visualize these . each of these large squares you can view as a whole . and this one on the left , i have it split into 100 smaller sections 'cause you notice we have 10 rows , and each row has 10 squares , so we have 100 squares here . so , each of these squares represents 1/100 of the whole . and so this number up here , 0.17 , we could view that as 17/100 . so let 's color in 17/100 . 17/100 is going to be , so that 's 1/100 , two , three , four , five , six , seven , eight , nine , 10/100 . and notice 10/100 , i filled in one out of the 10 rows , so this is the same thing as 1/10 . so that 's 10/100 , 11/100 , 12/100 , 13 , 14 , 15 , 16 , 17 . so , one way to visualize 0.17 , which is 17/100 , is the fraction of this whole that is filled in magenta . now what about 0.2 ? 0.2 is the same thing as 2/10 . so we could take our whole and divide it into 1/10 , divide it into 10 equal sections which we 've done here . notice each of these sections is equal to 10/100 . and that makes sense , 1/10 is equal to 10/100 . let 's fill in two of them now 'cause we 're dealing with 2/10 . so let 's fill in two of them . we have 1/10 and 2/10 . which of these is larger ? we see we 're filling in more of the whole when we 're doing 2/10 than when we 're doing 17/100 . and that makes sense because 2/10 is the same thing . 2/10 is the same thing . another way of me saying 2/10 is you could write it as 20/100 . 20/100 is greater than 17/100 . to get the 20/100 , you 'd have to fill in these three as well . notice when you fill out those three , you 're filling out the same fraction of your whole . so , which one 's larger ? 2/10 is . and how do we write that down ? when we write it in equality , we wan na open to the larger number . you want to open to that one . so we have 0.17 or 17/100 , is less than 2/10 . now , another way that we could 've tackled this , even without having to draw all of this , is we could 've just gone to the largest place value . so if you went to the one 's place , we have zero ones . and we have zero ones here , so that does n't help us . then you go to the next largest , so you go to the 1/10 place . here you have 1/10 , here you have 2/10 . so immediately , without even looking at what comes after this , this could be 0.17358 , it could keep going , but the bottom line is you have more 1/10 here than you have 1/10 here , which tells you that this one over here is going to be larger . so the general thing is , look at the largest place value . if one of the numbers has a larger digit in the largest place value , it 's gon na be larger . if they 're equal , go to the next larger place value . and you could keep going like that , but another way to think about it is just to visualize it , just like we did over here .
you want to open to that one . so we have 0.17 or 17/100 , is less than 2/10 . now , another way that we could 've tackled this , even without having to draw all of this , is we could 've just gone to the largest place value .
if 17 hundredths equal 0.17 then what does 17 tenths look like ?
the goal of this video is to try to compare these two quantities . i have 0.17 and i have 0.2 , and i wan na figure out which of these two is larger or maybe they 're equal , and i encourage you to pause the video to try to figure that out . one way to do it is to try to visualize these . each of these large squares you can view as a whole . and this one on the left , i have it split into 100 smaller sections 'cause you notice we have 10 rows , and each row has 10 squares , so we have 100 squares here . so , each of these squares represents 1/100 of the whole . and so this number up here , 0.17 , we could view that as 17/100 . so let 's color in 17/100 . 17/100 is going to be , so that 's 1/100 , two , three , four , five , six , seven , eight , nine , 10/100 . and notice 10/100 , i filled in one out of the 10 rows , so this is the same thing as 1/10 . so that 's 10/100 , 11/100 , 12/100 , 13 , 14 , 15 , 16 , 17 . so , one way to visualize 0.17 , which is 17/100 , is the fraction of this whole that is filled in magenta . now what about 0.2 ? 0.2 is the same thing as 2/10 . so we could take our whole and divide it into 1/10 , divide it into 10 equal sections which we 've done here . notice each of these sections is equal to 10/100 . and that makes sense , 1/10 is equal to 10/100 . let 's fill in two of them now 'cause we 're dealing with 2/10 . so let 's fill in two of them . we have 1/10 and 2/10 . which of these is larger ? we see we 're filling in more of the whole when we 're doing 2/10 than when we 're doing 17/100 . and that makes sense because 2/10 is the same thing . 2/10 is the same thing . another way of me saying 2/10 is you could write it as 20/100 . 20/100 is greater than 17/100 . to get the 20/100 , you 'd have to fill in these three as well . notice when you fill out those three , you 're filling out the same fraction of your whole . so , which one 's larger ? 2/10 is . and how do we write that down ? when we write it in equality , we wan na open to the larger number . you want to open to that one . so we have 0.17 or 17/100 , is less than 2/10 . now , another way that we could 've tackled this , even without having to draw all of this , is we could 've just gone to the largest place value . so if you went to the one 's place , we have zero ones . and we have zero ones here , so that does n't help us . then you go to the next largest , so you go to the 1/10 place . here you have 1/10 , here you have 2/10 . so immediately , without even looking at what comes after this , this could be 0.17358 , it could keep going , but the bottom line is you have more 1/10 here than you have 1/10 here , which tells you that this one over here is going to be larger . so the general thing is , look at the largest place value . if one of the numbers has a larger digit in the largest place value , it 's gon na be larger . if they 're equal , go to the next larger place value . and you could keep going like that , but another way to think about it is just to visualize it , just like we did over here .
so let 's fill in two of them . we have 1/10 and 2/10 . which of these is larger ?
are decimals remainders for dividing by 10 ?
the goal of this video is to try to compare these two quantities . i have 0.17 and i have 0.2 , and i wan na figure out which of these two is larger or maybe they 're equal , and i encourage you to pause the video to try to figure that out . one way to do it is to try to visualize these . each of these large squares you can view as a whole . and this one on the left , i have it split into 100 smaller sections 'cause you notice we have 10 rows , and each row has 10 squares , so we have 100 squares here . so , each of these squares represents 1/100 of the whole . and so this number up here , 0.17 , we could view that as 17/100 . so let 's color in 17/100 . 17/100 is going to be , so that 's 1/100 , two , three , four , five , six , seven , eight , nine , 10/100 . and notice 10/100 , i filled in one out of the 10 rows , so this is the same thing as 1/10 . so that 's 10/100 , 11/100 , 12/100 , 13 , 14 , 15 , 16 , 17 . so , one way to visualize 0.17 , which is 17/100 , is the fraction of this whole that is filled in magenta . now what about 0.2 ? 0.2 is the same thing as 2/10 . so we could take our whole and divide it into 1/10 , divide it into 10 equal sections which we 've done here . notice each of these sections is equal to 10/100 . and that makes sense , 1/10 is equal to 10/100 . let 's fill in two of them now 'cause we 're dealing with 2/10 . so let 's fill in two of them . we have 1/10 and 2/10 . which of these is larger ? we see we 're filling in more of the whole when we 're doing 2/10 than when we 're doing 17/100 . and that makes sense because 2/10 is the same thing . 2/10 is the same thing . another way of me saying 2/10 is you could write it as 20/100 . 20/100 is greater than 17/100 . to get the 20/100 , you 'd have to fill in these three as well . notice when you fill out those three , you 're filling out the same fraction of your whole . so , which one 's larger ? 2/10 is . and how do we write that down ? when we write it in equality , we wan na open to the larger number . you want to open to that one . so we have 0.17 or 17/100 , is less than 2/10 . now , another way that we could 've tackled this , even without having to draw all of this , is we could 've just gone to the largest place value . so if you went to the one 's place , we have zero ones . and we have zero ones here , so that does n't help us . then you go to the next largest , so you go to the 1/10 place . here you have 1/10 , here you have 2/10 . so immediately , without even looking at what comes after this , this could be 0.17358 , it could keep going , but the bottom line is you have more 1/10 here than you have 1/10 here , which tells you that this one over here is going to be larger . so the general thing is , look at the largest place value . if one of the numbers has a larger digit in the largest place value , it 's gon na be larger . if they 're equal , go to the next larger place value . and you could keep going like that , but another way to think about it is just to visualize it , just like we did over here .
and notice 10/100 , i filled in one out of the 10 rows , so this is the same thing as 1/10 . so that 's 10/100 , 11/100 , 12/100 , 13 , 14 , 15 , 16 , 17 . so , one way to visualize 0.17 , which is 17/100 , is the fraction of this whole that is filled in magenta .
what is 100 divided by 3 ?
the goal of this video is to try to compare these two quantities . i have 0.17 and i have 0.2 , and i wan na figure out which of these two is larger or maybe they 're equal , and i encourage you to pause the video to try to figure that out . one way to do it is to try to visualize these . each of these large squares you can view as a whole . and this one on the left , i have it split into 100 smaller sections 'cause you notice we have 10 rows , and each row has 10 squares , so we have 100 squares here . so , each of these squares represents 1/100 of the whole . and so this number up here , 0.17 , we could view that as 17/100 . so let 's color in 17/100 . 17/100 is going to be , so that 's 1/100 , two , three , four , five , six , seven , eight , nine , 10/100 . and notice 10/100 , i filled in one out of the 10 rows , so this is the same thing as 1/10 . so that 's 10/100 , 11/100 , 12/100 , 13 , 14 , 15 , 16 , 17 . so , one way to visualize 0.17 , which is 17/100 , is the fraction of this whole that is filled in magenta . now what about 0.2 ? 0.2 is the same thing as 2/10 . so we could take our whole and divide it into 1/10 , divide it into 10 equal sections which we 've done here . notice each of these sections is equal to 10/100 . and that makes sense , 1/10 is equal to 10/100 . let 's fill in two of them now 'cause we 're dealing with 2/10 . so let 's fill in two of them . we have 1/10 and 2/10 . which of these is larger ? we see we 're filling in more of the whole when we 're doing 2/10 than when we 're doing 17/100 . and that makes sense because 2/10 is the same thing . 2/10 is the same thing . another way of me saying 2/10 is you could write it as 20/100 . 20/100 is greater than 17/100 . to get the 20/100 , you 'd have to fill in these three as well . notice when you fill out those three , you 're filling out the same fraction of your whole . so , which one 's larger ? 2/10 is . and how do we write that down ? when we write it in equality , we wan na open to the larger number . you want to open to that one . so we have 0.17 or 17/100 , is less than 2/10 . now , another way that we could 've tackled this , even without having to draw all of this , is we could 've just gone to the largest place value . so if you went to the one 's place , we have zero ones . and we have zero ones here , so that does n't help us . then you go to the next largest , so you go to the 1/10 place . here you have 1/10 , here you have 2/10 . so immediately , without even looking at what comes after this , this could be 0.17358 , it could keep going , but the bottom line is you have more 1/10 here than you have 1/10 here , which tells you that this one over here is going to be larger . so the general thing is , look at the largest place value . if one of the numbers has a larger digit in the largest place value , it 's gon na be larger . if they 're equal , go to the next larger place value . and you could keep going like that , but another way to think about it is just to visualize it , just like we did over here .
you want to open to that one . so we have 0.17 or 17/100 , is less than 2/10 . now , another way that we could 've tackled this , even without having to draw all of this , is we could 've just gone to the largest place value .
how did you know to make 0.17 have 10x10 but the other have the squares ?
let 's apply everything we 've learned to an actual differential equation . instead of just taking laplace transforms and taking their inverse , let 's actually solve a problem . so let 's say that i have the second derivative of my function y plus 4 times my function y is equal to sine of t minus the unit step function 0 up until 2 pi of t times sine of t minus 2 pi . let 's solve this differential equation , an interpretation of it . and i actually do a whole playlist on interpretations of differential equations and how you model it , but you know , you can kind of view this is a forcing function . that it 's a weird forcing function of this being applied to some weight with , you know , this is the acceleration term , right ? the second derivative with respect to time is the acceleration . so the mass would be 1 whatever units , and then as a function of its position , this is probably some type of spring constant . anyway , i wo n't go there . i do n't want to waste your time with the interpretation of it , but let 's solve it . we can do more about interpretations later . so we 're going take the laplace transform of both sides of this equation . so what 's the laplace transform of the left-hand side ? so the laplace transform of the second derivative of y is just s squared , so now i 'm taking the laplace transform of just that . the laplace transform of s squared times the laplace transform of y minus -- lower the degree there once -- minus s times y of 0 minus y prime of 0 . so clearly , i must have to give you some initial conditions in order to do this properly . and then plus 4 times the laplace transform of y is equal to -- what 's the laplace transform of sine of t ? that should be second nature by now . it 's just 1 over s squared plus 1 . and then we have minus the laplace transform of this thing . and i 'll do a little side note here to figure out the laplace transform of this thing right here . and we know , i showed it to you a couple of videos ago , we showed that the laplace transform -- actually i could just write it out here . this is going to be the same thing as the laplace transform of sine of t , but we 're going to have to multiply it by e to the minus -- if you remember that last formula -- e to the minus cs , where c is 2 pi . actually , let me write that down . i decided to write it down , then i decided , oh , no , i do n't want to do this . but let me write that . so the laplace transform of the unit step function that goes up to c times some function shifted by c is equal to e to the minus cs times the laplace transform of just the original function times the laplace transform of f of t. so if we 're taking the laplace transform of this thing , our c is 2 pi . our f of t is just sine of t , right ? so then this is just going to be equal to -- if we just do this piece right here -- it 's going to be equal to e to the minus cs -- our c is 2 pi -- e to the minus 2 pi s times the laplace transform of f of t. f of t is just sine of t before we shifted . this is f of t minus 2 pi . so f of t is just going to be sine of t. so it 's going to be times 1 over s squared plus 1 . this is the laplace transform of sine of t. so let 's go back to where we had left off . so we 've taken the laplace transform of both sides of this equation . and clearly , i have some initial conditions here , so the problem must have given me some and i just forgot to write them down . so let 's see , the initial conditions i 'm given , and they are written kind of in the margin here , they tell us -- i 'll do it in orange , they tell us that y of 0 is equal to 0 , and y prime of 0 is equal to 0 . that makes the math easy . that 's 0 and that 's 0 . so let 's see if i can simplify my equation . so the left-hand side , let 's factor out the laplace transform . so let 's factor out this term and that term . so we get the laplace transform of y times this plus this times s squared plus 4 is equal to the right-hand side . and what 's the right-hand side ? we could simplify this . well , i 'll just write it out . i do n't want to do too many steps at once . it 's 1 over s squared plus 1 and then plus -- or minus actually , this is a minus -- minus the laplace transfer of this thing , which was e to the minus 2 pi s over s squared plus 1 . so if we divide both sides of this equation by the s squared plus 4 , then we get the laplace transform of y is equal to -- and actually , i can just merge these two . they 're the same denominator . so before i even divide by s squared plus 4 , that right-hand side will look like this . it will look like with a denominator of s squared plus 1 and you have a numerator of 1 minus e to the minus 2 pi s. and , of course , we 're dividing both sides of this equation by s squared plus 4 , so we 're going to have to stick that s squared plus 4 over here . now , we 're at the hard part . in order to figure out why , we have to take the inverse laplace transform of this thing . so how do we take the inverse laplace transform of this thing ? that 's where the hard part is always , you know , it makes solving the differential equation 's easy if you know the laplace transforms . so it looks like we 're going to have to do some partial fraction expansion . so let 's see if we can do that . so we can rewrite this equation right here . actually , let 's write it as this , because this 'll kind of simplify our work . let 's factor this whole thing out . so we 're going to write it as 1 minus e to the minus 2 pi s , all of that times -- i 'll do it in orange -- all of that times 1 over s squared plus 1 times s squared plus 4 . now , we need to do some partial fraction expansion to simplify this thing right here . we 're going to do this on the side . maybe i should do this over on the right here . this thing -- let me rewrite it -- 1 over s squared plus 1 times s squared plus 4 should be able to be rewritten as two separate fractions , s squared plus 1 and s squared plus 4 , with the numerators . this one would be as plus b . it 's going to have to have degree 1 , because this is degree 2 . here and then we 'd have cs plus d. and so when you add these two things up , you get as plus b times s squared plus 4 plus cs plus d times s squared plus 1 , all of that over the common denominator . we 've seen this story before . we just have to do some algebra here . as you can tell , these differential equations problems , they require a lot of stamina . you kind of just have to say i will keep moving forward and do the algebra that i need to do in order to get the answer . and you kind of have to get excited about that notion that you have all this algebra to do . so let 's figure it out . so this top can be simplified to as to the third plus bs squared plus 4as plus 4b . and then this one , you end up with cs to the third plus ds squared plus cs plus d. so when you add of these up together , you get -- and this is all the algebra that we have to do , for better , for worse -- a plus c over s to the third plus b plus d times s squared plus 4a plus c times s -- let 's scroll over a little bit -- plus 4b plus d. and now we just have to say , ok , all of this is equal to this thing up here . this is the numerator . we just simplified the numerator . this is the numerator . that 's the numerator right there . and all of this is going to be over your original s squared plus 1 times your s squared plus 4 . and we established that this thing should be -- let me just write this -- that 1 over s squared plus 1 times s squared plus 4 should equal this thing . and then you just pattern match on the coefficients . this is all just intense partial fraction expansion . and you say , look , a plus c is the coefficient of the s cubed terms . i do n't see any s cubed terms here . so a plus c must be equal to 0 . and then you see , ok , b plus d is the coefficient of the s squared terms . i do n't see any s squared terms there . so b plus d must be equal to 0 . 4a plus c , the coefficient of the s terms . i do n't see any s terms over here . so 4a plus c must be equal to 0 . and then we 're almost done . 4b plus d must be the constant terms . there is a constant term there . so 4b plus d is equal to 1 . so let 's see if we can do anything here . if we subtract this from that , we get minus 3a is equal to 0 , or a is equal to 0 . if a is equal to 0 , then c is equals to 0 . and let 's see what we can get here . if we subtract this from that , we get minus 3b . the d 's cancel out . it 's equal to minus 1 , or b is equal to 1/3 . and then , of course , we have d is equal to minus b , if you subtract b from both sides . so d is equal to 1/3 . so all of that work , and we actually have a pretty simple result . our equation , this thing here , can be rewritten as -- the a disappeared . it 's 1/3 over s squared plus 1 . b was the coefficient on the -- let me make it very clear . b was the coefficient on the -- or it was a term on top of the s squared plus 1 , so that 's why i 'm using b there . and then d is minus b , so d is minus 1 . so let me make sure i have that . b is 1/3 minus -- let me make sure i get that right . d is 1/3 . so , sorry , b as in boy is 1/3 , so d is minus 1/3 . so b , there 's a term on top of the s squared plus 1 . and then you have minus d over the minus 1/3 over s squared plus 4 . this takes a lot of stamina to record this video . i hope you appreciate it . ok , so let me rewrite everything , just so we can get back to the problem because when you take the partial fraction detour , you forget -- not even to speak of the problem , you forget what day it is . let 's see , so you get the laplace transform of y is equal to 1 minus e to the minus 2 pi s times what that mess that we just solved for , times -- and i 'll write it like this . 1/3 times 1 over s squared plus 1 minus 1/3 times -- actually , let me write it this way . because i have this s squared plus 4 , so i really want to have a 2 there . so i want to have a 2 in the numerator , so you want to have a 2 over s squared plus 4 . so if i put a 2 in the numerator there , i have to divide this by 2 as well . so let me change this to a 6 . minus 1/6 times 2 is minus 1/3 . so i did that just so i get this in the form of the laplace transform of sine of t. now , let 's see if there 's anything that i can do from here . this is an epic problem . i 'll be amazed if i do n't make a careless mistake while i do this . so we can rewrite everything . let 's see if we can simplify this . and by simplifying it , i 'm just going to make it longer . we can write the laplace transform of y is equal to -- i 'm just going to multiply the 1 out , and then i 'm going to multiply the e to the minus 2 pi s out . so if you multiply the 1 out , you get 1/3 times 1 over s squared plus 1 -- i 'm just multiplying the 1 out -- minus 1/6 -- these are all the 1 's times the 1 -- times 2 over s squared plus 4 . and then i 'm going to multiply the minus e. let me just switch colors , do the minus e. so then you get minus e to the minus 2 pi s over 3 times 1 over s squared plus 1 . and then the minus and the minus cancel out , so you get plus e to the minus 2 pi s over 6 times 2 over s squared plus 4 . now , taking the inverse laplace transform of these things are pretty straightforward . so let 's do that . let 's take the inverse laplace transform of the whole thing . and we get y is equal to the inverse laplace transform of this guy right here , is just 1/3 sine of t -- i do n't have to write a parentheses there -- sine of t , and then this is minus 1/6 times -- this is the laplace transform of sine of 2t . that 's that term right there . now , these are almost the same , but we have this little pesky character over here . we have this e to the minus 2 pi s. and there , we just have to remind ourselves -- i 'll write it here in the bottom . we just have to remind ourselves that the laplace transform of the unit step function -- i 'll put the pi there , just 2 pi times f of t minus 2 pi -- i should put as the step function of t -- is equal to e to the minus 2 pi s times the laplace transform of just -- or let me just write it this way -- times the laplace transform of f of t. so if we view f of t as just sine of t or sine of 2t , then we can kind of backwards pattern match . and we 'll have to shift it and multiply it by the unit step function . so i want to make that clear . if you did n't have this guy here , the inverse laplace transform of this guy would be the same thing as this guy . it 'd just be sine of t. the inverse laplace transform of this guy would be sine of 2t . but we have this pesky character here , which essentially , instead of having the inverse laplace transform just being our f of t , it 's going to be our f of t shifted by 2 pi times the unit step function , where it steps up at 2pi . so this is going to be minus 1/3 times the unit step function , where c is 2 pi of t times -- instead of sine of t -- sine of t minus 2pi . and then we 're almost done . i 'll do it in magenta to celebrate it . plus this very last term , which is 1/6 times the unit step function 2 pi of t , the unit step function that steps up at 2 pi times sine of -- and we have to be careful here . wherever we had a t before , we 're going to replace it with a t minus 2 pi . so sine of , instead of 2t , is going to be 2 times t minus 2 pi . and there you have it . we finally have solved our very hairy problem . we could take some time if we want to simplify this a little bit . in fact , we might as well . at the risk of making a careless mistake at the last moment , let me see if i can make any simplifications here . well , we could factor out this guy right here , but other than that , that seems about as simple as we can get . so this is our function of t that satisfies our otherwise simple-looking differential equation that we had up here . this looked fairly straightforward , but we got this big mess to actually satisfy that equation , given those initial conditions that we had initially .
so clearly , i must have to give you some initial conditions in order to do this properly . and then plus 4 times the laplace transform of y is equal to -- what 's the laplace transform of sine of t ? that should be second nature by now .
i understand the laplace transform process and solution , but should n't sal be adding the homogeneous solution as well ?
let 's apply everything we 've learned to an actual differential equation . instead of just taking laplace transforms and taking their inverse , let 's actually solve a problem . so let 's say that i have the second derivative of my function y plus 4 times my function y is equal to sine of t minus the unit step function 0 up until 2 pi of t times sine of t minus 2 pi . let 's solve this differential equation , an interpretation of it . and i actually do a whole playlist on interpretations of differential equations and how you model it , but you know , you can kind of view this is a forcing function . that it 's a weird forcing function of this being applied to some weight with , you know , this is the acceleration term , right ? the second derivative with respect to time is the acceleration . so the mass would be 1 whatever units , and then as a function of its position , this is probably some type of spring constant . anyway , i wo n't go there . i do n't want to waste your time with the interpretation of it , but let 's solve it . we can do more about interpretations later . so we 're going take the laplace transform of both sides of this equation . so what 's the laplace transform of the left-hand side ? so the laplace transform of the second derivative of y is just s squared , so now i 'm taking the laplace transform of just that . the laplace transform of s squared times the laplace transform of y minus -- lower the degree there once -- minus s times y of 0 minus y prime of 0 . so clearly , i must have to give you some initial conditions in order to do this properly . and then plus 4 times the laplace transform of y is equal to -- what 's the laplace transform of sine of t ? that should be second nature by now . it 's just 1 over s squared plus 1 . and then we have minus the laplace transform of this thing . and i 'll do a little side note here to figure out the laplace transform of this thing right here . and we know , i showed it to you a couple of videos ago , we showed that the laplace transform -- actually i could just write it out here . this is going to be the same thing as the laplace transform of sine of t , but we 're going to have to multiply it by e to the minus -- if you remember that last formula -- e to the minus cs , where c is 2 pi . actually , let me write that down . i decided to write it down , then i decided , oh , no , i do n't want to do this . but let me write that . so the laplace transform of the unit step function that goes up to c times some function shifted by c is equal to e to the minus cs times the laplace transform of just the original function times the laplace transform of f of t. so if we 're taking the laplace transform of this thing , our c is 2 pi . our f of t is just sine of t , right ? so then this is just going to be equal to -- if we just do this piece right here -- it 's going to be equal to e to the minus cs -- our c is 2 pi -- e to the minus 2 pi s times the laplace transform of f of t. f of t is just sine of t before we shifted . this is f of t minus 2 pi . so f of t is just going to be sine of t. so it 's going to be times 1 over s squared plus 1 . this is the laplace transform of sine of t. so let 's go back to where we had left off . so we 've taken the laplace transform of both sides of this equation . and clearly , i have some initial conditions here , so the problem must have given me some and i just forgot to write them down . so let 's see , the initial conditions i 'm given , and they are written kind of in the margin here , they tell us -- i 'll do it in orange , they tell us that y of 0 is equal to 0 , and y prime of 0 is equal to 0 . that makes the math easy . that 's 0 and that 's 0 . so let 's see if i can simplify my equation . so the left-hand side , let 's factor out the laplace transform . so let 's factor out this term and that term . so we get the laplace transform of y times this plus this times s squared plus 4 is equal to the right-hand side . and what 's the right-hand side ? we could simplify this . well , i 'll just write it out . i do n't want to do too many steps at once . it 's 1 over s squared plus 1 and then plus -- or minus actually , this is a minus -- minus the laplace transfer of this thing , which was e to the minus 2 pi s over s squared plus 1 . so if we divide both sides of this equation by the s squared plus 4 , then we get the laplace transform of y is equal to -- and actually , i can just merge these two . they 're the same denominator . so before i even divide by s squared plus 4 , that right-hand side will look like this . it will look like with a denominator of s squared plus 1 and you have a numerator of 1 minus e to the minus 2 pi s. and , of course , we 're dividing both sides of this equation by s squared plus 4 , so we 're going to have to stick that s squared plus 4 over here . now , we 're at the hard part . in order to figure out why , we have to take the inverse laplace transform of this thing . so how do we take the inverse laplace transform of this thing ? that 's where the hard part is always , you know , it makes solving the differential equation 's easy if you know the laplace transforms . so it looks like we 're going to have to do some partial fraction expansion . so let 's see if we can do that . so we can rewrite this equation right here . actually , let 's write it as this , because this 'll kind of simplify our work . let 's factor this whole thing out . so we 're going to write it as 1 minus e to the minus 2 pi s , all of that times -- i 'll do it in orange -- all of that times 1 over s squared plus 1 times s squared plus 4 . now , we need to do some partial fraction expansion to simplify this thing right here . we 're going to do this on the side . maybe i should do this over on the right here . this thing -- let me rewrite it -- 1 over s squared plus 1 times s squared plus 4 should be able to be rewritten as two separate fractions , s squared plus 1 and s squared plus 4 , with the numerators . this one would be as plus b . it 's going to have to have degree 1 , because this is degree 2 . here and then we 'd have cs plus d. and so when you add these two things up , you get as plus b times s squared plus 4 plus cs plus d times s squared plus 1 , all of that over the common denominator . we 've seen this story before . we just have to do some algebra here . as you can tell , these differential equations problems , they require a lot of stamina . you kind of just have to say i will keep moving forward and do the algebra that i need to do in order to get the answer . and you kind of have to get excited about that notion that you have all this algebra to do . so let 's figure it out . so this top can be simplified to as to the third plus bs squared plus 4as plus 4b . and then this one , you end up with cs to the third plus ds squared plus cs plus d. so when you add of these up together , you get -- and this is all the algebra that we have to do , for better , for worse -- a plus c over s to the third plus b plus d times s squared plus 4a plus c times s -- let 's scroll over a little bit -- plus 4b plus d. and now we just have to say , ok , all of this is equal to this thing up here . this is the numerator . we just simplified the numerator . this is the numerator . that 's the numerator right there . and all of this is going to be over your original s squared plus 1 times your s squared plus 4 . and we established that this thing should be -- let me just write this -- that 1 over s squared plus 1 times s squared plus 4 should equal this thing . and then you just pattern match on the coefficients . this is all just intense partial fraction expansion . and you say , look , a plus c is the coefficient of the s cubed terms . i do n't see any s cubed terms here . so a plus c must be equal to 0 . and then you see , ok , b plus d is the coefficient of the s squared terms . i do n't see any s squared terms there . so b plus d must be equal to 0 . 4a plus c , the coefficient of the s terms . i do n't see any s terms over here . so 4a plus c must be equal to 0 . and then we 're almost done . 4b plus d must be the constant terms . there is a constant term there . so 4b plus d is equal to 1 . so let 's see if we can do anything here . if we subtract this from that , we get minus 3a is equal to 0 , or a is equal to 0 . if a is equal to 0 , then c is equals to 0 . and let 's see what we can get here . if we subtract this from that , we get minus 3b . the d 's cancel out . it 's equal to minus 1 , or b is equal to 1/3 . and then , of course , we have d is equal to minus b , if you subtract b from both sides . so d is equal to 1/3 . so all of that work , and we actually have a pretty simple result . our equation , this thing here , can be rewritten as -- the a disappeared . it 's 1/3 over s squared plus 1 . b was the coefficient on the -- let me make it very clear . b was the coefficient on the -- or it was a term on top of the s squared plus 1 , so that 's why i 'm using b there . and then d is minus b , so d is minus 1 . so let me make sure i have that . b is 1/3 minus -- let me make sure i get that right . d is 1/3 . so , sorry , b as in boy is 1/3 , so d is minus 1/3 . so b , there 's a term on top of the s squared plus 1 . and then you have minus d over the minus 1/3 over s squared plus 4 . this takes a lot of stamina to record this video . i hope you appreciate it . ok , so let me rewrite everything , just so we can get back to the problem because when you take the partial fraction detour , you forget -- not even to speak of the problem , you forget what day it is . let 's see , so you get the laplace transform of y is equal to 1 minus e to the minus 2 pi s times what that mess that we just solved for , times -- and i 'll write it like this . 1/3 times 1 over s squared plus 1 minus 1/3 times -- actually , let me write it this way . because i have this s squared plus 4 , so i really want to have a 2 there . so i want to have a 2 in the numerator , so you want to have a 2 over s squared plus 4 . so if i put a 2 in the numerator there , i have to divide this by 2 as well . so let me change this to a 6 . minus 1/6 times 2 is minus 1/3 . so i did that just so i get this in the form of the laplace transform of sine of t. now , let 's see if there 's anything that i can do from here . this is an epic problem . i 'll be amazed if i do n't make a careless mistake while i do this . so we can rewrite everything . let 's see if we can simplify this . and by simplifying it , i 'm just going to make it longer . we can write the laplace transform of y is equal to -- i 'm just going to multiply the 1 out , and then i 'm going to multiply the e to the minus 2 pi s out . so if you multiply the 1 out , you get 1/3 times 1 over s squared plus 1 -- i 'm just multiplying the 1 out -- minus 1/6 -- these are all the 1 's times the 1 -- times 2 over s squared plus 4 . and then i 'm going to multiply the minus e. let me just switch colors , do the minus e. so then you get minus e to the minus 2 pi s over 3 times 1 over s squared plus 1 . and then the minus and the minus cancel out , so you get plus e to the minus 2 pi s over 6 times 2 over s squared plus 4 . now , taking the inverse laplace transform of these things are pretty straightforward . so let 's do that . let 's take the inverse laplace transform of the whole thing . and we get y is equal to the inverse laplace transform of this guy right here , is just 1/3 sine of t -- i do n't have to write a parentheses there -- sine of t , and then this is minus 1/6 times -- this is the laplace transform of sine of 2t . that 's that term right there . now , these are almost the same , but we have this little pesky character over here . we have this e to the minus 2 pi s. and there , we just have to remind ourselves -- i 'll write it here in the bottom . we just have to remind ourselves that the laplace transform of the unit step function -- i 'll put the pi there , just 2 pi times f of t minus 2 pi -- i should put as the step function of t -- is equal to e to the minus 2 pi s times the laplace transform of just -- or let me just write it this way -- times the laplace transform of f of t. so if we view f of t as just sine of t or sine of 2t , then we can kind of backwards pattern match . and we 'll have to shift it and multiply it by the unit step function . so i want to make that clear . if you did n't have this guy here , the inverse laplace transform of this guy would be the same thing as this guy . it 'd just be sine of t. the inverse laplace transform of this guy would be sine of 2t . but we have this pesky character here , which essentially , instead of having the inverse laplace transform just being our f of t , it 's going to be our f of t shifted by 2 pi times the unit step function , where it steps up at 2pi . so this is going to be minus 1/3 times the unit step function , where c is 2 pi of t times -- instead of sine of t -- sine of t minus 2pi . and then we 're almost done . i 'll do it in magenta to celebrate it . plus this very last term , which is 1/6 times the unit step function 2 pi of t , the unit step function that steps up at 2 pi times sine of -- and we have to be careful here . wherever we had a t before , we 're going to replace it with a t minus 2 pi . so sine of , instead of 2t , is going to be 2 times t minus 2 pi . and there you have it . we finally have solved our very hairy problem . we could take some time if we want to simplify this a little bit . in fact , we might as well . at the risk of making a careless mistake at the last moment , let me see if i can make any simplifications here . well , we could factor out this guy right here , but other than that , that seems about as simple as we can get . so this is our function of t that satisfies our otherwise simple-looking differential equation that we had up here . this looked fairly straightforward , but we got this big mess to actually satisfy that equation , given those initial conditions that we had initially .
let 's solve this differential equation , an interpretation of it . and i actually do a whole playlist on interpretations of differential equations and how you model it , but you know , you can kind of view this is a forcing function . that it 's a weird forcing function of this being applied to some weight with , you know , this is the acceleration term , right ? the second derivative with respect to time is the acceleration .
is the forcing function not allowing the solution to go past certain boundaries ?
let 's apply everything we 've learned to an actual differential equation . instead of just taking laplace transforms and taking their inverse , let 's actually solve a problem . so let 's say that i have the second derivative of my function y plus 4 times my function y is equal to sine of t minus the unit step function 0 up until 2 pi of t times sine of t minus 2 pi . let 's solve this differential equation , an interpretation of it . and i actually do a whole playlist on interpretations of differential equations and how you model it , but you know , you can kind of view this is a forcing function . that it 's a weird forcing function of this being applied to some weight with , you know , this is the acceleration term , right ? the second derivative with respect to time is the acceleration . so the mass would be 1 whatever units , and then as a function of its position , this is probably some type of spring constant . anyway , i wo n't go there . i do n't want to waste your time with the interpretation of it , but let 's solve it . we can do more about interpretations later . so we 're going take the laplace transform of both sides of this equation . so what 's the laplace transform of the left-hand side ? so the laplace transform of the second derivative of y is just s squared , so now i 'm taking the laplace transform of just that . the laplace transform of s squared times the laplace transform of y minus -- lower the degree there once -- minus s times y of 0 minus y prime of 0 . so clearly , i must have to give you some initial conditions in order to do this properly . and then plus 4 times the laplace transform of y is equal to -- what 's the laplace transform of sine of t ? that should be second nature by now . it 's just 1 over s squared plus 1 . and then we have minus the laplace transform of this thing . and i 'll do a little side note here to figure out the laplace transform of this thing right here . and we know , i showed it to you a couple of videos ago , we showed that the laplace transform -- actually i could just write it out here . this is going to be the same thing as the laplace transform of sine of t , but we 're going to have to multiply it by e to the minus -- if you remember that last formula -- e to the minus cs , where c is 2 pi . actually , let me write that down . i decided to write it down , then i decided , oh , no , i do n't want to do this . but let me write that . so the laplace transform of the unit step function that goes up to c times some function shifted by c is equal to e to the minus cs times the laplace transform of just the original function times the laplace transform of f of t. so if we 're taking the laplace transform of this thing , our c is 2 pi . our f of t is just sine of t , right ? so then this is just going to be equal to -- if we just do this piece right here -- it 's going to be equal to e to the minus cs -- our c is 2 pi -- e to the minus 2 pi s times the laplace transform of f of t. f of t is just sine of t before we shifted . this is f of t minus 2 pi . so f of t is just going to be sine of t. so it 's going to be times 1 over s squared plus 1 . this is the laplace transform of sine of t. so let 's go back to where we had left off . so we 've taken the laplace transform of both sides of this equation . and clearly , i have some initial conditions here , so the problem must have given me some and i just forgot to write them down . so let 's see , the initial conditions i 'm given , and they are written kind of in the margin here , they tell us -- i 'll do it in orange , they tell us that y of 0 is equal to 0 , and y prime of 0 is equal to 0 . that makes the math easy . that 's 0 and that 's 0 . so let 's see if i can simplify my equation . so the left-hand side , let 's factor out the laplace transform . so let 's factor out this term and that term . so we get the laplace transform of y times this plus this times s squared plus 4 is equal to the right-hand side . and what 's the right-hand side ? we could simplify this . well , i 'll just write it out . i do n't want to do too many steps at once . it 's 1 over s squared plus 1 and then plus -- or minus actually , this is a minus -- minus the laplace transfer of this thing , which was e to the minus 2 pi s over s squared plus 1 . so if we divide both sides of this equation by the s squared plus 4 , then we get the laplace transform of y is equal to -- and actually , i can just merge these two . they 're the same denominator . so before i even divide by s squared plus 4 , that right-hand side will look like this . it will look like with a denominator of s squared plus 1 and you have a numerator of 1 minus e to the minus 2 pi s. and , of course , we 're dividing both sides of this equation by s squared plus 4 , so we 're going to have to stick that s squared plus 4 over here . now , we 're at the hard part . in order to figure out why , we have to take the inverse laplace transform of this thing . so how do we take the inverse laplace transform of this thing ? that 's where the hard part is always , you know , it makes solving the differential equation 's easy if you know the laplace transforms . so it looks like we 're going to have to do some partial fraction expansion . so let 's see if we can do that . so we can rewrite this equation right here . actually , let 's write it as this , because this 'll kind of simplify our work . let 's factor this whole thing out . so we 're going to write it as 1 minus e to the minus 2 pi s , all of that times -- i 'll do it in orange -- all of that times 1 over s squared plus 1 times s squared plus 4 . now , we need to do some partial fraction expansion to simplify this thing right here . we 're going to do this on the side . maybe i should do this over on the right here . this thing -- let me rewrite it -- 1 over s squared plus 1 times s squared plus 4 should be able to be rewritten as two separate fractions , s squared plus 1 and s squared plus 4 , with the numerators . this one would be as plus b . it 's going to have to have degree 1 , because this is degree 2 . here and then we 'd have cs plus d. and so when you add these two things up , you get as plus b times s squared plus 4 plus cs plus d times s squared plus 1 , all of that over the common denominator . we 've seen this story before . we just have to do some algebra here . as you can tell , these differential equations problems , they require a lot of stamina . you kind of just have to say i will keep moving forward and do the algebra that i need to do in order to get the answer . and you kind of have to get excited about that notion that you have all this algebra to do . so let 's figure it out . so this top can be simplified to as to the third plus bs squared plus 4as plus 4b . and then this one , you end up with cs to the third plus ds squared plus cs plus d. so when you add of these up together , you get -- and this is all the algebra that we have to do , for better , for worse -- a plus c over s to the third plus b plus d times s squared plus 4a plus c times s -- let 's scroll over a little bit -- plus 4b plus d. and now we just have to say , ok , all of this is equal to this thing up here . this is the numerator . we just simplified the numerator . this is the numerator . that 's the numerator right there . and all of this is going to be over your original s squared plus 1 times your s squared plus 4 . and we established that this thing should be -- let me just write this -- that 1 over s squared plus 1 times s squared plus 4 should equal this thing . and then you just pattern match on the coefficients . this is all just intense partial fraction expansion . and you say , look , a plus c is the coefficient of the s cubed terms . i do n't see any s cubed terms here . so a plus c must be equal to 0 . and then you see , ok , b plus d is the coefficient of the s squared terms . i do n't see any s squared terms there . so b plus d must be equal to 0 . 4a plus c , the coefficient of the s terms . i do n't see any s terms over here . so 4a plus c must be equal to 0 . and then we 're almost done . 4b plus d must be the constant terms . there is a constant term there . so 4b plus d is equal to 1 . so let 's see if we can do anything here . if we subtract this from that , we get minus 3a is equal to 0 , or a is equal to 0 . if a is equal to 0 , then c is equals to 0 . and let 's see what we can get here . if we subtract this from that , we get minus 3b . the d 's cancel out . it 's equal to minus 1 , or b is equal to 1/3 . and then , of course , we have d is equal to minus b , if you subtract b from both sides . so d is equal to 1/3 . so all of that work , and we actually have a pretty simple result . our equation , this thing here , can be rewritten as -- the a disappeared . it 's 1/3 over s squared plus 1 . b was the coefficient on the -- let me make it very clear . b was the coefficient on the -- or it was a term on top of the s squared plus 1 , so that 's why i 'm using b there . and then d is minus b , so d is minus 1 . so let me make sure i have that . b is 1/3 minus -- let me make sure i get that right . d is 1/3 . so , sorry , b as in boy is 1/3 , so d is minus 1/3 . so b , there 's a term on top of the s squared plus 1 . and then you have minus d over the minus 1/3 over s squared plus 4 . this takes a lot of stamina to record this video . i hope you appreciate it . ok , so let me rewrite everything , just so we can get back to the problem because when you take the partial fraction detour , you forget -- not even to speak of the problem , you forget what day it is . let 's see , so you get the laplace transform of y is equal to 1 minus e to the minus 2 pi s times what that mess that we just solved for , times -- and i 'll write it like this . 1/3 times 1 over s squared plus 1 minus 1/3 times -- actually , let me write it this way . because i have this s squared plus 4 , so i really want to have a 2 there . so i want to have a 2 in the numerator , so you want to have a 2 over s squared plus 4 . so if i put a 2 in the numerator there , i have to divide this by 2 as well . so let me change this to a 6 . minus 1/6 times 2 is minus 1/3 . so i did that just so i get this in the form of the laplace transform of sine of t. now , let 's see if there 's anything that i can do from here . this is an epic problem . i 'll be amazed if i do n't make a careless mistake while i do this . so we can rewrite everything . let 's see if we can simplify this . and by simplifying it , i 'm just going to make it longer . we can write the laplace transform of y is equal to -- i 'm just going to multiply the 1 out , and then i 'm going to multiply the e to the minus 2 pi s out . so if you multiply the 1 out , you get 1/3 times 1 over s squared plus 1 -- i 'm just multiplying the 1 out -- minus 1/6 -- these are all the 1 's times the 1 -- times 2 over s squared plus 4 . and then i 'm going to multiply the minus e. let me just switch colors , do the minus e. so then you get minus e to the minus 2 pi s over 3 times 1 over s squared plus 1 . and then the minus and the minus cancel out , so you get plus e to the minus 2 pi s over 6 times 2 over s squared plus 4 . now , taking the inverse laplace transform of these things are pretty straightforward . so let 's do that . let 's take the inverse laplace transform of the whole thing . and we get y is equal to the inverse laplace transform of this guy right here , is just 1/3 sine of t -- i do n't have to write a parentheses there -- sine of t , and then this is minus 1/6 times -- this is the laplace transform of sine of 2t . that 's that term right there . now , these are almost the same , but we have this little pesky character over here . we have this e to the minus 2 pi s. and there , we just have to remind ourselves -- i 'll write it here in the bottom . we just have to remind ourselves that the laplace transform of the unit step function -- i 'll put the pi there , just 2 pi times f of t minus 2 pi -- i should put as the step function of t -- is equal to e to the minus 2 pi s times the laplace transform of just -- or let me just write it this way -- times the laplace transform of f of t. so if we view f of t as just sine of t or sine of 2t , then we can kind of backwards pattern match . and we 'll have to shift it and multiply it by the unit step function . so i want to make that clear . if you did n't have this guy here , the inverse laplace transform of this guy would be the same thing as this guy . it 'd just be sine of t. the inverse laplace transform of this guy would be sine of 2t . but we have this pesky character here , which essentially , instead of having the inverse laplace transform just being our f of t , it 's going to be our f of t shifted by 2 pi times the unit step function , where it steps up at 2pi . so this is going to be minus 1/3 times the unit step function , where c is 2 pi of t times -- instead of sine of t -- sine of t minus 2pi . and then we 're almost done . i 'll do it in magenta to celebrate it . plus this very last term , which is 1/6 times the unit step function 2 pi of t , the unit step function that steps up at 2 pi times sine of -- and we have to be careful here . wherever we had a t before , we 're going to replace it with a t minus 2 pi . so sine of , instead of 2t , is going to be 2 times t minus 2 pi . and there you have it . we finally have solved our very hairy problem . we could take some time if we want to simplify this a little bit . in fact , we might as well . at the risk of making a careless mistake at the last moment , let me see if i can make any simplifications here . well , we could factor out this guy right here , but other than that , that seems about as simple as we can get . so this is our function of t that satisfies our otherwise simple-looking differential equation that we had up here . this looked fairly straightforward , but we got this big mess to actually satisfy that equation , given those initial conditions that we had initially .
so let 's say that i have the second derivative of my function y plus 4 times my function y is equal to sine of t minus the unit step function 0 up until 2 pi of t times sine of t minus 2 pi . let 's solve this differential equation , an interpretation of it . and i actually do a whole playlist on interpretations of differential equations and how you model it , but you know , you can kind of view this is a forcing function .
can laplace transformation give general solution of differential equation ?
let 's apply everything we 've learned to an actual differential equation . instead of just taking laplace transforms and taking their inverse , let 's actually solve a problem . so let 's say that i have the second derivative of my function y plus 4 times my function y is equal to sine of t minus the unit step function 0 up until 2 pi of t times sine of t minus 2 pi . let 's solve this differential equation , an interpretation of it . and i actually do a whole playlist on interpretations of differential equations and how you model it , but you know , you can kind of view this is a forcing function . that it 's a weird forcing function of this being applied to some weight with , you know , this is the acceleration term , right ? the second derivative with respect to time is the acceleration . so the mass would be 1 whatever units , and then as a function of its position , this is probably some type of spring constant . anyway , i wo n't go there . i do n't want to waste your time with the interpretation of it , but let 's solve it . we can do more about interpretations later . so we 're going take the laplace transform of both sides of this equation . so what 's the laplace transform of the left-hand side ? so the laplace transform of the second derivative of y is just s squared , so now i 'm taking the laplace transform of just that . the laplace transform of s squared times the laplace transform of y minus -- lower the degree there once -- minus s times y of 0 minus y prime of 0 . so clearly , i must have to give you some initial conditions in order to do this properly . and then plus 4 times the laplace transform of y is equal to -- what 's the laplace transform of sine of t ? that should be second nature by now . it 's just 1 over s squared plus 1 . and then we have minus the laplace transform of this thing . and i 'll do a little side note here to figure out the laplace transform of this thing right here . and we know , i showed it to you a couple of videos ago , we showed that the laplace transform -- actually i could just write it out here . this is going to be the same thing as the laplace transform of sine of t , but we 're going to have to multiply it by e to the minus -- if you remember that last formula -- e to the minus cs , where c is 2 pi . actually , let me write that down . i decided to write it down , then i decided , oh , no , i do n't want to do this . but let me write that . so the laplace transform of the unit step function that goes up to c times some function shifted by c is equal to e to the minus cs times the laplace transform of just the original function times the laplace transform of f of t. so if we 're taking the laplace transform of this thing , our c is 2 pi . our f of t is just sine of t , right ? so then this is just going to be equal to -- if we just do this piece right here -- it 's going to be equal to e to the minus cs -- our c is 2 pi -- e to the minus 2 pi s times the laplace transform of f of t. f of t is just sine of t before we shifted . this is f of t minus 2 pi . so f of t is just going to be sine of t. so it 's going to be times 1 over s squared plus 1 . this is the laplace transform of sine of t. so let 's go back to where we had left off . so we 've taken the laplace transform of both sides of this equation . and clearly , i have some initial conditions here , so the problem must have given me some and i just forgot to write them down . so let 's see , the initial conditions i 'm given , and they are written kind of in the margin here , they tell us -- i 'll do it in orange , they tell us that y of 0 is equal to 0 , and y prime of 0 is equal to 0 . that makes the math easy . that 's 0 and that 's 0 . so let 's see if i can simplify my equation . so the left-hand side , let 's factor out the laplace transform . so let 's factor out this term and that term . so we get the laplace transform of y times this plus this times s squared plus 4 is equal to the right-hand side . and what 's the right-hand side ? we could simplify this . well , i 'll just write it out . i do n't want to do too many steps at once . it 's 1 over s squared plus 1 and then plus -- or minus actually , this is a minus -- minus the laplace transfer of this thing , which was e to the minus 2 pi s over s squared plus 1 . so if we divide both sides of this equation by the s squared plus 4 , then we get the laplace transform of y is equal to -- and actually , i can just merge these two . they 're the same denominator . so before i even divide by s squared plus 4 , that right-hand side will look like this . it will look like with a denominator of s squared plus 1 and you have a numerator of 1 minus e to the minus 2 pi s. and , of course , we 're dividing both sides of this equation by s squared plus 4 , so we 're going to have to stick that s squared plus 4 over here . now , we 're at the hard part . in order to figure out why , we have to take the inverse laplace transform of this thing . so how do we take the inverse laplace transform of this thing ? that 's where the hard part is always , you know , it makes solving the differential equation 's easy if you know the laplace transforms . so it looks like we 're going to have to do some partial fraction expansion . so let 's see if we can do that . so we can rewrite this equation right here . actually , let 's write it as this , because this 'll kind of simplify our work . let 's factor this whole thing out . so we 're going to write it as 1 minus e to the minus 2 pi s , all of that times -- i 'll do it in orange -- all of that times 1 over s squared plus 1 times s squared plus 4 . now , we need to do some partial fraction expansion to simplify this thing right here . we 're going to do this on the side . maybe i should do this over on the right here . this thing -- let me rewrite it -- 1 over s squared plus 1 times s squared plus 4 should be able to be rewritten as two separate fractions , s squared plus 1 and s squared plus 4 , with the numerators . this one would be as plus b . it 's going to have to have degree 1 , because this is degree 2 . here and then we 'd have cs plus d. and so when you add these two things up , you get as plus b times s squared plus 4 plus cs plus d times s squared plus 1 , all of that over the common denominator . we 've seen this story before . we just have to do some algebra here . as you can tell , these differential equations problems , they require a lot of stamina . you kind of just have to say i will keep moving forward and do the algebra that i need to do in order to get the answer . and you kind of have to get excited about that notion that you have all this algebra to do . so let 's figure it out . so this top can be simplified to as to the third plus bs squared plus 4as plus 4b . and then this one , you end up with cs to the third plus ds squared plus cs plus d. so when you add of these up together , you get -- and this is all the algebra that we have to do , for better , for worse -- a plus c over s to the third plus b plus d times s squared plus 4a plus c times s -- let 's scroll over a little bit -- plus 4b plus d. and now we just have to say , ok , all of this is equal to this thing up here . this is the numerator . we just simplified the numerator . this is the numerator . that 's the numerator right there . and all of this is going to be over your original s squared plus 1 times your s squared plus 4 . and we established that this thing should be -- let me just write this -- that 1 over s squared plus 1 times s squared plus 4 should equal this thing . and then you just pattern match on the coefficients . this is all just intense partial fraction expansion . and you say , look , a plus c is the coefficient of the s cubed terms . i do n't see any s cubed terms here . so a plus c must be equal to 0 . and then you see , ok , b plus d is the coefficient of the s squared terms . i do n't see any s squared terms there . so b plus d must be equal to 0 . 4a plus c , the coefficient of the s terms . i do n't see any s terms over here . so 4a plus c must be equal to 0 . and then we 're almost done . 4b plus d must be the constant terms . there is a constant term there . so 4b plus d is equal to 1 . so let 's see if we can do anything here . if we subtract this from that , we get minus 3a is equal to 0 , or a is equal to 0 . if a is equal to 0 , then c is equals to 0 . and let 's see what we can get here . if we subtract this from that , we get minus 3b . the d 's cancel out . it 's equal to minus 1 , or b is equal to 1/3 . and then , of course , we have d is equal to minus b , if you subtract b from both sides . so d is equal to 1/3 . so all of that work , and we actually have a pretty simple result . our equation , this thing here , can be rewritten as -- the a disappeared . it 's 1/3 over s squared plus 1 . b was the coefficient on the -- let me make it very clear . b was the coefficient on the -- or it was a term on top of the s squared plus 1 , so that 's why i 'm using b there . and then d is minus b , so d is minus 1 . so let me make sure i have that . b is 1/3 minus -- let me make sure i get that right . d is 1/3 . so , sorry , b as in boy is 1/3 , so d is minus 1/3 . so b , there 's a term on top of the s squared plus 1 . and then you have minus d over the minus 1/3 over s squared plus 4 . this takes a lot of stamina to record this video . i hope you appreciate it . ok , so let me rewrite everything , just so we can get back to the problem because when you take the partial fraction detour , you forget -- not even to speak of the problem , you forget what day it is . let 's see , so you get the laplace transform of y is equal to 1 minus e to the minus 2 pi s times what that mess that we just solved for , times -- and i 'll write it like this . 1/3 times 1 over s squared plus 1 minus 1/3 times -- actually , let me write it this way . because i have this s squared plus 4 , so i really want to have a 2 there . so i want to have a 2 in the numerator , so you want to have a 2 over s squared plus 4 . so if i put a 2 in the numerator there , i have to divide this by 2 as well . so let me change this to a 6 . minus 1/6 times 2 is minus 1/3 . so i did that just so i get this in the form of the laplace transform of sine of t. now , let 's see if there 's anything that i can do from here . this is an epic problem . i 'll be amazed if i do n't make a careless mistake while i do this . so we can rewrite everything . let 's see if we can simplify this . and by simplifying it , i 'm just going to make it longer . we can write the laplace transform of y is equal to -- i 'm just going to multiply the 1 out , and then i 'm going to multiply the e to the minus 2 pi s out . so if you multiply the 1 out , you get 1/3 times 1 over s squared plus 1 -- i 'm just multiplying the 1 out -- minus 1/6 -- these are all the 1 's times the 1 -- times 2 over s squared plus 4 . and then i 'm going to multiply the minus e. let me just switch colors , do the minus e. so then you get minus e to the minus 2 pi s over 3 times 1 over s squared plus 1 . and then the minus and the minus cancel out , so you get plus e to the minus 2 pi s over 6 times 2 over s squared plus 4 . now , taking the inverse laplace transform of these things are pretty straightforward . so let 's do that . let 's take the inverse laplace transform of the whole thing . and we get y is equal to the inverse laplace transform of this guy right here , is just 1/3 sine of t -- i do n't have to write a parentheses there -- sine of t , and then this is minus 1/6 times -- this is the laplace transform of sine of 2t . that 's that term right there . now , these are almost the same , but we have this little pesky character over here . we have this e to the minus 2 pi s. and there , we just have to remind ourselves -- i 'll write it here in the bottom . we just have to remind ourselves that the laplace transform of the unit step function -- i 'll put the pi there , just 2 pi times f of t minus 2 pi -- i should put as the step function of t -- is equal to e to the minus 2 pi s times the laplace transform of just -- or let me just write it this way -- times the laplace transform of f of t. so if we view f of t as just sine of t or sine of 2t , then we can kind of backwards pattern match . and we 'll have to shift it and multiply it by the unit step function . so i want to make that clear . if you did n't have this guy here , the inverse laplace transform of this guy would be the same thing as this guy . it 'd just be sine of t. the inverse laplace transform of this guy would be sine of 2t . but we have this pesky character here , which essentially , instead of having the inverse laplace transform just being our f of t , it 's going to be our f of t shifted by 2 pi times the unit step function , where it steps up at 2pi . so this is going to be minus 1/3 times the unit step function , where c is 2 pi of t times -- instead of sine of t -- sine of t minus 2pi . and then we 're almost done . i 'll do it in magenta to celebrate it . plus this very last term , which is 1/6 times the unit step function 2 pi of t , the unit step function that steps up at 2 pi times sine of -- and we have to be careful here . wherever we had a t before , we 're going to replace it with a t minus 2 pi . so sine of , instead of 2t , is going to be 2 times t minus 2 pi . and there you have it . we finally have solved our very hairy problem . we could take some time if we want to simplify this a little bit . in fact , we might as well . at the risk of making a careless mistake at the last moment , let me see if i can make any simplifications here . well , we could factor out this guy right here , but other than that , that seems about as simple as we can get . so this is our function of t that satisfies our otherwise simple-looking differential equation that we had up here . this looked fairly straightforward , but we got this big mess to actually satisfy that equation , given those initial conditions that we had initially .
so then this is just going to be equal to -- if we just do this piece right here -- it 's going to be equal to e to the minus cs -- our c is 2 pi -- e to the minus 2 pi s times the laplace transform of f of t. f of t is just sine of t before we shifted . this is f of t minus 2 pi . so f of t is just going to be sine of t. so it 's going to be times 1 over s squared plus 1 .
0 , should n't the f ( t ) in blue be f ( s ) instead ?
let 's apply everything we 've learned to an actual differential equation . instead of just taking laplace transforms and taking their inverse , let 's actually solve a problem . so let 's say that i have the second derivative of my function y plus 4 times my function y is equal to sine of t minus the unit step function 0 up until 2 pi of t times sine of t minus 2 pi . let 's solve this differential equation , an interpretation of it . and i actually do a whole playlist on interpretations of differential equations and how you model it , but you know , you can kind of view this is a forcing function . that it 's a weird forcing function of this being applied to some weight with , you know , this is the acceleration term , right ? the second derivative with respect to time is the acceleration . so the mass would be 1 whatever units , and then as a function of its position , this is probably some type of spring constant . anyway , i wo n't go there . i do n't want to waste your time with the interpretation of it , but let 's solve it . we can do more about interpretations later . so we 're going take the laplace transform of both sides of this equation . so what 's the laplace transform of the left-hand side ? so the laplace transform of the second derivative of y is just s squared , so now i 'm taking the laplace transform of just that . the laplace transform of s squared times the laplace transform of y minus -- lower the degree there once -- minus s times y of 0 minus y prime of 0 . so clearly , i must have to give you some initial conditions in order to do this properly . and then plus 4 times the laplace transform of y is equal to -- what 's the laplace transform of sine of t ? that should be second nature by now . it 's just 1 over s squared plus 1 . and then we have minus the laplace transform of this thing . and i 'll do a little side note here to figure out the laplace transform of this thing right here . and we know , i showed it to you a couple of videos ago , we showed that the laplace transform -- actually i could just write it out here . this is going to be the same thing as the laplace transform of sine of t , but we 're going to have to multiply it by e to the minus -- if you remember that last formula -- e to the minus cs , where c is 2 pi . actually , let me write that down . i decided to write it down , then i decided , oh , no , i do n't want to do this . but let me write that . so the laplace transform of the unit step function that goes up to c times some function shifted by c is equal to e to the minus cs times the laplace transform of just the original function times the laplace transform of f of t. so if we 're taking the laplace transform of this thing , our c is 2 pi . our f of t is just sine of t , right ? so then this is just going to be equal to -- if we just do this piece right here -- it 's going to be equal to e to the minus cs -- our c is 2 pi -- e to the minus 2 pi s times the laplace transform of f of t. f of t is just sine of t before we shifted . this is f of t minus 2 pi . so f of t is just going to be sine of t. so it 's going to be times 1 over s squared plus 1 . this is the laplace transform of sine of t. so let 's go back to where we had left off . so we 've taken the laplace transform of both sides of this equation . and clearly , i have some initial conditions here , so the problem must have given me some and i just forgot to write them down . so let 's see , the initial conditions i 'm given , and they are written kind of in the margin here , they tell us -- i 'll do it in orange , they tell us that y of 0 is equal to 0 , and y prime of 0 is equal to 0 . that makes the math easy . that 's 0 and that 's 0 . so let 's see if i can simplify my equation . so the left-hand side , let 's factor out the laplace transform . so let 's factor out this term and that term . so we get the laplace transform of y times this plus this times s squared plus 4 is equal to the right-hand side . and what 's the right-hand side ? we could simplify this . well , i 'll just write it out . i do n't want to do too many steps at once . it 's 1 over s squared plus 1 and then plus -- or minus actually , this is a minus -- minus the laplace transfer of this thing , which was e to the minus 2 pi s over s squared plus 1 . so if we divide both sides of this equation by the s squared plus 4 , then we get the laplace transform of y is equal to -- and actually , i can just merge these two . they 're the same denominator . so before i even divide by s squared plus 4 , that right-hand side will look like this . it will look like with a denominator of s squared plus 1 and you have a numerator of 1 minus e to the minus 2 pi s. and , of course , we 're dividing both sides of this equation by s squared plus 4 , so we 're going to have to stick that s squared plus 4 over here . now , we 're at the hard part . in order to figure out why , we have to take the inverse laplace transform of this thing . so how do we take the inverse laplace transform of this thing ? that 's where the hard part is always , you know , it makes solving the differential equation 's easy if you know the laplace transforms . so it looks like we 're going to have to do some partial fraction expansion . so let 's see if we can do that . so we can rewrite this equation right here . actually , let 's write it as this , because this 'll kind of simplify our work . let 's factor this whole thing out . so we 're going to write it as 1 minus e to the minus 2 pi s , all of that times -- i 'll do it in orange -- all of that times 1 over s squared plus 1 times s squared plus 4 . now , we need to do some partial fraction expansion to simplify this thing right here . we 're going to do this on the side . maybe i should do this over on the right here . this thing -- let me rewrite it -- 1 over s squared plus 1 times s squared plus 4 should be able to be rewritten as two separate fractions , s squared plus 1 and s squared plus 4 , with the numerators . this one would be as plus b . it 's going to have to have degree 1 , because this is degree 2 . here and then we 'd have cs plus d. and so when you add these two things up , you get as plus b times s squared plus 4 plus cs plus d times s squared plus 1 , all of that over the common denominator . we 've seen this story before . we just have to do some algebra here . as you can tell , these differential equations problems , they require a lot of stamina . you kind of just have to say i will keep moving forward and do the algebra that i need to do in order to get the answer . and you kind of have to get excited about that notion that you have all this algebra to do . so let 's figure it out . so this top can be simplified to as to the third plus bs squared plus 4as plus 4b . and then this one , you end up with cs to the third plus ds squared plus cs plus d. so when you add of these up together , you get -- and this is all the algebra that we have to do , for better , for worse -- a plus c over s to the third plus b plus d times s squared plus 4a plus c times s -- let 's scroll over a little bit -- plus 4b plus d. and now we just have to say , ok , all of this is equal to this thing up here . this is the numerator . we just simplified the numerator . this is the numerator . that 's the numerator right there . and all of this is going to be over your original s squared plus 1 times your s squared plus 4 . and we established that this thing should be -- let me just write this -- that 1 over s squared plus 1 times s squared plus 4 should equal this thing . and then you just pattern match on the coefficients . this is all just intense partial fraction expansion . and you say , look , a plus c is the coefficient of the s cubed terms . i do n't see any s cubed terms here . so a plus c must be equal to 0 . and then you see , ok , b plus d is the coefficient of the s squared terms . i do n't see any s squared terms there . so b plus d must be equal to 0 . 4a plus c , the coefficient of the s terms . i do n't see any s terms over here . so 4a plus c must be equal to 0 . and then we 're almost done . 4b plus d must be the constant terms . there is a constant term there . so 4b plus d is equal to 1 . so let 's see if we can do anything here . if we subtract this from that , we get minus 3a is equal to 0 , or a is equal to 0 . if a is equal to 0 , then c is equals to 0 . and let 's see what we can get here . if we subtract this from that , we get minus 3b . the d 's cancel out . it 's equal to minus 1 , or b is equal to 1/3 . and then , of course , we have d is equal to minus b , if you subtract b from both sides . so d is equal to 1/3 . so all of that work , and we actually have a pretty simple result . our equation , this thing here , can be rewritten as -- the a disappeared . it 's 1/3 over s squared plus 1 . b was the coefficient on the -- let me make it very clear . b was the coefficient on the -- or it was a term on top of the s squared plus 1 , so that 's why i 'm using b there . and then d is minus b , so d is minus 1 . so let me make sure i have that . b is 1/3 minus -- let me make sure i get that right . d is 1/3 . so , sorry , b as in boy is 1/3 , so d is minus 1/3 . so b , there 's a term on top of the s squared plus 1 . and then you have minus d over the minus 1/3 over s squared plus 4 . this takes a lot of stamina to record this video . i hope you appreciate it . ok , so let me rewrite everything , just so we can get back to the problem because when you take the partial fraction detour , you forget -- not even to speak of the problem , you forget what day it is . let 's see , so you get the laplace transform of y is equal to 1 minus e to the minus 2 pi s times what that mess that we just solved for , times -- and i 'll write it like this . 1/3 times 1 over s squared plus 1 minus 1/3 times -- actually , let me write it this way . because i have this s squared plus 4 , so i really want to have a 2 there . so i want to have a 2 in the numerator , so you want to have a 2 over s squared plus 4 . so if i put a 2 in the numerator there , i have to divide this by 2 as well . so let me change this to a 6 . minus 1/6 times 2 is minus 1/3 . so i did that just so i get this in the form of the laplace transform of sine of t. now , let 's see if there 's anything that i can do from here . this is an epic problem . i 'll be amazed if i do n't make a careless mistake while i do this . so we can rewrite everything . let 's see if we can simplify this . and by simplifying it , i 'm just going to make it longer . we can write the laplace transform of y is equal to -- i 'm just going to multiply the 1 out , and then i 'm going to multiply the e to the minus 2 pi s out . so if you multiply the 1 out , you get 1/3 times 1 over s squared plus 1 -- i 'm just multiplying the 1 out -- minus 1/6 -- these are all the 1 's times the 1 -- times 2 over s squared plus 4 . and then i 'm going to multiply the minus e. let me just switch colors , do the minus e. so then you get minus e to the minus 2 pi s over 3 times 1 over s squared plus 1 . and then the minus and the minus cancel out , so you get plus e to the minus 2 pi s over 6 times 2 over s squared plus 4 . now , taking the inverse laplace transform of these things are pretty straightforward . so let 's do that . let 's take the inverse laplace transform of the whole thing . and we get y is equal to the inverse laplace transform of this guy right here , is just 1/3 sine of t -- i do n't have to write a parentheses there -- sine of t , and then this is minus 1/6 times -- this is the laplace transform of sine of 2t . that 's that term right there . now , these are almost the same , but we have this little pesky character over here . we have this e to the minus 2 pi s. and there , we just have to remind ourselves -- i 'll write it here in the bottom . we just have to remind ourselves that the laplace transform of the unit step function -- i 'll put the pi there , just 2 pi times f of t minus 2 pi -- i should put as the step function of t -- is equal to e to the minus 2 pi s times the laplace transform of just -- or let me just write it this way -- times the laplace transform of f of t. so if we view f of t as just sine of t or sine of 2t , then we can kind of backwards pattern match . and we 'll have to shift it and multiply it by the unit step function . so i want to make that clear . if you did n't have this guy here , the inverse laplace transform of this guy would be the same thing as this guy . it 'd just be sine of t. the inverse laplace transform of this guy would be sine of 2t . but we have this pesky character here , which essentially , instead of having the inverse laplace transform just being our f of t , it 's going to be our f of t shifted by 2 pi times the unit step function , where it steps up at 2pi . so this is going to be minus 1/3 times the unit step function , where c is 2 pi of t times -- instead of sine of t -- sine of t minus 2pi . and then we 're almost done . i 'll do it in magenta to celebrate it . plus this very last term , which is 1/6 times the unit step function 2 pi of t , the unit step function that steps up at 2 pi times sine of -- and we have to be careful here . wherever we had a t before , we 're going to replace it with a t minus 2 pi . so sine of , instead of 2t , is going to be 2 times t minus 2 pi . and there you have it . we finally have solved our very hairy problem . we could take some time if we want to simplify this a little bit . in fact , we might as well . at the risk of making a careless mistake at the last moment , let me see if i can make any simplifications here . well , we could factor out this guy right here , but other than that , that seems about as simple as we can get . so this is our function of t that satisfies our otherwise simple-looking differential equation that we had up here . this looked fairly straightforward , but we got this big mess to actually satisfy that equation , given those initial conditions that we had initially .
instead of just taking laplace transforms and taking their inverse , let 's actually solve a problem . so let 's say that i have the second derivative of my function y plus 4 times my function y is equal to sine of t minus the unit step function 0 up until 2 pi of t times sine of t minus 2 pi . let 's solve this differential equation , an interpretation of it .
what is a periodic function ?
let 's apply everything we 've learned to an actual differential equation . instead of just taking laplace transforms and taking their inverse , let 's actually solve a problem . so let 's say that i have the second derivative of my function y plus 4 times my function y is equal to sine of t minus the unit step function 0 up until 2 pi of t times sine of t minus 2 pi . let 's solve this differential equation , an interpretation of it . and i actually do a whole playlist on interpretations of differential equations and how you model it , but you know , you can kind of view this is a forcing function . that it 's a weird forcing function of this being applied to some weight with , you know , this is the acceleration term , right ? the second derivative with respect to time is the acceleration . so the mass would be 1 whatever units , and then as a function of its position , this is probably some type of spring constant . anyway , i wo n't go there . i do n't want to waste your time with the interpretation of it , but let 's solve it . we can do more about interpretations later . so we 're going take the laplace transform of both sides of this equation . so what 's the laplace transform of the left-hand side ? so the laplace transform of the second derivative of y is just s squared , so now i 'm taking the laplace transform of just that . the laplace transform of s squared times the laplace transform of y minus -- lower the degree there once -- minus s times y of 0 minus y prime of 0 . so clearly , i must have to give you some initial conditions in order to do this properly . and then plus 4 times the laplace transform of y is equal to -- what 's the laplace transform of sine of t ? that should be second nature by now . it 's just 1 over s squared plus 1 . and then we have minus the laplace transform of this thing . and i 'll do a little side note here to figure out the laplace transform of this thing right here . and we know , i showed it to you a couple of videos ago , we showed that the laplace transform -- actually i could just write it out here . this is going to be the same thing as the laplace transform of sine of t , but we 're going to have to multiply it by e to the minus -- if you remember that last formula -- e to the minus cs , where c is 2 pi . actually , let me write that down . i decided to write it down , then i decided , oh , no , i do n't want to do this . but let me write that . so the laplace transform of the unit step function that goes up to c times some function shifted by c is equal to e to the minus cs times the laplace transform of just the original function times the laplace transform of f of t. so if we 're taking the laplace transform of this thing , our c is 2 pi . our f of t is just sine of t , right ? so then this is just going to be equal to -- if we just do this piece right here -- it 's going to be equal to e to the minus cs -- our c is 2 pi -- e to the minus 2 pi s times the laplace transform of f of t. f of t is just sine of t before we shifted . this is f of t minus 2 pi . so f of t is just going to be sine of t. so it 's going to be times 1 over s squared plus 1 . this is the laplace transform of sine of t. so let 's go back to where we had left off . so we 've taken the laplace transform of both sides of this equation . and clearly , i have some initial conditions here , so the problem must have given me some and i just forgot to write them down . so let 's see , the initial conditions i 'm given , and they are written kind of in the margin here , they tell us -- i 'll do it in orange , they tell us that y of 0 is equal to 0 , and y prime of 0 is equal to 0 . that makes the math easy . that 's 0 and that 's 0 . so let 's see if i can simplify my equation . so the left-hand side , let 's factor out the laplace transform . so let 's factor out this term and that term . so we get the laplace transform of y times this plus this times s squared plus 4 is equal to the right-hand side . and what 's the right-hand side ? we could simplify this . well , i 'll just write it out . i do n't want to do too many steps at once . it 's 1 over s squared plus 1 and then plus -- or minus actually , this is a minus -- minus the laplace transfer of this thing , which was e to the minus 2 pi s over s squared plus 1 . so if we divide both sides of this equation by the s squared plus 4 , then we get the laplace transform of y is equal to -- and actually , i can just merge these two . they 're the same denominator . so before i even divide by s squared plus 4 , that right-hand side will look like this . it will look like with a denominator of s squared plus 1 and you have a numerator of 1 minus e to the minus 2 pi s. and , of course , we 're dividing both sides of this equation by s squared plus 4 , so we 're going to have to stick that s squared plus 4 over here . now , we 're at the hard part . in order to figure out why , we have to take the inverse laplace transform of this thing . so how do we take the inverse laplace transform of this thing ? that 's where the hard part is always , you know , it makes solving the differential equation 's easy if you know the laplace transforms . so it looks like we 're going to have to do some partial fraction expansion . so let 's see if we can do that . so we can rewrite this equation right here . actually , let 's write it as this , because this 'll kind of simplify our work . let 's factor this whole thing out . so we 're going to write it as 1 minus e to the minus 2 pi s , all of that times -- i 'll do it in orange -- all of that times 1 over s squared plus 1 times s squared plus 4 . now , we need to do some partial fraction expansion to simplify this thing right here . we 're going to do this on the side . maybe i should do this over on the right here . this thing -- let me rewrite it -- 1 over s squared plus 1 times s squared plus 4 should be able to be rewritten as two separate fractions , s squared plus 1 and s squared plus 4 , with the numerators . this one would be as plus b . it 's going to have to have degree 1 , because this is degree 2 . here and then we 'd have cs plus d. and so when you add these two things up , you get as plus b times s squared plus 4 plus cs plus d times s squared plus 1 , all of that over the common denominator . we 've seen this story before . we just have to do some algebra here . as you can tell , these differential equations problems , they require a lot of stamina . you kind of just have to say i will keep moving forward and do the algebra that i need to do in order to get the answer . and you kind of have to get excited about that notion that you have all this algebra to do . so let 's figure it out . so this top can be simplified to as to the third plus bs squared plus 4as plus 4b . and then this one , you end up with cs to the third plus ds squared plus cs plus d. so when you add of these up together , you get -- and this is all the algebra that we have to do , for better , for worse -- a plus c over s to the third plus b plus d times s squared plus 4a plus c times s -- let 's scroll over a little bit -- plus 4b plus d. and now we just have to say , ok , all of this is equal to this thing up here . this is the numerator . we just simplified the numerator . this is the numerator . that 's the numerator right there . and all of this is going to be over your original s squared plus 1 times your s squared plus 4 . and we established that this thing should be -- let me just write this -- that 1 over s squared plus 1 times s squared plus 4 should equal this thing . and then you just pattern match on the coefficients . this is all just intense partial fraction expansion . and you say , look , a plus c is the coefficient of the s cubed terms . i do n't see any s cubed terms here . so a plus c must be equal to 0 . and then you see , ok , b plus d is the coefficient of the s squared terms . i do n't see any s squared terms there . so b plus d must be equal to 0 . 4a plus c , the coefficient of the s terms . i do n't see any s terms over here . so 4a plus c must be equal to 0 . and then we 're almost done . 4b plus d must be the constant terms . there is a constant term there . so 4b plus d is equal to 1 . so let 's see if we can do anything here . if we subtract this from that , we get minus 3a is equal to 0 , or a is equal to 0 . if a is equal to 0 , then c is equals to 0 . and let 's see what we can get here . if we subtract this from that , we get minus 3b . the d 's cancel out . it 's equal to minus 1 , or b is equal to 1/3 . and then , of course , we have d is equal to minus b , if you subtract b from both sides . so d is equal to 1/3 . so all of that work , and we actually have a pretty simple result . our equation , this thing here , can be rewritten as -- the a disappeared . it 's 1/3 over s squared plus 1 . b was the coefficient on the -- let me make it very clear . b was the coefficient on the -- or it was a term on top of the s squared plus 1 , so that 's why i 'm using b there . and then d is minus b , so d is minus 1 . so let me make sure i have that . b is 1/3 minus -- let me make sure i get that right . d is 1/3 . so , sorry , b as in boy is 1/3 , so d is minus 1/3 . so b , there 's a term on top of the s squared plus 1 . and then you have minus d over the minus 1/3 over s squared plus 4 . this takes a lot of stamina to record this video . i hope you appreciate it . ok , so let me rewrite everything , just so we can get back to the problem because when you take the partial fraction detour , you forget -- not even to speak of the problem , you forget what day it is . let 's see , so you get the laplace transform of y is equal to 1 minus e to the minus 2 pi s times what that mess that we just solved for , times -- and i 'll write it like this . 1/3 times 1 over s squared plus 1 minus 1/3 times -- actually , let me write it this way . because i have this s squared plus 4 , so i really want to have a 2 there . so i want to have a 2 in the numerator , so you want to have a 2 over s squared plus 4 . so if i put a 2 in the numerator there , i have to divide this by 2 as well . so let me change this to a 6 . minus 1/6 times 2 is minus 1/3 . so i did that just so i get this in the form of the laplace transform of sine of t. now , let 's see if there 's anything that i can do from here . this is an epic problem . i 'll be amazed if i do n't make a careless mistake while i do this . so we can rewrite everything . let 's see if we can simplify this . and by simplifying it , i 'm just going to make it longer . we can write the laplace transform of y is equal to -- i 'm just going to multiply the 1 out , and then i 'm going to multiply the e to the minus 2 pi s out . so if you multiply the 1 out , you get 1/3 times 1 over s squared plus 1 -- i 'm just multiplying the 1 out -- minus 1/6 -- these are all the 1 's times the 1 -- times 2 over s squared plus 4 . and then i 'm going to multiply the minus e. let me just switch colors , do the minus e. so then you get minus e to the minus 2 pi s over 3 times 1 over s squared plus 1 . and then the minus and the minus cancel out , so you get plus e to the minus 2 pi s over 6 times 2 over s squared plus 4 . now , taking the inverse laplace transform of these things are pretty straightforward . so let 's do that . let 's take the inverse laplace transform of the whole thing . and we get y is equal to the inverse laplace transform of this guy right here , is just 1/3 sine of t -- i do n't have to write a parentheses there -- sine of t , and then this is minus 1/6 times -- this is the laplace transform of sine of 2t . that 's that term right there . now , these are almost the same , but we have this little pesky character over here . we have this e to the minus 2 pi s. and there , we just have to remind ourselves -- i 'll write it here in the bottom . we just have to remind ourselves that the laplace transform of the unit step function -- i 'll put the pi there , just 2 pi times f of t minus 2 pi -- i should put as the step function of t -- is equal to e to the minus 2 pi s times the laplace transform of just -- or let me just write it this way -- times the laplace transform of f of t. so if we view f of t as just sine of t or sine of 2t , then we can kind of backwards pattern match . and we 'll have to shift it and multiply it by the unit step function . so i want to make that clear . if you did n't have this guy here , the inverse laplace transform of this guy would be the same thing as this guy . it 'd just be sine of t. the inverse laplace transform of this guy would be sine of 2t . but we have this pesky character here , which essentially , instead of having the inverse laplace transform just being our f of t , it 's going to be our f of t shifted by 2 pi times the unit step function , where it steps up at 2pi . so this is going to be minus 1/3 times the unit step function , where c is 2 pi of t times -- instead of sine of t -- sine of t minus 2pi . and then we 're almost done . i 'll do it in magenta to celebrate it . plus this very last term , which is 1/6 times the unit step function 2 pi of t , the unit step function that steps up at 2 pi times sine of -- and we have to be careful here . wherever we had a t before , we 're going to replace it with a t minus 2 pi . so sine of , instead of 2t , is going to be 2 times t minus 2 pi . and there you have it . we finally have solved our very hairy problem . we could take some time if we want to simplify this a little bit . in fact , we might as well . at the risk of making a careless mistake at the last moment , let me see if i can make any simplifications here . well , we could factor out this guy right here , but other than that , that seems about as simple as we can get . so this is our function of t that satisfies our otherwise simple-looking differential equation that we had up here . this looked fairly straightforward , but we got this big mess to actually satisfy that equation , given those initial conditions that we had initially .
so then this is just going to be equal to -- if we just do this piece right here -- it 's going to be equal to e to the minus cs -- our c is 2 pi -- e to the minus 2 pi s times the laplace transform of f of t. f of t is just sine of t before we shifted . this is f of t minus 2 pi . so f of t is just going to be sine of t. so it 's going to be times 1 over s squared plus 1 .
at 2*pi at any given point this term will equal zero right ?
let 's apply everything we 've learned to an actual differential equation . instead of just taking laplace transforms and taking their inverse , let 's actually solve a problem . so let 's say that i have the second derivative of my function y plus 4 times my function y is equal to sine of t minus the unit step function 0 up until 2 pi of t times sine of t minus 2 pi . let 's solve this differential equation , an interpretation of it . and i actually do a whole playlist on interpretations of differential equations and how you model it , but you know , you can kind of view this is a forcing function . that it 's a weird forcing function of this being applied to some weight with , you know , this is the acceleration term , right ? the second derivative with respect to time is the acceleration . so the mass would be 1 whatever units , and then as a function of its position , this is probably some type of spring constant . anyway , i wo n't go there . i do n't want to waste your time with the interpretation of it , but let 's solve it . we can do more about interpretations later . so we 're going take the laplace transform of both sides of this equation . so what 's the laplace transform of the left-hand side ? so the laplace transform of the second derivative of y is just s squared , so now i 'm taking the laplace transform of just that . the laplace transform of s squared times the laplace transform of y minus -- lower the degree there once -- minus s times y of 0 minus y prime of 0 . so clearly , i must have to give you some initial conditions in order to do this properly . and then plus 4 times the laplace transform of y is equal to -- what 's the laplace transform of sine of t ? that should be second nature by now . it 's just 1 over s squared plus 1 . and then we have minus the laplace transform of this thing . and i 'll do a little side note here to figure out the laplace transform of this thing right here . and we know , i showed it to you a couple of videos ago , we showed that the laplace transform -- actually i could just write it out here . this is going to be the same thing as the laplace transform of sine of t , but we 're going to have to multiply it by e to the minus -- if you remember that last formula -- e to the minus cs , where c is 2 pi . actually , let me write that down . i decided to write it down , then i decided , oh , no , i do n't want to do this . but let me write that . so the laplace transform of the unit step function that goes up to c times some function shifted by c is equal to e to the minus cs times the laplace transform of just the original function times the laplace transform of f of t. so if we 're taking the laplace transform of this thing , our c is 2 pi . our f of t is just sine of t , right ? so then this is just going to be equal to -- if we just do this piece right here -- it 's going to be equal to e to the minus cs -- our c is 2 pi -- e to the minus 2 pi s times the laplace transform of f of t. f of t is just sine of t before we shifted . this is f of t minus 2 pi . so f of t is just going to be sine of t. so it 's going to be times 1 over s squared plus 1 . this is the laplace transform of sine of t. so let 's go back to where we had left off . so we 've taken the laplace transform of both sides of this equation . and clearly , i have some initial conditions here , so the problem must have given me some and i just forgot to write them down . so let 's see , the initial conditions i 'm given , and they are written kind of in the margin here , they tell us -- i 'll do it in orange , they tell us that y of 0 is equal to 0 , and y prime of 0 is equal to 0 . that makes the math easy . that 's 0 and that 's 0 . so let 's see if i can simplify my equation . so the left-hand side , let 's factor out the laplace transform . so let 's factor out this term and that term . so we get the laplace transform of y times this plus this times s squared plus 4 is equal to the right-hand side . and what 's the right-hand side ? we could simplify this . well , i 'll just write it out . i do n't want to do too many steps at once . it 's 1 over s squared plus 1 and then plus -- or minus actually , this is a minus -- minus the laplace transfer of this thing , which was e to the minus 2 pi s over s squared plus 1 . so if we divide both sides of this equation by the s squared plus 4 , then we get the laplace transform of y is equal to -- and actually , i can just merge these two . they 're the same denominator . so before i even divide by s squared plus 4 , that right-hand side will look like this . it will look like with a denominator of s squared plus 1 and you have a numerator of 1 minus e to the minus 2 pi s. and , of course , we 're dividing both sides of this equation by s squared plus 4 , so we 're going to have to stick that s squared plus 4 over here . now , we 're at the hard part . in order to figure out why , we have to take the inverse laplace transform of this thing . so how do we take the inverse laplace transform of this thing ? that 's where the hard part is always , you know , it makes solving the differential equation 's easy if you know the laplace transforms . so it looks like we 're going to have to do some partial fraction expansion . so let 's see if we can do that . so we can rewrite this equation right here . actually , let 's write it as this , because this 'll kind of simplify our work . let 's factor this whole thing out . so we 're going to write it as 1 minus e to the minus 2 pi s , all of that times -- i 'll do it in orange -- all of that times 1 over s squared plus 1 times s squared plus 4 . now , we need to do some partial fraction expansion to simplify this thing right here . we 're going to do this on the side . maybe i should do this over on the right here . this thing -- let me rewrite it -- 1 over s squared plus 1 times s squared plus 4 should be able to be rewritten as two separate fractions , s squared plus 1 and s squared plus 4 , with the numerators . this one would be as plus b . it 's going to have to have degree 1 , because this is degree 2 . here and then we 'd have cs plus d. and so when you add these two things up , you get as plus b times s squared plus 4 plus cs plus d times s squared plus 1 , all of that over the common denominator . we 've seen this story before . we just have to do some algebra here . as you can tell , these differential equations problems , they require a lot of stamina . you kind of just have to say i will keep moving forward and do the algebra that i need to do in order to get the answer . and you kind of have to get excited about that notion that you have all this algebra to do . so let 's figure it out . so this top can be simplified to as to the third plus bs squared plus 4as plus 4b . and then this one , you end up with cs to the third plus ds squared plus cs plus d. so when you add of these up together , you get -- and this is all the algebra that we have to do , for better , for worse -- a plus c over s to the third plus b plus d times s squared plus 4a plus c times s -- let 's scroll over a little bit -- plus 4b plus d. and now we just have to say , ok , all of this is equal to this thing up here . this is the numerator . we just simplified the numerator . this is the numerator . that 's the numerator right there . and all of this is going to be over your original s squared plus 1 times your s squared plus 4 . and we established that this thing should be -- let me just write this -- that 1 over s squared plus 1 times s squared plus 4 should equal this thing . and then you just pattern match on the coefficients . this is all just intense partial fraction expansion . and you say , look , a plus c is the coefficient of the s cubed terms . i do n't see any s cubed terms here . so a plus c must be equal to 0 . and then you see , ok , b plus d is the coefficient of the s squared terms . i do n't see any s squared terms there . so b plus d must be equal to 0 . 4a plus c , the coefficient of the s terms . i do n't see any s terms over here . so 4a plus c must be equal to 0 . and then we 're almost done . 4b plus d must be the constant terms . there is a constant term there . so 4b plus d is equal to 1 . so let 's see if we can do anything here . if we subtract this from that , we get minus 3a is equal to 0 , or a is equal to 0 . if a is equal to 0 , then c is equals to 0 . and let 's see what we can get here . if we subtract this from that , we get minus 3b . the d 's cancel out . it 's equal to minus 1 , or b is equal to 1/3 . and then , of course , we have d is equal to minus b , if you subtract b from both sides . so d is equal to 1/3 . so all of that work , and we actually have a pretty simple result . our equation , this thing here , can be rewritten as -- the a disappeared . it 's 1/3 over s squared plus 1 . b was the coefficient on the -- let me make it very clear . b was the coefficient on the -- or it was a term on top of the s squared plus 1 , so that 's why i 'm using b there . and then d is minus b , so d is minus 1 . so let me make sure i have that . b is 1/3 minus -- let me make sure i get that right . d is 1/3 . so , sorry , b as in boy is 1/3 , so d is minus 1/3 . so b , there 's a term on top of the s squared plus 1 . and then you have minus d over the minus 1/3 over s squared plus 4 . this takes a lot of stamina to record this video . i hope you appreciate it . ok , so let me rewrite everything , just so we can get back to the problem because when you take the partial fraction detour , you forget -- not even to speak of the problem , you forget what day it is . let 's see , so you get the laplace transform of y is equal to 1 minus e to the minus 2 pi s times what that mess that we just solved for , times -- and i 'll write it like this . 1/3 times 1 over s squared plus 1 minus 1/3 times -- actually , let me write it this way . because i have this s squared plus 4 , so i really want to have a 2 there . so i want to have a 2 in the numerator , so you want to have a 2 over s squared plus 4 . so if i put a 2 in the numerator there , i have to divide this by 2 as well . so let me change this to a 6 . minus 1/6 times 2 is minus 1/3 . so i did that just so i get this in the form of the laplace transform of sine of t. now , let 's see if there 's anything that i can do from here . this is an epic problem . i 'll be amazed if i do n't make a careless mistake while i do this . so we can rewrite everything . let 's see if we can simplify this . and by simplifying it , i 'm just going to make it longer . we can write the laplace transform of y is equal to -- i 'm just going to multiply the 1 out , and then i 'm going to multiply the e to the minus 2 pi s out . so if you multiply the 1 out , you get 1/3 times 1 over s squared plus 1 -- i 'm just multiplying the 1 out -- minus 1/6 -- these are all the 1 's times the 1 -- times 2 over s squared plus 4 . and then i 'm going to multiply the minus e. let me just switch colors , do the minus e. so then you get minus e to the minus 2 pi s over 3 times 1 over s squared plus 1 . and then the minus and the minus cancel out , so you get plus e to the minus 2 pi s over 6 times 2 over s squared plus 4 . now , taking the inverse laplace transform of these things are pretty straightforward . so let 's do that . let 's take the inverse laplace transform of the whole thing . and we get y is equal to the inverse laplace transform of this guy right here , is just 1/3 sine of t -- i do n't have to write a parentheses there -- sine of t , and then this is minus 1/6 times -- this is the laplace transform of sine of 2t . that 's that term right there . now , these are almost the same , but we have this little pesky character over here . we have this e to the minus 2 pi s. and there , we just have to remind ourselves -- i 'll write it here in the bottom . we just have to remind ourselves that the laplace transform of the unit step function -- i 'll put the pi there , just 2 pi times f of t minus 2 pi -- i should put as the step function of t -- is equal to e to the minus 2 pi s times the laplace transform of just -- or let me just write it this way -- times the laplace transform of f of t. so if we view f of t as just sine of t or sine of 2t , then we can kind of backwards pattern match . and we 'll have to shift it and multiply it by the unit step function . so i want to make that clear . if you did n't have this guy here , the inverse laplace transform of this guy would be the same thing as this guy . it 'd just be sine of t. the inverse laplace transform of this guy would be sine of 2t . but we have this pesky character here , which essentially , instead of having the inverse laplace transform just being our f of t , it 's going to be our f of t shifted by 2 pi times the unit step function , where it steps up at 2pi . so this is going to be minus 1/3 times the unit step function , where c is 2 pi of t times -- instead of sine of t -- sine of t minus 2pi . and then we 're almost done . i 'll do it in magenta to celebrate it . plus this very last term , which is 1/6 times the unit step function 2 pi of t , the unit step function that steps up at 2 pi times sine of -- and we have to be careful here . wherever we had a t before , we 're going to replace it with a t minus 2 pi . so sine of , instead of 2t , is going to be 2 times t minus 2 pi . and there you have it . we finally have solved our very hairy problem . we could take some time if we want to simplify this a little bit . in fact , we might as well . at the risk of making a careless mistake at the last moment , let me see if i can make any simplifications here . well , we could factor out this guy right here , but other than that , that seems about as simple as we can get . so this is our function of t that satisfies our otherwise simple-looking differential equation that we had up here . this looked fairly straightforward , but we got this big mess to actually satisfy that equation , given those initial conditions that we had initially .
and then you just pattern match on the coefficients . this is all just intense partial fraction expansion . and you say , look , a plus c is the coefficient of the s cubed terms .
does it really matter that you use ( as +b ) and ( cs +d ) as the new numerators in the process of partial fraction decompositon when the numerator on the other side is merely an integer ?
let 's apply everything we 've learned to an actual differential equation . instead of just taking laplace transforms and taking their inverse , let 's actually solve a problem . so let 's say that i have the second derivative of my function y plus 4 times my function y is equal to sine of t minus the unit step function 0 up until 2 pi of t times sine of t minus 2 pi . let 's solve this differential equation , an interpretation of it . and i actually do a whole playlist on interpretations of differential equations and how you model it , but you know , you can kind of view this is a forcing function . that it 's a weird forcing function of this being applied to some weight with , you know , this is the acceleration term , right ? the second derivative with respect to time is the acceleration . so the mass would be 1 whatever units , and then as a function of its position , this is probably some type of spring constant . anyway , i wo n't go there . i do n't want to waste your time with the interpretation of it , but let 's solve it . we can do more about interpretations later . so we 're going take the laplace transform of both sides of this equation . so what 's the laplace transform of the left-hand side ? so the laplace transform of the second derivative of y is just s squared , so now i 'm taking the laplace transform of just that . the laplace transform of s squared times the laplace transform of y minus -- lower the degree there once -- minus s times y of 0 minus y prime of 0 . so clearly , i must have to give you some initial conditions in order to do this properly . and then plus 4 times the laplace transform of y is equal to -- what 's the laplace transform of sine of t ? that should be second nature by now . it 's just 1 over s squared plus 1 . and then we have minus the laplace transform of this thing . and i 'll do a little side note here to figure out the laplace transform of this thing right here . and we know , i showed it to you a couple of videos ago , we showed that the laplace transform -- actually i could just write it out here . this is going to be the same thing as the laplace transform of sine of t , but we 're going to have to multiply it by e to the minus -- if you remember that last formula -- e to the minus cs , where c is 2 pi . actually , let me write that down . i decided to write it down , then i decided , oh , no , i do n't want to do this . but let me write that . so the laplace transform of the unit step function that goes up to c times some function shifted by c is equal to e to the minus cs times the laplace transform of just the original function times the laplace transform of f of t. so if we 're taking the laplace transform of this thing , our c is 2 pi . our f of t is just sine of t , right ? so then this is just going to be equal to -- if we just do this piece right here -- it 's going to be equal to e to the minus cs -- our c is 2 pi -- e to the minus 2 pi s times the laplace transform of f of t. f of t is just sine of t before we shifted . this is f of t minus 2 pi . so f of t is just going to be sine of t. so it 's going to be times 1 over s squared plus 1 . this is the laplace transform of sine of t. so let 's go back to where we had left off . so we 've taken the laplace transform of both sides of this equation . and clearly , i have some initial conditions here , so the problem must have given me some and i just forgot to write them down . so let 's see , the initial conditions i 'm given , and they are written kind of in the margin here , they tell us -- i 'll do it in orange , they tell us that y of 0 is equal to 0 , and y prime of 0 is equal to 0 . that makes the math easy . that 's 0 and that 's 0 . so let 's see if i can simplify my equation . so the left-hand side , let 's factor out the laplace transform . so let 's factor out this term and that term . so we get the laplace transform of y times this plus this times s squared plus 4 is equal to the right-hand side . and what 's the right-hand side ? we could simplify this . well , i 'll just write it out . i do n't want to do too many steps at once . it 's 1 over s squared plus 1 and then plus -- or minus actually , this is a minus -- minus the laplace transfer of this thing , which was e to the minus 2 pi s over s squared plus 1 . so if we divide both sides of this equation by the s squared plus 4 , then we get the laplace transform of y is equal to -- and actually , i can just merge these two . they 're the same denominator . so before i even divide by s squared plus 4 , that right-hand side will look like this . it will look like with a denominator of s squared plus 1 and you have a numerator of 1 minus e to the minus 2 pi s. and , of course , we 're dividing both sides of this equation by s squared plus 4 , so we 're going to have to stick that s squared plus 4 over here . now , we 're at the hard part . in order to figure out why , we have to take the inverse laplace transform of this thing . so how do we take the inverse laplace transform of this thing ? that 's where the hard part is always , you know , it makes solving the differential equation 's easy if you know the laplace transforms . so it looks like we 're going to have to do some partial fraction expansion . so let 's see if we can do that . so we can rewrite this equation right here . actually , let 's write it as this , because this 'll kind of simplify our work . let 's factor this whole thing out . so we 're going to write it as 1 minus e to the minus 2 pi s , all of that times -- i 'll do it in orange -- all of that times 1 over s squared plus 1 times s squared plus 4 . now , we need to do some partial fraction expansion to simplify this thing right here . we 're going to do this on the side . maybe i should do this over on the right here . this thing -- let me rewrite it -- 1 over s squared plus 1 times s squared plus 4 should be able to be rewritten as two separate fractions , s squared plus 1 and s squared plus 4 , with the numerators . this one would be as plus b . it 's going to have to have degree 1 , because this is degree 2 . here and then we 'd have cs plus d. and so when you add these two things up , you get as plus b times s squared plus 4 plus cs plus d times s squared plus 1 , all of that over the common denominator . we 've seen this story before . we just have to do some algebra here . as you can tell , these differential equations problems , they require a lot of stamina . you kind of just have to say i will keep moving forward and do the algebra that i need to do in order to get the answer . and you kind of have to get excited about that notion that you have all this algebra to do . so let 's figure it out . so this top can be simplified to as to the third plus bs squared plus 4as plus 4b . and then this one , you end up with cs to the third plus ds squared plus cs plus d. so when you add of these up together , you get -- and this is all the algebra that we have to do , for better , for worse -- a plus c over s to the third plus b plus d times s squared plus 4a plus c times s -- let 's scroll over a little bit -- plus 4b plus d. and now we just have to say , ok , all of this is equal to this thing up here . this is the numerator . we just simplified the numerator . this is the numerator . that 's the numerator right there . and all of this is going to be over your original s squared plus 1 times your s squared plus 4 . and we established that this thing should be -- let me just write this -- that 1 over s squared plus 1 times s squared plus 4 should equal this thing . and then you just pattern match on the coefficients . this is all just intense partial fraction expansion . and you say , look , a plus c is the coefficient of the s cubed terms . i do n't see any s cubed terms here . so a plus c must be equal to 0 . and then you see , ok , b plus d is the coefficient of the s squared terms . i do n't see any s squared terms there . so b plus d must be equal to 0 . 4a plus c , the coefficient of the s terms . i do n't see any s terms over here . so 4a plus c must be equal to 0 . and then we 're almost done . 4b plus d must be the constant terms . there is a constant term there . so 4b plus d is equal to 1 . so let 's see if we can do anything here . if we subtract this from that , we get minus 3a is equal to 0 , or a is equal to 0 . if a is equal to 0 , then c is equals to 0 . and let 's see what we can get here . if we subtract this from that , we get minus 3b . the d 's cancel out . it 's equal to minus 1 , or b is equal to 1/3 . and then , of course , we have d is equal to minus b , if you subtract b from both sides . so d is equal to 1/3 . so all of that work , and we actually have a pretty simple result . our equation , this thing here , can be rewritten as -- the a disappeared . it 's 1/3 over s squared plus 1 . b was the coefficient on the -- let me make it very clear . b was the coefficient on the -- or it was a term on top of the s squared plus 1 , so that 's why i 'm using b there . and then d is minus b , so d is minus 1 . so let me make sure i have that . b is 1/3 minus -- let me make sure i get that right . d is 1/3 . so , sorry , b as in boy is 1/3 , so d is minus 1/3 . so b , there 's a term on top of the s squared plus 1 . and then you have minus d over the minus 1/3 over s squared plus 4 . this takes a lot of stamina to record this video . i hope you appreciate it . ok , so let me rewrite everything , just so we can get back to the problem because when you take the partial fraction detour , you forget -- not even to speak of the problem , you forget what day it is . let 's see , so you get the laplace transform of y is equal to 1 minus e to the minus 2 pi s times what that mess that we just solved for , times -- and i 'll write it like this . 1/3 times 1 over s squared plus 1 minus 1/3 times -- actually , let me write it this way . because i have this s squared plus 4 , so i really want to have a 2 there . so i want to have a 2 in the numerator , so you want to have a 2 over s squared plus 4 . so if i put a 2 in the numerator there , i have to divide this by 2 as well . so let me change this to a 6 . minus 1/6 times 2 is minus 1/3 . so i did that just so i get this in the form of the laplace transform of sine of t. now , let 's see if there 's anything that i can do from here . this is an epic problem . i 'll be amazed if i do n't make a careless mistake while i do this . so we can rewrite everything . let 's see if we can simplify this . and by simplifying it , i 'm just going to make it longer . we can write the laplace transform of y is equal to -- i 'm just going to multiply the 1 out , and then i 'm going to multiply the e to the minus 2 pi s out . so if you multiply the 1 out , you get 1/3 times 1 over s squared plus 1 -- i 'm just multiplying the 1 out -- minus 1/6 -- these are all the 1 's times the 1 -- times 2 over s squared plus 4 . and then i 'm going to multiply the minus e. let me just switch colors , do the minus e. so then you get minus e to the minus 2 pi s over 3 times 1 over s squared plus 1 . and then the minus and the minus cancel out , so you get plus e to the minus 2 pi s over 6 times 2 over s squared plus 4 . now , taking the inverse laplace transform of these things are pretty straightforward . so let 's do that . let 's take the inverse laplace transform of the whole thing . and we get y is equal to the inverse laplace transform of this guy right here , is just 1/3 sine of t -- i do n't have to write a parentheses there -- sine of t , and then this is minus 1/6 times -- this is the laplace transform of sine of 2t . that 's that term right there . now , these are almost the same , but we have this little pesky character over here . we have this e to the minus 2 pi s. and there , we just have to remind ourselves -- i 'll write it here in the bottom . we just have to remind ourselves that the laplace transform of the unit step function -- i 'll put the pi there , just 2 pi times f of t minus 2 pi -- i should put as the step function of t -- is equal to e to the minus 2 pi s times the laplace transform of just -- or let me just write it this way -- times the laplace transform of f of t. so if we view f of t as just sine of t or sine of 2t , then we can kind of backwards pattern match . and we 'll have to shift it and multiply it by the unit step function . so i want to make that clear . if you did n't have this guy here , the inverse laplace transform of this guy would be the same thing as this guy . it 'd just be sine of t. the inverse laplace transform of this guy would be sine of 2t . but we have this pesky character here , which essentially , instead of having the inverse laplace transform just being our f of t , it 's going to be our f of t shifted by 2 pi times the unit step function , where it steps up at 2pi . so this is going to be minus 1/3 times the unit step function , where c is 2 pi of t times -- instead of sine of t -- sine of t minus 2pi . and then we 're almost done . i 'll do it in magenta to celebrate it . plus this very last term , which is 1/6 times the unit step function 2 pi of t , the unit step function that steps up at 2 pi times sine of -- and we have to be careful here . wherever we had a t before , we 're going to replace it with a t minus 2 pi . so sine of , instead of 2t , is going to be 2 times t minus 2 pi . and there you have it . we finally have solved our very hairy problem . we could take some time if we want to simplify this a little bit . in fact , we might as well . at the risk of making a careless mistake at the last moment , let me see if i can make any simplifications here . well , we could factor out this guy right here , but other than that , that seems about as simple as we can get . so this is our function of t that satisfies our otherwise simple-looking differential equation that we had up here . this looked fairly straightforward , but we got this big mess to actually satisfy that equation , given those initial conditions that we had initially .
so let me change this to a 6 . minus 1/6 times 2 is minus 1/3 . so i did that just so i get this in the form of the laplace transform of sine of t. now , let 's see if there 's anything that i can do from here .
why we ca n't use a/s^2+1 + b/s^2+4 in this question ?
let 's apply everything we 've learned to an actual differential equation . instead of just taking laplace transforms and taking their inverse , let 's actually solve a problem . so let 's say that i have the second derivative of my function y plus 4 times my function y is equal to sine of t minus the unit step function 0 up until 2 pi of t times sine of t minus 2 pi . let 's solve this differential equation , an interpretation of it . and i actually do a whole playlist on interpretations of differential equations and how you model it , but you know , you can kind of view this is a forcing function . that it 's a weird forcing function of this being applied to some weight with , you know , this is the acceleration term , right ? the second derivative with respect to time is the acceleration . so the mass would be 1 whatever units , and then as a function of its position , this is probably some type of spring constant . anyway , i wo n't go there . i do n't want to waste your time with the interpretation of it , but let 's solve it . we can do more about interpretations later . so we 're going take the laplace transform of both sides of this equation . so what 's the laplace transform of the left-hand side ? so the laplace transform of the second derivative of y is just s squared , so now i 'm taking the laplace transform of just that . the laplace transform of s squared times the laplace transform of y minus -- lower the degree there once -- minus s times y of 0 minus y prime of 0 . so clearly , i must have to give you some initial conditions in order to do this properly . and then plus 4 times the laplace transform of y is equal to -- what 's the laplace transform of sine of t ? that should be second nature by now . it 's just 1 over s squared plus 1 . and then we have minus the laplace transform of this thing . and i 'll do a little side note here to figure out the laplace transform of this thing right here . and we know , i showed it to you a couple of videos ago , we showed that the laplace transform -- actually i could just write it out here . this is going to be the same thing as the laplace transform of sine of t , but we 're going to have to multiply it by e to the minus -- if you remember that last formula -- e to the minus cs , where c is 2 pi . actually , let me write that down . i decided to write it down , then i decided , oh , no , i do n't want to do this . but let me write that . so the laplace transform of the unit step function that goes up to c times some function shifted by c is equal to e to the minus cs times the laplace transform of just the original function times the laplace transform of f of t. so if we 're taking the laplace transform of this thing , our c is 2 pi . our f of t is just sine of t , right ? so then this is just going to be equal to -- if we just do this piece right here -- it 's going to be equal to e to the minus cs -- our c is 2 pi -- e to the minus 2 pi s times the laplace transform of f of t. f of t is just sine of t before we shifted . this is f of t minus 2 pi . so f of t is just going to be sine of t. so it 's going to be times 1 over s squared plus 1 . this is the laplace transform of sine of t. so let 's go back to where we had left off . so we 've taken the laplace transform of both sides of this equation . and clearly , i have some initial conditions here , so the problem must have given me some and i just forgot to write them down . so let 's see , the initial conditions i 'm given , and they are written kind of in the margin here , they tell us -- i 'll do it in orange , they tell us that y of 0 is equal to 0 , and y prime of 0 is equal to 0 . that makes the math easy . that 's 0 and that 's 0 . so let 's see if i can simplify my equation . so the left-hand side , let 's factor out the laplace transform . so let 's factor out this term and that term . so we get the laplace transform of y times this plus this times s squared plus 4 is equal to the right-hand side . and what 's the right-hand side ? we could simplify this . well , i 'll just write it out . i do n't want to do too many steps at once . it 's 1 over s squared plus 1 and then plus -- or minus actually , this is a minus -- minus the laplace transfer of this thing , which was e to the minus 2 pi s over s squared plus 1 . so if we divide both sides of this equation by the s squared plus 4 , then we get the laplace transform of y is equal to -- and actually , i can just merge these two . they 're the same denominator . so before i even divide by s squared plus 4 , that right-hand side will look like this . it will look like with a denominator of s squared plus 1 and you have a numerator of 1 minus e to the minus 2 pi s. and , of course , we 're dividing both sides of this equation by s squared plus 4 , so we 're going to have to stick that s squared plus 4 over here . now , we 're at the hard part . in order to figure out why , we have to take the inverse laplace transform of this thing . so how do we take the inverse laplace transform of this thing ? that 's where the hard part is always , you know , it makes solving the differential equation 's easy if you know the laplace transforms . so it looks like we 're going to have to do some partial fraction expansion . so let 's see if we can do that . so we can rewrite this equation right here . actually , let 's write it as this , because this 'll kind of simplify our work . let 's factor this whole thing out . so we 're going to write it as 1 minus e to the minus 2 pi s , all of that times -- i 'll do it in orange -- all of that times 1 over s squared plus 1 times s squared plus 4 . now , we need to do some partial fraction expansion to simplify this thing right here . we 're going to do this on the side . maybe i should do this over on the right here . this thing -- let me rewrite it -- 1 over s squared plus 1 times s squared plus 4 should be able to be rewritten as two separate fractions , s squared plus 1 and s squared plus 4 , with the numerators . this one would be as plus b . it 's going to have to have degree 1 , because this is degree 2 . here and then we 'd have cs plus d. and so when you add these two things up , you get as plus b times s squared plus 4 plus cs plus d times s squared plus 1 , all of that over the common denominator . we 've seen this story before . we just have to do some algebra here . as you can tell , these differential equations problems , they require a lot of stamina . you kind of just have to say i will keep moving forward and do the algebra that i need to do in order to get the answer . and you kind of have to get excited about that notion that you have all this algebra to do . so let 's figure it out . so this top can be simplified to as to the third plus bs squared plus 4as plus 4b . and then this one , you end up with cs to the third plus ds squared plus cs plus d. so when you add of these up together , you get -- and this is all the algebra that we have to do , for better , for worse -- a plus c over s to the third plus b plus d times s squared plus 4a plus c times s -- let 's scroll over a little bit -- plus 4b plus d. and now we just have to say , ok , all of this is equal to this thing up here . this is the numerator . we just simplified the numerator . this is the numerator . that 's the numerator right there . and all of this is going to be over your original s squared plus 1 times your s squared plus 4 . and we established that this thing should be -- let me just write this -- that 1 over s squared plus 1 times s squared plus 4 should equal this thing . and then you just pattern match on the coefficients . this is all just intense partial fraction expansion . and you say , look , a plus c is the coefficient of the s cubed terms . i do n't see any s cubed terms here . so a plus c must be equal to 0 . and then you see , ok , b plus d is the coefficient of the s squared terms . i do n't see any s squared terms there . so b plus d must be equal to 0 . 4a plus c , the coefficient of the s terms . i do n't see any s terms over here . so 4a plus c must be equal to 0 . and then we 're almost done . 4b plus d must be the constant terms . there is a constant term there . so 4b plus d is equal to 1 . so let 's see if we can do anything here . if we subtract this from that , we get minus 3a is equal to 0 , or a is equal to 0 . if a is equal to 0 , then c is equals to 0 . and let 's see what we can get here . if we subtract this from that , we get minus 3b . the d 's cancel out . it 's equal to minus 1 , or b is equal to 1/3 . and then , of course , we have d is equal to minus b , if you subtract b from both sides . so d is equal to 1/3 . so all of that work , and we actually have a pretty simple result . our equation , this thing here , can be rewritten as -- the a disappeared . it 's 1/3 over s squared plus 1 . b was the coefficient on the -- let me make it very clear . b was the coefficient on the -- or it was a term on top of the s squared plus 1 , so that 's why i 'm using b there . and then d is minus b , so d is minus 1 . so let me make sure i have that . b is 1/3 minus -- let me make sure i get that right . d is 1/3 . so , sorry , b as in boy is 1/3 , so d is minus 1/3 . so b , there 's a term on top of the s squared plus 1 . and then you have minus d over the minus 1/3 over s squared plus 4 . this takes a lot of stamina to record this video . i hope you appreciate it . ok , so let me rewrite everything , just so we can get back to the problem because when you take the partial fraction detour , you forget -- not even to speak of the problem , you forget what day it is . let 's see , so you get the laplace transform of y is equal to 1 minus e to the minus 2 pi s times what that mess that we just solved for , times -- and i 'll write it like this . 1/3 times 1 over s squared plus 1 minus 1/3 times -- actually , let me write it this way . because i have this s squared plus 4 , so i really want to have a 2 there . so i want to have a 2 in the numerator , so you want to have a 2 over s squared plus 4 . so if i put a 2 in the numerator there , i have to divide this by 2 as well . so let me change this to a 6 . minus 1/6 times 2 is minus 1/3 . so i did that just so i get this in the form of the laplace transform of sine of t. now , let 's see if there 's anything that i can do from here . this is an epic problem . i 'll be amazed if i do n't make a careless mistake while i do this . so we can rewrite everything . let 's see if we can simplify this . and by simplifying it , i 'm just going to make it longer . we can write the laplace transform of y is equal to -- i 'm just going to multiply the 1 out , and then i 'm going to multiply the e to the minus 2 pi s out . so if you multiply the 1 out , you get 1/3 times 1 over s squared plus 1 -- i 'm just multiplying the 1 out -- minus 1/6 -- these are all the 1 's times the 1 -- times 2 over s squared plus 4 . and then i 'm going to multiply the minus e. let me just switch colors , do the minus e. so then you get minus e to the minus 2 pi s over 3 times 1 over s squared plus 1 . and then the minus and the minus cancel out , so you get plus e to the minus 2 pi s over 6 times 2 over s squared plus 4 . now , taking the inverse laplace transform of these things are pretty straightforward . so let 's do that . let 's take the inverse laplace transform of the whole thing . and we get y is equal to the inverse laplace transform of this guy right here , is just 1/3 sine of t -- i do n't have to write a parentheses there -- sine of t , and then this is minus 1/6 times -- this is the laplace transform of sine of 2t . that 's that term right there . now , these are almost the same , but we have this little pesky character over here . we have this e to the minus 2 pi s. and there , we just have to remind ourselves -- i 'll write it here in the bottom . we just have to remind ourselves that the laplace transform of the unit step function -- i 'll put the pi there , just 2 pi times f of t minus 2 pi -- i should put as the step function of t -- is equal to e to the minus 2 pi s times the laplace transform of just -- or let me just write it this way -- times the laplace transform of f of t. so if we view f of t as just sine of t or sine of 2t , then we can kind of backwards pattern match . and we 'll have to shift it and multiply it by the unit step function . so i want to make that clear . if you did n't have this guy here , the inverse laplace transform of this guy would be the same thing as this guy . it 'd just be sine of t. the inverse laplace transform of this guy would be sine of 2t . but we have this pesky character here , which essentially , instead of having the inverse laplace transform just being our f of t , it 's going to be our f of t shifted by 2 pi times the unit step function , where it steps up at 2pi . so this is going to be minus 1/3 times the unit step function , where c is 2 pi of t times -- instead of sine of t -- sine of t minus 2pi . and then we 're almost done . i 'll do it in magenta to celebrate it . plus this very last term , which is 1/6 times the unit step function 2 pi of t , the unit step function that steps up at 2 pi times sine of -- and we have to be careful here . wherever we had a t before , we 're going to replace it with a t minus 2 pi . so sine of , instead of 2t , is going to be 2 times t minus 2 pi . and there you have it . we finally have solved our very hairy problem . we could take some time if we want to simplify this a little bit . in fact , we might as well . at the risk of making a careless mistake at the last moment , let me see if i can make any simplifications here . well , we could factor out this guy right here , but other than that , that seems about as simple as we can get . so this is our function of t that satisfies our otherwise simple-looking differential equation that we had up here . this looked fairly straightforward , but we got this big mess to actually satisfy that equation , given those initial conditions that we had initially .
if a is equal to 0 , then c is equals to 0 . and let 's see what we can get here . if we subtract this from that , we get minus 3b .
can we get a real life example using the dirac delta function ?
let 's apply everything we 've learned to an actual differential equation . instead of just taking laplace transforms and taking their inverse , let 's actually solve a problem . so let 's say that i have the second derivative of my function y plus 4 times my function y is equal to sine of t minus the unit step function 0 up until 2 pi of t times sine of t minus 2 pi . let 's solve this differential equation , an interpretation of it . and i actually do a whole playlist on interpretations of differential equations and how you model it , but you know , you can kind of view this is a forcing function . that it 's a weird forcing function of this being applied to some weight with , you know , this is the acceleration term , right ? the second derivative with respect to time is the acceleration . so the mass would be 1 whatever units , and then as a function of its position , this is probably some type of spring constant . anyway , i wo n't go there . i do n't want to waste your time with the interpretation of it , but let 's solve it . we can do more about interpretations later . so we 're going take the laplace transform of both sides of this equation . so what 's the laplace transform of the left-hand side ? so the laplace transform of the second derivative of y is just s squared , so now i 'm taking the laplace transform of just that . the laplace transform of s squared times the laplace transform of y minus -- lower the degree there once -- minus s times y of 0 minus y prime of 0 . so clearly , i must have to give you some initial conditions in order to do this properly . and then plus 4 times the laplace transform of y is equal to -- what 's the laplace transform of sine of t ? that should be second nature by now . it 's just 1 over s squared plus 1 . and then we have minus the laplace transform of this thing . and i 'll do a little side note here to figure out the laplace transform of this thing right here . and we know , i showed it to you a couple of videos ago , we showed that the laplace transform -- actually i could just write it out here . this is going to be the same thing as the laplace transform of sine of t , but we 're going to have to multiply it by e to the minus -- if you remember that last formula -- e to the minus cs , where c is 2 pi . actually , let me write that down . i decided to write it down , then i decided , oh , no , i do n't want to do this . but let me write that . so the laplace transform of the unit step function that goes up to c times some function shifted by c is equal to e to the minus cs times the laplace transform of just the original function times the laplace transform of f of t. so if we 're taking the laplace transform of this thing , our c is 2 pi . our f of t is just sine of t , right ? so then this is just going to be equal to -- if we just do this piece right here -- it 's going to be equal to e to the minus cs -- our c is 2 pi -- e to the minus 2 pi s times the laplace transform of f of t. f of t is just sine of t before we shifted . this is f of t minus 2 pi . so f of t is just going to be sine of t. so it 's going to be times 1 over s squared plus 1 . this is the laplace transform of sine of t. so let 's go back to where we had left off . so we 've taken the laplace transform of both sides of this equation . and clearly , i have some initial conditions here , so the problem must have given me some and i just forgot to write them down . so let 's see , the initial conditions i 'm given , and they are written kind of in the margin here , they tell us -- i 'll do it in orange , they tell us that y of 0 is equal to 0 , and y prime of 0 is equal to 0 . that makes the math easy . that 's 0 and that 's 0 . so let 's see if i can simplify my equation . so the left-hand side , let 's factor out the laplace transform . so let 's factor out this term and that term . so we get the laplace transform of y times this plus this times s squared plus 4 is equal to the right-hand side . and what 's the right-hand side ? we could simplify this . well , i 'll just write it out . i do n't want to do too many steps at once . it 's 1 over s squared plus 1 and then plus -- or minus actually , this is a minus -- minus the laplace transfer of this thing , which was e to the minus 2 pi s over s squared plus 1 . so if we divide both sides of this equation by the s squared plus 4 , then we get the laplace transform of y is equal to -- and actually , i can just merge these two . they 're the same denominator . so before i even divide by s squared plus 4 , that right-hand side will look like this . it will look like with a denominator of s squared plus 1 and you have a numerator of 1 minus e to the minus 2 pi s. and , of course , we 're dividing both sides of this equation by s squared plus 4 , so we 're going to have to stick that s squared plus 4 over here . now , we 're at the hard part . in order to figure out why , we have to take the inverse laplace transform of this thing . so how do we take the inverse laplace transform of this thing ? that 's where the hard part is always , you know , it makes solving the differential equation 's easy if you know the laplace transforms . so it looks like we 're going to have to do some partial fraction expansion . so let 's see if we can do that . so we can rewrite this equation right here . actually , let 's write it as this , because this 'll kind of simplify our work . let 's factor this whole thing out . so we 're going to write it as 1 minus e to the minus 2 pi s , all of that times -- i 'll do it in orange -- all of that times 1 over s squared plus 1 times s squared plus 4 . now , we need to do some partial fraction expansion to simplify this thing right here . we 're going to do this on the side . maybe i should do this over on the right here . this thing -- let me rewrite it -- 1 over s squared plus 1 times s squared plus 4 should be able to be rewritten as two separate fractions , s squared plus 1 and s squared plus 4 , with the numerators . this one would be as plus b . it 's going to have to have degree 1 , because this is degree 2 . here and then we 'd have cs plus d. and so when you add these two things up , you get as plus b times s squared plus 4 plus cs plus d times s squared plus 1 , all of that over the common denominator . we 've seen this story before . we just have to do some algebra here . as you can tell , these differential equations problems , they require a lot of stamina . you kind of just have to say i will keep moving forward and do the algebra that i need to do in order to get the answer . and you kind of have to get excited about that notion that you have all this algebra to do . so let 's figure it out . so this top can be simplified to as to the third plus bs squared plus 4as plus 4b . and then this one , you end up with cs to the third plus ds squared plus cs plus d. so when you add of these up together , you get -- and this is all the algebra that we have to do , for better , for worse -- a plus c over s to the third plus b plus d times s squared plus 4a plus c times s -- let 's scroll over a little bit -- plus 4b plus d. and now we just have to say , ok , all of this is equal to this thing up here . this is the numerator . we just simplified the numerator . this is the numerator . that 's the numerator right there . and all of this is going to be over your original s squared plus 1 times your s squared plus 4 . and we established that this thing should be -- let me just write this -- that 1 over s squared plus 1 times s squared plus 4 should equal this thing . and then you just pattern match on the coefficients . this is all just intense partial fraction expansion . and you say , look , a plus c is the coefficient of the s cubed terms . i do n't see any s cubed terms here . so a plus c must be equal to 0 . and then you see , ok , b plus d is the coefficient of the s squared terms . i do n't see any s squared terms there . so b plus d must be equal to 0 . 4a plus c , the coefficient of the s terms . i do n't see any s terms over here . so 4a plus c must be equal to 0 . and then we 're almost done . 4b plus d must be the constant terms . there is a constant term there . so 4b plus d is equal to 1 . so let 's see if we can do anything here . if we subtract this from that , we get minus 3a is equal to 0 , or a is equal to 0 . if a is equal to 0 , then c is equals to 0 . and let 's see what we can get here . if we subtract this from that , we get minus 3b . the d 's cancel out . it 's equal to minus 1 , or b is equal to 1/3 . and then , of course , we have d is equal to minus b , if you subtract b from both sides . so d is equal to 1/3 . so all of that work , and we actually have a pretty simple result . our equation , this thing here , can be rewritten as -- the a disappeared . it 's 1/3 over s squared plus 1 . b was the coefficient on the -- let me make it very clear . b was the coefficient on the -- or it was a term on top of the s squared plus 1 , so that 's why i 'm using b there . and then d is minus b , so d is minus 1 . so let me make sure i have that . b is 1/3 minus -- let me make sure i get that right . d is 1/3 . so , sorry , b as in boy is 1/3 , so d is minus 1/3 . so b , there 's a term on top of the s squared plus 1 . and then you have minus d over the minus 1/3 over s squared plus 4 . this takes a lot of stamina to record this video . i hope you appreciate it . ok , so let me rewrite everything , just so we can get back to the problem because when you take the partial fraction detour , you forget -- not even to speak of the problem , you forget what day it is . let 's see , so you get the laplace transform of y is equal to 1 minus e to the minus 2 pi s times what that mess that we just solved for , times -- and i 'll write it like this . 1/3 times 1 over s squared plus 1 minus 1/3 times -- actually , let me write it this way . because i have this s squared plus 4 , so i really want to have a 2 there . so i want to have a 2 in the numerator , so you want to have a 2 over s squared plus 4 . so if i put a 2 in the numerator there , i have to divide this by 2 as well . so let me change this to a 6 . minus 1/6 times 2 is minus 1/3 . so i did that just so i get this in the form of the laplace transform of sine of t. now , let 's see if there 's anything that i can do from here . this is an epic problem . i 'll be amazed if i do n't make a careless mistake while i do this . so we can rewrite everything . let 's see if we can simplify this . and by simplifying it , i 'm just going to make it longer . we can write the laplace transform of y is equal to -- i 'm just going to multiply the 1 out , and then i 'm going to multiply the e to the minus 2 pi s out . so if you multiply the 1 out , you get 1/3 times 1 over s squared plus 1 -- i 'm just multiplying the 1 out -- minus 1/6 -- these are all the 1 's times the 1 -- times 2 over s squared plus 4 . and then i 'm going to multiply the minus e. let me just switch colors , do the minus e. so then you get minus e to the minus 2 pi s over 3 times 1 over s squared plus 1 . and then the minus and the minus cancel out , so you get plus e to the minus 2 pi s over 6 times 2 over s squared plus 4 . now , taking the inverse laplace transform of these things are pretty straightforward . so let 's do that . let 's take the inverse laplace transform of the whole thing . and we get y is equal to the inverse laplace transform of this guy right here , is just 1/3 sine of t -- i do n't have to write a parentheses there -- sine of t , and then this is minus 1/6 times -- this is the laplace transform of sine of 2t . that 's that term right there . now , these are almost the same , but we have this little pesky character over here . we have this e to the minus 2 pi s. and there , we just have to remind ourselves -- i 'll write it here in the bottom . we just have to remind ourselves that the laplace transform of the unit step function -- i 'll put the pi there , just 2 pi times f of t minus 2 pi -- i should put as the step function of t -- is equal to e to the minus 2 pi s times the laplace transform of just -- or let me just write it this way -- times the laplace transform of f of t. so if we view f of t as just sine of t or sine of 2t , then we can kind of backwards pattern match . and we 'll have to shift it and multiply it by the unit step function . so i want to make that clear . if you did n't have this guy here , the inverse laplace transform of this guy would be the same thing as this guy . it 'd just be sine of t. the inverse laplace transform of this guy would be sine of 2t . but we have this pesky character here , which essentially , instead of having the inverse laplace transform just being our f of t , it 's going to be our f of t shifted by 2 pi times the unit step function , where it steps up at 2pi . so this is going to be minus 1/3 times the unit step function , where c is 2 pi of t times -- instead of sine of t -- sine of t minus 2pi . and then we 're almost done . i 'll do it in magenta to celebrate it . plus this very last term , which is 1/6 times the unit step function 2 pi of t , the unit step function that steps up at 2 pi times sine of -- and we have to be careful here . wherever we had a t before , we 're going to replace it with a t minus 2 pi . so sine of , instead of 2t , is going to be 2 times t minus 2 pi . and there you have it . we finally have solved our very hairy problem . we could take some time if we want to simplify this a little bit . in fact , we might as well . at the risk of making a careless mistake at the last moment , let me see if i can make any simplifications here . well , we could factor out this guy right here , but other than that , that seems about as simple as we can get . so this is our function of t that satisfies our otherwise simple-looking differential equation that we had up here . this looked fairly straightforward , but we got this big mess to actually satisfy that equation , given those initial conditions that we had initially .
let 's apply everything we 've learned to an actual differential equation . instead of just taking laplace transforms and taking their inverse , let 's actually solve a problem . so let 's say that i have the second derivative of my function y plus 4 times my function y is equal to sine of t minus the unit step function 0 up until 2 pi of t times sine of t minus 2 pi .
can we use laplace transforms for des which are n't initial value problems ?
let 's apply everything we 've learned to an actual differential equation . instead of just taking laplace transforms and taking their inverse , let 's actually solve a problem . so let 's say that i have the second derivative of my function y plus 4 times my function y is equal to sine of t minus the unit step function 0 up until 2 pi of t times sine of t minus 2 pi . let 's solve this differential equation , an interpretation of it . and i actually do a whole playlist on interpretations of differential equations and how you model it , but you know , you can kind of view this is a forcing function . that it 's a weird forcing function of this being applied to some weight with , you know , this is the acceleration term , right ? the second derivative with respect to time is the acceleration . so the mass would be 1 whatever units , and then as a function of its position , this is probably some type of spring constant . anyway , i wo n't go there . i do n't want to waste your time with the interpretation of it , but let 's solve it . we can do more about interpretations later . so we 're going take the laplace transform of both sides of this equation . so what 's the laplace transform of the left-hand side ? so the laplace transform of the second derivative of y is just s squared , so now i 'm taking the laplace transform of just that . the laplace transform of s squared times the laplace transform of y minus -- lower the degree there once -- minus s times y of 0 minus y prime of 0 . so clearly , i must have to give you some initial conditions in order to do this properly . and then plus 4 times the laplace transform of y is equal to -- what 's the laplace transform of sine of t ? that should be second nature by now . it 's just 1 over s squared plus 1 . and then we have minus the laplace transform of this thing . and i 'll do a little side note here to figure out the laplace transform of this thing right here . and we know , i showed it to you a couple of videos ago , we showed that the laplace transform -- actually i could just write it out here . this is going to be the same thing as the laplace transform of sine of t , but we 're going to have to multiply it by e to the minus -- if you remember that last formula -- e to the minus cs , where c is 2 pi . actually , let me write that down . i decided to write it down , then i decided , oh , no , i do n't want to do this . but let me write that . so the laplace transform of the unit step function that goes up to c times some function shifted by c is equal to e to the minus cs times the laplace transform of just the original function times the laplace transform of f of t. so if we 're taking the laplace transform of this thing , our c is 2 pi . our f of t is just sine of t , right ? so then this is just going to be equal to -- if we just do this piece right here -- it 's going to be equal to e to the minus cs -- our c is 2 pi -- e to the minus 2 pi s times the laplace transform of f of t. f of t is just sine of t before we shifted . this is f of t minus 2 pi . so f of t is just going to be sine of t. so it 's going to be times 1 over s squared plus 1 . this is the laplace transform of sine of t. so let 's go back to where we had left off . so we 've taken the laplace transform of both sides of this equation . and clearly , i have some initial conditions here , so the problem must have given me some and i just forgot to write them down . so let 's see , the initial conditions i 'm given , and they are written kind of in the margin here , they tell us -- i 'll do it in orange , they tell us that y of 0 is equal to 0 , and y prime of 0 is equal to 0 . that makes the math easy . that 's 0 and that 's 0 . so let 's see if i can simplify my equation . so the left-hand side , let 's factor out the laplace transform . so let 's factor out this term and that term . so we get the laplace transform of y times this plus this times s squared plus 4 is equal to the right-hand side . and what 's the right-hand side ? we could simplify this . well , i 'll just write it out . i do n't want to do too many steps at once . it 's 1 over s squared plus 1 and then plus -- or minus actually , this is a minus -- minus the laplace transfer of this thing , which was e to the minus 2 pi s over s squared plus 1 . so if we divide both sides of this equation by the s squared plus 4 , then we get the laplace transform of y is equal to -- and actually , i can just merge these two . they 're the same denominator . so before i even divide by s squared plus 4 , that right-hand side will look like this . it will look like with a denominator of s squared plus 1 and you have a numerator of 1 minus e to the minus 2 pi s. and , of course , we 're dividing both sides of this equation by s squared plus 4 , so we 're going to have to stick that s squared plus 4 over here . now , we 're at the hard part . in order to figure out why , we have to take the inverse laplace transform of this thing . so how do we take the inverse laplace transform of this thing ? that 's where the hard part is always , you know , it makes solving the differential equation 's easy if you know the laplace transforms . so it looks like we 're going to have to do some partial fraction expansion . so let 's see if we can do that . so we can rewrite this equation right here . actually , let 's write it as this , because this 'll kind of simplify our work . let 's factor this whole thing out . so we 're going to write it as 1 minus e to the minus 2 pi s , all of that times -- i 'll do it in orange -- all of that times 1 over s squared plus 1 times s squared plus 4 . now , we need to do some partial fraction expansion to simplify this thing right here . we 're going to do this on the side . maybe i should do this over on the right here . this thing -- let me rewrite it -- 1 over s squared plus 1 times s squared plus 4 should be able to be rewritten as two separate fractions , s squared plus 1 and s squared plus 4 , with the numerators . this one would be as plus b . it 's going to have to have degree 1 , because this is degree 2 . here and then we 'd have cs plus d. and so when you add these two things up , you get as plus b times s squared plus 4 plus cs plus d times s squared plus 1 , all of that over the common denominator . we 've seen this story before . we just have to do some algebra here . as you can tell , these differential equations problems , they require a lot of stamina . you kind of just have to say i will keep moving forward and do the algebra that i need to do in order to get the answer . and you kind of have to get excited about that notion that you have all this algebra to do . so let 's figure it out . so this top can be simplified to as to the third plus bs squared plus 4as plus 4b . and then this one , you end up with cs to the third plus ds squared plus cs plus d. so when you add of these up together , you get -- and this is all the algebra that we have to do , for better , for worse -- a plus c over s to the third plus b plus d times s squared plus 4a plus c times s -- let 's scroll over a little bit -- plus 4b plus d. and now we just have to say , ok , all of this is equal to this thing up here . this is the numerator . we just simplified the numerator . this is the numerator . that 's the numerator right there . and all of this is going to be over your original s squared plus 1 times your s squared plus 4 . and we established that this thing should be -- let me just write this -- that 1 over s squared plus 1 times s squared plus 4 should equal this thing . and then you just pattern match on the coefficients . this is all just intense partial fraction expansion . and you say , look , a plus c is the coefficient of the s cubed terms . i do n't see any s cubed terms here . so a plus c must be equal to 0 . and then you see , ok , b plus d is the coefficient of the s squared terms . i do n't see any s squared terms there . so b plus d must be equal to 0 . 4a plus c , the coefficient of the s terms . i do n't see any s terms over here . so 4a plus c must be equal to 0 . and then we 're almost done . 4b plus d must be the constant terms . there is a constant term there . so 4b plus d is equal to 1 . so let 's see if we can do anything here . if we subtract this from that , we get minus 3a is equal to 0 , or a is equal to 0 . if a is equal to 0 , then c is equals to 0 . and let 's see what we can get here . if we subtract this from that , we get minus 3b . the d 's cancel out . it 's equal to minus 1 , or b is equal to 1/3 . and then , of course , we have d is equal to minus b , if you subtract b from both sides . so d is equal to 1/3 . so all of that work , and we actually have a pretty simple result . our equation , this thing here , can be rewritten as -- the a disappeared . it 's 1/3 over s squared plus 1 . b was the coefficient on the -- let me make it very clear . b was the coefficient on the -- or it was a term on top of the s squared plus 1 , so that 's why i 'm using b there . and then d is minus b , so d is minus 1 . so let me make sure i have that . b is 1/3 minus -- let me make sure i get that right . d is 1/3 . so , sorry , b as in boy is 1/3 , so d is minus 1/3 . so b , there 's a term on top of the s squared plus 1 . and then you have minus d over the minus 1/3 over s squared plus 4 . this takes a lot of stamina to record this video . i hope you appreciate it . ok , so let me rewrite everything , just so we can get back to the problem because when you take the partial fraction detour , you forget -- not even to speak of the problem , you forget what day it is . let 's see , so you get the laplace transform of y is equal to 1 minus e to the minus 2 pi s times what that mess that we just solved for , times -- and i 'll write it like this . 1/3 times 1 over s squared plus 1 minus 1/3 times -- actually , let me write it this way . because i have this s squared plus 4 , so i really want to have a 2 there . so i want to have a 2 in the numerator , so you want to have a 2 over s squared plus 4 . so if i put a 2 in the numerator there , i have to divide this by 2 as well . so let me change this to a 6 . minus 1/6 times 2 is minus 1/3 . so i did that just so i get this in the form of the laplace transform of sine of t. now , let 's see if there 's anything that i can do from here . this is an epic problem . i 'll be amazed if i do n't make a careless mistake while i do this . so we can rewrite everything . let 's see if we can simplify this . and by simplifying it , i 'm just going to make it longer . we can write the laplace transform of y is equal to -- i 'm just going to multiply the 1 out , and then i 'm going to multiply the e to the minus 2 pi s out . so if you multiply the 1 out , you get 1/3 times 1 over s squared plus 1 -- i 'm just multiplying the 1 out -- minus 1/6 -- these are all the 1 's times the 1 -- times 2 over s squared plus 4 . and then i 'm going to multiply the minus e. let me just switch colors , do the minus e. so then you get minus e to the minus 2 pi s over 3 times 1 over s squared plus 1 . and then the minus and the minus cancel out , so you get plus e to the minus 2 pi s over 6 times 2 over s squared plus 4 . now , taking the inverse laplace transform of these things are pretty straightforward . so let 's do that . let 's take the inverse laplace transform of the whole thing . and we get y is equal to the inverse laplace transform of this guy right here , is just 1/3 sine of t -- i do n't have to write a parentheses there -- sine of t , and then this is minus 1/6 times -- this is the laplace transform of sine of 2t . that 's that term right there . now , these are almost the same , but we have this little pesky character over here . we have this e to the minus 2 pi s. and there , we just have to remind ourselves -- i 'll write it here in the bottom . we just have to remind ourselves that the laplace transform of the unit step function -- i 'll put the pi there , just 2 pi times f of t minus 2 pi -- i should put as the step function of t -- is equal to e to the minus 2 pi s times the laplace transform of just -- or let me just write it this way -- times the laplace transform of f of t. so if we view f of t as just sine of t or sine of 2t , then we can kind of backwards pattern match . and we 'll have to shift it and multiply it by the unit step function . so i want to make that clear . if you did n't have this guy here , the inverse laplace transform of this guy would be the same thing as this guy . it 'd just be sine of t. the inverse laplace transform of this guy would be sine of 2t . but we have this pesky character here , which essentially , instead of having the inverse laplace transform just being our f of t , it 's going to be our f of t shifted by 2 pi times the unit step function , where it steps up at 2pi . so this is going to be minus 1/3 times the unit step function , where c is 2 pi of t times -- instead of sine of t -- sine of t minus 2pi . and then we 're almost done . i 'll do it in magenta to celebrate it . plus this very last term , which is 1/6 times the unit step function 2 pi of t , the unit step function that steps up at 2 pi times sine of -- and we have to be careful here . wherever we had a t before , we 're going to replace it with a t minus 2 pi . so sine of , instead of 2t , is going to be 2 times t minus 2 pi . and there you have it . we finally have solved our very hairy problem . we could take some time if we want to simplify this a little bit . in fact , we might as well . at the risk of making a careless mistake at the last moment , let me see if i can make any simplifications here . well , we could factor out this guy right here , but other than that , that seems about as simple as we can get . so this is our function of t that satisfies our otherwise simple-looking differential equation that we had up here . this looked fairly straightforward , but we got this big mess to actually satisfy that equation , given those initial conditions that we had initially .
so d is equal to 1/3 . so all of that work , and we actually have a pretty simple result . our equation , this thing here , can be rewritten as -- the a disappeared .
`` '' arent b and d supposed to be negatives of each other ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule .
simply to indicate one additional bond is possible ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons .
boron in bf3 molecule the b has only 3 electrons in outermost shell and after bonding with flourine the b contains total 6 electrons but according to the octet rule the b should contain 8 ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in .
for c2h4 , why did it start out with one electron in the s2 orbital and and extra electron in the p orbital ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here .
when does sp2 hybridization take place ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond .
what about triple bonds with the central atoms , how many sigma bonds and pi bonds would there be ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule .
why are double bonds always made of one sigma and one pi bond ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here .
why would there be such a thing as pi bonds ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three .
why 2 electrons was not filled in 2s for full and then fill in 2p ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that .
jay tell us that boron has one unhybridised p orbital.how does he know that ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three .
but the sp2 hybrid orbital have only 3 electrons and hence will they be able to gain 3 more electrons ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that .
here what the left out p orbital does which is left unhybridised in ethene or ethylene ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule .
is there only one pi bond , or would that be showing two pi bonds ( one for each overlap ) ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character .
what happened to the double bond of the carbons in the structure ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react .
what is a `` lewis acid '' ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three .
i get that carbon can make four bonds in total , but why ca n't each carbon just make bonds with three other atoms and keep an unshared pair of electrons ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine .
for bf3 , why does b not have 4 single bonds ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero .
does n't it have to obey the octet rule and add a lone pair of electrons to it ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron .
while bonding for example in carbon the electron excites from where the energy comes to excite the electron ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create .
why does n't boron bonds with 4 flourines to gain its octet rather than bonding with 3 f ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that .
so , like the bf3 which used its empty p-orbitals to accept an electron pair to be lewis acid , will the c2h4 uses its half-filled unhybridized p-orbital to accept one electron and acts as lewis acid ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three .
do we only count sigma bonds to calculate the steric number ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule .
what does make the atom form a pi bond ... ... not all bonds are sigma ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now .
is ethylene an other name for ethene ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond .
in c2h4 , are the sigma bonds ionic bonds and the pi bonds covalent ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that .
what happens to the third p orbital ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here .
what is the difference between sp2 and sp3 hybridization ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four .
what is an electron density ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon .
why does n't carbon form 2 sigma bonds instead of one sigma and one pi ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in .
why do n't all the 4 orbitals become hybridized like in methane even though carbon still forms 4 bonds in both methane and ethene ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital .
7 , why does n't one of the two carbons in the ethylene compound spin around so that the two electrons in the unhybridized orbitals are farther from each other ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here .
will all tetrahedral shapes have a hybridization of sp3 , trigonal planar of sp2 , and linear of sp1 ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that .
how does i unhybridiesedp orbital efect the molecule as whole ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized .
why the number of lone pairs are taken as zero ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond .
however , when it bonds with flourine , what shape will flourine have ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital .
will it just have one of the three regular 'p ' shaped orbitals bonding to boron 's sp2 hybrid orbitals , or will flourine also become hybridized and have sp3 orbitals bonding to boron 's sp2 orbitals ( based on steric number i.e the 3 lone pairs+ 1 sigma bond ) ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example .
does only the least electronegative atom get hybridized ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here .
why do water molecules and ammonia molecules need hybridization ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron .
can we calculate steric no for coordination compounds ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital .
what makes a hybrid orbital ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here .
how does the pi bond contribute to no free rotation ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry .
how do we calculate angstorms ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals .
and does s stand for sigma bond and p for pi bond ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three .
how can i know wether a molecule will accept electrons or donate electrons ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital .
what 's the difference between sp3 hybrid orbitals and sp2 hybrid orbitals ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine .
why doesnt the element other than the central element like f in bf3 also get hybridized ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example .
why does only the central atom geta hybridized ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine .
in the bf3 example , are the fluorine 's orbitals hybridized ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon .
when do orbitals hybridize and when not ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now .
is there a difference between ethene and ethylene ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here .
how do scientists predict the 3d configuration of the entire molecule ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine .
onwards structure of bf3 is explained , we see that fluorine has a steric number of 4 does that mean fluorine is sp3 hybridized ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine .
is bf3 an exception to octet rule since b has only 6 electrons in bf3 ?
: in an earlier video , we saw that when carbon is bonded to four atoms , we have an sp3 hybridization with a tetrahedral geometry and an ideal bonding over 109.5 degrees . if you look at one of the carbons in ethenes , let 's say this carbon right here , we do n't see the same geometry . the geometry of the atoms around this carbon happens to be planar . actually , this entire molecule is planar . you could think about all this in a plane here . and the bond angles are close to 120 degrees . approximately , 120 degree bond angles and this carbon that i 've underlined here is bonded to only three atoms . a hydrogen , a hydrogen and a carbon and so we must need a different hybridization for each of the carbon 's presence in the ethylene molecule . we 're gon na start with our electron configurations over here , the excited stage . we have carbons four , valence electron represented . one , two , three and four . in the video on sp3 hybridization , we took all four of these orbitals and combined them to make four sp3 hybrid orbitals . in this case , we only have a carbon bonded to three atoms . we only need three of our orbitals . we 're going to promote the s orbital . we 're gon na promote the s orbital up and this time , we only need two of the p orbitals . we 're gon na take one of the p 's and then another one of the p 's here . that is gon na leave one of the our p orbitals unhybridized . each one of these orbitals has one electron and it 's like that . this is no longer an s orbital . this is an sp2 hybrid orbital . this is no longer a p orbital . this is an sp2 hybrid orbital and same with this one , an sp2 hybrid orbital . we call this sp2 hybridization . let me go and write this up here . and use a different color here . this is sp2 hybridization because we 're using one s orbital and two p orbitals to form our new hybrid orbitals . this carbon right here is sp2 hybridized and same with this carbon . notice that we left a p orbital untouched . we have a p orbital unhybridized like that . in terms of the shape of our new hybrid orbital , let 's go ahead and get some more space down here . we 're taking one s orbital . we know s orbitals are shaped like spheres . we 're taking two p orbitals . we know that a p orbital is shaped like a dumbbell . we 're gon na take these orbitals and hybridized them to form three sp2 hybrid orbitals and they have a bigger front lobe and a smaller back lobe here like that . once again , when we draw the pictures , we 're going to ignore the smaller back lobe . this gives us our sp2 hybrid orbitals . in terms of what percentage character , we have three orbitals that we 're taking here and one of them is an s orbital . one out of three , gives us 33 % s character in our new hybrid sp2 orbital and then we have two p orbitals . two out of three gives us 67 % p character . 33 % s character and 67 % p character . there 's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital and since the electron density in an s orbital is closer to the nucleus . we think about the electron density here being closer to the nucleus that means that we could think about this lobe right here being a little bit shorter with the electron density being closer to the nucleus and that 's gon na have an effect on the length of the bonds that we 're gon na be forming . let 's go ahead and draw the picture of the ethylene molecule now . we know that each of the carbons in ethylenes . just going back up here to emphasize the point . each of these carbons here is sp2 hybridized . that means each of those carbons is going to have three sp2 hybrid orbitals around it and once unhybridized p orbital . let 's go ahead and draw that . we have a carbon right here and this is an sp2 hybridized orbitals . we 're gon na draw in . there 's one sp2 hybrid orbital . here 's another sp2 hybrid orbital and here 's another one . then we go back up to here and we can see that each one of those orbitals . let me go ahead and mark this . each one of those sp2 hybrid orbitals has one electron in it . each one of these orbitals has one electron . i go back down here and i put in the one electron in each one of my orbitals like that . i know that each of those carbons is going to have an unhybridized p orbital here . an unhybridized p orbital with one electron too . let me go ahead and draw that in . i 'll go ahead and use a different color . we have our unhybridized p orbital like that and there 's one electron in our unhybridized p orbital . each of the carbons was sp2 hybridized . let me go ahead and draw the dot structure right here again so we can take a look at it . the dot structure for ethylene . let 's do the other carbon now . the carbon on the right is also sp2 hybridized . we can go ahead and draw in an sp2 hybrid orbital and there 's one electron in that orbital and then there 's another one with one electron and then here 's another one with one electron . this carbon being sp2 hybridized also has an unhybridized p orbital with one electron . go ahead and draw in that p orbital with its one electron . we also have some hydrogens . we have some hydrogens to think about here . each carbon is bonded to two hydrogens . let me go ahead and put in the hydrogens . the hydrogen has a valance electron in an unhybridized s orbital . i 'm going ahead and putting in the s orbital and the one valance electron from hydrogen like this . when we take a look at what we 've drawn here , we can see some head on overlap of orbitals , which we know from our earlier video is called a sigma bond . here 's the head on overlap of orbitals . that 's a sigma bond . here 's another head on overlap of orbitals . the carbon carbon bond , here 's also a head on overlap of orbitals and then we have these two over here . we have a total of five sigma bonds in our molecules . let me go ahead and write that over here . there are five sigma bonds . if i would try to find those on my dot structure this would be a sigma bond . this would be a sigma bond . one of these two is a sigma bond and then these over here . a total of five sigma bonds and then we have a new type of bonding . these unhybridized p orbitals can overlap side by side . up here and down here . we get side by side overlap of our p orbitals and this creates a pi bond . a pi bond , let me go ahead and write that here . a pi bond is side by side overlap . there is overlap above and below this sigma bond here and that 's going to prevent free rotation . when we 're looking at the example of ethane , we have free rotation about the sigma bond that connected the two carbons but because of this pi bond here , this pi bond is going to prevent rotations so we do n't get different confirmations of the ethylene molecules . no free rotation due to the pi bonds . when you 're looking at the dot structure , one of these bonds is the pi bonds , i 'm just gon na say it 's this one right here . if you have a double bond , one of those bonds , the sigma bond and one of those bonds is a pi bond . we have a total of one pi bond in the ethylene molecule . if you 're thinking about the distance between the two carbons , let me go ahead and use a different color for that . the distance between this carbon and this carbon . it turns out to be approximately 1.34 angstroms , which is shorter than the distance between the two carbons in the ethane molecule . remember for ethane , the distance was approximately 1.54 angstroms . a double bond is shorter than a single bond . one way to think about that is the increased s character . this increased s character means electron density is closer to the nucleus and that 's going to make this lobe a little bit shorter than before and that 's going to decrease the distance between these two carbon atoms here . 1.34 angstroms . let 's look at the dot structure again and see how we can analyze this using the concept of steric number . let me go ahead and redraw the dot structure . we have our carbon carbon double bond here and our hydrogens like that . if you 're approaching this situation using steric number remember to find the hybridization . we can use this concept . steric number is equal to the number of sigma bonds plus number of lone pairs of electrons . if my goal was to find the steric number for this carbon . i count up my number of sigma bonds . that 's one , two and then i know when i double bond one of those is sigma and one of those is pi . one of those is a sigma bond . a total of three sigma bonds . i have zero loan pairs of electrons around that carbon . three plus zero , gives me a steric number of three . i need three hybrid orbitals and we 've just seen in this video that three sp2 hybrid orbitals form if we 're dealing with sp2 hybridization . if we get a steric number of three , you 're gon na think about sp2 hybridization . one s orbital and two p orbitals hybridizing . that carbon is sp2 hybridized and of course , this one is too . both of them are sp2 hybridized . let 's do another example . let 's do boron trifluoride . bf3 . if you wan na draw the dot structure of bf3 , you would have boron and then you would surround it with your flourines here and you would have an octet of electrons around each flourine . i go ahead and put those in on my dot structure . if your goal is to figure out the hybridization of this boron here . what is the hybridization stage of this boron ? let 's use the concept of steric number . once again , let 's use steric number . find the hybridization of this boron . steric number is equal to number of sigma bonds . that 's one , two , three . three sigma bonds plus lone pairs of electrons . that 's zero . steric number of three tells us this boron is sp2 hybridized . this boron is gon na have three sp2 hybrid orbitals and one p orbital . one unhybridized p orbital . let 's go ahead and draw that . we have a boron here bonded to three flourines and also it 's going to have an unhybrized p orbital . now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three . when you 're thinking about the sp2 hybrid orbitals that you create . sp2 hybrid orbital , sp2 , sp2 and then one unhybridized p orbital right here . boron only has three valance electrons . let 's go ahead and put in those valance electrons . one , two and three . it does n't have any electrons in its unhybridized p orbital . over here when we look at the picture , this has an empty orbital and so boron can accept a pair of electrons . we 're thinking about its chemical behavior , one of the things that bf3 can do , the boron can accept an electron pair and function as a lewis acid . that 's one way in thinking about how hybridizational allows you to think about the structure and how something might react . this boron turns out to be sp2 hybridized . this boron here is sp2 hybridized and so we can also talk about the geometry of the molecule . it 's planar . around this boron , it 's planar and so therefore , your bond angles are 120 degrees . if you have boron right here and you 're thinking about a circle . a circle is 360 degrees . if you divide a 360 by 3 , you get 120 degrees for all of these bond angles . in the next video , we 'll look at sp hybridization .
now , remember when you are dealing with boron , it has one last valance electron and carbon . carbon have four valance electrons . boron has only three .
despite originally having space in its 2p^2 orbitals for 2 electrons of opposite spin ?