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3tl5gAwHj6w-002|It's called valence shell electron pair repulsion. |
3tl5gAwHj6w-003|That is, you have electron pairs and either bonds or lone pairs and they repel each other, and you'd like to distribute them in space as far apart as possible. |
3tl5gAwHj6w-004|So how do you do that? |
3tl5gAwHj6w-005|Well, you know the steric number is the number of lone pairs and the number of bonded atoms-- that tells you how many things you have to distribute in space. |
3tl5gAwHj6w-006|We can look at that for several examples. |
3tl5gAwHj6w-007|For instance, if we have six things distributed in space, here are six balloons distributed around the central point. |
3tl5gAwHj6w-008|And the balloons touching each other demonstrates the steric interaction, how everything is trying to spread itself out as far as possible in space. |
3tl5gAwHj6w-009|And what you get is this kind of arrangement. |
3tl5gAwHj6w-010|In this arrangement, all the positions are equivalent, even though I've got red and yellow balloons. |
3tl5gAwHj6w-011|The yellow balloons or the green balloons could be on top, each position is identical. |
3tl5gAwHj6w-012|But what you have here is an octahedral shape. |
3tl5gAwHj6w-013|An octahedral shape is six things arranged around a central point-- six vertices. |
3tl5gAwHj6w-018|What if I have five things? |
3tl5gAwHj6w-019|Well, with a balloon, it's easy to go from six to five. |
3tl5gAwHj6w-024|So here's a 90-degree bond angle between green and yellow, but the green greens are 120 degrees. |
3tl5gAwHj6w-028|We'll lose one more of our balloons. |
3tl5gAwHj6w-029|Four things. |
3tl5gAwHj6w-030|Now here's a tetrahedral shape. |
3tl5gAwHj6w-031|Notice that these four things don't naturally arrange themselves as a square plane. |
3tl5gAwHj6w-032|I could probably force them into a square planar configuration, but I have to hold them there. |
3tl5gAwHj6w-039|So you can see this VSEPR arrangement is better energetically, the electron pairs have bigger angles, more space between them. |
AV5nRtwehD0-000|Let's see if you can predict some bond orders. |
AV5nRtwehD0-001|Here we'll take nitrite going to nitrate-- NO2- to NO3-. |
AV5nRtwehD0-002|In that conversion, it's an oxidation. |
AV5nRtwehD0-008|We're looking at NO2- and NO3- and how the bond order changes when I do that oxidation reaction. |
AV5nRtwehD0-009|Now, the way to do this is to draw a Lewis dot structure and all the resonance structures for NO2- and NO3- and determine the bond order. |
AV5nRtwehD0-010|Now I know you can do that, so I'm going to show you a shortcut that also works. |
AV5nRtwehD0-011|What we're going to recognize is that NO2- is isoelectronic with something we already know-- ozone. |
AV5nRtwehD0-012|Ozone has 3 oxygen atoms, each has 6 valence electrons, so that's 18 electrons. |
AV5nRtwehD0-013|Ozone is an 18-electron system. |
AV5nRtwehD0-014|What about NO2-? |
AV5nRtwehD0-015|Well, let's look at the periodic table. |
AV5nRtwehD0-016|NO2-, here's oxygen with its 6 valence electrons. |
AV5nRtwehD0-017|Nitrogen with 5. |
AV5nRtwehD0-019|So in total, we have 6 electrons from each oxygen, that's 12, 5 from the nitrogen, that 17, 1 from the charge, that's 18. |
AV5nRtwehD0-020|So an 18-electron system. |
AV5nRtwehD0-022|So if I have the same number of atoms, the same number of electrons, they'll bond together in the same way. |
AV5nRtwehD0-025|We already know the carbonate bond order is 1 and 1/3. |
AV5nRtwehD0-026|So going from 1 and 1/2 to 1 and 1/3 in that oxidation reaction decreases the bond order. |
AV5nRtwehD0-027|The correct answer here-- decreases. |
BSnGqJsLCZA-000|Acid strength is dependent on a variety of factors. |
BSnGqJsLCZA-001|Certainly one of them is the strength of the bond holding the proton on the molecule. |
BSnGqJsLCZA-004|And that equilibrium constant is determined not just by enthalpic or bond strength considerations. |
BSnGqJsLCZA-005|It's also entropic considerations. |
BSnGqJsLCZA-006|Remember, K for this reaction depends on the standard state free energy difference between the products and reactants. |
BSnGqJsLCZA-007|In fact, we define weak acids as having a K less than 1 or a positive free energy difference. |
BSnGqJsLCZA-008|And strong acids as a K greater than 1 for this reaction or a spontaneous or negative standard state free energy difference. |
BSnGqJsLCZA-009|So it's an enthalpic and an entropic term. |
BSnGqJsLCZA-013|HF, it takes 543 kilojoules to break that bond. |
BSnGqJsLCZA-014|Where HBr, only 354 kilojoules to break that bond. |
BSnGqJsLCZA-018|In this case, depended upon the strengths of the bond holding the proton on the molecule. |
t-WkmDPN8Ws-000|Let's look at another physical equilibrium, the solid gas sublimation equilibrium for iodine. |
t-WkmDPN8Ws-001|I've written the reaction here, iodine solid going to iodine gas. |
t-WkmDPN8Ws-002|The question I have is what happens when I add more iodine solid to this equilibrium? |
t-WkmDPN8Ws-003|What happens to the intensity of the I2 gas? |
t-WkmDPN8Ws-011|We're talking about the sublimation of iodine going from the solid to the gas. |
t-WkmDPN8Ws-013|So an equilibrium that's dynamic, but from a macroscopic sense it appears static. |
t-WkmDPN8Ws-014|Let's look at the equilibrium expression if we write products over reactants. |
t-WkmDPN8Ws-015|For this we would have the partial pressure of the I2 gas, I2 solid, that's a pure solid. |
t-WkmDPN8Ws-016|And pure liquids and solids don't appear in our equilibrium expressions. |
t-WkmDPN8Ws-019|If it's independent of the solid, then as I add solid I shouldn't expect this equilibrium to change. |
t-WkmDPN8Ws-020|Doesn't appear in the equilibrium expression, the solid, so the solid doesn't affect the equilibrium expression. |
t-WkmDPN8Ws-021|Let's actually see if that happens. |
t-WkmDPN8Ws-022|What I have here is the solid and vapor in equilibrium. |
t-WkmDPN8Ws-031|The equilibrium constant is a function only of the vapor pressure. |
t-WkmDPN8Ws-035|But at the same time, I've added that more solid, that gives more surface area for the vapor that's there to condense back on the solid. |
t-WkmDPN8Ws-036|So I don't change the position of equilibrium. |
-DM5LRsQNbA-000|Let's look at the equilibrium between NO2 gas and N2O4 if we change the temperature. |
-DM5LRsQNbA-001|So given this equilibrium, what happens to the color intensity when I raise the temperature? |
-DM5LRsQNbA-002|Will it increase, decrease, or stay the same? |
-DM5LRsQNbA-009|We're talking about the equilibrium between NO2 and N2O4. |
-DM5LRsQNbA-010|NO2, a brown gas, N2O4, a clear gas. |
-DM5LRsQNbA-011|Now we know that this is an exothermic chemical reaction, so heat is a product. |
-DM5LRsQNbA-012|If we add heat, then we'd expect the equilibrium to shift back towards reactants to relieve that stress by Le Chatelier's principle. |
-DM5LRsQNbA-014|So let's see that happen. |
-DM5LRsQNbA-015|I have two flasks here, both with the N2O4 and the NO2 gas. |
-DM5LRsQNbA-016|The NO2 gas is the brown gas, the reactant. |
-DM5LRsQNbA-017|I'm going to heat one and cool the other to see the effect of temperature. |
-DM5LRsQNbA-018|We expect that the hotter, the higher temperature will be darker. |
-DM5LRsQNbA-019|Let's see that happen. |
-DM5LRsQNbA-026|So that's a beautiful effect. |
-DM5LRsQNbA-027|And in fact, it's common for equilibrium constants to behave like this. |
-DM5LRsQNbA-028|If you have an equilibrium constant for an exothermic reaction, the equilibrium constant will decrease with temperature. |
-DM5LRsQNbA-032|Now this is actually the equilibrium constant changing. |
-DM5LRsQNbA-033|So this is not a shift where q changes, it's actually k that changes. |
-DM5LRsQNbA-034|So I go to a k that smaller, a smaller k means bigger denominator, more reactants. |
-DM5LRsQNbA-035|The color increases, k gets smaller. |
rYgOBHeQyjA-000|Let's look at a titration where we dilute the solution and then titrate again. |
rYgOBHeQyjA-001|So I'm going to titrate the strong acid, nitric acid, HNO3, with the strong base, potassium hydroxide. |
rYgOBHeQyjA-002|I'm then going to dilute it and do the titration again. |
rYgOBHeQyjA-003|So shown here in yellow is the initial titration. |
rYgOBHeQyjA-004|I'm going to dilute by a factor of 10, just add water, titrate again. |
rYgOBHeQyjA-012|We're looking at two titrations, one where we've diluted by a factor of 10. |
rYgOBHeQyjA-013|If you dilute a strong acid solution by a factor of 10, you change the H3O+ concentration by a factor of 10, and you increase the pH by one unit. |
rYgOBHeQyjA-014|So the initial pH should be one unit higher than the undiluted. |
rYgOBHeQyjA-015|But in dilution, you're just adding water. |
rYgOBHeQyjA-016|You're not adding any more moles of acid. |
rYgOBHeQyjA-017|So the same number of moles of acid are there. |
rYgOBHeQyjA-018|So it requires the same number of moles of base to reach equivalence point. |
nyp9rT6pRXo-001|That's a fancy way of saying water absorbs energy as it boils. |
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