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|---|---|---|---|---|---|---|---|---|---|---|---|
1ldppgyi8
|
chemistry
|
chemical-bonding-and-molecular-structure
|
hybridization-and-vsepr-theory
|
<p>Match List I with List II</p>
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;}
.tg .tg-7btt{border-color:inherit;font-weight:bold;text-align:center;vertical-align:top}
.tg .tg-0pky{border-color:inherit;text-align:left;vertical-align:top}
</style>
<table class="tg" style="undefined;table-layout: fixed; width: 581px">
<colgroup>
<col style="width: 50px"/>
<col style="width: 223px"/>
<col style="width: 45px"/>
<col style="width: 263px"/>
</colgroup>
<thead>
<tr>
<th class="tg-7btt"></th>
<th class="tg-7btt">List I</th>
<th class="tg-7btt"></th>
<th class="tg-7btt">List II</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-0pky">A.</td>
<td class="tg-0pky">$$\mathrm{XeF_4}$$</td>
<td class="tg-0pky">I.</td>
<td class="tg-0pky">See-saw</td>
</tr>
<tr>
<td class="tg-0pky">B.</td>
<td class="tg-0pky">$$\mathrm{SF_4}$$</td>
<td class="tg-0pky">II.</td>
<td class="tg-0pky">Square-planar</td>
</tr>
<tr>
<td class="tg-0pky">C.</td>
<td class="tg-0pky">$$\mathrm{NH_{4}^{+}}$$</td>
<td class="tg-0pky">III.</td>
<td class="tg-0pky">Bent T-shaped</td>
</tr>
<tr>
<td class="tg-0pky">D.</td>
<td class="tg-0pky">$$\mathrm{BrF_3}$$</td>
<td class="tg-0pky">IV.</td>
<td class="tg-0pky">Tetrahedral</td>
</tr>
</tbody>
</table></p>
<p>Choose the correct answer from the options given below :</p>
|
[{"identifier": "A", "content": "A - II, B - I, C - III, D - IV"}, {"identifier": "B", "content": "A - II, B - I, C - IV, D - III"}, {"identifier": "C", "content": "A - IV, B - I, C - II, D - III"}, {"identifier": "D", "content": "A - IV, B - III, C - II, D - I"}]
|
["B"]
| null |
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1leh9wcqv/41039764-7736-413d-ace2-672e44c709ef/ae342970-b390-11ed-96ad-459254a85a5f/file-1leh9wcqw.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1leh9wcqv/41039764-7736-413d-ace2-672e44c709ef/ae342970-b390-11ed-96ad-459254a85a5f/file-1leh9wcqw.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 31st January Morning Shift Chemistry - Chemical Bonding & Molecular Structure Question 79 English Explanation 1">
<br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1leh9wwqk/a4055a2a-f9a3-4e3d-8fc8-a8effaaa39fb/bda58fc0-b390-11ed-adcf-7927667ba5c3/file-1leh9wwql.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1leh9wwqk/a4055a2a-f9a3-4e3d-8fc8-a8effaaa39fb/bda58fc0-b390-11ed-adcf-7927667ba5c3/file-1leh9wwql.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 31st January Morning Shift Chemistry - Chemical Bonding & Molecular Structure Question 79 English Explanation 2">
<br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1leh9xf0t/8ffea33d-56a4-4d21-951b-ce1fe5b0fa63/cbc56fd0-b390-11ed-adcf-7927667ba5c3/file-1leh9xf0u.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1leh9xf0t/8ffea33d-56a4-4d21-951b-ce1fe5b0fa63/cbc56fd0-b390-11ed-adcf-7927667ba5c3/file-1leh9xf0u.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 31st January Morning Shift Chemistry - Chemical Bonding & Molecular Structure Question 79 English Explanation 3">
<br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1leh9xuwm/b5a1d86b-8158-4ff0-904f-fbf4afe04b8c/d80a7560-b390-11ed-adcf-7927667ba5c3/file-1leh9xuwn.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1leh9xuwm/b5a1d86b-8158-4ff0-904f-fbf4afe04b8c/d80a7560-b390-11ed-adcf-7927667ba5c3/file-1leh9xuwn.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 31st January Morning Shift Chemistry - Chemical Bonding & Molecular Structure Question 79 English Explanation 4">
|
mcq
|
jee-main-2023-online-31st-january-morning-shift
| 873
|
1ldr4ho5x
|
chemistry
|
chemical-bonding-and-molecular-structure
|
hybridization-and-vsepr-theory
|
<p>Match List I with List II</p>
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;}
.tg .tg-7btt{border-color:inherit;font-weight:bold;text-align:center;vertical-align:top}
.tg .tg-0pky{border-color:inherit;text-align:left;vertical-align:top}
</style>
<table class="tg" style="undefined;table-layout: fixed; width: 607px">
<colgroup>
<col style="width: 50px"/>
<col style="width: 249px"/>
<col style="width: 45px"/>
<col style="width: 263px"/>
</colgroup>
<thead>
<tr>
<th class="tg-7btt"></th>
<th class="tg-7btt">List I<br/>(molecules/ions)</th>
<th class="tg-7btt"></th>
<th class="tg-7btt">List II<br/>(No. of lone pairs of e$$^-$$ on central atom)</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-0pky">A.</td>
<td class="tg-0pky">$$\mathrm{IF_7}$$</td>
<td class="tg-0pky">I.</td>
<td class="tg-0pky">Three</td>
</tr>
<tr>
<td class="tg-0pky">B.</td>
<td class="tg-0pky">$$\mathrm{ICl}$$$$_4^ - $$</td>
<td class="tg-0pky">II.</td>
<td class="tg-0pky">One</td>
</tr>
<tr>
<td class="tg-0pky">C.</td>
<td class="tg-0pky">$$\mathrm{XeF_6}$$</td>
<td class="tg-0pky">III.</td>
<td class="tg-0pky">Two</td>
</tr>
<tr>
<td class="tg-0pky">D.</td>
<td class="tg-0pky">$$\mathrm{XeF_2}$$</td>
<td class="tg-0pky">IV.</td>
<td class="tg-0pky">Zero</td>
</tr>
</tbody>
</table></p>
<p>Choose the correct answer from the options given below :</p>
|
[{"identifier": "A", "content": "A - II, B - I, C - IV, D - III"}, {"identifier": "B", "content": "A - IV, B - I, C - II, D - III"}, {"identifier": "C", "content": "A - IV, B - III, C - II, D - I"}, {"identifier": "D", "content": "A - II, B - III, C - IV, D - I"}]
|
["C"]
| null |
<p>(A) IF$$_7$$ $$-$$ 0 lone pairs</p>
<p>(B) ICI$$_4^ - $$ $$-$$ 2 lone pairs</p>
<p>(C) XeF$$_6$$ $$-$$ 1 lone pair</p>
<p>(D) XeF$$_2$$ $$-$$ 3 lone pairs</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lepzncy5/03ee6ab7-d140-4db8-bd2d-97f2c0c7d311/a6a17fd0-b85b-11ed-8195-4f3c56fa1eb5/file-1lepzncy6.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lepzncy5/03ee6ab7-d140-4db8-bd2d-97f2c0c7d311/a6a17fd0-b85b-11ed-8195-4f3c56fa1eb5/file-1lepzncy6.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 30th January Morning Shift Chemistry - Chemical Bonding & Molecular Structure Question 78 English Explanation"></p>
|
mcq
|
jee-main-2023-online-30th-january-morning-shift
| 874
|
1ldv10o5c
|
chemistry
|
chemical-bonding-and-molecular-structure
|
hybridization-and-vsepr-theory
|
<p>The total number of lone pairs of electrons on oxygen atoms of ozone is __________.</p>
|
[]
| null |
6
|
<p>Total no, of lone pairs on oxygen atoms = 6</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lebwzzd6/9bf8acbe-9e54-4d5f-a5e8-daffdbfcaa39/512da2a0-b09e-11ed-a1d6-7715f24e8c48/file-1lebwzzd7.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lebwzzd6/9bf8acbe-9e54-4d5f-a5e8-daffdbfcaa39/512da2a0-b09e-11ed-a1d6-7715f24e8c48/file-1lebwzzd7.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 25th January Morning Shift Chemistry - Chemical Bonding & Molecular Structure Question 71 English Explanation"></p>
|
integer
|
jee-main-2023-online-25th-january-morning-shift
| 875
|
lgnze5sm
|
chemistry
|
chemical-bonding-and-molecular-structure
|
hybridization-and-vsepr-theory
|
The number of $\mathrm{P}-\mathrm{O}-\mathrm{P}$ bonds in $\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{7},\left(\mathrm{HPO}_{3}\right)_{3}$ and $\mathrm{P}_{4} \mathrm{O}_{10}$ are respectively :
|
[{"identifier": "A", "content": "$1,3,6$"}, {"identifier": "B", "content": "$1,2,4$"}, {"identifier": "C", "content": "$0,3,4$"}, {"identifier": "D", "content": "$0,3,6$"}]
|
["A"]
| null |
Let's analyze each compound separately:
<br><br>
1. $\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{7}$: This compound is also known as pyrophosphoric acid. Its structure has two phosphorus atoms (P) connected by one oxygen atom (O), forming a P-O-P bond. There are no other P-O-P bonds in its structure.
Number of P-O-P bonds = 1
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lgrz4gh5/49778edb-2079-49bb-b3be-cf467d6d80be/9909f590-e10b-11ed-ad8b-a134a1ca2315/file-1lgrz4gh6.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lgrz4gh5/49778edb-2079-49bb-b3be-cf467d6d80be/9909f590-e10b-11ed-ad8b-a134a1ca2315/file-1lgrz4gh6.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 15th April Morning Shift Chemistry - Chemical Bonding & Molecular Structure Question 67 English Explanation 1">
<br><br>
2. $\left(\mathrm{HPO}_{3}\right)_{3}$: This compound is also known as cyclotriphosphoric acid. It has three phosphorus atoms and three oxygen atoms, arranged in a cyclic structure with each P atom connected to two O atoms, forming a six-membered ring. In this structure, there are three P-O-P bonds.
Number of P-O-P bonds = 3
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lgrz55iw/42f2a330-ff44-4349-8918-fcd23d610352/ac636b80-e10b-11ed-8f74-951e213f5826/file-1lgrz55ix.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lgrz55iw/42f2a330-ff44-4349-8918-fcd23d610352/ac636b80-e10b-11ed-8f74-951e213f5826/file-1lgrz55ix.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 15th April Morning Shift Chemistry - Chemical Bonding & Molecular Structure Question 67 English Explanation 2">
<br><br>
3. $\mathrm{P}_{4} \mathrm{O}_{10}$: This compound is also known as phosphorus pentoxide. Its structure consists of four phosphorus atoms (P) and ten oxygen atoms (O) connected in a cyclic arrangement. In the structure, each phosphorus atom is connected to four oxygen atoms, with two terminal oxygen atoms and two bridging oxygen atoms. There are six P-O-P bonds in the P<sub>4</sub>O<sub>10</sub> structure.
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lgrz5yu6/122fe8e7-9567-4d3f-af8c-59e722c2256d/c3083be0-e10b-11ed-8f74-951e213f5826/file-1lgrz5yu7.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lgrz5yu6/122fe8e7-9567-4d3f-af8c-59e722c2256d/c3083be0-e10b-11ed-8f74-951e213f5826/file-1lgrz5yu7.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 15th April Morning Shift Chemistry - Chemical Bonding & Molecular Structure Question 67 English Explanation 3">
<br><br>
So, the number of P-O-P bonds in $\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{7},\left(\mathrm{HPO}_{3}\right)_{3}$, and $\mathrm{P}_{4} \mathrm{O}_{10}$ are 1, 3, and 6, respectively.
|
mcq
|
jee-main-2023-online-15th-april-morning-shift
| 876
|
1lgq472j1
|
chemistry
|
chemical-bonding-and-molecular-structure
|
hybridization-and-vsepr-theory
|
$$\mathrm{ClF}_{5}$$ at room temperature is a:
|
[{"identifier": "A", "content": "Colourless liquid with square pyramidal geometry"}, {"identifier": "B", "content": "Colourless gas with square pyramidal geometry"}, {"identifier": "C", "content": "Colourless gas with trigonal bipyramidal geometry."}, {"identifier": "D", "content": "Colourless liquid with trigonal bipyramidal geometry"}]
|
["A"]
| null |
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lgtthcsj/24428c52-a29a-4693-b7e1-f2c2c9a5e1de/1a3c7530-e20f-11ed-9e52-3dd25100c100/file-1lgtthcsk.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lgtthcsj/24428c52-a29a-4693-b7e1-f2c2c9a5e1de/1a3c7530-e20f-11ed-9e52-3dd25100c100/file-1lgtthcsk.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 13th April Morning Shift Chemistry - Chemical Bonding & Molecular Structure Question 64 English Explanation"><br>$\mathrm{ClF}_5$ is actually a colorless liquid at room temperature with a square pyramidal geometry.
|
mcq
|
jee-main-2023-online-13th-april-morning-shift
| 877
|
1lgsz22g1
|
chemistry
|
chemical-bonding-and-molecular-structure
|
hybridization-and-vsepr-theory
|
<p>The maximum number of lone pairs of electrons on the central atom from the following species is ____________.</p>
<p>$$\mathrm{ClO}_{3}{ }^{-}, \mathrm{XeF}_{4}, \mathrm{SF}_{4}$$ and $$\mathrm{I}_{3}{ }^{-}$$</p>
|
[]
| null |
3
|
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1libj9fix/3a3c9e8e-e382-40b1-8b6a-179c32cfb663/75a6ac90-ff99-11ed-8d3c-6fd4c50a7427/file-1libj9fiy.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1libj9fix/3a3c9e8e-e382-40b1-8b6a-179c32cfb663/75a6ac90-ff99-11ed-8d3c-6fd4c50a7427/file-1libj9fiy.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 11th April Evening Shift Chemistry - Chemical Bonding & Molecular Structure Question 61 English Explanation">
|
integer
|
jee-main-2023-online-11th-april-evening-shift
| 878
|
1lguzfmzq
|
chemistry
|
chemical-bonding-and-molecular-structure
|
hybridization-and-vsepr-theory
|
<p>Match List - I with List - II:</p>
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;}
.tg .tg-baqh{text-align:center;vertical-align:top}
.tg .tg-amwm{font-weight:bold;text-align:center;vertical-align:top}
</style>
<table class="tg" style="undefined;table-layout: fixed; width: 554px">
<colgroup>
<col style="width: 49px"/>
<col style="width: 201px"/>
<col style="width: 54px"/>
<col style="width: 250px"/>
</colgroup>
<thead>
<tr>
<th class="tg-amwm"></th>
<th class="tg-amwm">List - I Species</th>
<th class="tg-amwm"></th>
<th class="tg-amwm">List - II Geometry/Shape</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-baqh">A.</td>
<td class="tg-baqh">$$\mathrm{H_3O^+}$$</td>
<td class="tg-baqh">I.</td>
<td class="tg-baqh">Tetrahedral</td>
</tr>
<tr>
<td class="tg-baqh">B.</td>
<td class="tg-baqh">Acetylide anion</td>
<td class="tg-baqh">II.</td>
<td class="tg-baqh">Linear</td>
</tr>
<tr>
<td class="tg-baqh">C.</td>
<td class="tg-baqh">$$\mathrm{NH_4^+}$$</td>
<td class="tg-baqh">III.</td>
<td class="tg-baqh">Pyramidal</td>
</tr>
<tr>
<td class="tg-baqh">D.</td>
<td class="tg-baqh">$$\mathrm{ClO_2^-}$$</td>
<td class="tg-baqh">IV.</td>
<td class="tg-baqh">Bent</td>
</tr>
</tbody>
</table></p>
<p>Choose the correct answer from the options given below:</p>
|
[{"identifier": "A", "content": "A-III, B-I, C-II, D-IV"}, {"identifier": "B", "content": "A-III, B-II, C-I, D-IV"}, {"identifier": "C", "content": "A-III, B-IV, C-I, D-II"}, {"identifier": "D", "content": "A-III, B-IV, C-II, D-I"}]
|
["B"]
| null |
<p>Let's consider each species in List I and determine their geometries/shapes.</p>
<p>A. $\mathrm{H_3O^+}$: This ion is formed by the addition of a proton to a water molecule. The central atom (O) is surrounded by three Hydrogen atoms and one lone pair, making its shape pyramidal.</p>
<p>B. Acetylide anion: The acetylide ion ($\mathrm{C_2H^-}$ or $\mathrm{C_2^{2-}}$) has a linear geometry as the three atoms are in a straight line.</p>
<p>C. $\mathrm{NH_4^+}$: The ammonium ion has a central nitrogen atom surrounded by four hydrogen atoms, giving it a tetrahedral geometry.</p>
<p>D. $\mathrm{ClO_2^-}$: The chlorite ion has a central chlorine atom surrounded by two oxygen atoms and one lone pair, which gives it a bent or V-shaped geometry.</p>
<p>Matching these to List II, we get:</p>
<p>A. $\mathrm{H_3O^+}$ - III. Pyramidal<br/><br/>
B. Acetylide anion - II. Linear<br/><br/>
C. $\mathrm{NH_4^+}$ - I. Tetrahedral<br/><br/>
D. $\mathrm{ClO_2^-}$ - IV. Bent</p>
|
mcq
|
jee-main-2023-online-11th-april-morning-shift
| 879
|
1lgvv5z7m
|
chemistry
|
chemical-bonding-and-molecular-structure
|
hybridization-and-vsepr-theory
|
<p>The number of molecules from the following which contain only two lone pair of electrons is ________</p>
<p>$$\mathrm{H}_{2} \mathrm{O}, \mathrm{N}_{2}, \mathrm{CO}, \mathrm{XeF}_{4}, \mathrm{NH}_{3}, \mathrm{NO}, \mathrm{CO}_{2}, \mathrm{~F}_{2}$$</p>
|
[]
| null |
3
|
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lixqco47/38985813-f5e4-4424-ba74-edb4b5dd5979/6de7f270-0bce-11ee-acbf-7f9d4e788425/file-6y3zli1lixqco48.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lixqco47/38985813-f5e4-4424-ba74-edb4b5dd5979/6de7f270-0bce-11ee-acbf-7f9d4e788425/file-6y3zli1lixqco48.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh; vertical-align: baseline;" alt="JEE Main 2023 (Online) 10th April Evening Shift Chemistry - Chemical Bonding & Molecular Structure Question 59 English Explanation 1">
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lixqd7pq/3939164a-df87-4ac3-8b2b-9c6f43cbad64/7d0badf0-0bce-11ee-acbf-7f9d4e788425/file-6y3zli1lixqd7pr.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lixqd7pq/3939164a-df87-4ac3-8b2b-9c6f43cbad64/7d0badf0-0bce-11ee-acbf-7f9d4e788425/file-6y3zli1lixqd7pr.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh; vertical-align: baseline;" alt="JEE Main 2023 (Online) 10th April Evening Shift Chemistry - Chemical Bonding & Molecular Structure Question 59 English Explanation 2">
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lixqe8wf/68ad2051-a3ab-467c-8364-82bcb3ca8fa9/99c531f0-0bce-11ee-acbf-7f9d4e788425/file-6y3zli1lixqe8wg.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lixqe8wf/68ad2051-a3ab-467c-8364-82bcb3ca8fa9/99c531f0-0bce-11ee-acbf-7f9d4e788425/file-6y3zli1lixqe8wg.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh; vertical-align: baseline;" alt="JEE Main 2023 (Online) 10th April Evening Shift Chemistry - Chemical Bonding & Molecular Structure Question 59 English Explanation 3"><br>
The number of molecules having only 2 lone pair of electrons $=3$<br><br>
Which are $\mathrm{H}_2 \mathrm{O} , \mathrm{N}_2$ and $\mathrm{CO}$<br><br>
<b>Note:-</b><br><br>
$\mathrm{XeF}_4$ have 2 lps on central atom, but we are asked about lone pair in molecule
|
integer
|
jee-main-2023-online-10th-april-evening-shift
| 880
|
1lgyfyt7u
|
chemistry
|
chemical-bonding-and-molecular-structure
|
hybridization-and-vsepr-theory
|
<p>The compound which does not exist is</p>
|
[{"identifier": "A", "content": "(NH$$_4$$)$$_2$$BeF$$_4$$"}, {"identifier": "B", "content": "PbEt$$_4$$"}, {"identifier": "C", "content": "BeH$$_2$$"}, {"identifier": "D", "content": "NaO$$_2$$"}]
|
["D"]
| null |
Sodium superoxide (NaO$_2$) is not a stable compound. It is not found in normal conditions due to its high reactivity.
|
mcq
|
jee-main-2023-online-10th-april-morning-shift
| 881
|
1lgyhijw5
|
chemistry
|
chemical-bonding-and-molecular-structure
|
hybridization-and-vsepr-theory
|
<p>The number of bent-shaped molecule/s from the following is __________</p>
<p>N$$_3^-$$, NO$$_2^-$$, I$$_3^-$$, O$$_3$$, SO$$_2$$</p>
|
[]
| null |
3
|
<p>A bent-shaped molecule has a molecular geometry with a central atom bonded to two other atoms and one or two pairs of non-bonding electrons. The VSEPR (Valence Shell Electron Pair Repulsion) theory helps us predict the shapes of molecules.</p>
<p>Let's consider the given molecules:</p>
<ol>
<li><p>N$_3^-$: The azide ion has a linear shape, not bent. The nitrogen in the middle is connected to two other nitrogens and there are no lone pairs on the central atom.</p>
</li><br/>
<li><p>NO$_2^-$: The nitrite ion has a bent shape. The central nitrogen atom is connected to two oxygen atoms and has one lone pair of electrons, giving it a bent geometry.</p>
</li><br/>
<li><p>I$_3^-$: The triiodide ion has a linear shape, not bent. The central iodine atom is connected to two other iodine atoms and there are no lone pairs on the central atom.</p>
</li><br/>
<li><p>O$_3$: Ozone has a bent shape. The central oxygen atom is connected to two other oxygen atoms and has one lone pair of electrons, giving it a bent geometry.</p>
</li><br/>
<li><p>SO$_2$: Sulfur dioxide has a bent shape. The central sulfur atom is connected to two oxygen atoms and has one lone pair of electrons, giving it a bent geometry.</p>
</li>
</ol>
<p>Therefore, there are 3 bent-shaped molecules: NO$_2^-$, O$_3$, and SO$_2$.</p>
|
integer
|
jee-main-2023-online-10th-april-morning-shift
| 882
|
1lgyhlpij
|
chemistry
|
chemical-bonding-and-molecular-structure
|
hybridization-and-vsepr-theory
|
<p>The sum of lone pairs present on the central atom of the interhalogen IF$$_5$$ and IF$$_7$$ is _________</p>
|
[]
| null |
1
|
<p>Interhalogen compounds are the substances that consist of two different halogens. The most common type of interhalogen compounds are binary, containing only two different elements.</p>
<p>In IF$_5$, iodine (I) is the central atom. Iodine has 7 valence electrons, 5 of which are used for bonding with the 5 fluorine (F) atoms. This leaves 2 electrons, or 1 lone pair on the iodine atom.</p>
<p>In IF$_7$, iodine (I) is the central atom again. Iodine has 7 valence electrons, all of which are used for bonding with the 7 fluorine (F) atoms. So, there are no lone pairs on the iodine atom.</p>
<p>Therefore, the sum of lone pairs present on the central atom of the interhalogens IF$_5$ and IF$_7$ is 1.</p>
|
integer
|
jee-main-2023-online-10th-april-morning-shift
| 883
|
1lh27wei6
|
chemistry
|
chemical-bonding-and-molecular-structure
|
hybridization-and-vsepr-theory
|
<p>Match List I with List II</p>
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;}
.tg .tg-c3ow{border-color:inherit;text-align:center;vertical-align:top}
.tg .tg-7btt{border-color:inherit;font-weight:bold;text-align:center;vertical-align:top}
.tg .tg-0pky{border-color:inherit;text-align:left;vertical-align:top}
</style>
<table class="tg" style="undefined;table-layout: fixed; width: 598px">
<colgroup>
<col style="width: 46px"/>
<col style="width: 240px"/>
<col style="width: 48px"/>
<col style="width: 264px"/>
</colgroup>
<thead>
<tr>
<th class="tg-7btt"></th>
<th class="tg-7btt">LIST I <br/>Oxide</th>
<th class="tg-7btt"></th>
<th class="tg-7btt">LIST II <br/>Type of bond</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-c3ow">A.</td>
<td class="tg-0pky">$$\mathrm{N_2O_4}$$</td>
<td class="tg-c3ow">I.</td>
<td class="tg-0pky">1 N = O bond</td>
</tr>
<tr>
<td class="tg-c3ow">B.</td>
<td class="tg-0pky">$$\mathrm{NO_2}$$</td>
<td class="tg-c3ow">II.</td>
<td class="tg-0pky">1 N $$-$$ O $$-$$ N bond</td>
</tr>
<tr>
<td class="tg-c3ow">C.</td>
<td class="tg-0pky">$$\mathrm{N_2O_5}$$</td>
<td class="tg-c3ow">III.</td>
<td class="tg-0pky">1 N $$-$$ N bond</td>
</tr>
<tr>
<td class="tg-c3ow">D.</td>
<td class="tg-0pky">$$\mathrm{N_2O}$$</td>
<td class="tg-c3ow">IV.</td>
<td class="tg-0pky">1 N=N / N $$\equiv$$ N bond</td>
</tr>
</tbody>
</table></p>
<p>Choose the correct answer from the options given below:</p>
|
[{"identifier": "A", "content": "A-II, B-I, C-III, D-IV"}, {"identifier": "B", "content": "A-III, B-I, C-II, D-IV"}, {"identifier": "C", "content": "A-III, B-I, C-IV, D-II"}, {"identifier": "D", "content": "A-II, B-IV, C-III, D-I"}]
|
["B"]
| null |
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1llm9poeo/b606cdcb-1706-473e-ad1b-625b32825c55/63727d00-40e5-11ee-8358-1796f8234e47/file-6y3zli1llm9poep.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1llm9poeo/b606cdcb-1706-473e-ad1b-625b32825c55/63727d00-40e5-11ee-8358-1796f8234e47/file-6y3zli1llm9poep.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 60vh;vertical-align: baseline" alt="JEE Main 2023 (Online) 6th April Morning Shift Chemistry - Chemical Bonding & Molecular Structure Question 52 English Explanation">
|
mcq
|
jee-main-2023-online-6th-april-morning-shift
| 885
|
1lh29lf3g
|
chemistry
|
chemical-bonding-and-molecular-structure
|
hybridization-and-vsepr-theory
|
<p>The number of species from the following which have square pyramidal structure is _________</p>
<p>$$\mathrm{PF}_{5}, \mathrm{BrF}_{4}^{-}, \mathrm{IF}_{5}, \mathrm{BrF}_{5}, \mathrm{XeOF}_{4}, \mathrm{ICl}_{4}^{-}$$</p>
|
[]
| null |
3
|
<p>A square pyramidal structure has five bonds and one lone pair, making a total of six electron pairs around the central atom. The geometry of such a molecule can be analyzed using the VSEPR theory.</p>
<ol>
<li>$\mathrm{PF}_{5}$: 5 bond pairs and 0 lone pairs, so its geometry is trigonal bipyramidal, not square pyramidal.</li>
<br/><li>$\mathrm{BrF}_{4}^{-}$: 4 bond pairs and 2 lone pairs, so its geometry is square planar, not square pyramidal.</li><br/>
<li>$\mathrm{IF}_{5}$: 5 bond pairs and 1 lone pair, so its geometry is square pyramidal.</li><br/>
<li>$\mathrm{BrF}_{5}$: 5 bond pairs and 1 lone pair, so its geometry is square pyramidal.</li>
<br/><li>$\mathrm{XeOF}_{4}$: 5 bond pairs and 1 lone pair, so its geometry is square pyramidal.</li><br/>
<li>$\mathrm{ICl}_{4}^{-}$: 4 bond pairs and 2 lone pairs, so its geometry is square planar, not square pyramidal.</li>
</ol>
<p>So the number of species from the given list that have a square pyramidal structure is 3.</p>
|
integer
|
jee-main-2023-online-6th-april-morning-shift
| 886
|
1lh32lx4j
|
chemistry
|
chemical-bonding-and-molecular-structure
|
hybridization-and-vsepr-theory
|
<p>The number of species having a square planar shape from the following is __________.</p>
<p>$$\mathrm{XeF}_{4}, \mathrm{SF}_{4}, \mathrm{SiF}_{4}, \mathrm{BF}_{4}^{-}, \mathrm{BrF}_{4}^{-},\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+},\left[\mathrm{FeCl}_{4}\right]^{2-},\left[\mathrm{PtCl}_{4}\right]^{2-}$$</p>
|
[]
| null |
4
|
$\mathrm{XeF}_4 \rightarrow$ Square planar<br/><br/>
$\mathrm{SF}_4 \rightarrow$ See saw<br/><br/>
$\mathrm{SiF}_4 \rightarrow$ Tetrahedral<br/><br/>
$\mathrm{BF}_4^{-} \rightarrow$ Tetrahedral<br/><br/>
$\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right]^{2+} \rightarrow$ Square planar<br/><br/>
$\left[\mathrm{FeCl}_4\right]^{2-} \rightarrow$ Tetrahedral<br/><br/>
$\left[\mathrm{PtCl}_4\right]^{2-} \rightarrow$ Square planar<br/><br/>
$\mathrm{BrF}_4^{-} \rightarrow$ Square planar<br/><br/>
So, 4 square planer shape compounds are present.
|
integer
|
jee-main-2023-online-6th-april-evening-shift
| 887
|
lsbn2unr
|
chemistry
|
chemical-bonding-and-molecular-structure
|
hybridization-and-vsepr-theory
|
The number of molecules/ion/s having trigonal bipyramidal shape is _______.
<br/><br/>$\mathrm{PF}_5, \mathrm{BrF}_5, \mathrm{PCl}_5,\left[\mathrm{Pt} \mathrm{Cl}_4\right]^{2-}, \mathrm{BF}_3, \mathrm{Fe}(\mathrm{CO})_5$
|
[]
| null |
3
|
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsfah88p/a66619bb-5ed5-4177-af56-d7795c400dad/68b8ce90-c7a4-11ee-a4c7-e152a9884fc5/file-6y3zli1lsfah88q.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsfah88p/a66619bb-5ed5-4177-af56-d7795c400dad/68b8ce90-c7a4-11ee-a4c7-e152a9884fc5/file-6y3zli1lsfah88q.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 1st February Morning Shift Chemistry - Chemical Bonding & Molecular Structure Question 45 English Explanation 1">
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsfah4os/d3171949-b1b9-4a09-9cf0-8f24491fe2a9/65fa23c0-c7a4-11ee-a4c7-e152a9884fc5/file-6y3zli1lsfah4ot.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsfah4os/d3171949-b1b9-4a09-9cf0-8f24491fe2a9/65fa23c0-c7a4-11ee-a4c7-e152a9884fc5/file-6y3zli1lsfah4ot.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 1st February Morning Shift Chemistry - Chemical Bonding & Molecular Structure Question 45 English Explanation 2">
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsfaivfp/bc6e091a-18aa-4fdc-a935-fd91f70e5aff/9672bd50-c7a4-11ee-a4c7-e152a9884fc5/file-6y3zli1lsfaivfq.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsfaivfp/bc6e091a-18aa-4fdc-a935-fd91f70e5aff/9672bd50-c7a4-11ee-a4c7-e152a9884fc5/file-6y3zli1lsfaivfq.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 1st February Morning Shift Chemistry - Chemical Bonding & Molecular Structure Question 45 English Explanation 3">
|
integer
|
jee-main-2024-online-1st-february-morning-shift
| 888
|
jaoe38c1lscs8hzs
|
chemistry
|
chemical-bonding-and-molecular-structure
|
hybridization-and-vsepr-theory
|
<p>The number of non-polar molecules from the following is _________. $$\mathrm{HF}, \mathrm{H}_2 \mathrm{O}, \mathrm{SO}_2, \mathrm{H}_2, \mathrm{CO}_2, \mathrm{CH}_4, \mathrm{NH}_3, \mathrm{HCl}, \mathrm{CHCl}_3, \mathrm{BF}_3$$</p>
|
[]
| null |
4
|
<p>To determine whether a molecule is polar or non-polar, we must consider the difference in electronegativity between the atoms and the symmetry of the molecule. Polar molecules occur when there is an electronegativity difference between the bonded atoms. Non-polar molecules either do not have any polar bonds or the polarities cancel each other out because of a symmetrical arrangement. Let's consider each molecule individually:</p>
<ul>
<li><b>HF</b> (Hydrogen fluoride): This molecule is polar due to the high electronegativity difference between hydrogen (H) and fluorine (F).</li>
<li><b>H<sub>2</sub>O</b> (Water): It is polar because of its bent shape and the difference in electronegativity between oxygen and hydrogen.</li>
<li><b>SO<sub>2</sub></b> (Sulfur dioxide): The molecule is non-linear and has polar bonds, making it a polar molecule.</li>
<li><b>H<sub>2</sub></b> (Hydrogen): This molecule is non-polar because it consists of two identical atoms which share electrons evenly.</li>
<li><b>CO<sub>2</sub></b> (Carbon dioxide): Even though C=O bonds are polar, the molecule is linear and the polarities cancel out, making CO<sub>2</sub> non-polar.</li>
<li><b>CH<sub>4</sub></b> (Methane): It is non-polar because the C-H bonds are evenly distributed in a tetrahedral shape, canceling out any dipole moments.</li>
<li><b>NH<sub>3</sub></b> (Ammonia): This molecule is polar; it has a trigonal pyramidal shape with a lone pair on nitrogen, creating a dipole moment.</li>
<li><b>HCl</b> (Hydrogen chloride): The molecule is polar due to the electronegativity difference between hydrogen and chlorine.</li>
<li><b>CHCl<sub>3</sub></b> (Chloroform): It has polar C-Cl bonds and is not symmetrical, resulting in a polar molecule.</li>
<li><b>BF<sub>3</sub></b> (Boron trifluoride): This molecule is non-polar despite having polar bonds, because the shape of the molecule is trigonal planar and the dipoles cancel out.</li>
</ul>
<p>Given this information, the non-polar molecules from the list are: $$\mathrm{H}_2, \mathrm{CO}_2, \mathrm{CH}_4, \mathrm{BF}_3$$.</p>
<p>Therefore, there are 4 non-polar molecules in the list given.</p>
|
integer
|
jee-main-2024-online-27th-january-evening-shift
| 889
|
jaoe38c1lsfk70hj
|
chemistry
|
chemical-bonding-and-molecular-structure
|
hybridization-and-vsepr-theory
|
<p>Number of compounds with one lone pair of electrons on central atom amongst following is _________.</p>
<p>$$\mathrm{O}_3, \mathrm{H}_2 \mathrm{O}, \mathrm{SF}_4, \mathrm{ClF}_3, \mathrm{NH}_3, \mathrm{BrF}_5, \mathrm{XeF}_4$$</p>
|
[]
| null |
4
|
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lt2oifv6/1f764780-52ba-411c-a9aa-a11f5170d5e2/512a7230-d481-11ee-ac6c-b973c46afe12/file-1lt2oifv7.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lt2oifv6/1f764780-52ba-411c-a9aa-a11f5170d5e2/512a7230-d481-11ee-ac6c-b973c46afe12/file-1lt2oifv7.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 29th January Morning Shift Chemistry - Chemical Bonding & Molecular Structure Question 34 English Explanation"></p>
|
integer
|
jee-main-2024-online-29th-january-morning-shift
| 891
|
1lsg7kbq3
|
chemistry
|
chemical-bonding-and-molecular-structure
|
hybridization-and-vsepr-theory
|
<p>The molecule / ion with square pyramidal shape is</p>
|
[{"identifier": "A", "content": "$$\\mathrm{PCl}_5$$\n"}, {"identifier": "B", "content": "$$\\left[\\mathrm{Ni}(\\mathrm{CN})_4\\right]^{2-}$$\n"}, {"identifier": "C", "content": "$$\\mathrm{PF}_5$$\n"}, {"identifier": "D", "content": "$$\\mathrm{BrF}_5$$"}]
|
["D"]
| null |
<p>$$\mathrm{BrF_5}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lspvcpn2/a7b05e75-c4d1-4eca-8add-63d4a0a5db32/e1529ee0-cd75-11ee-9be1-85ca54b9efcb/file-6y3zli1lspvcpn3.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lspvcpn2/a7b05e75-c4d1-4eca-8add-63d4a0a5db32/e1529ee0-cd75-11ee-9be1-85ca54b9efcb/file-6y3zli1lspvcpn3.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 30th January Evening Shift Chemistry - Chemical Bonding & Molecular Structure Question 32 English Explanation"></p>
|
mcq
|
jee-main-2024-online-30th-january-evening-shift
| 892
|
1lsgxnho5
|
chemistry
|
chemical-bonding-and-molecular-structure
|
hybridization-and-vsepr-theory
|
<p>Aluminium chloride in acidified aqueous solution forms an ion having geometry</p>
|
[{"identifier": "A", "content": "Square planar\n"}, {"identifier": "B", "content": "Octahedral\n"}, {"identifier": "C", "content": "Trigonal bipyramidal\n"}, {"identifier": "D", "content": "Tetrahedral"}]
|
["B"]
| null |
<p>$$\mathrm{AlCl}_3$$ in acidified aqueous solution forms octahedral geometry $$[\mathrm{Al}(\mathrm{H}_2 \mathrm{O})_6]^{3+}$$</p>
|
mcq
|
jee-main-2024-online-30th-january-morning-shift
| 893
|
1lsgy08ma
|
chemistry
|
chemical-bonding-and-molecular-structure
|
hybridization-and-vsepr-theory
|
<p>Match List I with List II.</p>
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;}
.tg .tg-baqh{text-align:center;vertical-align:top}
.tg .tg-amwm{font-weight:bold;text-align:center;vertical-align:top}
.tg .tg-0lax{text-align:left;vertical-align:top}
</style>
<table class="tg" style="undefined;table-layout: fixed; width: 551px">
<colgroup>
<col style="width: 74px"/>
<col style="width: 179px"/>
<col style="width: 82px"/>
<col style="width: 216px"/>
</colgroup>
<thead>
<tr>
<th class="tg-amwm"></th>
<th class="tg-amwm">List I<br/>Molecule</th>
<th class="tg-amwm"></th>
<th class="tg-amwm">List II<br/>Shape</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-baqh">(A)</td>
<td class="tg-0lax">$$\mathrm{BrF_5}$$</td>
<td class="tg-baqh">(I)</td>
<td class="tg-0lax">T-Shape</td>
</tr>
<tr>
<td class="tg-baqh">(B)</td>
<td class="tg-0lax">$$\mathrm{H_2O}$$</td>
<td class="tg-baqh">(II)</td>
<td class="tg-0lax">See saw</td>
</tr>
<tr>
<td class="tg-baqh">(C)</td>
<td class="tg-0lax">$$\mathrm{ClF_3}$$</td>
<td class="tg-baqh">(III)</td>
<td class="tg-0lax">Bent</td>
</tr>
<tr>
<td class="tg-baqh">(D)</td>
<td class="tg-0lax">$$\mathrm{SF_4}$$</td>
<td class="tg-baqh">(IV)</td>
<td class="tg-0lax">Square pyramidal</td>
</tr>
</tbody>
</table></p>
<p>Choose the correct answer from the options given below :</p>
|
[{"identifier": "A", "content": "(A)-(I), (B)-(II), (C)-(IV), (D)-(III)\n"}, {"identifier": "B", "content": "(A)-(IV), (B)-(III), (C)-(I), (D)-(II)\n"}, {"identifier": "C", "content": "(A)-(III), (B)-(IV), (C)-(I), (D)-(II)\n"}, {"identifier": "D", "content": "(A)-(II), (B)-(I), (C)-(III), (D)-(IV)"}]
|
["B"]
| null |
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsqmhidy/bdac6e90-3aaf-4cbe-9e2a-a19aaadc1c8f/fd40af60-cddf-11ee-a0d3-7b75c4537559/file-6y3zli1lsqmhidz.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsqmhidy/bdac6e90-3aaf-4cbe-9e2a-a19aaadc1c8f/fd40af60-cddf-11ee-a0d3-7b75c4537559/file-6y3zli1lsqmhidz.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 30th January Morning Shift Chemistry - Chemical Bonding & Molecular Structure Question 27 English Explanation 1">
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsqmhez4/c4faed98-bb86-4560-9e01-7085d854101f/fa9df100-cddf-11ee-a0d3-7b75c4537559/file-6y3zli1lsqmhez5.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsqmhez4/c4faed98-bb86-4560-9e01-7085d854101f/fa9df100-cddf-11ee-a0d3-7b75c4537559/file-6y3zli1lsqmhez5.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 30th January Morning Shift Chemistry - Chemical Bonding & Molecular Structure Question 27 English Explanation 2">
|
mcq
|
jee-main-2024-online-30th-january-morning-shift
| 894
|
luxzq9h9
|
chemistry
|
chemical-bonding-and-molecular-structure
|
hybridization-and-vsepr-theory
|
<p>The correct increasing order for bond angles among $$\mathrm{BF}_3, \mathrm{PF}_3$$ and $$\mathrm{ClF}_3$$ is :</p>
|
[{"identifier": "A", "content": "$$\\mathrm{BF}_3=\\mathrm{PF}_3<\\mathrm{ClF}_3$$\n"}, {"identifier": "B", "content": "$$\\mathrm{BF}_3<\\mathrm{PF}_3<\\mathrm{ClF}_3$$\n"}, {"identifier": "C", "content": "$$\\mathrm{ClF}_3<\\mathrm{PF}_3<\\mathrm{BF}_3$$\n"}, {"identifier": "D", "content": "$$\\mathrm{PF}_3<\\mathrm{BF}_3<\\mathrm{ClF}_3$$"}]
|
["C"]
| null |
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw2xkwii/4c270791-c55b-4947-92e4-4c0873ba2d87/6df13890-1009-11ef-bffa-cb0d0a489c35/file-1lw2xkwij.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw2xkwii/4c270791-c55b-4947-92e4-4c0873ba2d87/6df13890-1009-11ef-bffa-cb0d0a489c35/file-1lw2xkwij.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 9th April Evening Shift Chemistry - Chemical Bonding & Molecular Structure Question 25 English Explanation"></p>
<p>Order of bond angle is</p>
<p>$$\mathrm{ClF}_3<\mathrm{PF}_3<\mathrm{BF}_3$$</p>
|
mcq
|
jee-main-2024-online-9th-april-evening-shift
| 895
|
lv0vyqj5
|
chemistry
|
chemical-bonding-and-molecular-structure
|
hybridization-and-vsepr-theory
|
<p>Number of molecules/ions from the following in which the central atom is involved in $$\mathrm{sp}^3$$ hybridization is ________.</p>
<p>$$\mathrm{NO}_3^{-}, \mathrm{BCl}_3, \mathrm{ClO}_2^{-}, \mathrm{ClO}_3^{-}$$</p>
|
[{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "3"}]
|
["A"]
| null |
<p>To determine the number of molecules/ions in which the central atom is involved in $$\mathrm{sp}^3$$ hybridization, we must analyze the hybridization state for each central atom. Hybridization is typically determined by the number of sigma bonds and lone pairs on the central atom.</p>
<p>Let's evaluate each molecule/ion:</p>
<p><b>1. $$\mathrm{NO}_3^{-}$$</b>:</p>
<p>The central atom is nitrogen (N). In $$\mathrm{NO}_3^{-}$$, nitrogen forms three sigma bonds with oxygen atoms and has no lone pairs. Using the formula for hybridization:</p>
<p> Number of hybrid orbitals = number of sigma bonds + number of lone pairs</p>
<p>For $$\mathrm{NO}_3^{-}$$: $$3 + 0 = 3$$ hybrid orbitals. Therefore, nitrogen is $$\mathrm{sp}^2$$ hybridized.</p>
<p><b>2. $$\mathrm{BCl}_3$$</b>:</p>
<p>The central atom is boron (B). In $$\mathrm{BCl}_3$$, boron forms three sigma bonds with chlorine atoms and has no lone pairs. Using the same formula:</p>
<p> For $$\mathrm{BCl}_3$$: $$3 + 0 = 3$$ hybrid orbitals. Therefore, boron is $$\mathrm{sp}^2$$ hybridized.</p>
<p><b>3. $$\mathrm{ClO}_2^{-}$$</b>:</p>
<p>The central atom is chlorine (Cl). In $$\mathrm{ClO}_2^{-}$$, chlorine forms two sigma bonds with oxygen atoms and has two lone pairs. Therefore,:</p>
<p> For $$\mathrm{ClO}_2^{-}$$: $$2 + 2 = 4$$ hybrid orbitals. Thus, chlorine is $$\mathrm{sp}^3$$ hybridized.</p>
<p><b>4. $$\mathrm{ClO}_3^{-}$$</b>:</p>
<p>The central atom is chlorine (Cl). In $$\mathrm{ClO}_3^{-}$$, chlorine forms three sigma bonds with oxygen atoms and has one lone pair. Therefore,:</p>
<p> For $$\mathrm{ClO}_3^{-}$$: $$3 + 1 = 4$$ hybrid orbitals. Thus, chlorine is $$\mathrm{sp}^3$$ hybridized.</p>
<p>Based on the analysis, the molecules/ions $$\mathrm{ClO}_2^{-}$$ and $$\mathrm{ClO}_3^{-}$$ have their central atoms involved in $$\mathrm{sp}^3$$ hybridization.</p>
<p>Therefore, the total number is <strong>2</strong>.</p>
<p>Thus, the correct option is <strong>Option A</strong>.</p>
|
mcq
|
jee-main-2024-online-4th-april-morning-shift
| 897
|
lv2es2q1
|
chemistry
|
chemical-bonding-and-molecular-structure
|
hybridization-and-vsepr-theory
|
<p>The number of species from the following that have pyramidal geometry around the central atom is _________</p>
<p>$$\mathrm{S}_2 \mathrm{O}_3^{2-}, \mathrm{SO}_4^{2-}, \mathrm{SO}_3^{2-}, \mathrm{S}_2 \mathrm{O}_7^{2-}$$</p>
|
[{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "1"}]
|
["D"]
| null |
<p>$$\mathrm{SO}_3^{2-}$$ is the only species with pyramidal geometry.</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwhsvaob/e56e6438-5a66-40f1-a233-1345578a2fd3/c311a9b0-1836-11ef-9081-0ded966256d8/file-1lwhsvaoc.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwhsvaob/e56e6438-5a66-40f1-a233-1345578a2fd3/c311a9b0-1836-11ef-9081-0ded966256d8/file-1lwhsvaoc.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 4th April Evening Shift Chemistry - Chemical Bonding & Molecular Structure Question 16 English Explanation"></p>
|
mcq
|
jee-main-2024-online-4th-april-evening-shift
| 898
|
lv3xmasa
|
chemistry
|
chemical-bonding-and-molecular-structure
|
hybridization-and-vsepr-theory
|
<p>The shape of carbocation is :</p>
|
[{"identifier": "A", "content": "tetrahedral\n"}, {"identifier": "B", "content": "diagonal pyramidal\n"}, {"identifier": "C", "content": "diagonal\n"}, {"identifier": "D", "content": "trigonal planar"}]
|
["D"]
| null |
<p>Carbocation is $$s p^2$$ hybridised hence it's shape is trigonal planar</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw572fuz/7d920da9-b2af-44a2-8fce-743980ccd75b/191aa9b0-1148-11ef-b5fc-89359a20637a/file-1lw572fv0.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw572fuz/7d920da9-b2af-44a2-8fce-743980ccd75b/191aa9b0-1148-11ef-b5fc-89359a20637a/file-1lw572fv0.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 8th April Evening Shift Chemistry - Chemical Bonding & Molecular Structure Question 14 English Explanation"></p>
|
mcq
|
jee-main-2024-online-8th-april-evening-shift
| 899
|
lv5gsw7q
|
chemistry
|
chemical-bonding-and-molecular-structure
|
hybridization-and-vsepr-theory
|
<p>Match List I with List II</p>
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;}
.tg .tg-c3ow{border-color:inherit;text-align:center;vertical-align:top}
.tg .tg-7btt{border-color:inherit;font-weight:bold;text-align:center;vertical-align:top}
.tg .tg-0pky{border-color:inherit;text-align:left;vertical-align:top}
</style>
<table class="tg" style="undefined;table-layout: fixed; width: 664px">
<colgroup>
<col style="width: 75px"/>
<col style="width: 252px"/>
<col style="width: 75px"/>
<col style="width: 262px"/>
</colgroup>
<thead>
<tr>
<th class="tg-7btt"></th>
<th class="tg-7btt">LIST I<br/>(Molecule)</th>
<th class="tg-7btt"></th>
<th class="tg-7btt">LIST II<br/>(Shape)</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-c3ow">A.</td>
<td class="tg-0pky">$$\mathrm{NH_3}$$</td>
<td class="tg-c3ow">I.</td>
<td class="tg-0pky">Square pyramid</td>
</tr>
<tr>
<td class="tg-c3ow">B. </td>
<td class="tg-0pky">$$\mathrm{BrF_5}$$</td>
<td class="tg-c3ow">II.</td>
<td class="tg-0pky">Tetrahedral</td>
</tr>
<tr>
<td class="tg-c3ow">C.</td>
<td class="tg-0pky">$$\mathrm{PCl_5}$$</td>
<td class="tg-c3ow">III.</td>
<td class="tg-0pky">Trigonal pyramidal</td>
</tr>
<tr>
<td class="tg-c3ow">D.</td>
<td class="tg-0pky">$$\mathrm{CH_4}$$</td>
<td class="tg-c3ow">IV.</td>
<td class="tg-0pky">Trigonal bipyramidal</td>
</tr>
</tbody>
</table></p>
<p>Choose the correct answer from the options given below:</p>
|
[{"identifier": "A", "content": "A-II, B-IV, C-I, D-III\n"}, {"identifier": "B", "content": "A-III, B-I, C-IV, D-II\n"}, {"identifier": "C", "content": "A-IV, B-III, C-I, D-II\n"}, {"identifier": "D", "content": "A-III, B-IV, C-I, D-II"}]
|
["B"]
| null |
<p>To match the molecules in List I with their corresponding shapes in List II, we need to consider the electronic geometry and the VSEPR (Valence Shell Electron Pair Repulsion) theory. Here’s the detailed analysis:</p>
<p><b>$$\mathrm{NH_3}$$ (Ammonia)</b>: Ammonia has a central nitrogen atom bonded to three hydrogen atoms and has one lone pair of electrons. This leads to a trigonal pyramidal shape.</p>
<p><b>$$\mathrm{BrF_5}$$ (Bromine pentafluoride)</b>: Bromine pentafluoride has a central bromine atom bonded to five fluorine atoms and has one lone pair of electrons. This leads to a square pyramidal shape.</p>
<p><b>$$\mathrm{PCl_5}$$ (Phosphorus pentachloride)</b>: Phosphorus pentachloride has a central phosphorus atom bonded to five chlorine atoms and no lone pairs. This results in a trigonal bipyramidal shape.</p>
<p><b>$$\mathrm{CH_4}$$ (Methane)</b>: Methane has a central carbon atom bonded to four hydrogen atoms with no lone pairs, forming a tetrahedral structure.</p>
<p>Using the above analyses, we can match the molecules and shapes as follows:</p>
<ul>
<li>A ($$\mathrm{NH_3}$$) - III (Trigonal pyramidal)</li>
<li>B ($$\mathrm{BrF_5}$$) - I (Square pyramid)</li>
<li>C ($$\mathrm{PCl_5}$$) - IV (Trigonal bipyramidal)</li>
<li>D ($$\mathrm{CH_4}$$) - II (Tetrahedral)</li>
</ul>
<p>Therefore, the correct matching is option B:</p>
<p><b>Option B: A-III, B-I, C-IV, D-II</b></p>
|
mcq
|
jee-main-2024-online-8th-april-morning-shift
| 900
|
lv9s26z7
|
chemistry
|
chemical-bonding-and-molecular-structure
|
hybridization-and-vsepr-theory
|
<p>Match List I with List II.</p>
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;}
.tg .tg-c3ow{border-color:inherit;text-align:center;vertical-align:top}
.tg .tg-7btt{border-color:inherit;font-weight:bold;text-align:center;vertical-align:top}
.tg .tg-0pky{border-color:inherit;text-align:left;vertical-align:top}
</style>
<table class="tg" style="undefined;table-layout: fixed; width: 703px">
<colgroup>
<col style="width: 75px"/>
<col style="width: 252px"/>
<col style="width: 75px"/>
<col style="width: 301px"/>
</colgroup>
<thead>
<tr>
<th class="tg-7btt"></th>
<th class="tg-7btt">LIST I<br/></th>
<th class="tg-7btt"></th>
<th class="tg-7btt">LIST II<br/></th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-c3ow">A.</td>
<td class="tg-0pky">$$<br/>\mathrm{ICl}<br/>$$</td>
<td class="tg-c3ow">I.</td>
<td class="tg-0pky">T - shape</td>
</tr>
<tr>
<td class="tg-c3ow">B.</td>
<td class="tg-0pky">$$<br/>\mathrm{ICl}_3<br/>$$</td>
<td class="tg-c3ow">II.</td>
<td class="tg-0pky">Square pyramidal</td>
</tr>
<tr>
<td class="tg-c3ow">C.</td>
<td class="tg-0pky">$$<br/>\mathrm{ClF}_5<br/>$$</td>
<td class="tg-c3ow">III.</td>
<td class="tg-0pky">Pentagonal bipyramidal</td>
</tr>
<tr>
<td class="tg-c3ow">D.</td>
<td class="tg-0pky">$$<br/>\mathrm{IF}_7<br/>$$</td>
<td class="tg-c3ow">IV.</td>
<td class="tg-0pky">Linear</td>
</tr>
</tbody>
</table></p>
<p>Choose the correct answer from the options given below :</p>
|
[{"identifier": "A", "content": "(A)-(I), (B)-(IV), (C)-(III), (D)-(II)\n"}, {"identifier": "B", "content": "(A)-(IV), (B)-(I), (C)-(II), (D)-(III)\n"}, {"identifier": "C", "content": "(A)-(I), (B)-(III), (C)-(II), (D)-(IV)\n"}, {"identifier": "D", "content": "(A)-(IV), (B)-(III), (C)-(II), (D)-(I)"}]
|
["B"]
| null |
<p>$$\begin{aligned}
& \mathrm{ICl} \rightarrow s p^3 \rightarrow 1 \mathrm{bp}+3 \mathrm{lp} \rightarrow \text { Linear } \\
& \mathrm{ICl}_3 \rightarrow s p^3 d \rightarrow 3 \mathrm{bp}+2 \mathrm{lp} \rightarrow \mathrm{T} \text {-shape } \\
& \mathrm{CIF}_5 \rightarrow s p^3 d^2 \rightarrow 5 \mathrm{bp}+1 \mathrm{lp} \rightarrow \text { Square Pyramidal } \\
& \mathrm{IF}_7 \rightarrow s p^3 d^3 \rightarrow 7 \text { bp only } \rightarrow \text { Pentagonal bipyramidal } \\
& \text { (A)-(IV), (B)-(I), (C)-(II), D-(III) }
\end{aligned}$$</p>
|
mcq
|
jee-main-2024-online-5th-april-evening-shift
| 901
|
lvb2acem
|
chemistry
|
chemical-bonding-and-molecular-structure
|
hybridization-and-vsepr-theory
|
<p>Total number of species from the following with central atom utilising $$\mathrm{sp}^2$$ hybrid orbitals for bonding is ________.</p> <p>$$\mathrm{NH}_3, \mathrm{SO}_2, \mathrm{SiO}_2, \mathrm{BeCl}_2, \mathrm{C}_2 \mathrm{H}_2, \mathrm{C}_2 \mathrm{H}_4, \mathrm{BCl}_3, \mathrm{HCHO}, \mathrm{C}_6 \mathrm{H}_6, \mathrm{BF}_3, \mathrm{C}_2 \mathrm{H}_4 \mathrm{Cl}_2$$</p>
|
[]
| null |
6
|
<p>$$\mathrm{SO}_2, \mathrm{C}_2 \mathrm{H}_4, \mathrm{BCl}_3, \mathrm{HCHO}, \mathrm{C}_6 \mathrm{H}_6, \mathrm{BF}_3$$ are $$s p^2$$ hybridised central atom</p>
|
integer
|
jee-main-2024-online-6th-april-evening-shift
| 902
|
lvc57i5y
|
chemistry
|
chemical-bonding-and-molecular-structure
|
hybridization-and-vsepr-theory
|
<p>Match List I with List II</p>
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;}
.tg .tg-c3ow{border-color:inherit;text-align:center;vertical-align:top}
.tg .tg-7btt{border-color:inherit;font-weight:bold;text-align:center;vertical-align:top}
.tg .tg-0pky{border-color:inherit;text-align:left;vertical-align:top}
</style>
<table class="tg" style="undefined;table-layout: fixed; width: 690px">
<colgroup>
<col style="width: 75px"/>
<col style="width: 239px"/>
<col style="width: 75px"/>
<col style="width: 301px"/>
</colgroup>
<thead>
<tr>
<th class="tg-7btt"></th>
<th class="tg-7btt">LIST I<br/>(Compound/Species)</th>
<th class="tg-7btt"></th>
<th class="tg-7btt">LIST II<br/>(Shape/Geometry)</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-c3ow">A.</td>
<td class="tg-0pky">$$\mathrm{SF_4}$$</td>
<td class="tg-c3ow">I.</td>
<td class="tg-0pky">Tetrahedral</td>
</tr>
<tr>
<td class="tg-c3ow">B. </td>
<td class="tg-0pky">$$\mathrm{BrF_3}$$</td>
<td class="tg-c3ow">II.</td>
<td class="tg-0pky">Pyramidal</td>
</tr>
<tr>
<td class="tg-c3ow">C.</td>
<td class="tg-0pky">$$\mathrm{BrO_3^-}$$</td>
<td class="tg-c3ow">III.</td>
<td class="tg-0pky">See saw</td>
</tr>
<tr>
<td class="tg-c3ow">D.</td>
<td class="tg-0pky">$$\mathrm{NH_4^+}$$</td>
<td class="tg-c3ow">IV.</td>
<td class="tg-0pky">Bent T-Shape</td>
</tr>
</tbody>
</table></p>
<p>Choose the correct answer from the options given below:</p>
|
[{"identifier": "A", "content": "A-III, B-II, C-IV, D-I\n"}, {"identifier": "B", "content": "A-III, B-IV, C-II, D-I\n"}, {"identifier": "C", "content": "A-II, B-IV, C-III, D-I\n"}, {"identifier": "D", "content": "A-II, B-III, C-I, D-IV"}]
|
["B"]
| null |
<p>$$\begin{aligned}
& \text { A } \rightarrow \text { Sea-saw (III) } \\
& \text { B } \rightarrow \text { Bent T-shape (IV) } \\
& \text { C } \rightarrow \text { Pyramidal (II) } \\
& \text { D } \rightarrow \text { Tetrahedral (I) }
\end{aligned}$$</p>
|
mcq
|
jee-main-2024-online-6th-april-morning-shift
| 903
|
lvc588zk
|
chemistry
|
chemical-bonding-and-molecular-structure
|
hybridization-and-vsepr-theory
|
<p>Match List I with List II</p>
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;}
.tg .tg-c3ow{border-color:inherit;text-align:center;vertical-align:top}
.tg .tg-7btt{border-color:inherit;font-weight:bold;text-align:center;vertical-align:top}
.tg .tg-0pky{border-color:inherit;text-align:left;vertical-align:top}
</style>
<table class="tg" style="undefined;table-layout: fixed; width: 690px">
<colgroup>
<col style="width: 75px"/>
<col style="width: 239px"/>
<col style="width: 75px"/>
<col style="width: 301px"/>
</colgroup>
<thead>
<tr>
<th class="tg-7btt"></th>
<th class="tg-7btt">LIST I<br/>(Molecule/Species)</th>
<th class="tg-7btt"></th>
<th class="tg-7btt">LIST II<br/>(Property/Shape)</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-c3ow">A.</td>
<td class="tg-0pky">$$\mathrm{SO_2Cl_2}$$</td>
<td class="tg-c3ow">I.</td>
<td class="tg-0pky">Paramagnetic</td>
</tr>
<tr>
<td class="tg-c3ow">B. </td>
<td class="tg-0pky">$$\mathrm{NO}$$</td>
<td class="tg-c3ow">II.</td>
<td class="tg-0pky">Diamagnetic</td>
</tr>
<tr>
<td class="tg-c3ow">C.</td>
<td class="tg-0pky">$$\mathrm{NO_2^-}$$</td>
<td class="tg-c3ow">III.</td>
<td class="tg-0pky">Tetrahedral</td>
</tr>
<tr>
<td class="tg-c3ow">D.</td>
<td class="tg-0pky">$$\mathrm{I_3^-}$$</td>
<td class="tg-c3ow">IV.</td>
<td class="tg-0pky">Linear</td>
</tr>
</tbody>
</table></p>
<p>Choose the correct answer from the options given below:</p>
|
[{"identifier": "A", "content": "A-II, B-III, C-I, D-IV\n"}, {"identifier": "B", "content": "A-III, B-I, C-II, D-IV\n"}, {"identifier": "C", "content": "A-IV, B-I, C-III, D-II\n"}, {"identifier": "D", "content": "A-III, B-IV, C-II, D-I"}]
|
["B"]
| null |
<p>$$\begin{aligned}
& \mathrm{A} \rightarrow \text { Tetrahedral (III) } \\
& \mathrm{B} \rightarrow \text { Paramagnetic (I) } \\
& \mathrm{C} \rightarrow \text { Diamagnetic (II) } \\
& \mathrm{D} \rightarrow \text { Linear (IV) }
\end{aligned}$$</p>
|
mcq
|
jee-main-2024-online-6th-april-morning-shift
| 904
|
RduOhjDepCS6jqrz
|
chemistry
|
chemical-bonding-and-molecular-structure
|
hydrogen-bonding
|
Which of the following hydrogen bonds is the strongest?
|
[{"identifier": "A", "content": "O\u2212H\u2026\u2026.N"}, {"identifier": "B", "content": "F\u2212H\u2026\u2026.F "}, {"identifier": "C", "content": "O\u2212H\u2026\u2026.O "}, {"identifier": "D", "content": "O\u2212H\u2026\u2026.F "}]
|
["B"]
| null |
Among F, O and N, F is most electronegative so F pulls bond pair of electron in F - H towards itself and develops highly positive charge on H atom.
<br><br>This highly positive charged H atom creates stongest hydrogen bonding by taking lone pair of electron form electronegative atom F/N/O. Hence among the given options F – H ...... F is the strongest bond.
<br><br><b>Note :</b> F – H ...... N has strongest hydogen bonding among F – H ...... F, F – H ...... O and F – H ...... N because N is least electronegative among F, O and N and can easily donate lone pair of electron.
|
mcq
|
aieee-2007
| 905
|
InA3umoGpofamZ2t
|
chemistry
|
chemical-bonding-and-molecular-structure
|
hydrogen-bonding
|
The intermolecular interaction that is dependent on the inverse cube of distance between the molecule is:
|
[{"identifier": "A", "content": "ion-dipole interaction"}, {"identifier": "B", "content": "London force"}, {"identifier": "C", "content": "hydrogen bond"}, {"identifier": "D", "content": "ion-ion interaction"}]
|
["C"]
| null |
Hydrogen bond is a type of strong electrostatic dipole- dipole intersection and dependent on the inverse cube of distance between the molecular ion-dipole -
<br><br> interaction $$ \propto {1 \over {{r^3}}}.$$
|
mcq
|
jee-main-2015-offline
| 906
|
C8wM97uJad4kiBxUSAxvh
|
chemistry
|
chemical-bonding-and-molecular-structure
|
hydrogen-bonding
|
HF has highest boiling point among hydrogen
halides, because it has :
|
[{"identifier": "A", "content": "lowest dissociation enthalpy"}, {"identifier": "B", "content": "strongest hydrogen bonding"}, {"identifier": "C", "content": "lowest ionic character"}, {"identifier": "D", "content": "strongest van der Waals' interactions"}]
|
["B"]
| null |
Due to strong H-bonding
between HF molecules, HF has highest boiling point among the
hydrogen halides.
|
mcq
|
jee-main-2019-online-9th-april-evening-slot
| 907
|
2San0gkpPqtZNXmtsbjgy2xukfjfemc7
|
chemistry
|
chemical-bonding-and-molecular-structure
|
hydrogen-bonding
|
The potential energy curve for the H<sub>2</sub>
molecule as a function of internuclear distance is :
|
[{"identifier": "A", "content": "<img src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263495/exam_images/ebsqxtpz2rbzxfo0v6jf.webp\" style=\"max-width: 100%;height: auto;display: block;margin: 0 auto;\" loading=\"lazy\" alt=\"JEE Main 2020 (Online) 5th September Morning Slot Chemistry - Chemical Bonding & Molecular Structure Question 143 English Option 1\">"}, {"identifier": "B", "content": "<img src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267455/exam_images/g66jdpfvghuonbxnrsuu.webp\" style=\"max-width: 100%;height: auto;display: block;margin: 0 auto;\" loading=\"lazy\" alt=\"JEE Main 2020 (Online) 5th September Morning Slot Chemistry - Chemical Bonding & Molecular Structure Question 143 English Option 2\">"}, {"identifier": "C", "content": "<img src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267306/exam_images/gfgvygdqedrzud5q17zp.webp\" style=\"max-width: 100%;height: auto;display: block;margin: 0 auto;\" loading=\"lazy\" alt=\"JEE Main 2020 (Online) 5th September Morning Slot Chemistry - Chemical Bonding & Molecular Structure Question 143 English Option 3\">"}, {"identifier": "D", "content": "<img src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263645/exam_images/zbhmtcfaqoa0tnwfdquq.webp\" style=\"max-width: 100%;height: auto;display: block;margin: 0 auto;\" loading=\"lazy\" alt=\"JEE Main 2020 (Online) 5th September Morning Slot Chemistry - Chemical Bonding & Molecular Structure Question 143 English Option 4\">"}]
|
["C"]
| null |
Potential energy curve for H<sub>2</sub> molecules is
<img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267306/exam_images/gfgvygdqedrzud5q17zp.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 5th September Morning Slot Chemistry - Chemical Bonding & Molecular Structure Question 143 English Explanation">
|
mcq
|
jee-main-2020-online-5th-september-morning-slot
| 908
|
jrIpSPZ42eyUlUxyKk1klud2nda
|
chemistry
|
chemical-bonding-and-molecular-structure
|
hydrogen-bonding
|
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.<br/><br/>Assertion A : Dipole-dipole interactions are the only non-covalent interactions, resulting in hydrogen bond formation.<br/><br/>Reason R : Fluorine is the most electronegative element and hydrogen bonds in HF are symmetrical.<br/><br/>In the light of the above statements, choose the most appropriate answer from the options given below :
|
[{"identifier": "A", "content": "Both A and R are true and R is the correct explanation of A"}, {"identifier": "B", "content": "A is true but R is false"}, {"identifier": "C", "content": "A is false but R is true"}, {"identifier": "D", "content": "Both A and R are true but R is NOT the correct explanation of A"}]
|
["C"]
| null |
Dipole - Dipole are not only the interaction
responsible for hydrogen bond formation.
Ion-dipole can also be responsible for
hydrogen bond formation.
<br><br> F is most electronegative element and
anhydrous HF in solid phase has
symmetrical hydrogen bonding.
|
mcq
|
jee-main-2021-online-26th-february-morning-slot
| 909
|
1ktiha5vq
|
chemistry
|
chemical-bonding-and-molecular-structure
|
hydrogen-bonding
|
The number of hydrogen bonded water molecule(s) associated with stoichiometry CuSO<sub>4</sub>.5H<sub>2</sub>O is ____________.
|
[]
| null |
1
|
<img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264346/exam_images/uhagajtbrpgxxvbseqkp.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 31st August Morning Shift Chemistry - Chemical Bonding & Molecular Structure Question 115 English Explanation"><br><br>One hydrogen bonded H<sub>2</sub>O molecule
|
integer
|
jee-main-2021-online-31st-august-morning-shift
| 910
|
1l56wur7s
|
chemistry
|
chemical-bonding-and-molecular-structure
|
hydrogen-bonding
|
<p>The correct order of increasing intermolecular hydrogen bond strength is</p>
|
[{"identifier": "A", "content": "HCN < H<sub>2</sub>O < NH<sub>3</sub>"}, {"identifier": "B", "content": "HCN < CH<sub>4</sub> < NH<sub>3</sub>"}, {"identifier": "C", "content": "CH<sub>4</sub> < HCN < NH<sub>3</sub>"}, {"identifier": "D", "content": "CH<sub>4</sub> < NH<sub>3</sub> < HCN"}]
|
["C"]
| null |
Due to the high difference in electronegativity of H and
N the H-bond strength of NH<sub>3</sub> is highest. There is
no H-bond in CH<sub>4</sub>.<br/><br/>
CH<sub>4</sub> < HCN < NH<sub>3</sub>
|
mcq
|
jee-main-2022-online-27th-june-evening-shift
| 911
|
1ldyfls2h
|
chemistry
|
chemical-bonding-and-molecular-structure
|
hydrogen-bonding
|
<p>Decreasing order of the hydrogen bonding in following forms of water is correctly represented by</p>
<p>A. Liquid water</p>
<p>B. Ice</p>
<p>C. Impure water</p>
<p>Choose the correct answer from the options given below :</p>
|
[{"identifier": "A", "content": "A = B > C"}, {"identifier": "B", "content": "B > A > C"}, {"identifier": "C", "content": "A > B > C"}, {"identifier": "D", "content": "C > B > A"}]
|
["B"]
| null |
Extent of hydrogen bonding :
<br/><br/>
Ice $>$ liquid water $>$ impure water
<br/><br/>
- In ice, 4 molecules of $\mathrm{H}_{2} \mathrm{O}$ are connected to $\mathrm{H}_{2} \mathrm{O}$ molecule.
<br/><br/>
- Impure water will have less hydrogen bonding.
|
mcq
|
jee-main-2023-online-24th-january-morning-shift
| 912
|
1lh333die
|
chemistry
|
chemical-bonding-and-molecular-structure
|
hydrogen-bonding
|
<p>In an ice crystal, each water molecule is hydrogen bonded to ____________ neighbouring molecules.</p>
|
[]
| null |
4
|
In an ice crystal, each water molecule is hydrogen bonded to four neighbouring molecules.
|
integer
|
jee-main-2023-online-6th-april-evening-shift
| 913
|
lsap1fic
|
chemistry
|
chemical-bonding-and-molecular-structure
|
hydrogen-bonding
|
Select the compound from the following that will show intramolecular hydrogen bonding.
|
[{"identifier": "A", "content": "<img src=\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsar2vc6/4f0ca88f-8841-4a89-a07a-96469081fcba/695fec60-c525-11ee-94a9-d124b3795c4c/file-6y3zli1lsar2vc7.png?format=png\" data-orsrc=\"https://app-content.cdn.examgoal.net/image/6y3zli1lsar2vc6/4f0ca88f-8841-4a89-a07a-96469081fcba/695fec60-c525-11ee-94a9-d124b3795c4c/file-6y3zli1lsar2vc7.png\" loading=\"lazy\" style=\"max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline\" alt=\"JEE Main 2024 (Online) 1st February Evening Shift Chemistry - Chemical Bonding & Molecular Structure Question 48 English Option 1\">"}, {"identifier": "B", "content": "$\\mathrm{H}_2 \\mathrm{O}$"}, {"identifier": "C", "content": "$\\mathrm{C}_2 \\mathrm{H}_5 \\mathrm{OH}$"}, {"identifier": "D", "content": "$\\mathrm{NH}_3$"}]
|
["A"]
| null |
<p>Intramolecular hydrogen bonding occurs when a hydrogen atom is covalently bonded to a highly electronegative atom, like oxygen or nitrogen, and this hydrogen atom is also attracted to another electronegative atom within the same molecule. This phenomenon tends to happen when the molecule can form a six-membered or five-membered ring as a result of this bonding, which offers a stable configuration.</p>
<p>Let's consider each of the given options:</p>
<ul>
<li>Option A : The structure in this option contains an ortho-nitrophenol molecule, in which the hydroxyl group (-OH) and the nitro group (-NO2) are positioned on the benzene ring such that they are adjacent to each other. This allows for the formation of an intramolecular hydrogen bond between the hydrogen of the hydroxyl group and the oxygen of the nitro group. Hence, this compound can exhibit intramolecular hydrogen bonding.</li>
<br/><li>Option B : Water ($\mathrm{H}_2 \mathrm{O}$) is capable of intermolecular hydrogen bonding, but it cannot form intramolecular hydrogen bonds as it is a simple molecule with just one oxygen atom and no possibility to form a stable ring structure within a single molecule.</li>
<br/><li>Option C : Ethanol ($\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$) can form intermolecular hydrogen bonds through its hydroxyl group with other ethanol molecules or water molecules, but like water, it cannot form intramolecular hydrogen bonds because it does not have the correct geometry for the hydroxyl group to bond within the same molecule.</li>
<br/><li>Option D : Ammonia ($\mathrm{NH}_3$) is known for intermolecular hydrogen bonding with other ammonia molecules or with water molecules, but there are no hydrogen bond acceptors within the molecule that would allow for intramolecular hydrogen bonding.</li>
</ul>
<p>Therefore, the compound that can show intramolecular hydrogen bonding is in Option A, the ortho-nitrophenol molecule.</p>
|
mcq
|
jee-main-2024-online-1st-february-evening-shift
| 914
|
lsbmkq49
|
chemistry
|
chemical-bonding-and-molecular-structure
|
hydrogen-bonding
|
Given below are two statements: one is labelled as <b>Assertion (A</b>) and the other is labelled as <b>Reason (R)</b>.<br/><br/>
<b>Assertion (A)</b>: $\mathrm{PH}_3$ has lower boiling point than $\mathrm{NH}_3$.<br/><br/>
<b>Reason (R)</b> : In liquid state $\mathrm{NH}_3$ molecules are associated through vander Waal's forces, but $\mathrm{PH}_3$ molecules are associated through hydrogen bonding.
<br/><br/>
In the light of the above statements, choose the <b>most appropriate</b> answer from the options given below :
|
[{"identifier": "A", "content": "Both (A) and (R) are correct and (R) is the correct explanation of (A)"}, {"identifier": "B", "content": "(A) is not correct but (R) is correct"}, {"identifier": "C", "content": "(A) is correct but (R) is not correct"}, {"identifier": "D", "content": "Both $(\\mathbf{A})$ and $(\\mathbf{R})$ are correct but $(\\mathbf{R})$ is not the correct explanation of $(\\mathbf{A})$"}]
|
["C"]
| null |
<p>The correct answer is Option C: (A) is correct but (R) is not correct.</p>
<p><b>Assertion (A)</b>: $\mathrm{PH}_3$ has lower boiling point than $\mathrm{NH}_3$.</p>
<p>This assertion is true. The boiling point of ammonia ($\mathrm{NH}_3$) is higher than that of phosphine ($\mathrm{PH}_3$). Ammonia has a boiling point of about -33.34°C, whereas phosphine has a boiling point of about -87.7°C. The reason for the higher boiling point of ammonia as compared to phosphine lies in the strength and type of intermolecular forces present in the substances.</p>
<p><b>Reason (R)</b> : In liquid state $\mathrm{NH}_3$ molecules are associated through Vander Waals' forces, but $\mathrm{PH}_3$ molecules are associated through hydrogen bonding.</p>
<p>This reason is incorrect. Ammonia ($\mathrm{NH}_3$) can form hydrogen bonds due to the presence of a highly electronegative nitrogen atom bonded to hydrogen atoms. This allows $\mathrm{NH}_3$ molecules to strongly associate with each other through hydrogen bonding, which is a much stronger intermolecular force than Van der Waals forces. This hydrogen bonding is responsible for the relatively high boiling point of ammonia.</p>
<p>On the other hand, phosphine ($\mathrm{PH}_3$) does not form hydrogen bonds because the electronegativity difference between phosphorus and hydrogen is not significant enough to enable the formation of hydrogen bonds. Instead, phosphine molecules are associated mainly through weaker Van der Waals forces, leading to a lower boiling point when compared to ammonia.</p>
<p>In conclusion, the assertion that $\mathrm{PH}_3$ has a lower boiling point than $\mathrm{NH}_3$ is correct, due to the hydrogen bonding present in $\mathrm{NH}_3$ and absent in $\mathrm{PH}_3$. However, the reason provided is incorrect because it incorrectly states that $\mathrm{PH}_3$ forms hydrogen bonds and that $\mathrm{NH}_3$ is associated through Van der Waals forces. It is actually the other way around, so the correct answer is that (A) is true but (R) is false.</p>
|
mcq
|
jee-main-2024-online-1st-february-morning-shift
| 915
|
lv2erjdj
|
chemistry
|
chemical-bonding-and-molecular-structure
|
hydrogen-bonding
|
<p>The correct statement/s about Hydrogen bonding is/are</p>
<p>A. Hydrogen bonding exists when H is covalently bonded to the highly electro negative atom.</p>
<p>B. Intermolecular H bonding is present in $$o$$-nitro phenol</p>
<p>C. Intramolecular $$\mathrm{H}$$ bonding is present in HF.</p>
<p>D. The magnitude of $$\mathrm{H}$$ bonding depends on the physical state of the compound.</p>
<p>E. H-bonding has powerful effect on the structure and properties of compounds</p>
<p>Choose the correct answer from the options given below:</p>
|
[{"identifier": "A", "content": "A, B, C only\n"}, {"identifier": "B", "content": "A only\n"}, {"identifier": "C", "content": "A, D, E only\n"}, {"identifier": "D", "content": "A, B, D only"}]
|
["C"]
| null |
<p>In o-nitrophenol intra molecular hydrogen bonding
is present.</p>
<p>In HF intermolecular hydrogen bonding is present. </p>
<p>Other statements are correct except B and C. </p>
|
mcq
|
jee-main-2024-online-4th-april-evening-shift
| 916
|
lvb2ackq
|
chemistry
|
chemical-bonding-and-molecular-structure
|
hydrogen-bonding
|
<p>Consider the following reactions</p>
<p><img src="data:image/png;base64,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"/></p>
<p>The number of protons that do not involve in hydrogen bonding in the product B is _________.</p>
|
[]
| null |
12
|
<p>$$\mathrm{NiS + HN{O_3} + HCl\buildrel {} \over
\longrightarrow \mathop {NiC{l_2}}\limits_{(A)} + S + NO + {H_2}O}$$</p>
<p>$$\mathrm{\mathop {NiC{l_2}}\limits_{(A)} + N{H_4}OH +}$$ Dimethylgyoxime $$\mathrm{\buildrel {} \over
\longrightarrow \mathop {Ni{{(dmg)}_2}}\limits_{(B)} + N{H_4}Cl + {H_2}O}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwbmse3q/e8123551-b987-47da-b57a-5609937bd1cf/479e5060-14d2-11ef-8218-dfd8f7832166/file-1lwbmse3r.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwbmse3q/e8123551-b987-47da-b57a-5609937bd1cf/479e5060-14d2-11ef-8218-dfd8f7832166/file-1lwbmse3r.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 6th April Evening Shift Chemistry - Chemical Bonding & Molecular Structure Question 4 English Explanation"></p>
<p>Total 12 proton do not involve in H-bond</p>
|
integer
|
jee-main-2024-online-6th-april-evening-shift
| 917
|
rSpvxqnXjBtwo9bfkI1klsd0e3n
|
chemistry
|
chemical-bonding-and-molecular-structure
|
hydrolysis
|
Among the following, the number of halide(s) which is/are inert to hydrolysis is _________.<br/><br/>(A) BF<sub>3</sub><br/><br/>(B) SiCl<sub>4</sub><br/><br/>(C) PCl<sub>5</sub><br/><br/>(D) SF<sub>6</sub>
|
[]
| null |
1
|
BF<sub>3</sub> – Shows Partial hydrolysis
<br><br>SiCl<sub>4</sub> – Undergoes hydrolysis readily
<br><br>PCl<sub>5</sub> – Undergoes hydrolysis by addition–
elimination mechanism.
<br><br>SF<sub>6</sub> – Due to crowding Inert towards hydrolysis.
|
integer
|
jee-main-2021-online-25th-february-morning-slot
| 919
|
fq1DNfNhwProQ708
|
chemistry
|
chemical-bonding-and-molecular-structure
|
ionic-bond
|
Lattice energy of an ionic compounds depends upon
|
[{"identifier": "A", "content": "Charge on the ion only"}, {"identifier": "B", "content": "Size of the ion only "}, {"identifier": "C", "content": "Packing of ions only"}, {"identifier": "D", "content": "Charge on the ion and size of the ion"}]
|
["D"]
| null |
<p>The lattice energy of an ionic compound is a measure of the strength of the bonds in that ionic compound. Specifically, it is the energy required to separate one mole of an ionic solid into its constituent ions in the gaseous state. The lattice energy depends on several factors, primarily the charge on the ions and the size (or radius) of the ions.</p>
<p>1. <strong>Charge on the ions:</strong> The lattice energy is directly proportional to the product of the charges of the cation and the anion. Higher charges result in stronger electrostatic attraction between the ions, thereby increasing the lattice energy. This relationship can be derived from Coulomb's law, which states that the force of attraction between two charged particles is directly proportional to the product of their charges.</p>
<p><strong>Mathematically:</strong></p>
<p>$$U \propto \frac{q_1 \cdot q_2}{r}$$</p>
<p>where $U$ is the lattice energy, $q_1$ and $q_2$ are the charges on the ions, and $r$ is the distance between the centers of the ions.</p>
<p>2. <strong>Size of the ions:</strong> The lattice energy is inversely proportional to the distance between the ions, which depends on the sum of their ionic radii. Smaller ions can come closer together, increasing the electrostatic attractions between them, thereby increasing the lattice energy. Larger ions, being further apart, result in a lower lattice energy.</p>
<p>Considering these two factors, the correct choice is:</p>
<p><strong>Option D: Charge on the ion and size of the ion</strong></p>
|
mcq
|
aieee-2005
| 920
|
RWNDdbhUYKYBoil1
|
chemistry
|
chemical-bonding-and-molecular-structure
|
ionic-bond
|
The charge/size ratio of a cation determines its polarizing power. Which one of the following
sequences represents the increasing order of the polarizinig order of the polarizing power of the
cationic species, K<sup>+</sup>, Ca<sup>2+</sup>, Mg<sup>2+</sup>, Be<sup>2+</sup>?
|
[{"identifier": "A", "content": "Mg<sup>2+</sup> < Be<sup>2+</sup> < K<sup>+</sup> < Ca<sup>2+</sup>"}, {"identifier": "B", "content": "K<sup>+</sup> < Ca<sup>2+</sup> < Mg<sup>2+</sup> < Be<sup>2+</sup>"}, {"identifier": "C", "content": "Be<sup>2+</sup> < K<sup>+</sup> < Ca<sup>2+</sup> < Mg<sup>2+</sup>"}, {"identifier": "D", "content": "Ca<sup>2+</sup> < Mg<sup>2+</sup> < Be<sup>2+</sup> \n < K<sup>+</sup>"}]
|
["B"]
| null |
As charge/size ratio of a cation determines its polarizing power so high charge and small size of the cations increases
polarisation.
<br><br>As the size of the given cations decreases as
<br><br> K<sup>+</sup>
> Ca<sup>2+</sup> > Mg<sup>2+</sup> > Be<sup>2+</sup>
<br><br>Hence, polarising power decreases as
<br><br> K<sup>+</sup> < Ca<sup>2+</sup> < Mg<sup>2+</sup> < Be<sup>2+</sup>
|
mcq
|
aieee-2007
| 921
|
NlQwrJQxyidRbufT
|
chemistry
|
chemical-bonding-and-molecular-structure
|
ionic-bond
|
Among the following the maximum covalent character is shown by the compound :
|
[{"identifier": "A", "content": "SnCl<sub>2 </sub>"}, {"identifier": "B", "content": "AlCl<sub>3</sub> "}, {"identifier": "C", "content": "MgCl<sub>2 </sub>"}, {"identifier": "D", "content": "FeCl<sub>2</sub>"}]
|
["B"]
| null |
Charge of cation/Size of cation is called polarising power.
<br><br>$$ \therefore $$ (i) Polarising power $$ \propto $$ charge of cation
<br><br>(ii) Polarising power $$ \propto $$ <span style="display: inline-block;vertical-align: middle;">
<div style="text-align: center;border-bottom: 1px solid black;">1</div>
<div style="text-align: center;">size of cation</div>
</span>
<br><br>Here the AlCl<sub>3</sub>
is satisfying the above two conditions i.e., Al
is in +3 oxidation state and also has small size. So it has more
covalent character.
|
mcq
|
aieee-2011
| 922
|
EKsMjnfbWj3zbUo1ppjgy2xukeyf4obe
|
chemistry
|
chemical-bonding-and-molecular-structure
|
ionic-bond
|
Match the type of interaction in column A with
the distance dependence of their interaction
energy in column B
<br/><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;}
.tg .tg-c3ow{border-color:inherit;text-align:center;vertical-align:top}
.tg .tg-0pky{border-color:inherit;text-align:left;vertical-align:top}
</style>
<table class="tg">
<thead>
<tr>
<th class="tg-c3ow">A</th>
<th class="tg-c3ow">B</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-0pky">(i) ion-ion</td>
<td class="tg-0pky">(a) $${1 \over r}$$</td>
</tr>
<tr>
<td class="tg-0pky">(ii) dipole-dipole</td>
<td class="tg-0pky">(b) $${1 \over {{r^2}}}$$</td>
</tr>
<tr>
<td class="tg-0pky">(iii) London dispersion</td>
<td class="tg-0pky">(c) $${1 \over {{r^3}}}$$</td>
</tr>
<tr>
<td class="tg-0pky"></td>
<td class="tg-0pky">(d) $${1 \over {{r^6}}}$$</td>
</tr>
</tbody>
</table>
|
[{"identifier": "A", "content": "(I)-(a), (II)-(b), (III)-(d)"}, {"identifier": "B", "content": "(I)-(b), (II)-(d), (III)-(c)"}, {"identifier": "C", "content": "(I)-(a), (II)-(b), (III)-(c)"}, {"identifier": "D", "content": "(I)-(a), (II)-(c), (III)-(d)"}]
|
["D"]
| null |
Ion-ion interaction energy $$ \propto $$ $${1 \over r}$$
<br><br>Dipole-dipole interaction energy $$ \propto $$ $${1 \over {{r^3}}}$$
<br><br>London dispersion $$ \propto $$ $${1 \over {{r^6}}}$$
|
mcq
|
jee-main-2020-online-2nd-september-evening-slot
| 923
|
HFY7Xy1k19cqRCzppU1klru73v4
|
chemistry
|
chemical-bonding-and-molecular-structure
|
ionic-bond
|
The correct set from the following in which both pairs are in correct order of melting point is :
|
[{"identifier": "A", "content": "LiCl > LiF ; NaCl > MgO"}, {"identifier": "B", "content": "LiCl > LiF ; MgO > NaCl"}, {"identifier": "C", "content": "LiF > LiCl ; NaCl > MgO"}, {"identifier": "D", "content": "LiF > LiCl ; MgO > NaCl"}]
|
["D"]
| null |
<p>Correct option is i.e. LiF > LiCl; MgO > NaCl. Melting point is directly proportional to lattice energy. Lattice energy is the energy required to separate a mole of an ionic solid into gaseous ions. It depends upon charge of ions and size of ions.</p>
<p>$$M.P. \propto L.E. \propto {{Charge} \over {Size}}$$</p>
<p>$$\matrix{
{Li \to + 1} & {Li \to + 1} \cr
{F = - 1} & {Cl \to - 1} \cr
} $$</p>
<p>Both LiF and LiCl having same charge, so melting point will depend on size.</p>
<p>Larger the size of anion, lesser the lattice energy and hence, melting point order is LiF > LiCl.</p>
<p>Similarly, $$\matrix{
{MgO} & {NaCl} \cr
{Mg \to 2 + } & {Na \to 1 + } \cr
{O \to 2 - } & {Cl \to 1 - } \cr
} $$</p>
<p>MgO having + 2 charge which is greater than NaCl (+ 1) charge. So, greater the charge on the ions greater will be lattice energy and hence, melting point order is MgO > NaCl.</p>
|
mcq
|
jee-main-2021-online-24th-february-evening-slot
| 924
|
1lh04mf3g
|
chemistry
|
chemical-bonding-and-molecular-structure
|
ionic-bond
|
<p>The number of following factors which affect the percent covalent character of the ionic bond is _________</p>
<p>(A) Polarising power of cation</p>
<p>(B) Extent of distortion of anion</p>
<p>(C) Polarisability of the anion</p>
<p>(D) Polarising power of anion</p>
|
[]
| null |
3
|
<p>The covalent character of an ionic bond is largely determined by the polarization of the ions involved in the bond. Polarization refers to the distortion of the electron cloud of an anion by a cation. This distortion leads to a shift in electron density towards the cation, thereby increasing the covalent character of the bond.</p>
<p>Here's a breakdown of the options:</p>
<p>(A) Polarising power of cation: The greater the polarising power of a cation, the greater the distortion of the anion, and therefore, the greater the covalent character of the bond. Thus, this statement is correct.</p>
<p>(B) Extent of distortion of anion: The more an anion is distorted, the more the electron density is shifted towards the cation, and the greater the covalent character of the bond. This statement is also correct.</p>
<p>(C) Polarisability of the anion: The greater the polarisability of an anion, the more it can be distorted, and therefore, the greater the covalent character of the bond. Thus, this statement is also correct.</p>
<p>(D) Polarising power of anion: This statement is incorrect. It is not the polarising power of the anion, but the polarisability of the anion and the polarising power of the cation that influence the covalent character of the bond.</p>
<p>Therefore, three of the factors listed (A, B, C) affect the percent covalent character of the ionic bond.</p>
|
integer
|
jee-main-2023-online-8th-april-morning-shift
| 925
|
tRqcHUfZAks8ckvLOhjgy2xukfuqwaai
|
chemistry
|
chemical-bonding-and-molecular-structure
|
lewis-theory
|
The number of Cl = O bonds in perchloric acid
is, "________".
|
[]
| null |
3
|
The structure of perchloric acid is
<img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266578/exam_images/jyszcvabceqqemsdzkut.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 6th September Morning Slot Chemistry - Chemical Bonding & Molecular Structure Question 141 English Explanation">
<br>The number Cl = O bonds in HClO<sub>4</sub> is 3.
|
integer
|
jee-main-2020-online-6th-september-morning-slot
| 928
|
1ldst4bbt
|
chemistry
|
chemical-bonding-and-molecular-structure
|
lewis-theory
|
<p>The number of molecules or ions from the following, which do not have odd number of electrons are _________.</p>
<p>(A) NO$$_2$$</p>
<p>(B) ICl$$_4^ - $$</p>
<p>(C) BrF$$_3$$</p>
<p>(D) ClO$$_2$$</p>
<p>(E) NO$$_2^ + $$</p>
<p>(F) NO</p>
|
[]
| null |
3
|
Only $\mathrm{ICl}_4^{-}, \mathrm{BrF}_3$ and $\mathrm{NO}_2^{+}$have even number of electrons.
<br/><br/>$$
\begin{aligned}
& \mathrm{NO}_2 \Rightarrow 23 e^{-} ; \\\\
& \mathrm{ICl}_4^{-} \Rightarrow 122 e^{-} ; \\\\
& \mathrm{BrF}_3 \Rightarrow 62 e^{-} ; \\\\
& \mathrm{ClO}_2 \Rightarrow 33 e^{-} ; \\\\
& \mathrm{NO}_2^{+} \Rightarrow 22 e^{-} ; \\\\
& \mathrm{NO} \Rightarrow 15 e^{-}
\end{aligned}
$$
|
integer
|
jee-main-2023-online-29th-january-morning-shift
| 929
|
lv5gsw8c
|
chemistry
|
chemical-bonding-and-molecular-structure
|
lewis-theory
|
<p>Number of molecules from the following which are exceptions to octet rule is _________.</p>
<p>$$\mathrm{CO}_2, \mathrm{NO}_2, \mathrm{H}_2 \mathrm{SO}_4, \mathrm{BF}_3, \mathrm{CH}_4, \mathrm{SiF}_4, \mathrm{ClO}_2, \mathrm{PCl}_5, \mathrm{BeF}_2, \mathrm{C}_2 \mathrm{H}_6, \mathrm{CHCl}_3, \mathrm{CBr}_4$$</p>
|
[]
| null |
6
|
<p>$$\mathrm{NO}_2, \mathrm{H}_2 \mathrm{SO}_4, \mathrm{BF}_3, \mathrm{ClO}_2, \mathrm{PCl}_5, \mathrm{BeF}_2$$</p>
<p>These are exception of octet rule</p>
|
integer
|
jee-main-2024-online-8th-april-morning-shift
| 930
|
lv7v47s8
|
chemistry
|
chemical-bonding-and-molecular-structure
|
lewis-theory
|
<p>In the lewis dot structure for $$\mathrm{NO}_2^{-}$$, total number of valence electrons around nitrogen is _________.</p>
|
[]
| null |
8
|
<p>$$\text { Lewis dot structure of } \mathrm{NO}_2^{-} \text {is: }$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwh56zal/04a24314-4a60-43cf-8f6c-8ea5670b9184/2ce32dd0-17da-11ef-a722-71ab49de2419/file-1lwh56zam.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwh56zal/04a24314-4a60-43cf-8f6c-8ea5670b9184/2ce32dd0-17da-11ef-a722-71ab49de2419/file-1lwh56zam.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 5th April Morning Shift Chemistry - Chemical Bonding & Molecular Structure Question 10 English Explanation"></p>
<p>$$\text { Total number of valence } \mathrm{e}^{\ominus} \text { around } \mathrm{N}=8$$</p>
|
integer
|
jee-main-2024-online-5th-april-morning-shift
| 931
|
5ewmfVaZwhG6amZL
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
Which of the following are arranged in an increasing order of their bond strengths?
|
[{"identifier": "A", "content": "$$O_2^-$$ < O<sub>2</sub> < $$O_2^+$$ < $$O_2^{2-}$$"}, {"identifier": "B", "content": "$$O_2^{2-}$$ < $$O_2^-$$ < $$O_2$$ < $$O_2^{+}$$"}, {"identifier": "C", "content": "$$O_2^-$$ < $$O_2^{2-}$$ < $$O_2$$ < $$O_2^{+}$$"}, {"identifier": "D", "content": "$$O_2^{+}$$ < $$O_2$$ < $$O_2^-$$ < $$O_2^{2-}$$"}]
|
["B"]
| null |
<b><u>Note</u> :</b>
<br><br>(1) $$\,\,\,\,$$ Bond strength $$ \propto $$ Bond order
<br><br>(2) $$\,\,\,\,$$ Bond length $$ \propto $$ $${1 \over {Bond\,\,order}}$$
<br><br>(3) $$\,$$ Bond order $$ = {1 \over 2}$$ [N<sub>b </sub> $$-$$ N<sub>a</sub>]
<br><br>N<sub>b</sub> = No of electrons in bonding molecular orbital
<br><br>N<sub>a</sub> $$=$$ No of electrons in anti bonding molecular orbital
<br><br>(4) $$\,\,\,\,$$ upto 14 electrons, molecular orbital configuration is
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264441/exam_images/dyi05nwqikfdrpmfsriz.webp" loading="lazy" alt="AIEEE 2002 Chemistry - Chemical Bonding & Molecular Structure Question 222 English Explanation 1">
<br><br>Here N<sub>a</sub> = Anti bonding electron $$=$$ 4 and N<sub>b</sub> = 10
<br><br>(5) $$\,\,\,\,$$ After 14 electrons to 20 electrons molecular orbital configuration is - - -
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267718/exam_images/tivzvzeox1orkr2tnz5y.webp" loading="lazy" alt="AIEEE 2002 Chemistry - Chemical Bonding & Molecular Structure Question 222 English Explanation 2">
<br><br>Here N<sub>a</sub> = 10
<br><br>and N<sub>b</sub> = 10
<br><br>In O atom 8 electrons present, so in O<sub>2</sub>, 8 $$ \times $$ 2 = 16 electrons present.
<br><br>Then in $$O_2^ + $$ no of electrons = 15
<br><br>in $$O_2^ - $$ no of electrons = 17
<br><br>in $$O_2^{2 - }$$ no of electrons = 18
<br><br>$$\therefore\,\,\,\,$$ Molecular orbital configuration of O<sub>2</sub> (16 electrons) is
<br><br>$${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$$ $${\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,$$ $${\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}} = {\pi _{2p_y^2}}\,\,\pi _{2p_x^1}^ * \,\, = \pi _{2p_y^1}^ * $$
<br><br>$$\therefore\,\,\,\,$$N<sub>a</sub> = 6
<br><br>N<sub>b</sub> = 10
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 6} \right] = 2$$
<br><br>Molecular orbital configuration of O$$_2^ + $$ (15 electrons) is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^o}^ * $$
<br><br>$$\therefore\,\,\,\,$$ N<sub>b</sub> = 10
<br><br>N<sub>a</sub> = 5
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 5} \right]$$ = 2.5
<br><br>Molecular orbital configuration of $$O_2^ - $$ (17 electrons) is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^1}^ * $$
<br><br>$$\therefore\,\,\,\,$$ N<sub>b</sub> = 10
<br><br>N<sub>a</sub> = 7
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 7} \right]$$ = 1.5
<br><br>Molecular orbital configuration of O $$_2^{2 - }$$ (18 electrons) is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^2}^ * $$
<br><br>$$\therefore\,\,\,\,$$ N<sub>b</sub> = 10
<br><br>N<sub>a</sub> = 8
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}$$ [ 10 $$-$$ 8] = 1
<br><br>As Bond strength $$ \propto $$ Bond order so, correct order is
<br><br>$$O_2^{2 - } < O_2^ - < {O_2} < O_2^ + $$
|
mcq
|
aieee-2002
| 932
|
nZWgMrVC61t5mc6a
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
The bond order in NO is 2.5 while that in NO<sup>+</sup> is 3. Which of the following statements is true for these two species?
|
[{"identifier": "A", "content": "Bond length in NO<sup>+</sup> is greater than in NO "}, {"identifier": "B", "content": "Bond length is unpredictable"}, {"identifier": "C", "content": "Bond length in NO<sup>+</sup> in equal to that in NO"}, {"identifier": "D", "content": "Bond length in NO is greater than in NO<sup>+</sup>"}]
|
["D"]
| null |
Molecular orbital configuration of NO (15 electrons) is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^o}^ * $$
<br><br>$$\therefore\,\,\,\,$$ N<sub>b</sub> = 10
<br><br>N<sub>a</sub> = 5
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 5} \right]$$ = 2.5
<br><br>Similarly Molecular orbital configuration of NO<sup>+</sup> (14 electrons) is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,$$
<br><br>$$\therefore\,\,\,\,$$ N<sub>b</sub> = 10
<br><br>N<sub>a</sub> = 4
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 4} \right]$$ = 3
<br><br><b>Note :</b>
<br>(2) $$\,\,\,\,$$ Bond length $$ \propto $$ $${1 \over {Bond\,\,order}}$$
<br><br>So, bond length in NO > NO<sup>+</sup>
|
mcq
|
aieee-2004
| 933
|
78vvWsuL3joAOjJY
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
Which one of the following species is diamagnetic in nature?
|
[{"identifier": "A", "content": "$$He_2^+$$"}, {"identifier": "B", "content": "H<sub>2</sub>"}, {"identifier": "C", "content": "$$H_2^+$$"}, {"identifier": "D", "content": "$$H_2^-$$"}]
|
["B"]
| null |
<b>TIPS/Formulae :</b>
<br><br>A diamagnetic substance contains no unpaired electron.
<br><br>$${H_2}$$ is diamagnetic as it contains all paired electrons
<br><br>$$\mathop {{H_2} = \sigma _b^2}\limits_{\left( {diamagnetic} \right)} \,\,,\,\,\mathop {H_2^ + = \sigma _b^1,}\limits_{\left( {paramagnetic} \right)} \,\,\mathop {H_2^ - = \sigma _b^2,}\limits_{\left( {paramagnetic} \right)} $$
<br><br>$$\mathop {\sigma _a^{ * 1};He_2^ + }\limits_{\left( {paramagnetic} \right)} = \mathop {\sigma _b^2,\sigma _a^{ \ne 1}}\limits_{\left( {paramagnetic} \right)} $$
|
mcq
|
aieee-2005
| 934
|
l1VU0nBBIoUEbBGm
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
Which of the following molecules/ions does not contain unpaired electrons?
|
[{"identifier": "A", "content": "$$O_2^{2\u2212} $$"}, {"identifier": "B", "content": "B<sub>2</sub>"}, {"identifier": "C", "content": "$$N_2^+$$"}, {"identifier": "D", "content": "O<sub>2</sub>"}]
|
["A"]
| null |
(A) Molecular orbital configuration of O $$_2^{2 - }$$ (18 electrons) is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^2}^ * $$
<br><br>So O $$_2^{2 - }$$ has no unpaired electrons.
<br><br>(B) Molecular orbital configuration of B<sub>2</sub> (10 electrons) is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\pi _{2p_x^1}} = {\pi _{2p_y^1}}$$
<br><br>Here in B<sub>2</sub>, 2 unpaired electrons present.
<br><br>(C) Moleculer orbital configuration of $$N_2^{ + }$$ (13 electrons)
<br><br>= $${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^1}}$$
<br><br>Here in $$N_2^{ + }$$, 1 unpaired electron present.
<br><br>(D) Molecular orbital configuration of O<sub>2</sub> (16 electrons) is
<br><br>$${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$$ $${\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,$$ $${\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}} = {\pi _{2p_y^2}}\,\,\pi _{2p_x^1}^ * \,\, = \pi _{2p_y^1}^ * $$
<br><br>So O$$_2$$ has 2 unpaired electrons.
|
mcq
|
aieee-2006
| 935
|
v25tvfRhzo8LhY59
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
Which of the following species exhibits the diamagnetic behaviour?
|
[{"identifier": "A", "content": "$$O_2^{2\u2212}$$"}, {"identifier": "B", "content": "NO"}, {"identifier": "C", "content": "$$O_2^+$$"}, {"identifier": "D", "content": "O<sub>2</sub>"}]
|
["A"]
| null |
Those species which have unpaired electrons are called paramagnetic species.
<br><br>And those species which have no unpaired electrons are called diamagnetic species.
<br><br>(a) $$O_2^{2−}$$ has 18 electrons.
<br><br>Moleculer orbital configuration of $$O_2^{2−}$$ is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^2}^ * $$
<br><br>Here is no unpaired electron so it is diamagnetic.
<br><br>(b) NO has 15 electrons.
<br><br>Moleculer orbital configuration of NO is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^0}^ * $$
<br><br>Here is 1 unpaired electron, So it is Paramagnetic.
<br><br>(c) $$O_2$$ has 16 electrons.
<br><br>Moleculer orbital configuration of $$O_2$$ is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^1}^ * $$
<br><br>Here 2 unpaired electron present, so it is paramagnetic.
<br><br>(d) $$O_2^{+}$$ has 15 electrons.
<br><br>Moleculer orbital configuration of $$O_2^{+}$$ is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^0}^ * $$
<br><br>Here 1 unpaired electron present, so it is paramagnetic.
|
mcq
|
aieee-2007
| 936
|
Mk1EqYkRfu6bSrWr
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
In which of the following ionization processes, the bond order has increased and the magnetic
behaviour has changed?
|
[{"identifier": "A", "content": "$$C_2 \\to C_2^+$$"}, {"identifier": "B", "content": "$$N_2 \\to N_2^+$$"}, {"identifier": "C", "content": "$$NO \\to NO^+$$"}, {"identifier": "D", "content": "$$O_2 \\to O_2^+$$"}]
|
["C"]
| null |
(A) Moleculer orbital configuration of $$C_2$$ (12 electrons)
<br><br>= $${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}$$
<br><br>$$\therefore\,\,\,\,$$N<sub>a</sub> = 4
<br><br>N<sub>b</sub> = 8
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {8 - 4} \right] = 2$$
<br><br>Here no unpaired electron present, so it is diamagnetic.
<br><br>Moleculer orbital configuration of $$C_2^+$$ (11 electrons)
<br><br>= $${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^1}}$$
<br><br>$$\therefore\,\,\,\,$$N<sub>a</sub> = 4
<br><br>N<sub>b</sub> = 7
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {7 - 4} \right] = 1.5$$
<br><br>Here 1 unpaired electron present, so it is paramagnetic.
<br><br>(B) Moleculer orbital configuration of $$N_2$$ (14 electrons)
<br><br>= $${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}$$
<br><br>$$\therefore\,\,\,\,$$N<sub>a</sub> = 4
<br><br>N<sub>b</sub> = 10
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 4} \right] = 3$$
<br><br>Here no unpaired electron present, so it is diamagnetic.
<br><br>Moleculer orbital configuration of $$N_2^+$$ (13 electrons)
<br><br>= $${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^1}}$$
<br><br>$$\therefore\,\,\,\,$$N<sub>a</sub> = 4
<br><br>N<sub>b</sub> = 9
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {9 - 4} \right] = 2.5$$
<br><br>Here 1 unpaired electron present, so it is paramagnetic.
<br><br>(C) Moleculer orbital configuration of NO (15 electrons) is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^0}^ * $$
<br><br>$$\therefore\,\,\,\,$$N<sub>a</sub> = 5
<br><br>N<sub>b</sub> = 10
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 5} \right] = 2.5$$
<br><br>Here is 1 unpaired electron, So it is paramagnetic.
<br><br>Moleculer orbital configuration of NO<sup>+</sup> (14 electrons) is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^0}^ * = \,\,\pi _{2p_y^0}^ * $$
<br><br>$$\therefore\,\,\,\,$$N<sub>a</sub> = 4
<br><br>N<sub>b</sub> = 10
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 4} \right] = 3$$
<br><br>Here is no unpaired electron, So it is diamagnetic.
<br><br>(D) Molecular orbital configuration of O<sub>2</sub> (16 electrons) is
<br><br>$${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$$ $${\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,$$ $${\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}} = {\pi _{2p_y^2}}\,\,\pi _{2p_x^1}^ * \,\, = \pi _{2p_y^1}^ * $$
<br><br>$$\therefore\,\,\,\,$$N<sub>a</sub> = 6
<br><br>N<sub>b</sub> = 10
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 6} \right] = 2$$
<br><br>Here 2 unpaired electrons present, so it is paramagnetic.
<br><br>Molecular orbital configuration of O$$_2^ + $$ (15 electrons) is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^o}^ * $$
<br><br>$$\therefore\,\,\,\,$$ N<sub>b</sub> = 10
<br><br>N<sub>a</sub> = 5
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 5} \right]$$ = 2.5
<br><br>Here 1 unpaired electrons present, so it is also paramagnetic.
|
mcq
|
aieee-2007
| 937
|
hVhRBuRAwNW6A8B1
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
Which of the following pair of species have the same bond order?
|
[{"identifier": "A", "content": "CN<sup>-</sup> and NO<sup>+</sup>"}, {"identifier": "B", "content": "CN<sup>-</sup> and CN<sup>+</sup>"}, {"identifier": "C", "content": "$$O_2^-$$ and CN<sup>-</sup>"}, {"identifier": "D", "content": "NO<sup>+</sup> and CN<sup>+</sup>"}]
|
["A"]
| null |
Number of electron in
NO<sup>+</sup>
= number of electron in CN<sup>–</sup>
= 14 electrons.
<br><br>As both have same number of electrons so their bond order is equal.
<br><br>Moleculer orbital configuration of NO<sup>+</sup> (14 electrons) is
<br><br>= $${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^0}^ * = \,\,\pi _{2p_y^0}^ * $$
<br><br>$$ \therefore $$ B.O = $${1 \over 2}\left[ {10 - 4} \right]$$ = 3
<br><br>Moleculer orbital configuration of CN<sup>-</sup> (14 electrons) is
<br><br>= $${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}$$
<br><br>$$ \therefore $$ B.O = $${1 \over 2}\left[ {10 - 4} \right]$$ = 3
|
mcq
|
aieee-2008
| 938
|
Vnurx6F2t4VBIDLe
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
Using MO theory, predict which of the following species has the shortest bond length?
|
[{"identifier": "A", "content": "$$O_2^+$$"}, {"identifier": "B", "content": "$$O_2^-$$"}, {"identifier": "C", "content": "$$O_2^{2-}$$"}, {"identifier": "D", "content": "$$O_2^{2+}$$"}]
|
["D"]
| null |
<b><u>Note</u> :</b>
<br><br>(1) $$\,\,\,\,$$ Bond length $$ \propto $$ $${1 \over {Bond\,\,order}}$$
<br><br>(2) $$\,$$ Bond order $$ = {1 \over 2}$$ [N<sub>b </sub> $$-$$ N<sub>a</sub>]
<br><br>N<sub>b</sub> = No of electrons in bonding molecular orbital
<br><br>N<sub>a</sub> $$=$$ No of electrons in anti bonding molecular orbital
<br><br>(4) $$\,\,\,\,$$ upto 14 electrons, molecular orbital configuration is
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264441/exam_images/dyi05nwqikfdrpmfsriz.webp" loading="lazy" alt="AIEEE 2009 Chemistry - Chemical Bonding & Molecular Structure Question 202 English Explanation 1">
<br><br>Here N<sub>a</sub> = Anti bonding electron $$=$$ 4 and N<sub>b</sub> = 10
<br><br>(5) $$\,\,\,\,$$ After 14 electrons to 20 electrons molecular orbital configuration is - - -
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267718/exam_images/tivzvzeox1orkr2tnz5y.webp" loading="lazy" alt="AIEEE 2009 Chemistry - Chemical Bonding & Molecular Structure Question 202 English Explanation 2">
<br><br>Here N<sub>a</sub> = 10
<br><br>and N<sub>b</sub> = 10
<br><br>In O atom 8 electrons present, so in O<sub>2</sub>, 8 $$ \times $$ 2 = 16 electrons present.
<br><br>Then in $$O_2^ + $$ no of electrons = 15
<br><br>in $$O_2^ - $$ no of electrons = 17
<br><br>in $$O_2^{2 - }$$ no of electrons = 18
<br><br>$$\therefore\,\,\,\,$$ Molecular orbital configuration of O<sub>2</sub> (16 electrons) is
<br><br>$${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$$ $${\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,$$ $${\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}} = {\pi _{2p_y^2}}\,\,\pi _{2p_x^1}^ * \,\, = \pi _{2p_y^1}^ * $$
<br><br>$$\therefore\,\,\,\,$$N<sub>a</sub> = 6
<br><br>N<sub>b</sub> = 10
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 6} \right] = 2$$
<br><br>Molecular orbital configuration of O$$_2^ + $$ (15 electrons) is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^o}^ * $$
<br><br>$$\therefore\,\,\,\,$$ N<sub>b</sub> = 10
<br><br>N<sub>a</sub> = 5
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 5} \right]$$ = 2.5
<br><br>Molecular orbital configuration of O$$_2^ {2+ }$$ (14 electrons) is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^0}^ * \, = \,\pi _{2p_y^o}^ * $$
<br><br>$$\therefore\,\,\,\,$$ N<sub>b</sub> = 10
<br><br>N<sub>a</sub> = 4
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 4} \right]$$ = 3
<br><br>Molecular orbital configuration of $$O_2^ - $$ (17 electrons) is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^1}^ * $$
<br><br>$$\therefore\,\,\,\,$$ N<sub>b</sub> = 10
<br><br>N<sub>a</sub> = 7
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 7} \right]$$ = 1.5
<br><br>Molecular orbital configuration of O $$_2^{2 - }$$ (18 electrons) is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^2}^ * $$
<br><br>$$\therefore\,\,\,\,$$ N<sub>b</sub> = 10
<br><br>N<sub>a</sub> = 8
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}$$ [ 10 $$-$$ 8] = 1
<br><br>As Bond length $$ \propto $$ $${1 \over {Bond\,\,order}}$$
<br><br>So $$O_2^{2+}$$ has the shortest bond length.
|
mcq
|
aieee-2009
| 939
|
IVnAUloaajyf6Ciz
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
Which one of the following molecules is expected to exhibit diamagnetic behaviour?
|
[{"identifier": "A", "content": "N<sub>2</sub>"}, {"identifier": "B", "content": "O<sub>2</sub>"}, {"identifier": "C", "content": "S<sub>2</sub>"}, {"identifier": "D", "content": "C<sub>2</sub>"}]
| null | null |
Those species which have unpaired electrons are called paramagnetic species.
<br><br>And those species which have no unpaired electrons are called diamagnetic species.
<br><br>(A) $$N_2$$ has 14 electrons.
<br><br>Moleculer orbital configuration of $$N_2$$
<br><br>= $${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}$$
<br><br>Here no unpaired electron present, so it is diamagnetic.
<br><br>(B) $$O_2$$ has 16 electrons.
<br><br>Moleculer orbital configuration of $$O_2$$ is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^1}^ * $$
<br><br>Here 2 unpaired electrons present, so it is paramagnetic.
<br><br>(C) $$S_2$$ has 32 electrons.
<br><br>It also has 2 unpaired electrons like $$O_2$$, so it is paramagnetic.
<br><br>(D) $$C_2$$ has 12 electrons.
<br><br>Moleculer orbital configuration of $$C_2$$
<br><br>= $${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}$$
<br><br>Here no unpaired electron present, so it is diamagnetic.
|
mcqm
|
jee-main-2013-offline
| 940
|
e9ZZ6hVkVRrr20He
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
In which of the following pairs of molecules/ions, both the species are not likely to exist?
|
[{"identifier": "A", "content": "$$H_2^+$$ , $$He_2^{2-}$$"}, {"identifier": "B", "content": "$$H_2^-$$ , $$He_2^{2-}$$"}, {"identifier": "C", "content": "$$H_2^{2+}$$ , $$He_2$$"}, {"identifier": "D", "content": "$$H_2^-$$ , $$He_2^{2+}$$"}]
|
["C"]
| null |
$$H_2^{2+}$$ , $$He_2$$
<br><br>Both have zero bond order. Thus, they do not exist.
|
mcq
|
jee-main-2013-offline
| 941
|
71vVjiJV8KEZUIQD
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
Stability of the species Li<sub>2</sub>, $$Li_2^−$$ and $$Li_2^+$$ increases in the order of:
|
[{"identifier": "A", "content": "Li<sub>2</sub> < $$Li_2^+$$ < $$Li_2^-$$"}, {"identifier": "B", "content": "$$Li_2^+$$ < $$Li_2^-$$ < Li<sub>2</sub>"}, {"identifier": "C", "content": "$$Li_2^-$$ <$$Li_2^+$$ < Li<sub>2</sub>"}, {"identifier": "D", "content": "$$Li_2^-$$ < Li<sub>2</sub> < $$Li_2^+$$"}]
|
["C"]
| null |
Li<sub>2</sub> = $${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$$ $${\sigma _{2{s^2}}} \,$$
<br><br>$$ \therefore $$ Bond order = $${1 \over 2}\left( {4 - 2} \right)$$ = 1
<br><br>$$Li_2^+$$ = $${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$$ $${\sigma _{2{s^1}}} \,$$
<br><br>$$ \therefore $$ Bond order = $${1 \over 2}\left( {3 - 2} \right)$$ = 0.5
<br><br>$$Li_2^-$$ = $${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$$ $${\sigma _{2{s^2}}} \,\,\sigma _{1{s^1}}^ * \,$$
<br><br>$$ \therefore $$ Bond order = $${1 \over 2}\left( {4 - 3} \right)$$ = 0.5
<br><br>The bond order of $$Li_2^+$$ and $$Li_2^-$$
is same but $$Li_2^+$$
is more stable
than $$Li_2^-$$ because $$Li_2^+$$
is smaller in size and
has 2 electrons in antibonding orbitals whereas $$Li_2^-$$ has 3 electrons in antibonding orbitals.
Hence $$Li_2^+$$
is more stable than $$Li_2^-$$.
|
mcq
|
jee-main-2013-offline
| 942
|
2I94DYV587XfHvOi
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
Which of the following species is <b>not</b> paramagnetic?
|
[{"identifier": "A", "content": "CO"}, {"identifier": "B", "content": "O<sub>2</sub>"}, {"identifier": "C", "content": "B<sub>2</sub>"}, {"identifier": "D", "content": "NO"}]
|
["A"]
| null |
Those species which have unpaired electrons are called paramagnetic species.
<br><br>(a) CO has 14 electrons.
<br><br>Moleculer orbital configuration of CO is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\pi _{2p_x^2}} =\,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}$$
<br><br>Here is no unpaired electron so it is diamagnetic.
<br><br>(b) $$O_2$$ has 16 electrons.
<br><br>Moleculer orbital configuration of $$O_2$$ is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^1}^ * $$
<br><br>Here 2 unpaired electron present, so it is paramagnetic.
<br><br>(c) B<sub>2</sub> has 10 electrons.
<br><br>Molecular orbital configuration of B<sub>2</sub> is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\pi _{2p_x^1}} = {\pi _{2p_y^1}}$$
<br><br>Here two unpaired electrons present. So it is paramagnetic.
<br><br>(d) NO has 15 electrons.
<br><br>Moleculer orbital configuration of NO is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^0}^ * $$
<br><br>Here is 1 unpaired electron, So it is Paramagnetic.
|
mcq
|
jee-main-2017-offline
| 943
|
YFVl81zZz9l5GTXPCmqsx
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
Which of the following is paramagnetic ?
|
[{"identifier": "A", "content": "NO<sup>+</sup> "}, {"identifier": "B", "content": "CO"}, {"identifier": "C", "content": "$$O_2^{2 - }$$"}, {"identifier": "D", "content": "B<sub>2</sub>"}]
|
["D"]
| null |
Those species which have unpaired electrons are called paramagnetic species.
<br><br>(a) NO<sup>+</sup> has 14 electrons.
<br><br>Moleculer orbital configuration of NO<sup>+</sup> is
<br><br>$${\sigma _{1{s^2}}}$$ $$\sigma _{1{s^2}}^ * $$ $${\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,\, = \,\,{\pi _{2p_y^2}}$$
<br><br>Here is no unpaired electron, So it is Diamagnetic.
<br><br>(b) CO has 14 electrons.
<br><br>Moleculer orbital configuration of CO is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\pi _{2p_x^2}} =\,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}$$
<br><br>Here is no unpaired electron so it is diamagnetic.
<br><br>(c) $$O_2^{2 - }$$ has 18 electrons.
<br><br>Moleculer orbital configuration of $$O_2^{2 - }$$ is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * = \,\,\pi _{2p_y^2}^ * $$
<br><br>Here also no unpaired electron present, so it is diamagnetic.
<br><br>(d) B<sub>2</sub> has 10 electrons.
<br><br>Molecular orbital configuration of B<sub>2</sub> is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\pi _{2p_x^1}} = {\pi _{2p_y^1}}$$
<br><br>Here two unpaired electrons present.
<br><br>So it is paramagnetic.
|
mcq
|
jee-main-2017-online-8th-april-morning-slot
| 944
|
xbbpwAyMOufn5c54
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
According to molecular orbital theory, which of the following will not be a viable molecule?
|
[{"identifier": "A", "content": "$${\\rm H}e_2^{2 + }$$ "}, {"identifier": "B", "content": "$${\\rm H}e_2^{ + }$$"}, {"identifier": "C", "content": "$${\\rm H}_2^{- }$$"}, {"identifier": "D", "content": "$${\\rm H}_2^{2 - }$$"}]
|
["D"]
| null |
<b><u>Note</u> :</b>
<br><br>According to molecules orbital theory, when a molecule have bond order = 0 then that molecule does not exist.
<br><br>(a)$$\,\,\,$$ Configuration of $$He_2^{2 + }$$ (2 electrons) is = $${\sigma _{1{s^2}}}$$
<br><br>$$\therefore\,\,\,$$ Bond order = $${1 \over 2}$$ (2 $$-$$0) = 1
<br><br>(b)$$\,\,\,$$ Configuration of $$He_2^ + $$ (3 electrons) is = $${\sigma _{1{s^2}}}$$ $$\sigma _{1{s^1}}^ * $$
<br><br>$$\therefore\,\,\,$$ Bond order = $${1 \over 2}$$ (2 $$-$$1) = 0.5
<br><br>(c) $$\,\,\,$$ Configuration of $$H_2^ - $$ (3 electrons) is = $${\sigma _{1{s^2}}}$$ $$\sigma _{1{s^1}}^ * $$
<br><br>$$\therefore$$ Bond order = $${1 \over 2}$$ (2 $$-$$ 1) = 0.5
<br><br>(d) $$\therefore\,\,\,$$ Configuration of $$He_2^{2 - }$$ is = $${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * $$
<br><br>$$\therefore\,\,\,$$ Bond order = $$ = {1 \over 2}$$ (2 $$-$$ 2) = 0
<br><br>$$\therefore\,\,\,$$ $$H_2^{2 - }$$ is not a viable molecule.
|
mcq
|
jee-main-2018-offline
| 947
|
5QPKgmMT1jSaJIB9R9HXC
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
According to molecular orbital theory, which of the following is true with respect to Li<sub>2</sub><sup>+</sup> and Li<sub>2</sub><sup>$$-$$</sup> ?
|
[{"identifier": "A", "content": "$$Li_2^ + $$ is unstable and $$Li_2^ - $$ is stable"}, {"identifier": "B", "content": "$$Li_2^ + $$ is stable and $$Li_2^ - $$ unstable"}, {"identifier": "C", "content": "Both are stable "}, {"identifier": "D", "content": "Both are unstable "}]
|
["C"]
| null |
$$Li_2^ + $$ (5 electrons) = $${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^1}}}$$
<br><br>$$Li_2^ - $$ (7 electrons) = $${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}$$ $$\,\sigma _{2{s^1}}^ * \,$$
<br><br>$$ \therefore $$ Bond order of $$Li_2^ + $$ = $${{3 - 2} \over 2}$$ = 0.5
<br><br>Bond order of $$Li_2^ - $$ = $${{4 - 3} \over 2}$$ = 0.5
<br><br>As both $$Li_2^ + $$ and $$Li_2^ - $$ has non-zero bond order, so both are stable.
|
mcq
|
jee-main-2019-online-9th-january-morning-slot
| 948
|
tajJ8St48Q1lcW1qdYfUS
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
Among the following species, the diamagnetic
molecule is
|
[{"identifier": "A", "content": "CO"}, {"identifier": "B", "content": "B<sub>2</sub>"}, {"identifier": "C", "content": "O<sub>2</sub>"}, {"identifier": "D", "content": "NO"}]
|
["A"]
| null |
Those species which have unpaired electrons are called paramagnetic species.
<br><br>And those species which have no unpaired electrons are called diamagnetic species.
<br><br>(a) CO has 14 electrons.
<br><br>Moleculer orbital configuration of CO is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\pi _{2p_x^2}} =\,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}$$
<br><br>Here is no unpaired electron so it is diamagnetic.
<br><br>(b) B<sub>2</sub> has 10 electrons.
<br><br>Molecular orbital configuration of B<sub>2</sub> is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\pi _{2p_x^1}} = {\pi _{2p_y^1}}$$
<br><br>Here two unpaired electrons present. So it is paramagnetic.
<br><br>(c) $$O_2$$ has 16 electrons.
<br><br>Moleculer orbital configuration of $$O_2$$ is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^1}^ * $$
<br><br>Here 2 unpaired electron present, so it is paramagnetic.
<br><br>(d) NO has 15 electrons.
<br><br>Moleculer orbital configuration of NO is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^0}^ * $$
<br><br>Here is 1 unpaired electron, So it is Paramagnetic.
|
mcq
|
jee-main-2019-online-9th-april-evening-slot
| 950
|
4kAdZGcAFWDgiyauTrdhC
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
Among the following, the molecule expected
to be stabilized by anion formation is :<br/>
C<sub>2</sub>, O<sub>2</sub>, NO, F<sub>2</sub>
|
[{"identifier": "A", "content": "C<sub>2</sub>"}, {"identifier": "B", "content": "NO"}, {"identifier": "C", "content": "O<sub>2</sub>"}, {"identifier": "D", "content": "F<sub>2</sub>"}]
|
["A"]
| null |
C<sub>2</sub> has 12 electrons.
<br><br>Moleculer orbital configuration of C<sub>2</sub> is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\pi _{2p_x^2}} =\,{\pi _{2p_y^2}}$$
<br><br>Bond order of C<sub>2</sub> = $${{8 - 4} \over 2}$$ = 2
<br><br>C<sub>2</sub><sup>-</sup> has 13 electrons.
<br><br>Moleculer orbital configuration of C<sub>2</sub><sup>-</sup> is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\pi _{2p_x^2}} =\,{\pi _{2p_y^2}}$$ $${\sigma _{2p_z^1}}$$
<br><br>Bond order of C<sub>2</sub><sup>-</sup> = $${{9 - 4} \over 2}$$ = 2.5
<br><br>As we know, higher is the bond order more will be the stability of the molecule.
<br><br>So we can say, stability of C<sub>2</sub> increases when it forms C<sub>2</sub><sup>-</sup> ion.
|
mcq
|
jee-main-2019-online-9th-april-morning-slot
| 951
|
jyTyBEbXoLWjxRE6s2Jln
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
Two pi and half sigma bonds are present in :
|
[{"identifier": "A", "content": "O<sub>2</sub>"}, {"identifier": "B", "content": "N$$_2^ + $$"}, {"identifier": "C", "content": "O$$_2^ + $$"}, {"identifier": "D", "content": "N<sub>2</sub>"}]
|
["B"]
| null |
Two pi and half sigma bonds are presents in molecule with bond order 2.5.
<br><br>Moleculer orbital configuration of $$N_2^+$$ (13 electrons)
<br><br>= $${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^1}}$$
<br><br>Bond order = $${1 \over 2}\left( {9 - 4} \right)$$ = 2.5
|
mcq
|
jee-main-2019-online-10th-january-morning-slot
| 953
|
rZmnCDvb0RMeec8w5Dkta
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
In which of the following process, the bond order has increased and paramagnetic character has charged to diamagnetic ?
|
[{"identifier": "A", "content": "NO $$ \\to $$ NO<sup>+</sup>"}, {"identifier": "B", "content": "N<sub>2</sub> $$ \\to $$ N<sub>2</sub><sup>+</sup>"}, {"identifier": "C", "content": "O<sub>2</sub> $$ \\to $$ O<sub>2</sub><sup>+</sup>"}, {"identifier": "D", "content": "O<sub>2</sub> $$ \\to $$ O<sub>2</sub><sup>2$$-$$</sup>"}]
|
["A"]
| null |
Molecular orbital configuration of NO (15 electrons) is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^o}^ * $$
<br><br>$$\therefore\,\,\,\,$$ N<sub>b</sub> = 10
<br><br>N<sub>a</sub> = 5
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 5} \right]$$ = 2.5
<br><br>And in NO one unpaired electron is present , so it is paramagnetic.
<br><br>Similarly Molecular orbital configuration of NO<sup>+</sup> (14 electrons) is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,$$
<br><br>$$\therefore\,\,\,\,$$ N<sub>b</sub> = 10
<br><br>N<sub>a</sub> = 4
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 4} \right]$$ = 3
<br><br>And in NO<sup>+</sup> no unpaired electron is present , so it is diamagnetic.
|
mcq
|
jee-main-2019-online-9th-january-evening-slot
| 954
|
39uduN98fcC5Wjng4F7k9k2k5epf47d
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
The bond order and the magnetic characteristics of CN<sup>-</sup>
are :
|
[{"identifier": "A", "content": "3, paramagnetic"}, {"identifier": "B", "content": "$$2{1 \\over 2}$$, paramagnetic"}, {"identifier": "C", "content": "3, diamagnetic"}, {"identifier": "D", "content": "$$2{1 \\over 2}$$, diamagnetic"}]
|
["C"]
| null |
CN<sup>-</sup> has 14 electrons.
<br><br>Moleculer orbital configuration of CN<sup>-</sup> is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\pi _{2p_x^2}} =\,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}$$
<br><br>Here is no unpaired electron so it is diamagnetic.
<br>N<sub>b</sub> = 10
<br><br>N<sub>a</sub> = 4
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 4} \right]$$ = 3
|
mcq
|
jee-main-2020-online-7th-january-evening-slot
| 955
|
OJ4HXN2l6O25EZHBDR7k9k2k5icjff5
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
If the magnetic moment of a dioxygen species
is 1.73 B.M, it may be :
|
[{"identifier": "A", "content": "$$O_2^ - $$ or $$O_2^ + $$"}, {"identifier": "B", "content": "O<sub>2</sub>, $$O_2^ - $$ or $$O_2^ + $$"}, {"identifier": "C", "content": "O<sub>2</sub> or $$O_2^ + $$"}, {"identifier": "D", "content": "O<sub>2</sub> or $$O_2^ - $$"}]
|
["A"]
| null |
Magnetic moment = 1.73 BM
<br><br>$$ \therefore $$ Unpaired electron = 1
<br><br>(1) $$O_2$$ has 16 electrons.
<br><br>Moleculer orbital configuration of $$O_2$$ is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^1}^ * $$
<br><br>Here 2 unpaired electron present.
<br><br>(2) $$O_2^{+}$$ has 15 electrons.
<br><br>Moleculer orbital configuration of $$O_2^{+}$$ is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^0}^ * $$
<br><br>Here 1 unpaired electron present.
<br><br>(3) $$O_2^{-}$$ has 17 electrons.
<br><br>Moleculer orbital configuration of $$O_2^{-}$$ is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * = \,\,\pi _{2p_y^1}^ * $$
<br><br>Here 1 unpaired electron present.
<br><br>Hence $$O_2^{-}$$ & $$O_2^{+}$$
have one unpaired electron.
|
mcq
|
jee-main-2020-online-9th-january-morning-slot
| 956
|
mHQPHyo2fQemHaqz1Fjgy2xukf2b5jgn
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
Of the species, NO, NO<sup>+</sup>, NO<sup>2+</sup> and NO<sup>-</sup>
, the one with minimum bond strength is :
|
[{"identifier": "A", "content": "NO<sup>\u2013</sup>"}, {"identifier": "B", "content": "NO"}, {"identifier": "C", "content": "NO<sup>+</sup>"}, {"identifier": "D", "content": "NO<sup>2+</sup>"}]
|
["A"]
| null |
<b><u>Note</u> :</b>
<br><br>(1) $$\,\,\,\,$$ Bond strength $$ \propto $$ Bond order
<br><br>(2) $$\,\,\,\,$$ Bond length $$ \propto $$ $${1 \over {Bond\,\,order}}$$
<br><br>(3) $$\,$$ Bond order $$ = {1 \over 2}$$ [N<sub>b </sub> $$-$$ N<sub>a</sub>]
<br><br>N<sub>b</sub> = No of electrons in bonding molecular orbital
<br><br>N<sub>a</sub> $$=$$ No of electrons in anti bonding molecular orbital
<br><br>(4) $$\,\,\,\,$$ upto 14 electrons, molecular orbital configuration is
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264441/exam_images/dyi05nwqikfdrpmfsriz.webp" loading="lazy" alt="JEE Main 2020 (Online) 3rd September Morning Slot Chemistry - Chemical Bonding & Molecular Structure Question 147 English Explanation 1">
<br><br>Here N<sub>a</sub> = Anti bonding electron $$=$$ 4 and N<sub>b</sub> = 10
<br><br>(5) $$\,\,\,\,$$ After 14 electrons to 20 electrons molecular orbital configuration is - - -
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267718/exam_images/tivzvzeox1orkr2tnz5y.webp" loading="lazy" alt="JEE Main 2020 (Online) 3rd September Morning Slot Chemistry - Chemical Bonding & Molecular Structure Question 147 English Explanation 2">
<br><br>Molecular orbital configuration of NO (15 electrons) is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^o}^ * $$
<br><br>$$\therefore\,\,\,\,$$ N<sub>b</sub> = 10
<br><br>N<sub>a</sub> = 5
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 5} \right]$$ = 2.5
<br><br>Similarly Molecular orbital configuration of NO<sup>+</sup> (14 electrons) is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,$$
<br><br>$$\therefore\,\,\,\,$$ N<sub>b</sub> = 10
<br><br>N<sub>a</sub> = 4
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 4} \right]$$ = 3
<br><br>Similarly Molecular orbital configuration of NO<sup>2+</sup> (13 electrons) is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^1}}\,$$
<br><br>$$\therefore\,\,\,\,$$ N<sub>b</sub> = 9
<br><br>N<sub>a</sub> = 4
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {9 - 4} \right]$$ = 2.5
<br><br>Molecular orbital configuration of NO<sup>-</sup> (16 electrons) is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^1}^ * $$
<br><br>$$\therefore\,\,\,\,$$ N<sub>b</sub> = 10
<br><br>N<sub>a</sub> = 6
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 6} \right]$$ = 2
<br><br>As Bond strength $$ \propto $$ Bond order
<br><br>$$ \therefore $$ NO<sup>–</sup> will have minimum bond strength.
|
mcq
|
jee-main-2020-online-3rd-september-morning-slot
| 957
|
LWpA0sfVy1vp0ZMoaD1kls92xym
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
According to molecular orbital theory, the species among the following that does not exist is :
|
[{"identifier": "A", "content": "$${O_2}^{2 - }$$"}, {"identifier": "B", "content": "$$B{e_2}$$"}, {"identifier": "C", "content": "$$H{e_2}^ - $$"}, {"identifier": "D", "content": "$$H{e_2}^ + $$"}]
|
["B"]
| null |
<b><u>Note</u> :</b>
<br><br>According to molecules orbital theory, when a molecule have bond order = 0 then that molecule does not exist.
<br><br>We know, Bond order $$ = {1 \over 2}$$ [N<sub>b </sub> $$-$$ N<sub>a</sub>]
<br><br>N<sub>b</sub> = No of electrons in bonding molecular orbital
<br><br>N<sub>a</sub> $$=$$ No of electrons in anti bonding molecular orbital
<br><br>(4) $$\,\,\,\,$$ upto 14 electrons, molecular orbital configuration is
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264441/exam_images/dyi05nwqikfdrpmfsriz.webp" loading="lazy" alt="JEE Main 2021 (Online) 25th February Morning Shift Chemistry - Chemical Bonding & Molecular Structure Question 136 English Explanation 1">
<br><br>Here N<sub>a</sub> = Anti bonding electron $$=$$ 4 and N<sub>b</sub> = 10
<br><br>(5) $$\,\,\,\,$$ After 14 electrons to 20 electrons molecular orbital configuration is - - -
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267718/exam_images/tivzvzeox1orkr2tnz5y.webp" loading="lazy" alt="JEE Main 2021 (Online) 25th February Morning Shift Chemistry - Chemical Bonding & Molecular Structure Question 136 English Explanation 2">
<br><br>Here N<sub>a</sub> = 10
<br><br>and N<sub>b</sub> = 10
<br><br>Molecular orbital configuration of O $$_2^{2 - }$$ (18 electrons) is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^2}^ * $$
<br><br>$$\therefore\,\,\,\,$$ N<sub>b</sub> = 10
<br><br>N<sub>a</sub> = 8
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}$$ [ 10 $$-$$ 8] = 1
<br><br>$$\,\,\,$$ Configuration of $$He_2^ + $$ (3 electrons) is = $${\sigma _{1{s^2}}}\,\,\sigma _{1{s^1}}^ * $$
<br><br>$$\therefore\,\,\,$$ Bond order = $${1 \over 2}$$ (2 $$-$$1) = 0.5
<br><br>$$\,\,\,$$ Configuration of $$He_2^ - $$ (5 electrons) is = $${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * {\sigma _{2{s^1}}}\,$$
<br><br>$$\therefore\,\,\,$$ Bond order = $${1 \over 2}$$ (3 $$-$$2) = 0.5
<br><br>$$\,\,\,$$ Configuration of $$Be_2 $$ (4 electrons) is = $${\sigma _{1{s^2}}}$$ $$\sigma _{1{s^2}}^ * $$
<br><br>$$\therefore$$ Bond order = $${1 \over 2}$$ (2 $$-$$ 2) = 0
|
mcq
|
jee-main-2021-online-25th-february-morning-slot
| 958
|
OJX2Tfv6pbCnjqppz81klurauoc
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
Match List - I with List - II.<br/><br/><table>
<thead>
<tr>
<th></th>
<th>List - I (Molecule)</th>
<th></th>
<th>List - II (Bond order)</th>
</tr>
</thead>
<tbody>
<tr>
<td>(a)</td>
<td>$$N{e_2}$$</td>
<td>(i)</td>
<td>1</td>
</tr>
<tr>
<td>(b)</td>
<td>$${N_2}$$</td>
<td>(ii)</td>
<td>2</td>
</tr>
<tr>
<td>(c)</td>
<td>$${F_2}$$</td>
<td>(iii)</td>
<td>0</td>
</tr>
<tr>
<td>(d)</td>
<td>$${O_2}$$</td>
<td>(iv)</td>
<td>3</td>
</tr>
</tbody>
</table><br/>Choose the correct answer from the options given below :
|
[{"identifier": "A", "content": "(a) $$ \\to $$ (i), (b) $$ \\to $$ (ii), (c) $$ \\to $$ (iii), (d) $$ \\to $$ (iv)"}, {"identifier": "B", "content": "(a) $$ \\to $$ (iv), (b) $$ \\to $$ (iii), (c) $$ \\to $$ (ii), (d) $$ \\to $$ (i)"}, {"identifier": "C", "content": "(a) $$ \\to $$ (iii), (b) $$ \\to $$ (iv), (c) $$ \\to $$ (i), (d) $$ \\to $$ (ii)"}, {"identifier": "D", "content": "(a) $$ \\to $$ (ii), (b) $$ \\to $$ (i), (c) $$ \\to $$ (iv), (d) $$ \\to $$ (iii)"}]
|
["C"]
| null |
Bond order $$ = {1 \over 2}$$ [N<sub>b </sub> $$-$$ N<sub>a</sub>]
<br><br>N<sub>b</sub> = No of electrons in bonding molecular orbital
<br><br>N<sub>a</sub> $$=$$ No of electrons in anti bonding molecular orbital
<br><br>(4) $$\,\,\,\,$$ upto 14 electrons, molecular orbital configuration is
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264441/exam_images/dyi05nwqikfdrpmfsriz.webp" loading="lazy" alt="JEE Main 2021 (Online) 26th February Evening Shift Chemistry - Chemical Bonding & Molecular Structure Question 133 English Explanation 1">
<br><br>Here N<sub>a</sub> = Anti bonding electron $$=$$ 4 and N<sub>b</sub> = 10
<br><br>(5) $$\,\,\,\,$$ After 14 electrons to 20 electrons molecular orbital configuration is - - -
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267718/exam_images/tivzvzeox1orkr2tnz5y.webp" loading="lazy" alt="JEE Main 2021 (Online) 26th February Evening Shift Chemistry - Chemical Bonding & Molecular Structure Question 133 English Explanation 2">
<br><br>Here N<sub>a</sub> = 10
<br><br>and N<sub>b</sub> = 10
<br><br>(a) Molecular orbital configuration of Ne<sub>2</sub> (20 electrons) is
<br><br>$${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$$ $${\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,$$ $${\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}} = {\pi _{2p_y^2}}\,\,\pi _{2p_x^2}^ * \,\, = \pi _{2p_y^2}^ * $$$$\sigma _{2p_z^2}^*$$
<br><br>$$\therefore\,\,\,\,$$N<sub>a</sub> = 10
<br><br>N<sub>b</sub> = 10
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 10} \right] = 0$$
<br><br>(Note : All inert gases has BO = 0 and it does not exist as molecule. Here Ne is also an inert gas.)
<br><br>(b) Moleculer orbital configuration of $$N_2$$ (14 electrons) is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}$$
<br><br>$$\therefore\,\,\,\,$$N<sub>a</sub> = 4
<br><br>N<sub>b</sub> = 10
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 4} \right] = 3$$
<br><br>(d) Molecular orbital configuration of F<sub>2</sub> (18 electrons) is
<br><br>$${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$$ $${\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,$$ $${\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}} = {\pi _{2p_y^2}}\,\,\pi _{2p_x^2}^ * \,\, = \pi _{2p_y^2}^ * $$
<br><br>$$\therefore\,\,\,\,$$N<sub>a</sub> = 8
<br><br>N<sub>b</sub> = 10
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 8} \right] = 1$$
<br><br>(d) Molecular orbital configuration of O<sub>2</sub> (16 electrons) is
<br><br>$${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$$ $${\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,$$ $${\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}} = {\pi _{2p_y^2}}\,\,\pi _{2p_x^1}^ * \,\, = \pi _{2p_y^1}^ * $$
<br><br>$$\therefore\,\,\,\,$$N<sub>a</sub> = 6
<br><br>N<sub>b</sub> = 10
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 6} \right] = 2$$
|
mcq
|
jee-main-2021-online-26th-february-evening-slot
| 959
|
9ZObK2CT7396w4lhw01kmlnkmj6
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
AX is a covalent diatomic molecule where A and X are second row elements of periodic table. Based on Molecular orbital theory, the bond order of AX is 2.5. The total number of electrons in AX is __________. (Round off to the Nearest Integer).
|
[]
| null |
15
|
The compound AX is NO its bond order is 2.5 and it has total 15 electrons.
<br><br><b>Note :</b> Total number of electrons equal to 13 will
also have the 2.5 bond order. But in this case
neutral diatomic molecule will not be possible.
|
integer
|
jee-main-2021-online-18th-march-morning-shift
| 960
|
1krx4sr8p
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
In the following the correct bond order sequence is :
|
[{"identifier": "A", "content": "$$O_2^{2 - } > O_2^ + > O_2^ - > {O_2}$$"}, {"identifier": "B", "content": "$$O_2^ + > O_2^ - > O_2^{2 - } > {O_2}$$"}, {"identifier": "C", "content": "$$O_2^ + > {O_2} > O_2^ - > O_2^{2 - }$$"}, {"identifier": "D", "content": "$${O_2} > O_2^ - > O_2^{2 - } > O_2^ + $$"}]
|
["C"]
| null |
<b><u>Note</u> :</b>
<br><br>(1) $$\,\,\,\,$$ Bond strength $$ \propto $$ Bond order
<br><br>(2) $$\,\,\,\,$$ Bond length $$ \propto $$ $${1 \over {Bond\,\,order}}$$
<br><br>(3) $$\,$$ Bond order $$ = {1 \over 2}$$ [N<sub>b </sub> $$-$$ N<sub>a</sub>]
<br><br>N<sub>b</sub> = No of electrons in bonding molecular orbital
<br><br>N<sub>a</sub> $$=$$ No of electrons in anti bonding molecular orbital
<br><br>(4) $$\,\,\,\,$$ upto 14 electrons, molecular orbital configuration is
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264441/exam_images/dyi05nwqikfdrpmfsriz.webp" loading="lazy" alt="JEE Main 2021 (Online) 25th July Evening Shift Chemistry - Chemical Bonding & Molecular Structure Question 123 English Explanation 1">
<br><br>Here N<sub>a</sub> = Anti bonding electron $$=$$ 4 and N<sub>b</sub> = 10
<br><br>(5) $$\,\,\,\,$$ After 14 electrons to 20 electrons molecular orbital configuration is - - -
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267718/exam_images/tivzvzeox1orkr2tnz5y.webp" loading="lazy" alt="JEE Main 2021 (Online) 25th July Evening Shift Chemistry - Chemical Bonding & Molecular Structure Question 123 English Explanation 2">
<br><br>Here N<sub>a</sub> = 10
<br><br>and N<sub>b</sub> = 10
<br><br>In O atom 8 electrons present, so in O<sub>2</sub>, 8 $$ \times $$ 2 = 16 electrons present.
<br><br>Then in $$O_2^ + $$ no of electrons = 15
<br><br>in $$O_2^ - $$ no of electrons = 17
<br><br>in $$O_2^{2 - }$$ no of electrons = 18
<br><br>$$\therefore\,\,\,\,$$ Molecular orbital configuration of O<sub>2</sub> (16 electrons) is
<br><br>$${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$$ $${\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,$$ $${\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}} = {\pi _{2p_y^2}}\,\,\pi _{2p_x^1}^ * \,\, = \pi _{2p_y^1}^ * $$
<br><br>$$\therefore\,\,\,\,$$N<sub>a</sub> = 6
<br><br>N<sub>b</sub> = 10
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 6} \right] = 2$$
<br><br>Molecular orbital configuration of O$$_2^ + $$ (15 electrons) is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^o}^ * $$
<br><br>$$\therefore\,\,\,\,$$ N<sub>b</sub> = 10
<br><br>N<sub>a</sub> = 5
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 5} \right]$$ = 2.5
<br><br>Molecular orbital configuration of $$O_2^ - $$ (17 electrons) is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^1}^ * $$
<br><br>$$\therefore\,\,\,\,$$ N<sub>b</sub> = 10
<br><br>N<sub>a</sub> = 7
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 7} \right]$$ = 1.5
<br><br>Molecular orbital configuration of O $$_2^{2 - }$$ (18 electrons) is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^2}^ * $$
<br><br>$$\therefore\,\,\,\,$$ N<sub>b</sub> = 10
<br><br>N<sub>a</sub> = 8
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}$$ [ 10 $$-$$ 8] = 1
<br><br>As Bond strength $$ \propto $$ Bond order so, correct order is
<br><br>$$O_2^{2 - } < O_2^ - < {O_2} < O_2^ + $$
|
mcq
|
jee-main-2021-online-25th-july-evening-shift
| 961
|
1krz3371p
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
The difference between bond orders of CO and NO$$^ \oplus $$ is $${x \over 2}$$ where x = _____________. (Round off to the Nearest Integer)
|
[]
| null |
0
|
Bond order of CO = 3<br><br>Bond order of NO<sup>+</sup> = 3<br><br>Difference = 0 = $${x \over 2}$$<br><br>$$ \Rightarrow $$ x = 0
<br><br><b><u>Note</u> :</b>
<br><br>(1) $$\,$$ Bond order $$ = {1 \over 2}$$ [N<sub>b </sub> $$-$$ N<sub>a</sub>]
<br><br>N<sub>b</sub> = No of electrons in bending molecular orbital
<br><br>N<sub>a</sub> $$=$$ No of electrons in anti bonding molecular orbital
<br><br>(4) $$\,\,\,\,$$ upto 14 electrons, molecular orbital configuration is
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264441/exam_images/dyi05nwqikfdrpmfsriz.webp" loading="lazy" alt="JEE Main 2021 (Online) 27th July Morning Shift Chemistry - Chemical Bonding & Molecular Structure Question 122 English Explanation 1">
<br><br>Here N<sub>a</sub> = Anti bonding electron $$=$$ 4 and N<sub>b</sub> = 10
<br><br>(5) $$\,\,\,\,$$ After 14 electrons to 20 electrons molecular orbital configuration is - - -
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267718/exam_images/tivzvzeox1orkr2tnz5y.webp" loading="lazy" alt="JEE Main 2021 (Online) 27th July Morning Shift Chemistry - Chemical Bonding & Molecular Structure Question 122 English Explanation 2">
<br><br>Here N<sub>a</sub> = 10
<br><br>and N<sub>b</sub> = 10
<br><br>(A) CO has 14 electrons.
<br><br>Moleculer orbital configuration of CO is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\pi _{2p_x^2}} =\,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}$$
<br><br>$$\therefore$$ N<sub>b</sub> = 10
<br><br>N<sub>a</sub> = 4
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}$$ [ 10 $$-$$ 4] = 3
<br><br>(B) NO<sup>+</sup> has 14 electrons.
<br><br>Moleculer orbital configuration of NO<sup>+</sup> is
<br><br>$${\sigma _{1{s^2}}}$$ $$\sigma _{1{s^2}}^ * $$ $${\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,\, = \,\,{\pi _{2p_y^2}}$$
<br><br>$$\therefore$$ N<sub>b</sub> = 10
<br><br>N<sub>a</sub> = 4
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}$$ [ 10 $$-$$ 4] = 3
|
integer
|
jee-main-2021-online-27th-july-morning-shift
| 962
|
1ks1jnmhr
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
The total number of electrons in all bonding molecular orbitals of $$O_2^{2 - }$$ is ______________.<br/><br/>(Round off to the nearest integer)
|
[]
| null |
10
|
N<sub>b</sub> = No of electrons in bonding molecular orbital
<br><br>N<sub>a</sub> $$=$$ No of electrons in anti bonding molecular orbital
<br><br>(1) $$\,\,\,\,$$ upto 14 electrons, molecular orbital configuration is
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264441/exam_images/dyi05nwqikfdrpmfsriz.webp" loading="lazy" alt="JEE Main 2021 (Online) 27th July Evening Shift Chemistry - Chemical Bonding & Molecular Structure Question 121 English Explanation 1">
<br><br>Here N<sub>a</sub> = Anti bonding electron $$=$$ 4 and N<sub>b</sub> = 10
<br><br>(2) $$\,\,\,\,$$ After 14 electrons to 20 electrons molecular orbital configuration is - - -
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267718/exam_images/tivzvzeox1orkr2tnz5y.webp" loading="lazy" alt="JEE Main 2021 (Online) 27th July Evening Shift Chemistry - Chemical Bonding & Molecular Structure Question 121 English Explanation 2">
<br><br>Here N<sub>a</sub> = 10
<br><br>and N<sub>b</sub> = 10
<br><br>In O atom 8 electrons present, so in O<sub>2</sub>, 8 $$ \times $$ 2 = 16 electrons present.
<br><br>Then in $$O_2^ + $$ no of electrons = 15
<br><br>in $$O_2^ - $$ no of electrons = 17
<br><br>in $$O_2^{2 - }$$ no of electrons = 18
<br><br>Molecular orbital configuration of O $$_2^{2 - }$$ (18 electrons) is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^2}^ * $$
<br><br>$$\therefore\,\,\,\,$$ N<sub>b</sub> = 10
<br><br>and N<sub>a</sub> = 8
|
integer
|
jee-main-2021-online-27th-july-evening-shift
| 963
|
1ktcnvdv8
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
The bond order and magnetic behaviour of $$O_2^ - $$ ion are respectively :
|
[{"identifier": "A", "content": "1.5 and paramagnetic "}, {"identifier": "B", "content": "1.5 and diamagnetic"}, {"identifier": "C", "content": "2 and diamagnetic"}, {"identifier": "D", "content": "1 and paramagnetic "}]
|
["A"]
| null |
$$O_2^ - = {({\sigma _{1s}})^2}{(\sigma _{1s}^*)^2}{({\sigma _{2s}})^2}{(\sigma _{2s}^*)^2}{({\sigma _{2{p_z}}})^2}$$$$\left( {\pi _{2{p_x}}^2 = \pi _{2{p_y}}^2} \right)\left( {\pi _{2{p_x}}^{*2} = \pi _{2{p_y}}^{*1}} \right)$$<br><br>Bond order = $${{10 - 7} \over 2} = 1.5$$ <br><br>and paramagnetic.
|
mcq
|
jee-main-2021-online-26th-august-evening-shift
| 964
|
1ktebop1v
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
Match items of List-I with those of List-II :<br/><br/><table>
<thead>
<tr>
<th></th>
<th>List - I<br/>(Property)</th>
<th></th>
<th>List - II<br/>(Example)</th>
</tr>
</thead>
<tbody>
<tr>
<td>(a)</td>
<td>Diamagnetism</td>
<td>(i)</td>
<td>MnO</td>
</tr>
<tr>
<td>(b)</td>
<td>Ferrimagnetism</td>
<td>(ii)</td>
<td>$${O_2}$$</td>
</tr>
<tr>
<td>(c)</td>
<td>Paramagnetism</td>
<td>(iii)</td>
<td>NaCl</td>
</tr>
<tr>
<td>(d)</td>
<td>Antiferromagnetism</td>
<td>(iv)</td>
<td>$$F{e_3}{O_4}$$</td>
</tr>
</tbody>
</table><br/><br/>Choose the most appropriate answer from the options given below :
|
[{"identifier": "A", "content": "(a)-(ii), (b)-(i), (c)-(iii), (d)-(iv)"}, {"identifier": "B", "content": "(a)-(i), (b)-(iii), (c)-(iv), (d)-(ii)"}, {"identifier": "C", "content": "(a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)"}, {"identifier": "D", "content": "(a)-(iv), (b)-(ii), (c)-(i), (d)-(iii)"}]
|
["C"]
| null |
<p> A. NaCl is diamagnetic because all electrons are paired in Na<sup>+</sup> and in Cl<sup>−</sup> . So, it shows diamagnetism. </p>
<p> B. Fe<sub>3</sub>O<sub>4</sub> is ferrimagnetic because of the presence of the unequal alignment of magnetic moment in opposite direction. </p>
<p> C. O<sub>2</sub> molecule has two unpaired electrons. So, it is paramagnetic. </p>
<p> D. In MnO, the orientation of the electrons of Mn and O is such that they cancel their effects, hence it is antiferromagnetic.</p>
|
mcq
|
jee-main-2021-online-27th-august-morning-shift
| 965
|
1ktn2xa2s
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
The spin-only magnetic moment value of $$B_2^ + $$ species is _____________ $$\times$$ 10<sup>$$-$$2</sup> BM. (Nearest integer) [Given : $$\sqrt 3 $$ = 1.73]
|
[]
| null |
173
|
$$B_2^ + \Rightarrow \sigma _{1s}^2\sigma _{1s}^{*2}\sigma _{2s}^2\sigma _{2s}^{*2}\pi _{2py}^1 \simeq \pi _{2pz}^0$$<br><br> It has one
unpaired electron.<br><br>Spin - only magnetic moment = $$\mu
$$
<br><br>= $$\sqrt {n\left( {n + 1} \right)} $$
<br><br>n = Number of unpaired electrons
<br><br>$$= \sqrt {1(1 + 2)} = \sqrt 3 $$ BM<br><br>= 1.73 BM<br><br>= 1.73 $$\times$$ 10<sup>$$-$$2</sup> BM
|
integer
|
jee-main-2021-online-1st-september-evening-shift
| 968
|
1l58di4xa
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
<p>Consider the ions/molecule</p>
<p>O$$_2^ + $$, O<sub>2</sub>, O$$_2^ - $$, O$$_2^ {2-} $$</p>
<p>For increasing bond order the correct option is :</p>
|
[{"identifier": "A", "content": "O$$_2^ {2-} $$ < O$$_2^ - $$ < O<sub>2</sub> < O$$_2^ + $$"}, {"identifier": "B", "content": "O$$_2^ - $$ < O$$_2^ {2-} $$ < O<sub>2</sub> < O$$_2^ + $$"}, {"identifier": "C", "content": "O$$_2^ - $$ < O$$_2^ {2-} $$ < O$$_2^ + $$ < O<sub>2</sub>"}, {"identifier": "D", "content": "O$$_2^ - $$ < O$$_2^ + $$ < O$$_2^ {2-} $$ < O<sub>2</sub>"}]
|
["A"]
| null |
<b><u>Note</u> :</b>
<br><br>(1) $$\,\,\,\,$$ Bond strength $$ \propto $$ Bond order
<br><br>(2) $$\,\,\,\,$$ Bond length $$ \propto $$ $${1 \over {Bond\,\,order}}$$
<br><br>(3) $$\,$$ Bond order $$ = {1 \over 2}$$ [N<sub>b </sub> $$-$$ N<sub>a</sub>]
<br><br>N<sub>b</sub> = Number of electrons in bonding molecular orbital
<br><br>N<sub>a</sub> $$=$$ Number of electrons in anti bonding molecular orbital
<br><br>(4) $$\,\,\,\,$$ upto 14 electrons, molecular orbital configuration is
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264441/exam_images/dyi05nwqikfdrpmfsriz.webp" loading="lazy" alt="JEE Main 2022 (Online) 26th June Morning Shift Chemistry - Chemical Bonding & Molecular Structure Question 104 English Explanation 1">
<br><br>Here N<sub>a</sub> = Anti bonding electron $$=$$ 4 and N<sub>b</sub> = 10
<br><br>(5) $$\,\,\,\,$$ After 14 electrons to 20 electrons molecular orbital configuration is - - -
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267718/exam_images/tivzvzeox1orkr2tnz5y.webp" loading="lazy" alt="JEE Main 2022 (Online) 26th June Morning Shift Chemistry - Chemical Bonding & Molecular Structure Question 104 English Explanation 2">
<br><br>Here N<sub>a</sub> = 10
<br><br>and N<sub>b</sub> = 10
<br><br>In O atom 8 electrons present, so in O<sub>2</sub>, 8 $$ \times $$ 2 = 16 electrons present.
<br><br>Then in $$O_2^ + $$ no of electrons = 15
<br><br>in $$O_2^ - $$ no of electrons = 17
<br><br>in $$O_2^{2 - }$$ no of electrons = 18
<br><br>$$\therefore\,\,\,\,$$ Molecular orbital configuration of O<sub>2</sub> (16 electrons) is
<br><br>$${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$$ $${\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,$$ $${\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}} = {\pi _{2p_y^2}}\,\,\pi _{2p_x^1}^ * \,\, = \pi _{2p_y^1}^ * $$
<br><br>$$\therefore\,\,\,\,$$N<sub>a</sub> = 6
<br><br>N<sub>b</sub> = 10
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 6} \right] = 2$$
<br><br>Molecular orbital configuration of O$$_2^ + $$ (15 electrons) is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^o}^ * $$
<br><br>$$\therefore\,\,\,\,$$ N<sub>b</sub> = 10
<br><br>N<sub>a</sub> = 5
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 5} \right]$$ = 2.5
<br><br>Molecular orbital configuration of $$O_2^ - $$ (17 electrons) is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^1}^ * $$
<br><br>$$\therefore\,\,\,\,$$ N<sub>b</sub> = 10
<br><br>N<sub>a</sub> = 7
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 7} \right]$$ = 1.5
<br><br>Molecular orbital configuration of O $$_2^{2 - }$$ (18 electrons) is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^2}^ * $$
<br><br>$$\therefore\,\,\,\,$$ N<sub>b</sub> = 10
<br><br>N<sub>a</sub> = 8
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}$$ [ 10 $$-$$ 8] = 1
<br><br> So, correct order of Bond order is
<br><br>$$O_2^{2 - } < O_2^ - < {O_2} < O_2^ + $$
|
mcq
|
jee-main-2022-online-26th-june-morning-shift
| 969
|
1l5alkri6
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
<p>Bonding in which of the following diatomic molecule(s) become(s) stronger, on the basis of MO Theory, by removal of an electron?</p>
<p>(A) NO</p>
<p>(B) N<sub>2</sub></p>
<p>(C) O<sub>2</sub></p>
<p>(D) C<sub>2</sub></p>
<p>(E) B<sub>2</sub></p>
<p>Choose the most appropriate answer from the options given below :</p>
|
[{"identifier": "A", "content": "(A), (B), (C) only"}, {"identifier": "B", "content": "(B), (C), (E) only"}, {"identifier": "C", "content": "(A), (C) only"}, {"identifier": "D", "content": "(D) only"}]
|
["C"]
| null |
If an electron is removed from the anti-bonding
orbital, then it will tend to increase the bond order.
The HOMO in NO and O<sub>2</sub> is antibonding molecular
orbital .<br/><br/>
Hence, in NO and O<sub>2</sub> bond order will increase on
loss of electron.
|
mcq
|
jee-main-2022-online-25th-june-morning-shift
| 970
|
1l5bcwle0
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
<p>The correct order of bond orders of $${C_2}^{2 - }$$, $${N_2}^{2 - }$$ and $${O_2}^{2 - }$$</p> is, respectively
|
[{"identifier": "A", "content": "$${C_2}^{2 - }$$ < $${N_2}^{2 - }$$ < $${O_2}^{2 - }$$"}, {"identifier": "B", "content": "$${O_2}^{2 - }$$ < $${N_2}^{2 - }$$ < $${C_2}^{2 - }$$"}, {"identifier": "C", "content": "$${C_2}^{2 - }$$ < $${O_2}^{2 - }$$ < $${N_2}^{2 - }$$"}, {"identifier": "D", "content": "$${N_2}^{2 - }$$ < $${C_2}^{2 - }$$ < $${O_2}^{2 - }$$"}]
|
["B"]
| null |
<b><u>Note</u> :</b>
<br><br>(1) $$\,$$ Bond order $$ = {1 \over 2}$$ [N<sub>b </sub> $$-$$ N<sub>a</sub>]
<br><br>N<sub>b</sub> = No of electrons in bonding molecular orbital
<br><br>N<sub>a</sub> $$=$$ No of electrons in anti bonding molecular orbital
<br><br>(2) $$\,\,\,\,$$ upto 14 electrons, molecular orbital configuration is
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264441/exam_images/dyi05nwqikfdrpmfsriz.webp" loading="lazy" alt="JEE Main 2022 (Online) 24th June Evening Shift Chemistry - Chemical Bonding & Molecular Structure Question 99 English Explanation 1">
<br><br>Here N<sub>a</sub> = Anti bonding electron $$=$$ 4 and N<sub>b</sub> = 10
<br><br>(3) $$\,\,\,\,$$ After 14 electrons to 20 electrons molecular orbital configuration is - - -
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267718/exam_images/tivzvzeox1orkr2tnz5y.webp" loading="lazy" alt="JEE Main 2022 (Online) 24th June Evening Shift Chemistry - Chemical Bonding & Molecular Structure Question 99 English Explanation 2">
<br><br>Here N<sub>a</sub> = 10
<br><br>and N<sub>b</sub> = 10
<br><br>In O atom 8 electrons present, so in O<sub>2</sub>, 8 $$ \times $$ 2 = 16 electrons present.
<br><br>in $$O_2^{2 - }$$ no of electrons = 18
<br><br>(A) Molecular orbital configuration of O $$_2^{2 - }$$ (18 electrons) is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^2}^ * $$
<br><br>$$\therefore\,\,\,\,$$ N<sub>b</sub> = 10
<br><br>N<sub>a</sub> = 8
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}$$ [ 10 $$-$$ 8] = 1
<br><br>(B) $$C_2^{2 - }$$ has 14 electrons.
<br><br>Moleculer orbital configuration of $$C_2^{2 - }$$ is
<br><br> $${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}$$
<br><br>$$\therefore\,\,\,\,$$N<sub>a</sub> = 4
<br><br>N<sub>b</sub> = 10
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 4} \right] = 3$$
<br><br>(C) $$N_2^{2 - }$$ has 16 electrons.
<br><br>Moleculer orbital configuration of $$N_2^{2 - }$$ is
<br><br> $${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^1}^ *$$
<br><br>$$\therefore\,\,\,\,$$N<sub>a</sub> = 6
<br><br>N<sub>b</sub> = 10
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {10 - 6} \right] = 2$$
<br><br>The correct order of bond orders of $${C_2}^{2 - }$$, $${N_2}^{2 - }$$ and $${O_2}^{2 - }$$
<br><br>$${O_2}^{2 - }$$ < $${N_2}^{2 - }$$ < $${C_2}^{2 - }$$
|
mcq
|
jee-main-2022-online-24th-june-evening-shift
| 971
|
1l6e1w620
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
<p>Among the following species</p>
<p>$$\mathrm{N}_{2}, \mathrm{~N}_{2}^{+}, \mathrm{N}_{2}^{-}, \mathrm{N}_{2}^{2-}, \mathrm{O}_{2}, \mathrm{O}_{2}^{+}, \mathrm{O}_{2}^{-}, \mathrm{O}_{2}^{2-}$$</p>
<p>the number of species showing diamagnesim is _______________.</p>
|
[]
| null |
2
|
Those species which have unpaired electrons are called paramagnetic species.
<br><br>And those species which have no unpaired electrons are called diamagnetic species.
<br><br>(1) $$N_2$$ has 14 electrons.
<br><br>Moleculer orbital configuration of $$N_2$$
<br><br>= $${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}$$
<br><br>Here no unpaired electron present, so it is diamagnetic.
<br><br>(2) Moleculer orbital configuration of $$N_2^{ + }$$ (13 electrons)
<br><br>= $${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^1}}$$
<br><br>Here in $$N_2^{ + }$$, 1 unpaired electron present, so it is paramagnetic.
<br><br>(3) $$\mathrm{N}_{2}^{2-}$$ has 16 electrons.
<br><br>Moleculer orbital configuration of $$\mathrm{N}_{2}^{2-}$$ is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^1}^ * $$
<br><br>Here 2 unpaired electron present, so it is paramagnetic.
<br><br>(4) $$\mathrm{N}_{2}^{-}$$ has 15 electrons.
<br><br>Moleculer orbital configuration of $$\mathrm{N}_{2}^{-}$$ is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^0}^ * $$
<br><br>Here 1 unpaired electron present, so it is paramagnetic.
<br><br>(a) $$O_2^{2−}$$ has 18 electrons.
<br><br>Moleculer orbital configuration of $$O_2^{2−}$$ is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^2}^ * $$
<br><br>Here is no unpaired electron so it is diamagnetic.
<br><br>(b) $$O_2^{−}$$ has 17 electrons.
<br><br>Moleculer orbital configuration of $$O_2^{2−}$$ is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^1}^ * $$
<br><br>Here 1 unpaired electron present, so it is paramagnetic.
<br><br>(c) $$O_2$$ has 16 electrons.
<br><br>Moleculer orbital configuration of $$O_2$$ is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^1}^ * $$
<br><br>Here 2 unpaired electron present, so it is paramagnetic.
<br><br>(d) $$O_2^{+}$$ has 15 electrons.
<br><br>Moleculer orbital configuration of $$O_2^{+}$$ is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^0}^ * $$
<br><br>Here 1 unpaired electron present, so it is paramagnetic.
|
integer
|
jee-main-2022-online-25th-july-morning-shift
| 972
|
1l6jn1ls9
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
<p>According to MO theory, number of species/ions from the following having identical bond order is ________.</p>
<p>$$\mathrm{CN}^{-}, \mathrm{NO}^{+}, \mathrm{O}_{2}, \mathrm{O}_{2}^{+}, \mathrm{O}_{2}^{2+}$$</p>
|
[]
| null |
3
|
$\mathrm{CN}^{-}, \mathrm{NO}^{+}$and $\mathrm{O}_{2}^{2+}$ have bond order of 3
<br/><br/>
$\mathrm{O}_{2}$ has bond order of 2
<br/><br/>
$\mathrm{O}_{2}^{+}$ has bond order of $2.5$
<br/><br/>
$\therefore 3$ species have similar bond order.
|
integer
|
jee-main-2022-online-27th-july-morning-shift
| 974
|
1ldscb4vt
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
<p>According to MO theory the bond orders for $$\mathrm{O}$$$$_2^{2 - }$$, $$\mathrm{CO}$$ and $$\mathrm{NO^+}$$ respectively, are</p>
|
[{"identifier": "A", "content": "1, 3 and 3"}, {"identifier": "B", "content": "2, 3 and 3"}, {"identifier": "C", "content": "1, 2 and 3"}, {"identifier": "D", "content": "1, 3 and 2"}]
|
["A"]
| null |
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;}
.tg .tg-baqh{text-align:center;vertical-align:top}
</style>
<table class="tg" style="undefined;table-layout: fixed; width: 227px">
<colgroup>
<col style="width: 113px">
<col style="width: 114px">
</colgroup>
<thead>
<tr>
<th class="tg-baqh">Species</th>
<th class="tg-baqh">B.O.</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-baqh">O$$_2^{2 - }$$</td>
<td class="tg-baqh">1</td>
</tr>
<tr>
<td class="tg-baqh">CO</td>
<td class="tg-baqh">3</td>
</tr>
<tr>
<td class="tg-baqh">NO$$^ \oplus $$</td>
<td class="tg-baqh">3</td>
</tr>
</tbody>
</table></p>
<b><u>Note</u> :</b>
<br><br>(1) $$\,$$ Bond order $$ = {1 \over 2}$$ [N<sub>b </sub> $$-$$ N<sub>a</sub>]
<br><br>N<sub>b</sub> = No of electrons in bending molecular orbital
<br><br>N<sub>a</sub> $$=$$ No of electrons in anti bonding molecular orbital
<br><br>(4) $$\,\,\,\,$$ upto 14 electrons, molecular orbital configuration is
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264441/exam_images/dyi05nwqikfdrpmfsriz.webp" loading="lazy" alt="JEE Main 2023 (Online) 29th January Evening Shift Chemistry - Chemical Bonding & Molecular Structure Question 76 English Explanation 1">
<br><br>Here N<sub>a</sub> = Anti bonding electron $$=$$ 4 and N<sub>b</sub> = 10
<br><br>(5) $$\,\,\,\,$$ After 14 electrons to 20 electrons molecular orbital configuration is - - -
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267718/exam_images/tivzvzeox1orkr2tnz5y.webp" loading="lazy" alt="JEE Main 2023 (Online) 29th January Evening Shift Chemistry - Chemical Bonding & Molecular Structure Question 76 English Explanation 2">
<br><br>Here N<sub>a</sub> = 10
<br><br>and N<sub>b</sub> = 10
<br><br>(A) In O atom 8 electrons present, so in O<sub>2</sub>, 8 $$ \times $$ 2 = 16 electrons present.
<br><br>Then in $$O_2^{2 - }$$ no of electrons = 18
<br><br>$$ \therefore $$ Molecular orbital configuration of O $$_2^{2 - }$$ (18 electrons) is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^2}^ * $$
<br><br>$$\therefore$$ N<sub>b</sub> = 10
<br><br>N<sub>a</sub> = 8
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}$$ [ 10 $$-$$ 8] = 1
<br><br>(B) CO has 14 electrons.
<br><br>Moleculer orbital configuration of CO is
<br><br>$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\pi _{2p_x^2}} =\,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}$$
<br><br>$$\therefore$$ N<sub>b</sub> = 10
<br><br>N<sub>a</sub> = 4
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}$$ [ 10 $$-$$ 4] = 3
<br><br>(C) NO<sup>+</sup> has 14 electrons.
<br><br>Moleculer orbital configuration of NO<sup>+</sup> is
<br><br>$${\sigma _{1{s^2}}}$$ $$\sigma _{1{s^2}}^ * $$ $${\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,\, = \,\,{\pi _{2p_y^2}}$$
<br><br>$$\therefore$$ N<sub>b</sub> = 10
<br><br>N<sub>a</sub> = 4
<br><br>$$\therefore\,\,\,\,$$ BO = $${1 \over 2}$$ [ 10 $$-$$ 4] = 3
|
mcq
|
jee-main-2023-online-29th-january-evening-shift
| 976
|
1ldwsjb3e
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
<p>What is the number of unpaired electron(s) in the highest occupied molecular orbital of the following species : $$\mathrm{{N_2};N_2^ + ;{O_2};O_2^ + }$$ ?</p>
|
[{"identifier": "A", "content": "0, 1, 0, 1"}, {"identifier": "B", "content": "2, 1, 0, 1"}, {"identifier": "C", "content": "0, 1, 2, 1"}, {"identifier": "D", "content": "2, 1, 2, 1"}]
|
["C"]
| null |
<style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;width:100%}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;}
.tg .tg-c3ow{border-color:inherit;text-align:center;vertical-align:top}
</style>
<table class="tg">
<thead>
<tr>
<th class="tg-c3ow">Molecule</th>
<th class="tg-c3ow">$\begin{gathered}\text { No. of unpaired electron in } \\\text { highest occupied } \\\text { molecular orbital }\end{gathered}$</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-c3ow">$\mathrm{N}_{2}$</td>
<td class="tg-c3ow">0</td>
</tr>
<tr>
<td class="tg-c3ow">$\mathrm{N}_{2}^{\oplus}$</td>
<td class="tg-c3ow">1</td>
</tr>
<tr>
<td class="tg-c3ow">$\mathrm{O}_{2}$</td>
<td class="tg-c3ow">2</td>
</tr>
<tr>
<td class="tg-c3ow">$\mathrm{O}_{2}^{\oplus}$</td>
<td class="tg-c3ow">1</td>
</tr>
</tbody>
</table>
|
mcq
|
jee-main-2023-online-24th-january-evening-shift
| 978
|
1lgq3vwp7
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
<p>In which of the following processes, the bond order increases and paramagnetic character changes to diamagnetic one ?</p>
|
[{"identifier": "A", "content": "$$\\mathrm{O}_{2} \\rightarrow \\mathrm{O}_{2}^{2-}$$"}, {"identifier": "B", "content": "$$\\mathrm{N}_{2} \\rightarrow \\mathrm{N}_{2}^{+}$$"}, {"identifier": "C", "content": "$$\\mathrm{NO} \\rightarrow \\mathrm{NO}^{+}$$"}, {"identifier": "D", "content": "$$\\mathrm{O}_{2} \\rightarrow \\mathrm{O}_{2}^{+}$$"}]
|
["C"]
| null |
Let's analyze each option:
<br/><br/>
Option A: $\mathrm{O}_{2} \rightarrow \mathrm{O}_{2}^{2-}$<br/><br/>
The bond order of $\mathrm{O}_{2}$ is 2 and it is paramagnetic. When two electrons are added to form $\mathrm{O}_{2}^{2-}$, the bond order becomes 1 (decreases) and it becomes diamagnetic. This option does not meet the required conditions.
<br/><br/>
Option B: $\mathrm{N}_{2} \rightarrow \mathrm{N}_{2}^{+}$<br/><br/>
The bond order of $\mathrm{N}_{2}$ is 3 and it is diamagnetic. When one electron is removed to form $\mathrm{N}_{2}^{+}$, the bond order becomes 2.5 (decreases) and it remains diamagnetic. This option does not meet the required conditions.
<br/><br/>
Option C: $\mathrm{NO} \rightarrow \mathrm{NO}^{+}$<br/><br/>
The bond order of $\mathrm{NO}$ is 2.5 and it is paramagnetic. When one electron is removed to form $\mathrm{NO}^{+}$, the bond order becomes 3 (increases) and it becomes diamagnetic. This option meets the required conditions.
<br/><br/>
Option D: $\mathrm{O}_{2} \rightarrow \mathrm{O}_{2}^{+}$<br/><br/>
The bond order of $\mathrm{O}_{2}$ is 2 and it is paramagnetic. When one electron is removed to form $\mathrm{O}_{2}^{+}$, the bond order becomes 2.5 (increases) and it remains paramagnetic. This option does not meet the required conditions.
<br/><br/>
Therefore, the correct answer is $\mathrm{NO} \rightarrow \mathrm{NO}^{+}$.
|
mcq
|
jee-main-2023-online-13th-april-morning-shift
| 979
|
1lgrkbmlp
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
<p>The bond order and magnetic property of acetylide ion are same as that of</p>
|
[{"identifier": "A", "content": "$$\\mathrm{O}_{2}^{+}$$"}, {"identifier": "B", "content": "$$\\mathrm{O}_{2}^{-}$$"}, {"identifier": "C", "content": "$$\\mathrm{N}_{2}^{+}$$"}, {"identifier": "D", "content": "$$\\mathrm{NO}^{+}$$"}]
|
["D"]
| null |
<p>The acetylide ion has the formula $$\mathrm{C}_{2}^{2-}$$. To determine its bond order, we need to first write the molecular orbital (MO) diagram for this ion. </p>
<p>The MO diagram for $$\mathrm{C}_{2}^{2-}$$ is:</p>
<p>$$\mathrm{σ_{1s}^{2}}\mathrm{σ_{1s}^{2}}\mathrm{σ_{2s}^{2}}\mathrm{σ_{2s}^{2}}\mathrm{π_{2p}^{4}}$$</p>
<p>The bond order is given by the difference between the number of bonding electrons and the number of antibonding electrons, divided by 2. In this case, there are 4 bonding electrons and 2 antibonding electrons, so the bond order is:</p>
<p>$$\frac{4-2}{2}=1$$</p>
<p>Therefore, the bond order of $$\mathrm{C}_{2}^{2-}$$ is 1.</p>
<p>Now, we need to determine the magnetic property of $$\mathrm{C}_{2}^{2-}$$. Since the bond order is 1, we know that the ion has a single unpaired electron. Therefore, it is paramagnetic.</p>
<p>Among the given options, the only ion with a bond order of 1 and paramagnetic property is $$\mathrm{NO}^{+}$$.</p>
|
mcq
|
jee-main-2023-online-12th-april-morning-shift
| 980
|
lsaowhvy
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
Given below are two statements :<br/><br/>
<b>Statement (I)</b> : A $\pi$ bonding MO has lower electron density above and below the inter-nuclear axis.
<br/><br/>
<b>Statement (II)</b> : The $\pi^*$ antibonding MO has a node between the nuclei.<br/><br/>
In the light of the above statements, choose the <b>most appropriate</b> answer from the options given below :
|
[{"identifier": "A", "content": "Both Statement I and Statement II are true"}, {"identifier": "B", "content": "Both Statement I and Statement II are false"}, {"identifier": "C", "content": "Statement I is true but Statement II is false"}, {"identifier": "D", "content": "Statement I is false but Statement II is true"}]
|
["D"]
| null |
<p>Let's analyze both statements:</p>
<p><b>Statement (I)</b>: "A $\pi$ bonding MO has lower electron density above and below the inter-nuclear axis."</p>
<p>This statement is false. In molecular orbital (MO) theory, a pi bond ($\pi$ bond) is formed by the sideways overlap of p-orbitals from two adjacent atoms. The characteristic electron density of a $\pi$ bonding molecular orbital is concentrated above and below the inter-nuclear axis, not lower. These regions of electron density are where the p-orbitals overlap and electrons are likely to be found. This concentration of electron density above and below the inter-nuclear axis is what allows the $\pi$ bond to provide additional stability to the molecule, alongside any sigma ($\sigma$) bonds that may be present.</p>
<p><b>Statement (II)</b>: "The $\pi^*$ antibonding MO has a node between the nuclei."</p>
<p>This statement is true. In a $\pi^*$ (pi-star) antibonding molecular orbital, there is indeed a nodal plane between the nuclei where the probability of finding electrons is essentially zero. This node arises from the out-of-phase combination of the p-orbitals from adjacent atoms, which results in destructive interference and a region of zero electron density in the bonding region. The presence of this nodal plane is what makes the $\pi^*$ orbital antibonding—the electrons in this orbital actually serve to destabilize the bond between the two atoms.</p>
<p>Therefore, the correct answer is:</p>
<p>Option D: Statement I is false but Statement II is true.</p>
|
mcq
|
jee-main-2024-online-1st-february-evening-shift
| 981
|
jaoe38c1lsc6u3er
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
<p>Sum of bond order of CO and NO$$^+$$ is ________.</p>
|
[]
| null |
6
|
<p>$$\begin{array}{lcl}
\mathrm{CO} \Rightarrow & \overline{\mathrm{C}} \equiv \stackrel{+}{\mathrm{O}} & : \mathrm{BO}=3 \\
\mathrm{NO}^{+} \Rightarrow & \mathrm{N} \equiv \mathrm{O}^{+} & : \mathrm{BO}=3
\end{array}$$</p>
|
integer
|
jee-main-2024-online-27th-january-morning-shift
| 982
|
jaoe38c1lse7imu1
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
<p>The linear combination of atomic orbitals to form molecular orbitals takes place only when the combining atomic orbitals</p>
<p>A. have the same energy</p>
<p>B. have the minimum overlap</p>
<p>C. have same symmetry about the molecular axis</p>
<p>D. have different symmetry about the molecular axis</p>
<p>Choose the most appropriate from the options given below:</p>
|
[{"identifier": "A", "content": "B, C, D only"}, {"identifier": "B", "content": "A, B, C only"}, {"identifier": "C", "content": "B and D only"}, {"identifier": "D", "content": "A and C only"}]
|
["D"]
| null |
<p>* Molecular orbital should have maximum overlap</p>
<p>* Symmetry about the molecular axis should be similar</p>
|
mcq
|
jee-main-2024-online-31st-january-morning-shift
| 983
|
jaoe38c1lsfk4s5c
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
<p>The number of species from the following which are paramagnetic and with bond order equal to one is _________.</p>
<p>$$\mathrm{H}_2, \mathrm{He}_2^{+}, \mathrm{O}_2^{+}, \mathrm{N}_2^{2-}, \mathrm{O}_2^{2-}, \mathrm{F}_2, \mathrm{Ne}_2^{+}, \mathrm{B}_2$$</p>
|
[]
| null |
1
|
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;}
.tg .tg-baqh{text-align:center;vertical-align:top}
.tg .tg-amwm{font-weight:bold;text-align:center;vertical-align:top}
</style>
<table class="tg" style="undefined;table-layout: fixed; width: 551px">
<colgroup>
<col style="width: 158px">
<col style="width: 210px">
<col style="width: 183px">
</colgroup>
<thead>
<tr>
<th class="tg-amwm">Sol.</th>
<th class="tg-amwm">Magnetic behaviour</th>
<th class="tg-amwm">Bond order</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-baqh">$$\mathrm{H}_2$$</td>
<td class="tg-baqh">Diamagnetic</td>
<td class="tg-baqh">$$1$$</td>
</tr>
<tr>
<td class="tg-baqh">$$\mathrm{He_2^+}$$</td>
<td class="tg-baqh">Paramagnetic</td>
<td class="tg-baqh">$$0.5$$</td>
</tr>
<tr>
<td class="tg-baqh">$$\mathrm{O_2^+}$$</td>
<td class="tg-baqh">Paramagnetic</td>
<td class="tg-baqh">$$2.5$$</td>
</tr>
<tr>
<td class="tg-baqh">$$\mathrm{N_2^{2-}}$$</td>
<td class="tg-baqh">Paramagnetic</td>
<td class="tg-baqh">$$2$$</td>
</tr>
<tr>
<td class="tg-baqh">$$\mathrm{O_2^{2-}}$$</td>
<td class="tg-baqh">Diamagnetic</td>
<td class="tg-baqh">$$1$$</td>
</tr>
<tr>
<td class="tg-baqh">$$\mathrm{F_2}$$</td>
<td class="tg-baqh">Diamagnetic</td>
<td class="tg-baqh">$$1$$</td>
</tr>
<tr>
<td class="tg-baqh">$$\mathrm{Ne_2^+}$$</td>
<td class="tg-baqh">Paramagnetic</td>
<td class="tg-baqh">$$0.5$$</td>
</tr>
<tr>
<td class="tg-baqh">$$\mathrm{B_2}$$</td>
<td class="tg-baqh">Paramagnetic</td>
<td class="tg-baqh">$$1$$</td>
</tr>
</tbody>
</table></p>
|
integer
|
jee-main-2024-online-29th-january-morning-shift
| 984
|
jaoe38c1lsfplid6
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
<p>The total number of anti bonding molecular orbitals, formed from $$2 s$$ and $$2 p$$ atomic orbitals in a diatomic molecule is _______.</p>
|
[]
| null |
4
|
<p>Antibonding molecular orbital from $$2 \mathrm{~s}=1$$</p>
<p>Antibonding molecular orbital from $$2 p=3$$</p>
<p>Total $$=4$$</p>
|
integer
|
jee-main-2024-online-29th-january-evening-shift
| 985
|
1lsgyk5tt
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
<p>The total number of molecular orbitals formed from $$2 \mathrm{s}$$ and $$2 \mathrm{p}$$ atomic orbitals of a diatomic molecule is __________.</p>
|
[]
| null |
8
|
<p>Two molecular orbitals $$\sigma 2 \mathrm{s}$$ and $$\sigma * 2 \mathrm{s}$$.</p>
<p>Six molecular orbitals $$\sigma 2 \mathrm{p}_z$$ and $$\sigma * 2 \mathrm{p}_{\mathrm{z}}$$.</p>
<p>$$\pi 2 \mathrm{p}_{\mathrm{x}}, \pi 2 \mathrm{p}_{\mathrm{y}}$$ and $$\pi * 2 \mathrm{p}_{\mathrm{x}}, \pi^* 2 \mathrm{p}_{\mathrm{y}}$$</p>
|
integer
|
jee-main-2024-online-30th-january-morning-shift
| 986
|
luy1mwmf
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
<p>Total number of electrons present in $$\left(\pi^*\right)$$ molecular orbitals of $$\mathrm{O}_2, \mathrm{O}_2^{+}$$ and $$\mathrm{O}_2^{-}$$ is ________.</p>
|
[]
| null |
6
|
<p>$$\begin{aligned}
& \mathrm{O}_2(16 \mathrm{e}):\left(\sigma_{1 \mathrm{~s}}\right)^2\left(\sigma_{1 \mathrm{~s}}^*\right)^2\left(\sigma_{2 \mathrm{~s}}\right)^2\left(\sigma_{2 \mathrm{~s}}^*\right)^2 \\
& \left(\sigma_{2 \mathrm{p}}\right)^2\left[\left(\pi_{2 \mathrm{p}}\right)^2=\left(\pi_{2 \mathrm{p}}\right)^2\right],\left[\left(\pi_{2 \mathrm{p}}^*\right)^1=\left(\pi_{2 \mathrm{p}}^*\right)^1\right]
\end{aligned}$$</p>
<p>Number of $$\mathrm{e}^{-}$$ present in $$\left(\pi^*\right)$$ of $$\mathrm{O}_2=2$$</p>
<p>Number of $$\mathrm{e}^{-}$$ present in $$\left(\pi^*\right)$$ of $$\mathrm{O}_2^{+}=1$$</p>
Number of $$\mathrm{e}^{-}$$ present in $$\left(\pi^*\right)$$ of $$\mathrm{O}_2^{-}=3$$</p>
<p>So total $$\mathrm{e}^{-}$$ in $$\left(\pi^*\right)=2+1+3=6$$</p>
|
integer
|
jee-main-2024-online-9th-april-evening-shift
| 987
|
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