Chemical Bonding and Molecular Structure MCQs for UPSC Prelims

108 practice questions covering Chemical Bonding and Molecular Structure, organised into 1 topics. 108 questions include a written explanation. Pick a topic below, or try the samples first.

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Sample questions

Q1
medium

In the ozone molecule the central oxygen atom forms one single bond and one double bond and carries one lone pair. What is the formal charge on that central atom?

  1. A $-1$
  2. B $+1$
  3. C $0$
  4. D $+2$
Show answer and explanation

Correct answer: B - $+1$

Formal charge is the number of valence electrons minus the non-bonding electrons minus half the bonding electrons. The central oxygen has $6$ valence electrons, $2$ non-bonding electrons from its single lone pair, and $6$ bonding electrons from one single and one double bond, so the formal charge is $6 - 2 - 3 = +1$. Counting the lone pair as one electron instead of two gives $+2$, while $-1$ belongs to the singly bonded terminal oxygen.

Q2
hard

Assertion (A): $\text{NF}_3$ has a larger dipole moment than $\text{NH}_3$. Reason (R): In $\text{NH}_3$ the bond dipoles and the dipole of the nitrogen lone pair point the same way, while in $\text{NF}_3$ they point in opposing directions.

  1. A Both A and R true, R explains A
  2. B Both A and R true, R does not explain A
  3. C A true, R false
  4. D A false, R true
Show answer and explanation

Correct answer: D - A false, R true

A is false. The dipole moment of $\text{NH}_3$ is about $1.47\ \text{D}$ against only about $0.24\ \text{D}$ for $\text{NF}_3$, so ammonia is by far the more polar. R is true and explains why: in $\text{NF}_3$ the $\text{N-F}$ dipoles pull electron density away from nitrogen and oppose the lone-pair dipole, so the two contributions largely cancel.

Q3
medium

How does the electron cloud of a sigma bond differ from that of a pi bond?

  1. A The pi cloud is symmetrical about the internuclear axis, while the sigma cloud has a nodal plane containing that axis
  2. B The sigma cloud is symmetrical about the internuclear axis, while the pi cloud lies above and below that axis with a nodal plane containing it
  3. C Both clouds have a nodal plane containing the internuclear axis, but the sigma cloud lies closer to the nuclei
  4. D Both clouds are symmetrical about the internuclear axis, but the pi cloud is the larger of the two
Show answer and explanation

Correct answer: B - The sigma cloud is symmetrical about the internuclear axis, while the pi cloud lies above and below that axis with a nodal plane containing it

A sigma bond is formed by head-on overlap, so its electron cloud is symmetrical about the line joining the nuclei and can be rotated about that line without change. A pi bond comes from sideways overlap of parallel orbitals, so its cloud sits in two regions above and below a nodal plane that contains the internuclear axis. Swapping these descriptions, or giving both bonds the same symmetry, misstates how the orbitals overlap.

Q4
medium

Assertion (A): In the oxygen molecule the $\sigma 2p_z$ molecular orbital lies higher in energy than the $\pi 2p_x$ and $\pi 2p_y$ orbitals. Reason (R): In molecules up to nitrogen, mixing of the $2s$ and $2p$ orbitals raises the $\sigma 2p_z$ orbital above the $\pi 2p$ pair.

  1. A Both A and R are true, and R is the correct explanation of A
  2. B Both A and R are true, but R is not the correct explanation of A
  3. C A is true, but R is false
  4. D A is false, but R is true
Show answer and explanation

Correct answer: D - A is false, but R is true

The reason states the $s$-$p$ mixing correctly: for $\text{B}_2$, $\text{C}_2$ and $\text{N}_2$ the mixing pushes $\sigma 2p_z$ above the degenerate $\pi 2p$ pair. Oxygen lies past that point, because its $2s$ and $2p$ levels are far enough apart for the mixing to be negligible, so in $\text{O}_2$ the $\sigma 2p_z$ orbital lies below the $\pi 2p$ orbitals. The assertion is therefore false while the reason is true.

Q5
hard

Assertion (A): The bond order of $\text{N}_2$ is greater than that of $\text{N}_2^+$. Reason (R): $\text{N}_2^+$ is formed by removing an electron from an antibonding molecular orbital of $\text{N}_2$.

  1. A Both A and R are true, and R explains A.
  2. B Both A and R are true, but R does not explain A.
  3. C A is true, but R is false.
  4. D A is false, but R is true.
Show answer and explanation

Correct answer: C - A is true, but R is false.

A is true: $\text{N}_2$ has a bond order of $3$ while $\text{N}_2^+$ has $2.5$. R is false. The highest occupied orbital of $\text{N}_2$ is the bonding $\sigma 2p_z$, so ionisation removes a bonding electron, not an antibonding one. Losing a bonding electron is exactly what lowers the bond order by one half; removing an antibonding electron would have raised it.

Q6
easy

Ice floats on water because:

  1. A Water molecules in ice repel each other, increasing volume.
  2. B Ice has ionic bonds that make it lighter than water.
  3. C The molecular weight of water decreases upon freezing.
  4. D Ice is less dense than water due to hydrogen bonding forming a lattice.
Show answer and explanation

Correct answer: D - Ice is less dense than water due to hydrogen bonding forming a lattice.

Ice is less dense than water because its hydrogen bonds form an open hexagonal lattice that occupies more space, making it less dense. Ice does not have ionic bonds, and freezing does not change the molecular weight of water. Repulsion is not a factor here.

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