molecular-orbital-theory-and-hydrogen-bonding MCQs for UPSC Prelims

27 practice questions on molecular-orbital-theory-and-hydrogen-bonding from the Chemical Bonding and Molecular Structure section of the UPSC Prelims syllabus. 27 come with a written explanation. Try the sample set below - the answer stays hidden until you ask for it.

6 Easy 15 Medium 6 Hard

Sample questions

Q1
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.

Q2
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.

Q3
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.

Q4
medium

Why does water (H2O) have a higher boiling point than hydrogen sulfide (H2S)?

  1. A H2S has stronger van der Waals forces than H2O, raising its boiling point.
  2. B The larger molecular weight of H2S increases its boiling point over H2O.
  3. C Water forms stronger hydrogen bonds than H2S, which lacks hydrogen bonding.
  4. D Water's covalent bonds are weaker than those in H2S, requiring more energy to break.
Show answer and explanation

Correct answer: C - Water forms stronger hydrogen bonds than H2S, which lacks hydrogen bonding.

Water (H2O) has a higher boiling point because it forms strong hydrogen bonds due to the presence of highly electronegative oxygen, which is not the case for H2S. The misconception that H2S has stronger forces due to mass is incorrect because the type of intermolecular force (hydrogen bonding vs van der Waals forces) is more significant than mass in this context.

Q5
medium

The bond order of O2+ is:

  1. A 2
  2. B 2.5
  3. C 3
  4. D 1.5
Show answer and explanation

Correct answer: B - 2.5

For O2+, the bond order is calculated as \( BO = \frac{(N_b - N_a)}{2} \). With 10 bonding and 7 antibonding electrons, the bond order is \( 2.5 \). Other options are incorrect calculations of this bond order.

Q6
hard

In o-nitrophenol and p-nitrophenol, which type of hydrogen bonding is predominant, respectively?

  1. A Intramolecular in both o- and p-nitrophenol
  2. B Intramolecular in o-nitrophenol and intermolecular in p-nitrophenol
  3. C Intermolecular in both o- and p-nitrophenol
  4. D Intermolecular in o-nitrophenol and intramolecular in p-nitrophenol
Show answer and explanation

Correct answer: B - Intramolecular in o-nitrophenol and intermolecular in p-nitrophenol

In o-nitrophenol, the proximity of the nitro and hydroxyl groups allows intramolecular hydrogen bonding, while in p-nitrophenol, hydrogen bonding is more likely intermolecular. Students might confuse the spatial arrangement and bonding types.

Q7
easy

How does an antibonding molecular orbital differ from the bonding molecular orbital formed from the same two atomic orbitals?

  1. A It lies lower in energy and concentrates electron density between the nuclei
  2. B It lies higher in energy and has a node between the nuclei, so electrons in it weaken the bond
  3. C It has exactly the same energy as the bonding orbital but takes electrons of opposite spin
  4. D It can hold four electrons, twice as many as the bonding orbital
Show answer and explanation

Correct answer: B - It lies higher in energy and has a node between the nuclei, so electrons in it weaken the bond

Subtractive combination of two atomic orbitals gives the antibonding molecular orbital: it lies higher in energy than either parent orbital and carries a node between the nuclei, so electrons placed in it pull the atoms apart. The bonding orbital is the lower-energy additive combination that builds electron density up between the nuclei. The two are not equal in energy, and every molecular orbital holds a maximum of two electrons.

Q8
hard

Assertion (A): $\text{HF}$ has a lower boiling point than $\text{H}_2\text{O}$. Reason (R): An aqueous solution of $\text{HF}$ behaves as a weak acid, whereas pure water is neutral.

  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: B - Both A and R are true, but R does not explain A.

Both statements are true. $\text{HF}$ boils at about $293\ \text{K}$ and water at $373\ \text{K}$: a water molecule has two hydrogen atoms and two lone pairs and so builds a three-dimensional hydrogen bonded network, whereas $\text{HF}$ has a single hydrogen and can only form chains. R is a true statement about proton transfer in solution, a separate property that says nothing about how tightly the molecules hold one another, so it does not explain A.

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