Molecular Orbital Theory and Hydrogen Bonding — NEET Chemistry MCQs with Solutions
Free NEET Chemistry Molecular Orbital Theory and Hydrogen Bonding MCQs with step-by-step solutions (75 questions). Part of Chemical Bonding. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Molecular Orbital Theory and Hydrogen Bonding · easy
Which of the following molecule(s) show paramagnetic behaviour?
A. O₂ B. N₂ C. F₂ D. S₂ E. Cl₂
Choose the correct answer from the options given below:
A. A and C only
B. A and E only
C. B only
D. A and D only ✓ Correct
Solution: O₂ (16 e⁻) has two unpaired electrons in the degenerate $\pi^*2p$ orbitals, so it is paramagnetic. S₂ is the third-period analogue of O₂ with the same valence configuration, so it is also paramagnetic. N₂, F₂ and Cl₂ have all electrons paired (diamagnetic). Hence A and D only.
Q2 — Molecular Orbital Theory and Hydrogen Bonding · medium
The correct statement(s) about hydrogen bonding is/are:
A. Hydrogen bonding exists when H is covalently bonded to a highly electronegative atom.
B. Intermolecular H-bonding is present in o-nitrophenol.
C. Intramolecular H-bonding is present in HF.
D. The magnitude of H-bonding depends on the physical state of the compound.
E. H-bonding has a powerful effect on the structure and properties of compounds.
Choose the correct answer from the options given below:
A. A, D, E only ✓ Correct
B. A, B, D only
C. A, B, C only
D. A only
Solution: A, D and E are correct statements about hydrogen bonding. B is wrong: o-nitrophenol shows INTRAmolecular H-bonding (the OH and NO₂ groups are adjacent). C is wrong: HF shows INTERmolecular H-bonding (zig-zag chains); a single HF molecule cannot H-bond within itself.
Q3 — Molecular Orbital Theory and Hydrogen Bonding · easy
When $\psi_A$ and $\psi_B$ are the wave functions of atomic orbitals, then the antibonding molecular orbital $\sigma^*$ is represented by
A. $\psi_A + \psi_B$
B. $\psi_A - \psi_B$ ✓ Correct
C. $\psi_A + 2\psi_B$
D. $\psi_A - 2\psi_B$
Solution: By the LCAO method, the constructive combination $\psi_A + \psi_B$ gives the bonding MO ($\sigma$), while the destructive combination $\psi_A - \psi_B$ gives the antibonding MO ($\sigma^*$).
Q4 — Molecular Orbital Theory and Hydrogen Bonding · easy
The linear combination of atomic orbitals to form molecular orbitals takes place only when the combining atomic orbitals
(A) have the same energy
(B) have the minimum overlap
(C) have the same symmetry about the molecular axis
(D) have different symmetry about the molecular axis
Choose the most appropriate answer from the options given below:
A. (A), (B), (C) only
B. (B), (C), (D) only
C. (A) and (C) only ✓ Correct
D. (B) and (D) only
Solution: For effective LCAO the combining atomic orbitals must have the same (or nearly the same) energy and the same symmetry about the molecular axis, and they must overlap to the MAXIMUM extent. So (A) and (C) are correct; (B) and (D) are wrong.
Q5 — Molecular Orbital Theory and Hydrogen Bonding · medium
The bond order and magnetic property of the acetylide ion (C₂²⁻) are the same as those of
A. N₂⁺
B. NO⁺ ✓ Correct
C. O₂⁻
D. O₂⁺
Solution: C₂²⁻ has 14 electrons: bond order = (10 − 4)/2 = 3, diamagnetic. NO⁺ (14 e⁻) also has bond order 3 and is diamagnetic. N₂⁺ (BO 2.5, paramagnetic), O₂⁻ (BO 1.5, paramagnetic) and O₂⁺ (BO 2.5, paramagnetic) do not match.
Q6 — Molecular Orbital Theory and Hydrogen Bonding · medium
In which of the following processes does the bond order increase and the paramagnetic character change to diamagnetic?
A. O₂ → O₂⁺
B. NO → NO⁺ ✓ Correct
C. N₂ → N₂⁺
D. O₂ → O₂²⁻
Solution: NO (15 e⁻): BO 2.5, one unpaired $\pi^*$ electron (paramagnetic). Removing it gives NO⁺ (14 e⁻): BO 3, diamagnetic — bond order rises AND paramagnetic → diamagnetic. O₂ → O₂⁺ raises BO (2 → 2.5) but O₂⁺ is still paramagnetic; N₂ → N₂⁺ lowers BO (3 → 2.5); O₂ → O₂²⁻ lowers BO (2 → 1).
Q7 — Molecular Orbital Theory and Hydrogen Bonding · medium
Consider the following statements:
(A) The NF₃ molecule has a trigonal planar structure.
(B) The bond length of N₂ is shorter than that of O₂.
(C) Isoelectronic molecules or ions have identical bond order.
(D) The dipole moment of H₂S is higher than that of the water molecule.
Choose the correct answer from the options given below:
A. (C) and (D) are correct
B. (B) and (C) are correct ✓ Correct
C. (A) and (B) are correct
D. (A) and (D) are correct
Solution: (A) is wrong: NF₃ is pyramidal (one lone pair on N). (B) is correct: N₂ has bond order 3 vs 2 for O₂, so N₂ has the shorter bond. (C) is correct: isoelectronic species (e.g. N₂, CO, NO⁺ — all 14 e⁻) have the same bond order (3). (D) is wrong: water (μ ≈ 1.85 D) has a higher dipole moment than H₂S (μ ≈ 0.95 D) because O is more electronegative and the H–O–H geometry gives a larger resultant.
Q8 — Molecular Orbital Theory and Hydrogen Bonding · medium
What is the number of unpaired electron(s) in the highest occupied molecular orbital of the following species: N₂; N₂⁻; O₂; O₂⁻ ?
A. 0, 1, 2, 1 ✓ Correct
B. 2, 1, 2, 1
C. 0, 1, 0, 1
D. 2, 1, 0, 1
Solution: N₂ (14 e⁻): HOMO is the filled $\sigma 2p_z$ → 0 unpaired. N₂⁻ (15 e⁻): extra electron enters $\pi^*2p$ → 1 unpaired. O₂ (16 e⁻): HOMO is $\pi^*2p$ with one electron in each degenerate orbital → 2 unpaired. O₂⁻ (17 e⁻): $\pi^*2p^3$ → 1 unpaired. Hence 0, 1, 2, 1.
Q9 — Molecular Orbital Theory and Hydrogen Bonding · medium
The magnetic behaviours of Li₂O, Na₂O₂ and KO₂, respectively, are
A. diamagnetic, paramagnetic and diamagnetic
B. diamagnetic, diamagnetic and paramagnetic ✓ Correct
C. paramagnetic, paramagnetic and diamagnetic
D. paramagnetic, diamagnetic and paramagnetic
Solution: Li₂O contains the oxide ion O²⁻ (all paired) → diamagnetic. Na₂O₂ contains the peroxide ion O₂²⁻ (18 e⁻, $\pi^*2p$ completely filled) → diamagnetic. KO₂ contains the superoxide ion O₂⁻ (17 e⁻, one unpaired $\pi^*$ electron) → paramagnetic.
Q10 — Molecular Orbital Theory and Hydrogen Bonding · easy
According to MO theory, the bond orders of O₂²⁻, CO and NO⁺, respectively, are
A. 2, 3 and 3
B. 1, 3 and 3 ✓ Correct
C. 1, 3 and 2
D. 1, 2 and 3
Solution: O₂²⁻ (18 e⁻): BO = (10 − 8)/2 = 1. CO (14 e⁻): BO = (10 − 4)/2 = 3. NO⁺ (14 e⁻, isoelectronic with CO): BO = 3. Hence 1, 3 and 3.
Q11 — Molecular Orbital Theory and Hydrogen Bonding · medium
The correct order of bond orders of C₂²⁻, N₂²⁻ and O₂²⁻ is, respectively,
A. C₂²⁻ < N₂²⁻ < O₂²⁻
B. O₂²⁻ < N₂²⁻ < C₂²⁻ ✓ Correct
C. C₂²⁻ < O₂²⁻ < N₂²⁻
D. N₂²⁻ < C₂²⁻ < O₂²⁻
Solution: C₂²⁻ has 14 e⁻: BO = (10 − 4)/2 = 3. N₂²⁻ has 16 e⁻: BO = (10 − 6)/2 = 2. O₂²⁻ has 18 e⁻: BO = (10 − 8)/2 = 1. So O₂²⁻ (1) < N₂²⁻ (2) < C₂²⁻ (3).
Q12 — Molecular Orbital Theory and Hydrogen Bonding · medium
Bonding in which of the following diatomic molecule(s) becomes stronger, on the basis of MO theory, by removal of an electron?
A. NO B. N₂ C. O₂ D. C₂ E. B₂
Choose the most appropriate answer from the options given below:
A. A, B, C only
B. B, C, E only
C. A, C only ✓ Correct
D. D only
Solution: Removing an electron strengthens the bond only if it comes from an ANTIBONDING orbital. NO (BO 2.5 → NO⁺ BO 3, electron removed from $\pi^*$) and O₂ (BO 2 → O₂⁺ BO 2.5, from $\pi^*$) get stronger. N₂ (3 → 2.5), C₂ (2 → 1.5) and B₂ (1 → 0.5) lose a BONDING electron and get weaker. Hence A and C only.
Q13 — Molecular Orbital Theory and Hydrogen Bonding · easy
The bond order and magnetic behaviour of the O₂⁻ ion are, respectively,
A. 1.5 and diamagnetic
B. 1.5 and paramagnetic ✓ Correct
C. 2 and diamagnetic
D. 1 and paramagnetic
Solution: O₂⁻ (superoxide) has 17 electrons: BO = (10 − 7)/2 = 1.5, with three electrons in the degenerate $\pi^*2p$ orbitals leaving one unpaired → paramagnetic.
Q14 — Molecular Orbital Theory and Hydrogen Bonding · easy
According to molecular orbital theory, the species among the following that does NOT exist is
A. Be₂ ✓ Correct
B. He₂⁺
C. O₂²⁻
D. He₂²⁺
Solution: Be₂ (8 e⁻): $\sigma 1s^2\,\sigma^*1s^2\,\sigma 2s^2\,\sigma^*2s^2$ → BO = (4 − 4)/2 = 0, so it does not exist. He₂⁺ has BO 0.5, He₂²⁺ has BO 1 and O₂²⁻ has BO 1 — all non-zero.
Q15 — Molecular Orbital Theory and Hydrogen Bonding · medium
Of the species NO, NO⁺, NO²⁺ and NO⁻, the one with the MINIMUM bond strength is
A. NO⁺
B. NO
C. NO²⁺
D. NO⁻ ✓ Correct
Solution: Bond orders: NO (15 e⁻) = 2.5; NO⁺ (14 e⁻) = 3; NO²⁺ (13 e⁻) = 2.5 (an electron is removed from the bonding $\sigma 2p_z$); NO⁻ (16 e⁻) = 2 (extra electron in $\pi^*$). Lowest bond order → weakest bond → NO⁻.
Q16 — Molecular Orbital Theory and Hydrogen Bonding · easy
The bond order and the magnetic characteristic of CN⁻ are
A. 3, diamagnetic ✓ Correct
B. 3, paramagnetic
C. 2½, paramagnetic
D. 2½, diamagnetic
Solution: CN⁻ has 14 electrons (isoelectronic with N₂ and CO): BO = (10 − 4)/2 = 3 with all electrons paired → diamagnetic.
Q17 — Molecular Orbital Theory and Hydrogen Bonding · medium
If the magnetic moment of a dioxygen species is 1.73 B.M., it may be
A. O₂, O₂⁻ or O₂⁺
B. O₂⁺ or O₂⁻ ✓ Correct
C. O₂ or O₂⁺
D. O₂ or O₂⁻
Solution: μ = 1.73 B.M. corresponds to exactly one unpaired electron ($\sqrt{n(n+2)} = \sqrt{3}$). O₂⁺ (15 e⁻, $\pi^*2p^1$) and O₂⁻ (17 e⁻, $\pi^*2p^3$) each have one unpaired electron. O₂ itself has two unpaired electrons (μ ≈ 2.83 B.M.), so any option containing O₂ is ruled out.
Q18 — Molecular Orbital Theory and Hydrogen Bonding · medium
Among the following molecules/ions — C₂²⁻, N₂²⁻, O₂²⁻, O₂ — which one is diamagnetic and has the shortest bond length?
A. C₂²⁻ ✓ Correct
B. O₂
C. N₂²⁻
D. O₂²⁻
Solution: C₂²⁻ (14 e⁻): BO 3, all paired → diamagnetic, and the highest bond order gives the shortest bond. N₂²⁻ (16 e⁻): BO 2 but has two unpaired $\pi^*$ electrons (paramagnetic). O₂ (BO 2) is paramagnetic. O₂²⁻ (BO 1) is diamagnetic but has the longest bond.
Q19 — Molecular Orbital Theory and Hydrogen Bonding · medium
Among the following, the molecule expected to be stabilized by anion formation is:
C₂, O₂, NO, F₂
A. NO
B. O₂
C. F₂
D. C₂ ✓ Correct
Solution: Adding an electron stabilizes a molecule only if it enters a BONDING orbital. In C₂ (12 e⁻) the next available orbital is the bonding $\sigma 2p_z$, so C₂⁻ has BO 2.5 > 2. For O₂ (2 → 1.5), NO (2.5 → 2) and F₂ (1 → 0.5) the added electron enters an antibonding $\pi^*$/$\sigma^*$ orbital and lowers the bond order.
Q20 — Molecular Orbital Theory and Hydrogen Bonding · easy
Among the following species, the diamagnetic molecule is
A. B₂
B. NO
C. O₂
D. CO ✓ Correct
Solution: CO (14 e⁻) has all electrons paired (BO 3) → diamagnetic. B₂ has two unpaired $\pi 2p$ electrons, O₂ has two unpaired $\pi^*2p$ electrons, and NO has one unpaired $\pi^*$ electron — all paramagnetic.
Q21 — Molecular Orbital Theory and Hydrogen Bonding · easy
During the change of O₂ to O₂⁻, the incoming electron goes to the orbital
A. $\sigma^*2p_z$
B. $\pi 2p_y$
C. $\pi^*2p_x$ ✓ Correct
D. $\pi 2p_x$
Solution: O₂ (16 e⁻) ends its filling with $\pi^*2p_x^1\,\pi^*2p_y^1$. The 17th electron of O₂⁻ pairs up in one of these half-filled antibonding orbitals, i.e. it enters $\pi^*2p_x$ (the $\sigma^*2p_z$ lies higher in energy; the bonding $\pi 2p$ orbitals are already full).
Q22 — Molecular Orbital Theory and Hydrogen Bonding · medium
According to molecular orbital theory, which of the following is true with respect to Li₂⁺ and Li₂⁻ ?
A. Both are stable ✓ Correct
B. Both are unstable
C. Li₂⁺ is unstable and Li₂⁻ is stable
D. Li₂⁺ is stable and Li₂⁻ is unstable
Solution: Li₂⁺ (5 e⁻): BO = (3 − 2)/2 = 0.5. Li₂⁻ (7 e⁻): BO = (4 − 3)/2 = 0.5. Both have a positive (non-zero) bond order, so both exist and are stable (though less stable than Li₂, BO 1).
Q23 — Molecular Orbital Theory and Hydrogen Bonding · easy
In which of the following processes has the bond order increased and the paramagnetic character changed to diamagnetic?
A. O₂ → O₂⁺
B. O₂ → O₂²⁻
C. N₂ → N₂⁺
D. NO → NO⁺ ✓ Correct
Solution: NO (BO 2.5, one unpaired $\pi^*$ electron, paramagnetic) → NO⁺ (BO 3, diamagnetic): bond order increases and the species becomes diamagnetic. O₂ → O₂⁺ increases BO but stays paramagnetic; O₂ → O₂²⁻ decreases BO (2 → 1); N₂ → N₂⁺ decreases BO (3 → 2.5).
Q24 — Molecular Orbital Theory and Hydrogen Bonding · hard
Two π bonds and half a σ bond are present in
A. O₂
B. O₂⁺
C. N₂
D. N₂⁺ ✓ Correct
Solution: N₂⁺ (13 e⁻): ... $\pi 2p_x^2\,\pi 2p_y^2\,\sigma 2p_z^1$ — the two full $\pi$ bonding orbitals give two π bonds and the singly occupied $\sigma 2p_z$ gives half a σ bond (total BO 2.5). O₂⁺ also has BO 2.5 but as one full σ bond plus 1.5 π bonds; N₂ has one σ and two π; O₂ has one σ and one π (net).
Q25 — Molecular Orbital Theory and Hydrogen Bonding · easy
According to molecular orbital theory, which of the following will NOT be a viable molecule?
A. He₂²⁺
B. He₂⁺
C. H₂⁻
D. H₂²⁻ ✓ Correct
Solution: H₂²⁻ (4 e⁻): $\sigma 1s^2\,\sigma^*1s^2$ → BO = (2 − 2)/2 = 0, so it is not viable. He₂²⁺ has BO 1, He₂⁺ has BO 0.5 and H₂⁻ has BO 0.5 — all non-zero.
Q26 — Molecular Orbital Theory and Hydrogen Bonding · medium
In the molecular orbital diagram for the molecular ion N₂⁺, the number of electrons in the $\sigma 2p_z$ molecular orbital is
A. 3
B. 1 ✓ Correct
C. 0
D. 2
Solution: N₂⁺ has 13 electrons. With s–p mixing (valid for N₂), the order is ... $\pi 2p_x^2\,\pi 2p_y^2\,\sigma 2p_z$; after filling 12 electrons up through the π set, the 13th and last electron sits alone in $\sigma 2p_z$ → 1 electron.
Q27 — Molecular Orbital Theory and Hydrogen Bonding · easy
Which of the following species is NOT paramagnetic?
A. O₂
B. B₂
C. NO
D. CO ✓ Correct
Solution: CO (14 e⁻) has all electrons paired → diamagnetic. O₂ (two unpaired $\pi^*$ electrons), B₂ (two unpaired $\pi 2p$ electrons) and NO (one unpaired $\pi^*$ electron) are all paramagnetic.
Q28 — Molecular Orbital Theory and Hydrogen Bonding · easy
Which of the following is paramagnetic?
A. CO
B. O₂²⁻
C. NO⁺
D. B₂ ✓ Correct
Solution: B₂ (10 e⁻): the last two electrons singly occupy the degenerate $\pi 2p_x$ and $\pi 2p_y$ orbitals → two unpaired electrons, paramagnetic. CO and NO⁺ (14 e⁻ each, BO 3) and O₂²⁻ (18 e⁻, $\pi^*$ fully filled) are all diamagnetic.
Q29 — Molecular Orbital Theory and Hydrogen Bonding · medium
Which one of the following properties is NOT shown by NO?
A. Its bond order is 2.5
B. It is diamagnetic in the gaseous state ✓ Correct
C. It is a neutral oxide
D. It combines with oxygen to form nitrogen dioxide
Solution: NO has 15 electrons with one unpaired $\pi^*$ electron, so it is PARAMAGNETIC in the gaseous state (it becomes diamagnetic only on dimerising to N₂O₂ in the liquid/solid state). Its bond order is (10 − 5)/2 = 2.5, it is a neutral oxide, and 2NO + O₂ → 2NO₂.
Q30 — Molecular Orbital Theory and Hydrogen Bonding · hard
Assuming that 2s–2p mixing is NOT operative, the paramagnetic species among the following is
A. Be₂
B. B₂
C. C₂ ✓ Correct
D. N₂
Solution: Without s–p mixing, $\sigma 2p_z$ lies BELOW the $\pi 2p$ set. C₂ (12 e⁻) then fills ... $\sigma 2p_z^2\,\pi 2p_x^1\,\pi 2p_y^1$ — two unpaired electrons → paramagnetic. B₂ (10 e⁻) becomes ... $\sigma 2p_z^2$ (diamagnetic, the reverse of the real situation), Be₂ ends at $\sigma^*2s^2$ (diamagnetic) and N₂ is fully paired either way.