Molecular Orbital Theory - Its important features. LCAOs, types of molecular orbitals (bonding, antibonding), sigma and pi-bonds, molecular orbital electronic configurations of homonuclear diatomic molecules, the concept of bond order, bond length, and bond energy. Elementary idea of metallic bonding. Hydrogen bonding and its applications.: NEET MCQs with Answers

15 free questions on Molecular Orbital Theory - Its important features. LCAOs, types of molecular orbitals (bonding, antibonding), sigma and pi-bonds, molecular orbital electronic configurations of homonuclear diatomic molecules, the concept of bond order, bond length, and bond energy. Elementary idea of metallic bonding. Hydrogen bonding and its applications. for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

Practise all 210 Molecular Orbital Theory - Its important features. LCAOs, types of molecular orbitals (bonding, antibonding), sigma and pi-bonds, molecular orbital electronic configurations of homonuclear diatomic molecules, the concept of bond order, bond length, and bond energy. Elementary idea of metallic bonding. Hydrogen bonding and its applications. questions free Take a free timed test

Question 1

Previous-year question
Arrange the following compounds in increasing order of their solubility in chloroform: NaCl, CH₃OH, cyclohexane, CH₃CN. [Re-NEET 2024]
  1. NaCl < CH₃CN < CH₃OH < Cyclohexane
  2. CH₃OH < CH₃CN < NaCl < Cyclohexane
  3. NaCl < CH₃OH < CH₃CN < Cyclohexane
  4. Cyclohexane < CH₃CN < CH₃OH < NaCl
Show answer
Answer: (A) NaCl < CH₃CN < CH₃OH < Cyclohexane
Explanation: Chloroform is a weakly polar solvent, so solubility increases as solute polarity decreases ('like dissolves like'). Ionic NaCl is least soluble; CH₃CN (dipole moment ≈ 3.9 D) is more polar than CH₃OH (≈ 1.7 D), so it is less soluble; non-polar cyclohexane is most soluble. Hence NaCl < CH₃CN < CH₃OH < cyclohexane.

Question 2

Previous-year question
Which one of the following molecules is paramagnetic? [NEET 2024 (Sawaimadhopur)]
  1. Li₂
  2. C₂
  3. O₂
  4. H₂
Show answer
Answer: (C) O₂
Explanation: According to MO theory, O₂ has two unpaired electrons in the degenerate π*2pₓ and π*2p_y orbitals, so it is paramagnetic. Li₂, C₂ and H₂ have all their electrons paired and are diamagnetic.

Question 3

Previous-year question
Intermolecular forces are the attractive and repulsive forces acting between interacting particles. Which of the following are included among them? [NEET 2023]
A. Dipole–dipole forces
B. Dipole–induced dipole forces
C. Hydrogen bonding
D. Covalent bonding
E. Dispersion forces
  1. A, B, C, D are correct
  2. A, B, C, E are correct
  3. A, C, D, E are correct
  4. B, C, D, E are correct
Show answer
Answer: (B) A, B, C, E are correct
Explanation: Intermolecular forces include dipole–dipole forces, dipole–induced dipole forces, dispersion (London) forces and hydrogen bonding. A covalent bond is an intramolecular bond holding atoms together within a molecule, so D is excluded; the answer is A, B, C, E.

Question 4

Previous-year question
Identify the correct decreasing order of bond enthalpy for the species given below. [NEET 2022 (Overseas)]
  1. N₂ > O₂ > O₂²⁻ > O₂⁻
  2. N₂ > O₂ > O₂⁻ > O₂²⁻
  3. O₂ > N₂ > O₂⁻ > O₂²⁻
  4. O₂²⁻ > O₂⁻ > O₂ > N₂
Show answer
Answer: (B) N₂ > O₂ > O₂⁻ > O₂²⁻
Explanation: Bond enthalpy increases with bond order. Bond orders from MO theory: N₂ = 3, O₂ = 2, O₂⁻ = 1.5 and O₂²⁻ = 1. Therefore the bond enthalpy order is N₂ > O₂ > O₂⁻ > O₂²⁻.

Question 5

Previous-year question
Choose the correct order of boiling points of the following hydrides. [NEET 2022 (Overseas)]
  1. BiH₃ > SbH₃ > NH₃ > AsH₃ > PH₃
  2. NH₃ > PH₃ > AsH₃ > SbH₃ > BiH₃
  3. PH₃ > NH₃ > AsH₃ > SbH₃ > BiH₃
  4. AsH₃ > PH₃ > NH₃ > SbH₃ > BiH₃
Show answer
Answer: (A) BiH₃ > SbH₃ > NH₃ > AsH₃ > PH₃
Explanation: From PH₃ to BiH₃ the boiling point rises with molar mass because van der Waals (dispersion) forces increase. NH₃ boils abnormally high (≈ 238.5 K) because of intermolecular hydrogen bonding, placing it above AsH₃ and PH₃ but below the heavy SbH₃ and BiH₃. Hence BiH₃ > SbH₃ > NH₃ > AsH₃ > PH₃.

Question 6

Previous-year question
Identify the incorrect statement among the following. [NEET 2022]
  1. C₂ molecule has four electrons in its two degenerate π molecular orbitals.
  2. H₂⁺ ion has one electron.
  3. O₂⁺ ion is diamagnetic.
  4. The bond orders of O₂⁺, O₂, O₂⁻ and O₂²⁻ are 2.5, 2, 1.5 and 1, respectively.
Show answer
Answer: (C) O₂⁺ ion is diamagnetic.
Explanation: O₂⁺ (15 electrons) has one unpaired electron in a π*2p orbital, so it is paramagnetic, not diamagnetic; the statement calling it diamagnetic is the incorrect one. C₂ has four electrons in the degenerate π2pₓ and π2p_y orbitals, H₂⁺ has a single electron, and the bond orders of O₂⁺, O₂, O₂⁻ and O₂²⁻ are indeed 2.5, 2, 1.5 and 1.

Question 7

Previous-year question
Choose the correct order of bond order for the species given in the options. [AIPMT 2015]
  1. O₂²⁺ < O₂⁻ < O₂⁺
  2. O₂⁺ < O₂⁻ < O₂²⁺
  3. O₂⁻ < O₂⁺ < O₂²⁺
  4. O₂²⁺ < O₂⁺ < O₂⁻
Show answer
Answer: (C) O₂⁻ < O₂⁺ < O₂²⁺
Explanation: Bond order = ½(Nb − Na). O₂²⁺ (14 electrons) has bond order 3, O₂⁺ (15 electrons) has 2.5 and O₂⁻ (17 electrons) has 1.5. Therefore the order is O₂⁻ (1.5) < O₂⁺ (2.5) < O₂²⁺ (3).
Doing well? Create a free account to take timed tests on this topic and see your weak areas.

Question 8

Previous-year question
Identify a molecule which does not exist. [NEET 2020]
  1. Li₂
  2. C₂
  3. O₂
  4. He₂
Show answer
Answer: (D) He₂
Explanation: In He₂ the two bonding electrons in σ1s are exactly cancelled by the two antibonding electrons in σ*1s, giving a bond order of zero. The molecule therefore does not exist, whereas Li₂, C₂ and O₂ all have positive bond orders.

Question 9

Previous-year question
Which of the following diatomic molecular species has only π bonds according to molecular orbital theory? [NEET 2019]
  1. O₂
  2. N₂
  3. C₂
  4. Be₂
Show answer
Answer: (C) C₂
Explanation: For C₂ the eight valence electrons fill σ2s², σ*2s², π2p_x², π2p_y². The σ2s and σ*2s contributions cancel, so the bond order of 2 comes entirely from the two filled π orbitals — C₂ has two π bonds and no net σ bond.

Question 10

Previous-year question
The manganate and permanganate ions are tetrahedral, due to [NEET 2019]
  1. the π-bonding involves overlap of p-orbitals of oxygen with d-orbitals of manganese
  2. there is no π-bonding
  3. the π-bonding involves overlap of p-orbitals of oxygen with p-orbitals of manganese
  4. the π-bonding involves overlap of d-orbitals of oxygen with d-orbitals of manganese
Show answer
Answer: (A) the π-bonding involves overlap of p-orbitals of oxygen with d-orbitals of manganese
Explanation: In MnO₄²⁻ and MnO₄⁻ the manganese uses sp³ hybrid orbitals for the sigma framework, which fixes the tetrahedral shape, while additional pπ–dπ bonding arises from overlap of filled 2p orbitals of oxygen with the vacant 3d orbitals of manganese.

Question 11

Previous-year question
Consider the species CN⁺, CN⁻, NO and CN. Which one of these will have the highest bond order? [NEET 2018]
  1. NO
  2. CN⁻
  3. CN
  4. CN⁺
Show answer
Answer: (B) CN⁻
Explanation: CN⁻ has 14 valence electrons, the same as N₂, giving a bond order of 3. CN has 13 (bond order 2.5), CN⁺ has 12 (bond order 2) and NO has 15 (bond order 2.5). Hence CN⁻ has the highest bond order.

Question 12

Previous-year question
Which of the following pairs of species have the same bond order? [NEET 2017]
  1. O₂, NO⁺
  2. CN⁻, CO
  3. N₂, O₂⁻
  4. CO, NO
Show answer
Answer: (B) CN⁻, CO
Explanation: Both CN⁻ and CO have 14 valence electrons, exactly as N₂ does, so both are isoelectronic with nitrogen and have a bond order of 3. In the other pairs the two species differ in their electron counts and therefore in bond order.

Question 13

Previous-year question
The decreasing order of stability of O₂, O₂⁻, O₂⁺ and O₂²⁻ is [AIPMT 2015]
  1. O₂⁺ > O₂ > O₂⁻ > O₂²⁻
  2. O₂²⁻ > O₂⁻ > O₂ > O₂⁺
  3. O₂ > O₂⁺ > O₂²⁻ > O₂⁻
  4. O₂⁻ > O₂²⁻ > O₂⁺ > O₂
Show answer
Answer: (A) O₂⁺ > O₂ > O₂⁻ > O₂²⁻
Explanation: Stability follows bond order: O₂⁺ = 2.5, O₂ = 2, O₂⁻ = 1.5 and O₂²⁻ = 1. The species with the highest bond order is the most stable, giving O₂⁺ > O₂ > O₂⁻ > O₂²⁻.

Question 14

Previous-year question
The correct bond order in the following species is: [AIPMT 2015]
  1. O₂²⁺ < O₂⁻ < O₂⁺
  2. O₂⁻ < O₂²⁺ < O₂⁺
  3. O₂⁻ < O₂ < O₂²⁺
  4. O₂²⁺ < O₂ < O₂⁻
Show answer
Answer: (C) O₂⁻ < O₂ < O₂²⁺
Explanation: Bond orders are O₂⁻ = 1.5, O₂ = 2 and O₂²⁺ = 3, since removing electrons from the antibonding π* orbitals steadily raises the bond order. Hence O₂⁻ < O₂ < O₂²⁺.

Question 15

Previous-year question
Be²⁺ is isoelectronic with which of the following ions? [AIPMT 2014]
  1. H⁺
  2. Li⁺
  3. Na⁺
  4. Mg²⁺
Show answer
Answer: (B) Li⁺
Explanation: Be²⁺ has 4 − 2 = 2 electrons and Li⁺ has 3 − 1 = 2 electrons, so the two ions are isoelectronic. H⁺ has none, while Na⁺ and Mg²⁺ each have 10.

Want more Molecular Orbital Theory - Its important features. LCAOs, types of molecular orbitals (bonding, antibonding), sigma and pi-bonds, molecular orbital electronic configurations of homonuclear diatomic molecules, the concept of bond order, bond length, and bond energy. Elementary idea of metallic bonding. Hydrogen bonding and its applications. questions?

E-GYAN NIDHI has 210 questions on this topic and 63,000+ in total, all free: timed CBT tests, NEET marking and instant analysis.

Register free Free Sunday Mock Test

More topics in CHEMICAL BONDING AND MOLECULAR STRUCTURE

↑