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Chemical Bonding and Molecular Structure: questions with answers

Class 11 Chemistry, Chapter 4 (NCERT). 15 questions by concept. Open “Show answer” for the answer and why.

How to teach this chapter · Common misconceptions & worked visual for this chapter

Hydrogen bonding: cause, types, and consequences

Question 1

Which hydrogen-bond arrangement reduces intermolecular association and explains the greater steam volatility of o-nitrophenol?

  1. (a)Intramolecular O–H···O attraction between separate molecules
  2. (b)Maximum intermolecular hydrogen bonding in the vapour
  3. (c)The covalent O–H bond within each molecule
  4. (d)Intramolecular O–H···O attraction within one molecule
Show answer

(d) Intramolecular O–H···O attraction within one molecule

The ortho arrangement permits an internal O–H···O contact, while the para arrangement favours association between molecules.

Question 2

In the chapter’s comparison of physical states, how does the extent of hydrogen bonding rank?

  1. (a)Liquid > solid > gas
  2. (b)Gas > liquid > solid
  3. (c)Solid > liquid > gas
  4. (d)Solid = liquid = gas
Show answer

(c) Solid > liquid > gas

The chapter ranks hydrogen bonding greatest in solids and least in gases. Persistent close molecular association favours hydrogen bonding in condensed phases.

Kössel–Lewis bonding, Lewis structures, formal charge, and octet limits

Question 3

How many valence electrons belong in the Lewis electron budget of NO₂⁻?

  1. (a)17
  2. (b)24
  3. (c)16
  4. (d)18
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(d) 18

Nitrogen supplies 5 valence electrons and two oxygen atoms supply 12. The negative charge adds one, giving 18.

Question 4

A Lewis structure for NH₄⁺ contains four single-bond lines. How many bonding electrons do those lines represent?

  1. (a)4 bonding electrons
  2. (b)12 bonding electrons
  3. (c)8 bonding electrons
  4. (d)10 bonding electrons
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(c) 8 bonding electrons

Each single-bond line represents two electrons, so four lines represent eight bonding electrons.

Ionic bond formation and lattice enthalpy

Question 5

Producing M⁺ costs 500 kJ mol⁻¹; adding an electron to X(g) costs 40 kJ mol⁻¹. Forming MX(s) from these ions releases 800 kJ mol⁻¹. What is the net enthalpy?

  1. (a)+260 kJ mol⁻¹
  2. (b)−340 kJ mol⁻¹
  3. (c)−260 kJ mol⁻¹
  4. (d)+1340 kJ mol⁻¹
Show answer

(c) −260 kJ mol⁻¹

The stated steps give 500 + 40 − 800 = −260 kJ mol⁻¹. Electron gain is endothermic in this example.

Question 6

Separating one mole of NaCl(s) into gaseous ions requires 788 kJ. What follows for the reverse lattice-formation step?

  1. (a)Enthalpy change: −788 kJ mol⁻¹
  2. (b)Energetic favourability guaranteed by completed octets alone
  3. (c)Enthalpy change: +788 kJ mol⁻¹
  4. (d)Enthalpy change: 0 kJ mol⁻¹
Show answer

(a) Enthalpy change: −788 kJ mol⁻¹

The chapter defines lattice enthalpy for separating the solid into gaseous ions; reversing the process reverses the sign.

Bond parameters, bond order trends, and resonance

Question 7

The two successive O–H dissociation enthalpies of gaseous H₂O are 502 and 427 kJ mol⁻¹. What is the mean O–H bond enthalpy?

  1. (a)502 kJ mol⁻¹
  2. (b)427 kJ mol⁻¹
  3. (c)464.5 kJ mol⁻¹
  4. (d)929 kJ mol⁻¹
Show answer

(c) 464.5 kJ mol⁻¹

The mean is (502 + 427)/2 = 464.5 kJ mol⁻¹. Successive cleavage changes the remaining species, so the two dissociation energies differ.

Question 8

How do axial and equatorial P–Cl bond lengths in gaseous PCl₅ compare, and where are those lengths measured?

  1. (a)Axial longer; empty gap between atomic edges
  2. (b)Axial shorter; nucleus to nucleus
  3. (c)All five equal; nucleus to nucleus
  4. (d)Axial longer; nucleus to nucleus
Show answer

(d) Axial longer; nucleus to nucleus

Bond length is the equilibrium distance between nuclei. Axial P–Cl bonds are longer because they experience more right-angle repulsions than equatorial bonds.

Bond polarity, molecular dipole moment, and polarisation

Question 9

Two cations have the same charge, but M⁺ is smaller than N⁺. With the same anion, which bond is expected to have greater covalent character?

  1. (a)Both compounds have equal covalent character
  2. (b)The compound containing M⁺
  3. (c)The compound containing N⁺
  4. (d)Neither compound has any covalent character
Show answer

(b) The compound containing M⁺

For equal charge, smaller size means greater charge density and stronger distortion of the anion.

Question 10

When Li⁺ distorts the F⁻ electron cloud in LiF, how does the bond character change?

  1. (a)Covalent character decreases
  2. (b)The bond becomes completely covalent
  3. (c)The bond remains completely ionic
  4. (d)Partial covalent character increases
Show answer

(d) Partial covalent character increases

Ionic and covalent labels are ideal limits; polarisation produces partial covalent character.

Hybridisation, molecular geometry, and multiple bonding

Question 11

How many σ and π bonds are present in C₂H₄?

  1. (a)1 σ bond and 5 π bonds
  2. (b)4 σ bonds and 1 π bond
  3. (c)5 σ bonds and 1 π bond
  4. (d)6 σ bonds and 0 π bonds
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(c) 5 σ bonds and 1 π bond

Ethene contains four C–H σ bonds and one C–C σ bond; the remaining p orbitals form one π bond.

VSEPR electron-domain geometry and molecular shape

Question 12

What are the electron geometry and molecular shape of AB₃E?

  1. (a)Tetrahedral electron geometry and trigonal-pyramidal molecular shape
  2. (b)Trigonal-planar shape because only the three B atoms are counted
  3. (c)Trigonal-pyramidal electron geometry and the same molecular shape by definition
  4. (d)Five domains because a multiple bond counts twice
Show answer

(a) Tetrahedral electron geometry and trigonal-pyramidal molecular shape

Three bond pairs plus one lone pair make four tetrahedral domains; only the three bonded atoms define the pyramidal shape.

Valence-bond overlap, potential energy, and σ/π bonds

Question 13

Gaseous H₂ has its minimum potential energy at a nuclear separation of 74 pm. What happens when the nuclei are forced closer?

  1. (a)Potential energy falls continuously toward zero separation
  2. (b)Potential energy rises
  3. (c)Potential energy first falls, then stays constant
  4. (d)Potential energy remains constant
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(b) Potential energy rises

Below the equilibrium separation, repulsive effects outweigh the extra attraction and potential energy rises.

MO electron configurations, bond order, stability, and magnetism

Question 14

A species has 10 bonding and 6 antibonding electrons. How does its bond order change when one additional electron enters an antibonding orbital?

  1. (a)2 → 1.5
  2. (b)5 → 5
  3. (c)2 → 1
  4. (d)2 → 2.5
Show answer

(a) 2 → 1.5

Initially, bond order is (10 − 6)/2 = 2. Adding an antibonding electron changes it to (10 − 7)/2 = 1.5.

Molecular-orbital formation, symmetry, and energy ordering

Question 15

For 2p_z and 2p_x orbitals on different atoms, with z as the internuclear axis, what is their net overlap?

  1. (a)Zero through cancellation of opposite-sign contributions
  2. (b)Maximum head-on overlap
  3. (c)Nonzero whenever the orbital energies match
  4. (d)Maximum sideways overlap
Show answer

(a) Zero through cancellation of opposite-sign contributions

The two atomic orbitals have incompatible symmetry about the molecular axis. Positive and negative overlap contributions cancel.

Common misconception in Chemical Bonding and Molecular Structure

Open full misconception guide →

Do polar bonds always give a polar molecule?

Wrong

CO₂ is polar because its C=O bonds are polar.

Why it feels right. I check the bonds one at a time.

Right

The net molecular dipole depends on both bond dipoles and geometry.

Worked example

Equal bond dipoles point in opposite directions along the molecule. The net dipole is zero.
  1. Draw the two equal C=O dipole arrows in linear CO₂.
  2. They point in opposite directions and cancel.