Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure – Extra Questions with Answers

Electron transfer forming ionic bonds, and the tetrahedral shape and sp³ hybridization of methane, are among the recurring ideas tested in this set. Other questions look at coordinate bonds and how molecular orbital theory explains bonding through bonding and antibonding orbitals.

Last Updated: September 23, 2026

Very Short Answer Questions (1 mark)

Q1. What type of bond is formed by electron transfer?
Ans: Ionic (electrovalent) bond.

Q2. What is the shape of a methane (CH₄) molecule?
Ans: Tetrahedral.

Q3. What is the hybridization of carbon in CH₄?
Ans: sp³.

Q4. What is a coordinate (dative) bond?
Ans: A covalent bond where both shared electrons come from the same atom.

Q5. Name the theory that explains bonding using bonding and antibonding molecular orbitals.
Ans: Molecular orbital theory.

Short Answer Questions (2–3 marks)

Q6. Draw the Lewis structure of water (H₂O) and state its molecular shape.
Ans: Oxygen has 2 bond pairs (to H atoms) and 2 lone pairs; shape is bent/angular (due to lone pair repulsion) with bond angle ≈104.5°.

Q7. Explain why NH₃ has a pyramidal shape rather than a perfect tetrahedral shape, despite nitrogen being sp³ hybridized.
Ans: Nitrogen in NH₃ has 3 bond pairs and 1 lone pair. The lone pair exerts greater repulsion than bond pairs, pushing the three N-H bonds closer together, distorting the shape from ideal tetrahedral (109.5°) to pyramidal (≈107°).

Q8. Calculate the bond order of N₂ using molecular orbital theory (10 bonding, 4 antibonding electrons).
Ans: Bond order=½(10−4)=3 (triple bond), consistent with N₂’s known triple bond.

Higher-Order Thinking / Application Questions

Q9. Using molecular orbital theory, explain why O₂ is paramagnetic (has unpaired electrons) even though its Lewis structure suggests all electrons should be paired, and describe how this illustrates a limitation of the simple Lewis/valence bond approach.
Ans: The Lewis structure of O₂ shows a double bond with all electrons apparently paired, predicting O₂ should be diamagnetic. However, molecular orbital theory reveals that when filling the molecular orbitals of O₂ in order of increasing energy, the last two electrons occupy two degenerate (equal energy) π* antibonding orbitals singly (following Hund’s rule), rather than pairing up in one orbital. This gives O₂ two unpaired electrons, correctly predicting its observed paramagnetism (attraction to a magnetic field), something the simple Lewis dot structure approach completely fails to predict, demonstrating that molecular orbital theory provides a more accurate picture of electron distribution in molecules like O₂ where subtle orbital-filling effects matter.

Q10. Explain, using the concept of hydrogen bonding, why water has an unusually high boiling point (100°C) compared to H₂S (boiling point around −60°C), even though sulfur is larger and might be expected to have stronger van der Waals forces.
Ans: Oxygen is significantly more electronegative than sulfur and much smaller in size, which allows water molecules to form strong hydrogen bonds between the highly polarized O-H bonds of one molecule and the lone pairs on the oxygen of a neighbouring molecule. These hydrogen bonds, though individually weaker than covalent bonds, are considerably stronger than ordinary van der Waals (dispersion) forces and create an extensive, dynamic hydrogen-bonded network throughout liquid water, requiring significantly more thermal energy to break in order to vaporize the liquid. H₂S, by contrast, involves the less electronegative and larger sulfur atom, which cannot form effective hydrogen bonds (S-H bonds are much less polarized), so H₂S molecules are held together only by weaker van der Waals forces, resulting in a much lower boiling point despite sulfur’s larger size (which would otherwise be expected to increase van der Waals attraction).

Frequently Asked Questions

How would you predict the shape of a methane molecule using VSEPR theory?
Methane has 4 bond pairs and no lone pairs around carbon, so VSEPR theory predicts a tetrahedral shape with bond angles of 109.5 degrees.

How would you determine the hybridization of the central atom in ammonia?
Nitrogen in ammonia has 3 bond pairs and 1 lone pair, giving sp3 hybridization.

Written by Satish

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