Take timed practice tests on Chemical Bonding for JEE Main and JEE Advanced with session-wise drills, score review, and explanation-led revision.
Six 20-question timed sessions plus a 60-question chapter module. Each item is original and reframed for copyright safety.
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1. In a Lewis structure, a single bond represents:
Explanation: A single covalent bond = one bonding pair = 2 electrons (one from each atom, or both from one in a coordinate bond).
2. The octet rule states that atoms tend to form bonds to achieve:
Explanation: The octet rule: most atoms achieve stability by surrounding themselves with 8 valence electrons (like noble gases). Exceptions: H (duet), BF₃ (sextet), PCl₅ (expanded octet).
3. The bond order of N₂ is:
Explanation: N₂ has a triple bond (one σ + two π). Bond order = 3. This explains the very high bond dissociation energy and short bond length.
4. The shape of BeCl₂ (2 bond pairs, 0 lone pairs) according to VSEPR is:
Explanation: 2 bond pairs with no lone pairs → linear geometry (180°). VSEPR predicts minimum electron-pair repulsion.
5. The shape of NH₃ (3 bond pairs, 1 lone pair) is:
Explanation: NH₃: 4 electron pairs (3 bonding + 1 lone). Electron geometry = tetrahedral, but molecular shape (ignoring lone pair) = trigonal pyramidal. Bond angle ≈ 107°.
6. The hybridisation of carbon in methane (CH₄) is:
Explanation: CH₄: C forms 4 bonds with H. Steric number = 4 → sp³ hybridisation. Tetrahedral shape with 109.5° bond angles.
7. The hybridisation of the central atom in SF₆ is:
Explanation: SF₆: S has 6 bonds, no lone pairs. Steric number = 6 → sp³d² hybridisation. Octahedral shape.
8. Which of the following has sp² hybridisation of the central atom?
Explanation: BF₃: B has 3 bonds, 0 lone pairs. Steric number = 3 → sp² hybridisation. Trigonal planar (120°).
9. The shape of PCl₅ is:
Explanation: PCl₅: 5 bond pairs, 0 lone pairs. Steric number = 5 → sp³d. Shape: trigonal bipyramidal (3 equatorial at 120°, 2 axial at 90°).
10. Which of the following has zero dipole moment despite polar bonds?
Explanation: BF₃ is trigonal planar — the three B−F dipoles cancel by symmetry. Zero net dipole moment despite polar bonds.
11. Bond polarity is highest in:
Explanation: H−F has the highest polarity because fluorine has the highest electronegativity (4.0 on Pauling scale), creating the greatest electronegativity difference.
12. The formal charge on nitrogen in NH₄⁺ is:
Explanation: N in NH₄⁺: valence e = 5, nonbonding e = 0, bonding e = 8. FC = 5 − 0 − 8/2 = 5 − 4 = +1.
13. In the resonance structures of SO₃, the S−O bond order is:
Explanation: SO₃ has 3 equivalent resonance structures. Each structure has one double bond and two single bonds. Average bond order = (2+1+1)/3 = 4/3.
14. The bond angle in H₂O (2 bond pairs, 2 lone pairs) is approximately:
Explanation: H₂O: 4 electron pairs. Tetrahedral electron geometry but bent molecular shape. Two lone pairs (stronger repulsion) compress the bond angle from 109.5° to 104.5°.
15. The bond order of O₂ by molecular orbital theory is:
Explanation: O₂ MO configuration: σ1s²σ*1s²σ2s²σ*2s²σ2p²π2p⁴π*2p². Bond order = (bonding − antibonding)/2 = (10−6)/2 = 2. O₂ is paramagnetic (2 unpaired electrons in π*).
16. Which molecule is paramagnetic according to MOT?
Explanation: O₂ has 2 unpaired electrons in the degenerate π*2p orbitals. It is paramagnetic. N₂, F₂, H₂ are all diamagnetic.
17. The bond order of He₂ by MOT is:
Explanation: He₂: σ1s² σ*1s². Bond order = (2−2)/2 = 0. He₂ does not exist as a stable molecule.
18. Lattice energy of an ionic compound is defined as:
Explanation: Lattice energy = energy absorbed when 1 mol of ionic crystal is completely dissociated into gaseous ions at infinite separation. A large lattice energy implies a more stable crystal.
19. Among N−N, N=N and N≡N, the shortest bond is:
Explanation: Higher bond order → shorter bond length and stronger bond. Triple bond
20. Which of the following molecules is polar?
Explanation: H₂O is bent (not linear), so the two O−H bond dipoles do not cancel → net dipole moment. CO₂ (linear), CCl₄ (tetrahedral), BF₃ (trigonal planar) are all symmetric → zero dipole moment.
21. Hydrogen bonding occurs between F, O, N because these atoms are:
Explanation: H-bonding requires: (1) H bonded to a small, highly electronegative atom (F, O, N), (2) lone pair on acceptor atom. The small size allows close approach for electrostatic interaction.
22. The bond formed when BF₃ accepts a lone pair from NH₃ is:
Explanation: When both electrons of a bond come from one atom (donor) to another (acceptor), it is a coordinate (dative) covalent bond. NH₃ donates its N lone pair to B in BF₃.
23. In PCl₅, the phosphorus forms an expanded octet because:
Explanation: Period 3 and beyond elements (P, S, Cl, etc.) can expand their octet by involving empty d-orbitals in bonding. Period 2 elements (N, O, F) cannot because they lack accessible d-orbitals.
24. SO₂ has a bent shape. The hybridisation of S in SO₂ is:
Explanation: S in SO₂: 2 bonding domains + 1 lone pair = steric number 3 → sp² hybridisation. Bent shape with lone pair in one sp² orbital. Bond angle ≈ 119°.
25. BF₃ is Lewis acid because:
Explanation: BF₃: B has only 6 electrons (incomplete octet). It is electron-deficient and accepts lone pairs from Lewis bases → Lewis acid.
26. The shape of XeF₄ (4 bond pairs, 2 lone pairs) is:
Explanation: XeF₄: 6 electron pairs (4 bond + 2 lone). Octahedral electron geometry. Two lone pairs occupy axial positions opposite each other → molecular shape is square planar.
27. Which species is most stable (highest bond order) among NO, NO⁺, NO⁻?
Explanation: NO: bond order = 2.5. NO⁺ (loses antibonding e from π*): bond order = 3. NO⁻ (gains e in π*): bond order = 2. NO⁺ is isoelectronic with N₂ and most stable.
28. Bond angle order for NH₃, NF₃, PH₃ is:
Explanation: NH₃ ≈ 107°. NF₃ ≈ 102° (F withdraws electrons by induction, reducing bonding pair repulsion, compressing angle). PH₃ ≈ 93° (P uses mostly unhybridised p-orbitals, smaller angle). Order: NH₃ > NF₃ > PH₃.
29. N₂O is a linear molecule with the structure N=N=O. The formal charge on the central N is:
Explanation: N=N=O: Central N has 4 bonds (double bond to each neighbour), 0 lone pairs. FC = 5 − 0 − 8/2 = 5 − 4 = +1.
30. Among ortho, meta and para nitrotoluene, the molecule with the largest dipole moment is:
Explanation: In ortho-nitrotoluene, the NO₂ and CH₃ groups are adjacent. Both CH₃ (+I) and NO₂ (−M) contributions add in ortho due to spatial proximity and mutual reinforcement. Ortho isomer generally has the largest dipole moment in this series.
31. Among NaF, NaCl, NaBr, NaI, the lattice energy decreases in the order:
Explanation: Lattice energy ∝ charge product / distance. As the anion size increases (F NaCl > NaBr > NaI.
32. HF has a higher boiling point than HCl because:
Explanation: F is the most electronegative element → H−F bonds are highly polar → strong intermolecular H-bonding in HF. HCl has no significant H-bonding. Hence HF has an anomalously high boiling point.
33. The bond lengths in the carbonate ion CO₃²⁻ are all equal because:
Explanation: CO₃²⁻ has 3 equivalent resonance structures. The delocalised electrons make all C−O bonds identical with bond order = 4/3, intermediate between single and double bond.
34. The number of lone pairs on Cl in Cl₂ is:
Explanation: Each Cl in Cl₂ has 3 lone pairs (6 non-bonding electrons) and 1 bonding pair. So 3 on each Cl (= 6 total on each Cl—wait: 3 lone pairs per Cl atom). Answer: 3 on each Cl.
35. The hybridisation of carbon in CO₂ is:
Explanation: CO₂: O=C=O. C has 2 double bonds (treated as 2 bonding domains, no lone pairs). Steric number = 2 → sp hybridisation. Linear shape.
36. The structure of I₃⁻ (3 lone pairs on central I, 2 bonding domains) is:
Explanation: I₃⁻: central I has 2 I−I bonds + 3 lone pairs = 5 electron pairs → sp³d, trigonal bipyramidal electron geometry. The 3 lone pairs occupy equatorial positions → 2 axial bonding positions → linear shape.
37. The enthalpy of atomisation of Cl₂ (g) is equal to:
Explanation: Cl₂ → 2Cl(g), ΔH = bond dissociation energy = enthalpy of atomisation of Cl₂ (since breaking one mole of Cl₂ gives 2 mol Cl atoms). Per mole Cl₂: ΔHatom = BDE of Cl₂.
38. Which species does NOT exist according to MOT (bond order = 0)?
Explanation: He₂: σ1s² σ*1s². Bond order = 0. Does not exist. He₂⁺: σ1s² σ*1s¹; BO = 0.5, exists. H₂⁻: σ1s² σ*1s¹; BO = 0.5, exists. H₂: BO = 1, exists.