Chemical Bonding and Molecular Structure — JEE Main practice

39 questions

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Sample questions with solutions

Q1 · 2026

Given below are two statements :

Statement I :

The number of species among $\mathrm{SF}_4, \mathrm{NH}_4^{+},\left[\mathrm{NiCl}_4\right]^{2-}, \mathrm{XeF}_4,\left[\mathrm{PtCl}_4\right]^{2-}, \mathrm{SeF}_4$ and $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ , that have tetrahedral geometry is 3 .

Statement II :

In the set [ $\mathrm{NO}_2, \mathrm{BeH}_2, \mathrm{BF}_3, \mathrm{AlCl}_3$ ], all the molecules have incomplete octet around central atom.

In the light of the above statements, choose the correct answer from the options given below:

A.
Both Statement I and Statement II are true
B.
Statement I is false but Statement II is true
C.
Statement I is true but Statement II is false
D.
Both Statement I and Statement II are false

Answer: B

For Statement I, check each species' actual geometry: SF₄ (see-saw), NH₄⁺ (tetrahedral), $[\mathrm{NiCl}_4]^{2-}$ (tetrahedral, since Ni²⁺ with weak-field Cl⁻ ligands stays d⁸ high-spin tetrahedral), XeF₄ (square planar), $[\mathrm{PtCl}_4]^{2-}$ (square planar, d⁸ with strong crystal field), SeF₄ (see-saw), $[\mathrm{Ni(CN)}_4]^{2-}$ (square planar, since CN⁻ is a strong-field ligand causing pairing).
Count only the truly tetrahedral species: NH₄⁺ and $[\mathrm{NiCl}_4]^{2-}$ , giving a count of 2, not 3 as claimed.
Since the actual count is 2, Statement I is false.
For Statement II, check each molecule for incomplete octet: NO₂ (odd-electron molecule, N cannot complete octet), BeH₂ (only 4 electrons around Be), BF₃ (only 6 electrons around B), AlCl₃ (only 6 electrons around Al in its monomeric form).
Since all four molecules indeed have incomplete octets around their central atoms, Statement II is true.

Hence, Statement I is false but Statement II is true, so the answer is option B.

Q2 · 2026

Given below are two statements :

Statement I : The correct order in terms of bond dissociation enthalpy is Cl $_2$ > Br $_2$ > F $_2$ > I $_2$ .

Statement II : The correct trend in the covalent character of the metal halides is [SnCl $_4$ > SnCl $_2$ ], [PbCl $_4$ > PbCl $_2$ ] and [UF $_4$ > UF $_6$ ].

In the light of the above statements, choose the correct answer from the options given below :

A.
Statement I is true but Statement II is false
B.
Both Statement I and Statement II are true
C.
Both Statement I and Statement II are false
D.
Statement I is false but Statement II is true

Answer: A

For Statement I, note that bond dissociation enthalpy generally decreases down a group as bond length increases, but F₂ is a well-known exception due to strong lone pair-lone pair repulsion between the small fluorine atoms, weakening its bond. Given this exception, the actual order is $D(\mathrm{Cl}_2) > D(\mathrm{Br}_2) > D(\mathrm{F}_2) > D(\mathrm{I}_2)$ , matching the statement, so Statement I is true.
For Statement II, apply Fajans' rule: covalent character increases with higher cation charge and smaller cation size (greater polarizing power), so check each comparison. Given $\mathrm{SnCl}_4$ vs $\mathrm{SnCl}_2$ : $Sn^{4+}$ is smaller and more highly charged than $Sn^{2+}$ , so $\mathrm{SnCl}_4$ is more covalent — this part is correct. Similarly $\mathrm{PbCl}_4 > \mathrm{PbCl}_2$ in covalent character — correct.
However for $\mathrm{UF}_4$ vs $\mathrm{UF}_6$ : since $U^{6+}$ has a higher charge than $U^{4+}$ , it should have greater polarizing power, meaning $\mathrm{UF}_6$ should actually be MORE covalent than $\mathrm{UF}_4$ — the opposite of what the statement claims.
Since this last comparison in Statement II is incorrect, Statement II is false overall.

Hence, Statement I is true but Statement II is false, so the answer is option A.

Q3 · 2026

The correct increasing order of C–H(A), C–O(B), C=O(C) and C≡N(D) bonds in terms of covalent bond length is :

A.
D < C < B < A
B.
A < D < C < B
C.
D < C < A < B
D.
A < B < C < D

Answer: B

Since bond length decreases as bond order increases, and single bonds with larger atoms are generally longer, compare each bond type systematically. Given C–H is a single bond but with a very small hydrogen atom, making it notably short.
Compare C≡N (triple bond), which should be the shortest overall due to the highest bond order among these bonds.
Compare C=O (double bond), which is shorter than any single bond but longer than the triple bond.
Compare C–O (single bond with the larger oxygen atom), making it the longest of the four.
Combine these comparisons into the overall increasing order. Hence,

A(\text{C–H}) < D(\text{C}\equiv\text{N}) < C(\text{C=O}) < B(\text{C–O})

Hence, the increasing order is A < D < C < B, so the answer is option B.

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