Chemical Bonding and Molecular Structure JEE Main previous year questions with solutions

5 solved JEE Main questions on Chemical Bonding and Molecular Structure, free to read — no sign-in needed. The full chapter has 312 questions; sign in to attempt the remaining 307 in the exam simulator.

1. Q1JEE Main 2026 (06 Apr, Shift 2)Covalent and Co-ordinate Bonding
   Given below are two statements : Statement I : The number of compounds among SO$_2$, SO$_3$, SF$_4$, SF$_6$ and H$_2$S in which sulphur does not obey the Octet rule is $3$. Statement II : Among [H$_2$O, ClF$_3$, SF$_4$], [NH$_3$, BrF$_5$, SF$_4$], [BrF$_5$, ClF$_3$, XeF$_4$] and [XeF$_4$, ClF$_3$, H$_2$O], the number of sets in which all the molecules have one lone pair of electrons on the central atom is $1$. In the light of the above statements, choose the correct answer from the options given below :
   1. A.Both Statement I and Statement II are true
   2. B.Both Statement I and Statement II are false
   3. C.Statement I is true but Statement II is false
   4. D.Statement I is false but Statement II is true
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   Answer: (D)

   Statement I: Number of compounds in which sulphur does not follow the octet rule (from $SO_2$, $SO_3$, $SF_4$, $SF_6$ and $H_2S$). $SO_2$: The formal-charge-minimised structure $O=S=O$ places $10$ electrons around sulphur (expanded octet). $SO_3$: With three $S=O$ double bonds, sulphur has $12$ electrons around it (expanded octet). $SF_4$: $4$ bond pairs $+ 1$ lone pair $= 10$ electrons around $S$ (expanded octet). $SF_6$: $6$ bond pairs $= 12$ electrons around $S$ (expanded octet). $H_2S$: $2$ bond pairs $+ 2$ lone pairs $= 8$ electrons around $S$ (obeys the octet rule). Depending on whether formal-charge-minimised structures are used for $SO_2$ and $SO_3$, either $2$ ($SF_4, SF_6$) or $4$ ($SO_2, SO_3, SF_4, SF_6$) species violate the octet rule, but the number is never $3$. Hence, Statement I is false. Statement II: Among the given sets, exactly one contains only molecules having one lone pair on the central atom. Number of lone pairs on the central atom, $\text{LP} = \dfrac{\text{valence electrons of central atom} - \text{number of bonded monovalent atoms}}{2}$: $H_2O$: $\dfrac{6 - 2}{2} = 2$ lone pairs $ClF_3$: $\dfrac{7 - 3}{2} = 2$ lone pairs $SF_4$: $\dfrac{6 - 4}{2} = 1$ lone pair $NH_3$: $\dfrac{5 - 3}{2} = 1$ lone pair $BrF_5$: $\dfrac{7 - 5}{2} = 1$ lone pair $XeF_4$: $\dfrac{8 - 4}{2} = 2$ lone pairs Checking the sets: $\{H_2O\ (2),\ ClF_3\ (2),\ SF_4\ (1)\}$: not all have $1$ lone pair. $\{NH_3\ (1),\ BrF_5\ (1),\ SF_4\ (1)\}$: all have exactly $1$ lone pair. $\{BrF_5\ (1),\ ClF_3\ (2),\ XeF_4\ (2)\}$: not all have $1$ lone pair. $\{XeF_4\ (2),\ ClF_3\ (2),\ H_2O\ (2)\}$: not all have $1$ lone pair. Exactly one set satisfies the condition, so Statement II is true. Hence, Statement I is false but Statement II is true, which corresponds to option $(4)$.
2. Q2JEE Main 2025 (22 Jan, Shift 2)Dipole moment
   <p>Arrange the following compounds in increasing order of their dipole moment : $\mathrm{HBr}, \mathrm{H}_2 \mathrm{~S}, \mathrm{NF}_3$ and $\mathrm{CHCl}_3$</p>
   1. A.$\mathrm{H}_2 \mathrm{~S} \lt \mathrm{HBr} \lt \mathrm{NF}_3 \lt \mathrm{CHCl}_3$
   2. B.<p>$\mathrm{NF}_3 \lt \mathrm{HBr} \lt \mathrm{H}_2 \mathrm{~S} \lt \mathrm{CHCl}_3$</p>
   3. C.$\mathrm{HBr} \lt \mathrm{H}_2 \mathrm{~S} \lt \mathrm{NF}_3 \lt \mathrm{CHCl}_3$
   4. D.<p>$\mathrm{CHCl}_3 \lt \mathrm{NF}_3 \lt \mathrm{HBr} \lt \mathrm{H}_2 \mathrm{~S}$</p>
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   Answer: (B)

   <p>$\begin{aligned} & \mu_{\mathrm{HBr}}=0.78 \mathrm{D} \\ & \mu_{\mathrm{H}_2 \mathrm{~S}}=0.95 \mathrm{D} \\ & \mu_{\mathrm{NF}_3}=0.24 \mathrm{D} \\ & \mu_{\mathrm{CHCl}_3}=1.01 \mathrm{D} \end{aligned}$ Hence dipole moment of $\mathrm{NF}_3 \lt \mathrm{HBr} \lt \mathrm{H}_2 \mathrm{~S} \lt \mathrm{CHCl}_3$</p>
3. Q3JEE Main 2024 (08 Apr, Shift 1)Back Bonding
   Given below are two statements: Statement I: $\mathrm{N}\left(\mathrm{CH}_3\right)_3$ and $\mathrm{P}\left(\mathrm{CH}_3\right)_3$ can act as ligands to form transition metal complexes. Statement II: As N and P are from same group, the nature of bonding of $\mathrm{N}\left(\mathrm{CH}_3\right)_3$ and $\mathrm{P}\left(\mathrm{CH}_3\right)_3$ is always same with transition metals. In the light of the above statements, choose the most appropriate answer from the options given below:
   1. A.Statement I is correct but Statement II is incorrect.
   2. B.Statement I is incorrect but Statement II is correct.
   3. C.Both Statement I and Statement II are correct.
   4. D.Both Statement I and Statement II are incorrect.
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   Answer: (A)

   $\mathrm{N}\left(\mathrm{CH}_3\right)_3$ and $\mathrm{P}\left(\mathrm{CH}_3\right)_3$ both are Lewis base and acts as ligand, However, $\mathrm{P}\left(\mathrm{CH}_3\right)_3$ has a $\pi$-acceptor character.
4. Q4JEE Main 2023 (06 Apr, Shift 2)Hybridisation and VSEPR Theory
   The number of species having a square planar shape from the following is ${XeF}_{4},{SF}_{4},{SiF}_{4},{{BF}_{4}}^{-},{{BrF}_{4}}^{-},{\left[Cu{\left({NH}_{3}\right)}_{4}\right]}^{2+},{\left[{FeCl}_{4}\right]}^{2-},{\left[{PtCl}_{4}\right]}^{2-}$
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   Answer: 4

   ${XeF}_{4}$ is ${sp}^{3}{d}^{2}$ hybridised with two lone pairs of electrons, hence, it is square planar. ${SF}_{4}$ is ${sp}^{3}d$ hybridised with one lone pairs of electrons, hence, it is See saw. ${SiF}_{4}$ is ${sp}^{3}$ hybridised with zero lone pairs of electrons, hence, it is Tetrahedral ${{BF}_{4}}^{-}$ is ${sp}^{3}$ hybridised with zero lone pairs of electrons, hence, it is Tetrahedral $[Cu({NH}_{3}{)}_{4}]{}^{2+}$is ${dsp}^{2}$ hybridised and is Square planar $[{FeCl}_{4}]{}^{2–}$ is ${sp}^{3}$ hybridised and is Tetrahedral $[{PtCl}_{4}]{}^{2–}$is ${dsp}^{2}$ hybridised and is Square planar ${{BrF}_{4}}^{-}$ is ${sp}^{3}{d}^{2}$ with Square planar geometry. So, $4$ square planer shape compounds are present.
5. Q5JEE Main 2022 (29 Jun, Shift 1)Characteristics of covalent compound
   Arrange the following in the decreasing order of their covalent character : (A) $LiCl$ (B) $NaCl$ (C) $KCl$ (D) $CsCl$ Choose the most appropriate answer from the options given below
   1. A.$\left(A\right)\gt \left(C\right)\gt \left(B\right)\gt \left(D\right)$
   2. B.$\left(B\right)\gt \left(A\right)\gt \left(C\right)\gt \left(D\right)$
   3. C.$\left(A\right)\gt \left(B\right)\gt \left(C\right)\gt \left(D\right)$
   4. D.$\left(A\right)\gt \left(B\right)\gt \left(D\right)\gt \left(C\right)$
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   Answer: (C)

   Lithium chloride is the alkali metal chloride which is most covalent in nature. This is due to the small size and high polarising power of lithium. That is, smaller the cation, greater is the polarising power and hence, greater is the covalent character. So order of covalent character will be $CsCl\lt RbCl\lt NaCl\lt LiCl$