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 307 questions; sign in to attempt the remaining 302 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
   Show answer & solution

   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>
   Show answer & solution

   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 (09 Apr, Shift 2)Hybridisation and VSEPR Theory
   The incorrect statement regarding ethyne is
   1. A.The $\mathrm{C}-\mathrm{C}$ bonds in ethyne is shorter than that in ethene
   2. B.Both carbons are sp hybridised
   3. C.Ethyne is linear
   4. D.The corbon - carbon bonds in ethyne is weaker than that in ethene
   Show answer & solution

   Answer: (D)

   The carbon-carbon bonds in ethyne is stronger than that in ethene. $(\mathrm{H}-\mathrm{C} \equiv \mathrm{C}-\mathrm{H})$ Ethyne is linear and carbon atoms are SP hybridised.
4. Q4JEE Main 2023 (13 Apr, Shift 1)Molecular Orbital Theory
   In which of the following processes, the bond order increases and paramagnetic character changes to diamagnetic one?
   1. A.${O}_{2}\rightarrow {O}_{2}^{+}$
   2. B.${O}_{2}\rightarrow {O}_{2}^{2-}$
   3. C.$NO\rightarrow {NO}^{+}$
   4. D.${N}_{2}\rightarrow {N}_{2}^{+}$
   Show answer & solution

   Answer: (C)

   Process Change in magnetic nature Bonad order change ${N}_{2}\rightarrow {N}_{2}^{+}$ Dia $\rightarrow$ Para $3\rightarrow 2.5$ $NO\rightarrow {NO}^{+}$ Para $\rightarrow$Dia $2.5\rightarrow 3$ ${O}_{2}\rightarrow {O}_{2}^{-2}$ Para$\rightarrow$ Dia $2\rightarrow 1$ ${O}_{2}\rightarrow {O}_{2}^{+}$ Para $\rightarrow$ Para $2\rightarrow 2.5$
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.$\left(A\right)\gt \left(B\right)\gt \left(D\right)\gt \left(C\right)$
   Show answer & solution

   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 $CsCl\lt RbCl\lt NaCl\lt LiCl$