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

5 solved JEE Advanced questions on Chemical Bonding and Molecular Structure, free to read — no sign-in needed. The full chapter has 34 questions; sign in to attempt the remaining 29 in the exam simulator.

1. Q1JEE Advanced Adv 2026 (Paper 2)
   The correct order of ONO bond angle in the given species is
   1. A.$\text{NO}_2^+ \lt \text{NO}_2 \lt \text{NO}_3^- \lt \text{NO}_2^-$
   2. B.$\text{NO}_2^- \lt \text{NO}_3^- \lt \text{NO}_2 \lt \text{NO}_2^+$
   3. C.$\text{NO}_3^- \lt \text{NO}_2^- \lt \text{NO}_2 \lt \text{NO}_2^+$
   4. D.$\text{NO}_2^- \lt \text{NO}_3^- \lt \text{NO}_2^+ \lt \text{NO}_2$
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   Answer: (B)

   For $\text{NO}_2^+$, the central nitrogen atom is $sp$ hybridized with no lone pairs, resulting in a linear geometry and a bond angle of $180^\circ$. For $\text{NO}_2$, the central nitrogen atom is $sp^2$ hybridized with one unpaired electron. The repulsion from a single electron is less than that from a bond pair, so the bond angle opens up to be greater than $120^\circ$ (approximately $134^\circ$). For $\text{NO}_3^-$, the central nitrogen atom is $sp^2$ hybridized with no lone pairs, resulting in a trigonal planar geometry and a bond angle of exactly $120^\circ$. For $\text{NO}_2^-$, the central nitrogen atom is $sp^2$ hybridized with one lone pair. The lone pair-bond pair repulsion is greater than bond pair-bond pair repulsion, compressing the bond angle to less than $120^\circ$ (approximately $115^\circ$). Therefore, the correct order of O-N-O bond angles is $\text{NO}_2^- \lt \text{NO}_3^- \lt \text{NO}_2 \lt \text{NO}_2^+$. Answer: $\text{NO}_2^- \lt \text{NO}_3^- \lt \text{NO}_2 \lt \text{NO}_2^+$
2. Q2JEE Advanced Adv 2026 (Paper 1)
   Consider the following species: $\text{SOCl}_2$, $\text{XeOF}_4$, $\text{ClF}_3$, $\text{ClF}_5$, $\text{XeF}_5^+$, $\text{SO}_3^{2-}$, $\text{XeF}_3^+$, $\text{SF}_4$ List-I contains different molecular shapes and List-II contains total number of species with the same molecular shapes from the given species. Match each entry in List-I with the appropriate entry in List-II and choose the correct option. List-IList-II(P) See-saw(1) one(Q) T-Shaped(2) two(R) Trigonal Planar(3) three(S) Square Pyramidal(4) four(5) zero
   1. A.P $\rightarrow$ 1; Q $\rightarrow$ 2; R $\rightarrow$ 5; S $\rightarrow$ 3
   2. B.P $\rightarrow$ 5; Q $\rightarrow$ 4; R $\rightarrow$ 2; S $\rightarrow$ 3
   3. C.P $\rightarrow$ 3; Q $\rightarrow$ 2; R $\rightarrow$ 1; S $\rightarrow$ 4
   4. D.P $\rightarrow$ 1; Q $\rightarrow$ 3; R $\rightarrow$ 5; S $\rightarrow$ 4
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   Answer: (A)

   Let us determine the hybridization and shape of each given species by finding the number of bond pairs (bp) and lone pairs (lp) on the central atom: $\text{SOCl}_2$: Central atom S has 6 valence electrons. It forms 3 $\sigma$ bonds (one with O, two with Cl) and has 1 lone pair. Steric number = 4 ($sp^3$). Shape is Trigonal Pyramidal. $\text{XeOF}_4$: Central atom Xe has 8 valence electrons. It forms 5 $\sigma$ bonds (one with O, four with F) and has 1 lone pair. Steric number = 6 ($sp^3d^2$). Shape is Square Pyramidal. $\text{ClF}_3$: Central atom Cl has 7 valence electrons. It forms 3 $\sigma$ bonds with F and has 2 lone pairs. Steric number = 5 ($sp^3d$). Shape is T-shaped. $\text{ClF}_5$: Central atom Cl has 7 valence electrons. It forms 5 $\sigma$ bonds with F and has 1 lone pair. Steric number = 6 ($sp^3d^2$). Shape is Square Pyramidal. $\text{XeF}_5^+$: Central atom Xe has 8 valence electrons. The positive charge leaves 7 valence electrons. It forms 5 $\sigma$ bonds with F and has 1 lone pair. Steric number = 6 ($sp^3d^2$). Shape is Square Pyramidal. $\text{SO}_3^{2-}$: Central atom S has 6 valence electrons. The $-2$ charge gives 8 valence electrons. It forms 3 $\sigma$ bonds with O and has 1 lone pair. Steric number = 4 ($sp^3$). Shape is Trigonal Pyramidal. $\text{XeF}_3^+$: Central atom Xe has 8 valence electrons. The positive charge leaves 7 valence electrons. It forms 3 $\sigma$ bonds with F and has 2 lone pairs. Steric number = 5 ($sp^3d$). Shape is T-shaped. $\text{SF}_4$: Central atom S has 6 valence electrons. It forms 4 $\sigma$ bonds with F and has 1 lone pair. Steric number = 5 ($sp^3d$). Shape is See-saw. Counting the species for each shape: (P) See-saw: $1$ ($\text{SF}_4$) (Q) T-Shaped: $2$ ($\text{ClF}_3$, $\text{XeF}_3^+$) (R) Trigonal Planar: $0$ (None) (S) Square Pyramidal: $3$ ($\text{XeOF}_4$, $\text{ClF}_5$, $\text{XeF}_5^+$) Matching with List-II: P $\rightarrow$ 1 Q $\rightarrow$ 2 R $\rightarrow$ 5 S $\rightarrow$ 3 Answer: P $\rightarrow$ 1; Q $\rightarrow$ 2; R $\rightarrow$ 5; S $\rightarrow$ 3
3. Q3JEE Advanced Adv 2026 (Paper 1)
   The correct order of dipole moments for the given species is
   1. A.$\text{BF}_3 = \text{NH}_4^+ \lt \text{NF}_3 \lt \text{NH}_3$
   2. B.$\text{BF}_3 \lt \text{NH}_4^+ \lt \text{NF}_3 \lt \text{NH}_3$
   3. C.$\text{NH}_4^+ \lt \text{BF}_3 \lt \text{NH}_3 \lt \text{NF}_3$
   4. D.$\text{BF}_3 \lt \text{NH}_4^+ \lt \text{NH}_3 \lt \text{NF}_3$
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   Answer: (A)

   The dipole moment of a molecule depends on its geometry and the vector sum of its individual bond dipoles. $\text{BF}_3$ has a symmetrical trigonal planar geometry. The three B-F bond dipoles cancel each other out completely, resulting in a net dipole moment of zero ($\mu = 0$). $\text{NH}_4^+$ has a symmetrical tetrahedral geometry. The four N-H bond dipoles cancel each other out, resulting in a net dipole moment of zero ($\mu = 0$). Thus, $\mu(\text{BF}_3) = \mu(\text{NH}_4^+) = 0$. Both $\text{NH}_3$ and $\text{NF}_3$ have a trigonal pyramidal geometry with one lone pair on the central nitrogen atom. In $\text{NH}_3$, nitrogen is more electronegative than hydrogen. The N-H bond dipoles point towards the nitrogen atom, which is in the same direction as the orbital dipole of the lone pair. These dipoles reinforce each other, giving a higher net dipole moment. In $\text{NF}_3$, fluorine is more electronegative than nitrogen. The N-F bond dipoles point away from the nitrogen atom, which is in the opposite direction to the orbital dipole of the lone pair. These dipoles partially cancel each other out, resulting in a lower net dipole moment. Therefore, $\mu(\text{NF}_3) \lt \mu(\text{NH}_3)$. The correct order of dipole moments is $\text{BF}_3 = \text{NH}_4^+ \lt \text{NF}_3 \lt \text{NH}_3$. Answer: $\text{BF}_3 = \text{NH}_4^+ \lt \text{NF}_3 \lt \text{NH}_3$
4. Q4JEE Advanced Adv 2025 (Paper 2)
   The correct statements (s) about intermolecular forces is(are)
   1. A.The potential energy between two point charges approaches zero more rapidly than the potential energy between a point dipole and a point charge as the distance between them approaches infinity.
   2. B.The average potential energy of two rotating polar molecules that are separated by a distance $r$ has $1 / r^3$ dependence.
   3. C.The dipole-induced dipole average interaction energy is independent of temperature.
   4. D.Nonpolar molecules attract one another even though neither has a permanent dipole moment.
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   Answer: C,D

   (i) Ion - Ion $\rightarrow$ Interaction energy $\propto \frac{1}{r}$ Ion - dipole $\rightarrow$ Interaction energy $\propto \frac{1}{r^2}$ Ion - dipole Interaction energy approaches zero more rapidly as ' $r$ ' increases (ii) Rotating Polar molecules $\rightarrow$ Interaction energy $\propto \frac{1}{\mathrm{r}^6}$. (iii) Dipole - induced dipole forces are independent of temperature. (iv) Non-polar species show London dispersion forces.
5. Q5JEE Advanced Adv 2025 (Paper 1)
   Regarding the molecular orbital (MO) energy levels for homonuclear diatomic molecules, the INCORRECT statement(s) is(are)
   1. A.$\text { Bond order of } \mathrm{Ne}_2 \text { is zero. }$
   2. B.$\text { The highest occupied molecular orbital (HOMO) of } F_2 \text { is } \sigma \text {-type. }$
   3. C.$\text { Bond energy of } \mathrm{O}_2^{+} \text {is smaller than the bond energy of } \mathrm{O}_2 \text {. }$
   4. D.$\text { Bond length of } \mathrm{Li}_2 \text { is larger than the bond length of } \mathrm{B}_2$
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   Answer: B,C

   (i) $\mathrm{Ne}_2 \Rightarrow\left(\sigma 1 \mathrm{~s}^2\right)\left(\sigma^* 1 \mathrm{~s}^2\right)\left(\sigma 2 \mathrm{~s}^2\right)\left(\sigma^* 2 \mathrm{~s}^2\right)\left(\sigma 2 \mathrm{p}_{\mathrm{z}}^2\right)\left(\pi 2 \mathrm{p}_{\mathrm{x}}^2=\pi 2 \mathrm{p}_{\mathrm{y}}^2\right)\left(\pi^* 2 \mathrm{p}_{\mathrm{x}}^2=\pi^* 2 \mathrm{p}_{\mathrm{y}}^2\right)\left(\sigma^* 2 \mathrm{p}_{\mathrm{z}}^2\right)$ B.O. $=\frac{6-6}{2}=0$ (ii) $\mathrm{F}_2 \Rightarrow\left(\sigma 1 \mathrm{~s}^2\right)\left(\sigma^* 1 \mathrm{~s}^2\right)\left(\sigma 2 \mathrm{~s}^2\right)\left(\sigma^* 2 \mathrm{~s}^2\right)\left(\sigma 2 \mathrm{p}_{\mathrm{z}}^2\right)\left(\pi 2 \mathrm{p}_{\mathrm{x}}^2=\pi 2 \mathrm{p}_{\mathrm{y}}^2\right)\left(\pi^* 2 \mathrm{p}_{\mathrm{x}}^2=\pi^* 2 \mathrm{p}_{\mathrm{y}}^2\right)$ (iii) $\mathrm{O}_2^{\oplus} \Rightarrow\left(\sigma 1 s^2\right)\left(\sigma^* 1 s^2\right)\left(\sigma 2 s^2\right)\left(\sigma^* 2 s^2\right)\left(\sigma 2 p_z^2\right)\left(\pi 2 p_x^2=\pi 2 p_y^2\right)\left(\pi^* 2 p_x^1=\pi^* 2 p_y\right)$ B.O. $=\frac{6-1}{2}=2.5$ $\mathrm{O}_2 \Rightarrow\left(\sigma 1 s^2\right)\left(\sigma^* 1 s^2\right)\left(\sigma 2 s^2\right)\left(\sigma^* 2 s^2\right)\left(\sigma 2 p_{\mathrm{z}}^2\right)\left(\pi 2 p_{\mathrm{x}}^2=\pi 2 p_{\mathrm{y}}^2\right)\left(\pi^* 2 p_{\mathrm{x}}^1=\pi^* 2 p_{\mathrm{y}}^1\right)$ B.O. $\frac{6-2}{2}=2$ (Bond order increases, Bond strength increases) (iv) Size of atom increases, Bond length increases Size of Li > B So, Bond length of $\mathrm{Li}_2\gt \mathrm{B}_2$