Chemical Bonding Previous Year Questions (JEE Main 2026)
Solved JEE Main 2026 previous year questions on Chemical Bonding with step-by-step solutions covering Lewis structures, VSEPR shapes, hybridization, molecular orbital theory and bond length order.
A curated set of JEE Main 2026 previous-year questions on Chemical Bonding, each solved step by step so you can check both the final answer and the reasoning.
Solutions are AI-generated and pending review.
Solution
Total valence electrons:
$$\underbrace{8}_{\text{Xe}} + \underbrace{6\times 6}_{6\,\text{O}} + \underbrace{4}_{\text{charge}} = 48\ e^-$$In $\text{XeO}_6^{4-}$ the central Xe is bonded to six oxygen atoms in an octahedral arrangement. Each Xe–O linkage contributes one $\sigma$ bond, so there are $6$ $\sigma$ bond-pairs around Xe.
The steric number of Xe is $6$ (six $\sigma$ bonds, no lone pair), giving a regular octahedral geometry, so Xe carries zero lone pairs.
$$\sigma\text{ bond-pairs} + \text{lone pairs} = 6 + 0 = 6$$Answer: 6
Solution
Two species have a “similar” Lewis dot structure only when they are isoelectronic (same number of valence electrons) and therefore share the same bonding pattern, lone-pair count, and geometry. Count the valence electrons for each pair:
- A $[\text{SO}_3^{2-}, \text{CO}_3^{2-}]$: $\text{CO}_3^{2-}$ has $4+3(6)+2 = 24$ valence electrons — trigonal planar ($sp^2$), central C carries no lone pair. $\text{SO}_3^{2-}$ has $6+3(6)+2 = 26$ valence electrons — pyramidal ($sp^3$), central S carries one lone pair. Different electron count, geometry and lone-pair count — NOT similar.
- B $[\text{O}_2^{2-}, \text{F}_2]$: both have $14$ valence electrons — one single bond and three lone pairs on each atom. Isoelectronic — similar.
- C $[\text{CN}^-, \text{CO}]$: both have $10$ valence electrons — a triple bond and one lone pair on each atom. Isoelectronic — similar.
- D $[\text{NH}_3, \text{H}_3\text{O}^+]$: both have $8$ valence electrons — pyramidal with one lone pair on the central atom. Isoelectronic — similar.
- E $[\text{MnO}_4^{2-}, \text{CrO}_4^{2-}]$: both are tetrahedral $+6$ oxoanions, but they are not isoelectronic — $\text{CrO}_4^{2-}$ has a $d^0$ centre while $\text{MnO}_4^{2-}$ has one extra electron ($d^1$). The electron distributions of the two Lewis structures therefore differ — NOT similar.
Only pairs A and E fail to share a similar Lewis dot structure.
Answer: B (A and E)
Solution
Using molecular orbital theory, the bond order is:
$$O_2^+ : 2.5,\quad O_2 : 2.0,\quad O_2^- : 1.5,\quad O_2^{2-} : 1.0$$Higher bond order $\Rightarrow$ shorter bond, so bond length increases as bond order falls:
$$O_2^+ < O_2 < O_2^- < O_2^{2-}\quad\checkmark\ (\textbf{Statement I is true})$$Number of unpaired electrons (from the $\pi^*$ occupation):
$$O_2 : 2,\quad O_2^+ : 1,\quad O_2^- : 1,\quad O_2^{2-} : 0$$Statement II claims $O_2 > O_2^+ > O_2^- > O_2^{2-}$, i.e. $2 > 1 > 1 > 0$. Since $O_2^+$ and $O_2^-$ both have $1$ unpaired electron, the strict inequality $O_2^+ > O_2^-$ is wrong.
$$\Rightarrow \textbf{Statement II is false}$$Answer: C
Solution
Longer bond $\Leftrightarrow$ more single-bond (lower bond-order) character in the C–O linkage.
- (y) acetone: an essentially pure, localized $\mathrm{C{=}O}$ double bond (no lengthening resonance). Shortest.
- (z) acetate: the two C–O bonds are equivalent by resonance with bond order $1.5$, so the bond is longer than a pure double bond. Intermediate.
- (x) tropone: resonance transfers electron density from $\mathrm{C{=}O}$ into the ring to generate the aromatic tropylium cation ($\mathrm{C{-}O^{-}}$ with a $6\pi$ aromatic ring). This gives the C–O bond the greatest single-bond character. Longest.
Answer: D
Solution
$SF_4$ has steric number $5$ ($4$ bond pairs $+ 1$ lone pair, $AX_4E$, $sp^3d$) $\Rightarrow$ see-saw shape.
Check each option:
- A. $BrF_4^{-}$: $4$ bonds $+ 2$ lone pairs ($AX_4E_2$) $\Rightarrow$ square planar. No.
- B. $CH_4$: $4$ bonds, no lone pair ($AX_4$) $\Rightarrow$ tetrahedral. No.
- C. $IF_4^{+}$: I contributes $7$ electrons, $-1$ for the charge, $4$ bonds $\Rightarrow$ $(7-1-4)/2 = 1$ lone pair, $AX_4E$ $\Rightarrow$ see-saw. Yes.
- D. $XeF_4$: $4$ bonds $+ 2$ lone pairs $\Rightarrow$ square planar. No.
- E. $XeO_2F_2$: Xe has $2$ $=O$, $2$ $-F$ and $1$ lone pair ($AX_4E$) $\Rightarrow$ see-saw. Yes.
Species isostructural with $SF_4$: C and E.
Answer: B
Solution
Statement I — electrons around S:
- $SO_2$: expanded octet ($10\,e^-$) — does not obey.
- $SO_3$: $12\,e^-$ — does not obey.
- $SF_4$: $10\,e^-$ — does not obey.
- $SF_6$: $12\,e^-$ — does not obey.
- $H_2S$: $2$ bonds $+ 2$ lone pairs $= 8\,e^-$ — obeys.
So S does not obey the octet in $4$ compounds, not $3$. $\Rightarrow$ Statement I is false.
Statement II — lone pairs on the central atom:
$$H_2O:2,\ \ NH_3:1,\ \ H_3O^+\text{—},\ \ ClF_3:2,\ \ SF_4:1,\ \ BrF_5:1,\ \ XeF_4:2$$- $[H_2O, ClF_3, SF_4]$: $2,2,1$ — not all one.
- $[NH_3, BrF_5, SF_4]$: $1,1,1$ — all one. $\checkmark$
- $[BrF_5, ClF_3, XeF_4]$: $1,2,2$ — not all one.
- $[XeF_4, ClF_3, H_2O]$: $2,2,2$ — not all one.
Exactly $1$ set qualifies. $\Rightarrow$ Statement II is true.
Answer: D
Solution
$sp^3d$ hybridization corresponds to a steric number (SN) of $5$.
| Species | Bond pairs | Lone pairs | SN | Hybridization |
|---|---|---|---|---|
| $BrF_5$ | 5 | 1 | 6 | $sp^3d^2$ |
| $XeF_5^-$ | 5 | 2 | 7 | $sp^3d^3$ |
| $BF_4^-$ | 4 | 0 | 4 | $sp^3$ |
| $ICl_4^-$ | 4 | 2 | 6 | $sp^3d^2$ |
| $XeF_4$ | 4 | 2 | 6 | $sp^3d^2$ |
| $SF_4$ | 4 | 1 | 5 | $sp^3d$ ✓ |
| $NH_4^+$ | 4 | 0 | 4 | $sp^3$ |
| $ClF_3$ | 3 | 2 | 5 | $sp^3d$ ✓ |
| $XeF_2$ | 2 | 3 | 5 | $sp^3d$ ✓ |
| $ICl_2^-$ | 2 | 3 | 5 | $sp^3d$ ✓ |
Species with $sp^3d$: $SF_4,\ ClF_3,\ XeF_2,\ ICl_2^-$ $\Rightarrow 4$.
Answer: 4
Solution
The covalent bond length is the internuclear A–B distance, which is the sum of the two covalent radii:
$$d_{A-B} = r_A + r_B$$The total length of the AB molecule spans from the outer edge of A to the outer edge of B:
$$L = r_A + d_{A-B} + r_B = r_A + (r_A + r_B) + r_B = 2(r_A + r_B)$$$$\Rightarrow\ \text{bond length} = (r_A + r_B),\qquad \text{total length} = 2(r_A + r_B)$$Answer: A
Solution
Autoionization: $\ 2\,BrF_3 \rightleftharpoons BrF_2^{+} + BrF_4^{-}$.
Cation $BrF_2^{+}$: Br has $7$ electrons, $-1$ for the positive charge, $2$ used in bonding:
$$\text{lone pairs} = \frac{7 - 1 - 2}{2} = 2,\quad SN = 2 + 2 = 4\ (AX_2E_2)\ \Rightarrow\ \textbf{bent}$$Anion $BrF_4^{-}$: Br has $7$ electrons, $+1$ for the negative charge, $4$ used in bonding:
$$\text{lone pairs} = \frac{7 + 1 - 4}{2} = 2,\quad SN = 4 + 2 = 6\ (AX_4E_2)\ \Rightarrow\ \textbf{square planar}$$Answer: A