Chemical Bonding and Molecular Structure JEE Main PYQs (2020–2026)

Solved Chemical Bonding and Molecular Structure PYQs

3 recent questions with answers. The full set of 198 is free to practise.

1. JEE Main 2026MediumVSEPR Theory & Shapes/Geometry of Molecules
   Q1

   Bromine trifluoride autoionizes to form ${BrF}_2^{\oplus }$ and ${BrF}_4^{\ominus }$ . The shapes of the cation and anion are respectively $\_\_\_\_$ and $\_\_\_\_$ .

   • Abent, square planar
   • Blinear, square planar
   • Cbent, see-saw
   • Dlinear, tetrahedral

   Show answer & solution

   Correct answer: (A)

   First, let us recall the autoionization of bromine trifluoride:
   
   $$2Br{F}_3\rightleftharpoons Br{F}_2^{+}+Br{F}_4^{-}$$
   
   We have to determine the shapes of the cation and the anion.
   
   Central atom: Br
   
   Valence electrons of Br = 7
   
   Each F contributes one bond → 2 Br–F bonds = 2 pairs of bonding electrons
   
   So, total electrons around Br for shape determination:
   
   $$7-2(\text{by bonding})-1(\text{positive charge})=4$$
   
   These 4 electrons form 2 lone pairs on Br.
   
   Hence, the total number of electron pairs around Br =
   
   2 (bonding) + 2 (lone pairs) = 4 electron domains.
   
   According to VSEPR theory:
   
   4 electron pairs → tetrahedral electron geometry
   
   With 2 bonding and 2 lone pairs → bent shape
   
   Thus, $Br{F}_2^{+}$ is bent.
   
   Central atom: Br
   
   Valence electrons of Br = 7
   
   Each F contributes one bond → 4 Br–F bonds = 4 pairs of bonding electrons
   
   Additional one electron due to negative charge → +1
   
   Hence total electrons on central bromine:
   
   $$7+1-4=4$$
   
   which means 2 lone pairs on Br.
   
   Thus, the total number of electron pairs =
   
   4 (bonding) + 2 (lone pairs) = 6 electron domains.
   
   According to VSEPR theory:
   
   6 electron domains → octahedral electron geometry
   
   With 4 bonding and 2 lone pairs → square planar shape
   
   Shapes are:
   
   $Br{F}_2^{+}$ → bent
   
   $Br{F}_4^{-}$ → square planar
   
   Correct Option: A) bent, square planar
2. JEE Main 2026MediumMolecular Orbital Theory (MOT): Homonuclear & Heteronuclear Diatomics
   Q2

   Pair of species among the following having same bond order as well as paramagnetic character will be-

   • A$O_2^{-},N_2^{+}$
   • B$O_2^{+},N_2^{2-}$
   • C$O_2^{-},N_2^{-}$
   • D$O_2^{+},N_2^{-}$

   Show answer & solution

   Correct answer: (D)

   $$\begin{array}{ccc}\text{ Species }& \begin{array}{c}\text{ Bond }\\ \text{ order }\end{array}& \begin{array}{c}\text{ Magnetic }\\ \text{ Nature }\end{array}\\ O_2^{+}& 2.5& \text{ Paramagnetic }\\ O_2^{-}& 1.5& \text{ Paramagnetic }\\ O_2^{+}& 2.5& \text{ Paramagnetic }\\ N_2^{-}& 2.5& \text{ Paramagnetic }\\ N_2^{2-}& 2& \text{ Paramagnetic }\\ & & \end{array}$$
3. JEE Main 2026MediumHybridization involving s, p, d Orbitals
   Q3

   Two p-block elements X and Y form fluorides of the type ${EF}_3$ . The fluoride compound ${XF}_3$ is a Lewis acid and ${YF}_3$ is a Lewis base. The hybridizations of the central atoms of ${XF}_3$ and ${YF}_3$ respectively are

   • A${sp}^2$ and ${sp}^3$
   • BBoth ${sp}^3$
   • CBoth ${sp}^2$
   • D${sp}^3$ and ${sp}^2$

   Show answer & solution

   Correct answer: (A)

   ${XF}_3$ acts as a Lewis acid, so it must have an incomplete octet (it can accept an electron pair).
   
   A common p-block fluoride with formula ${EF}_3$ that is a Lewis acid is ${BF}_3$ .
   
   In ${BF}_3$ , boron forms three $\sigma$ bonds with fluorine and has no lone pair, so it has only 6 electrons around it. Therefore, it is electron-deficient and behaves as a Lewis acid.
   
   With three bond pairs around boron, the hybridization is ${sp}^2$ .
   
   ${YF}_3$ acts as a Lewis base, so it must have a lone pair to donate.
   
   A common p-block fluoride with formula ${EF}_3$ that is a Lewis base is ${NF}_3$ .
   
   In ${NF}_3$ , nitrogen forms three $\sigma$ bonds with fluorine and has one lone pair, so it can donate that lone pair and behave as a Lewis base.
   
   With four electron pairs (3 bond pairs + 1 lone pair) around nitrogen, the hybridization is ${sp}^3$ .
   
   So, ${XF}_3={BF}_3$ has hybridization ${sp}^2$ and ${YF}_3={NF}_3$ has hybridization ${sp}^3$ .
   
   $$\begin{array}{rl}& {XF}_3={BF}_3;{sp}^2\\ & {YF}_3={NF}_3;{sp}^3\end{array}$$

Topic-wise question count

Where JEE Main has asked from Chemical Bonding and Molecular Structure, most frequent first.

• VSEPR Theory & Shapes/Geometry of Molecules59
• Molecular Orbital Theory (MOT): Homonuclear & Heteronuclear Diatomics46
• Covalent Bond, Dipole Moment & Fajan's Rules37
• Lewis Dot Structures, Formal Charge & Octet Rule27
• Hybridization involving s, p, d Orbitals15
• Hydrogen Bonding (Intermolecular and Intramolecular)11
• Ionic Bond, Lattice Enthalpy & Born-Haber Cycle2
• Valence Bond Theory (VBT) & Orbital Overlap1

Year-wise Chemical Bonding and Molecular Structure questions in JEE Main

Number of questions from this chapter in each year's paper(s).