JEE Main 2025 Jan 28 Shift 1, Chemistry Q6: Valence Shell Electron Pair Repulsion (VSEPR) Theory

JEE Main2025Jan 28, Shift 1Chemistry

Consider ' $n$ ' is the number of lone pair of electrons present in the equatorial position of the most stable structure of ClF $_{3}$ . The ions from the following with ' $n$ ' number of unpaired electrons are :

A. V $^{3 +}$
B. Ti $^{3 +}$
C. Cu $^{2 +}$
D. Ni $^{2 +}$
E. Ti $^{2 +}$

Choose the correct answer from the options given below :

A
A and C only
B
A, D and E only
C
B and C only
D
B and D only

Solution

ClF $_{3}$ has the central Cl with 3 bond pairs and 2 lone pairs (AX $_{3}$ E $_{2}$ , T-shaped). In the trigonal bipyramidal electron geometry, both lone pairs sit in the equatorial plane to minimise repulsion. So $n = 2$ .

Count the $d$ -electrons (and hence unpaired electrons) for each cation:

Ion	Configuration	Unpaired e $^{-}$
V $^{3 +}$	[Ar] $3 d^{2}$	2
Ti $^{3 +}$	[Ar] $3 d^{1}$	1
Cu $^{2 +}$	[Ar] $3 d^{9}$	1
Ni $^{2 +}$	[Ar] $3 d^{8}$	2
Ti $^{2 +}$	[Ar] $3 d^{2}$	2

Ions with exactly 2 unpaired electrons: V $^{3 +}$ , Ni $^{2 +}$ , Ti $^{2 +}$ , that is A, D and E.

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JEE Main 2025 Jan 28 Shift 1, Chemistry Q6: Valence Shell Electron Pair Repulsion (VSEPR) Theory

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