Consider ' $n$ ' is the number of lone pair of electrons present in the equatorial position of the most…

Consider ' $n$ ' is the number of lone pair of electrons present in the equatorial position of the most stable structure of $\mathrm{ClF}_3$. The ions from the following with ' $n$ ' number of unpaired electrons are
A. $\mathrm{V}^{3+}$
B. $\mathrm{Ti}^{3+}$
C. $\mathrm{Cu}^{2+}$
D. $\mathrm{Ni}^{2+}$
E. $\mathrm{Ti}^{2+}$
Choose the correct answer from the options given below:
  1. A and C Only
  2. A, D and E Only
  3. B and D Only
  4. B and C Only

Solution

$\mathrm{ClF}_3$

$\mathrm{n}=2($ No of lone pair present in equitorial plane $)$ (Unpaired $\mathrm{e}^{-}$)
(A) $\mathrm{V}^{+3}:[\mathrm{Ar}] 3 \mathrm{~d}^2 \quad 2$
(B) $\mathrm{Ti}^{3+}:[\mathrm{Ar}] 3 \mathrm{~d}^1 \quad 1$
(C) $\mathrm{Cu}^{+2}:[\mathrm{Ar}] 3 \mathrm{~d}^9 \quad 1$
(D) $\mathrm{Ni}^{+2}:[\mathrm{Ar}] 3 \mathrm{~d}^8 \quad 2$
(E) $\mathrm{Ti}^{+2}:[\mathrm{Ar}] 3 \mathrm{~d}^2 \quad 2$

Asked in: JEE Main 2025 (28 Jan Shift 1)

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