JEE MAIN - Chemistry (2025 - 28th January Morning Shift - No. 14)
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:
Explanation
_28th_January_Morning_Shift_en_14_1.png)
$\mathrm{n}=2($ No of lone pair present in equitorial plane)
$$\begin{array}{lc} & \text { (Unpaired } \mathrm{e}^{-} \text {) } \\ \text { (A) } \mathrm{V}^{+3}:[\mathrm{Ar}] 3 \mathrm{~d}^2 & 2 \\ \text { (B) } \mathrm{Ti}^{3+}:[\mathrm{Ar}] 3 \mathrm{~d}^1 & 1 \\ \text { (C) } \mathrm{Cu}^{+2}:[\mathrm{Ar}] 3 \mathrm{~d}^9 & 1 \\ \text { (D) } \mathrm{Ni}^{+2}:[\mathrm{Ar}] 3 \mathrm{~d}^8 & 2 \\ \text { (E) } \mathrm{Ti}^{+2}:[\mathrm{Ar}] 3 \mathrm{~d}^2 & 2 \end{array}$$
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