Number of complexes from the following with even number of unpaired " \(\mathrm{d}\) " electrons is…

1. 2
2. 1
3. 4
4. 5

Solution

$\begin{aligned} & {\left[\mathrm{V}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+} \rightarrow \mathrm{d}^2 \mathrm{sp}^3} \\ & { }_{23} \mathrm{~V}:-[\mathrm{Ar}] 3 \mathrm{~d}^3 4 \mathrm{~s}^2 \\ & \left.\mathrm{~V}^{+3}:-[\mathrm{Ar}] 3 \mathrm{~d}^2, \mathrm{n}=2 \text { (even number of unpaired } \mathrm{e}^{-}\right) \\ & {\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+} \rightarrow \mathrm{sp}^3 \mathrm{~d}^2} \\ & { }_{24} \mathrm{Cr}:-[\mathrm{Ar}] 3 \mathrm{~d}^5 4 \mathrm{~s}^1 \\ & \left.\mathrm{Cr}^{+2}:-[\mathrm{Ar}] 3 \mathrm{~d}^4, \mathrm{n}=4 \text { (even number of unpaired } \mathrm{e}^{-}\right)\end{aligned}$
$\begin{aligned} & {\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+} \rightarrow \mathrm{sp}^3 \mathrm{~d}^2} \\ & \mathrm{Fe}^{3+}:-[\mathrm{Ar}] 3 \mathrm{~d}^5 4 \mathrm{~s}^0 \\ & \mathrm{n}=5\left(\text { odd number of unpaired } \mathrm{e}^{-}\right) \\ & {\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+} \rightarrow \mathrm{sp}^3 \mathrm{~d}^2} \\ & \mathrm{Ni}:-[\mathrm{Ar}] 3 \mathrm{~d}^8 4 \mathrm{~s}^2 \\ & \left.\mathrm{Ni}^{+3}:-[\mathrm{Ar}] 3 \mathrm{~d}^7, \mathrm{n}=3 \text { (odd number of unpaired } \mathrm{e}^{-}\right) \\ & {\left[\mathrm{Cu}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+} \rightarrow \mathrm{sp}^3 \mathrm{~d}^2} \\ & \mathrm{Cu}:-[\mathrm{Ar}] 3 \mathrm{~d}^9 4 \mathrm{~s}^0 \\ & \left.\mathrm{n}=1 \text { (odd number of unpaired } \mathrm{e}^{-}\right)\end{aligned}$

Asked in: JEE Main 2024 (04 Apr Shift 1)

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