The $d$ electron configuration of $\mathrm{Cr}^{2+}$, $\mathrm{Mn}^{2+}, \mathrm{Fe}^{2+}$ and…

The $d$ electron configuration of $\mathrm{Cr}^{2+}$, $\mathrm{Mn}^{2+}, \mathrm{Fe}^{2+}$ and $\mathrm{Ni}^{2+}$ are $3 \mathrm{~d}^4, 3 \mathrm{~d}^5, 3 \mathrm{~d}^6$ and $3 \mathrm{~d}^8$ respectively. Which one of the following aqua complexes will exhibit the minimum paramagnetic behaviour? (At. No. $\mathrm{Cr}=24, \mathrm{Mn}=25, \mathrm{Fe}=26, \mathrm{Ni}=$ 28)
  1. $\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$
  2. $\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$
  3. $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$
  4. $\left[\mathrm{Mn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$

Solution

$\begin{aligned} & \mathrm{Cr}^{++}\left(3 \mathrm{~d}^4\right)=\begin{array}{|l|l|l|l|l|} \hline \upharpoonleft & \upharpoonleft & \upharpoonleft & \upharpoonleft & \\ \hline \end{array} \\ & =4 \text { unpaired } e^{-} \\ & \operatorname{Mn}^{++}\left(3 \mathrm{~d}^4\right)=\begin{array}{|l|l|l|l|l|} \hline \upharpoonleft & \upharpoonleft & \upharpoonleft & \upharpoonleft & \upharpoonleft \\ \hline \end{array} \\ & =5 \text { unpaired } e^{-} \\ & \mathrm{Fe}^{++}\left(3 \mathrm{~d}^6\right)=\begin{array}{|l|l|l|l|l|} \hline \upharpoonleft \downharpoonright & \upharpoonleft & \upharpoonleft & \upharpoonleft & \upharpoonleft \\ \hline \end{array} \\ & =4 \text { unpaired } e^{-} \\ & \mathrm{Ni}^{++}\left(3 \mathrm{~d}^8\right)=\begin{array}{|l|l|l|l|l|} \hline \upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright & \upharpoonleft \downharpoonright & \upharpoonleft & \upharpoonleft \\ \hline \end{array} \\ & =2 \text { unpaired } e^{-} \end{aligned}$ As $\mathrm{Ni}^{++}$ has minimum no. of unpaired $e^{-}$ thus this is least paramagnetic.

Asked in: NEET 2007

Practice more Coordination Compounds questions on Aicharya