Chemistry › Coordination Compounds › Werner's theory and Valence Bond Theory
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)
$\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ $\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ $\left[\mathrm{Mn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$
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
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