Coordination Compounds - Result Question 58
####65. The $d$ electron configurations of $Cr^{2+}, Mn^{2+}, Fe^{2+}$ and $Ni^{2+}$ are $3 d^{4}, 3 d^{5}, 3 d^{6}$ and $3 d^{8}$ respectively. Which one of the following aqua complexes will exhibit the minimum paramagnetic behaviour?
[2007]
(a) $[Fe(H_2 O)_6]^{2+}$
(b) $[Ni(H_2 O)_6]^{2+}$
(c) $[Cr(H_2 O)_6]^{2+}$
(d) $[Mn(H_2 O)_6]^{2+}$
(At. No. $Cr=24, Mn=25, Fe=26, Ni=28$ )
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Solution:
- (b) Lesser is the number of unpaired electrons smaller will be the paramagnetic behaviour. As $Cr^{2+}, Mn^{2+}, Fe^{2+} & Ni^{2+}$ contains.
$=4$ unpaired $e^{-}$.
$Mn^{2+}(3 d^{5})=$\begin{tabular}{|l|l|l|l|l|} \hline 1 & 1 & 1 & 1 & 1 \\ \hline \end{tabular}
$=5$ unpaired $e^{-}$.
$Fe^{2+}(3 d^{6})=$\begin{tabular}{|l|l|l|l|l|} \hline 1 & 1 & 1 & 1 & 1 \\ \hline \end{tabular}
$=4$ unpaired $e^{-}$.
$Ni^{2+}(3 d^{8})=$\begin{tabular}{|l|l|l|l|l|} \hline $\mathbb{1}$ & $\mathbb{1}$ & $\mathbb{1}$ & 1 & 1 \\ \hline \end{tabular}
$=2$ unpaired $e^{-}$.
As $Ni^{2+}$ has minimum no. of unpaired $e^{-}$thus this is least paramagnetic.
