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Amongst $\mathrm{Ni}(\mathrm{CO})_{4},\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ and $\left[\mathrm{NiCl}_{4}^{2-}\right]$
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Verified Answer
The correct answer is:
$\mathrm{Ni}(\mathrm{CO})_{4}$ and $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ are diamagnetic but $\mathrm{NiCl}_{4}^{2-}$ is paramagnetic
The electronic configuration of $\mathrm{Ni}$ is
$$
\begin{array}{l}
\mathrm{Ni}(28)=[\mathrm{Ar}] 3 \mathrm{~d}^{8}, 4 \mathrm{~s}^{2} \\
\mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{8}
\end{array}
$$
Both $\mathrm{Ni}$ and $\mathrm{Ni}^{2+}$ have two unpaired electrons.
$\mathrm{CO}$ and $\mathrm{CN}^{-}$are strong field ligands and thus unpaired electrons get paired. Hence, $\mathrm{Ni}(\mathrm{CO})_{4}$ and $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ are diamagnetic. $\mathrm{Cl}^{-}$is a weak field ligand hence, no pairing of $\mathrm{e}^{-}$will take place. hence $\mathrm{NiCl}_{4}^{2-}$ is paramagnetic.
$$
\begin{array}{l}
\mathrm{Ni}(28)=[\mathrm{Ar}] 3 \mathrm{~d}^{8}, 4 \mathrm{~s}^{2} \\
\mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{8}
\end{array}
$$
Both $\mathrm{Ni}$ and $\mathrm{Ni}^{2+}$ have two unpaired electrons.
$\mathrm{CO}$ and $\mathrm{CN}^{-}$are strong field ligands and thus unpaired electrons get paired. Hence, $\mathrm{Ni}(\mathrm{CO})_{4}$ and $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ are diamagnetic. $\mathrm{Cl}^{-}$is a weak field ligand hence, no pairing of $\mathrm{e}^{-}$will take place. hence $\mathrm{NiCl}_{4}^{2-}$ is paramagnetic.
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