Coordination Compounds 2 Question 46
####48. Amongst $Ni(CO) _4,\left[Ni(CN) _4\right]^{2-}$ and $NiCl _4^{2-}$
(1991, 1M)
(a) $Ni(CO) _4$ and $NiCl _4^{2-}$ are diamagnetic and $\left[Ni(CN) _4\right]^{2-}$ is paramagnetic
(b) $\left[NiCl _4\right]^{2-}$ and $\left[Ni(CN) _4\right]^{2-}$ are diamagnetic and $Ni(CO) _4$ is paramagnetic
(c) $Ni(CO) _4$ and $\left[Ni(CN) _4\right]^{2-}$ are diamagnetic and $\left[NiCl _4\right]^{2-}$ is paramagnetic
(d) $Ni(CO) _4$ is diamagnetic and $\left[NiCl _4\right]^{2-}$ and $\left[Ni(CN) _4\right]^{2-}$ are paramagnetic
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Solution:
- In $Ni(CO) _4, Ni$ has $3 d^{10}$-configuration, diamagnetic. In $\left.Ni(CN) _4\right]^{2-}$, $Ni$ has $3 d^{8}$-configuration but due to strong ligand field, all the $d$-electrons are spin paired giving $d s p^{2}$-hybridisation, diamagnetic.
$\operatorname{In}\left[NiCl _4\right]^{2-}$, Ni has $3 d^{8}$-configuration and there is two unpaired electrons (weak chloride ligand do not pair up $d$-electrons), hence paramagnetic.
