Question

Amongst $Ni(CO{)}_4$, $\left[Ni\right(CN{)}_4{]}^{2-}$ and $NiC{l}_4^{2-}$:

• A. $Ni(CO{)}_4$, $\left[Ni\right(CN{)}_4{]}^{2-}$ are diamagnetic and $NiC{l}_4^{2-}$ is paramagnetic
• B. $Ni(CO{)}_4$, $\left[Ni\right(CN{)}_4{]}^{2-}$ are $ds{p}^2$ hybridised and $NiC{l}_4^{2-}$ is $s{p}^3$ hybridised
• C. $\left[Ni\right(CN{)}_4{]}^{2-}$ is square planar and $NiC{l}_4^{2-}$, $Ni(CO{)}_4$ are tetrahedral
• D. $Ni(CO{)}_4$, $\left[Ni\right(CN{)}_4{]}^{2-}$ are paramagnetic and $NiC{l}_4^{2-}$ is diamagnetic

Answer 1

Answer: (A), (C)

The correct options are
A $Ni(CO{)}_4$, $\left[Ni\right(CN{)}_4{]}^{2-}$ are diamagnetic and $NiC{l}_4^{2-}$ is paramagnetic
C $\left[Ni\right(CN{)}_4{]}^{2-}$ is square planar and $NiC{l}_4^{2-}$, $Ni(CO{)}_4$ are tetrahedral

In $Ni(CO{)}_4$, $Ni$ has $3d^{10}$ configuration, therefore it is diamagnetic in nature and has a tetrahedral structure and is $s{p}^3$ hybridised.
In $\left[Ni\right(CN{)}_4{]}^{2-}$, $Ni$ has $3d^8$ configuration but due to $C{N}^{-}$ being a strong field ligand, all the electrons are spin paired giving $ds{p}^2$ hybridisation, square planar structure and is diamagnetic.
​In $NiC{l}_4^{2-}$, $Ni$ has $3d^8$ configuration but due to $C{l}^{-}$ being a weak field ligand, electrons will not pair giving $s{p}^3$ hybridised tetrahedral structure and is paramagnetic.
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