Question

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

• A. $Ni(CO{)}_4$ and $NiC{l}_4^{2-}$ are diamagnetic and $[Ni(CN{)}_4{]}^{2-}$ is paramagnetic
• B. $[NiC{l}_4{]}^{2-}$ and $[Ni(CN{)}_4{]}^{2-}$ are diamagnetic and $Ni(CO{)}_4$ is paramagnetic
• C. $Ni(CO{)}_4$ and $[Ni(CN{)}_4{]}^{2-}$ are diamagnetic and $[NiC{l}_4{]}^{2-}$ is paramagnetic
• D. $Ni(CO{)}_4$ is diamagnetic and $[NiC{l}_4{]}^{2-}$ and $[Ni(CN{)}_4{]}^{2-}$ are paramagnetic

Answer 1

Answer: (C)

In $Ni(CO{)}_4,Ni$ has $3d^{10}$ -configuration, diamagnetic. In $Ni(CN{)}_4{]}^{2-},Ni$ has $3d^8$ -configuration but due to strong ligand field, all the ($d$-electrons are spin paired giving $ds{p}^2$ -hybridisation diamagnetic.
In $[NiC{l}_4{]}^{2-},Ni$ has $3d^8$ -configuration and there is two unpaired electrons (weak chloride ligand do not pair up $d$ - electrons), hence paramagnetic.