Question

What is the geometry of $\left[Ni\right(CN{)}_4{]}^{2-},[NiC{l}_4{]}^{2-}$ and $\left[Ni\right(CO{)}_4]$?

• A. $[NiC{l}_4{]}^{2-}$ is square planar and $\left[Ni\right(CN{)}_4{]}^{2-},\left[Ni\right(CO{)}_4]$ are tetrahedral
• B. $\left[Ni\right(CO{)}_4]$ is square planar and $\left[Ni\right(CN{)}_4{]}^{2-},[NiC{l}_4{]}^{2-}$ are tetrahedral
• C. $\left[Ni\right(CN{)}_4{]}^{2-}$ is square planar and $[NiC{l}_4{]}^{2-},[Ni\left(CO{)}_4\right]$ are tetrahedral
• D. None of the above

Answer 1

The correct option is C $\left[Ni\right(CN{)}_4{]}^{2-}$ is square planar and

$[NiC{l}_4{]}^{2-},[Ni\left(CO{)}_4\right]$ are tetrahedral

$\left[Ni\right(CN{)}_4{]}^{2-}$ is a square planar geometry formed by $ds{p}^2$ hybridisation and not tetrahedral by sp3.

$[NiC{l}_4{]}^{2-}$, there is $N{i}^{2+}$ ion, However, in the presence of weak field $C{l}^{-}$ ligands, NO pairing of d-electrons occurs. Therefore, $N{i}^{2+}$ undergoes $s{p}^3$ hybridization to make bonds with $C{l}^{-}$ ligands in tetrahedral geometry

$\left[Ni\right(CO{)}_4]$ is also tetrahedral because of $s{p}^3$

Hence C is correct