Question

The geometry of $\left[Ni\right(CO{)}_4]$ and $[NiC{l}_4{]}^{2-}$ are respectively:

• A. Both are square planar
• B. Tetrahedral and square planar
• C. Both are tetrahedral
• D. Square planar and tetrahedral

Answer 1

The correct option is C Both are tetrahedral

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

Using VBT, in the case of $\left[Ni\right(CO{)}_4]$,
${}_{28}Ni\to \left[Ar\right]3d^{8}4s^2$

Since, $CO$ (carbonyl) is a strong and neutral ligand
$\therefore$ it pairs up `3d` orbitals to give $\left[Ar\right]3d^{10}4s^o$ configuration. $4s$ and $4p$ orbitals participate in hybridisation shows $s{p}^3$ hybridisation.

Using VBT, in the case of $[NiC{l}_4{]}^{2-}$,
$N{i}^{2+}\to \left[Ar\right]3d^{8}4s^0$
$C{l}^{-}$ is a weak ligand.
$\therefore$ no pairing occurs and it will show $s{p}^3$ hybridisation and thus the geometry will be tetrahedral.