Question

Which of the following complexes will have a tetrahedral shape?

• A. $[PdC{l}_4{]}^{2-}$
• B. $\left[Pd\right(CN{)}_4{]}^{2-}$
• C. $\left[Ni\right(CN{)}_4{]}^{2-}$
• D. $[NiC{l}_4{]}^{2-}$

Answer 1

The correct option is D $[NiC{l}_4{]}^{2-}$

In case of 3d series, central metal atom of C.N. 4

has tetrahedral shape for weak field ligands & square planar for strong field ligands.

For Pd both ligands act as strong field.

$[NiC{l}_4{]}^{2-}=$ $C{l}^{-}\to$ Weak field ligand.

$\Rightarrow N{i}^{2+}\Rightarrow \left[Ar\right]3d^8$

As the ligand is weak field ligands the complex has tetrahedral geometry of $s{p}^3$ hybridization.