Question

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

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

Answer 1

The correct option is D Tetrahedral and square planar

In the case of $\left[Ni\right(CO{)}_4]$ the Ni possess the electronic configuration of $3d^{8}4s^2$ but due to the presence of strong field ligand like CO the electrons in 4s orbitals gets paired up with unpaired electrons in 3d orbital and hence forms a outer orbital complex of $s{p}^3$ i.e tetrahedral. In the case of $[PdC{l}_4{]}^{2-}$ the 4d orbital of $P{d}^{2+}$ is diffused and therefore forms strong overlap with the orbitals of Cl and this leads to a larger splitting in the d orbitals and hence it forms square planar complex. Therefore the correct option is D.