Question

Match the complex in List-I with the hybridisation of the central ion given in List-II.

Answer 1

(A) $Ni\left(C{O}_4\right)$: It has a coordination number of $4$ as four atoms of ligands are present, so hybridisation is $s{p}^3$.

(B) $\left[Ni\right(CN{)}_4{]}^{2-}$: It has a coordination number of $4$ as four atoms of ligands is present. So, hybridisation is $s{p}^3$.

(C) $\left[Fe\right(CN{)}_6{]}^{4-}$: It has a coordination number of $6$, so an octahedral complex will be formed with a hybridsation of $d^{2}s{p}^3$ as cyanide $\left(C{N}^{-}\right)$ is a strong field ligand so it will form a inner orbital complex.

(D) $[Mn{F}_6{]}^{4-}$: Here, the ligand is a weak field ligand and the coordination number is $6$. So, the complex will be octahedral but an outer complex with hybridisation of $s{p}^3{d}^2$.