Question

Match list-I (Complexes) with list - II (Hybridization of Central Atom) and select the correct answer using the codes given below the lists :

	List - I		List - II
A.	$\left[Ni\right(CO{)}_4]$	1.	$s{p}^3$
B.	$\left[Ni\right(CN{)}_4{]}^{2-}$	2.	$ds{p}^2$
C.	$\left[Fe\right(CN{)}_6{]}^{4-}$	3.	$s{p}^3{d}^2$
D.	$[Mn{F}_6{]}^{4-}$	4.	$d^{2}s{p}^3$
		5.	$s{p}^{2}d$

• A. A-1, B-3, C-2, D-4
• B. A-5, B-2, C-4, D-3
• C. A-5, B-3, C-2, D-4
• D. A-1, B-2, C-4, D-3

Answer 1

The correct option is D A-1, B-2, C-4, D-3

Ni atom in $\left[Ni\right(CO{)}_4]$ is $s{p}^3$ hybridized. This results in a tetrahedral geometry.

Ni atom in $\left[Ni\right(CN{)}_4{]}^{2-}$ is $ds{p}^2$ hybridized. This results in square planar geometry. Here it is showing $ds{p}^2$ hybridization because here $C{N}^{-}$ is strong field ligand hence it forms an inner complex.

Fe atom in $Fe(CN{)}_6{]}^{4-}$ is $d^{2}s{p}^3$ hybridized. This results in an octahedral geometry. Here it is showing $d^{2}s{p}^3$ hybridization because here again cyanide group is present which is strong field ligand which forms inner complexes.

Mn atom in $[Mn{F}_6{]}^{4-}$ is $s{p}^3{d}^2$ hybridized. This results in octahedral geometry.

Therefore, the correct option is D.