Question

Match the column and choose the correct option.

	List-I (Complex)		List-II (Geometry and Magnetic behaviour)
(P)	$[Ni(CN)_4]^{2-}$	(1)	Octahedral (inner)
(Q)	$[MnF_6]^{4-}$	(2)	Paramagnetic
(R)	$[Fe(CN)_6]^{3-}$	(3)	Square planar
(S)	$[Cr(NH_3)_6]^{3+}$	(4)	Octahedral (outer)
		(5)	Diamagnetic

Answer 1

(P)-3,5; (Q)-4,2; (R)-1,2; (S)-1,2

Answer 2

1. $[Ni(CN)_4]^{2-}$: Ni is +2 ($3d^8$). CN$^-$ is a strong field ligand. This leads to $dsp^2$ hybridization, resulting in a square planar geometry. The $3d^8$ configuration with a strong field ligand results in all electrons being paired, making it diamagnetic. (P) matches with (3) and (5).

2. $[MnF_6]^{4-}$: Mn is +2 ($3d^5$). F$^-$ is a weak field ligand. This leads to $sp^3d^2$ hybridization, resulting in an outer octahedral geometry. The $3d^5$ configuration with a weak field ligand has 5 unpaired electrons, making it paramagnetic. (Q) matches with (4) and (2).

3. $[Fe(CN)_6]^{3-}$: Fe is +3 ($3d^5$). CN$^-$ is a strong field ligand. This leads to $d^2sp^3$ hybridization, resulting in an inner octahedral geometry. The $3d^5$ configuration with a strong field ligand has 1 unpaired electron, making it paramagnetic. (R) matches with (1) and (2).

4. $[Cr(NH_3)_6]^{3+}$: Cr is +3 ($3d^3$). NH$_3$ is a moderately strong field ligand. This leads to $d^2sp^3$ hybridization, resulting in an inner octahedral geometry. The $3d^3$ configuration has 3 unpaired electrons, making it paramagnetic. (S) matches with (1) and (2).