Match List-I with List-II, Coordination Compounds-121

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Solution

Q.
Match List-I with List-II [2024]

List-I (Complex ion): List-II (Spin only magnetic moment in B.M.)

(A) ${[Cr {({NH}_{3})}_{6}]}^{3 +}$ (I) 4.90

(B) ${[{NiCl}_{4}]}^{2 -}$ (II) 3.87

(C) ${[{CoF}_{6}]}^{3 -}$ (III) 0.0

(D) ${[Ni {(CN)}_{4}]}^{2 -}$ (IV) 2.83

Choose the correct answer from the options given below:

1 (A)-(I), (B)-(IV), (C)-(II), (D)-(III)

2 (A)-(II), (B)-(III), (C)-(I), (D)-(IV)

3 (A)-(II), (B)-(IV), (C)-(I), (D)-(III)

4 (A)-(IV), (B)-(III), (C)-(II), (D)-(I)

Ans.
(3)

(A) ${[Cr {({NH}_{3})}_{6}]}^{3 +}$

Oxidation number of Cr = +3

${Cr}^{3 +} = {}_{18}{[Ar] 3 d^{3}}$

As inner d orbitals are available, it undergoes $d^{2} s p^{3}$ hybridization.

Orbital diagram of ${Cr}^{3 +}$ in presence of ${NH}_{3}$

It has 3 unpaired electrons. Spin only magnetic moment $(μ) = \sqrt{n (n + 2)} B M = \sqrt{3 (3 + 2)} B M = 3.87 B M$

(B) ${[{NiCl}_{4}]}^{2 -}$

Oxidation number of Ni = +2

${Ni}^{2 +} = {}_{18}{[Ar] 3 d^{8}}$

${Cl}^{-}$ is weak field ligand hence does not cause pairing of $3 d^{8}$ electrons, ${Ni}^{2 +}$ undergoes $s p^{3}$ hybridization and lone pairs from ${Cl}^{-}$ are accommodated in $s p^{3}$ orbitals.

Orbital diagram of ${Ni}^{2 +}$ in presence of ${Cl}^{-}$

As it contains 2 unpaired electrons. Spin only magnetic moment $(μ) = \sqrt{n (n + 2)} B M = \sqrt{2 (2 + 2)} B M = 2.82 B M$

(C) ${[{CoF}_{6}]}^{3 -}$

Oxidation number of Co = +3

${Co}^{3 +} = {}_{18}{[Ar] 3 d^{6}}$

$F^{-}$ is a weak field ligand. It cannot pair up d electrons. Thus, 4d orbitals are used for $s p^{3} d^{2}$ hybridization. Six $e^{-}$ pairs from six $F^{-}$ are accommodated in these six $s p^{3} d^{2}$ orbitals.

Orbital diagram of ${Co}^{3 +}$ in presence of $F^{-}$

It has 4 unpaired electrons. Spin only magnetic moment $(μ) = \sqrt{n (n + 2)} B M = \sqrt{4 (4 + 2)} B M = 4.9 B M$

(D) ${[Ni {(CN)}_{4}]}^{2 -}$

Oxidation number of Ni = +2

${Ni}^{2 +} = {}_{18}{[Ar] 3 d^{8}}$

${CN}^{-}$ is strong field ligand hence causes pairing of $3 d^{8}$ electrons, then causes $d s p^{2}$ hybridization using 3d, 4s, and 4p orbitals.

Orbital diagram of $N i^{2 +}$ in presence of $C N^{-}$

It has no unpaired electron. Spin only magnetic moment $(μ) = \sqrt{n (n + 2)} B M = \sqrt{0 (0 + 2)} B M = 0 B M$

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