Match List I with List II, Coordination Compounds - 1

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Solution

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

List I List II

A. $K_{2} [Ni {(CN)}_{4}]$ I. $s p^{3}$

B. $[Ni {(CO)}_{4}]$ II. $s p^{3} d^{2}$

C. $[Co {({NH}_{3})}_{6}] {Cl}_{3}$ III. $d s p^{2}$

D. ${Na}_{3} [{CoF}_{6}]$ IV. $d^{2} s p^{3}$

Choose the correct answer from the options given below:

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

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

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

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

Ans.
(4)

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

Oxidation number of Ni = +2

${CN}^{-}$ 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.

B. $[Ni {(CO)}_{4}]$ :

oxidation number of Ni = 0

CO is a strong field ligand. It shifts 4s electrons to 3d orbital to give $3 d^{10}$ configuration.

4s orbital becomes vacant. Ni then undergoes $s p^{3}$ hybridization and lone pairs from CO are accommodated in $s p^{3}$ orbitals.

C. ${[Co {({NH}_{3})}_{6}]}^{3 +}$

Oxidation number of Co = +3

${NH}_{3}$ is a strong field ligand. It pairs up d electrons. After pairing d electrons, two 3d orbitals are vacant. ${Co}^{3 +}$ undergoes $d^{2} {sp}^{3}$ hybridisation by using two 3d orbitals. Six $e^{-}$ pairs from six ${NH}_{3}$ are accommodated in these six $d^{2} {sp}^{3}$ orbitals.

D. ${[{CoF}_{6}]}^{3 -}$

Oxidation number of Co = +3

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

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