QJEEAdv21Ch2bdb: JEE Advanced 2021 Chemistry - Class 12 JEEAdv21Ch2bdb Solution: The pair(s) of complexes wherein both exhibit tetrahedral…

(a) ${[FeC{l_4}]^ - }$

Atomic number of Fe = 26

Oxidation number of Fe = x + 4 ( $-$ 1) = $-$ 1

$\Rightarrow$ x = 3

Electronic configuration of Fe = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s²

Electronic configuration of Fe³⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s⁰

Cl^- is a weak field ligand, so no pairing of electrons occurs.


Hybridisation is sp³

Geometry is tetrahedral.

${[Fe{(CO)_4}]^{2 - }}$ Oxidation number of Fe = x + 4(0) = - 2 $\Rightarrow$ x = - 2

Electronic configuration of Fe^2- = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁸ 4s²



CO is a strong field ligand and cause electrons of 3d orbitals to pair up and shift electrons of 4s to 3d orbital.



One 4s and three 4p orbitals undergoes hybridisation to form four sp³ hybrid orbitals which arrange into a tetrahedral geometry.



Each of the four sp³ hybrid orbitals accept a pair of electrons from carbonyl ligand.



Geometry is tetrahedral.

(b) ${[Co{(CO)_4}]^ - }$ Atomic number of Co = 27

Oxidation number of Co = x + 4(0) = $-$ 1 $\Rightarrow$ x = $-$ 1

Electronic configuration of Co = 1s²2s²2p⁶3s²3p⁶3d⁷4s²

Electronic configuration of Co^- = 1s²2s²2p⁶3s²3p⁶3d⁸4s²

CO is a strong field ligand, so pairing of electrons occurs.


Hybridisation is sp³

Geometry is tetrahedral.

${[CoC{l_4}]^{2 - }}$

Oxidation number of Co = x + 4( $-$ 1) = $-$ 2

$\Rightarrow$ x = + 2

Electronic configuration of Co²⁺ = 1s²2s²2p⁶3s²3p⁶3d⁷4s⁰

Cl^- is a weak field ligand, so pairing of electrons do not occur.


Hybridisation is sp³

Geometry is tetrahedral

(c) $[Ni{(CO)_4}]$ Oxidation number of Ni = x + 4(0) = 0 $\Rightarrow$ x = 0

Atomic number of Ni = 28

Electronic configuration of Ni = 1s²2s²2p⁶3s²3p⁶3d⁸4s²

CO is a strong field ligand, so pairing of electrons occur.


Hybridisation is sp³

Geometry is tetrahedral.

${[Ni{(CN)_4}]^{2 - }}$

Oxidation number of Ni = x + 4( $-$ 1) = $-$ 2 $\Rightarrow$ x = + 2

Electronic configuration of Ni²⁺ = 1s²2s²2p⁶3s²3p⁶3d⁸4s⁰

CN is a strong field ligand, so pairing of electrons occur.


Hybridisation is dsp².

Geometry is square planar.

(d) ${[Cu{(py)_4}]^ + }$

Oxidation number of Cu = x + 4(0) = + 1 $\Rightarrow$ x = + 1

Atomic number of Cu = 29

Electronic configuration of Cu = 1s²2s²2p⁶3s²3p⁶3d¹⁰4s¹

Electronic configuration of Cu⁺ = 1s²2s²2p⁶3s²3p⁶3d¹⁰4s⁰


Hybridisation is sp³

Geometry is tetrahedral.

${[Cu{(CN)_4}]^{3 - }}$

Oxidation number of Cu = x + 4( $-$ 1) = $-$ 3 $\Rightarrow$ x = + 1


Hybridisation is sp³

Geometry is tetrahedral.