QJEEM21Ch7421: JEE Main 2021 Chemistry - Class 12 JEEM21Ch7421 Solution: According to the valence bond theory the hybridization of…

According to VBT i.e. valence bond theory,

Electronic configuration of Ni = [Ar]3d⁸4s².

(a) NiCl₂ . 6H₂O

NiCl₂ . 6H₂O $\rightarrow$ NiCl₂ + 6H₂O

Oxidation number of Ni(x) = x + 2( $-$ 1) = 0; where x, $-$ 1 and 2 are the oxidation number of Ni, oxidation number of Cl and number of Cl atoms respectively.

NiCl₂ $\Rightarrow$ x + ( $-$ 2) = 0 $\Rightarrow$ x = 2

Electronic configuration of Ni²⁺ = [Ar]3d⁸4s⁰

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

For C.N. = 6


(b) K₂[Ni(CN)₄]

K₂[Ni(CN)₄] $\rightarrow$ 2K⁺ + [Ni(CN)₄]^{2 $-$}

x + 4( $-$ 1) $-$ ( $-$ 2) = 0; where x, 4, $-$ 1 and $-$ 2 are the oxidation number of Ni, number of CN ligands, charge on one CN and charge on complex.

[Ni(CN)₄]^{2 $-$} $\Rightarrow$ x $-$ 4 + 2 = 0 $\Rightarrow$ x = +2

Electronic configuration of Ni²⁺ $\Rightarrow$ [Ar]3d⁸4s⁰

CN ^$-$ is a strong field ligand. So, pairing of electrons occur. For C.N. = 4


(c) Ni(CO)₄

CO is neutral and strong field ligand. So, pairing of electrons occur.

Oxidation number of Ni is zero

Electronic configuration of Ni = [Ar]3d¹⁰4s⁰

For C.N. = 4


(d) Na₂[NiCl₄]

Na₂[NiCl₄] $\rightarrow$ 2Na + [NiCl₄]^{2 $-$}

x + 4( $-$ 1) $-$ ( $-$ 2) = 0; where x, 4, $-$ 1 and $-$ 2 are the oxidation number of Ni, number of CN ligands, charge on one CN and charge on complex.

[NiCl₄]^{2 $-$} $\Rightarrow$ x $-$ 4 + 2 = 0 $\Rightarrow$ x = +2

Electronic configuration of Ni²⁺ = [Ar]3d⁸4s⁰

For C.N. = 4


Hence, only K₂[Ni(CN)₄] has dsp² hybridization.