(A) In $[Ni(CO)_4]$,the oxidation state of $Ni$ is $0$. The electronic configuration of $Ni$ $(Z=28)$ is $[Ar] 3d^8 4s^2$. Since $CO$ is a strong fiel

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(A) In $[Ni(CO)_4]$,the oxidation state of $Ni$ is $0$. The electronic configuration of $Ni$ $(Z=28)$ is $[Ar] 3d^8 4s^2$. Since $CO$ is a strong field ligand,it causes pairing of electrons in the $3d$ orbital. The $4s$ electrons shift to the $3d$ orbital,resulting in a $3d^{10}$ configuration. The hybridization is $sp^3$,which corresponds to a tetrahedral geometry.

Class 12 Chemistry Coordination Compounds Hybridisation and Geometry

Medium

What is the geometry of $[Ni(CO)_4]$?

A
Tetrahedral
B
Square planar
C
Octahedral
D
Trigonal bipyramidal

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Coordination Compounds

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Hybridisation and Geometry

Which one of the following is wrongly matched?

EasyKCET 2010

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Which ion has tetrahedral geometry:

Medium

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How many different types of $O-B-O$ angles are there in sodium peroxoborate?

Difficult

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Match the following based on valence bond theory $(VBT)$.
Hybridisation Geometry Complex structure
$(A) \ sp^3$ $(i) \ \text{Square planar}$ $(p) \ [Fe(CN)_6]^{3-}$
$(B) \ d^2sp^3$ $(ii) \ \text{Tetrahedral}$ $(q) \ [ZnCl_4]^{2-}$
$(C) \ dsp^2$ $(iii) \ \text{Octahedral}$ $(r) \ [Ni(NH_3)_4]^{2+}$
- $(iv) \ \text{Linear}$ $(s) \ [Ag(CN)_2]^-$

Hybridisation	Geometry	Complex structure
$(A) \ sp^3$	$(i) \ \text{Square planar}$	$(p) \ [Fe(CN)_6]^{3-}$
$(B) \ d^2sp^3$	$(ii) \ \text{Tetrahedral}$	$(q) \ [ZnCl_4]^{2-}$
$(C) \ dsp^2$	$(iii) \ \text{Octahedral}$	$(r) \ [Ni(NH_3)_4]^{2+}$
-	$(iv) \ \text{Linear}$	$(s) \ [Ag(CN)_2]^-$