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(D) The central metal atom is $Ni$ with atomic number $28$. The electronic configuration of $Ni$ is $[Ar] 3d^8 4s^2$.In the complex ion $[Ni(CN)_4]^{2

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(D) The central metal atom is $Ni$ with atomic number $28$. The electronic configuration of $Ni$ is $[Ar] 3d^8 4s^2$.In the complex ion $[Ni(CN)_4]^{2-}$,the oxidation state of $Ni$ is $x + 4(-1) = -2$,which gives $x = +2$.Thus,$Ni^{2+}$ has the configuration $[Ar] 3d^8$.$CN^-$ is a strong field ligand,which causes pairing of electrons in the $3d$ orbitals.After pairing,all $8$ electrons occupy the $d_{xy}, d_{yz}, d_{xz},$ and $d_{z^2}$ orbitals in pairs,leaving no unpaired electrons.Therefore,the number of unpaired electrons is $0$.

Coordination number and type of hybridisation	Distribution of hybrid orbitals in space
$a. 4, sp^3$	$i. \text{trigonal bipyramidal}$
$b. 4, dsp^2$	$ii. \text{octahedral}$
$c. 5, sp^3d$	$iii. \text{tetrahedral}$
$d. 6, d^2sp^3$	$iv. \text{square planar}$

How many number of unpaired electrons are there in the complex ion $[Ni(CN)_4]^{2-}$?

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