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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 $0$.

List-$I$ (Hybridisation)	List-$II$ (Compound/ion)
$A. sp^3d$	$I. [PtCl_4]^{2-}$
$B. sp^3d^2$	$II. SF_6$
$C. dsp^2$	$III. BCl_3$
$D. dsp^3$	$IV. PCl_5$
	$V. ClF_3$

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

Similar Questions

The complex ion which has no $d$ electrons in the central metal atom is

Match the following:
List-$I$ (Hybridisation) List-$II$ (Compound/ion)
$A. sp^3d$ $I. [PtCl_4]^{2-}$
$B. sp^3d^2$ $II. SF_6$
$C. dsp^2$ $III. BCl_3$
$D. dsp^3$ $IV. PCl_5$
$V. ClF_3$

The correct match is:

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Which of the following configurations will show an outer orbital octahedral complex?

What is the hybridization of the complex $[Mn(CN)_6]^{3-}$?

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