The magnetic moment of [NiCl_4]^{2-} is .... | vedclass

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(C) In $[NiCl_4]^{2-}$,the oxidation state of $Ni$ is $+2$. The atomic number of $Ni$ is $28$,so the electronic configuration of $Ni^{2+}$ is $[Ar] 3d

Vedclass · Accepted Answer

$2.82$

Vedclass · Answer

(C) In $[NiCl_4]^{2-}$,the oxidation state of $Ni$ is $+2$. The atomic number of $Ni$ is $28$,so the electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$. $Cl^-$ is a weak field ligand,so it does not cause pairing of electrons in the $3d$ orbitals. Thus,there are $2$ unpaired electrons in the $3d$ orbitals. The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ B.M.$,where $n$ is the number of unpaired electrons. Substituting $n = 2$,we get $\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.82 \ B.M.$

The magnetic moment of $[NiCl_4]^{2-}$ is .... $B.M.$ (spin-only).

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