The magnetic moment (spin-only) of [NiCl_4]^{ | vedclass

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(A) In $[NiCl_4]^{2-}$,the oxidation state of $Ni$ is $+2$.The electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$.Since $Cl^-$ is a weak field ligan

Vedclass · Accepted Answer

$2.82$

Vedclass · Answer

(A) In $[NiCl_4]^{2-}$,the oxidation state of $Ni$ is $+2$.The electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8$.Since $Cl^-$ is a weak field ligand,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 $\mu = \sqrt{n(n+2)} \ B.M.$Substituting $n = 2$,we get $\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.82 \ B.M.$

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

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