The magnetic moment (spin only) of [NiCl_{4}] | vedclass

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(C) In $[NiCl_{4}]^{2-}$,the oxidation state of $Ni$ is $+2$.$Ni^{2+}$ has the electronic configuration $[Ar] 3d^{8}$.Since $Cl^{-}$ is a weak field l

Vedclass · Accepted Answer

$2.82$

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(C) In $[NiCl_{4}]^{2-}$,the oxidation state of $Ni$ is $+2$.$Ni^{2+}$ has the electronic configuration $[Ar] 3d^{8}$.Since $Cl^{-}$ is a weak field ligand,it does not cause pairing of electrons in the $3d$ orbitals.Thus,there are $n = 2$ unpaired electrons.The spin-only magnetic moment is calculated as $\mu = \sqrt{n(n+2)} \ B.M.$$\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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