The magnetic moment (spin only) of $[N i C l_{4}]^{2 -}$ is :

1.82 BM

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5.46 BM

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2.82 BM

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1.41 BM

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Solution

The correct option is C 2.82 BM
$C l^{-}$ is a weak field ligand and therefore $d^{8}$ ion will have two unpaired electron.
$μ = \sqrt{n (n + 2)}$ $= \sqrt{24}$ $= 2.82 B . M .$

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