The spin only magnetic moment value of Cr(CO) | vedclass

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(D) The central metal atom in $Cr(CO)_6$ is $Cr$ in its zero oxidation state $(Cr^0)$.The electronic configuration of $Cr^0$ is $[Ar] 3d^5 4s^1$.$CO$

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(D) The central metal atom in $Cr(CO)_6$ is $Cr$ in its zero oxidation state $(Cr^0)$.The electronic configuration of $Cr^0$ is $[Ar] 3d^5 4s^1$.$CO$ is a strong field ligand,which causes pairing of all electrons in the $3d$ and $4s$ orbitals.After pairing,there are no unpaired electrons $(n = 0)$.The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)} \ BM$.Substituting $n = 0$,we get $\mu = \sqrt{0(0+2)} = 0 \ BM$.

The spin only magnetic moment value of $Cr(CO)_6$ is (in $BM$)

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