The spin only magnetic moment value (in Bohr | vedclass

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(A) In $Cr(CO)_6$, the central metal atom $Cr$ is in the $0$ oxidation state.The ground state electronic configuration of $Cr$ is $[Ar] 3d^5 4s^1$.$CO

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$0$

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(A) In $Cr(CO)_6$, the central metal atom $Cr$ is in the $0$ oxidation state.The ground state electronic configuration of $Cr$ is $[Ar] 3d^5 4s^1$.$CO$ is a strong field ligand, which causes the pairing of all $6$ electrons ($5$ from $3d$ and $1$ from $4s$) into the $3d$ orbitals.As a result, the number of unpaired electrons $(n)$ is $0$.The spin-only magnetic moment $(\mu)$ 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 (in Bohr magneton units) of $Cr(CO)_6$ is

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