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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(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 $\mu = \sqrt{0(0+2)} = 0 \ BM$.

List-$I$ (Complex ion)	List-$II$ (Calculated spin only magnetic moment $(BM)$)
$A$. $[Cr(H_2O)_6]^{2+}$	$I$. $3.87$
$B$. $[Co(H_2O)_6]^{2+}$	$II$. $5.92$
$C$. $[Cu(H_2O)_6]^{2+}$	$III$. $4.90$
$D$. $[Mn(H_2O)_6]^{2+}$	$IV$. $1.73$

The spin only magnetic moment value (in Bohr magneton units) of $Cr(CO)_6$ is

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