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Q.
The spin only magnetic moment value (in Bohr magneton units) of $Cr ( CO )_{6}$ is
Coordination Compounds
Solution:
In $Cr ( CO )_{6}, CO$ is a neutral ligand, so $Cr$ is in zero oxidation state, but since it is a strong field ligand, it causes pairing of all the electrons.
$\therefore \,\, n\left(\right.$ No. of unpaired $\left.e^{-}\right)=0 \Rightarrow \mu=\sqrt{n(n+2)}=0$
