Question

The spin only magnetic moment value (in $$B.M.$$ unit) of $Cr(CO)_{6}$ is

• A. $0$
• B. $2.84$
• C. $4.90$
• D. $5.92$

Answer 1

Answer: (A)

$0$

Answer 2

In $Cr(CO)_{B}$, $Cr$ is present as $Cr^{0}$ $Cr$ or $Cr^{0}=[Ar] 3d^{5} \, 4S^{-1}$ $CO$ being strong field ligand pair up the unpaired electrons of $Cr$ In this condition, number of unpaired electrons, $n=0$ Magnetic moment, $\mu= \sqrt{n\left(n+2\right)}$ $=\sqrt{0\left(0+2\right)}$ $=0 BM$