Question

What is the spin only magnetic moment value in (Bohr Magneton units) of Cr(CO)6?

Answer 1

In Cr(CO)6, the central Cr atom is d2sp3 hybridised. Hence all the electrons in its valence shell is paired. So it has net magnetic moment zero (0 BM).

The electron configuration is [Ar]3d54s1.We have to accomodate the 6 Ligands and the fact that CO is a strong ligand.

This results in d2sp3 hybridization. Therefore, there are no unpaired electrons in Cr(CO)6. Hence n=0

And the spin only magnetic moment is also 0.

Answer 2

Answer:

$\mathrm{CO}$ exists a strong ligand, all the six electrons of the valence shell of $\mathrm{Cr}$ stand paired and spin only magnetic moment $=0$.

Explanation:

The Bohr magneton μB​ exists as a physical constant and the natural unit for describing the magnetic moment of an electron generated by either its orbital or spin angular momentum.

In $C r(C O)_{B}, C r$is present as $C r^{0}$

$C r$ or$C r^{0}=[A r] 3 d^{5} 4 S^{-1}$

$C O$being strong field ligand pairs up the unpaired electrons of $C r$

In this condition,

number of unpaired electrons, $n=0$

Magnetic moment

$\mu=\sqrt{n(n+2)}$

$&=\sqrt{0(0+2)}$

$&=0 B M$.

$\mathrm{CO}$ exists a strong ligand, all the six electrons of the valence shell  $\mathrm{Cr}$ stand paired and spin only magnetic moment $=0$.

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