Chemistry, asked by kalyankv4035, 1 year ago

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

Answers

Answered by Anonymous
11

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.

Answered by nafibarli789
0

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 $=0$.

#SPJ2

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