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What is the spin only magnetic moment value in (Bohr Magneton units) of Cr(CO)$_6$?

$\begin {array} {1 1} 0 \\ 2.84 \\ 4.90 \\ 5.92 \end {array}$

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Answer: 0
The electron configuration is $[Ar]3d^54s^1$.We have to accomodate the 6 Ligands and the fact that CO is a strong ligand.
This results in $d^2sp^3$ hybridization. Therefore, there are no unpaired electrons in $Cr(CO)_6$. Hence $n = 0$
And the spin only magnetic moment is also 0.
answered Jan 30, 2014 by sreemathi.v
edited Aug 5, 2014 by balaji.thirumalai

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