# The spin only magnetic moment value (in Bohr magneton units) of Cr(CO)_"6" is?

Aug 10, 2017

A big fat zero $\text{BM}$.

The spin-only magnetic moment is given by:

${\mu}_{S} = 2.0023 \sqrt{S \left(S + 1\right)}$

where $g = 2.0023$ is the gyromagnetic ratio and $S$ is the total spin of all unpaired electrons in the system. If there are none... then ${\mu}_{S} = 0$.

Spin-only means we ignore the total orbital angular momentum $L = | {\sum}_{i} {m}_{l , i} |$ for the $i$th electrons.

By conservation of charge, "Cr"("CO")_6 has a $\text{Cr}$ atom in its $0$ oxidation state.

For transition metal complexes, the ligand orbitals belong primarily to the ligands, and the metal's orbitals belong primarily to the metal, because the interacting atoms are going to have significantly different electronegativities.

Hence, the unpaired electrons found (if any) are based on the metal oxidation state.

$\text{Cr} \left(0\right)$ brought in $\boldsymbol{6}$ post-noble-gas-core electrons, i.e. $5 \times 3 d + 1 \times 4 s = 6$.

The $\text{CO}$ ligands are $\sigma$ donors (which is obvious, as they must make bonds!) AND $\pi$ acceptors (which is not so obvious)...

...so they are strong-field ligands, which promote low-spin octahedral complexes (large ${\Delta}_{o}$) by lowering the ${t}_{2 g}$ orbital energies (the $\sigma$ donor factor raises the ${e}_{g}^{\text{*}}$ orbital energies). So, we have a low-spin ${d}^{6}$ complex, which has no unpaired electrons.

${\Delta}_{o} \left\{\begin{matrix}{\text{ "" "" "" "bar(color(white)(uarr darr))" "bar(color(white)(uarr darr))" "stackrel(e_g)(" ") \\ " " \\ " " \\ " "" "ul(uarr darr)" "ul(uarr darr)" "ul(uarr darr)" }}_{{t}_{2 g}}\end{matrix}\right.$

Thus, the spin-only magnetic moment is $\boldsymbol{0}$. Don't believe me? Here's some proof of ZERO unpaired electrons: