# A particular compound of vanadium show a magnetic moment of 3.86BM the oxidation state of vanadium in the compound is( Z=23)?

Jul 25, 2017

${\text{V}}^{+ 2}$, probably (such as in ${\text{VCl}}_{2}$), but it could potentially be ${\text{V}}^{0}$, depending on the particular compound in question. It would not be an issue if the compound is octahedral.

It's not entirely clear what kind of magnetic moment this is. Is it spin-only, or based on spin and orbital angular momentum? I'm guessing spin-only, because it is reasonable to solve.

(We would have needed to know how many unpaired electrons there were in the first place, which we should not yet know, to know how to get the total magnetic moment, if given.)

The magnetic moment is given by (Miessler et al.):

$\mu = 2.0023 \sqrt{S \left(S + 1\right) + \frac{1}{4} L \left(L + 1\right)}$

where:

• $S = {\sum}_{i} | {m}_{s , i} |$ is the total spin of all the electrons in the compound. From this you could find the number of unpaired electrons only (but you will have no idea how many paired electrons!).
• $L = {\sum}_{i} | {m}_{l , i} |$ is the total orbital angular momentum of the orbitals in which each electron resides, and is found by summing the magnetic quantum number for each orbital, for each electron in that orbital, then taking the absolute value of the sum.
• $g = 2.0023$ is the gyromagnetic ratio.

But the spin-only magnetic moment is given by

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

If ${\mu}_{S} = \text{3.86 BM}$ is correct for the spin-only magnetic moment, then...

${\left(\frac{3.86}{2.0023}\right)}^{2} = S \left(S + 1\right) = {S}^{2} + S$

Thus, we have a quadratic equation to solve:

${S}^{2} + S - 3.7163 = 0$

Solve this to obtain $S = 1.492 \approx 1.5$ as the physically-reasonable total spin. This means we have:

$S = | \left(- \frac{1}{2}\right) + \left(- \frac{1}{2}\right) + \left(- \frac{1}{2}\right) | = | \left(+ \frac{1}{2}\right) + \left(+ \frac{1}{2}\right) + \left(+ \frac{1}{2}\right) | = \frac{3}{2}$

$= 1.5 \approx 1.492$,

or $\boldsymbol{3}$ unpaired $3 d$ electrons. However, we do not know what compound this was.

This would be an issue. Since vanadium has no more than $5$ valence electrons, i.e.

$\text{V} : \left[A r\right] 3 {d}^{3} 4 {s}^{2}$,

there is no other way for there to be $3$ unpaired electrons than if it were $\textcolor{b l u e}{{\text{V}}^{+ 2}}$ (which lost two $4 s$ electrons to the surrounding ligands' orbitals), except maybe as ${\text{V}}^{0}$.

It could be either one, but is more likely the $+ 2$.

• If any electrons were lost to the ligands, then it would have to be ${\text{V}}^{+ 2}$ (such as in ${\text{VCl}}_{2}$).
• If all the ligands are neutral and the complex is neutral as well, it could be ${\text{V}}^{0}$. However, "V"("CO")_6 is an example with not $3$ unpaired electrons, but $1$.