# How many unpaired electrons are in a transition metal complex that has a spin-only magnetic moment of sqrt(15)?

Aug 9, 2016

I got $3$ unpaired electrons.

The spin-only magnetic moment is written as:

$\setminus m a t h b f \left({\mu}_{S} = 2.00023 \sqrt{S \left(S + 1\right)}\right)$

where:

• $g = 2.00023$ is the gyromagnetic ratio.
• $S$ is the total spin of all electrons in the atom. Paired electrons contribute $0$ to $S$ because $+ \frac{1}{2} + \left(- \frac{1}{2}\right) = 0$.

Hence, solving for $S$ allows us to determine the number of unpaired electrons.

There would also be the contribution by the orbital magnetic moment, ${\mu}_{L}$, but to find the unpaired electrons in the ion of a relatively light transition metal (i.e. first/second row transition metals), it's sufficiently accurate to use ${\mu}_{S}$.

For example, for ${\text{Fe}}^{3 +}$, a ${d}^{5}$ metal, the calculated ${\mu}_{S} = 2.00023 \sqrt{\frac{5}{2} \left(\frac{5}{2} + 1\right)} = 5.92$, whereas the observed (including spin and orbital magnetic moments) ${\mu}_{S + L} \approx 5.9$, which is pretty excellent agreement.

What we have is ${\mu}_{S} \approx \sqrt{15}$ bohr magnetons. Therefore:

$\sqrt{15} = 2.00023 \sqrt{S \left(S + 1\right)}$

$15 = {2.00023}^{2} \left(S \left(S + 1\right)\right)$

$\frac{15}{2.00023} ^ 2 = {S}^{2} + S$

$0 = {S}^{2} + S - \frac{15}{2.00023} ^ 2$

Solve the quadratic formula to get:

$\textcolor{b l u e}{S \approx 1.5 \implies \frac{3}{2}}$

Therefore, the total spin $S$ is $\frac{3}{2}$, the number of unpaired electrons is $\setminus m a t h b f \left(3\right)$, and the atomic ion could be either a ${d}^{3}$ or ${d}^{7}$ configuration, as you would see here:

${d}^{3}$:

$\underline{\uparrow \textcolor{w h i t e}{\downarrow}} \text{ " ul(uarr color(white)(darr)) " " ul(uarr color(white)(darr)) " " ul( color(white)(uarrdarr)) " } \underline{\textcolor{w h i t e}{\uparrow \downarrow}}$

${d}^{7}$:

$\underline{\uparrow \downarrow} \text{ " ul(uarr darr) " " ul(uarr color(white)(darr)) " " ul(uarr color(white)(darr)) " } \underline{\uparrow \textcolor{w h i t e}{\downarrow}}$