# Question 3b3d4

Mar 12, 2017

The spin-only magnetic moment is most applicable to light-enough transition metals with minimal orbital contribution to $\mu$ that it accurately corresponds to the electronic structure. Then, we would have something like:

$\boldsymbol{{\mu}_{S} = 2.00023 \sqrt{S \left(S + 1\right)}}$

where $S$ is the total spin in the system. The electron configurations of these ions are:

$\left[A r\right] 3 {d}^{9}$, ${\text{Cu}}^{2 +}$
$\left[A r\right] 3 {d}^{8}$, ${\text{Ni}}^{2 +}$
$\left[A r\right] 3 {d}^{6}$, ${\text{Co}}^{3 +}$
$\left[A r\right] 3 {d}^{6}$, ${\text{Fe}}^{2 +}$

Write out the orbital diagrams.

$3 {d}^{9}$:

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

$3 {d}^{8}$:

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

$3 {d}^{6}$:

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

One can clearly see the increasing number of unpaired electrons going from ${\text{Cu}}^{2 +}$ to ${\text{Fe}}^{2 +}$. ${\mu}_{S}$ for each of these configurations is:

$\textcolor{g r e e n}{{\mu}_{S , 3 {d}^{9}}} = {\mu}_{S , {\text{Cu}}^{2 +}} = 2.00023 \sqrt{\frac{1}{2} \left(\frac{1}{2} + 1\right)} = \textcolor{g r e e n}{1.732}$

$\textcolor{g r e e n}{{\mu}_{S , 3 {d}^{8}}} = {\mu}_{S , {\text{Ni}}^{2 +}} = 2.00023 \sqrt{\frac{2}{2} \left(\frac{2}{2} + 1\right)} = \textcolor{g r e e n}{2.829}$

$\textcolor{g r e e n}{{\mu}_{S , 3 {d}^{6}}} = {\mu}_{S , {\text{Co"^(3+),"Fe}}^{2 +}} = 2.00023 \sqrt{\frac{4}{2} \left(\frac{4}{2} + 1\right)} = \textcolor{g r e e n}{4.900}$

Alternatively, ${\mu}_{S + L}$, the spin-and-orbital magnetic moment, is:

$\boldsymbol{{\mu}_{S + L} = 2.00023 \sqrt{S \left(S + 1\right) + \frac{1}{4} L \left(L + 1\right)}}$

where $L$ is the total orbital quantum number for each unpaired electron. For each of the above configurations, we would have the same increasing pattern with the number of unpaired electrons:

$\textcolor{g r e e n}{{\mu}_{S + l , {\text{Cu}}^{2 +}}} = 2.00023 \sqrt{\frac{1}{2} \left(\frac{1}{2} + 1\right) + \frac{1}{4} \cdot 2 \left(2 + 1\right)} = \textcolor{g r e e n}{3.000}$

$\textcolor{g r e e n}{{\mu}_{S + l , {\text{Ni}}^{2 +}}} = 2.00023 \sqrt{\frac{2}{2} \left(\frac{2}{2} + 1\right) + \frac{1}{4} \cdot 3 \left(3 + 1\right)} = \textcolor{g r e e n}{4.473}$

$\textcolor{g r e e n}{{\mu}_{S + l , {\text{Co"^(3+),"Fe}}^{2 +}}} = 2.00023 \sqrt{\frac{4}{2} \left(\frac{4}{2} + 1\right) + \frac{1}{4} \cdot 2 \left(2 + 1\right)} = \textcolor{g r e e n}{5.478}$

For instance, $L$ for $3 {d}^{6}$ is $2 + 1 + 0 + \left(- 1\right) = 2$, and for $3 {d}^{8}$ is $2 + 1 = 3$.

Do note that the observed magnetic moments for each of these cations are $1.7 - 2.2$, $2.8 - 4.0$, ~5.4#, and $5.1 - 5.5$, respectively (Inorganic Chemistry, Miessler et al., pg. 360).

So, ${\mu}_{S}$ has the better agreement for ${\text{Cu}}^{2 +}$ and ${\text{Ni}}^{2 +}$, but ${\mu}_{S + L}$ has better agreement for ${\text{Co}}^{3 +}$ and ${\text{Fe}}^{2 +}$.