# Why is "Sn"^"2+" diamagnetic?

Mar 30, 2016

If you look at the periodic table, you should see that the electron configuration for ${\text{Sn}}^{0}$ is

$\textcolor{g r e e n}{\left[K r\right] 4 {d}^{10} 5 {s}^{2} 5 {p}^{2}}$,

so its valence shell has ten $4 d$, two $5 s$, and two $5 p$ electrons.

Upon ionization, $\text{Sn}$ will lose two of its highest-energy electrons to form ${\text{Sn}}^{2 +}$.

The highest-energy electrons here are the $5 p$ electrons (by about $\text{8.55 eV}$ above the energy of the $5 s$ electrons), so we now achieve an ionic electron configuration of

$\textcolor{b l u e}{\left[K r\right] 4 {d}^{10} 5 {s}^{2}}$.

Since the $4 d$ and $5 s$ subshells are entirely filled, there are no unpaired electrons and ${\text{Sn}}^{2 +}$ by definition is diamagnetic.