# Question #0c88c

May 2, 2015

The only paramagnetic one is selenium.

#### Explanation:

A paramagnetic species will have an electron configuration that shows unpaired electrons. Unpaired electrons are what will cause the respective atom (or ion) will be attracted to a magnetic field.

If the opposite is true, i.e. a species has no unpaired electrons in its electron configuration, than that species will be diamagnetic.

So, all you have to do is write the electron configurations (I'll use noble gas shorthand notation) for all the species listed.

Start with the chlorine ion, $C {l}^{-}$. Since neutral chlorine has 17 electrons, the anion will have 18, one electron more. This will bring the chlorine ion to the electron configuration of argon, a noble gas.

${\text{Cl}}^{-} : \left[A r\right]$

Since a noble gas configuration has no unpaired electrons, it will result in a diamagnetic species.

$A r : 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6}$

The same idea applies for the barium and scandium cations as well. Neutral barium is

$\text{Ba} : \left[X e\right] 6 {s}^{2}$

If you remove the outermost two electrons, you'll get the noble gas configuration of xenon. Once again, this will result in a diamagnetic species.

$\text{Ba"^(2+)} : \left[X e\right]$

Likewise,

$\text{Sc} : \left[A r\right] 4 {s}^{2} 3 {d}^{1}$ $\to$ ${\text{Sc}}^{3 +} : \left[A r\right]$ $\to$ diamagnetic species.

This brings us to selenium, $\text{Se}$. Its electron configuration looks like this

$\text{Se} : \left[A r\right] 4 {s}^{2} 3 {d}^{10} 4 {p}^{4}$

Now, you'd be tempted to say that selenium has no unpaired electrons either, but you'd be wrong. Here's why

Selenium actually has 2 unpaired electrons in the 4p-orbital, which makes it paramagnetic.