# Question 81fd8

Oct 11, 2015

It would carry a $\left(2 +\right)$ charge.

#### Explanation:

Beryllium, $\text{Be}$, is located in period 2, group 2 of the periodic table, and has an atomic number equal to $4$.

This means that a neutral beryllium atom will have a total of 4 electrons surrounding its nucleus.

Now, out of these four electrons, $2$ are core electrons and $2$ are valence electrons. The electron configuration of beryllium - the noble gas shorthand notation - looks like this

"Be": ["He"] 2s^2

So, if beryllium were to give up its valence electrons, what charge would you expect the resulting cation to have?

Well, since the atom would have to give up two electrons, both from the 2s-orbital, the charge of the cation would be $\left(2 +\right)$. You can describe the ionization of the beryllium atom as taking place in two steps

${\text{Be" -> "Be}}^{+} + 1 {e}^{-}$

The electron configuration of the ${\text{Be}}^{+}$ cation would be

"Be"^(+): ["He"] 2s^1

and

${\text{Be"^(+) -> "Be}}^{2 +} + 1 {e}^{-}$

with the electron configuration of the cation being

"Be"^(2+): ["He"]#

Now, beryllium doesn't actually form cations because its valence electrons are held very, very tightly by the nucleus.

Beryllium has a very small atomic radius, which implies that its electrons are very close to the nucleus. Moreover, the two valence electrons are not effectively screened by the two core electrons, so beryllium does not form cations.

This is why the question was given in the form "If it were to give up its valence electrons", not when it gives up its valence electrons.