# Question #fbdfe

Dec 29, 2014

The difference in ionization energy between $B$ and $B e$ can be explained by detailing their respective electron configurations:

$B e$: $1 {s}^{2} 2 {s}^{2}$
$B$: $1 {s}^{2} 2 {s}^{2} 2 {p}^{1}$

Notice that $B e$ has a stable electron configuration, which means that more energy is required to remove an electron from it. On the other hand, $B$'s ionization energy is smaller because of the electron present in its p-orbital.

Since p-orbitals are higher in energy than s-orbital, therefore less stable, it is easier for an electron to be removed from a p-orbital. Electrons located in p-orbitals are further away from the nucleus, which means that they can withstand its effective nuclear charge better $\iff$ more easy to remove from the atom.

Losing an electron from its outermost shell will make $B$ more stable, since it will be lower in energy.