# Why is the ionization energy of beryllium, Z=4, GREATER than the ionization energy of boron, Z=5?

Aug 27, 2017

Consider the electronic configuration of each element.......

#### Explanation:

And we consider the ELEMENT not the molecule, inasmuch as this is how ionization enthalpies are defined......

${\text{Atom(g)"+Deltararr"Atom(g)}}^{+} + {e}^{-}$

For $B e$ we gots $Z = 4$, $1 {s}^{2} 2 {s}^{2}$

For $B$ we gots $Z = 5$, $1 {s}^{2} 2 {s}^{2} 2 {p}^{1}$

The ionization of boron involves removal of a $\text{p-electron}$, which has NO electron density at the nuclear core, and thus should be easier to remove than a $s$ electron, which can lie closer to the nuclear core.

For $N$ we gots $Z = 7$, $1 {s}^{2} 2 {s}^{2} 2 {p}^{3}$

For $O$ we gots $Z = 8$, $1 {s}^{2} 2 {s}^{2} 2 {p}^{4}$

For nitrogen, we have a half-filled $\text{p-shell}$ which is energetically stabilized with respect to $\text{Hund's rule}$ of maximum multiplicity. For the oxygen atom, even tho it has $Z = 8$, the electronic configuration of the cation is SLIGHTLY stabilized by Hund's rule. None of this treatment is any substitute for reading the relevant section of your text.