# Explain this exception to the general trend of ionization energy in terms of electron arrangements and attraction/repulsion?

## The trend for ionization energy is a general increase from left to right across a period. However, phosphorus (P) is found to have a higher first ionization energy value than sulfur (S).

Jan 8, 2018

Well, it is for the same reason nitrogen atom has a higher first ionization energy than oxygen atom.

Compare their valence electron configurations:

$\text{N}$: $2 {s}^{2} 2 {p}^{3}$
$\text{O}$: $2 {s}^{2} 2 {p}^{4}$

$\text{P}$: $3 {s}^{2} 3 {p}^{3}$
$\text{S}$: $3 {s}^{2} 3 {p}^{4}$

In either case, we compare the following... take for example, $\text{P}$ vs. $\text{S}$:

${\underbrace{\underline{\textcolor{b l u e}{\uparrow} \textcolor{w h i t e}{\downarrow}} \text{ "ul(uarr color(white)(darr))" } \underline{\uparrow \textcolor{w h i t e}{\downarrow}}}}_{3 p , P}$

${\underbrace{\underline{\textcolor{b l u e}{\uparrow \downarrow}} \text{ "ul(uarr color(white)(darr))" } \underline{\uparrow \textcolor{w h i t e}{\downarrow}}}}_{3 p , S}$

Since the $3 p$ electrons in sulfur (that sulfur would lose) are paired, sulfur has more electron repulsion in those orbitals than phosphorus does, so it takes less energy input to remove an electron from sulfur.

Hence, since the ionization occurs more easily, the ionization energy is smaller.