# Arrange the following atoms in order of decreasing ionization energy? "Cl", "S", "Se"

Aug 28, 2017

$\overbrace{\text{IE"_("Cl"))^("12.968 eV") > overbrace("IE"_("S"))^("10.360 eV") > overbrace("IE"_("Se"))^("9.752 eV}}$

Data from NIST.

You should refer back to the periodic table trends for:

• increasing atomic radii due to new quantum levels
• decreasing atomic radii due to increased effective nuclear charge

The former is a vertical trend and the latter a horizontal trend. They are gone into more detail here. In short:

color(white)(""^(color(black)"atomic radii increase") color(black)(darr){(stackrel(color(black)("atomic radii decrease"))(overbrace(stackrel(" ")stackrel(" ")color(black)"Li"" "color(black)"Be"" "color(black)(cdots)" "" "color(black)"F"" ")^(color(black)(->)))),(color(black)"Na"),(color(black)(vdots)),(color(black)"Fr") :})

Knowing that, we refer to the periodic table:

Since $\text{Cl}$ is to the upper-most-right, and $\text{Se}$ is to the lower-most-left, $\text{Cl}$ is smallest and $\text{Se}$ is the largest in atomic radius. $\text{S}$ is intermediate in atomic radius. That is,

${r}_{\text{Se") > r_("S") > r_("Cl}}$

The smallest atom holds onto its valence electrons most tightly.

Therefore, $\text{Cl}$ is hardest to ionize, and $\text{Se}$ is easiest to ionize. As ionization energy is large for atoms that are difficult to ionize, we have:

$\textcolor{b l u e}{\overline{\underline{| \stackrel{\text{ ")(" ""IE"_("Cl") > "IE"_("S") > "IE"_("Se")" }}{|}}}}$