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

#overbrace("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 #"Cl"# is to the upper-most-right, and #"Se"# is to the lower-most-left, #"Cl"# is smallest and #"Se"# is the largest in atomic radius. #"S"# is intermediate in atomic radius. That is,

#r_("Se") > r_("S") > r_("Cl")#

The smallest atom holds onto its valence electrons most tightly.

Therefore, #"Cl"# is hardest to ionize, and #"Se"# is easiest to ionize. As ionization energy is large for atoms that are difficult to ionize, we have:

#color(blue)(barul(|stackrel(" ")(" ""IE"_("Cl") > "IE"_("S") > "IE"_("Se")" ")|))#