How does "ionization energy" evolve in terms of the Periodic Table?

Jul 19, 2017

Well, ionization energy INCREASES across a Period.......

$\text{order of decreasing ionization energy}$
$\stackrel{\rightarrow}{F \left(Z = 9\right) , O \left(Z = 8\right) , N \left(Z = 7\right) , B \left(Z = 5\right) , B e \left(Z = 4\right)}$

Explanation:

Well, ionization energy INCREASES across a Period, a ROW of the Periodic Table, from left to right as we face the Table.

And ionization energy DECREASES down a Group.

Why so? Well, there is a contest between (i) nuclear charge, $Z$, the number of positive charges in the nucleus, which naturally tends to increase the energy of ionization, and (ii) shielding by other electrons. Now it is a fact that INCOMPLETE valence shells shield the nuclear charge VERY INEFFECTIVELY.

And thus we can account for measurable ionization energies, i.e. the energy associated with the following equation....

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

......on the basis of periodicity. Elements to the right of the Periodic Table (as we face it!) should have higher ionization enthalpies. As chemists, as physical scientists, we should interrogate the data.

This is listed in $\text{eV}$ but the Periodic trend, to which I have alluded, is clear.

And after all that we gots..............

$B e \left(Z = 4\right) , B \left(Z = 5\right) , N \left(Z = 7\right) , O \left(Z = 8\right) , \mathmr{and} F \left(Z = 9\right) .$

But that's round the wrong way in the order of INCREASING first ionization energy. And so......

$\text{order of decreasing ionization energy}$
$\stackrel{\rightarrow}{F \left(Z = 9\right) , O \left(Z = 8\right) , N \left(Z = 7\right) , B \left(Z = 5\right) , B e \left(Z = 4\right)}$

Clearly, this follows $Z$, the atomic number.......