What is the order of the second ionization energies for elements in the second row of the Periodic Table?

What is the order of the second ionization energies for elements in the second row of the Periodic Table?

Answer:
WARNING! Long answer! The order is \(Be < C < B < N < F < O < Ne < Li.\)

Explanation:
The second ionization energy \(\left(\mathrm{IE}_{2}\right)\) is the energy required to remove an electron from a \(1+\) cation in the gaseous state.

\(\mathrm{X}^{+}(\mathrm{g}) \rightarrow \mathrm{X}^{2-}(\mathrm{g})+\mathrm{e}^{-}\)

Just like the first ionization energy, \(\left(\mathrm{IE}_{2}\right)\) is affected by size, effective nuclear charge, and electron configuration.

We would expect second ionization energies to increase from left to right as the ionic size decreases.

Here’s a table listing the electron configurations of the ions involved.

And here is a plot of the ionization energies.

We notice three things:

1. \(\mathrm{Li}\) has the highest \(\mathrm{IE}_{2},\) because to remove the second electron we must break the stable \(1 \mathrm{~s}^{2}\) noble gas shell.
2. \(B\) has a greater \(\mathrm{IE}_{2},\) than \(C.\) This is probably due to the extra stability of the \(\mathrm{s}^{2}\) subshell in the \(\mathrm{B}^{+}\) ion.
3. \(O\) has a greater \(\mathrm{IE}_{2},\) than \(F.\) The \(\mathrm{F}^{+}\) ion has a \(\mathrm{p}^{4}\) configuration in which electronic repulsions raise the energy and decrease the \(\mathrm{IE}_{2}.\)

Thus, the order of second ionization energies is

\(Be < C < B < N < F < O < Ne < Li.\)