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.
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:
- \(\mathrm{Li}\) has the highest \(\mathrm{IE}_{2},\) because to remove the second electron we must break the stable \(1 \mathrm{~s}^{2}\) noble gas shell.
- \(B\) 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.
- \(O\) 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