Periodic trend in difference of energy between the s and p orbitals

Why does the difference of energy between the 2s and 2p orbitals of the second period elements increase with increasing atomic number? Does this difference increases by moving down a group, e.g. is the difference between 3s and 3p bigger than 2p and 2p and why?

Atomic Orbital Ionization Energies, eV $$ \begin{array}{lccccc} \hline \text{Atom} & \mathrm{1s} & \mathrm{2s} & \mathrm{2p} & \mathrm{3s} & \mathrm{3p} \\ \hline \ce{H} & 13.6 & & & & \\ \ce{He} & 24.5 & & & & \\ \ce{Li} & & 5.45 & & & \\ \ce{Be} & & 9.30 & & & \\ \ce{B} & & 14.0 & 8.30 & & \\ \ce{C} & & 19.5 & 10.7 & & \\ \ce{N} & & 25.5 & 13.1 & & \\ \ce{O} & & 32.3 & 15.9 & & \\ \ce{F} & & 46.4 & 18.7 & & \\ \ce{Ne} & & 48.5 & 21.5 & & \\ \ce{Na} & & & & 5.21 & \\ \ce{Mg} & & & & 7.68 & \\ \ce{Al} & & & & 11.3 & 5.95 \\ \ce{Si} & & & & 15.0 & 7.81 \\ \ce{P} & & & & 18.7 & 10.2 \\ \ce{S} & & & & 20.7 & 11.7 \\ \ce{Cl} & & & & 25.3 & 13.8 \\ \ce{Ar} & & & & 29.2 & 15.9 \\ \hline \end{array} $$

Data source: www.colby.edu

Answer

Let's start with neon, the attractive force of 10 protons, screened to varying degrees by the 9 other electrons, must be overcome in order to remove the outermost p electron, this requires 21.5 eV. There are only 9 protons holding the outermost electron in fluorine in place, again screening plays a role, but with only 9 protons it is easier to remove fluorine's outermost electron (18.7 eV). This trend of decreasing IP continues for the remaining elements with 2p electrons as we move to the left across this row of elements.

The corresponding 2s IP's are much higher. For example, the neon 2s IP=48.5 eV. To measure this IP, we have already removed neon's 6 outermost 2p electrons. The 2s electron is still held in place by 10 protons, but now only 3 other electrons remain to screen these protons. Consequently it is much more difficult to remove one of neon's inner 2s electrons. As we move to the left across this row of elements looking at the 2s IP's we see a trend similar to what we saw with the 2p IP's, but 2 factors are now involved. It get's easier to remove the 2s electron as the number of protons decreases (reduced nuclear attraction) and as the number of already removed outer 2p electrons becomes smaller (more screening).

The same trends are observed in the n=3 series, but the IP's are lower to start with and differences between successive ionizations are compressed because the number of protons, and therefore the number of screening electrons, has increased by 8.

Everytime we add a proton to the nucleus, we also add an electron to keep the charge balanced. In the case of neon, that 10th electron does not feel the full attraction of the 10 protons - the other electrons screen this last electron from the nucleus. The bigger and bigger we make our nucleus, the more electrons there are between the last electron we added and the nucleus, therefore that last electron feels less and less of a full proton's attraction, as the number of electrons between it and the nucleus increases.

Here is a good example of compression. The first IP of neon is 21.5 ev, the IP to pull out a fifth electron (after the first 4 electrons have already been removed) is 126.2 eV - about 6 times more energy is required. Now consider radon, its first IP is only 10.7 eV. That electron in radon has so many more electrons between it and the nucleus (76 more to be exact) that screen that last electron from the nucleus, that it is much easier to remove that outermost electron in radon. The fifth IP for radon is 55 eV - only about 5 times more energy (less than 6) is required here, that is compression.

Here is a link to a nice, thorough discussion on ionization energies and their trends that you may find of interest.

Finally here is a pictorial representation of what we have been discussing.

Note how the spacing between the lines decreases as we move to higher atomic number; also note that the slope of the lines tends to increase as the atomic numbers increases. Both of these observations illustrate what we have been discussing. As the atomic number increases there are more and more electrons between the outermost electron and the nucleus making it easier to remove that last electron and making the both the absolute and relative differences between successive ionizations (e.g. 1st IP and 2nd IP, 1st IP and 5th IP, etc.) smaller.