Trend in ionisation potential in period 2

Carbon has valence electronic configuration of $2s^2$ $2p^2$. Beryllium has configuration $2s^2$.So carbon should have less ionisation potential than Beryllium as it is easier to remove electrons from $2p$ subshell than fully filled $2s$ subshell. 


But I found that in JD Lee it is mentioned that carbon has more I.P. than beryllium. How can it be explained?




[![IP trends, group II](https://i.stack.imgur.com/t8AJS.png)](https://i.stack.imgur.com/t8AJS.png)


Emperically speaking, Be is on the "metallic" side of the table. Lower IE than non-metals is obviously expected. 


A quick glance at the $IE\_1$ trends for period II tells us that like the OP noted, Be certainly has an abnormally high $IE\_1$, deviating significantly from the expected plot, due to the fully filled configuration.


However, Carbon (Z=6) has a higher nuclear charge(=more protons) than Be (Z=4), a smaller radius and hence a stronger hold on its valence electrons. 


Even Nitrogen displays an exception, due to half-filled $2p$ subshell, as evident in the plot.


Just because of a tiny bump in the plot, expecting N and Be to have higher IE than all the elements of the periodic table is clearly an exaggeration.


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These periodic trends in 1st ionisation energy (I.E.) are always the result of more than one factor changing. The most important factors are:


1. The effective nuclear charge (Zeff) that the valence electron being removed experiences (roughly equal, for s and p blocks, to the number of valence electrons)
2. The potential energy of the sublevel (1s, 2s, 2p etc) that the electron is being removed from
3. Any electron-electron repulsion that the electron being removed experiences in its orbital
[![enter image description here](https://i.stack.imgur.com/jltnC.png)](https://i.stack.imgur.com/jltnC.png)
As we move left to right across the period, factor 1 increases and this dominates the overall shape shown in the graph. A drop occurs from Be to B as we move to the higher potential energy 2p orbital (so the electron is easier to remove) and this is more important than the increasing Zeff from 2+ to 3+. As we move to C, the Zeff increases to 4+, so it has a higher I.E. than B (same sublevel, 2p).


The reason that C's I.E. is higher than Be also is because the change in Zeff from Be to C is +2 to +4 (it roughly doubles) and this is more important than the difference in potential energy between the 2s and 2p sublevels, so factor 1 takes over again, over factor 2.


I haven't discussed factor 3. Anytime an electron is removed from a paired orbital, it is a little **easier** (lower I.E.) than exactly the same situation (same Zeff, same sublevel potential energy) from an unpaired orbital, due to electron-electron repulsion within the paired orbital. It is a small factor between Be and C (Be is a paired 2s, C is an unpaired 2p) but it is very minor compared to factors 1 and 2 changing between these species, so C is still harder to remove.


It is tempting to say that a filled sublevel, or orbital, is "stable" = higher I.E. This statement comes not from the fact that the sublevel is full (these paired electrons are actually easier to remove than you would expect, due to electron-electron repulsion) but because filled sublevels always correspond to the highest Zeff for that sublevel. 


You can employ these arguments to explain all the trends in the simple (s and p block) 2nd and 3rd periods. Things get much more complicated in other parts of the table! 


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