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Which requires more ionization energy: $\ce{Ne}$ or $\ce{Ne+}?$
It seems to me like it should be neon because of noble gas configuration, but the answer given is $\ce{Ne+}.$ Does this anything to do with size of the atom shrinking?
Is this because cationic counterparts always have more ionization energy than their neutral counterparts because it becomes successively more difficult to remove electrons from atoms on successive removals?
Does this anything to do with size of the atom shrinking?
Yes, precisely; this is one way of putting it. Another way is to say that because one electron has been removed, the remaining electrons are less shielded from the nuclear charge: therefore, the effective nuclear charge increases, and the remaining electrons are harder to remove.
Is this because cationic counterparts always have more ionization energy than their neutral counterparts because it becomes successively more difficult to remove electrons from atoms on successive removals?
Yes. This is true for every element in the Periodic Table, with no exceptions, and it doesn't matter what the electronic configuration is.
If all else were equal, then it would indeed be harder to remove an electron from a noble gas configuration than a non-noble gas configuration. However, that is a very big if! Between $\ce{Ne}$ and $\ce{Ne+}$, the electron configurations are indeed different, but all else is not equal: as discussed previously, the effective nuclear charge in $\ce{Ne+}$ is greater than that in $\ce{Ne}$.
If the second ionization potential were less than the first, a certain amount of (limited) ionization would doubly ionize half of the neons and leave the other half unionized. You could never observe a significant concentration of singly ionized neon; two singles would collide to give a neutral and a double.
Although neon is far more complex than a hydrogen, you could consider Ne+ and Ne++ to be positive attractors for an electron. The attraction of Ne++ would be about twice as great as from Ne+ (21.5 eV vs 41 eV, CRC Handbook).
What is interesting is that the next 6 ionizations require between 63.5 and 239.1 eV, but the last two require 1195.8 and 1362.2 eV. Big jump in energy required to get the last 2 s-electrons off!
