I understand that the 1st ionization energy gets bigger for elements along a period from left to right and along a group from down to up.

But why is the 1st ionization energy of $\ce{Na+}$ bigger than $\ce{Ne}$?

I would have guessed the opposite.

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    $\begingroup$ $\ce{Na+}$ has more protons than $\ce{Ne}$, so there would be a greater pull on the electrons towards the nucleus, meaning that you would have to supply more energy to remove the electron. Remember, $\ce{Na+}$ has a noble-gas configuration and is already very stable! $\endgroup$ – DHMO Jan 17 '17 at 15:20
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    $\begingroup$ $\ce{Na+}$ is the same as $\ce{Ne}$, only with more protons. Guess what does that mean for electrons. $\endgroup$ – Ivan Neretin Jan 17 '17 at 15:21

A sodium cation $\ce{Na+}$ and a neutral neon atom $\ce{Ne}$ are isoelectronic species; meaning they have the same number of electrons and also the same electronic configurations.

Yet, the two have different nuclei: most prominently, sodium always has an additional proton when compared to neon. Therefore, only one of the isoelectronic species (neon) is neutral, the other is positively charged.

It may now become obvious why the ionisation enthalpy of $\ce{Na+}$ is much larger than that of neon. Different ways to express this all boil down to the fact that it is harder to remove a negatively charged electron from a positively charged species than it is to remove one from a neutral species.


$\ce{Na+}$ and $\ce{Ne}$ are isoelectronic species and size of cationic species is smaller among isoelectronic species. Ionization energy increases with decrease in size.


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