According to my textbook, fluorine has a pretty high (compared to elements like sodium) first ionization energy. But why does it have such a high requirement to ionize? After all, it actively seeks out atoms to bond with, even tearing through already existing bonds in its drive to do so.

I would expect that fluorine would have a very low ionization energy, as it is soooooooo close to having a full octet (just like sodium). Yet unlike sodium, it has a higher first ionization energy. Why is that?

(My guess is that fluorine's protons' affinity for electrons is so great that it simply "overpowers" the octet rule.)

Apparently, neon can also be ionized. Doesn't that mean it can bond with cations? But it is unusually stable, so why doesn't it?

I think there is an underlying law behind this that can explain these, please help!

EDIT: Fluorine has now been explained, but why can neon be ionized? After all, if you give it a positive charge by taking away electrons, won't it be able to bond to anions? But in reality, it doesn't. What is going on?

  • $\begingroup$ ionisation is losing of electron - if it's energy is high enough you won't find a chemical way to do it $\endgroup$
    – Mithoron
    Commented Jul 17, 2015 at 16:39
  • $\begingroup$ Ionization can be also the addition of an electron, right? After all when fluorine is satisfied its charge is a -1. $\endgroup$ Commented Jul 17, 2015 at 16:44
  • $\begingroup$ Yes but than you don't talk about ionization energy but electron affinity - you can ionise atom using only some energy to make a cation not anion. Maybe I'll try to make answer. $\endgroup$
    – Mithoron
    Commented Jul 17, 2015 at 17:18
  • $\begingroup$ Indeed, if it really wants to add an electron to become an anion, it is really unlikely to want to easily give one up to become a cation. Therefore, one can determine that, no, ionization energy does not uniquely determine how reactive an element is (or, perhaps, it is not single-valued - both low and high ionization energies would be reactive, just in different ways). $\endgroup$
    – Jon Custer
    Commented Jul 17, 2015 at 17:47

1 Answer 1


This comes about as a consequence of the effective nuclear charge on the atom or Z. eff, and the size of the atom. Z effective if you are not familiar is a concept usually touched upon in chem 2 as a periodic trend. It is a measure of the charge felt by the outer most electron. It is calculated by accounting for the electrons on the inner shells essentially canceling out or shielding the outer electron from feeling all of the charge from the positive nucleus. As you could expect this make the electron easier or harder to take away causing ionization depending on the Zeff. A small atomic radius causes the energy barrier to be higher to remove an electron, because the electron is closer to the nucleus. If the atom is larger it will be easier to remove the outer electron because it's further away from the nucleus. As you move along the periods of the periodic table from left to right and low to high, the effective nuclear charge increases. As you move along the periodic table from top down and right to left, the atomic radius increases.

We find that fluorine is at the top right of the periodic table, where the effective nuclear charge is high, and the radius of the atom is small. This means that it has a small condensed positive charge permeating through the inner shells. This causes the reactivity of fluorine to other atoms due to it's electron affinity from the unshielded nucleus. It is hard to ionize compared to sodium because of its smaller than sodium, and it's z eff is higher. These both cause the electron to be attracted to the nucleus more, causing a higher energy barrier to take it away.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.