# Why does chlorine have a higher electron affinity than fluorine?

Since fluorine has its valence electrons in the n=2 energy level, and since chlorine has its valence electrons in the n=3 energy level, one would initially expect that an electron rushing towards fluorine would release more energy, as it would land in the n=2 energy level, whereas in chlorine, the electron would land only in the n=3 energy level, and would then not release as much energy. Thus, one would expect fluorine to have a greater electron affinity than chlorine. However, why is it that chlorine has a higher electron affinity (349 kJ/mol) than fluorine (328.165 kJ/mol)?

Fluorine, though higher than chlorine in the periodic table, has a very small atomic size. This makes the fluoride anion so formed unstable (highly reactive) due to a very high charge/mass ratio. Also, fluorine has no d-orbitals, which limits its atomic size. As a result, fluorine has an electron affinity less than that of chlorine.

See this.

• I don't think d orbitals contribute to atomic size, since they're always only populated in an internal shell. Calculated atomic radii decrease continuously across a period, and they decrease a little slower in the d-block because the inner d orbitals shield the valence s/p electrons from the nuclear charge better than adding electrons to the outer shell. If anything, the presence of a populated d subshell actually helps contract the atom, allowing it to reach a higher effective nuclear charge than expected in a period with no d subshell. Commented Nov 12, 2013 at 11:22
• ....but chlorine doesn't have any d-orbitals either? Commented Oct 29, 2016 at 16:49
• @user3932000 Chlorine have vacant d-orbitals. Since empty orbitals are omitted when writing electronic configuration, it is written as $1s^2 2s^2 2p^6 3s^2 3p^5$. Complete electronic configuration would be $1s^2 2s^2 2p^6 3s^2 3p^5 3d^0$.
– ashu
Commented Nov 6, 2016 at 14:42
• @ashu You could say the same for fluorine and say fluorine also has vacant d-orbitals, since its configuration would then be 1s2 2s2 2p5 3s0 3p0 4s0 3d0. Commented Nov 6, 2016 at 17:39
• @ashu There is absolutely no reason to consider vacant d orbitals in partly filled electron shells. You don’t consider f orbitals for bromine either. In fact, while this answer does start off well I have to give it a $-1$ for invoking a concept that does not help and is introduced incorrectly.
– Jan
Commented Sep 29, 2017 at 8:52

The electron being gained by fluorine would be taken in to a much smaller 2p orbital and requires more electron coupling energy than that of much larger 3p orbital of chlorine. Therefore, energy released during the electron gaining process of fluorine is less than that of chlorine.

It isn't just chlorine versus fluorine. In the preceding group sulfur neats oxygen despite sulfur's reputation as a relatively low-electronegativity nonmetal (this is one reason sulfides of electropositive metals show a high degree of ionic character), and the third period element actually wins all the way back to Group 13 (aluminum beats boron).

The old Bohr model of the hydrogen atom provides an explanation. In this model the $$n=1$$ shell is only half as far from the nucleus as the $$n=2$$ shell, whereas this ratio is 2/3 between $$n=2$$ and $$n=3$$, 3/4 between $$n=3$$ and $$n=4$$, and so on approaching 1 as we move to more distant adjacent shells. In the multielectronic case this difference between distance ratios is amplified by the $$n=1$$ shell being exposed to essentially the full nuclear charge while tending to shield higher shells.

So in, let us say, boron, where the valence shell is $$n=2$$ and the shekl immediately below is $$n=1$$, the latter is buried so deep inside the former that is provides essentiall full shielding; whereas in aluminum the $$n=2$$ "valence minus one" shell is not so deeply buried and thus is less effective as a first line of defense for the $$n=3$$ valence shell. And so it goes for silicon versus carbon and onwards, ultimately to chlorine versus fluorine.