# Why is fluorine the most electronegative atom?

It seems related to the atomic size but hydrogen has a smaller atomic size than fluorine. Why is fluorine the most electronegative atom?

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One reason is hydrogen has one shell and one proton, as compared to fluorine with 2 shells and 9 protons. – Satwik Pasani Feb 27 '14 at 10:15
It's interesting to note that if you define electronegativity as the "strength" with which the valence electrons are attracted to the nucleus (for a certain definition of strength), then the noble gasses also have very high electronegativities. In particular, neon becomes the element with the highest electronegativity, since it has the same number of shells as fluorine while having an extra proton! It's kind of odd that neon should be more electronegative than fluorine, because we often associate extreme electronegativity with high reactivity, but it makes a lot of sense. – Nicolau Saker Neto Feb 27 '14 at 12:08

Fluorine is the most electronegative element because the definition of electronegativity makes it so. The electronengativity scales are defined based on experimentally determined properties of the elements. Fluorine has appropriate values for all of the common scales to ensure it has the highest electronegativity.

The Pauling scale, which is the first electronegativity scale proposed, relates the difference in electronegativity $\Delta\chi_{A,B}$ between two atoms $(A\ \& \ B)$ is related to the bond dissociation energies $(E_d)$ of the diatomic species $AA,\ BB,\ \& \ AB$:

$$\Delta \chi_{A,B}=|\chi_A - \chi_B| = (\text{eV})^{-1/2}\sqrt{E_d(AB)-[E_d(AA)+E_d(BB)]/2}$$

Since only differences can be determined, one element (hydrogen) was chosen to have an arbitrarily set value. On the Pauling Scale Hydrogen has a value of $2.20$. Since the $\ce{H-F}$ bond is much stronger $(E_d=586\ \text{kJ/mol})$ than the average of the $\ce{H-H}$ bond $(E_d=436\ \text{kJ/mol})$ and the $\ce{F-F}$ bond $(E_d=157\ \text{kJ/mol})$. Data from here. In general, since fluorine forms stronger bonds to all other elements than it does to itself, fluorine will have the highest electronegativity.

The Mulliken scale relates electronegativity to two other properties: electron affinity $E_{ea}$ and ionization energy $E_i$:

$$\chi_A = \dfrac{E_{ea}(A)+E_i(A)}{2}$$

Fluorine has a high electron affinity (easy to gain an electron) and a high ionization energy (hard to lose an electron).

There are other scales, including the Allred-Rochow scale, based on effective nuclear charge and covalent radius, the Sanderson scale based on atomic size, and the Allen scale based on the average energy of the valence electrons.

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