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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$\begingroup$ One reason is hydrogen has one shell and one proton, as compared to fluorine with 2 shells and 9 protons. $\endgroup$– stochastic13Commented Feb 27, 2014 at 10:15
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6$\begingroup$ 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. $\endgroup$– Nicolau Saker NetoCommented Feb 27, 2014 at 12:08
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$\begingroup$ Electronegativity is the tendency of an atom to attract a shared pair of electrons (not the valence electrons ) towards itself $\endgroup$– Adnan AL-AmlehCommented Sep 1, 2018 at 5:25
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3 Answers
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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 $E_i$:
$$\chi_A = \frac{E_{ea}(A)+E_i(A)}{2}$$
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.
answered Feb 27, 2014 at 11:54
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Fluorine has a higher electronegatevity if compared with other than elements in the periodic tables due to:
- atomic size is decreased.
- nuclear charge increase due to gain two electrons.
- attraction force between nucleus and adding electron increase so electronegatevity increase too.
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It seems that (although I am no chemist) that the valence electrons in the 2nd shell have a higher priority than any other shell, if the 2nd shell is the valence shell. The closer it gets to being full, we know that electronegativity increases. With Fluorine only needing one more electron to be complete. The big battle for second place is interesting. Oxygen only needs 2 electrons in its valence shell. Chlorine only needs one, but in its 3rd shell. But we know Oxygen wins out by a slim margin.
