# Why is fluorine one of the most electronegative atoms even though it doesn't have the largest effective nuclear charge?

Why doesn't electronegativity increase as the effective nuclear charge increases? Most atoms with a higher atomic number than fluorine have a higher effective nuclear charge. However, flourine is a more electronegative atom.

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).