Why is fluorine more electronegative than sodium?

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Fluorine is more electronegative than sodium due to the increase in effective nuclear charge experienced by the outermost electrons as one moves across the periodic table. While both elements have electrons in their outer shells, fluorine's electrons are less shielded from the nuclear charge, leading to a stronger attraction to the nucleus. This phenomenon is explained by Slater's rules, which illustrate the inefficiency of electron shielding in the same shell. The Allred-Rochow electronegativity scale effectively correlates with the Pauli scale, providing a quantitative measure of this trend.

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I've learned the trends in electronegativity, but why is fluorine more electronegative than sodium? What is the real reasoning behind it?
 
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When going from sodium to fluorine, both the nuclear charge and the electrons in the outmost shell (i.e. electrons with approximately the same distance from the nucleus on average) increases. These electrons in the same shell are not very efficient in screening each other from the nuclear charge as about 50% of the electrons are farther outside than a given electron under consideration (See Slater's rules: http://en.wikipedia.org/wiki/Slater's_rules). Hence the effective nuclear charge seen by the outmost electrons increases continuously and the electrons become more tightly bound leading to increased electronegativity.
This physical argument forms the basis for the definition of the Allred-Rochow electronegativity scale which correlates well with the Pauli scale:
http://en.wikipedia.org/wiki/Electronegativity
 

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