What causes an element to be more electronegative than others? (Besides the electrons getting further from the positively charged nucleus and obtaining a noble gas configuration)

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What causes an element to be more electronegative than others? (Besides the electrons getting further from the positively charged nucleus and obtaining a noble gas configuration)

Chemistry
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Electronegativity is described as an element's desire to attract electrons from other elements to attain the same electron configuration of a noble gas. |dw:1440296674221:dw| Covalent bonds are “characterized by the sharing of pairs of electrons between atoms, or between atoms and other covalent bonds”. With , the ability of elements to attract shared electrons, the more electronegative an element is the more energy it pulls itself toward that same element. For example, C-F F has the most electronegative , so therefore electrons gets to be pulled towards it. |dw:1440297051000:dw|
Electronegativity is a measure of the attraction of nuclei towards electrons in a covalent bond, it decreases with increasing nuclear radius, and it decreases by shielding of inner electrons. The pattern of electronegativity shown by Zale is caused by the balance between the number of protons (Z), the number of electrons (e) and the distance (radius) from the nucleus. Compare the values between O, F and Cl. |dw:1440299119025:dw| Valence e in F feel a greater force than those in O because there are more protons causing a greater "pull", similarly, valence e in Cl feel less force than in F even if there are more protons, this is because the (electromagnetic) force between charged particles decreases in an inverse square law - coulomb's law, \(\sf F=k\dfrac{q_1q_2}{r^2}\) Well compare O and F because they have the same amount of shielding (inner electrons covering up nuclear charge). The force experienced by a valence electron in an oxygen atom: \(F_O=k\dfrac{8e*1e}{(60~pm)^2}\approx k~0.0022 e^2\) The force experienced by a valence electron in an oxygen atom: \(F_F=k\dfrac{9e*1e}{(60~pm)^2}\approx k~0.0036e^2\) You can see that there is a greater pull of valence e by the F nuclei. radii: https://en.wikipedia.org/wiki/Atomic_radii_of_the_elements_(data_page) Coulomb's law: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elefor.html
I meant to write 50 pm for F's radius, the force is correct, it's just a typo.

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@aaronq What does pm stand for when you wrote the radius of oxygen in 60 pm? Is it a unit of measure of some sort?
picometers, \(\sf 1~pm=1.0*10^{-12}~m\)
@aaronq @Zale101 Thank you for taking the time to answer my question.

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