# How many valence electrons does F have?

May 21, 2018

$7$
$\left[H e\right] 2 {s}^{2} \setminus 2 {p}^{5}$ or $1 {s}^{2} \setminus 2 {s}^{2} \setminus 2 {p}^{5}$
Therefore, its valence shell is $n = 2$, and so it contains $2 + 5 = 7$ valence electrons. Since it is one electron away from achieving an electron configuration of neon, a noble gas, would you expect fluorine to have high electronegativity or not?