Question

Calculate the electronegativity of fluorine from the following data:
$E_{H-H}=104.2$ kcal mol${}^{-1}$
$E_{F-F}=36.6$ kcal mol${}^{-1}$
$E_{H-F}=134.6$ kcal mol${}^{-1}$
$X_H=2.1$
Express your answer to the nearest possible integral.

Answer 1

Answer: 4

Let $X_H$ and $X_F$ be the electronegativity of $H$ and $F$, then
$X_H\sim X_F=0.208{\left[E_{H-F}{\left(E_{H-H}\times E_{F-F}\right)}^{1/2}\right]}^{1/2}$
$X_H\sim X_F=0.208{\left[134.6-{\left(104.2\times 36.6\right)}^{1/2}\right]}^{1/2}$
$X_H\sim X_F=1.78$ and $X_H<X_F$
since, $X_H=2.1$
$X_F=2.1+1.78=3.88=4$