Question

Calculate the electronegativity of fluorine from the following data:
Bond energy of H-H bond $E_{H-H}=104.2kcalmo{l}^{-1}$
Bond energy of F-F bond$E_{F-F}=36.6kcalmo{l}^{-1}$
Bond energy of H-F bond $E_{H-F}=134.6kcalmo{l}^{-l}$
Electronegativity of Hydrogen $X_H=\mathrm{2.1.}$

• A. 3.87
• B. 2.76
• C. -3.87
• D. -2.76

Answer 1

The correct option is A 3.87

Let the electronegativity of fluorine be $X_F$. Applying Pauling equation,
$X_F-X_H=0.208[E_{H-F}-(E_{F-F}\times E_{H-H}{)}^{0.5}{]}^{0.5}$
In this equation, dissociation energies are taken in $kcalmo{l}^{-1}$
$X_F-2.1=0.208[134.6-(104.2\times 36.6{)}^{0.5}{]}^{0.5}$
$X_F$ = 3.87