Calculate the electronegativity of fluorine from the following data:
`E_(H-H) = 104.4 kcal mol^(-1), E_(F-F) = 36.6 kcal mol^(-1)`
`E_(H-F) = 134.3 kcal mol^(-1), chi_(H) = 2.1`

Explanation

Let the electronegativity of fluorine be `chi_(F)`.
Applying Pauling's equation
`chi_(F)-chi_(H) = 0.208 [F_(H-F)-(E_(F-F) xx E_(H-H))^(1//2)]^(1//2)`
In this equation, dissociation energies are taken in `kcal mol^(-1)`
`chi_(F) -2.1 = 0.208[134.6-(104.2 xx 36.6)^(1//2)]^(1//2)`
`chi_(F) = 3.87`