Covalent Bonding

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Question

The electronegativity of fluorine from the following data is
$E_{H - H} = 104 .2 {kcalmol}^{- 1}, E_{F - F} = 36 .6 {kcalmol}^{- 1}, E_{H - F} = 134 .6 kcalmol^{- 1}$
(Round off to nearest integer)

Difficult

Solution

We know that electronegativity of H = 2.1. Since bond energy is given in ${k cal mol}^{-}$ , so :
$\begin{array}{l} X_{F} - X_{H} = 0.208 {[{BE}_{H - F} - {\{{BE}_{H - H} \times {BE}_{F - F}\}}^{1 / 2}]}^{1 / 2} \\ X_{F} - 2.1 = 0.208 {[134.6 - {104.2 \times 36.6}^{1 / 2}]}^{1 / 2} \\ X_{F} = 2.1 + 0.208 \end{array}$
$\begin{array}{l} {[134.6 - (3813.72)^{1 / 2} i.e., 61.76]}^{1 / 2} \\ = 2.1 + 0.208 \times (72.84)^{1 / 2} \\ = 2.1 + (0.208 \times 8.53) = 2.1 + 1.77 \\ = 3.87 \end{array}$

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