The electronegativity of fluorine obtained from the following data is $x$ .
$E_{H - H} = 104.2 k c a l m o l^{- 1};$
$E_{F - F} = 36.6$ kcal $m o l^{- 1}$
$E_{H - F} = 134.6 k c a l m o l^{- 1}$
Electronegativity of hydrogen $= 2.1$
Find the value of $x$ to the nearest integer.

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Solution

$χ_{H} - χ_{F} = 0.208 \sqrt{E_{H - F} - 0.5 (E_{H - H} + E_{F - F})} = 0.208 \sqrt{(134.6) - 0.5 (104.2 + 36.6)} = 1.76$
$χ_{F} = χ_{H} + 1.76 = 2.1 + 1.76 = 3.76 = x$

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Similar questions

E_{H - H} = 104.2 {kcalmol}^{- 1}, E_{F - F} = 36.6 {kcalmol}^{- 1},

E_{H - F} = 134.6 {kcalmol}^{- 1}, X_{H} = 2.1

E_{H - H} = 104.2 k c a l m o l^{- 1}

E_{F - F} = 36.6 k c a l m o l^{- 1}

E_{H - F} = 134.6 k c a l m o l^{- l}

X_{H} = 2.1.

E_{H - H} = 104.2 k c a l m o l^{- 1}

E_{F - F} = 36.6 k c a l m o l^{- 1}

E_{H - F} = 134.6 k c a l m o l^{- l}

X_{H} = 2.1.

E_{H - H} = 104.2 k c a l

{m o l}^{- 1}

E_{F - F} = 36.6 k c a l

{m o l}^{- 1}

E_{H - F} = 134.6 k c a l

{m o l}^{- 1}

X_{H} = 2.1

E_{H - H} = 104.2 {kcalmol}^{- 1}, E_{F - F} = 36.6 {kcalmol}^{- 1},

E_{H - F} = 134.6 {kcalmol}^{- 1}, X_{H} = 2.1

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