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Scales of Electronegativity

Calculate the...

Question

Calculate the electronegativity (10 time) of fluorine on Pauling scale form the following data:
$E_{H - H} = 104.2 k c a l$ ${m o l}^{- 1}$
$E_{F - F} = 36.6 k c a l$ ${m o l}^{- 1}$
$E_{H - F} = 134.6 k c a l$ ${m o l}^{- 1}$
Given= $X_{H} = 2.1$

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Solution

$X_{H} - X_{F} = 0.208 [E_{H - F} - 0.5 (E H - H + E_{F - F})]^{0.5}$
$X_{H} - X_{F} = 0.208 [134.6 - 0.5 (104.2 + 36.6)]^{0.5} = 1.67$
$X_{F} = X_{H} - 1.67 = 2.1 - 1.67 \approx 0.4$

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Similar questions

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.

x

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

E_{F - F} = 36.6

m o l^{- 1}

E_{H - F} = 134.6 k c a l m o l^{- 1}

= 2.1

x

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

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

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