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Electronegativity

Using the fol...

Question

Using the following data, calculate the electronegativity of fluorine.
$E_{H - H} = 104.2 {kcalmol}^{- 1}, E_{F - F} = 36.6 {kcalmol}^{- 1},$
$E_{H - F} = 134.6 {kcalmol}^{- 1}, X_{H} = 2.1$

3.87

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1.77

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2.67

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4.87

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Solution

The correct option is A 3.87
$Δ_{H - F} = E_{H - F} - \sqrt{(E_{F - F} - E_{H - H}} = 134.6 - \sqrt{36.6 \times 104.2} = 134.6 - 61.7 = 72.2$
$(X_{F} - X_{H}) = 0.208 \sqrt{Δ_{H - F}} k c a l m o l^{- 1} = 0.208 \sqrt{72.5} = 1.77 X_{F} = X_{H} + 1.77 = 2.1 + 1.77 = 3.87$

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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}^{- 1}

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 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.

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