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Scales of Electronegativity

If ∈ C - C =3...

Question

If $\in_{C - C} = 348 {kJmol}^{- 1}, \in_{C - H} = 412 {kJmol}^{- 1} and \in_{H - H} = - 436 {kJmol}^{- 1}$ , the Pauling electronegativity of $C$ is about ______. (Given: electronegativity of $H$ = 2.1)

1.64

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1.8

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2.58

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2.91

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Solution

The correct option is C 2.58
$Δ_{C - H} = E_{C - H} - \sqrt{(E_{C - C} - E_{H - H}} = 412 - \sqrt{348 \times 436} = 412 - 389.5 = 22.5$
$(X_{C} - X_{H}) = 0.107 \sqrt{Δ_{C - H}} k J m o l^{- 1} = 0.102 \sqrt{22.5} = 0.48 X_{C} = X_{H} + 0.48 = 2.1 + 0.48 = 2.58$

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Similar questions

E_{C - C} = 348 {kJ mol}^{- 1}, E_{C - H} = 412 {kJ mol}^{- 1} and E_{H - H} = 436 {kJ mol}^{- 1}

C

H

\in_{C - C} = 348 {kJmol}^{- 1}, \in_{C - H} = 412 {kJmol}^{- 1} and \in_{H - H} = - 436 {kJmol}^{- 1}

C

H

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

E_{C - C} = 83.1 k c a l m o l^{- 1}

E_{C - H} = 98.8 k c a l m o l^{- l}

X_{H} = 2.1.

Which carbon atom will show minimum electronegativity -

H $^{1} C$ $\equiv$ $^{2} C$ - C $^{3} H$ - C $^{4} H$ - C $^{5} H$

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

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