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Scales of Electronegativity

If ∈ C-C = 34...

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

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

C

H

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

C

H

C_{2} H_{6} \to C_{2} H_{4} + H_{2}

Δ_{r} H =

{kJ mol}^{- 1}

{kJ mol}^{- 1}

Δ_{f} H^{o}

N_{2} H_{4} (g)

N - N = 159 {kJ mol}^{- 1}; H - H = 436 {kJ mol}^{- 1}

N \equiv N = 941 {kJ mol}^{- 1}; N - H = 398 {kJ mol}^{- 1}

ϵ_{N \equiv N} = 945 {kJ mol}^{- 1}, ϵ_{H - H} = 436 {kJ mol}^{- 1}

ϵ_{(N - H)} = 391 {kJ mol}^{- 1}

N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g)

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