The bond dissociation energy of H - H, C - C, and C - H bonds respectively are 104.2, 83.1 and 98.8 Kcal/ mole. The electro negativity of e is

2.53

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2.51

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2.50

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Solution

The correct option is A
2.53

$△$ = $E_{C - H}$ - $\sqrt{E_{H - H} \times E_{C - C}}$

= 98.8 - ${[(104.2) \times (83.1)]}^{\frac{1}{2}}$ = 5.75 kcal.

( $X_{C - H}$ ) E.N of C-H bond = 0.218 $\sqrt{△}$ = 0.218 ${(5.75)}^{\frac{1}{2}}$

=0.43

$X_{C}$ = 0.43 + $X_{H}$ = 0.43 + 2.1 2.53

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Similar questions

The bond dissociation energy of H - H, C - C, and C - H bonds respectively are 104.2, 83.1 and 98.8 kcal/ mole. The electronegativity of C is?

The bond dissociation energy of H-H, C-C,C-H bonds respectively are 104.2, 83.1 and 98.8kcal/mole . The electronegative of c is?

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.

H - H, C - C

C - H

104.2, 83.1

98.8

m o l^{- 1}

The increasing order of bond dissociation energies of C - C, C - H and H - H bonds is

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