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## Q. 20.14

Using average bond energies from Table A.9 of the appendix, calculate the difference in electronegativity between H and F.

 Table A.9            Average Bond Energies Bond $Bond Energy/kJ mol^{−1}$ Br−Br 193 C−C 343 C=C 615 C≡C 812 C−Cl 326 C−F 490 C−H 416 C−N 290 C≡N 891 C−O 351 C=O 724 $C=O (in CO_{2})$ 799 C≡O 1046 Cl−Cl 244 F−F 158 H−Br 366 H−Cl 432 H−H 436 H−F 568 H−I 298 H−N 391 H−S 367 N−N 160 N≡N 946 N−O 176 O−H 464 O−O 144 $O=O (in O_{2})$ 498 From F. T.Wall, Chemical Thermodynamics, 3rd ed.,W. H. Freeman,San Francisco, 1974, p. 63.

## Verified Solution

We omit the units on the factors:

$\left|X_{H}-X_{Li} \right|=(0.102)[568 − (1/2)(436 + 158)]^{1/2}=1.7$          (Table value _ 1.9)