MHT CET Chemistry question
Calculate the bond enthalpy of Xe F bond as given in the equation,
\mathrm{XeF_4(g) \to Xe^{+}(g) + F^{-}(g) + F_2(g) + F(g)};\ \Delta_rH = 292\ \mathrm{kcal\,mol^{-1}}
Ionisation energy of Xe kcal/mol
Bond energy (F F) kcal/mol
Electron affinity of F kcal/mol
Answer: (B)
Solution
Break 4 XeF bonds (), ionise Xe (), form F (electron gain, ) and form F from 2 F (): $292 = 4x +
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