What is the bond enthalpy of Xe–F bond? array r XeF _ 4 Xe ^ + ( g )+ F ^ - ( g )+ F _ 2…
Chemistry · JEE Advanced · NTA Exams — Chemical Thermodynamics
What is the bond enthalpy of Xe–F bond?
\(\begin{array}{r} \mathrm{XeF}_{4} \rightarrow \mathrm{Xe}^{+}(\mathrm{g})+\mathrm{F}^{-}(\mathrm{g})+\mathrm{F}_{2}(\mathrm{~g})+\mathrm{F}(\mathrm{~g}) ; \\ \Delta_{1} \mathrm{H}=292 \mathrm{kcal} / \mathrm{mol} \end{array}\)
Given: Ionization energy of Xe = 279 kcal/mol
B.E. (F–F) = 38 kcal/mol,
Electron affinity of F = 85 kcal/mol
\(\begin{array}{r} \mathrm{XeF}_{4} \rightarrow \mathrm{Xe}^{+}(\mathrm{g})+\mathrm{F}^{-}(\mathrm{g})+\mathrm{F}_{2}(\mathrm{~g})+\mathrm{F}(\mathrm{~g}) ; \\ \Delta_{1} \mathrm{H}=292 \mathrm{kcal} / \mathrm{mol} \end{array}\)
Given: Ionization energy of Xe = 279 kcal/mol
B.E. (F–F) = 38 kcal/mol,
Electron affinity of F = 85 kcal/mol
- 24 kcal/mol
- 34 kcal/mol
- 8.5 kcal/mol
- None of these
Answer
(B) 34 kcal/mol
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