What is the bond enthalpy of Xe-F bond? `XeF_(4)(g) rarr Xe^(+) (g) + F^(-)(g) + F_(2) (g) + F(g), Delta_(f)H=292`kcal/mol. Given that I.E of Xenon =279k Cal/mole, B.E of `F_(2)= 38` k Cal/mol. E.A of F=85k Cal/mole.

Options

A. 24k Cal/mol
B. 34 k Cal/mol
C. 8.5 k Cal/mol
D. 142.12 k Cal/mol

Answer

B, D

Explanation

Reactions for the given data are as follows
`XeF_(4(g)) rarr Xe_((g))^(+) + F_((g))^(-) + F_(2(g)) + F_((g)) , Delta H_(1)` …(1)
`Xe_((g)) rarr Xe^(+) + e^(-), Delta H_(2)` …(2)
`F_(2(g)) rarr 2F_((g)) , Delta H_(3)` …(3)
`F_((g)) + e^(-) rArr F_((g))^(-), Delta H_(4)` ...(4)
From the above reactions we will giet the following reaction.
`Xe_((g)) + 2F_(2(g)) rarr XeF_(4(g))`... (5)
eq.(5) = eq.(2) + eq.(3) + eq(4)- eq.(1)
`Delta H_(5) = 279 + 38- 85 - 292`
`= -60 = 2` (B. E of `F_(2)`) - 4 (B.E of `Xe- F`)
`=60 = 2(38) - 4(Xe - F)`
(Xe-F) Bond energy = 34 K.Cal