What is the bond enthalpy of `Xe-F` bond ?
`XeF_(4)(g)rarrXe^(+)(g)+F^(-)(g)+F_(2)(g)+F(g)," "Delta_(r)H=292"kcal/mol"`
Given : Ionization energy of `Xe=279"kcal/mol"`
`B.E.(F-F) = 30"kcal/mol"`, Electron affinity of F = 85 kcal/mol

What is the bond enthalpy of Xe-F bond if Ionization energy of Xe = 279 kcal/mol B.E.(F-F) = 38 kcal/mol, electron affinity of F = 85 kcal//mol

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.

Calculate the enthalpy change for the following reaction: XeF_(4) rarr Xe^(o+) +F^(Theta) +F_(2) +F . The average Xe-F bond energy is 34 kcal mol^(-1) , first IE of Xe is 279 kcal mol^(-1), EA of F is 85 kcal mol^(-1) and bond dissociation enegry of F_(2) is 38 kcal mol^(-1)

Calculate the enthalpy change for the following reaction: XeF_(4) rarr Xe^(o+) +F^(Theta) +F_(2) +F . The average Xe-F bond energy is 34 kcal mol^(-1) , first IE of Xe is 279 kcal mol^(-1), EA of F is 85 kcal mol^(-1) and bond dissociation enegry of F_(2) is 38 kcal mol^(-1)

Calculate the bond energy of S-F bond in SF_(6),S(g),F(g) are -1100, 275, 80 kJ*mol^(-1) respectively.

In XeF_(4) the bond angle F-Xe-F is