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)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 ? 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

The average Xe-F bond energy is 34 kcal/mol, first I.E of Xe is 279 kcal/mol, electron affinity of F is 85 kcal/mol & bond dissocitation energy of F_(2) si 38 kcal/mol. Then, the enthalpy change of the reaction XeF_(4)rarrXe^(+)+F^(-)+F_(2)+F will be

The average Xe-F bond energy is 34Kcal//mol , first I.E . Of Xe is 279Kcal//mol , electron affinity of F is 85Kcal//mol . Then, the enthalpy chnage for the reaction XeF_(4)rarrXe^(+)+F^(-)+F_(2)+F will be

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

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.

The lattice enthalpy of KI will be, if the enthalpy of (I) Delta_f H^- (KI) = -78.0 kcal mol^-1 (II) Ionisation energy of K to K^+ is 4.0 eV (III) Dissociation energy of I_2 is 28.0 kcal mol^-1 (IV) Sublimation energy of K is 20.0 kcal mol^-1 (V) Electron gain enthalpy for I to I^- is -70.0 kcal mol^-1 (VI) Sublimation energy of I_2 is 14.0 kcal mol^-1 (1 eV =23. 0 kcal mol^-1)