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Ionisation energy of F^(ɵ) is 320 kJ mol...

Ionisation energy of `F^(ɵ)` is `320 kJ mol^(-1)`. The electron gain enthalpy of fluorine would be

A

`-320 kJ mol^(-1)`

B

`-160 k J mol^(-1)`

C

`320 kJ mol^(-1)`

D

`160 k J mol^(-1)`

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The correct Answer is:
A
EA of F (g) is opposite to `IE_(1)` of `F^(-) (g)`
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Ionisation of energy F^(ɵ) is 320 kJ mol^(-1) . The electronic gain enthalpy of fluorine would be

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Calculate the lattice enthalpy of KCl from the following data by Born- Haber's Cycle. Enthalpy of sublimation of K = 89 kJ mol^(–1) Enthalpy of dissociation of Cl = 244 kJ mol^(–1) Ionization enthalpy of potassium = 425 kJ mol^(–1) Electron gain enthalpy of chlorine = –355 kJ mol^(–1) Enthalpy of formation of KCl = –438 kJ mol^(-1)

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Use the following data to calculate Delta_("lattice")H^(@) for NaBr. Delta_("sub")H^(@) for sodium metal = 108.4 kJ "mol"^(-1) .Ionization enthalpy of sodium = 496 kJ "mol"^(-1) Electron gain enthalpy of bromine = -325 kJ "mol"^(-1) .Bond dissociation enthalpy of bromine = 192 kJ "mol"^(-1) . Delta_(f)^(H^(@)) for NaBr (s) = -360.1 kJ "mol"^(-1) .

Given Based on data provided, the value of electron gain enthalpy of fluorine would be

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