MCQ
The ionization enthalpy of $Na ^{+}$ formation from $Na _{( g )}$ is $495.8\, kJ\, mol ^{-1},$ while the electron gain enthalpy of $Br$ is $-325.0\, kJ\, mol ^{-1}$. Given the lattice enthalpy of $NaBr$ is $-728.4\, kJ\, mol ^{-1}$. The energy for the formation of NaBr ionic solid is $(-)$ ........... $\times 10^{-1} kJ \,mol ^{-1}$
- A$5581$
- B$4856$
- C$5596$
- ✓$5576$
