lf the ionisation energy of Na is 495.8 kJ mol^-1, the electron affinity of Cl is 349.4 kJ mol^-1 and lattice energy is 776 kJ mol^-1, the true statement for the formation of ionic bond
in NaCI is

$$\begin{gathered} \mathrm{Na}\left(\mathrm{g}\right)+495.8⟶{\mathrm{Na}}^{+}\left(\mathrm{g}\right)+{\mathrm{e}}^{-} . . . \left(\mathrm{i}\right) \\ \mathrm{Cl}\left(\mathrm{g}\right)+{\mathrm{e}}^{-}⟶{\mathrm{Cl}}^{-}\left(\mathrm{g}\right)+349.4 \\ {\mathrm{Na}}^{+}\left(\mathrm{g}\right)+{\mathrm{Cl}}^{-}\left(\mathrm{g}\right)⟶\mathrm{NaCl}\left(\mathrm{s}\right)+776 \mathrm{kJ} {\mathrm{mol}}^{-1} \end{gathered}$$

Total energy released per mole of NaCl formed = 776 + 349.4 = 1125.4 kJ

Total energy absorbed in step (i) = 495.8 kJ

Net energy released = 629.6 kJ

Thus, a stable ionic bond is formed between NaCl.