Q.CaO and NaCl have the same crystal structure and approximately the same radii. If U is the lattice energy of NaCl, the approximate lattice energy of CaO is

When the electronegativity difference between two combining atoms is LARGE, the lower-electronegativity atom does not share its valence electrons but transfers them COMPLETELY to the higher-electronegativity atom. Both atoms end up with a noble-gas-like electron configuration -- the donor as a cation, the acceptor as an anion -- and the resulting oppositely-charged ions are held together purely by electrostatic (Coulombic) attraction. This is the ionic or electrovalent bond.

Worked example — potassium chloride. Potassium ([Ar]4s¹) loses its lone valence electron to become K⁺ ([Ar]); chlorine ([Ne]3s²3p⁵) accepts that electron to become Cl⁻ ([Ar]). Considered in isolation, this electron transfer is NOT energetically favourable: forming K⁺(g) costs +418.81 kJ mol⁻¹ (potassium's ionisation energy), while forming Cl⁻(g) releases only −348.56 kJ mol⁻¹ (chlorine's electron-gain enthalpy) -- a net cost of +70.25 kJ mol⁻¹ for the free gaseous ions. What makes solid KCl form anyway is the further step of the free ions condensing into a crystal lattice, releasing a further 718 kJ mol⁻¹ of lattice energy -- easily outweighing the earlier endothermic step and making the overall process strongly favourable. This two-step accounting (ionisation/electron-gain cost versus lattice-energy payoff) is the general pattern behind every ionic compound's stability, and is formalised later (competitive exams) as a Born-Haber cycle. …