Q.Statement (I): Electrovalent compounds possess high lattice enthalpy, high melting and boiling points.
Statement (II): Due to strong coulombic forces of attraction between the oppositely charged ions, electrovalent compounds exist mostly as hard crystalline solids.

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. …