JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Solid State

A formula of ${\mathrm{A}}_{0.95}\mathrm{O}$ shows the metal sites are deficient: for each oxide ion there is slightly less than one metal ion, and electrical neutrality is restored by oxidising some ${\mathrm{A}}^{2+}$ to ${\mathrm{A}}^{3+}$. Removing one ${\mathrm{A}}^{2+}$ leaves a cation vacancy carrying a net shortfall of two positive charges, and this is balanced by promoting two neighbouring ${\mathrm{A}}^{2+}$ ions to ${\mathrm{A}}^{3+}$, so every vacancy is paired with two trivalent ions while the oxide lattice stays complete. Among the three lattices drawn, only the one marked A shows cation vacancies together with the matching pair of ${\mathrm{A}}^{3+}$ ions, so it alone represents the metal-deficient oxide correctly.