JEE Main · 2026 · 06 Apr 2026 · Shift 1 · p-Block Elements

The standard electrode potential $E_{\mathrm{A}{\mathrm{l}}^{3+}/Al}^{\circ }=-1.66\mathrm{V}$ is negative while $E_{\mathrm{T}{\mathrm{l}}^{3+}/Tl}^{\circ }$ is positive, so aluminium is oxidised far more readily than thallium and is the more electropositive of the two; Statement-I is correct.

For boron the sum of the first three ionisation enthalpies is very large ($\approx 6900\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$), so the energy needed to make ${\mathrm{B}}^{3+}$ cannot be repaid by lattice or hydration energy and boron bonds covalently; aluminium, with much smaller ionisation enthalpies, readily forms $\mathrm{A}{\mathrm{l}}^{3+}$. Statement-II is correct.

Both statements are true.