Statements Analysis on Electropositivity and Ionization Enthalpies of Group 13 Elements

To determine the correctness of the given statements, let us analyze them individually:

1. Analysis of Statement I:

   • The standard electrode potential ($E^\circ$) measures the tendency of an element to be reduced. A more negative reduction potential value implies a greater tendency for oxidation (loss of electrons) and hence higher electropositive character.
   • For aluminium, the standard reduction potential is negative ($E^\circ_{\text{Al}^{3+}/\text{Al}} = -1.66\text{ V}$), which shows its strong tendency to lose electrons and act as a electropositive metal.
   • For thallium, the standard reduction potential is positive ($E^\circ_{\text{Tl}^{3+}/\text{Tl}} = +1.26\text{ V}$), indicating that it is less electropositive due to poor shielding of inner $4f$ and $5d$ orbitals, as well as the inert pair effect.
   • Hence, aluminium is indeed more electropositive than thallium, making Statement I true.

2. Analysis of Statement II:

   • Due to the extremely small size of the boron atom, the sum of its first three ionization enthalpies ($\sum_{i=1}^{3} \Delta_i H$) is exceptionally high ($\sim 6887\text{ kJ mol}^{-1}$). This prevents boron from losing three electrons to form the $\text{B}^{3+}$ cation, leading it to form only covalent compounds.
   • On the other hand, the sum of the first three ionization enthalpies of aluminium is significantly lower ($\sim 5137\text{ kJ mol}^{-1}$), which is easily compensated for by hydration or lattice energies, allowing aluminium to form $\text{Al}^{3+}$ ions.
   • Thus, Statement II is true.

Since both Statement I and Statement II are correct, the correct option is A (Both Statement I and Statement II are true).