JEE Main 2026 — Classification of Elements and Periodicity in Properties Question with Solution

The electronegativity of Ge on the Pauling scale is $1.8$. In Group 13, the electronegativity values are B ($2.0$), Al ($1.5$), Ga ($1.6$), In ($1.7$), and Tl ($1.8$). Thus, the element 'E' is Thallium (Tl).

Due to the inert pair effect, the $+1$ oxidation state is much more stable than the $+3$ oxidation state for the heavier elements of Group 13, especially Thallium. Therefore, $T{l}^{+}$ is more stable than $T{l}^{3+}$. This makes statement C incorrect.

Since $T{l}^{3+}$ is relatively unstable, it readily accepts two electrons to form the more stable $T{l}^{+}$ ion ($T{l}^{3+}+2e^{-}\to T{l}^{+}$). Because it undergoes reduction easily, $T{l}^{3+}$ acts as a strong oxidizing agent. This makes statement A incorrect and statement B correct.

The standard electrode potential for the reduction of $T{l}^{3+}$ to Tl ($E_{T{l}^{3+}/Tl}^{\circ }$) is $+1.26\text{ V}$, which is a positive value. This makes statement D correct.

Therefore, only statements B and D are correct.