Chemistry Q58 – JEE-MAIN 2026

Home Chemistry Periodicity & ElementsQuestion #507

The electronegativity of a group 13 element 'E' is same as that of Ge (on Pauling scale and upto one decimal point). The CORRECT statements about

E^{3 +}

are A. It can act as a reducing agent. B. It can act as an oxidizing agent. C.

E^{3 +}

is more stable than

E^{+}

. D. The standard electrode potential for

E^{3 +} / E

is positive. Choose the correct answer from the options given below:

Clue 1 of 3: Identifying the Element

Use the given electronegativity value to identify the group 13 element 'E'. Consider the trends in electronegativity down the group and potential anomalies for heavier elements.

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Ninja Strategy•Redox and Stability Logic

Instantly eliminate any option with statement A (E³⁺ as a reducing agent) or statement C (E³⁺ more stable than E⁺ for a heavy element), which leaves only one possible answer.

🧠 Full Solution Path

Step 1: Identify the Element 'E'✦ Active

The electronegativity of Germanium (Ge) on the Pauling scale is approximately 2.0. We need to find a group 13 element with a similar value. The electronegativity values for group 13 are B (2.0), Al (1.6), Ga (1.8), In (1.8), Tl (1.8-2.0). Due to relativistic effects, the electronegativity of Thallium (Tl) is anomalously high, close to 2.0. The question's focus on the relative stability of $E^{3 +}$ and $E^{+}$ ions is a strong indicator of the inert pair effect, which is most prominent in the heaviest element of the group. Therefore, E is Thallium (Tl).

Step 2: Evaluate the Statements for Thallium (Tl)○ Expand

Let's analyze each statement with E = Tl: A. **It can act as a reducing agent.** This is incorrect. $T l^{3 +}$ is in the highest common oxidation state for a group 13 element and cannot be oxidized further. B. **It can act as an oxidizing agent.** This is correct. $T l^{3 +}$ can be readily reduced to the more stable $T l^{+}$ state. The high positive standard reduction potential for $T l^{3 +} / T l^{+}$ ( $+ 1.25$ V) confirms it is a strong oxidizing agent. C. ** $E^{3 +}$ is more stable than $E^{+}$ .** This is incorrect. Due to the inert pair effect, the +1 oxidation state ( $T l^{+}$ ) is significantly more stable than the +3 oxidation state ( $T l^{3 +}$ ). D. **The standard electrode potential for $E^{3 +} / E$ is positive.** This is correct. The standard reduction potential for the reaction $T l^{3 +} (a q) + 3 e^{-} \to T l (s)$ is $E^{\circ} = + 0.72$ V.

Step 3: Determine the Correct Option○ Expand

Based on the analysis, statements B and D are correct. Therefore, the correct option is the one that includes 'B and D Only'.

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Chemistry Question 58 – JEE-MAIN 2026

Clue 1 of 3: Identifying the Element

🧠 Full Solution Path