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Chemistry Question 59 – JEE-MAIN 2026

Given below are two statements : Statement I: The second ionization enthalpy of B, Al and Ga is in the order of B > Al > Ga. Statement II: The correct order in terms of first ionization enthalpy is Si < Ge < Pb < Sn. In the light of the above statements, choose the correct answer from the options given below :

Ionization enthalpy generally decreases down a group due to increasing atomic size and shielding, but exceptions occur due to poor shielding by d and f electrons (e.g., d-block contraction, lanthanide contraction).

Step 1: Analyze Statement I (Second Ionization Enthalpy of Group 13)✦ Active

Statement I claims the order of second ionization enthalpy (IE2) for B, Al, and Ga is B > Al > Ga. The elements are Boron (B), Aluminum (Al), and Gallium (Ga). Their actual IE2 values are B (2427 kJ/mol), Al (1817 kJ/mol), and Ga (1979 kJ/mol). The correct order is B > Ga > Al. This deviation from the simple decreasing trend (B > Al > Ga) is due to the poor shielding of 3d10 electrons in Ga, which increases its effective nuclear charge, making it harder to remove the second electron from Ga+ compared to Al+. Therefore, Statement I is false.

Step 2: Analyze Statement II (First Ionization Enthalpy of Group 14)○ Expand

Statement II claims the order of first ionization enthalpy (IE1) for Si, Ge, Pb, and Sn is Si < Ge < Pb < Sn. The elements are Silicon (Si), Germanium (Ge), Tin (Sn), and Lead (Pb). Their actual IE1 values are Si (786 kJ/mol), Ge (762 kJ/mol), Sn (708 kJ/mol), and Pb (715 kJ/mol). The correct order is Si > Ge > Sn < Pb. This trend is influenced by the d-block contraction (Ge, Sn) and f-block contraction (Pb), which cause an increase in effective nuclear charge, particularly for Pb, making its IE1 higher than Sn. Therefore, Statement II is false.

Step 3: Combine Conclusions○ Expand

Since both Statement I and Statement II are false, the correct option is the one stating that both statements are false.

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