Chemistry · Periodicity of Elements and Periodic Properties

JEE Main 2026 — 24 January, Morning Shift — Question 62

Given below are two statements : one is labelled as Statement I and the other is labelled as Statement II :

Statement I: $\mathrm{K}>\mathrm{Mg}>\mathrm{Al}>\mathrm{B}$ is the correct order in terms of metallic character.

Statement II: Atomic radius is always greater than the ionic radius for any element.

In the light of the above statements, choose the correct answer from the options given below:

Option A:
Both Statement I and Statement II are true
Option B:
Both Statement I and Statement II are false
Option C:
Statement I is false but Statement II is true
Option D:
Statement I is true but Statement II is false
Correct

Answer: D

Step-by-step solution

Statement I is true: Metallic character increases down a group and decreases across a period. $\mathrm{K}$ (group 1, period 4) is most metallic. Between $\mathrm{Mg}$ and $\mathrm{Al}$ in period 3, metallic character decreases from left to right. Thus, $\mathrm{Mg > Al}$ and $\mathrm{B}$ (period 2, group 13) is least metallic. So, correct order is $\mathrm{K > Mg > Al > B}$ .

Statement II is false: For cations, ionic radius is smaller than atomic radius. But for anions, ionic radius is larger than atomic radius. Hence atomic radius is not always greater than ionic radius.

Therefore, correct option is D.

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 24 January, Morning Shift
Subject: Chemistry
Chapter: Periodicity of Elements and Periodic Properties
Topic: Periodicity in Physical Properties (shielding and sizes)