Chemistry Q60 – JEE-MAIN 2026

Home Chemistry d- & f-Block ElementsQuestion #363

Given below are two statements : Statement I: The number of pairs among

[T i^{4 +}, V^{2 +}], [V^{2 +}, M n^{2 +}], [M n^{2 +}, F e^{3 +}]

and

[V^{2 +}, C r^{2 +}]

in which both ions are coloured is 3. Statement II: The number of pairs among

[L a^{3 +}, Y b^{2 +}], [L u^{3 +}, C e^{4 +}]

and

[A c^{3 +}, L r^{3 +}]

ions in which both are diamagnetic is 3. In the light of the above statements, choose the correct from the options given below :

Clue 1 of 3: Electronic Configuration and Properties

Recall the relationship between electronic configuration and properties like color and magnetism for transition metal and f-block ions.

🧠 Full Solution Path

Step 1: Analyze Statement I for colored ions✦ Active

An ion is colored if it has unpaired d-electrons. Let's determine the electronic configuration for each ion and check for unpaired electrons:

- $T i^{4 +}$ : $3 d^{0}$ (colorless)

- $V^{2 +}$ : $3 d^{3}$ (colored, 3 unpaired electrons)

- $M n^{2 +}$ : $3 d^{5}$ (colored, 5 unpaired electrons)

- $F e^{3 +}$ : $3 d^{5}$ (colored, 5 unpaired electrons)

- $C r^{2 +}$ : $3 d^{4}$ (colored, 4 unpaired electrons)

Now, let's check the given pairs:

- $[T i^{4 +}, V^{2 +}]$ : $T i^{4 +}$ is colorless. (Not both colored)

- $[V^{2 +}, M n^{2 +}]$ : Both are colored. (1st pair)

- $[M n^{2 +}, F e^{3 +}]$ : Both are colored. (2nd pair)

- $[V^{2 +}, C r^{2 +}]$ : Both are colored. (3rd pair)

The number of pairs in which both ions are colored is 3. Therefore, Statement I is correct.

Step 2: Analyze Statement II for diamagnetic ions○ Expand

An ion is diamagnetic if all its electrons are paired (i.e., $d^{0}$ , $d^{10}$ , $f^{0}$ , or $f^{14}$ configurations). Let's determine the electronic configuration for each ion:

- $L a^{3 +}$ : $[X e] 4 f^{0}$ (diamagnetic)

- $Y b^{2 +}$ : $[X e] 4 f^{14}$ (diamagnetic)

- $L u^{3 +}$ : $[X e] 4 f^{14}$ (diamagnetic)

- $C e^{4 +}$ : $[X e] 4 f^{0}$ (diamagnetic)

- $A c^{3 +}$ : $[R n] 5 f^{0}$ (diamagnetic)

- $L r^{3 +}$ : $[R n] 5 f^{14}$ (diamagnetic)

Now, let's check the given pairs:

- $[L a^{3 +}, Y b^{2 +}]$ : Both are diamagnetic. (1st pair)

- $[L u^{3 +}, C e^{4 +}]$ : Both are diamagnetic. (2nd pair)

- $[A c^{3 +}, L r^{3 +}]$ : Both are diamagnetic. (3rd pair)

The number of pairs in which both ions are diamagnetic is 3. Therefore, Statement II is correct.

Step 3: Conclusion○ Expand

Since both Statement I and Statement II are correct, the correct option is 'Both Statement I and Statement II are correct'.

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Chemistry Question 60 – JEE-MAIN 2026

Clue 1 of 3: Electronic Configuration and Properties

🧠 Full Solution Path