JEE Main 2026 Apr 8 Shift 2, Chemistry Q10: d-Block Elements

JEE Main2026Apr 8, Shift 2Chemistry

Given below are two statements :

Statement I : The number of pairs among $[Ti^{4 +}, V^{2 +}]$ , $[V^{2 +}, Mn^{2 +}]$ , $[Mn^{2 +}, Fe^{3 +}]$ and $[V^{2 +}, Cr^{2 +}]$ in which both ions are coloured is $3$ .

Statement II : The number of pairs among $[La^{3 +}, Yb^{2 +}]$ , $[Lu^{3 +}, Ce^{4 +}]$ and $[Ac^{3 +}, Lr^{3 +}]$ ions in which both are diamagnetic is $3$ .

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

A
Both Statement I and Statement II are correct
B
Both Statement I and Statement II are incorrect
C
Statement I is correct but Statement II is incorrect
D
Statement I is incorrect but Statement II is correct

Solution

Statement I. Colour in $d$ -block ions usually arises from $d$ – $d$ transitions, which require partially-filled $d$ orbitals.

Ti $^{4 +}$ ( $3 d^{0}$ ) — colourless. V $^{2 +}$ ( $3 d^{3}$ ), Mn $^{2 +}$ ( $3 d^{5}$ ), Fe $^{3 +}$ ( $3 d^{5}$ ), Cr $^{2 +}$ ( $3 d^{4}$ ) — all coloured.

Pairs where both ions are coloured: [V $^{2 +}$ , Mn $^{2 +}$ ], [Mn $^{2 +}$ , Fe $^{3 +}$ ], [V $^{2 +}$ , Cr $^{2 +}$ ] → $3$ pairs. The [Ti $^{4 +}$ , V $^{2 +}$ ] pair is disqualified by Ti $^{4 +}$ . Statement I is correct.

Statement II. Check $f$ -configurations: La $^{3 +}$ ( $4 f^{0}$ ), Yb $^{2 +}$ ( $4 f^{14}$ ), Lu $^{3 +}$ ( $4 f^{14}$ ), Ce $^{4 +}$ ( $4 f^{0}$ ), Ac $^{3 +}$ ( $5 f^{0}$ ), Lr $^{3 +}$ ( $5 f^{14}$ ).

All listed ions have either empty or completely filled $f$ subshells, so all are diamagnetic. All $3$ pairs satisfy the condition. Statement II is correct.

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JEE Main 2026 Apr 8 Shift 2, Chemistry Q10: d-Block Elements

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