JEE Main 2026 — Apr 4, Shift 2 — Chemistry solution

JEE Main2026Apr 4, Shift 2Chemistry

Consider $∣ x ∣$ is the difference in oxidation states of Mn in highest manganese fluoride and highest manganese oxide. The ions with $∣ x ∣$ number of unpaired electrons from the following are:

A. Sc $^{3 +}$ B. Zn $^{2 +}$ C. V $^{2 +}$ D. Fe $^{2 +}$ E. Co $^{2 +}$

Choose the correct answer from the options given below:

A
A and B Only
B
C, D and E Only
C
C and E Only
D
B and E Only

Solution

Highest oxide of Mn is Mn $_{2}$ O $_{7}$ (Mn in +7); highest fluoride is MnF $_{4}$ (Mn in +4). Difference $∣ x ∣ = 7 - 4 = 3$ .

Need ions with exactly 3 unpaired electrons.

Sc $^{3 +}$ : [Ar] — 0 unpaired.

Zn $^{2 +}$ : [Ar] 3d $^{10}$ — 0 unpaired.

V $^{2 +}$ : [Ar] 3d $^{3}$ — 3 unpaired. ✓

Fe $^{2 +}$ : [Ar] 3d $^{6}$ — 4 unpaired.

Co $^{2 +}$ : [Ar] 3d $^{7}$ — 3 unpaired. ✓

$∴$ C and E only.

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