JEE Main 2026 — Jan 22, Shift 1 — Chemistry solution

JEE Main2026Jan 22, Shift 1Chemistry

Consider the transition metal ions $M n^{3 +}$ , $C r^{3 +}$ , $F e^{3 +}$ and $C o^{3 +}$ and all form low spin octahedral complexes. The correct decreasing order of unpaired electrons in their respective d-orbitals of the complexes is

A
$C r^{3 +} > F e^{3 +} > C o^{3 +} > M n^{3 +}$
B
$M n^{3 +} > F e^{3 +} > C o^{3 +} > C r^{3 +}$
C
$F e^{3 +} > C o^{3 +} > M n^{3 +} > C r^{3 +}$
D
$C r^{3 +} > M n^{3 +} > F e^{3 +} > C o^{3 +}$

Solution

Count unpaired electrons in each low-spin octahedral d-configuration. With strong-field splitting, electrons fill $t_{2 g}$ before $e_{g}$ .

$C r^{3 +}$ — $d^{3}$ : $t_{2 g}^{1, 1, 1} e_{g}^{0}$ , 3 unpaired.

$M n^{3 +}$ — $d^{4}$ LS: $t_{2 g}^{2, 1, 1} e_{g}^{0}$ , 2 unpaired.

$F e^{3 +}$ — $d^{5}$ LS: $t_{2 g}^{2, 2, 1} e_{g}^{0}$ , 1 unpaired.

$C o^{3 +}$ — $d^{6}$ LS: $t_{2 g}^{2, 2, 2} e_{g}^{0}$ , 0 unpaired.

Decreasing order: $C r^{3 +} > M n^{3 +} > F e^{3 +} > C o^{3 +}$ .

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