Chemistry · Coordination Compounds

JEE Main 2026 — 22 January, Morning Shift — Question 49

Consider the transition metal ions $\mathrm{Mn}^{3+}, \mathrm{Cr}^{3+}, \mathrm{Fe}^{3+}$ and $\mathrm{Co}^{3+}$ and all form low spin octahedral complexes. The correct decreasing order of unpaired electrons in their respective d-orbitals of the complexes is

Option A:
$\mathrm{Cr}^{3+}>\mathrm{Fe}^{3+}>\mathrm{Co}^{3+}>\mathrm{Mn}^{3+}$
Option B:
$\mathrm{Mn}^{3+}>\mathrm{Fe}^{3+}>\mathrm{Co}^{3+}>\mathrm{Cr}^{3+}$
Option C:
$\mathrm{Fe}^{3+}>\mathrm{Co}^{3+}>\mathrm{Mn}^{3+}>\mathrm{Cr}^{3+}$
Option D:
$\mathrm{Cr}^{3+}>\mathrm{Mn}^{3+}>\mathrm{Fe}^{3+}>\mathrm{Co}^{3+}$
Correct

Answer: D

Step-by-step solution

$\mathrm{Co}^{3+} \longrightarrow 3 \mathrm{~d}^{6} \Rightarrow \mathrm{t}_{2 \mathrm{~g}}^{2,2,2} \mathrm{e}_{\mathrm{g}}^{0,0}$ , unpaired electron $=0$ $\mathrm{Fe}^{3+} \rightarrow 3 \mathrm{~d}^{5} \Rightarrow \mathrm{t}_{2 \mathrm{~g}}^{2,2,1} \mathrm{e}_{\mathrm{g}}^{0,0}$ unpaired electron $=1$ $\mathrm{Cr}^{3+} \rightarrow 3 \mathrm{~d}^{3} \Rightarrow \mathrm{t}_{2 \mathrm{~g}}^{1,1,1} \mathrm{e}_{\mathrm{g}}^{0,0}$ unpaired electron $=3$ $\mathrm{Mn}^{3+} \rightarrow 3 \mathrm{~d}^{4} \Rightarrow \mathrm{t}_{2 \mathrm{~g}}^{2,1,1} \mathrm{e}_{\mathrm{g}}^{0,0}$ unpaired electron $=2$

Answer key and solution verified before publishing.

Practise Coordination Compounds

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2026
Paper: 22 January, Morning Shift
Subject: Chemistry
Chapter: Coordination Compounds
Topic: Fundamentals of Co-ordination Compounds