Consider the transition metal ions Mn^3+, Cr^3+, Fe^3+ and Co^3+… — Solution | JEE Chemistry

MCQEasyJEE 2026Crystal Field Theory

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

A
$Cr^{3+} > Mn^{3+} > Fe^{3+} > Co^{3+}$
B
$Fe^{3+} > Co^{3+} > Mn^{3+} > Cr^{3+}$
C
$Mn^{3+} > Fe^{3+} > Co^{3+} > Cr^{3+}$
D
$Cr^{3+} > Fe^{3+} > Co^{3+} > Mn^{3+}$

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The ions are $Cr^{3+}$ , $Mn^{3+}$ , $Fe^{3+}$ and $Co^{3+}$ , and all form low spin octahedral complexes.

Find: The correct decreasing order of unpaired electrons in their respective $d$ -orbitals.

In low spin octahedral complexes, electrons pair in the $t_{2g}$ set before occupying the $e_g$ set.

First determine the $d$ -electron count of each ion:

\text{Cr}^{3+} : d^3, \quad \text{Mn}^{3+} : d^4, \quad \text{Fe}^{3+} : d^5, \quad \text{Co}^{3+} : d^6

Now count the unpaired electrons for the low spin case:

\text{Cr}^{3+} (d^3) : 3 \text{ unpaired}

\text{Mn}^{3+} (d^4) : 2 \text{ unpaired}

\text{Fe}^{3+} (d^5) : 1 \text{ unpaired}

\text{Co}^{3+} (d^6) : 0 \text{ unpaired}

Therefore, the decreasing order is:

\text{Cr}^{3+} > \text{Mn}^{3+} > \text{Fe}^{3+} > \text{Co}^{3+}

So, the correct option is A.

Common mistakes

Assuming all octahedral complexes are high spin. This is wrong because the question explicitly states low spin octahedral complexes. Always fill the $t_{2g}$ orbitals first with pairing before placing electrons in $e_g$ .
Using the neutral atom configuration instead of the ion configuration. This is wrong because unpaired electrons must be counted for $Cr^{3+}$ , $Mn^{3+}$ , $Fe^{3+}$ and $Co^{3+}$ , not for the neutral atoms. First convert each metal to its correct $d^n$ state.
Counting only total electrons and not the number of unpaired electrons. This is wrong because the order depends on unpaired electrons in the crystal field arrangement, not on the total number of $d$ -electrons. Draw or mentally fill the low spin octahedral splitting pattern before comparing.

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Step-by-step solution

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