JEE Main 2026 — Chemistry Inorganic Chemistry

2026mcqmedium

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

Official previous-year question

Held on 22 Jan 2026 · Verified 6 Jul 2026.

Options

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

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Solution

For low spin octahedral complexes, electrons fill the $t_{2g}$ orbitals first before pairing or moving to $e_g$ orbitals. The configuration depends on the number of d-electrons.

For $Cr^{3+}$ : Electronic configuration is $3d^3$ . In an octahedral field, the configuration is $t_{2g}^3 e_g^0$ . Number of unpaired electrons ( $n$ ) = 3.

For $Mn^{3+}$ : Electronic configuration is $3d^4$ . For low spin, the configuration is $t_{2g}^4 e_g^0$ . Number of unpaired electrons ( $n$ ) = 2.

For $Fe^{3+}$ : Electronic configuration is $3d^5$ . For low spin, the configuration is $t_{2g}^5 e_g^0$ . Number of unpaired electrons ( $n$ ) = 1.

For $Co^{3+}$ : Electronic configuration is $3d^6$ . For low spin, the configuration is $t_{2g}^6 e_g^0$ . Number of unpaired electrons ( $n$ ) = 0.

Comparing the number of unpaired electrons:

$Cr^{3+} (3) > Mn^{3+} (2) > Fe^{3+} (1) > Co^{3+} (0)$ .

The correct decreasing order is $Cr^{3+} > Mn^{3+} > Fe^{3+} > Co^{3+}$ .

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Options

$\mathrm{Mn}^{3+}>\mathrm{Fe}^{3+}>\mathrm{Co}^{3+}>\mathrm{Cr}^{3+}$

$\mathrm{Cr}^{3+}>\mathrm{Mn}^{3+}>\mathrm{Fe}^{3+}>\mathrm{Co}^{3+}$

$\mathrm{Fe}^{3+}>\mathrm{Co}^{3+}>\mathrm{Mn}^{3+}>\mathrm{Cr}^{3+}$

$\mathrm{Cr}^{3+}>\mathrm{Fe}^{3+}>\mathrm{Co}^{3+}>\mathrm{Mn}^{3+}$

Solution

For low spin octahedral complexes, electrons fill the $t_{2g}$ orbitals first before pairing or moving to $e_g$ orbitals. The configuration depends on the number of d-electrons.

For $Cr^{3+}$ : Electronic configuration is $3d^3$ . In an octahedral field, the configuration is $t_{2g}^3 e_g^0$ . Number of unpaired electrons ( $n$ ) = 3.

For $Mn^{3+}$ : Electronic configuration is $3d^4$ . For low spin, the configuration is $t_{2g}^4 e_g^0$ . Number of unpaired electrons ( $n$ ) = 2.

For $Fe^{3+}$ : Electronic configuration is $3d^5$ . For low spin, the configuration is $t_{2g}^5 e_g^0$ . Number of unpaired electrons ( $n$ ) = 1.

For $Co^{3+}$ : Electronic configuration is $3d^6$ . For low spin, the configuration is $t_{2g}^6 e_g^0$ . Number of unpaired electrons ( $n$ ) = 0.

Comparing the number of unpaired electrons:

$Cr^{3+} (3) > Mn^{3+} (2) > Fe^{3+} (1) > Co^{3+} (0)$ .

The correct decreasing order is $Cr^{3+} > Mn^{3+} > Fe^{3+} > Co^{3+}$ .

JEE Main 2026 — Chemistry Inorganic Chemistry

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JEE Main 2026 — Chemistry Inorganic Chemistry

Options

Did you get this right?

Solution

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JEE Main Inorganic Chemistry in other years

More JEE Main Inorganic Chemistry questions

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