JEE Main 2026 — Chemistry Inorganic Chemistry

2026integermedium

Identify the metal ions among $\mathrm{Co}^{2+}, \mathrm{Ni}^{2+}, \mathrm{Fe}^{2+}, \mathrm{V}^{3+}$ and $\mathrm{Ti}^{2+}$ having a spin-only magnetic moment value more than 3.0 BM. The sum of unpaired electrons present in the high spin octahedral complexes formed by those metal ions is $\_\_\_\_$ .

Official previous-year question

Held on 21 Jan 2026 · Verified 6 Jul 2026.

Did you get this right?

Solution

The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons.

For $\mu > 3.0$ BM, $n$ must be $\ge 3$ (since for $n=2$ , $\mu = \sqrt{8} \approx 2.83$ BM and for $n=3$ , $\mu = \sqrt{15} \approx 3.87$ BM).

Electronic configurations and unpaired electrons ( $n$ ) in high spin octahedral complexes:

Metal Ion	Configuration	$n$	$\mu$ (BM)
$Co^{2+}$	$[Ar] 3d^7$	3	3.87
$Ni^{2+}$	$[Ar] 3d^8$	2	2.83
$Fe^{2+}$	$[Ar] 3d^6$	4	4.90
$V^{3+}$	$[Ar] 3d^2$	2	2.83
$Ti^{2+}$	$[Ar] 3d^2$	2	2.83

The metal ions with $\mu > 3.0$ BM are $Co^{2+}$ ( $n=3$ ) and $Fe^{2+}$ ( $n=4$ ).

The sum of unpaired electrons for these ions is $3 + 4 = 7$ .

Your note

JEE Main Inorganic Chemistry in other years

Inorganic Chemistry 2025 Inorganic Chemistry 2024 Inorganic Chemistry 2023 Inorganic Chemistry 2022 Inorganic Chemistry 2021 Inorganic Chemistry 2020 Inorganic Chemistry 2019 Inorganic Chemistry 2018 Inorganic Chemistry 2017 Inorganic Chemistry 2016 Inorganic Chemistry 2015 Inorganic Chemistry 2014

Keep practicing Inorganic Chemistry

Work through every JEE Main Inorganic Chemistry PYQ, year by year.

Browse all Chemistry topics

JEE Main 2026 — Chemistry Inorganic Chemistry

2026integermedium

Official previous-year question

Held on 21 Jan 2026 · Verified 6 Jul 2026.

Did you get this right?

Solution

The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons.

For $\mu > 3.0$ BM, $n$ must be $\ge 3$ (since for $n=2$ , $\mu = \sqrt{8} \approx 2.83$ BM and for $n=3$ , $\mu = \sqrt{15} \approx 3.87$ BM).

Electronic configurations and unpaired electrons ( $n$ ) in high spin octahedral complexes:

Metal Ion	Configuration	$n$	$\mu$ (BM)
$Co^{2+}$	$[Ar] 3d^7$	3	3.87
$Ni^{2+}$	$[Ar] 3d^8$	2	2.83
$Fe^{2+}$	$[Ar] 3d^6$	4	4.90
$V^{3+}$	$[Ar] 3d^2$	2	2.83
$Ti^{2+}$	$[Ar] 3d^2$	2	2.83

The metal ions with $\mu > 3.0$ BM are $Co^{2+}$ ( $n=3$ ) and $Fe^{2+}$ ( $n=4$ ).

The sum of unpaired electrons for these ions is $3 + 4 = 7$ .

Your note

JEE Main Inorganic Chemistry in other years

Keep practicing Inorganic Chemistry

Work through every JEE Main Inorganic Chemistry PYQ, year by year.

Browse all Chemistry topics

JEE Main 2026 — Chemistry Inorganic Chemistry

Did you get this right?

Solution

Your note

JEE Main Inorganic Chemistry in other years

More JEE Main Inorganic Chemistry questions

Other Chemistry topics for JEE Main

Physical Chemistry

Organic Chemistry

Keep practicing Inorganic Chemistry

JEE Main 2026 — Chemistry Inorganic Chemistry

Did you get this right?

Solution

Your note

JEE Main Inorganic Chemistry in other years

More JEE Main Inorganic Chemistry questions

Other Chemistry topics for JEE Main

Physical Chemistry

Organic Chemistry

Keep practicing Inorganic Chemistry