JEE Main 2026 — Chemistry Inorganic Chemistry

2026mcqhard

The correct increasing order of spin-only magnetic moment values of the complex ions $\left[\mathrm{MnBr}_{4}\right]^{2-}(\mathrm{A}),\left[\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}(\mathrm{B}),\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}(\mathrm{C})$ and $\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}(\mathrm{D})$ is :

Official previous-year question

Held on 28 Jan 2026 · Verified 6 Jul 2026.

Options

A
$\mathrm{C}=\mathrm{D} < \mathrm{B} < \mathrm{A}$
B
$\mathrm{C} < \mathrm{B} < \mathrm{D} < \mathrm{A}$
C
$\mathrm{A}=\mathrm{B} < \mathrm{D} < \mathrm{C}$
D
$A=B < C < D$

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Solution

Calculate spin-only magnetic moment using $\mu_s = \sqrt{n(n+2)}$ where $n$ is the number of unpaired electrons.

For $[\text{MnBr}_4]^{2-}$ (A): Mn(II) in tetrahedral environment is high spin with 5 unpaired electrons. $\mu_s = \sqrt{5(7)} = \sqrt{35} \approx 5.92$ BM.

For $[\text{Cu}(\text{H}_2\text{O})_6]^{2+}$ (B): Cu(II) has 1 unpaired electron. $\mu_s = \sqrt{1(3)} = \sqrt{3} \approx 1.73$ BM.

For $[\text{Ni}(\text{CN})_4]^{2-}$ (C): Ni(II) in square planar geometry is low spin with 0 unpaired electrons. $\mu_s = 0$ BM.

For $[\text{Ni}(\text{H}_2\text{O})_6]^{2+}$ (D): Ni(II) in octahedral environment is high spin with 2 unpaired electrons. $\mu_s = \sqrt{2(4)} = \sqrt{8} \approx 2.83$ BM.

Increasing order: C (0) < B (1.73) < D (2.83) < A (5.92)

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Solution

Calculate spin-only magnetic moment using $\mu_s = \sqrt{n(n+2)}$ where $n$ is the number of unpaired electrons.

For $[\text{MnBr}_4]^{2-}$ (A): Mn(II) in tetrahedral environment is high spin with 5 unpaired electrons. $\mu_s = \sqrt{5(7)} = \sqrt{35} \approx 5.92$ BM.

For $[\text{Cu}(\text{H}_2\text{O})_6]^{2+}$ (B): Cu(II) has 1 unpaired electron. $\mu_s = \sqrt{1(3)} = \sqrt{3} \approx 1.73$ BM.

For $[\text{Ni}(\text{CN})_4]^{2-}$ (C): Ni(II) in square planar geometry is low spin with 0 unpaired electrons. $\mu_s = 0$ BM.

For $[\text{Ni}(\text{H}_2\text{O})_6]^{2+}$ (D): Ni(II) in octahedral environment is high spin with 2 unpaired electrons. $\mu_s = \sqrt{2(4)} = \sqrt{8} \approx 2.83$ BM.

Increasing order: C (0) < B (1.73) < D (2.83) < A (5.92)

JEE Main 2026 — Chemistry Inorganic Chemistry

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JEE Main 2026 — Chemistry Inorganic Chemistry

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Solution

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