JEE Main 2026 — Chemistry Inorganic Chemistry

2026mcqhard

Given below are two statements:

Statement I: Hybridisation, shape and spin only magnetic moment of $\mathrm{K}_{3}\left[\mathrm{Co}\left(\mathrm{CO}_{3}\right)_{3}\right]$ is $\mathrm{sp}^{3} \mathrm{~d}^{2}$ , octahedral and 4.9 BM respectively.

Statement II: Geometry, hybridisation and spin only magnetic moment values $(\mathrm{BM})$ of the ions $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-},\left[\mathrm{MnBr}_{4}\right]^{2-}$ and $\left[\mathrm{CoF}_{6}\right]^{3-}$ respectively are square planar, tetrahedral, octahedral; $\mathrm{dsp}^{2}, \mathrm{sp}^{3}, \mathrm{sp}^{3} \mathrm{~d}^{2}$ and $0,5.9,4.9$ .

In the light of the above statements, choose the correct answer from the options given below

Official previous-year question

Held on 24 Jan 2026 · Verified 6 Jul 2026.

Options

A
Statement I is true but Statement II is false
B
Both Statement I and Statement II are false
C
Both Statement I and Statement II are true
D
Statement I is false but Statement II is true

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Solution

In Statement I, for $K_3[Co(CO_3)_3]$ , Cobalt is in $+3$ oxidation state ( $Co^{3+}: 3d^6$ ). Carbonate ( $CO_3^{2-}$ ) is a weak field ligand. Thus, no pairing occurs, resulting in 4 unpaired electrons. Hybridisation is $sp^3d^2$ , shape is octahedral, and $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.9$ BM. Statement I is true.

In Statement II:

For $[Ni(CN)_4]^{2-}$ , $Ni^{2+}$ is $3d^8$ . $CN^-$ is a strong field ligand, causing pairing. Hybridisation is $dsp^2$ , geometry is square planar, and $\mu = 0$ BM.

For $[MnBr_4]^{2-}$ , $Mn^{2+}$ is $3d^5$ . $Br^-$ is a weak field ligand. Hybridisation is $sp^3$ , geometry is tetrahedral, and $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.9$ BM.

For $[CoF_6]^{3-}$ , $Co^{3+}$ is $3d^6$ . $F^-$ is a weak field ligand. Hybridisation is $sp^3d^2$ , geometry is octahedral, and $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.9$ BM.

Statement II is true.

Both statements are true.

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Solution

In Statement II:

For $[Ni(CN)_4]^{2-}$ , $Ni^{2+}$ is $3d^8$ . $CN^-$ is a strong field ligand, causing pairing. Hybridisation is $dsp^2$ , geometry is square planar, and $\mu = 0$ BM.

For $[MnBr_4]^{2-}$ , $Mn^{2+}$ is $3d^5$ . $Br^-$ is a weak field ligand. Hybridisation is $sp^3$ , geometry is tetrahedral, and $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.9$ BM.

For $[CoF_6]^{3-}$ , $Co^{3+}$ is $3d^6$ . $F^-$ is a weak field ligand. Hybridisation is $sp^3d^2$ , geometry is octahedral, and $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.9$ BM.

Statement II is true.

Both statements are true.

JEE Main 2026 — Chemistry Inorganic Chemistry

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JEE Main 2026 — Chemistry Inorganic Chemistry

Options

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Solution

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