1. It is sp³d² hybridised and tetrahedral 2. It is d²sp³ hybridised and octahedral 3. It is dsp² hybridised and square planar 4. It is sp³d² hybridised and octahedral

Answer: 2. It is d²sp³ hybridised and octahedral Key idea: cyanide is a strong-field ligand, which pairs the d electrons. In [Mn(CN)₆]³⁻ , Mn is +3 with 3d⁴ . With the strong ligand, the electrons pair up in two d orbitals ( t_2g⁴ ), freeing two 3d, one 4s and three 4p orbitals. Hybridisation is d²sp³ , geometry octahedral (inner orbital, low spin complex). sp³d² is the outer-orbital case, found with weak-field ligands.

Topic: Valence bond approach· Coordination Compounds

Pick out the correct statement with respect to [Mn(CN)_6]^3- :

1. It is \(sp^3d^2\) hybridised and tetrahedral

2. It is \(d^2sp^3\) hybridised and octahedral

3. It is \(dsp^2\) hybridised and square planar

4. It is \(sp^3d^2\) hybridised and octahedral

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Answer: 2. It is \(d^2sp^3\) hybridised and octahedral

Key idea: cyanide is a strong-field ligand, which pairs the d electrons.

In \([Mn(CN)_6]^{3-}\), Mn is +3 with \(3d^4\). With the strong ligand, the electrons pair up in two d orbitals (\(t_{2g}^4\)), freeing two 3d, one 4s and three 4p orbitals. Hybridisation is \(d^2sp^3\), geometry octahedral (inner orbital, low spin complex). \(sp^3d^2\) is the outer-orbital case, found with weak-field ligands.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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Pick out the correct statement with respect to [Mn(CN)_6]^3- :

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