Question:
In which of the following complex ion, the central metal ion is in a state of \(sp^3d^2\) hybridisation?
In which of the following complex ion, the central metal ion is in a state of \(sp^3d^2\) hybridisation?
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Weak field ligands (F\(^-\), Cl\(^-\), H\(_2\)O) favour outer orbital (\(sp^3d^2\)) complexes.
Updated On: Apr 20, 2026
- \([\mathrm{CoF}_6]^{3-}\)
- \([\mathrm{Co}(\mathrm{NH}_3)_6]^{3+}\)
- \([\mathrm{Fe}(\mathrm{CN})_6]^{3-}\)
- \([\mathrm{Cr}(\mathrm{NH}_3)_6]^{3+}\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
\(sp^3d^2\) hybridisation gives outer orbital (high spin) octahedral complexes. Weak field ligands (F\(^-\)) cause this.
Step 2: Detailed Explanation:
(A) Co\(^{3+}\) in [CoF\(_6\)]\(^{3-}\): F\(^-\) is weak field ligand, so electrons do not pair, hybridisation = \(sp^3d^2\) (outer orbital).
(B) [Co(NH\(_3\))\(_6\)]\(^{3+}\): NH\(_3\) is strong field, hybridisation = \(d^2sp^3\) (inner orbital).
(C) [Fe(CN)\(_6\)]\(^{3-}\): CN\(^-\) is strong field, hybridisation = \(d^2sp^3\).
(D) [Cr(NH\(_3\))\(_6\)]\(^{3+}\): NH\(_3\) is strong field for Cr\(^{3+}\), hybridisation = \(d^2sp^3\).
Step 3: Final Answer:
\([\mathrm{CoF}_6]^{3-}\)
\(sp^3d^2\) hybridisation gives outer orbital (high spin) octahedral complexes. Weak field ligands (F\(^-\)) cause this.
Step 2: Detailed Explanation:
(A) Co\(^{3+}\) in [CoF\(_6\)]\(^{3-}\): F\(^-\) is weak field ligand, so electrons do not pair, hybridisation = \(sp^3d^2\) (outer orbital).
(B) [Co(NH\(_3\))\(_6\)]\(^{3+}\): NH\(_3\) is strong field, hybridisation = \(d^2sp^3\) (inner orbital).
(C) [Fe(CN)\(_6\)]\(^{3-}\): CN\(^-\) is strong field, hybridisation = \(d^2sp^3\).
(D) [Cr(NH\(_3\))\(_6\)]\(^{3+}\): NH\(_3\) is strong field for Cr\(^{3+}\), hybridisation = \(d^2sp^3\).
Step 3: Final Answer:
\([\mathrm{CoF}_6]^{3-}\)
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