Question:
Identify the geometry of a complex compound \([\text{CoF}_{6}]^{-3}\).
Identify the geometry of a complex compound \([\text{CoF}_{6}]^{-3}\).
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Find the oxidation state of cobalt, then the hybridisation given a weak field ligand.
Updated On: Nov 12, 2026
- Square planar
- Tetrahedral
- Octahedral
- Trigonal pyramidal
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept
The geometry depends on the hybridisation of the central metal ion, which is decided by the ligand field strength.
Step 2: Calculation
In \([\text{CoF}_6]^{3-}\), \(x+6(-1)=-3\), so cobalt is \(\text{Co}^{3+}\) with \(3d^{6}\).
\(\text{F}^-\) is a weak field ligand, so no pairing of \(3d\) electrons occurs.
Step 3: Hybridisation
The outer \(4s\), three \(4p\) and two \(4d\) orbitals are used, giving \(sp^{3}d^{2}\) hybridisation (high spin).
Six ligands with \(sp^{3}d^{2}\) hybridisation sit at the corners of an octahedron.
Final Answer:
The complex is octahedral. \[ \boxed{\text{(C) }\text{Octahedral}} \]
The geometry depends on the hybridisation of the central metal ion, which is decided by the ligand field strength.
Step 2: Calculation
In \([\text{CoF}_6]^{3-}\), \(x+6(-1)=-3\), so cobalt is \(\text{Co}^{3+}\) with \(3d^{6}\).
\(\text{F}^-\) is a weak field ligand, so no pairing of \(3d\) electrons occurs.
Step 3: Hybridisation
The outer \(4s\), three \(4p\) and two \(4d\) orbitals are used, giving \(sp^{3}d^{2}\) hybridisation (high spin).
Six ligands with \(sp^{3}d^{2}\) hybridisation sit at the corners of an octahedron.
Final Answer:
The complex is octahedral. \[ \boxed{\text{(C) }\text{Octahedral}} \]
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