In the complexes `[SbF_(5)]^(2-),sp^(3)d` hybridisation is present. Geometry of the complex is:

Options

A. Square pyramidal
B. Square bipyramidal
C. Tetrahedral
D. square planar

Answer

To determine the geometry of the complex \([SbF_5]^{2-}\) with \(sp^3d\) hybridization, we can follow these steps:

### Step 1: Identify the central atom and its oxidation state
- The central atom in the complex is Antimony (Sb).
- In the complex \([SbF_5]^{2-}\), we can deduce the oxidation state of Sb. Fluorine (F) has an oxidation state of -1, and there are five fluorine atoms contributing a total of -5. Therefore, the oxidation state of Sb can be calculated as follows:
\[ x + 5(-1) = -2 \implies x - 5 = -2 \implies x = +3 \]
- Thus, the oxidation state of Sb in \([SbF_5]^{2-}\) is +3.

### Step 2: Determine the number of ligands
- The complex has five fluorine ligands coordinated to the central antimony atom.

### Step 3: Identify the hybridization
- The hybridization can be determined by the number of ligands and lone pairs around the central atom.
- For five ligands, the hybridization is \(sp^3d\). This indicates that one \(s\), three \(p\), and one \(d\) orbital are involved in hybridization.

### Step 4: Determine the geometry based on hybridization
- The geometry associated with \(sp^3d\) hybridization is square pyramidal.
- In square pyramidal geometry, there are five ligands: four ligands occupy the corners of a square base, and one ligand is positioned above the center of the square.

### Step 5: Conclusion
- Therefore, the geometry of the complex \([SbF_5]^{2-}\) with \(sp^3d\) hybridization is square pyramidal.

### Final Answer
The geometry of the complex \([SbF_5]^{2-}\) is square pyramidal.

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