`SbF_5` react with `XeF_4` to form an adduct. The shapes of cation and anion in the adduct are respectively:

Options

A. Square planar, trigonal bipyramidal
B. T-shaped, octahedral
C. Square pyramidal, octahedral
D. Square planar, octahedral

Explanation

To determine the shapes of the cation and anion formed when SbF5 reacts with XeF4, we can follow these steps:

### Step 1: Identify the Reactants and Products
- The reactants are **SbF5** (antimony pentafluoride) and **XeF4** (xenon tetrafluoride).
- When these two react, they form a cation **[XeF3]+** and an anion **[SbF6]−**.

### Step 2: Determine the Shape of the Cation [XeF3]+
- **Xenon (Xe)** is the central atom in the cation [XeF3]+.
- Xenon has 8 valence electrons. Since it has a positive charge, it effectively has 7 valence electrons (8 - 1 = 7).
- In [XeF3]+, xenon is bonded to 3 fluorine atoms and has 2 lone pairs of electrons.
- The arrangement of 3 bond pairs and 2 lone pairs leads to a **T-shaped** molecular geometry.

### Step 3: Determine the Shape of the Anion [SbF6]−
- **Antimony (Sb)** is the central atom in the anion [SbF6]−.
- Antimony has 5 valence electrons. With a negative charge, it has 6 effective valence electrons (5 + 1 = 6).
- In [SbF6]−, antimony is bonded to 6 fluorine atoms.
- The arrangement of 6 bond pairs with no lone pairs leads to an **octahedral** molecular geometry.

### Step 4: Conclusion
- The shapes of the cation and anion in the adduct formed from the reaction of SbF5 and XeF4 are:
- Cation: **T-shaped**
- Anion: **Octahedral**

Thus, the correct answer is **(b) T-shaped, octahedral**.

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