The shape of `XeF_(5)^(-)` will be

Options

A. square pyramidal
B. trigonal bipyramidal
C. planar
D. pentagonal bipyramidal

Explanation

To determine the shape of the ion \( \text{XeF}_5^{-} \), we can follow these steps:

### Step 1: Determine the Valence Electrons
- **Xenon (Xe)** has 8 valence electrons.
- Each **Fluorine (F)** atom contributes 1 electron, and since there are 5 fluorine atoms, they contribute a total of \( 5 \times 1 = 5 \) electrons.
- The negative charge of the ion adds 1 additional electron.

**Total Calculation:**
\[ \text{Total electrons} = \text{Valence electrons of Xe} + \text{Electrons from F} + \text{Charge} = 8 + 5 + 1 = 14 \]

### Step 2: Determine the Hybridization
- The total number of electrons (14) corresponds to 7 electron pairs.
- The hybridization for 7 electron pairs is \( \text{sp}^3\text{d}^3 \).

### Step 3: Determine the Geometry
- With \( \text{sp}^3\text{d}^3 \) hybridization, the geometry is typically **pentagonal bipyramidal**.
- However, we need to account for lone pairs. In \( \text{XeF}_5^{-} \), there are 5 bond pairs (from the 5 fluorine atoms) and 2 lone pairs (since we have 7 total pairs).

### Step 4: Determine the Shape
- The presence of 2 lone pairs will affect the shape. The lone pairs will occupy equatorial positions to minimize repulsion.
- Therefore, the shape of \( \text{XeF}_5^{-} \) will be **pentagonal planar**.

### Final Answer
The shape of \( \text{XeF}_5^{-} \) is **pentagonal planar**.

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