The shape of `XeF_3^+` is

Options

A. Trigonal planar
B. Pyramidal
C. Bent T-shape
D. See - saw

Answer

To determine the shape of the molecule \( \text{XeF}_3^+ \), we will follow these steps:

### Step 1: Determine the Valence Electrons
- **Xenon (Xe)** has 8 valence electrons (Group 18).
- **Fluorine (F)** has 7 valence electrons. Since there are 3 fluorine atoms, the total contribution from fluorine is \( 3 \times 7 = 21 \) electrons.
- The positive charge (+1) indicates that one electron is lost. Therefore, we subtract 1 from the total count.

**Total valence electrons calculation:**
\[ \text{Total} = 8 (\text{from Xe}) + 21 (\text{from 3 F}) - 1 (\text{due to +1 charge}) = 28 \text{ valence electrons} \]

### Step 2: Draw the Lewis Structure
- Place xenon in the center and surround it with three fluorine atoms.
- Connect each fluorine atom to xenon with a single bond (each bond uses 2 electrons).
- After forming 3 bonds (6 electrons used), we have \( 28 - 6 = 22 \) electrons remaining.
- Place 3 lone pairs (6 electrons) on each fluorine to complete their octets.
- The remaining \( 22 - 18 = 4 \) electrons will be placed as 2 lone pairs on xenon.

### Step 3: Count Electron Groups
- The central atom (xenon) is surrounded by:
- 3 bonding pairs (from the Xe-F bonds)
- 2 lone pairs
- This gives a total of \( 3 + 2 = 5 \) electron groups.

### Step 4: Determine Electron Geometry
- With 5 electron groups, the electron geometry is **trigonal bipyramidal**.

### Step 5: Determine Molecular Shape
- The presence of 2 lone pairs will affect the molecular shape. Lone pairs occupy equatorial positions to minimize repulsion.
- The remaining bonding pairs will form a **T-shaped** molecular geometry.

### Step 6: Conclusion
- Therefore, the shape of \( \text{XeF}_3^+ \) is **T-shaped**.

### Final Answer
The shape of \( \text{XeF}_3^+ \) is T-shaped.

---