The shape of `XeF_6` is

Options

A. Pentagonal bipyramidal
B. Square planar
C. Octahedral
D. Distrorted octahedral

Answer

To determine the shape of \( \text{XeF}_6 \), we can follow these steps:

### Step 1: Identify the central atom and its valence electrons
The central atom in \( \text{XeF}_6 \) is xenon (Xe). Xenon is a noble gas and has 8 valence electrons.

### Step 2: Count the number of fluorine atoms and their contribution
Each fluorine (F) atom contributes 1 electron, and since there are 6 fluorine atoms, they contribute a total of 6 electrons.

### Step 3: Calculate the total number of electrons around the central atom
The total number of electrons around the xenon atom is:
\[ \text{Valence electrons of Xe} + \text{Electrons from F} = 8 + 6 = 14 \text{ electrons} \]

### Step 4: Determine the number of bonding pairs and lone pairs
Each bond between xenon and fluorine uses 2 electrons. Since there are 6 fluorine atoms, we have:
\[ \text{Bonding pairs} = \frac{6 \text{ bonds} \times 2 \text{ electrons/bond}}{2} = 6 \text{ bonding pairs} \]
Now, we have 14 total electrons and 12 electrons used in bonding (6 bonds), leaving us with:
\[ \text{Lone pairs} = \frac{14 - 12}{2} = 1 \text{ lone pair} \]

### Step 5: Determine the geometry using VSEPR theory
According to VSEPR (Valence Shell Electron Pair Repulsion) theory, the arrangement of electron pairs around the central atom determines the molecular geometry. With 6 bonding pairs and 1 lone pair, the arrangement is based on an octahedral geometry.

### Step 6: Identify the shape of the molecule
The presence of the lone pair distorts the ideal octahedral shape. Therefore, the shape of \( \text{XeF}_6 \) is a distorted octahedral geometry.

### Final Answer
The shape of \( \text{XeF}_6 \) is **distorted octahedral**.

---