The structure of `XeOF_(4)` is

Options

A. Trigonal bipyramidal
B. Square pyramidal
C. Pentagonal planar
D. Octahedral

Answer

To determine the structure of \( \text{XeOF}_4 \), we will follow these steps:

### Step 1: Determine the Valence Electrons
- **Xenon (Xe)** has 8 valence electrons.
- **Oxygen (O)** has 6 valence electrons.
- **Fluorine (F)** has 7 valence electrons, and since there are 4 fluorine atoms, they contribute \( 4 \times 7 = 28 \) valence electrons.

**Total Valence Electrons Calculation:**
\[ \text{Total} = 8 (\text{Xe}) + 6 (\text{O}) + 28 (4 \times \text{F}) = 42 \text{ valence electrons} \]

### Step 2: Determine the Central Atom and Bonds
- The central atom in this molecule is xenon (Xe).
- Xenon will form a double bond with oxygen and single bonds with the four fluorine atoms.

### Step 3: Draw the Lewis Structure
- Start by placing the double bond between xenon and oxygen.
- Then, connect the four fluorine atoms to xenon with single bonds.
- After forming these bonds, we need to account for the remaining electrons.

**Bonding:**
- 1 double bond with oxygen uses 4 electrons.
- 4 single bonds with fluorine use 4 electrons (1 for each bond).

**Total Electrons Used:**
\[ 4 (\text{double bond}) + 4 (\text{single bonds}) = 8 \text{ electrons} \]

**Remaining Electrons:**
\[ 42 - 8 = 34 \text{ electrons} \]

### Step 4: Assign Lone Pairs
- After forming the bonds, the remaining electrons will be assigned as lone pairs.
- Since xenon can expand its octet, it will have one lone pair left after forming the bonds.

### Step 5: Determine Hybridization
- The hybridization of xenon in this case can be determined by counting the regions of electron density around it.
- There are 5 regions of electron density (4 single bonds with fluorine and 1 double bond with oxygen).
- This corresponds to \( \text{sp}^3\text{d}^2 \) hybridization.

### Step 6: Determine the Molecular Geometry
- With \( \text{sp}^3\text{d}^2 \) hybridization and one lone pair, the molecular geometry will be square pyramidal.

### Final Answer
The structure of \( \text{XeOF}_4 \) is **square pyramidal**.

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