Draw the structure of `XeF_4` molecule.

Options

A.
B.
C.
D.

Explanation

To draw the structure of the `XeF_4` molecule, follow these steps:

### Step 1: Determine the Valence Electrons
Xenon (Xe) is in Group 18 and has 8 valence electrons. Each fluorine (F) atom is in Group 17 and has 7 valence electrons. Since there are 4 fluorine atoms, the total number of valence electrons contributed by fluorine is \(4 \times 7 = 28\). Therefore, the total number of valence electrons in `XeF_4` is:
\[ 8 \, (\text{from Xe}) + 28 \, (\text{from 4 F}) = 36 \, \text{valence electrons} \]

### Step 2: Draw the Lewis Structure
- Place xenon in the center and the four fluorine atoms around it.
- Connect each fluorine atom to xenon with a single bond. This uses \(4 \times 2 = 8\) electrons (2 electrons per bond).
- After forming the bonds, we have \(36 - 8 = 28\) electrons left.

### Step 3: Distribute Remaining Electrons
- Each fluorine needs 6 more electrons to complete its octet (3 lone pairs). So, for 4 fluorine atoms, we need \(4 \times 6 = 24\) electrons.
- Place 6 electrons (3 lone pairs) around each fluorine atom. This uses up all 24 remaining electrons, leaving xenon with 2 lone pairs.

### Step 4: Determine the Molecular Geometry
- The molecular geometry of `XeF_4` can be determined by considering the positions of the bonding pairs and lone pairs.
- Xenon has 4 bonding pairs (from the 4 Xe-F bonds) and 2 lone pairs.
- The arrangement of 4 bonding pairs and 2 lone pairs corresponds to a square planar geometry.

### Step 5: Draw the Final Structure
- In the final structure, the fluorine atoms are positioned at the corners of a square, and the lone pairs are located above and below the plane of the square.

### Final Structure
```
F
|
F -- Xe -- F
|
F
```
This represents a square planar arrangement with two lone pairs above and below the plane.