Noble gases have compleately filled valance shall i.e. `ns^(2)np^(6)` exceps He (i.e) .Noble gases are monoomic under normal conductions .Law bolding point of the ligher noble gases are due to weak van dor wads forces between the atoms and abance of any interature imaractions `Xe` reacts with `F_(2)` so give a sourceof fouoxide mently `XeF_(2),XeF_(4),XeF_(4),XeF_(3)` on complete hydrolyes gives `XeFe_(3)`,
Structure of `XeF_(4)` is

Options

A. Linear
B. Square plater
C. Tetrahedral
D. Pyramidel

Explanation

To determine the structure of xenon tetrafluoride (XeF₄), we will follow a systematic approach based on the principles of hybridization and molecular geometry.

### Step 1: Identify the central atom and its valence electrons
- The central atom in XeF₄ is xenon (Xe).
- Xenon has the electronic configuration of [Kr] 5s² 5p⁶, which means it has 8 valence electrons.

**Hint:** Remember that noble gases have a full valence shell, which contributes to their stability.

### Step 2: Count the surrounding atoms
- In XeF₄, there are 4 fluorine (F) atoms surrounding the xenon atom. Each fluorine atom contributes 1 electron to form a bond with xenon.

**Hint:** Monovalent atoms like fluorine only contribute one bond each.

### Step 3: Use the hybridization formula
The formula for calculating the number of hybrid orbitals is:
\[ \text{Number of hybrid orbitals} = \frac{1}{2} \left( \text{Valence electrons on central atom} + \text{Number of surrounding monovalent atoms} - \text{Charge} \right) \]
- Valence electrons on xenon = 8
- Number of surrounding monovalent atoms (F) = 4
- Charge = 0 (since there is no charge on the molecule)

Substituting these values into the formula:
\[ \text{Number of hybrid orbitals} = \frac{1}{2} (8 + 4 - 0) = \frac{1}{2} (12) = 6 \]

**Hint:** The charge affects the count of electrons; if there were a positive charge, you would subtract, and if negative, you would add.

### Step 4: Determine the hybridization
- Since we have 6 hybrid orbitals, the hybridization is sp³d².

**Hint:** The hybridization tells us about the geometry of the molecule.

### Step 5: Account for lone pairs
- In XeF₄, xenon forms 4 bonds with fluorine, using 4 of its valence electrons. This leaves 4 electrons (or 2 lone pairs) on xenon.

**Hint:** Lone pairs can influence the shape of the molecule.

### Step 6: Determine the molecular geometry
- With 4 bond pairs and 2 lone pairs, the arrangement of the electron pairs around the xenon atom is octahedral. However, the presence of 2 lone pairs will push the bond pairs down, resulting in a square planar shape.

**Hint:** The arrangement of lone pairs and bond pairs determines the final shape of the molecule.

### Conclusion
The structure of xenon tetrafluoride (XeF₄) is square planar.

**Final Answer:** The structure of XeF₄ is square planar.

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