The comcept of redistribution of energy in different orbitals of an atom associated with different energies to give new orbitals of equal (or somethimes it may be non-equal) energy oriented in space in definite directions is called hybridization and formed new orbitals are called hybrid orbitals. The bonds formed by such orbitals are called hybrid bonds. The process of mixing of orbitals itself requires some energy. Thus, some additional energy, is needed for the hybridisation (mixing) of atomic orbitals.
Q. In neutral moleule `XeO_(n_(1)) F_(n_(2))`, central atom has no lone pair and ratio of `(n_(2))/(n_(1))` is two, then which of the following orbitals does not participate in bonding (`n_(1) and n_(2)` are natural numbers):

Options

A. `d_(x^(2))`
B. `p_(x)`
C. `d_(x^(2)-y^(2))`
D. None of these

Explanation

To solve the question regarding the hybridization and bonding in the molecule \( \text{XeO}_{n_1} \text{F}_{n_2} \) where the central atom has no lone pairs and the ratio of \( \frac{n_2}{n_1} = 2 \), we can follow these steps:

### Step 1: Determine the Valence Electrons of Xenon
Xenon (Xe) is a noble gas and has 8 valence electrons. The electron configuration of xenon is \( [Kr] 4d^{10} 5s^2 5p^6 \).

**Hint:** Remember that noble gases have a complete octet, which is crucial for understanding their bonding behavior.

### Step 2: Analyze the Ratio of \( n_2 \) to \( n_1 \)
Given that \( \frac{n_2}{n_1} = 2 \), we can express \( n_2 \) in terms of \( n_1 \):
- Let \( n_1 = 2 \) (for oxygen)
- Then \( n_2 = 4 \) (for fluorine)

**Hint:** The ratio helps us determine how many atoms of each element are bonded to the central atom.

### Step 3: Determine the Bonding Structure
With \( n_1 = 2 \) and \( n_2 = 4 \), the molecule can be represented as \( \text{XeO}_2\text{F}_4 \). In this structure:
- There are 2 double bonds with oxygen (each oxygen forms a double bond with xenon).
- There are 4 single bonds with fluorine.

**Hint:** Visualize the bonding to ensure that the total number of bonds corresponds to the number of valence electrons available.

### Step 4: Calculate the Steric Number
The steric number is calculated as the number of sigma bonds plus the number of lone pairs. In this case:
- Number of sigma bonds = 6 (2 from \( \text{O} \) and 4 from \( \text{F} \))
- Number of lone pairs = 0 (as stated in the question)

Thus, the steric number = 6.

**Hint:** The steric number helps determine the hybridization of the central atom.

### Step 5: Determine the Hybridization
With a steric number of 6, the hybridization of xenon in this molecule is \( \text{sp}^3\text{d}^2 \). This hybridization involves:
- 3 p orbitals and 2 d orbitals.

**Hint:** Knowing the hybridization can help identify which orbitals are involved in bonding.

### Step 6: Identify the Orbitals Involved in Bonding
The orbitals involved in \( \text{sp}^3\text{d}^2 \) hybridization are:
- \( p_x, p_y, p_z \) (the three p orbitals)
- \( d_{x^2-y^2} \) and \( d_{z^2} \) (the two d orbitals)

**Hint:** List the orbitals that participate in bonding to find out which one does not.

### Step 7: Identify the Orbital That Does Not Participate
From the \( \text{sp}^3\text{d}^2 \) hybridization, the orbital \( d_{x^2-y^2} \) does not participate in bonding as it is not involved in the formation of the bonds in this molecule.

**Final Answer:** The orbital that does not participate in bonding is \( d_{x^2-y^2} \).

### Summary
1. Determine valence electrons of xenon.
2. Analyze the ratio of \( n_2 \) to \( n_1 \).
3. Determine the bonding structure.
4. Calculate the steric number.
5. Determine the hybridization.
6. Identify the orbitals involved in bonding.
7. Identify the orbital that does not participate.