Number of non-bonding electron pair of Xe in `XeF_(4)` and `XeF_(2)` respectively is

Options

A. 6,4
B. 3,2
C. 2,3
D. 1, 2

Explanation

To determine the number of non-bonding electron pairs of xenon (Xe) in the compounds \(XeF_4\) and \(XeF_2\), we can follow these steps:

### Step 1: Determine the Valence Electrons of Xenon
Xenon (Xe) is a noble gas and has 8 valence electrons.

### Step 2: Determine the Total Number of Valence Electrons in Each Compound
1. For \(XeF_4\):
- Fluorine (F) has 7 valence electrons and there are 4 fluorine atoms.
- Total valence electrons from fluorine = \(4 \times 7 = 28\).
- Total valence electrons in \(XeF_4 = 8 (from Xe) + 28 (from F) = 36\).

2. For \(XeF_2\):
- Total valence electrons from fluorine = \(2 \times 7 = 14\).
- Total valence electrons in \(XeF_2 = 8 (from Xe) + 14 (from F) = 22\).

### Step 3: Determine the Hybridization and Bonding
1. For \(XeF_4\):
- The hybridization is \(sp^3d^2\) because there are 6 electron pairs (4 bond pairs and 2 lone pairs).
- Bonding pairs = 4 (from 4 \(Xe-F\) bonds).
- Non-bonding pairs = Total pairs - Bonding pairs = 6 - 4 = 2.

2. For \(XeF_2\):
- The hybridization is \(sp^3\) because there are 5 electron pairs (2 bond pairs and 3 lone pairs).
- Bonding pairs = 2 (from 2 \(Xe-F\) bonds).
- Non-bonding pairs = Total pairs - Bonding pairs = 5 - 2 = 3.

### Step 4: Conclusion
- The number of non-bonding electron pairs in \(XeF_4\) is 2.
- The number of non-bonding electron pairs in \(XeF_2\) is 3.

Thus, the answer is:
- Non-bonding electron pairs in \(XeF_4\) = 2
- Non-bonding electron pairs in \(XeF_2\) = 3

### Final Answer:
The number of non-bonding electron pairs of Xe in \(XeF_4\) and \(XeF_2\) respectively is 2 and 3.

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