What is the total number of lons pair of electron present on `Xe` in `XeF_(2)` ?

Explanation

To determine the total number of lone pairs of electrons present on xenon (Xe) in xenon difluoride (XeF₂), 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: Analyze the Bonding in XeF₂
In XeF₂, xenon forms bonds with two fluorine (F) atoms. Each fluorine atom requires one electron to complete its octet, which means xenon will share 2 of its valence electrons (one with each fluorine).

### Step 3: Calculate Remaining Electrons
After forming bonds with the two fluorine atoms, xenon will have:
- Initial valence electrons: 8
- Electrons used in bonding: 2
- Remaining electrons = 8 - 2 = 6

### Step 4: Identify Lone Pairs
The remaining 6 electrons will be present as lone pairs. Since each lone pair consists of 2 electrons, we can calculate the number of lone pairs:
- Number of lone pairs = Remaining electrons / 2 = 6 / 2 = 3 lone pairs

### Step 5: Conclusion
Thus, the total number of lone pairs of electrons present on xenon in XeF₂ is 3.

### Final Answer
**The total number of lone pairs of electrons present on xenon in XeF₂ is 3.**

---