Find the number of lone pairs of electrons preent in `OF_(2)` .

Explanation

To find the number of lone pairs of electrons present in OF₂, we can follow these steps:

### Step 1: Determine the electronic configuration of oxygen and fluorine.
- Oxygen (O) has an atomic number of 6. Its electronic configuration is 1s² 2s² 2p⁴.
- Fluorine (F) has an atomic number of 9. Its electronic configuration is 1s² 2s² 2p⁵.

### Step 2: Identify the valence electrons.
- Oxygen has 6 electrons in total, with 2 in the 1s orbital and 4 in the 2s and 2p orbitals. Therefore, it has 6 - 2 = 4 valence electrons.
- Fluorine has 7 electrons in total, with 2 in the 1s orbital and 5 in the 2s and 2p orbitals. Therefore, it has 7 - 2 = 5 valence electrons.

### Step 3: Determine how many electrons are needed for each atom to complete their octet.
- Oxygen needs 2 more electrons to complete its octet (it has 6 electrons and needs 8).
- Each fluorine atom needs 1 electron to complete its octet (each has 7 electrons and needs 8).

### Step 4: Draw the Lewis structure of OF₂.
- In the Lewis structure, oxygen will be the central atom. It will share one electron with each of the two fluorine atoms.
- This means that oxygen will form two single bonds (one with each fluorine).
- After forming these bonds, oxygen will have 2 lone pairs of electrons remaining (4 original valence electrons - 2 shared electrons = 2 lone pairs).
- Each fluorine atom will have 3 lone pairs remaining (5 original valence electrons - 1 shared electron = 4 lone pairs).

### Step 5: Count the total number of lone pairs.
- Oxygen has 2 lone pairs.
- Each fluorine has 3 lone pairs, and since there are 2 fluorine atoms, that totals to 3 x 2 = 6 lone pairs from fluorine.
- Therefore, the total number of lone pairs in OF₂ is 2 (from oxygen) + 6 (from fluorine) = 8 lone pairs.

### Final Answer:
The total number of lone pairs of electrons present in OF₂ is **8**.

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