Question

In OF\(_2\), the number of bond pairs and lone pairs of electrons are respectively:

• A. 2, 6
• B. 2, 8
• C. 2, 9
• D. 2, 10

Hint:

OF\(_2\): 2 bond pairs, 8 lone pairs (3 on each F, 2 on O)

Answer 1

The Correct Option is B

Molecule: OF\(_2\)
- Central atom = Oxygen (6 valence electrons)
- Fluorine atoms = 2 × 7 = 14 electrons
- Total valence electrons = 6 + 14 = 20
Bond pairs:
- 2 O–F bonds → 2 bond pairs = 4 electrons
Lone pairs:
- Remaining = \(20 - 4 = 16\) electrons → 8 lone pairs
Lone pair distribution:
- 3 lone pairs on each F → 6
- 2 lone pairs on O → 2
→ Total = 8 lone pairs

Answer 2

In OF₂, the number of bond pairs and lone pairs of electrons are respectively:

Step 1: Determine the total number of valence electrons:
- Oxygen (O) has 6 valence electrons
- Each fluorine (F) has 7 valence electrons
Total electrons in OF₂:
\[ 6\ (\text{O}) + 2 \times 7\ (\text{F}) = 6 + 14 = 20\ \text{valence electrons} \]

Step 2: Draw the Lewis structure:
- Oxygen is the central atom
- It forms a single bond with each fluorine atom
- Each bond contains 1 bond pair (2 electrons)

Step 3: Count the bond pairs:
- Two O–F single bonds → 2 bond pairs

Step 4: Distribute remaining electrons as lone pairs:
- 2 bonds = 4 electrons used
- Remaining = 20 – 4 = 16 electrons
- Each fluorine gets 3 lone pairs (6 electrons), and oxygen gets 2 lone pairs (4 electrons)
Total lone pairs = 3 (on F₁) + 3 (on F₂) + 2 (on O) = 8 lone pairs

Step 5: Final count:
- Bond pairs = 2
- Lone pairs = 8

Final Answer:
\[ \boxed{2,\ 8} \]