To determine the number of bond pairs and lone pairs of electrons in ${\mathrm{OF}}_2$, we first draw its Lewis structure.

Step 1: Calculate total valence electrons.
Oxygen (Group 16) has 6 valence electrons.
Fluorine (Group 17) has 7 valence electrons.
Total valence electrons = $6+2\times 7=6+14=20$ electrons.

Step 2: Determine the central atom and draw single bonds.
Oxygen is less electronegative than fluorine, so it is the central atom. Connect the two fluorine atoms to the oxygen atom with single bonds.
$\mathrm{F}-\mathrm{O}-\mathrm{F}$
These two single bonds account for $2\times 2=4$ electrons. Remaining electrons = $20-4=16$ electrons.

Step 3: Distribute remaining electrons as lone pairs.
First, complete the octets of the terminal fluorine atoms. Each fluorine needs 6 more electrons (3 lone pairs).
$2\times 6=12$ electrons are used for lone pairs on fluorine atoms.
Remaining electrons = $16-12=4$ electrons.

Step 4: Place remaining electrons on the central atom.
The 4 remaining electrons are placed as lone pairs on the central oxygen atom. This gives oxygen 2 lone pairs.

Step 5: Count bond pairs and lone pairs.
From the Lewis structure, there are 2 single bonds (O-F), so there are 2 bond pairs.
There are 3 lone pairs on each fluorine atom (total $2\times 3=6$ lone pairs) and 2 lone pairs on the oxygen atom. So, total lone pairs = $6+2=8$.

Option Analysis:

• A) 2,6: Incorrect. While there are 2 bond pairs, there are 8 lone pairs, not 6.
• B) 2,8: Correct. There are 2 bond pairs and 8 lone pairs in ${\mathrm{OF}}_2$.
• C) 2,10: Incorrect. There are 2 bond pairs, but only 8 lone pairs.
• D) 1,9: Incorrect. There are 2 bond pairs, not 1, and 8 lone pairs, not 9.

Correct Answer: (B)