Write the Lewis structure of $\mathrm{OF_2}$

Oxygen is the central atom: $\mathrm{F - O - F}$

• Oxygen has 6 valence electrons.
• Each fluorine has 7 valence electrons.
• Total valence electrons: $6 + 2\times 7 = 20$

Two $\mathrm{O-F}$ bonds use 4 electrons, leaving 16 electrons.

Each fluorine completes octet with 3 lone pairs each, using 12 electrons. Remaining 4 electrons stay on oxygen as 2 lone pairs.

So, oxygen has 2 lone pairs.

Therefore, A is true.

Determine shape and bond angle

Around oxygen, there are 4 electron pairs:

• 2 bond pairs
• 2 lone pairs

So the electron pair geometry is tetrahedral, but the molecular shape is bent / V-shaped.

Hence, D is true and E is false.

Compare bond angle with $104.5^\circ$

In $\mathrm{H_2O}$, the bond angle is about $104.5^\circ$.

In $\mathrm{OF_2}$, fluorine is highly electronegative and pulls bond pair electrons away from oxygen. This reduces bond pair-bond pair repulsion near oxygen, so the bond angle becomes smaller than $104.5^\circ$.

Experimentally, $\angle \mathrm{FOF} \approx 103^\circ$.

Therefore, B is true.

Find oxidation state of oxygen

Fluorine always has oxidation state $-1$.

Let oxidation state of oxygen be $x$: $x + 2(-1) = 0$ $x - 2 = 0$ $x = +2$

So oxygen is $+2$, not $-2$.

Therefore, C is false.

Evaluate all statements

• A: True
• B: True
• C: False
• D: True
• E: False

Hence the correct set is: $\boxed{\text{A, B, D only}}$

Match with options

This corresponds to Option B.