For OF2 molecule consider the following: (A) Number of lone pairs on oxygen is 2. (B) FOF angle is less than 104.5°.…

🧠 $\ce{OF2}$ (oxygen difluoride): F is more electronegative than O, so O acts as the central atom. O has 6 valence electrons, forms 2 bonds to F, leaving 4e = 2 lone pairs.

(A) "O has 2 lone pairs" — Correct ✓

(B) "FOF angle < 104.5°" — Correct ✓. F is more electronegative than H, so in $\ce{OF2}$ the bonding pairs are pulled further toward F than in $\ce{H2O}$. This increases LP–BP repulsion relative to BP–BP, compressing the F–O–F angle below the H–O–H angle of 104.5°. (Measured: ~103.2°)

(C) "Oxidation state of O is −2" — Incorrect ✗. F is more electronegative than O, so F has O.S. = −1. Using $\ce{OF2}$: OS(O) + 2(−1) = 0 → OS(O) = +2, not −2.

(D) "Molecule is bent V-shaped" — Correct ✓. 2 bonds + 2 LP → tetrahedral electron geometry → bent molecular shape.

(E) "Molecular geometry is linear" — Incorrect ✗. The presence of 2 lone pairs makes it bent, not linear.

$\boxed{\text{Answer: (d) A, B, D only}}$