(C₇H₅O₂)₂ →(hv) [X] → 2C₆H₅ + 2CO₂
Consider the above reaction and identify the intermediate 'X'

Solution

Chemistry ma’am · 4 steps

Light breaks the weak O–O bond evenly, so each half keeps one electron: you get a radical, not an ion.

1. Name the reactant. (C₇H₅O₂)₂ is benzoyl peroxide, C₆H₅COO-OOCC₆H₅. Its weakest link is the O-O single bond in the middle.
2. Homolysis by light. Light (hν) splits a bond homolytically: each oxygen takes back one of the two shared electrons. No charges form, so the two fragments are two identical benzoyloxy radicals.

   C₆H₅COO-OOCC₆H₅ →(hν) 2 C₆H₅COO• → 2 C₆H₅• + 2 CO₂

3. Check the radical. C₆H₅COO• has 28 + 5 + 12 = 45 valence electrons, an odd count. The unpaired electron sits on the single-bonded oxygen, next to its two lone pairs. Losing neutral CO₂ then leaves the neutral phenyl radical C₆H₅•, which matches the products.
4. So. X is the benzoyloxy radical C₆H₅COO•, drawn in option (d) with one lone electron and two lone pairs on oxygen.

The idea to keep. Light or heat on a peroxide means homolysis of the weak O–O bond, so expect radicals, never ions. Check that each fragment loses only neutral molecules to reach the products.

Answer from the NTA final answer key. Solution by PrepWiser.