Which of the following converts benzenediazonium chloride into phenol?

The correct answer is Option (1) → Allowing its aqueous solution to warm up to 283 K

Benzenediazonium chloride is stable only in the cold. When its solution is allowed to warm up to 283 K, the salt is hydrolysed by water: the diazonium group leaves as nitrogen gas and an –OH group takes its place.

    \( \displaystyle \ce{C6H5N2^+Cl^- + H2O -> C6H5OH + N2 ^ + HCl} \)

Why the other options are incorrect

Option (2) Treating its solution with potassium iodide — Incorrect. The iodide ion replaces the diazonium group and gives iodobenzene:

    \( \displaystyle \ce{C6H5N2^+Cl^- + KI -> C6H5I + KCl + N2 ^} \)

Option (3) Treating it with fluoroboric acid and heating the precipitate — Incorrect. Benzenediazonium fluoroborate is precipitated, and on heating it gives fluorobenzene:

    \( \displaystyle \ce{C6H5N2^+BF4^-} \xrightarrow{\Delta} \ce{C6H5F + BF3 + N2 ^} \)

Option (4) Treating its solution with Cu₂Cl₂/HCl — Incorrect. This is the Sandmeyer reaction: in the presence of Cu(I) the chloride ion replaces the diazonium group, giving chlorobenzene:

    \( \displaystyle \ce{C6H5N2^+Cl^-} \xrightarrow{\ce{Cu2Cl2/HCl}} \ce{C6H5Cl + N2 ^} \)

NCERT (2026-27 edition):
9.9 Chemical Reactions – “5. Replacement by hydroxyl group: If the temperature of the diazonium salt solution is allowed to rise upto 283 K, the salt gets hydrolysed to phenol.”