Given below are two statements :

Statement-I : Benzenediazonium salt is prepared by the reaction of aniline with nitrous acid at 273 – 278 K. It decomposes easily in the dry state.
Statement-II : Insertion of iodine into the benzene ring is difficult and hence iodobenzene is prepared through the reaction of benzenediazonium salt with KI.
In the light of the above statements, choose the most appropriate answer from the options given below :

**Statement I** – Formation and stability of benzenediazonium salt

1. Aniline (C₆H₅NH₂) reacts with nitrous acid (HNO₂) generated in situ from NaNO₂ and HCl.
2. The reaction is carried out at low temperature (273 – 278 K) to avoid rapid decomposition.
3. The product is benzenediazonium chloride (C₆H₅N₂⁺ Cl⁻).
4. In the dry solid state the diazonium ion is highly unstable; it readily loses N₂ gas and gives phenol, bromobenzene (with Br⁻), etc.
5. Therefore the statement that benzenediazonium salt is prepared from aniline and nitrous acid at 273 – 278 K and that it decomposes easily when dry is accurate.

**Statement II** – Preparation of iodobenzene

1. Direct electrophilic substitution of iodine into the benzene ring is difficult because I₂ is a relatively weak electrophile and the C–I bond formation is not favored under ordinary halogenation conditions.
2. A practical method is the Sand ‑ Meyer reaction: the benzenediazonium salt is treated with potassium iodide (KI).
3. The diazonium group is replaced by an iodine atom, giving iodobenzene (C₆H₅I) and nitrogen gas as by‑product.
4. This route circumvents the problematic direct iodination and is widely used in organic synthesis.
5. Hence the statement that iodobenzene is prepared through the reaction of benzenediazonium salt with KI is correct.

**Conclusion** – Both statements are true, so the appropriate answer is “Both Statement I and Statement II are correct.”