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:

Let us analyze the given statements one by one:

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.

Aniline reacts with nitrous acid ($HN{O}_2$, which is prepared in situ from sodium nitrite and a mineral acid like hydrochloric acid) at a low temperature of 273-278 K (0-5 °C) to form benzenediazonium chloride. This reaction is known as diazotization. Benzenediazonium salts are stable only in solution at low temperatures. In the dry state, they are highly unstable and decompose easily, sometimes explosively. Therefore, Statement I is correct.

Statement II: Insertion of iodine into the benzene ring is difficult and hence iodobenzene is prepared through the reaction of benzenediazonium salt with KI.

Direct iodination of benzene with iodine ($I_2$) is a reversible reaction because the hydroiodic acid ($HI$) formed during the reaction is a strong reducing agent that reduces iodobenzene back to benzene. Therefore, direct iodination is difficult and requires the presence of strong oxidizing agents (like $HN{O}_3$ or $HI{O}_4$) to remove $HI$. Alternatively, iodobenzene is easily prepared by heating a solution of benzenediazonium salt with potassium iodide (KI). Thus, Statement II is also correct.

Therefore, both Statement I and Statement II are correct.