Question

Which of the following statement is true for the Benzenediazonium fluoroborate?

• A. When it heated with \( \mathrm{NaNO_2} \) in presence of Cu, it gives aniline.
• B. On heating, it does not decompose to yield fluorobenzene.
• C. It is water insoluble and stable at room temperature.
• D. It is water soluble and unstable at room temperature.

Hint:

Remember benzenediazonium fluoroborate through the Balz-Schiemann reaction: it is stable, can be isolated, and on heating it gives fluorobenzene.

Answer 1

The Correct Option is C

Step 1: Understand the nature of benzenediazonium fluoroborate.
Benzenediazonium fluoroborate is a diazonium salt represented as \( \mathrm{C_6H_5N_2^+BF_4^-} \). It is a special diazonium salt because it is comparatively more stable than ordinary benzenediazonium chloride or bromide. This increased stability is due to the presence of the fluoroborate ion.
Step 2: Recall its important property.
Unlike many diazonium salts, benzenediazonium fluoroborate is water insoluble and can be isolated in solid form. It is also stable at room temperature. On heating, it undergoes decomposition to give fluorobenzene. This reaction is known as the Balz-Schiemann reaction.
\[ \mathrm{C_6H_5N_2^+BF_4^- \xrightarrow{\Delta} C_6H_5F + BF_3 + N_2} \] Step 3: Analyze the options.

• (A) Incorrect. This statement is not related to the normal behavior of benzenediazonium fluoroborate, and it does not give aniline in this way.
• (B) Incorrect. On heating, benzenediazonium fluoroborate does decompose to yield fluorobenzene.
• (C) Correct. It is water insoluble and stable at room temperature.
• (D) Incorrect. This is opposite to the actual property.

Step 4: Conclusion.
Therefore, the true statement about benzenediazonium fluoroborate is that it is water insoluble and stable at room temperature.
Final Answer:It is water insoluble and stable at room temperature.