Which of the following is correct about benzenediazonium fluoroborate?

The correct answer is Option (2) → It is water-insoluble and stable at room temperature

Benzenediazonium fluoroborate, \( \ce{C6H5N2^+BF4^-} \), differs from benzenediazonium chloride: the chloride dissolves readily in water and decomposes in the dry state, but the fluoroborate is insoluble in water and stable at room temperature. This is why it precipitates when fluoroboric acid is added to a diazonium chloride solution.

Why the other options are incorrect

Option (1) It is readily soluble in water at room temperature — Incorrect. Ready solubility in water describes benzenediazonium chloride; the fluoroborate is water insoluble.

Option (3) It is a coloured solid — Incorrect. Benzenediazonium chloride is a colourless crystalline solid, and the fluoroborate has the same diazonium ion with a colourless \( \ce{BF4^-} \) ion in place of \( \ce{Cl^-} \), so it is not a coloured solid either.

Option (4) It decomposes easily in the dry state — Incorrect. This is a property of benzenediazonium chloride; the fluoroborate is stable at room temperature.

NCERT (2026-27 edition):
9.8 Physical Properties – “… Benzenediazonium fluoroborate is water insoluble and stable at room temperature.”