Q3. What happens if a solution of benzenediazonium chloride is allowed to warm to 283 K?
It is hydrolysed to phenol, with loss of nitrogen: C6H5N2+Cl− + H2O → C6H5OH + N2 + HCl (NCERT page 276).
Q4. Which diazonium salt in NCERT is stable at room temperature?
Benzenediazonium fluoroborate: NCERT says it “is water insoluble and stable at room temperature” (page 275). Stable here is a comparison with the chloride; handle any isolated diazonium salt only under supervision (our safety advice).
Q5. What colour is benzenediazonium chloride, and is it a solid or a liquid?
It is a colourless crystalline solid (NCERT page 275).
Q6. Fluoroboric acid is added to a cold solution of 0.0150 mol of benzenediazonium chloride, and 2.50 g of dry benzenediazonium fluoroborate is collected. What is the percentage yield? (C = 12, H = 1, N = 14, B = 10.8, F = 19.)
One mole of chloride gives at most one mole of C6H5N2+BF4− (M = 191.8 g mol−1, Example 2): 0.0150 × 191.8 = 2.877 g. The yield is 2.50/2.877 × 100 = 86.9%, about 87%. This is a theoretical yield calculation for an assumed pure sample, not a laboratory instruction: NCERT’s “stable at room temperature” does not mean a diazonium salt can be safely dried (our safety advice).