Question

Question 64
An element A burns with golden flame in air. It reacts with another element B, atomic number 17 to give a product C. An aqueous solution of product C on electrolysis gives a compound D and liberates hydrogen. Identify A, B, C and D. Also write down the equations for the reactions involved.

Answer 1

1. When sodium is exposed to air it burns out with a golden flame. Hence, A is $\text{Na}$.

2. The element having atomic number 17 is chlorine. Hence B is ${\text{Cl}}_2$.

3. When sodium reacts with chlorine, it forms sodium chloride $\left(\text{NaCl}\right)$ which is an ionic compound. The reaction involved is:

$2\text{Na(s)}+{\text{Cl}}_2\text{(g)}\to 2\text{NaCl(s)}$

Hence, C is $\text{NaCl}$.

4. On electrolysis, $\text{NaCl}$ solution forms a sodium compound i.e. sodium hydroxide and releases chlorine and hydrogen gas. The reaction involved is:

$2\text{NaCl (aq)}+2{\text{H}}_2\text{O (l)}\to 2\text{NaOH (aq)}+{\text{Cl}}_2\text{(g)}+{\text{H}}_2\text{(g)}$

Hence, D is $\text{NaOH}$.