Question

An element A which is a part of common salt and kept under kerosene reacts with another element B of atomic number 17 to give a product C. When an aqueous solution of product C is electrolysed then a compound D is formed and two gases are liberated.

1. What are A and B?
2. Identify C and D.
3. What will be the action of C on litmus solution? Why?
4. State whether element B is a solid, liquid or gas at room temperature.
5. Write formula of the compound formed when element B reacts with an element E having atomic number 5.

Answer 1

1. Element A is sodium, and element B is chlorine. Sodium is kept under kerosene because it is highly reactive. The element with atomic number 17 is chlorine. Thus, element A is sodium and element B is chlorine.
2. When element A (sodium) reacts with element B (chlorine), it forms a product C, sodium chloride (NaCl).
   \[\ce{2Na + Cl2 -> 2NaCl}\]
   When an aqueous solution of sodium chloride is electrolysed, compound D, sodium hydroxide (NaOH), is formed along with the evolution of hydrogen and chlorine gases.
   \[\ce{2NaCl_{(aq)} + 2H2O_{(l)} ->[Electrolysis] 2NaOH_{(aq)} + H2_{(g)} + Cl2_{(g)}}\]
3. Compound C is sodium chloride. Sodium chloride has no action on litmus because it is a neutral salt formed from a strong acid (HCl) and a strong base (NaOH), so it does not hydrolyse in water.
4. Element B is chlorine. It is in a gaseous state at room temperature.
5. The element E with atomic number 5 is boron. When element B, chlorine, reacts with element E, boron, it forms a compound, boron trichloride. The formula of the compound formed is BCl₃.
   \[\ce{2B + 3Cl2 -> 2BCl3}\]