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

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Solution

(a) 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.

(b) When element A, sodium, reacts with element B, chlorine, it forms a product C, sodium chloride, NaCl.
$2 N a + C l_{2} \to 2 N a C l$
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.

$N a C l (a q) e l e c t r o l y s i s - ------ \to N a O H (s) + H_{2} (g) + C l_{2} (g)$

(c) Compound C is sodium chloride. It has no action on litmus solution. Because sodium chloride is a salt, it is neutral in nature and does not show any action with litmus solution.

(d) Element B is chlorine. It is in the gaseous state at room temperature.

(e) 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 $B C l_{3}$ .

$2 B + 3 C l_{2} \to 2 B C l_{3}$

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