Question

Write the equations for reaction of chlorine with the following:
(a) Cold and dilute $NaOH$
(b) Excess $N{H_3}$
(c) $Ca{(OH)_2}$
(d) $N{a_2}{S_2}{O_3}$

Answer 1

Hint: We know that chlorine has an atomic number of seventeen. Chlorine is an extremely reactive gas and it is an oxidising agent. It has the highest electron affinity. It is the third highest electronegative element in the modern periodic table after fluorine and oxygen. Chlorine is a P Block element.

Complete step by step solution:
As in the question we are given to react chlorine with four other different reactants. We will try to solve one by one. So,

(a) When we react chlorine with Cold and dilute $NaOH$, the chemical equation we will get is $2NaOH(aq) + C{l_2}(g) \to NaCl(aq) + NaOCl(aq) + {H_2}O$. Here we can see that we will get sodium chloride, water and sodium chlorate as products of the reaction.

(b) When we react chlorine with Excess $N{H_3}$, the chemical equation we will get is $8N{H_3} + C{l_2} \to 6N{H_4}Cl + {N_2}$. Here we can see that we will get nitrogen gas and ammonium chloride as products of the reaction.

(c) When we react chlorine with $Ca{(OH)_2}$, the chemical equation we will get is $Ca{(OH)_2} + C{l_2} \to CaOC{l_2} + {H_2}O$. Here we can see that we will get Calcium Hypochlorite and water as products of the reaction.

(d) When we react chlorine and $N{a_2}{S_2}{O_3}$, the chemical equation we will get is $N{a_2}{S_2}{O_3} + C{l_2} + {H_2}O \to N{a_2}S{O_4} + 2HCl + S$. Here we can see that we will get hydrochloric acid,sodium sulphate and sulphur as products of the reaction.

Note: Always remember that chlorine is a very reactive element. It has high electronegativity. It is a yellow green gas. It is a halogen. Remember the following reactions of chlorine because these are very important. $2NaOH(aq) + C{l_2}(g) \to NaCl(aq) + NaOCl(aq) + {H_2}O$, $8N{H_3} + C{l_2} \to 6N{H_4}Cl + {N_2}$,$Ca{(OH)_2} + C{l_2} \to CaOC{l_2} + {H_2}O$,$N{a_2}{S_2}{O_3} + C{l_2} + {H_2}O \to N{a_2}S{O_4} + 2HCl + S$.