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When chlorine water is exposed to sunlight, \[{{\text{O}}_2}\] is liberated. Hence:
A.Hydrogen has little affinity to \[{{\text{O}}_2}\]
B.Hydrogen has more affinity to \[{{\text{O}}_2}\]
C.Hydrogen has little affinity to \[{\text{C}}{{\text{l}}_2}\]
D.It is a reducing agent.

Last updated date: 22nd Jun 2024
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Hint: To write the reaction, one of the products is given to us. The other one can be found out by balancing the atoms. The other product form will tell us about the affinity of hydrogen towards the respective element.

Complete step by step answer:
Let us first write the reaction when chlorine water reacts with the oxygen in presence of sunlight. Chlorine exists as diatomic gas with the formula \[{\text{C}}{{\text{l}}_2}\]. Water is a binary compound made up of hydrogen and oxygen with a molecular formula \[{{\text{H}}_2}{\text{O}}\]. Oxygen is stable in a diatomic state, so hydrogen and chlorine will combine with each other to form a product which is hydrochloric acid. The reaction will be:
\[{\text{C}}{{\text{l}}_2} + {{\text{H}}_2}{\text{O}} \to 2{\text{HCl}} + \dfrac{1}{2}{{\text{O}}_2}\]
Now since the above reaction is happening then hydrogen must be getting attached with the chlorine leaving the oxygen atom. Hence it is clear from here that hydrogen is more likely to form bonds with chlorine than with oxygen. One of the reasons is low electronegativity of chlorine than oxygen. Chlorine easily gives electrons to hydrogen and the formation of ionic bonds is there. Hence hydrogen has more affinity towards the chlorine.

Hence, the correct option is B.

Chlorine is actually acting as an oxidizing agent and not a reducing agent. Addition of hydrogen is termed as reduction. In the above question hydrogen is being attached to chlorine so reduction of chlorine is taking place. The once who themselves get reduced, oxidize others, because oxidation and reduction are complementary to each other. Hence chlorine acts as an oxidizing agent.