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Among the following which is the strongest oxidizing agent?
a) \[B{{r}_{2}}\]
b) \[{{I}_{2}}\]
c) \[C{{l}_{2}}\]
d) \[{{F}_{2}}\]

Answer
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Hint: So, for checking the strength of oxidizing agents we have to check their tendency to transfer oxygen or to gain electrons in a redox chemical reaction.

Complete step by step answer:
In this question molecules of all halogen atoms are given. We know that oxidizing agents (also called an oxidizer or oxidant) is a chemical compound that readily transfers oxygen atoms or a substance that gains electrons in a redox chemical reaction. An element having a higher tendency to get reduced or to accept an electron is a strong oxidizing agent.

The Halogen which has a higher value of standard reduction potential will be the strongest oxidizing agent. Oxidizing agent oxidizes other elements and reduces itself. Fluorine is the most electronegative element because electronegativity decreases and size increases on moving down the group and also electron gain tendency decreases down the group. Hence, it gets reduced readily into \[{{F}^{-}}\] ion and is the strongest oxidizing agent.
Standard reduction potential value of these four halogen molecule is in the following order \[{{F}_{2}}\] >\[C{{l}_{2}}\]>\[B{{r}_{2}}\]>\[{{I}_{2}}\]
So, from the above explanation we can say that fluorine is the strongest oxidizing agent.

Then the correct answer is option “D”.

Note: All the halogens have a strong tendency to accept electrons. Therefore halogens act as strong oxidising agents and their oxidising power decreases from fluorine to iodine.