Question

Which of the following is the strongest oxidizing agent?

• A. F
• B. Br
• C. Cl
• D. I

Answer 1

The correct option is A

F

Explanation of the correct option:

(A)$\mathrm{F}$.

Oxidizing agent

“An oxidizing agent is a chemical species that tends to oxidize other substances, i.e. cause an increase in the oxidation state of the substance by making it lose electrons.”

• $\mathrm{F}$ is the most electronegative element and it can gain electrons easily to attain its stable state.
• The bond enthalpy $\mathrm{F}$ is low, so it has a high reduction potential.
• Because of its highest reduction potential $\mathrm{F}$ is the most oxidizing agent.

Explanation of the incorrect options:

(B)$\mathrm{Br}$.

• $\mathrm{Br}$ is considered a halogen compound and it has a larger atomic radius.
• Because of that the electrons present in the outer shells of it feel less the nuclear force of attraction and so easily release electrons, so it is known as a reducing agent.
• It is a weak oxidizing agent.

(C)$\mathrm{Cl}$.

• $\mathrm{Cl}$ is an electronegative element.
• It needs one electron in its valence shell to occupy the one empty space,
• It can easily take electrons from other atoms, so they are known as an oxidizing agent.
• But the atomic radius $\mathrm{Cl}$ is more than $\mathrm{F}$, so the $\mathrm{Cl}$ is a less oxidizing agent than $\mathrm{F}$.

(D)$\mathrm{I}$

• $\mathrm{I}$ has a larger atomic radius than all of the halogens.
• So the electrons present in the pouter shells feel the less nuclear force of attraction.
• So they can easily lose electrons to form iodine molecules and $\mathrm{I}$ is the weaker oxidizing agents and strong reducing agent.

The arrangement of the given halogen in the increasing order of their oxidizing power:

$\mathrm{I}$<$\mathrm{Br}$<$\mathrm{Cl}$<$\mathrm{F}$

Hence, option (A) is correct., $\mathrm{F}$ is the strongest oxidizing agent.