Question

The correct order of the oxidizing power of halogens is:
A. $B{r_2} < C{l_2} < {F_2} < {I_2}$
B. ${F_2} < {I_2} < B{r_2} < C{l_2}$
C. ${I_2} < B{r_2} < C{l_2} < {F_2}$
D. ${I_2} > B{r_2} > C{l_2} > {F_2}$

Answer 1

The atom who gains electrons and causes another atom to lose electrons is called an oxidizing agent and its tendency to gain electrons is known as oxidizing power. The oxidising agents have the highest possible oxidation state.

Complete step by step answer: Since all halogen atoms have a strong tendency to accept an electron due to its electronic configuration i.e., there is the only requirement of one electron to complete their octet. Hence they act as a strong oxidizing agent. However their oxidizing power decreases from ${F_2}{\text{ to }}{I_2}$. Since fluorine is the strongest oxidizing agent amongst halogen, it will oxidize all other halide ions to the corresponding halogen in solution or dry state.
${F_2} + 2{X^ - } \to 2{F^ - } + {X_2}{\text{ }}(X = C{l^ - },B{r^ - },{I^ - })$
Similarly, $C{l_2}$ will oxide $B{r^ - }{\text{ and }}{I^ - }$ ions while bromide ion oxidizes iodide ions from their solution. In general, a halogen of a lower atomic number oxidizes halide ions of higher atomic number.
Hence the correct order of the oxidizing power of halogen is ${I_2} < B{r_2} < C{l_2} < {F_2}$.
So, the correct answer is “Option C”.

Additional Information: Alternatively, the oxidizing power of halogens can be compared based on their standard electrode potential.
${F_2} + 2{e^ - } \to 2{F^ - };{\text{ }}{{\text{E}}^{\text{o}}} = + 2.87{\text{V}}$
C{l_2} + 2{e^ - } \to 2C{l^ - };{\text{ }}{{\text{E}}^{\text{^\circ }}}{\text{ = + 1}}{\text{.36V}}
B{r_2} + 2{e^ - } \to 2B{r^ - };{\text{ }}{{\text{E}}^{\text{^\circ }}}{\text{ = + 1}}{\text{.09V}}
{I_2} + 2{e^ - } \to 2{I^ - };{\text{ }}{{\text{E}}^{\text{^\circ }}}{\text{ = + 0}}{\text{.54V}}
As we see the electrode potential of ${F_2}/{F^ - }$ is the maximum while that of ${I_2}/{I^ - }$ is the minimum i.e., ${F_2}$ reduced most easily but ${I_2}$ is least easily. In other word ${F_2}$ is the strongest oxidizing agent and ${I_2}$ is the weakest oxidizing agent

Note: The relative oxidizing power of halogens can further be illustrated by their reaction with water. ${F_2}$ oxidises water to oxygen and ozone whereas chlorine and bromine react with water to form corresponding hydrohalic and hydrohalic acid. Also, the reaction of iodine with water is non-spontaneous.