WAEC Chemistry 2021 Paper

Concept:In Group VII, oxidizing power decreases down the group: $\mathrm{F_2 > Cl_2 > Br_2 > I_2}$. A halogen can oxidize the halide ion of an element below it in the group.Explanation:Oxidizing strength is the tendency to gain electrons and be reduced. This tendency is greatest for fluorine and decreases as atomic size increases down the group.Therefore, $\mathrm{F_2}$ is the strongest oxidizing agent, while $\mathrm{I_2}$ is the weakest.Option A is wrong because $\mathrm{Br_2}$ is below chlorine, so it cannot oxidize $\mathrm{Cl^-}$.Option B is wrong because $\mathrm{F_2}$ has the greatest tendency to be reduced, not the least.Option D is wrong because $\mathrm{I_2}$ is a weaker oxidizing agent than $\mathrm{F_2}$.Since chlorine is above iodine, $\mathrm{Cl_2}$ can oxidize $\mathrm{I^-}$ to iodine:$$\mathrm{Cl_2 + 2I^- \rightarrow 2Cl^- + I_2}$$Thus, the correct statement is option C.Answer:C. $\mathrm{Cl_2}$ will oxidize $\mathrm{I^-}$