Question

The oxidizing power of the halogens varies in the order

• A. > < >
• B. < > >
• C. > > >
• D.

Answer 1

The correct option is C > > >

Flourine is an extremely powerful oxidizing agent. This has more to do with its anomalous nature owing to very high electronegativity and small atomic radius. The $F-F$ bond is surprisingly weak due to the lone pair - bond pair ( and lone pair - lone pair) repulsions. This bond being weak, is liable to be broken and Fluorine being very electronegative tends to grab electrons from almost any available source. This explains why Fluorine is extremely oxidizing.

As for Chlorine, Bromine and Iodine, as we go down the group, the oxidizing nature of the element decreases. For example, Chlorine can oxidize Bromide or Iodide ions to their respective elements in solution. Similarly, Bromine can oxidize iodide ion (but not Chloride ion) to Iodine. Hence, the oxidizing power of $C{l}_2>B{r}_2>I_2$. Fluorine directly reacts with water so we cannot fit it into the same trend. But then, Fluorine is the most powerful oxidizing element.
Hence the trend is $F_2>C{l}_2>B{r}_2>I_2$.