The given reaction illustrates the direct iodination of...

Direct iodination of aromatic compounds is a reversible process because the byproduct, $HI$, is a strong reducing agent. It reduces the formed iodoarene back to the starting arene.

$Ar-H + I_2 \rightleftharpoons Ar-I + HI$

To drive the equilibrium in the forward direction, the $HI$ produced must be consumed. This is achieved by carrying out the reaction in the presence of an oxidizing agent such as $HIO_3$, $HIO_4$, or $HNO_3$.

The oxidizing agent oxidizes $HI$ to $I_2$, thereby preventing the reverse reaction and shifting the equilibrium to the right:

$7HI + HIO_4 \rightarrow 4I_2 + 4H_2O$

Among the given options, $HIO_4$ is the only suitable oxidizing agent.

Answer: $HIO_4$