The product formed on oxidation of \(I^–\) with \(MnO_4^–\) in alkaline medium is:

\(IO_3^–\)

The correct answer is option 4. \(IO_3^–\).

The product formed on oxidation of \(I^–\) with \(MnO_4^–\) in alkaline medium is (4) \(IO_3^–\).

In alkaline medium, \(MnO_4^–\) is reduced to \(MnO_2\), and \(I^–\) is oxidized to \(IO_3^–\). The reaction is as follows

\(2MnO_4^– + I^– + H_2O \longrightarrow 2MnO_2 + IO_3^– + 2OH^–\)

The oxidation state of iodine changes from –1 to +5 in this reaction. Therefore, the product is \(IO_3^–\).

The other options are incorrect:

Option (1), \(IO_4^–\), is incorrect because the oxidation state of iodine in \(IO_4^–\) is +7.

Option (2), \(I_2\), is incorrect because \(I_2\) is not formed in this reaction.

Option (3), \(IO^–\), is incorrect because the oxidation state of iodine in \(IO^–\) is +1.