In alkaline medium, $$MnO_4^-$$ oxidises $$I^-$$ to

We need to determine the product when $$MnO_4^-$$ oxidises $$I^-$$ in alkaline medium.

Understand the oxidising power of permanganate in alkaline medium.

In alkaline medium, permanganate ion ($$MnO_4^-$$) is a strong oxidising agent. It is reduced from Mn(+7) to Mn(+6) (as $$MnO_4^{2-}$$) or to MnO$$_2$$ depending on conditions.

Determine the oxidation product of iodide.

In alkaline medium, $$MnO_4^-$$ being a strong enough oxidiser can take iodide ($$I^-$$, oxidation state -1) all the way up to iodate ($$IO_3^-$$, oxidation state +5). The reaction is:

$$2MnO_4^- + I^- + H_2O \rightarrow 2MnO_2 + IO_3^- + 2OH^-$$

Verify the oxidation states.

- Iodine goes from -1 in $$I^-$$ to +5 in $$IO_3^-$$ (loss of 6 electrons)

- Manganese goes from +7 in $$MnO_4^-$$ to +4 in $$MnO_2$$ (gain of 3 electrons each, 6 total for 2 atoms)

Electrons are balanced.

The correct answer is Option (4): $$IO_3^-$$.