Hydrogen peroxide reacts with iodine in basic medium to give:

First, recall that hydrogen peroxide $$\mathrm{H_2O_2}$$ can behave both as an oxidising agent and as a reducing agent. In a basic medium, when it meets elemental iodine $$\mathrm{I_2}$$, it prefers to act as a reducing agent.

To understand what is formed, we analyse the redox process through the half-reaction method.

For hydrogen peroxide as a reductant, we write the oxidation half-reaction. The well-known relation is:

$$\mathrm{H_2O_2 \;\rightarrow\; O_2 + 2\,H^+ + 2\,e^-}$$

Because we are in basic (alkaline) medium, every $$\mathrm{H^+}$$ must be neutralised by adding an equal number of $$\mathrm{OH^-}$$ ions to both sides. Thus we add $$2\,\mathrm{OH^-}$$:

$$\mathrm{H_2O_2 + 2\,OH^- \;\rightarrow\; O_2 + 2\,H_2O + 2\,e^-}$$

Now we write the reduction half-reaction for iodine. Elemental iodine in its molecular form $$\mathrm{I_2}$$ accepts electrons to become iodide $$\mathrm{I^-}$$. The basic reduction relation is:

$$\mathrm{I_2 + 2\,e^- \;\rightarrow\; 2\,I^-}$$

Both half-reactions involve exactly $$2$$ electrons, so they already balance electrically and no scaling is needed. We now add them term by term:

$$\bigl(\mathrm{H_2O_2 + 2\,OH^- \;\rightarrow\; O_2 + 2\,H_2O + 2\,e^-}\bigr)$$

$$\bigl(\mathrm{I_2 + 2\,e^- \;\rightarrow\; 2\,I^-}\bigr)$$

Adding and cancelling the $$2\,e^-$$ on opposite sides, we obtain the overall ionic equation:

$$\mathrm{H_2O_2 + I_2 + 2\,OH^- \;\rightarrow\; 2\,I^- + 2\,H_2O + O_2}$$

The only iodine-containing species on the product side is $$\mathrm{I^-}$$, i.e. iodide ion.

Hence, the correct answer is Option C.