Concentrated $$HNO_3$$ reacts with Iodine to give

We need to determine the products formed when concentrated $$HNO_3$$ reacts with iodine ($$I_2$$).

Concentrated nitric acid is a strong oxidizing agent. When it reacts with iodine, it oxidizes iodine to its highest common oxide acid form. Iodine (oxidation state 0) gets oxidized to iodic acid $$HIO_3$$ (where iodine has an oxidation state of +5).

The balanced chemical equation for this reaction is:

$$I_2 + 10HNO_3 \text{(conc.)} \rightarrow 2HIO_3 + 10NO_2 + 4H_2O$$

Let us verify the balance. On the left side: 2 iodine atoms, 10 nitrogen atoms, 10 hydrogen atoms, and 30 oxygen atoms. On the right side: 2 iodine atoms (in $$2HIO_3$$), 10 nitrogen atoms (in $$10NO_2$$), 10 hydrogen atoms (2 in $$2HIO_3$$ and 8 in $$4H_2O$$), and oxygen: 6 (in $$2HIO_3$$) + 20 (in $$10NO_2$$) + 4 (in $$4H_2O$$) = 30 oxygen atoms. The equation is balanced.

The nitrogen in $$HNO_3$$ (oxidation state +5) is reduced to $$NO_2$$ (oxidation state +4), confirming that $$HNO_3$$ acts as the oxidizing agent while $$I_2$$ is oxidized.

So the products are $$HIO_3$$, $$NO_2$$, and $$H_2O$$.

Hence, the correct answer is Option C.