Iodine reacts with concentrated HN O _ 3 to yield Y along with...

${I}_{2} + 10HN{O}_{3} \to 2HI{O}_{3} + 10N{O}_{2} + 4{H}_{2}O$

Iodine gets oxidized to $I{O}_{3}^{-}$ when it reacts with an oxidizing agent \left(HN{O}_{3}\right) .$The oxidation number of$I$in product will be<br /><br />Oxidation:${I}_{2} - I{O}_{3}^{-}$<br /><br />$- 1 = x + 3 \times \left(- 2\right)

$- 1 = x - 6$

x = - 1 + 6 = + 5

Reduction: concentrated HNO₃ gives NO₂.