In alkaline medium, $I^{-}$is oxidised to ' X ' by $\mathrm{KMnO}_4$. The oxidation state of iodine in ' X ' is:

In alkaline medium, the reaction between permanganate ion and iodide ion is given by the balanced chemical equation:

2MnO$_{4}^{-}$ + I$^{-}$ + H$_{2}$O $\rightarrow$ 2MnO$_{2}$ + IO$_{3}^{-}$ + 2OH$^{-}$

The product 'X' formed by the oxidation of I$^{-}$ is the iodate ion, IO$_{3}^{-}$.

To find the oxidation state of iodine in IO$_{3}^{-}$, let the oxidation state be $x$.

$x + 3(-2) = -1$

$x - 6 = -1$

$x = +5$

Therefore, the oxidation state of iodine in 'X' is +5.

Answer: +5