Alkaline KMnO4 oxidises Iodide to a particular product (A). Determine the oxidation state of Iodine in compound (A).

To determine the oxidation state of iodine in the product formed when alkaline potassium permanganate (${\mathrm{KMnO}}_4$) oxidizes iodide ions ($I^{-}$), let us look at the chemical reaction in detail.

In an alkaline or faint/mildly basic medium, potassium permanganate (${\mathrm{KMnO}}_4$) acts as a strong oxidizing agent. It oxidizes iodide ($I^{-}$) to iodate ions (${\mathrm{IO}}_3^{-}$).
The balanced chemical equation for this redox reaction is:
$2{\mathrm{MnO}}_4^{-}+I^{-}+H_{2}O\to 2{\mathrm{MnO}}_2+{\mathrm{IO}}_3^{-}+2{\mathrm{OH}}^{-}$

From this reaction, the compound (A) containing iodine is the iodate ion, ${\mathrm{IO}}_3^{-}$.

Now, let us calculate the oxidation state of iodine in the iodate ion (${\mathrm{IO}}_3^{-}$):
Let the oxidation state of iodine ($I$) be $x$.
We know that the oxidation state of oxygen ($O$) is typically -2.
Since the overall charge on the iodate ion is -1, we can write the algebraic equation:
$x+3(-2)=-1$
$x-6=-1$
$x=-1+6$
$x=+5$

Thus, the oxidation state of iodine in compound (A) is +5.