When iodine reacts with a hot and concentrated solution of sodium hydroxide, the products obtained are

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$I^{-} + I O^{-}$

$I^{-} + I O_{3}^{-}$

$I^{-} + I_{2} O_{9}^{(} 4 -)$

answer is B.

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Detailed Solution

We therefore discover sodium iodide and sodium iodate as the results of the iodine's reaction with hot NaOH, with water acting as a by-product. Here the correct by product are,

$I^{-} + I O_{3}^{-}$

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