Question

Iodine reacts with hot $NaOH$ solution giving the products as:
(A) $NaI$
(B) $NaI{O_3}$
(C) Both A and B
(D) None of these.

Answer 1

Hint : Here, it is asked that if iodine is set to react with hot sodium hydroxide then what will be the product we will have after the completion of reaction. First understand what will happen when we set the reaction to occur in the vessel having iodine and add the hot $NaOH$ to it. Write that reaction and we will be able to find the answer to this question.

Complete Step By Step Answer:
Let us first form the chemical reaction that will occur between the iodine and sodium hydroxide such that:
${I_2} + NaO{H_{\left( {hot} \right)}}\; \to \;NaI + NaI{O_3} + {H_2}O$
So when the iodine is set to react with hot $NaOH$ then we find the products as sodium iodide and sodium iodate with water as the by-product.
The more balanced reaction of the above reaction is as given below:
$3{I_2} + 6NaO{H_{\left( {hot} \right)}}\; \to \;5NaI + NaI{O_3} + 3{H_2}O$
And this reaction is a reduction reaction.
So the correct answer is that we obtain two products namely $NaI$ (sodium iodide) and $NaI{O_3}$ (sodium iodate).
The correct option is C.

Note :
Also remember that if the iodine is set to react with cold sodium hydroxide then the product will be sodium iodide and sodium hypo iodate. These both reactions give sodium iodide but if with hot sodium hydroxide then it will give sodium iodate along with sodium it. And if cold then sodium hypo iodate with sodium iodide. This difference should be noted and understood.