One mole of acidified $K_{2} C r_{2} O_{7}$ on reaction with excess $K I$ will liberate_________mole(s)of $I_{2}$ .

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Solution

The correct option is A 3
The reaction between K $_{2} C r_{2} O_{7}$ and exces $K I$ is given as:

$K_{2} C r_{2} O_{7} + 7 H_{2} S O_{4} + 6 K I \to 4 K_{2} S O_{4} + C r_{2} (S O_{4})_{3} + 7 H_{2} O + 3 I_{2}$

From the above reaction, we can see that one mole of $K_{2} C r_{2} O_{7}$ gives 3 moles of $I_{2}$ on reaction with $K I$ .

Acidified K $_{2} C r_{2} O_{7}$ is an oxidising agent. It oxidises $I^{-}$ to $I_{2}$ and itself gets reduced.

Hence, option $A$ is correct.

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