What would happen when a solution of potassium chromate is treated with an excess of dilute nitric acid?

C r^{3 +}

and

C r_{2} O_{7}^{2 -}

are formed

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C r_{2} O_{7}^{2 -}

and

H_{2} O

are formed

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C r_{2} O_{4}^{2 -}

is reduced to +3 state of Cr

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C r_{2} O_{4}^{2 -}

oxidised to +7 state of Cr

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Solution

The correct option is B $C r_{2} O_{7}^{2 -}$ and $H_{2} O$ are formed
With dilute nitric acid the reaction between $K_{2} C r O_{4}$ and dil. $H N O_{3}$ :
$2 K_{2} C r O_{4} + 2 H N O_{3} (d i l) \to K_{2} C r_{2} O_{7} 2 K N O_{3} + H_{2} O$
Therefore; $C r O_{4}^{2 -} + 2 H N O_{3} (d i l) \to C r_{2} O_{7}^{2 -} + 2 N O_{3}^{-} + H_{2} O$
Dilute $H N O_{3}$ is an oxidising agent. In this reaction oxidation no. of Cr changes from $+ 3$ to $+ 6$ . So, dilute $H N O_{3}$ oxidises $K_{2} C r O_{4}$ to $K_{2} C r_{2} O_{7}$

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