In a reaction, $4$ mol of electrons are transferred to one mol of $H N O_{3}$ when it acts as an oxidant. The possible reduction product is :

0.5 m o l N_{2}

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0.5 m o l N_{2} O

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1 m o l N O_{2}

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1 m o l N H_{3}

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Solution

The correct option is B $0.5 m o l N_{2} O$
As it accepts $4$ mol of electrons, oxidation state of N in product is $5 - 4 = 1$ . Therefore, product is $N_{2} O$ , where oxidation state of N is $+ 1$ .

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