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Oxidation Number Method

When NaNO2 ...

Question

When $N a N O_{2}$ is treated with $N a$ , which of the following is/are formed?

A

N a N O_{3}

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B

N a_{2} O

and

N_{2}

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C

N a_{2} O_{2}

and

N_{2}

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D

N a_{2} O

and

N a_{3} N

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Solution

The correct option is B $N a_{2} O$ and $N_{2}$
When $N a N O_{2}$ is treated with $N a$ , $N a_{2} O$ and $N_{2}$ are formed. The reaction is as follows:
$2 N a N O_{2} + 6 N a (excess) \to 4 N a_{2} O + N_{2}$

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