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  4. Oxidation number of nitrogen in NaNO2 is

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Q. Oxidation number of nitrogen in $NaNO_{2}$ is

Redox Reactions

Solution:

Let the oxidation number of N in $NaNO_{2}$ be x $+ 1 + x + ( - 2) \times 2 = 0$ $1 + x - 4 = 0;x = + 3$