The oxidation state of nitrogen in $N a {N O}_{3}$ is :

+ 3

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+ 5

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- 4

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- 5

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Solution

The correct option is B $+ 5$
The overall charge is zero (because it is not an ion). Since sodium is a 1A family member, the charge is +1. The charge of oxygen is almost always -2.
Hence, $1 + O S o f N + (3 \times - 2) = 0$
So, the oxidation of nitrogen in that problem is +5.

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Similar questions

Suggest a list of the substances where carbon can exhibit oxidation states from –4 to +4 and nitrogen from –3 to +5.

N H_{4} N O_{3}

Oxidation states of $P$ in $H_{4} P_{2} O_{6}, H_{4} P_{2} O_{6}, H_{4} P_{2} O_{7}$ are respectively:

N H_{4} N O_{3}

N

+ 1, + 2, + 3, + 4

+ 5

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