The compound in which the oxidation state of nitrogen is + 3 is

To find the oxidation state of nitrogen in each compound, use the rule that oxygen has an oxidation state of \(-2\) and the sum of oxidation states in a neutral compound is zero.

For a compound of the form N\(_2\)O\(_x\):

\[2(\text{oxidation state of N}) + x(-2) = 0\]

\[\text{oxidation state of N} = \frac{2x}{2} = x\]

Wait - more precisely: \(\text{oxidation state of N} = \frac{2x}{2} = +x\). Let me calculate each:

• N\(_2\)O\(_5\): \(2n + 5(-2) = 0 \Rightarrow n = +5\)
• N\(_2\)O\(_3\): \(2n + 3(-2) = 0 \Rightarrow n = +3\) ✔
• N\(_2\)O\(_4\): \(2n + 4(-2) = 0 \Rightarrow n = +4\)
• N\(_2\)O: \(2n + 1(-2) = 0 \Rightarrow n = +1\)

The compound in which nitrogen has an oxidation state of +3 is N\(_2\)O\(_3\) (dinitrogen trioxide).