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

To find the oxidation state of nitrogen in these compounds, we use the rule that the sum of oxidation states in a neutral compound is zero, and the oxidation state of oxygen is typically -2.

Step-by-Step Analysis

Let the oxidation state of nitrogen be $x$.

A. N$_2$O$_5$ $2(x) + 5(-2) = 0$ $2x - 10 = 0$ $2x = 10$ $x = +5$

B. N$_2$O$_3$ $2(x) + 3(-2) = 0$ $2x - 6 = 0$ $2x = 6$ $x = +3$

C. N$_2$O$_4$ $2(x) + 4(-2) = 0$ $2x - 8 = 0$ $2x = 8$ $x = +4$

D. N$_2$O $2(x) + 1(-2) = 0$ $2x - 2 = 0$ $2x = 2$ $x = +1$

Conclusion

Comparing our results, the compound where nitrogen has an oxidation state of +3 is N$_2$O$_3$.

Exam Tip

When calculating oxidation states for oxides of nitrogen, remember that nitrogen can exhibit multiple oxidation states (+1 to +5). A quick trick is to look at the subscript of the oxygen: in $N_2O_y$, the oxidation state of nitrogen is simply $+y$. For example, in $N_2O_3$, $y=3$, so the state is $+3$. This works perfectly for these specific nitrogen oxides!