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(D) To find the oxidation state of nitrogen $(N)$ in each compound:$1$. In $N_2O$: $2x + (-2) = 0 \implies x = +1$$2$. In $NO$: $x + (-2) = 0 \implies

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$N_2O < NO < N_2O_3 < NO_2$

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(D) To find the oxidation state of nitrogen $(N)$ in each compound:$1$. In $N_2O$: $2x + (-2) = 0 \implies x = +1$$2$. In $NO$: $x + (-2) = 0 \implies x = +2$$3$. In $N_2O_3$: $2x + 3(-2) = 0 \implies 2x = 6 \implies x = +3$$4$. In $NO_2$: $x + 2(-2) = 0 \implies x = +4$Comparing the oxidation states: $+1 Therefore,the correct order is $N_2O < NO < N_2O_3 < NO_2$.

The correct order of the oxidation states of nitrogen in $NO$,$N_2O$,$NO_2$,and $N_2O_3$ is:

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