(B) Let the oxidation state of nitrogen be $x$. In $N_2O$,$2x + (-2) = 0$,so $2x = 2$,which gives $x = +1$. In $NO$,$x + (-2) = 0$,so $x = +2$. In $NH

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(B) Let the oxidation state of nitrogen be $x$. In $N_2O$,$2x + (-2) = 0$,so $2x = 2$,which gives $x = +1$. In $NO$,$x + (-2) = 0$,so $x = +2$. In $NH_2OH$,$x + 2(+1) + (-2) + (+1) = 0$,so $x + 1 = 0$,which gives $x = -1$. In $N_2H_4$,$2x + 4(+1) = 0$,so $2x = -4$,which gives $x = -2$. Therefore,the correct option is $(B)$.