Let the oxidation state of nitrogen in the given compounds be x. (a) N2H4 2(x)+(+1)4=02x=−4 So, x=−2. (b) NH3 x+(+1)3=0 So, x=−3 (c) N3H (x)3+(+1)=03x=−1 So, x=−1/3 (d) NH2OH x+(+1)2+(−2)+(+1)=0x+2−2+1=0x+1=0 So, x=−1 Thus, oxidation state of nitrogen is highest in N3H.