The oxidiation number of nitrogen in NH2OH is:
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x+2(+1)+(−2)+(+1)=0⇒x+2(+1)+(−2)+(+1)=0
x+2−2+1=0⇒x+2−2+1=0
x=−1
x+2−2+1=0⇒x+2−2+1=0
x=−1
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