One mole of N_2H_4 loses 10 moles of electron | vedclass

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(C) In $N_2H_4$,the oxidation state of $H$ is $+1$. Let the oxidation state of $N$ be $x$. $2x + 4(+1) = 0 \implies 2x = -4 \implies x = -2$. Total ox

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$+3$

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(C) In $N_2H_4$,the oxidation state of $H$ is $+1$. Let the oxidation state of $N$ be $x$. $2x + 4(+1) = 0 \implies 2x = -4 \implies x = -2$. Total oxidation number of two $N$ atoms in $N_2H_4$ is $-4$. When $N_2H_4$ loses $10$ moles of electrons,the total oxidation number of the two $N$ atoms increases by $10$. New total oxidation number of two $N$ atoms $= -4 + 10 = +6$. Since there are two $N$ atoms in the new compound $Z$,the oxidation state of each $N$ atom is $\frac{+6}{2} = +3$.

One mole of $N_2H_4$ loses $10$ moles of electrons to form a new compound $Z$. Assuming that all the nitrogen atoms appear in the new compound,what is the oxidation state of nitrogen in $Z$? (There is no change in the oxidation state of hydrogen.)

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