One mole of N_2H_4 loses 10 moles of electro | vedclass

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(C) The initial oxidation state of $N$ in $N_2H_4$ is calculated as: $2x + 4(+1) = 0 \implies 2x = -4 \implies x = -2$. Since $1 \ mole$ of $N_2H_4$ l

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$+3$

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(C) The initial oxidation state of $N$ in $N_2H_4$ is calculated as: $2x + 4(+1) = 0 \implies 2x = -4 \implies x = -2$. Since $1 \ mole$ of $N_2H_4$ loses $10 \ moles$ of electrons, the total increase in oxidation number is $10$. Let the oxidation state of $N$ in $Z$ be $y$. Since there are $2$ nitrogen atoms, the total oxidation state of $N$ in $Z$ is $2y$. The change in oxidation state is $2y - (-4) = +10$. $2y + 4 = 10 \implies 2y = 6 \implies y = +3$. Therefore, the oxidation state of nitrogen in $Z$ is $+3$.

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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