One mole of N_2H_4 loses 10 mol of electrons | vedclass

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

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$+3$

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(A) In $N_2H_4$,the oxidation state of $H$ is $+1$. Let the oxidation state of $N$ be $x$. $2x + 4(+1) = 0 \implies 2x = -4 \implies x = -2$. When $1 \ mol$ of $N_2H_4$ loses $10 \ mol$ of electrons,the total oxidation state increases by $10$. Let the oxidation state of $N$ in $Y$ be $a$. Since there are $2$ atoms of $N$,the total change is $2(a - (-2)) = 10$. $2(a + 2) = 10 $a + $2$ = $5$ $a = +3$.

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

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