One mole of N_2H_4 loses 10 moles of electro | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$+3$

Vedclass · Answer

(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.)

Similar Questions

In acidic medium,$H_2O_2$ changes $Cr_2O_7^{2-}$ to $CrO_5$ which has two peroxy linkages. The oxidation state of $Cr$ in $CrO_5$ is

The oxidation number of carbon in graphite is ....... .

The oxidation number of $H$ in $KH$,$MgH_2$,and $NaOH$ are respectively

What is the oxidation number of sulfur in $Na_2S_4O_6$?

The oxidation states of $S$ in $H_2S_2O_7$ are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module