One mole of \({N}_{{{2}}}{H}_{{{4}}}\) loses ten moles of electrons to form a new compound \({A}\). Assuming that all the nitrogen appears in the new compound, what is the oxidation state of nitrogen in \({A}\)? (There is no change in the oxidation state of hydrogen.)

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One mole of \({N}_{{{2}}}{H}_{{{4}}}\) loses ten moles of electrons to form a new compound \({A}\). Assuming that all the nitrogen appears in the new compound, what is the oxidation state of nitrogen in \({A}\)? (There is no change in the oxidation state of hydrogen.)

Answer: 3

\({N}_{{{2}}}{H}_{{{4}}}\rightarrow{\left({X}\right)}+{10}{e}\)

\(\because{X}\) contains all \({N}\)-atoms

\(\therefore{\left({N}^{{{2}-}}\right)}_{{{2}}}\rightarrow{\left({2}{N}\right)}^{{{a}}}+{10}{e}\)

Therefore \({2}{a}-{\left(-{4}\right)}={10}\Rightarrow{a}=+{3}\)

\({N}_{{{2}}}{H}_{{{4}}}\rightarrow{\left({X}\right)}+{10}{e}\)

\(\because{X}\) contains all \({N}\)-atoms

\(\therefore{\left({N}^{{{2}-}}\right)}_{{{2}}}\rightarrow{\left({2}{N}\right)}^{{{a}}}+{10}{e}\)

Therefore \({2}{a}-{\left(-{4}\right)}={10}\Rightarrow{a}=+{3}\)

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