JEE CHEMISTRY MCQ: REDOX REACTIONS

1 Hard

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Application +4 | -1

A sample of

2.5 \mathrm{~mol}

of hydrazine

\left(\mathrm{N}_2 \mathrm{H}_4\right)

loses

25 \mathrm{~mol}

of electrons on being converted to a new compound

\mathrm{X}.

Assuming that there is no loss of nitrogen in the formation of the new compound, what is the oxidation number of nitrogen in compound

\mathrm{X}?

View answer and explanation

Correct option: C

+3

Option C (+3) is correct because the oxidation number of nitrogen in the new compound X is calculated based on the electron loss and the initial oxidation state of nitrogen in hydrazine.

More Questions from REDOX REACTIONS

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The oxidation number of

\mathrm{H}

in sodium hydride is A reagent, which can decrease the oxidation number of an element, is called Consider the following decomposition reaction

2 \mathrm{KClO}_3 \longrightarrow 2 \mathrm{KCl}+3 \mathrm{O}_2

In the above reaction, Assertion (A):

\mathrm{HNO}_3

acts as oxidizing agent.

Reason (R): Oxidation no. of nitrogen is +5, no increase in oxidation no. beyond +5 can occur. The oxidation no. of

\mathrm{HNO}_3

can only decrease. Assertion :- Bromide ion is serving as a reducing agent in the reaction.

\mathrm{MnO}_4^{-}(\mathrm{aq})+\mathrm{Br}^{-}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{MnO}_2(\mathrm{aq})+\mathrm{BrO}_3^{-}(\mathrm{aq})+2 \mathrm{OH}^{-}(\mathrm{aq})

Reason :- Oxidation number of

\mathrm{Br}

increases from

-1

+5.

A substance, that by its sharp colour change indicates the completion of reaction is known as :