Nitrogen Disproportionation in the Reaction

Steps:

• Calculate oxidation state of N in NO₂: oxygen is -2, so N is +4 (since 2*(-2) + N = 0).
• In products, N in HNO₃ is +5 (H +1, 3O -6, total 0 so N +5); N in HNO₂ is +3 (H +1, 2O -4, total 0 so N +3).
• Nitrogen changes from +4 to both +5 (oxidation) and +3 (reduction) in the same reaction.
• This matches the definition of disproportionation, where one element is simultaneously oxidized and reduced.

Why B is correct:

• Disproportionation occurs when the same element undergoes both oxidation and reduction, as N does here from +4 to +5 and +3.

Why the others are wrong:

• A: Products HNO₃ and HNO₂ do not further disproportionate; the reaction shows N in NO₂ disproportionating.
• C: Hydrogen remains +1 in H₂O and both products, so its oxidation number is unchanged.
• D: Water provides H and O but does not oxidize N; N self-oxidizes/reduces via disproportionation.

Final answer: B