Redox balancing using oxidation states and mole ratio

Steps:

• Oxidation state of N in [NH3OH]+ is -1 (O = -2, 4H = +4, total charge +1).
• Each Fe3+ gains 1e– to form Fe2+; 2 Fe3+ require 2e– from 1 [NH3OH]+.
• N must oxidize by losing 2e– per [NH3OH]+, changing from -1 to +1.
• N2O forms from 2 [NH3OH]+ (total 4e– lost, as each N averages +1), matching 2:4 ratio or 1:2.

Why B is correct:

• In N2O, two N atoms total +2 oxidation state (O = -2); from 2 [NH3OH]+ at -1 each, total change is 4e– lost, reducing 4 Fe3+ for 1:2 ratio.

Why the others are wrong:

• A. NH3: N oxidizes to -3 (reduction, not oxidation as required).
• C. NO: N to +2 loses 3e– per [NH3OH]+, giving 1:3 ratio.
• D. N2: Two [NH3OH]+ to N2 (each N to 0) loses 2e– total, giving 1:1 ratio.

Final answer: B