Chemistry · Redox Equilibrium · NEET
Use the rule that H is +1 and O is -2, and the whole molecule is neutral. In NH3: N + 3(+1) = 0, so N = -3. In NO: N + (-2) = 0, so N = +2. In NO2: N + 2(-2) = 0, so N = +4. In HNO3: (+1) + N + 3(-2) = 0, so N = +5. So the sequence is -3, +2, +4, +5.
Nitrogen is oxidised throughout. Its oxidation number only goes up: from -3 in ammonia to +5 in nitric acid. Since oxidation means an increase in oxidation number (loss of electrons, OIL RIG), every stage of the Ostwald process is an oxidation of nitrogen. The oxygen from air is the oxidising agent.
From -3 (in NH3) to +5 (in HNO3) is a total increase of 8 units. This means each nitrogen atom loses 8 electrons across the whole process. NEET may ask for either the total change (8) or a single-step change, so read the question carefully.
Step 1: 4NH3 + 5O2 gives 4NO + 6H2O (N goes -3 to +2, catalysed by Pt/Rh gauze). Step 2: 2NO + O2 gives 2NO2 (N goes +2 to +4). Step 3: 3NO2 + H2O gives 2HNO3 + NO (N goes +4 to +5 in HNO3, but drops to +2 in NO). Step 3 is a disproportionation of NO2.
In 3NO2 + H2O giving 2HNO3 + NO, the same element (nitrogen, all starting at +4 in NO2) ends up at two different oxidation states: +5 in HNO3 (oxidised) and +2 in NO (reduced). When one species is both oxidised and reduced at the same time, it is called disproportionation.
The correct order of N-compounds in their decreasing order of oxidation states is
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Oxygen from air is the oxidising agent. It gains electrons (its oxidation state drops toward -2 in the oxides and water formed) while it forces nitrogen to lose electrons and rise in oxidation state. So oxygen is reduced and nitrogen is oxidised.
A platinum-rhodium (Pt-Rh) gauze at about 500 degrees Celsius catalyses the oxidation of ammonia to nitric oxide (NO). The catalyst speeds up the reaction but does not change the oxidation-state values of nitrogen.
Yes, only in the final step. When NO2 reacts with water (3NO2 + H2O gives 2HNO3 + NO), some nitrogen at +4 rises to +5 while some falls to +2 in NO. That NO is recycled back into the process, so overall nitrogen still ends up oxidised to +5 in HNO3.
Because a single industrial process shows nitrogen passing through many oxidation states (-3, +2, +4, +5) and includes both simple oxidation and a disproportionation. It lets examiners test oxidation-number assignment, redox identification, and the -3-to-+5 range change all at once.
