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Six oxides of nitrogen turn up in inorganic chemistry: N₂O, NO, N₂O₃, NO₂, N₂O₄ and N₂O₅. Students usually try to memorise six structures. A better route is to work out each oxidation state, count the electrons, and let those two numbers explain the shapes, colours and the way NO₂ pairs up. This post does that, with measured bond data, and starts by telling you which exam actually asks for it.
Which exams ask for oxides of nitrogen?
The JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus cover groups 13 to 18 only as a "general introduction": electronic configuration, trends down groups and across periods, and the unique behaviour of the first element. Neither names the oxides. The JEE (Advanced) 2026 brochure does: its group 15 entry ends with "oxides of nitrogen and oxoacids of phosphorus".
The textbook position has also changed. The current NCERT Class 12 Chemistry Part I contents run from Solutions to Coordination Compounds with no p-block unit. The older edition's Unit 7, The p-Block Elements (2022-23, archived copy), had a full section on oxides of nitrogen with Tables 7.3 and 7.4, and it is still the clearest school-level summary. So: if you are preparing for JEE Advanced, learn this properly; for JEE Main and NEET, it supports the "first element is different" questions but is not named. Check the current year's documents.
What is nitrogen's oxidation state in each oxide?
Take oxygen as −2 and solve for x:
N₂O: 2x − 2 = 0, so x = +1.
NO: x − 2 = 0, so x = +2.
N₂O₃: 2x − 6 = 0, so x = +3.
NO₂ and N₂O₄: x − 4 = 0, so x = +4.
N₂O₅: 2x − 10 = 0, so x = +5.
The old NCERT Table 7.3 lists the same states, and their appearance: N₂O and NO colourless gases, N₂O₃ a blue solid, NO₂ a brown gas, N₂O₄ and N₂O₅ colourless. It classes N₂O and NO as neutral oxides and the rest as acidic.
Which oxides are odd-electron molecules?
Count valence electrons: nitrogen brings 5, each oxygen 6.
NO: 5 + 6 = 11. Odd.
NO₂: 5 + 12 = 17. Odd.
N₂O: 10 + 6 = 16. N₂O₃: 10 + 18 = 28. N₂O₄: 34. N₂O₅: 40. All even.
An odd count means one unpaired electron, so NO and NO₂ are paramagnetic. The old NCERT text answers "Why does NO₂ dimerise?" in one line: it "contains odd number of valence electrons" and on dimerisation becomes stable N₂O₄ "with even number of electrons". PubChem describes nitrogen dioxide as a brown gas "in equilibrium with N2O4", and dinitrogen tetroxide as colourless. The old NCERT table writes the equilibrium as 2NO₂ ⇌ N₂O₄, with cooling favouring N₂O₄, so a colder flask is paler.
What do the measured structures look like?
The NIST Computational Chemistry Comparison and Benchmark Database lists experimental geometries:
N₂O: linear, N–N 1.128 Å and N–O 1.184 Å, with the oxygen at one end. It is not O–N–O, a common slip.
NO: N–O 1.154 Å.
NO₂: bent, N–O 1.193 Å, angle 134.1°.
N₂O₄: two NO₂ units joined N to N, N–N 1.782 Å, O–N–O 135.4°.
N₂O₃: an NO group joined to an NO₂ group through N–N, 1.864 Å.
For N₂O₅, the old NCERT Table 7.4 shows a planar O₂N–O–NO₂ structure, two NO₂ units bridged by one oxygen, with nitrogen's covalency four.
Worked example: why NO₂ splits off so easily
Compare the N–N bonds. In N₂O₄ it is 1.782 Å. In hydrazine, H₂N–NH₂, which has an ordinary N–N single bond, NIST lists 1.446 Å. The N₂O₄ bond is longer by 1.782 − 1.446 = 0.336 Å, about 23%. A much longer bond is a much weaker one, which fits the easy, reversible split 2NO₂ ⇌ N₂O₄ that you see as a colour change with temperature. The N–N bond in N₂O₃ is longer still, 1.864 Å; the old NCERT table makes N₂O₃ from NO and N₂O₄ at 250 K, a low temperature.
The bond angle in NO₂ tells a second story. A bent molecule with a lone pair on the central atom usually closes slightly below 120°. NO₂ has only one non-bonding electron on nitrogen, which repels the bonding pairs less than a full lone pair would, so the angle opens out to 134.1°.
If the shape reasoning is rusty, our guide to VSEPR and molecular shapes has the method, and our post on p-block groups 15 to 17 covers the trends JEE Main and NEET do ask for.
Which mistakes cost marks?
Drawing N₂O as O–N–O. It is N–N–O.
Calling NO₂ and N₂O₄ different oxidation states. Both are +4.
Writing that N₂O₅ has nitrogen with covalency five. It is four; nitrogen has no d orbitals to expand its valence shell.
Calling all nitrogen oxides acidic. N₂O and NO are neutral.
Forgetting the radicals. NO and NO₂ have odd electron counts.
A practice routine
Write the six formulas and, from memory, each oxidation state and electron count.
Draw one Lewis structure for each and check that every atom except the odd-electron nitrogens has an octet.
Explain, in two sentences, why NO₂ is brown and N₂O₄ is colourless, and what cooling the flask does.
If you are taking JEE Advanced, add one preparation reaction for each oxide from the old NCERT Table 7.3.
Quiz yourself again in three days; Dunlosky and colleagues (2013) rated spaced practice as high utility.
For a quick lesson on any idea here, from oxidation numbers to paramagnetism, the Fermi Concept Library covers every concept in physics, chemistry, biology, mathematics and computing, each with its own lesson.
Today, write the oxidation state and electron count for all six oxides without looking, then check them against this page.
