On this page
Draw the Lewis structure of oxygen and you get O=O with every electron paired. Yet experiments show that oxygen is paramagnetic: it is drawn into a magnetic field, which needs unpaired electrons. The Lewis picture cannot explain that, and molecular orbital (MO) theory can. That is why JEE likes to ask for the bond order and magnetic nature of O2, N2 and their ions. This post gives you one filling method, worked answers for six species, and the slips that cost marks.
What does the syllabus ask about MO theory?
The JEE (Main) 2026 syllabus lists, under Chemical Bonding, "Molecular Orbital Theory - Its important features, LCAOs, types of molecular orbitals (bonding, antibonding), sigma and pi-bonds, molecular orbital electronic configurations of homonuclear diatomic molecules, the concept of bond order, bond length and bond energy." The NCERT Class 11 chapter on bonding covers the same ground. Check the syllabus for your exam year.
The method itself earned a Nobel: the 1966 Nobel Prize in Chemistry went to Robert Mulliken for his work on chemical bonds and the electronic structure of molecules by the molecular orbital method.
Which order do the molecular orbitals fill in?
When two atoms come together, their atomic orbitals combine into bonding orbitals (lower energy) and antibonding orbitals (marked with a star, higher energy). For second-row diatomics there are two energy orders, and choosing the wrong one is the most common error.
For O2 and F2: σ1s, σ*1s, σ2s, σ*2s, σ2pz, then π2px = π2py, then π*2px = π*2py, then σ*2pz.
For B2, C2 and N2: the same, except the two π2p orbitals come below σ2pz. NCERT notes this is what experiment shows for molecules such as B2, C2 and N2.
Fill the orbitals lowest first, put one electron in each of a pair of equal-energy orbitals before pairing (Hund's rule), and count.
Bond order = (electrons in bonding orbitals - electrons in antibonding orbitals) / 2.
The 1s electrons cancel (two bonding, two antibonding), so you can count only valence electrons and get the same answer.
Worked example: N2 and O2
N2. Each nitrogen has 5 valence electrons, so N2 has 10. Using the B2 to N2 order: σ2s², σ*2s², π2px² π2py², σ2pz². Bonding electrons: 2 + 4 + 2 = 8. Antibonding: 2. Bond order = (8 - 2) / 2 = 3. Every electron is paired, so N2 is diamagnetic.
O2. Each oxygen has 6 valence electrons, so O2 has 12. Using the O2 order: σ2s², σ*2s², σ2pz², π2px² π2py², π*2px¹ π*2py¹. Bonding: 2 + 2 + 4 = 8. Antibonding: 2 + 2 = 4. Bond order = (8 - 4) / 2 = 2. The last two electrons sit alone in the two π* orbitals, so O2 has two unpaired electrons and is paramagnetic. NCERT notes that this prediction matches experiment.
The bond data in the NCERT chapter fit these bond orders. N2 has a bond length of 109 pm and a bond enthalpy of 946 kJ/mol. O2 has 121 pm and 498 kJ/mol. Higher bond order, shorter and stronger bond.
How do the oxygen ions compare?
This is a favourite JEE ordering question. Start from O2 (12 valence electrons) and add or remove electrons from the π* orbitals.
O2+ (11 electrons): one π* electron. Bond order (8 - 3) / 2 = 2.5. One unpaired electron, paramagnetic.
O2 (12): bond order 2, two unpaired, paramagnetic.
O2- (superoxide, 13): bond order (8 - 5) / 2 = 1.5. One unpaired, paramagnetic.
O2 2- (peroxide, 14): bond order (8 - 6) / 2 = 1. All paired, diamagnetic.
So bond order runs O2+ > O2 > O2- > O2 2-, and bond length runs the other way. NCERT itself lists O2 2- with F2 as having bond order 1.
For nitrogen, removing an electron from N2 takes it from the bonding σ2pz orbital, so N2+ has bond order (7 - 2) / 2 = 2.5 and one unpaired electron. Notice the contrast: removing an electron weakens N2 but strengthens O2, because in O2 the electron comes from an antibonding orbital.
Where do students go wrong?
A study of college students by Jenkins and Shoopman (2019) found that most could construct MO diagrams correctly, but many struggled to use and interpret them. There were clear gaps between drawing the diagram and saying what it meant for stability and reactivity. In JEE terms, drawing is not the hard part. Reading off bond order, magnetism and trends is.
The slips we see most:
Using the O2 order for N2. For B2, C2 and N2, π2p fills before σ2pz. With the wrong order B2 would come out diamagnetic, when it is paramagnetic.
Forgetting Hund's rule in the π* pair, which hides the two unpaired electrons in O2.
Removing the electron from the wrong orbital when forming a positive ion. Always remove from the highest occupied orbital.
Counting total and valence electrons in the same problem. Pick one and stay with it.
Assuming bond order predicts magnetism. C2 has bond order 2 like O2 but is diamagnetic, as NCERT notes.
A practice routine for MO questions
Write both filling orders from memory until you can do it in under a minute.
For each of B2, C2, N2, O2, F2 and Ne2, write the configuration, bond order and magnetism. Ne2 should come out at bond order zero.
Do the oxygen series and the nitrogen series (N2, N2+, N2-) and rank them by bond length.
Mix in five previous-year questions and say out loud which orbital each added or removed electron goes to.
If the atomic side is shaky, such as Hund's rule or the order of atomic orbitals, start with our post on quantum numbers and electron configuration. When you want to check your reasoning on a full question, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Today, rank O2+, O2, O2- and O2 2- by bond length without looking, then check each against the electron count.
