How to Write the Molecular Orbital (MO) Electronic Configuration
Chemistry · Chemical Bonding · NEET
To write a molecular orbital (MO) configuration, first count the total electrons in the molecule. Then fill them one by one into molecular orbitals following the correct energy order, from lowest energy to highest, obeying Aufbau, Pauli and Hund's rules — exactly like you fill atomic orbitals. Memory hook: "Count, Order, Fill" — count the electrons, pick the right energy order, then fill up.
The three-step method applied to N2: count 14 electrons, use the light-molecule energy order (π2p before σ2p_z), then fill from the bottom up. Reading the filled orbitals gives bond order 3 and shows all electrons are paired (diamagnetic).
Your doubts, answered
What are the exact steps to write an MO configuration?
Three steps. Step 1: count the TOTAL number of electrons in the molecule (add the electrons of both atoms; for ions add or remove electrons for charge). Step 2: choose the correct energy order of molecular orbitals. Step 3: fill electrons one at a time from the lowest-energy MO upward, putting 2 electrons max per MO (Pauli), and singly first in equal-energy orbitals (Hund). Example, N2 has 14 electrons: σ1s² σ*1s² σ2s² σ*2s² (π2p_x² = π2p_y²) σ2p_z². Remember: Count, Order, Fill.
Which energy order should I use — the N2 order or the O2 order?
There are TWO orders because of s-p mixing. For B2, C2 and N2 (light molecules, up to 14 electrons), the order is: σ1s < σ*1s < σ2s < σ*2s < (π2p_x = π2p_y) < σ2p_z < (π*2p_x = π*2p_y) < σ*2p_z — here the two π orbitals come BEFORE σ2p_z. For O2, F2 and Ne2 (heavier), σ2p_z comes BEFORE the two π orbitals. Rule of thumb: molecules with total electrons ≤ 14 use the first order; O2 and beyond use the second.
Do I include the inner 1s electrons?
Yes, for a full configuration you write σ1s² and σ*1s² too. But σ1s and σ*1s cancel each other (one bonding + one antibonding pair), so they do not change the bond order. In NEET you can write them as KK (like the noble-gas core shortcut) or write them out fully — both are accepted. What matters is you counted ALL electrons.
How do I count electrons for an ion like O2^- or CN^-?
Start from the neutral atoms' total, then adjust for charge. A negative charge means ADD that many electrons; a positive charge means REMOVE electrons. Example: O2 has 16 electrons, so O2^- has 17 and O2^+ has 15. CN neutral has 6+7=13 electrons, so CN^- has 14 electrons (same as N2, bond order 3). This is why NEET loves ion questions — one extra electron changes the bond order.
After I write the configuration, how do I get the bond order?
Use bond order = ½(N_b − N_a), where N_b = number of electrons in bonding MOs (σ2s, σ2p_z, π2p, plus σ1s) and N_a = number in antibonding MOs (the starred ones σ*, π*). For N2: bonding = 10, antibonding = 4, so bond order = ½(10−4) = 3. Higher bond order means shorter, stronger bond.
How do I know if the molecule is paramagnetic from the configuration?
Look at the highest filled orbitals. If ANY molecular orbital has an unpaired (single) electron, the molecule is paramagnetic. If all electrons are paired, it is diamagnetic. Example: O2 has configuration ...(π2p_x² π2p_y²)(π*2p_x¹ π*2p_y¹) — two unpaired electrons in the π* orbitals, so O2 is paramagnetic. This is the single most tested MO fact in NEET.
⚠️ The NEET trap ✗ Using the N2 energy order (π before σ2p_z) for O2, so you place electrons in the wrong orbitals and think O2 is diamagnetic. ✓ O2 uses the OTHER order: σ2p_z fills before the π2p orbitals. O2 ends with two unpaired electrons in π*2p, so O2 is PARAMAGNETIC. Bond order = 2. 🧠 O2 breaks the pattern: for O2, F2, Ne2 the σ2p goes BEFORE π2p. That single switch is why O2 is paramagnetic — the classic NEET catch.
Real NEET questions
NEET 2019
Which of the following is paramagnetic?
A · N2
B · H2
C · Li2
D · O2 ✓
Solution: Write the MO configuration and look for unpaired electrons. O2 has 16 electrons: KK σ2s² σ*2s² σ2p_z² (π2p_x² π2p_y²)(π*2p_x¹ π*2p_y¹). The two π* electrons stay unpaired (Hund's rule), so O2 is paramagnetic. N2, H2 and Li2 have all electrons paired, so they are diamagnetic. Answer: O2.
NEET 2018
Consider the species CN^+, CN^-, NO and CN. Which of these will have the highest bond order?
A · CN^+
B · CN^- ✓
C · NO
D · CN
Solution: Count electrons, write the config, then use bond order = ½(N_b − N_a). CN^- has 6+7+1 = 14 electrons, same as N2, giving bond order 3. CN has 13 (bond order 2.5), CN^+ has 12 (2.0), NO has 15 (2.5). Highest is CN^-. Answer: CN^-.
NEET 2023
The correct order of energies of the molecular orbitals of the N2 molecule is:
Solution: N2 has 14 electrons, so it uses the light-molecule order where the two π2p orbitals fill BEFORE σ2p_z (because of s-p mixing). That gives option A. Option B is the O2/F2 order. Knowing which energy order to use is the first step of writing any MO configuration. Answer: A.
Solved Chemical Bonding NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Is the MO filling order the same for every molecule?
No. Molecules up to 14 total electrons (B2, C2, N2) fill the two π2p orbitals before σ2p_z. Heavier molecules (O2, F2, Ne2) fill σ2p_z first. Pick the right order before you fill.
Why do we count total electrons and not just valence electrons?
A full MO configuration includes inner electrons (σ1s, σ*1s). They cancel and do not affect bond order, so for speed you can write them as KK, but you still must count them so your total is correct.
What is the quickest way to get bond order from the configuration?
Bond order = ½(bonding electrons − antibonding electrons). Bonding orbitals are σ2s, σ2p, π2p; antibonding are the starred ones. Higher bond order = stronger, shorter bond.
How do I decide diamagnetic or paramagnetic?
Read the configuration. Any single (unpaired) electron in any MO means paramagnetic. All electrons paired means diamagnetic. O2 (two unpaired π* electrons) is the famous paramagnetic example.
