Login Free Test Start Mock

Chemistry · Chemical Bonding and Molecular Structure

Molecular Orbital Theory

Molecular Orbital (MO) theory, developed by Hund and Mulliken in 1932, treats the electrons of a molecule as belonging to the molecule as a whole rather than to individual atoms. Following NCERT Class 11 Unit 4 (Sections 4.7–4.8), this note builds the MO picture from the linear combination of atomic orbitals through to electronic configurations, bond order, and the magnetic behaviour of $\ce{O2}$. For NEET, this is one of the most reliably tested ideas in the chapter, with bond-order and MO-ordering questions appearing almost every year.

Why MO Theory Was Needed

Valence bond theory pictures a covalent bond as a localised pair of electrons shared between two atoms. It accounts well for shapes and directionality, but it stumbles on a famous fact: liquid oxygen is attracted to a magnet. A Lewis structure of $\ce{O=O}$ shows every electron paired, which would make the molecule diamagnetic. MO theory was the framework that resolved this, and it does so as a natural consequence of how the orbitals fill.

The central shift in viewpoint is this: in a molecule the electrons occupy molecular orbitals that span two or more nuclei, just as the electrons of an atom occupy atomic orbitals around a single nucleus. An atomic orbital is monocentric; a molecular orbital is polycentric. The bonding picture then follows the same machinery you already know — the aufbau principle, Pauli's exclusion principle, and Hund's rule — applied to a new set of orbitals.

Salient feature of MO theoryConsequence
Electrons reside in molecular orbitals belonging to the whole moleculeBonding is delocalised, not strictly pairwise
Atomic orbitals of comparable energy and proper symmetry combineOnly matched orbitals mix (1s with 1s, 2pz with 2pz)
Number of MOs formed equals number of combining atomic orbitalsTwo AOs give one bonding and one antibonding MO
Bonding MO has lower energy; antibonding MO has higher energyFilling bonding MOs stabilises the molecule
MOs filled by aufbau, Pauli and Hund's rulesDegenerate orbitals fill singly first (key for $\ce{O2}$)

Forming MOs by LCAO

The Schrödinger equation cannot be solved exactly for a many-electron molecule, so MO theory uses an approximation called the Linear Combination of Atomic Orbitals (LCAO). The wave functions of the atomic orbitals are added and subtracted to build the molecular orbitals. Take the hydrogen molecule, with atoms A and B each carrying one electron in a 1s orbital described by $\psi_A$ and $\psi_B$.

Two combinations are possible, and they generate two molecular orbitals:

$$\sigma = \psi_A + \psi_B \qquad \sigma^{*} = \psi_A - \psi_B$$

The additive combination produces the bonding molecular orbital $\sigma$; the subtractive combination produces the antibonding molecular orbital $\sigma^{*}$. Qualitatively, the bonding MO arises from constructive interference of the two electron waves, while the antibonding MO arises from destructive interference. This is the wave-mechanical origin of the energy split that drives all the predictions below.

Bonding vs Antibonding Orbitals

In a bonding MO the electron density is concentrated between the two nuclei. This shielding reduces internuclear repulsion and pulls the nuclei together, so a bonding MO always lies lower in energy than either parent atomic orbital. In an antibonding MO the electron density is pushed away from the region between the nuclei; there is a nodal plane between them where the density is zero, so the nuclei repel each other strongly and the orbital lies higher in energy than the parent atomic orbitals.

A point examiners like to test: the energy raised on the antibonding orbital exactly balances the energy lowered on the bonding orbital. The total energy of the two molecular orbitals equals the total energy of the two original atomic orbitals — energy is redistributed, not created.

PropertyBonding MO (σ, π)Antibonding MO (σ*, π*)
Formed byAddition of wave functionsSubtraction of wave functions
InterferenceConstructiveDestructive
Electron density between nucleiHighLow — nodal plane present
Energy vs parent AOsLowerHigher
Effect on moleculeStabilisesDestabilises

Conditions and Types of MOs

Atomic orbitals do not combine indiscriminately. Three conditions must hold simultaneously for an effective linear combination, and NEET frequently probes the first two.

ConditionMeaningExample
Comparable energyCombining orbitals must have the same or nearly the same energy1s combines with 1s, not with 2s
Same symmetry about the bond axisTaking z as the molecular axis, only orbitals of matching symmetry mix2pz combines with 2pz, not with 2px or 2py
Maximum overlapThe greater the overlap, the greater the electron density between nucleiBetter overlap gives a stronger bond

Molecular orbitals are named by their symmetry about the bond axis. A sigma (σ) MO is symmetrical about the internuclear axis: head-on combination of two 1s orbitals gives $\sigma 1s$ and $\sigma^{*}1s$, and combination of two $2p_z$ orbitals (lying along the axis) gives $\sigma 2p_z$ and $\sigma^{*}2p_z$. A pi (π) MO is not symmetrical about the axis: sideways combination of $2p_x$ or $2p_y$ orbitals gives $\pi 2p_x$, $\pi 2p_y$ and their antibonding partners, with electron density in lobes above and below the axis.

The MO Energy-Level Diagram

For a second-row homonuclear diatomic, the eight valence atomic orbitals (2s and 2p on each atom) combine to give eight molecular orbitals: $\sigma 2s$, $\sigma^{*}2s$, $\sigma 2p_z$, $\sigma^{*}2p_z$, and the degenerate pairs $\pi 2p_x = \pi 2p_y$ and $\pi^{*}2p_x = \pi^{*}2p_y$. The diagram below is the standard schematic for $\ce{N2}$ — the representative diatomic in which the $\pi$ bonding orbitals lie below $\sigma 2p_z$.

Figure 1 Energy Atom A (N) Molecular orbitals Atom B (N) 2p 2s 2p 2s σ*2p z π*2p x π*2p y σ2p z π2p x π2p y σ*2s σ2s

MO energy-level diagram for $\ce{N2}$. Atomic 2s and 2p orbitals (purple) combine into bonding MOs (teal) and antibonding MOs (coral). For $\ce{N2}$ the $\pi 2p$ pair lies below $\sigma 2p_z$. The ten valence electrons fill $\sigma 2s$, $\sigma^{*}2s$, the two $\pi 2p$ orbitals, and $\sigma 2p_z$, giving a bond order of 3.

For $\ce{O2}$ and $\ce{F2}$ the relative positions of the $\pi 2p$ and $\sigma 2p_z$ orbitals are swapped. The next section explains exactly where and why this switch happens, since it is the single most common MO trap on NEET.

→
Build on this

See how the MO picture plays out across the second period in Homonuclear Diatomic Bonding.

The B2/C2/N2 vs O2/F2 Switch

The energy ordering of the molecular orbitals is fixed by experiment (spectroscopic data), and it is not the same for every second-row diatomic. For $\ce{O2}$, $\ce{F2}$ (and $\ce{Ne2}$), the $\sigma 2p_z$ orbital lies below the degenerate $\pi 2p$ orbitals:

$$\sigma 1s < \sigma^{*}1s < \sigma 2s < \sigma^{*}2s < \sigma 2p_z < (\pi 2p_x = \pi 2p_y) < (\pi^{*}2p_x = \pi^{*}2p_y) < \sigma^{*}2p_z$$

But for the lighter molecules $\ce{Li2}$, $\ce{Be2}$, $\ce{B2}$, $\ce{C2}$ and $\ce{N2}$, the order changes: the $\pi 2p$ pair drops below $\sigma 2p_z$.

$$\sigma 1s < \sigma^{*}1s < \sigma 2s < \sigma^{*}2s < (\pi 2p_x = \pi 2p_y) < \sigma 2p_z < (\pi^{*}2p_x = \pi^{*}2p_y) < \sigma^{*}2p_z$$

NEET Trap

The ordering flips between N₂ and O₂

The defining difference is the position of $\sigma 2p_z$. For $\ce{B2}$, $\ce{C2}$, $\ce{N2}$ the $\sigma 2p_z$ orbital is higher than $\pi 2p_x = \pi 2p_y$ (because 2s–2p mixing pushes it up). For $\ce{O2}$, $\ce{F2}$, $\ce{Ne2}$ the mixing is negligible and $\sigma 2p_z$ sits below the $\pi$ orbitals. NEET 2023 Q.64 asked precisely for the $\ce{N2}$ order and rewarded the $\pi 2p_x = \pi 2p_y$ then $\sigma 2p_z$ sequence.

Memory hook: up to and including N₂, π comes before σ2p₂ z. From O₂ onward, σ2p₂ z comes first.

Bond Order and What It Predicts

Bond order is the quantitative payoff of the whole theory. If $N_b$ is the number of electrons in bonding orbitals and $N_a$ the number in antibonding orbitals, then:

$$\text{Bond order} = \frac{1}{2}\,(N_b - N_a)$$

A positive bond order ($N_b > N_a$) means a stable molecule; a zero or negative bond order means the molecule is unstable and does not exist. Integral values of 1, 2 and 3 correspond to single, double and triple bonds. Two further correlations matter for NEET: bond length decreases as bond order increases, and bond dissociation enthalpy increases as bond order increases.

Worked Example

Compute the bond order of $\ce{N2}$, $\ce{O2}$ and $\ce{He2}$.

$\ce{N2}$ (14 e): $N_b = 10$, $N_a = 4$, so bond order $= \tfrac{1}{2}(10 - 4) = 3$ — a triple bond, the strongest of the three.

$\ce{O2}$ (16 e): $N_b = 10$, $N_a = 6$, so bond order $= \tfrac{1}{2}(10 - 6) = 2$ — a double bond.

$\ce{He2}$ (4 e): configuration $(\sigma 1s)^2(\sigma^{*}1s)^2$, so $N_b = 2$, $N_a = 2$ and bond order $= \tfrac{1}{2}(2 - 2) = 0$. With no net bonding, $\ce{He2}$ does not exist.

Because bond order depends only on the electron count and the orbital ordering, isoelectronic species share the same bond order. $\ce{CO}$, $\ce{CN-}$ and $\ce{NO+}$ are all isoelectronic with $\ce{N2}$ (14 valence-shell electrons in the standard count) and all carry bond order 3 — a relationship NEET tests repeatedly.

Configurations of Diatomic Molecules

Writing the configuration is mechanical once the ordering is fixed: fill the MOs by aufbau, pair only after each degenerate orbital has one electron (Hund's rule), and respect Pauli. The closed K shell $(\sigma 1s)^2(\sigma^{*}1s)^2$ is often abbreviated KK.

MoleculeElectronsValence MO configurationBond orderMagnetic nature
$\ce{H2}$2$(\sigma 1s)^2$1Diamagnetic
$\ce{He2}$4$(\sigma 1s)^2(\sigma^{*}1s)^2$0Does not exist
$\ce{Li2}$6$KK(\sigma 2s)^2$1Diamagnetic
$\ce{C2}$12$KK(\sigma 2s)^2(\sigma^{*}2s)^2(\pi 2p_x^2 = \pi 2p_y^2)$2Diamagnetic
$\ce{N2}$14$KK(\sigma 2s)^2(\sigma^{*}2s)^2(\pi 2p_x^2 = \pi 2p_y^2)(\sigma 2p_z)^2$3Diamagnetic
$\ce{O2}$16$KK(\sigma 2s)^2(\sigma^{*}2s)^2(\sigma 2p_z)^2(\pi 2p_x^2 = \pi 2p_y^2)(\pi^{*}2p_x^1 = \pi^{*}2p_y^1)$2Paramagnetic

Note the two ordering conventions in action: $\ce{C2}$ and $\ce{N2}$ fill the $\pi 2p$ orbitals before $\sigma 2p_z$, whereas $\ce{O2}$ fills $\sigma 2p_z$ first. A striking detail is that the double bond in $\ce{C2}$ is made of two pi bonds (four electrons in two $\pi$ MOs, with no net $\sigma 2p$ contribution), unlike the usual one-sigma-one-pi double bond.

NEET Trap

Bond order zero does not mean "weakly stable"

NEET 2025 Q.47 paired the false claim that a diatomic with zero bond order is "quite stable" with a second false claim that bond length increases with bond order. Both are wrong: zero bond order means the molecule is unstable, and bond length decreases as bond order increases. Read such statement-pairs literally and check each clause against the rules.

Higher bond order ⇒ shorter bond, higher bond enthalpy, greater stability.

Magnetic Behaviour

The magnetic nature follows directly from the configuration. If every molecular orbital is doubly occupied, all electron spins are paired and the substance is diamagnetic (weakly repelled by a magnetic field). If one or more MOs are singly occupied, the unpaired electron(s) make the substance paramagnetic (attracted by a magnetic field).

Figure 2 Highest occupied orbitals of O₂ π*2p x π*2p y Two unpaired electrons, parallel spin → paramagnetic

In $\ce{O2}$ the last two electrons enter the degenerate $\pi^{*}2p_x$ and $\pi^{*}2p_y$ orbitals singly with parallel spins, obeying Hund's rule. The resulting two unpaired electrons make $\ce{O2}$ paramagnetic — the observation that MO theory explains and Lewis structures cannot.

This is the headline triumph of MO theory: it predicts, from first principles, that oxygen is paramagnetic, in exact agreement with experiment. The same reasoning lets you classify any diatomic by inspecting whether its frontier orbitals are singly or doubly filled.

Quick Recap

Molecular Orbital Theory in one screen

  • MOs are built by LCAO: addition gives a bonding MO ($\sigma$, $\pi$); subtraction gives an antibonding MO ($\sigma^{*}$, $\pi^{*}$).
  • Two AOs give two MOs; the bonding MO is lower in energy, the antibonding MO higher, and total energy is conserved.
  • AOs combine only with matched energy, matched symmetry, and maximum overlap.
  • Ordering switch: for $\ce{B2}$, $\ce{C2}$, $\ce{N2}$ the $\pi 2p$ pair lies below $\sigma 2p_z$; for $\ce{O2}$, $\ce{F2}$, $\ce{Ne2}$, $\sigma 2p_z$ lies below the $\pi 2p$ pair.
  • Bond order $= \tfrac{1}{2}(N_b - N_a)$; positive means stable, zero or negative means non-existent.
  • Higher bond order means shorter bond and higher bond enthalpy; isoelectronic species share bond order.
  • Doubly filled MOs → diamagnetic; one or more singly filled MOs → paramagnetic (as in $\ce{O2}$).

NEET PYQ Snapshot — Molecular Orbital Theory

Real NEET questions on MO ordering, bond order, stability and magnetic nature (2018–2025).

NEET 2023 · Q.64

The correct order of energies of molecular orbitals of $\ce{N2}$ molecule, is —

  1. $\sigma 1s < \sigma^{*}1s < \sigma 2s < \sigma^{*}2s < (\pi 2p_x = \pi 2p_y) < (\pi^{*}2p_x = \pi^{*}2p_y) < \sigma 2p_z < \sigma^{*}2p_z$
  2. $\sigma 1s < \sigma^{*}1s < \sigma 2s < \sigma^{*}2s < (\pi 2p_x = \pi 2p_y) < \sigma 2p_z < (\pi^{*}2p_x = \pi^{*}2p_y) < \sigma^{*}2p_z$
  3. $\sigma 1s < \sigma^{*}1s < \sigma 2s < \sigma^{*}2s < \sigma 2p_z < (\pi 2p_x = \pi 2p_y) < (\pi^{*}2p_x = \pi^{*}2p_y) < \sigma^{*}2p_z$
  4. $\sigma 1s < \sigma^{*}1s < \sigma 2s < \sigma^{*}2s < \sigma 2p_z < \sigma^{*}2p_z < (\pi 2p_x = \pi 2p_y) < (\pi^{*}2p_x = \pi^{*}2p_y)$
Answer: (2)

For $\ce{B2}$, $\ce{C2}$, $\ce{N2}$ the degenerate $\pi 2p$ orbitals lie below $\sigma 2p_z$ because of 2s–2p mixing, giving the order in option (2).

NEET 2025 · Q.47

Two statements are given. Statement I: A hypothetical diatomic molecule with bond order zero is quite stable. Statement II: As bond order increases, the bond length increases. Choose the most appropriate answer.

  1. Statement I is false but Statement II is true
  2. Both Statement I and Statement II are true
  3. Both Statement I and Statement II are false
  4. Statement I is true but Statement II is false
Answer: (3)

Zero bond order means an unstable molecule, so Statement I is false. Bond length decreases as bond order increases, so Statement II is also false. Both statements are false.

NEET 2018 · Q.61

Consider the species $\ce{CN+}$, $\ce{CN-}$, $\ce{NO}$ and $\ce{CN}$. Which one of these will have the highest bond order?

  1. $\ce{NO}$
  2. $\ce{CN-}$
  3. $\ce{CN+}$
  4. $\ce{CN}$
Answer: (2)

Bond orders: $\ce{NO} = 2.5$, $\ce{CN+} = 2.0$, $\ce{CN} = 2.5$, $\ce{CN-} = 3.0$. $\ce{CN-}$ is isoelectronic with $\ce{N2}$ ($N_b - N_a = 6$), so its bond order is $\tfrac{1}{2}(6) = 3$ — the highest.

NEET 2020 · Q.151

Identify a molecule which does not exist.

  1. $\ce{Li2}$
  2. $\ce{C2}$
  3. $\ce{O2}$
  4. $\ce{He2}$
Answer: (4)

$\ce{He2}$ has four electrons in $(\sigma 1s)^2(\sigma^{*}1s)^2$, giving bond order $\tfrac{1}{2}(2-2) = 0$. With no net bonding the molecule does not exist.

NEET 2017 · Q.6

Which one of the following pairs of species has the same bond order?

  1. $\ce{N2}$, $\ce{O2^2-}$
  2. $\ce{CO}$, $\ce{NO}$
  3. $\ce{O2}$, $\ce{NO+}$
  4. $\ce{CN-}$, $\ce{CO}$
Answer: (4)

$\ce{CO}$ and $\ce{CN-}$ are isoelectronic, each with 14 electrons in the standard count, and both have bond order 3.

FAQs — Molecular Orbital Theory

Quick answers to the questions NEET aspirants ask most about MO theory.

What is the difference between a bonding and an antibonding molecular orbital?

A bonding molecular orbital is formed by the additive (constructive) combination of atomic orbital wave functions, so electron density is concentrated between the nuclei; it has lower energy than the parent atomic orbitals and stabilises the molecule. An antibonding molecular orbital is formed by the subtractive (destructive) combination, has a nodal plane between the nuclei, carries higher energy than the parent orbitals, and destabilises the molecule. The total energy of the two molecular orbitals equals the total energy of the two original atomic orbitals.

How do you calculate bond order in molecular orbital theory?

Bond order equals one half the difference between the number of electrons in bonding molecular orbitals (Nb) and the number in antibonding molecular orbitals (Na), that is bond order = (Nb − Na)/2. A positive bond order means a stable molecule, while a zero or negative bond order means the molecule is unstable and does not exist. Bond order also indicates bond type: values of 1, 2 and 3 correspond to single, double and triple bonds.

Why is the MO energy ordering different for N2 compared to O2?

For lighter second-row molecules up to N2 (that is Li2, Be2, B2, C2, N2), 2s–2p mixing pushes the sigma 2pz orbital above the pi 2px and pi 2py orbitals, giving the order with the pi orbitals filled before sigma 2pz. For O2, F2 and Ne2 this mixing is small, so sigma 2pz lies below the pi 2px and pi 2py orbitals. The dividing line falls between N2 and O2.

Why is O2 paramagnetic according to molecular orbital theory?

In O2 the last two electrons occupy the two degenerate antibonding orbitals π*2px and π*2py singly, in line with Hund's rule. Because the molecule contains two unpaired electrons it is attracted by a magnetic field and is therefore paramagnetic. MO theory predicts this correctly, whereas the simple Lewis or valence-bond picture of O=O wrongly suggests all electrons are paired.

Why does the He2 molecule not exist?

Helium has the configuration 1s2, so He2 would have four electrons filling both σ1s and σ*1s, giving the configuration (σ1s)2(σ*1s)2. The number of bonding and antibonding electrons is equal, so the bond order is ½(2 − 2) = 0. A zero bond order means there is no net bonding, so the He2 molecule is unstable and does not exist.

What conditions must atomic orbitals satisfy to combine into molecular orbitals?

Three conditions must be met. The combining atomic orbitals must have the same or nearly the same energy, so a 1s orbital combines with another 1s but not with a 2s. They must have the same symmetry about the molecular axis, so a 2pz orbital combines with another 2pz but not with a 2px or 2py. Finally, the orbitals must overlap to the maximum possible extent, since greater overlap gives greater electron density between the nuclei.

Related Topics
Chemical Bonding and Molecular Structure Back to the full chapter hub. Homonuclear Diatomic Bonding MO configurations across the second period. Valence Bond Theory The localised-bond model MO theory complements. Hybridisation How atomic orbitals mix to set molecular geometry. Bond Parameters Bond length, bond enthalpy and bond order links.