Mot — Concepts, Formulas & Examples

Molecular orbital theory — bonding in diatomic molecules.

CBSE JEE-MAIN NEET • 11 min read
Tags Mot

Molecular orbital (MO) theory explains bonding by combining atomic orbitals to form molecular orbitals that extend over the whole molecule. CBSE Class 11 and NEET test MO diagrams of diatomic molecules and the concept of bond order.

Core Concepts

Basic idea

When atomic orbitals combine, they form molecular orbitals — bonding (lower energy, between nuclei) and antibonding (higher energy, a node between nuclei). Electrons fill MOs from lowest to highest energy.

The key principle: nn atomic orbitals combine to form nn molecular orbitals. So two 1s orbitals from two H atoms give one σ1s\sigma_{1s} (bonding) and one σ1s∗\sigma^*_{1s} (antibonding) molecular orbital.

Rules for filling MOs:

  1. Aufbau principle: Fill lowest energy MO first
  2. Pauli exclusion: Maximum 2 electrons per MO, with opposite spins
  3. Hund’s rule: For degenerate MOs (same energy), fill each singly before pairing

Bonding and antibonding orbitals

Bonding MOs (sigma, pi) are labelled without a star. Antibonding MOs (sigma*, pi*) are labelled with a star. Electrons in bonding stabilise the molecule; in antibonding, destabilise it.

Sigma orbitals: Formed by head-on overlap of atomic orbitals. Cylindrically symmetric about the internuclear axis. σ1s\sigma_{1s}, σ2s\sigma_{2s}, σ2pz\sigma_{2p_z} are all sigma type.

Pi orbitals: Formed by sideways overlap of p orbitals perpendicular to the internuclear axis. π2px\pi_{2p_x} and π2py\pi_{2p_y} come in degenerate pairs.

Antibonding orbitals have a nodal plane between the two nuclei. They are higher in energy than the parent atomic orbitals. Electrons in antibonding orbitals weaken the bond.

Bond order

Bond order=Bonding electrons−Antibonding electrons2\text{Bond order} = \frac{\text{Bonding electrons} - \text{Antibonding electrons}}{2}

Higher bond order = stronger bond = shorter bond length. Zero bond order means no stable molecule forms.
Bond orderBond typeExampleBond lengthBond energy
1SingleH2_2, F2_2LongestWeakest
2DoubleO2_2MediumMedium
3TripleN2_2ShortestStrongest
0No bondHe2_2——

Energy level ordering

For O2_2, F2_2, Ne2_2 (Z > 7):

σ1s<σ1s∗<σ2s<σ2s∗<σ2pz<π2px=π2py<π2px∗=π2py∗<σ2pz∗\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}

For B2_2, C2_2, N2_2 (Z ≤\leq 7):

σ1s<σ1s∗<σ2s<σ2s∗<π2px=π2py<σ2pz<π2px∗=π2py∗<σ2pz∗\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}

The key difference: for lighter molecules (B2_2, C2_2, N2_2), the σ2pz\sigma_{2p_z} is HIGHER in energy than the π2p\pi_{2p} orbitals. For heavier molecules (O2_2, F2_2), σ2pz\sigma_{2p_z} is LOWER. This swap happens because of s-p mixing in lighter elements.

MO diagram of O2_2

Filling order gives 10 bonding and 6 antibonding electrons. Bond order = (10-6)/2 = 2. Two unpaired electrons in pi* orbitals make O2_2 paramagnetic — a MO theory triumph, since valence bond theory predicted it would be diamagnetic.

Electronic configuration of O2_2 (16 electrons):

(σ1s)2(σ1s∗)2(σ2s)2(σ2s∗)2(σ2pz)2(π2px)2(π2py)2(π2px∗)1(π2py∗)1(\sigma_{1s})^2 (\sigma^*_{1s})^2 (\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

The last two electrons go into the degenerate π∗\pi^* orbitals with parallel spins (Hund’s rule). These two unpaired electrons make O2_2 paramagnetic — liquid oxygen is attracted to a magnet.

MO diagram of N2_2

14 electrons total. Bond order = 3, giving a very strong triple bond. All electrons paired, so N2_2 is diamagnetic. This is why atmospheric nitrogen is so unreactive.

Electronic configuration of N2_2:

(σ1s)2(σ1s∗)2(σ2s)2(σ2s∗)2(π2px)2(π2py)2(σ2pz)2(\sigma_{1s})^2 (\sigma^*_{1s})^2 (\sigma_{2s})^2 (\sigma^*_{2s})^2 (\pi_{2p_x})^2 (\pi_{2p_y})^2 (\sigma_{2p_z})^2

Note: For N2_2, π2p\pi_{2p} fills before σ2pz\sigma_{2p_z} (the lighter-element ordering).

Bonding electrons = 10, antibonding = 4. Bond order = (10-4)/2 = 3. All electrons paired → diamagnetic.

MO configurations of important species

SpeciesTotal e−e^-Bond orderMagnetic natureStable?
H2_221DiamagneticYes
He2_240—No
He2+_2^+30.5ParamagneticYes (weakly)
Li2_261DiamagneticYes
B2_2101ParamagneticYes
C2_2122DiamagneticYes
N2_2143DiamagneticYes
O2_2162ParamagneticYes
O2−_2^-171.5ParamagneticYes
O22−_2^{2-}181DiamagneticYes
F2_2181DiamagneticYes
Ne2_2200—No

NEET and JEE often ask: “Which of the following is paramagnetic?” or “Arrange in order of bond length.” Use the table above. Remember: B2_2 and O2_2 are paramagnetic (this is the exam favourite). Higher bond order = shorter bond length.

Comparison with VBT

FeatureValence Bond TheoryMO Theory
BondingOverlap of atomic orbitalsCombination into molecular orbitals
ElectronsLocalised between two atomsDelocalised over entire molecule
Paramagnetism of O2_2Cannot explainCorrectly predicts
Bond order fractionsNot possiblePossible (e.g., 0.5 for He2+_2^+)
ResonanceNeeds multiple structuresSingle description

Worked Examples

4 electrons total, 2 in bonding sigma, 2 in antibonding sigma*. Bond order = (2-2)/2 = 0. No net bonding, no stable molecule. Helium remains monatomic.

MO diagram places two electrons in degenerate pi* orbitals. Hund’s rule puts them in different orbitals with parallel spins. Unpaired electrons make O2_2 paramagnetic — liquid O2_2 is attracted to a magnet.

O2−_2^- has 17 electrons (one more than O2_2). The extra electron goes into π2px∗\pi^*_{2p_x} (now 2 electrons there, 1 in π2py∗\pi^*_{2p_y}).

Bonding = 10, antibonding = 7. Bond order = (10-7)/2 = 1.5.

One unpaired electron → paramagnetic. Bond is weaker and longer than O2_2.

Ne has 10 electrons, so Ne2_2 has 20. Filling all MOs: bonding = 10, antibonding = 10. Bond order = 0. Ne2_2 does not exist — just like He2_2, both noble gas diatomics are unstable.

N2_2 has bond order 3 — the highest among homonuclear diatomics. Bond dissociation energy = 945 kJ/mol. This explains why breaking the N≡\equivN bond requires enormous energy (Haber process needs 450°C and 200 atm), and why N2_2 is so chemically inert.

Solved Problems (Exam Style)

Problem 1 (JEE Main pattern): Arrange in order of increasing bond length: N2_2, O2_2, F2_2.

N2_2: BO = 3. O2_2: BO = 2. F2_2: BO = 1. Higher bond order = shorter bond. N2_2 < O2_2 < F2_2 (increasing bond length).

Problem 2 (NEET pattern): Which of the following is paramagnetic — N2_2, O2_2, F2_2, C2_2?

N2_2: all paired → diamagnetic. O2_2: 2 unpaired in π∗\pi^* → paramagnetic. F2_2: all paired → diamagnetic. C2_2: all paired → diamagnetic. Answer: O2_2

Problem 3 (JEE Main): Calculate the bond order of CO.

CO has 14 electrons (6 from C + 8 from O). It is isoelectronic with N2_2, so it uses the same MO ordering.

Configuration same as N2_2: bonding = 10, antibonding = 4. Bond order = (10-4)/2 = 3.

CO has a triple bond — it is one of the strongest diatomic bonds, which is why CO is a very stable (and toxic) gas.

Common Mistakes

Saying O2_2 has only single bond. Bond order is 2.

Confusing sigma and pi orbitals. Sigma has end-on overlap; pi has sideways overlap with a node in the plane containing the nuclei.

Writing that N2_2 is paramagnetic. It is diamagnetic because all electrons are paired.

Using the same MO energy ordering for all diatomics. The ordering changes at O2_2 — for Z ≤\leq 7 (B, C, N), π2p\pi_{2p} is below σ2pz\sigma_{2p_z}. For Z > 7 (O, F), σ2pz\sigma_{2p_z} is below π2p\pi_{2p}. Getting this wrong changes the magnetic prediction.

Forgetting that bond order can be fractional. He2+_2^+ has bond order 0.5 and does exist (weakly). Fractional bond orders are a unique prediction of MO theory that VBT cannot make.

Exam Weightage and Revision

This topic is a repeat performer in board papers and entrance exams. NEET typically asks one to two questions on the core mechanisms, CBSE boards give three to six marks, and state PMT papers often include a diagram-based long answer. The PYQs cluster around a small set of facts — lock those and you clear the topic.

JEE Main 2024 asked about bond order of O2−_2^-. NEET 2023 tested paramagnetism of O2_2. JEE Main 2023 had a question on isoelectronic species (CO and N2_2). MOT gives 1-2 questions in JEE every year.

When a question gives a scenario, identify the core mechanism first, then match it to the concepts above. Most wrong answers come from reading the scenario too quickly.

Draw one MO diagram for O2_2 with electron fillings marked. That diagram answers most PYQs on MOT.

Practice Questions

Q1. Why does B2_2 have a different MO ordering than O2_2?

In lighter elements (Z ≤\leq 7), the 2s and 2p atomic orbitals are close enough in energy that they interact (s-p mixing). This pushes the σ2pz\sigma_{2p_z} orbital above the π2p\pi_{2p} orbitals. In heavier elements (O, F), the 2s-2p energy gap is large enough that s-p mixing is negligible, and σ2pz\sigma_{2p_z} stays below π2p\pi_{2p}.

Q2. What is the bond order of NO?

NO has 15 electrons (7 from N + 8 from O). Using the O2_2-type ordering: bonding = 10, antibonding = 5. Bond order = (10-5)/2 = 2.5. One unpaired electron in π∗\pi^* makes NO paramagnetic.

Q3. Arrange O2_2, O2−_2^-, O22−_2^{2-}, O2+_2^+ in order of increasing bond length.

Bond orders: O2+_2^+ = 2.5, O2_2 = 2, O2−_2^- = 1.5, O22−_2^{2-} = 1. Higher bond order = shorter bond. O2+_2^+ < O2_2 < O2−_2^- < O22−_2^{2-} (increasing bond length).

Q4. Why is CO a stronger bond than N2_2 despite both having bond order 3?

Both have bond order 3, but CO has a slightly stronger bond (1072 kJ/mol vs 945 kJ/mol for N2_2). This is because in CO, the electronegativity difference between C and O causes the bonding electrons to be more concentrated between the nuclei, leading to stronger bonding. The ionic contribution strengthens the overall bond.

Q5. Can MO theory explain why He2+_2^+ exists but He2_2 does not?

He2_2 has 4 electrons: 2 bonding + 2 antibonding = bond order 0. No net bonding. He2+_2^+ has 3 electrons: 2 bonding + 1 antibonding = bond order 0.5. A fractional but positive bond order means a weak bond exists. He2+_2^+ has been detected experimentally. VBT cannot explain this because it does not allow fractional bond orders.

FAQs

What is the difference between an orbit and an orbital in the context of MOT? An orbit (Bohr model) is a fixed circular path. An orbital (quantum model) is a mathematical function describing the probability of finding an electron in a region of space. Molecular orbitals are orbitals that extend over the entire molecule, not just one atom.

Why does MO theory predict O2_2 is paramagnetic while VBT does not? VBT draws O2_2 with a double bond (O=O) where all electrons are paired, predicting diamagnetism. MO theory fills molecular orbitals according to Hund’s rule and naturally places two electrons unpaired in degenerate π∗\pi^* orbitals, correctly predicting paramagnetism. This was one of the major triumphs of MO theory.

What are isoelectronic species? Species with the same number of electrons (and therefore the same MO configuration). N2_2 and CO are isoelectronic (14 electrons each). O2_2 and F22+_2^{2+} are isoelectronic. Isoelectronic species have the same bond order and similar properties.

Is MO theory better than VBT? MO theory is more general and can explain properties like paramagnetism and fractional bond orders that VBT cannot. However, VBT is simpler and more intuitive for describing molecular geometry (hybridisation). In practice, chemists use both theories where each works best.

MO theory is the modern view of bonding. It predicts real behaviour like the paramagnetism of O2_2 that simpler theories miss.

Practice Questions

Molecular orbital theory — bonding, antibonding, bond order calculation

hard

Molecular orbital theory — bonding, antibonding, bond order calculation

hard