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MHT-CET Chemistry · Chemical Bonding and Molecular Structure

Molecular Orbital Theory and Bond Order

Molecular orbital theory fills electrons into bonding and antibonding molecular orbitals; bond order = half of (bonding electrons minus antibonding electrons), and it fixes a molecule's stability, bond length and magnetic behaviour.

All Chemical Bonding and Molecular Structure notesMHT-CET Chemistry strategy

Why this matters

One of the most bankable subtopics in MHT-CET Chemical Bonding — almost every PYQ is a direct plug-in: write the MO configuration of a small diatomic (or its ion), count the bonding and antibonding electrons, and read off the bond order, magnetic nature or a bond-length ordering. The recurring traps are always the same: count TOTAL electrons including the charge on an ion, remember O2 is paramagnetic (two unpaired electrons), and note that ions can carry a fractional bond order. Learn the filling order plus the bond-order formula cold and you can attempt every question here on sight.

Concept 1 of 4

Molecular orbitals and the filling order

Intuition

When two atomic orbitals overlap they combine into two molecular orbitals: a low-energy bonding MO (electrons here glue the atoms together) and a high-energy antibonding MO, marked with a star (electrons here pull the atoms apart). Electrons fill these MOs from lowest energy upward, and for the second-period diatomics one pair of levels swaps depending on which atoms are involved.

Definition

Molecular orbital basics:

  • Overlap of two atomic orbitals gives one bonding MO (lower energy: σ, π\sigma,\ \pi) and one antibonding MO (higher energy: σ∗, π∗\sigma^*,\ \pi^*).
  • Electrons fill MOs following the Aufbau principle, Pauli exclusion and Hund's rule — lowest energy first, one electron per degenerate orbital before pairing.
  • **Order for B2, C2, N2\text{B}_2,\ \text{C}_2,\ \text{N}_2** (and lighter): σ1s<σ∗1s<σ2s<σ∗2s<π2px=π2py<σ2pz<π∗2px=π∗2py<σ∗2pz\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.
  • **Order swaps for O2, F2, Ne2\text{O}_2,\ \text{F}_2,\ \text{Ne}_2**: here σ2pz\sigma 2p_z drops below the π2p\pi 2p pair, so the sequence is …σ∗2s<σ2pz<π2px=π2py<π∗2px=π∗2py<σ∗2pz\ldots \sigma^*2s < \sigma 2p_z < \pi 2p_x = \pi 2p_y < \pi^*2p_x = \pi^*2p_y < \sigma^*2p_z.
  • First count the total number of electrons (add or subtract for an ion's charge), then fill.

Worked example

Write the molecular orbital configuration of the N2 molecule and state how many electrons occupy antibonding orbitals.
  1. Total electrons in N2\text{N}_2 = 7 + 7 = 14. Use the N2\text{N}_2 order (π2p\pi 2p below σ2pz\sigma 2p_z).
  2. Configuration: (σ1s)2(σ∗1s)2(σ2s)2(σ∗2s)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.
  3. Antibonding orbitals occupied are σ∗1s\sigma^*1s and σ∗2s\sigma^*2s: 2+2=42 + 2 = 4 electrons. The π∗2p\pi^*2p orbitals are empty.
Answer:N2 has 4 electrons in antibonding orbitals.
Practice this conceptself-check · 5 quick reps

Try it yourself

Which molecule holds the maximum number of electrons in antibonding molecular orbitals: Li2, N2, O2 or F2?

Practice — Level 1 (5 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    How many electrons in antibonding orbitals of N2?
  2. 2.
    How many electrons in bonding orbitals of O2?
  3. 3.
    Which pair of MOs swaps order for O2 and F2 versus N2?
  4. 4.
    Bonding and antibonding electrons in F2 respectively?
  5. 5.
    Total electrons to fill for the O2 molecule?

From the bank · past-year question

Example 1Chemical Bonding and Molecular StructureMODERATE
What is the number of electrons in bonding molecular orbitals and antibonding molecular orbitals respectively in F2F_2 molecule?

[Q76 · 25 April Shift I · 2025]

Count TOTAL electrons, and adjust for an ion's charge

Always start from the total electron count. N2\text{N}_2 has 14, O2\text{O}_2 has 16 — but an ion shifts this: N2+\text{N}_2^+ has 13 (one removed), O2−\text{O}_2^- has 17 (one added). Filling the wrong number of electrons is the single biggest source of wrong bond orders here.

Only σ* and π* orbitals count as antibonding

When asked for antibonding electrons in N2\text{N}_2, count only σ∗1s\sigma^*1s and σ∗2s\sigma^*2s (= 4). The π∗2p\pi^*2p orbitals are empty in N2\text{N}_2, so do not add them; and never count the bonding σ\sigma/π\pi electrons here.
Drill 3 more on molecular orbitals and the filling order

Concept 2 of 4

Bond order from the MO configuration

Intuition

Bond order tells you how many net bonds hold the two atoms together. Electrons in bonding MOs pull the atoms in; electrons in antibonding MOs push them apart. Subtract one from the other and halve it — that net count is the bond order. For ions it can come out as a half-integer, which is perfectly normal.

Definition

Bond order in molecular orbital theory:

  • Bond order =12(Nb−Na)= \tfrac12(N_b - N_a), where NbN_b = electrons in bonding MOs and NaN_a = electrons in antibonding MOs.
  • Bond order 0 means the molecule does not exist (e.g. hypothetical He2\text{He}_2); positive bond order means a stable molecule.
  • Standard values: H2=1\text{H}_2 = 1, Li2=1\text{Li}_2 = 1, N2=3\text{N}_2 = 3, O2=2\text{O}_2 = 2, F2=1\text{F}_2 = 1, CO=3\text{CO} = 3 (isoelectronic with N2\text{N}_2).
  • Ions give fractional bond orders: N2+=2.5\text{N}_2^+ = 2.5, O2+=2.5\text{O}_2^+ = 2.5, O2−=1.5\text{O}_2^- = 1.5, O22−=1\text{O}_2^{2-} = 1.

Bond order

Bond order=12(Nb−Na)\text{Bond order} = \tfrac{1}{2}\left(N_b - N_a\right)
  • N_bnumber of electrons in bonding molecular orbitals
  • N_anumber of electrons in antibonding molecular orbitals

Worked example

What is the bond order of the F2 molecule?
  1. Total electrons in F2\text{F}_2 = 9 + 9 = 18. Use the F2\text{F}_2 filling order.
  2. Nb=10N_b = 10 (σ1s2 σ2s2 σ2p2 π2p4\sigma1s^2\,\sigma2s^2\,\sigma2p^2\,\pi2p^4), Na=8N_a = 8 (σ∗1s2 σ∗2s2 π∗2p4\sigma^*1s^2\,\sigma^*2s^2\,\pi^*2p^4).
  3. Bond order =12(10−8)=12(2)=1= \tfrac12(10 - 8) = \tfrac12(2) = 1.
Answer:Bond order of F2 = 1 (a single bond).
Practice this conceptself-check · 6 quick reps

Try it yourself

Find the bond order of the N2+ ion.

Practice — Level 1 (6 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    Bond order of F2?
  2. 2.
    Bond order of CO?
  3. 3.
    Bond order of N2+?
  4. 4.
    Bond order of O2?
  5. 5.
    Which molecule has bond order 2: N2, H2, O2 or F2?
  6. 6.
    A molecule with bond order 0 — does it exist?

From the bank · past-year question

Example 2Chemical Bonding and Molecular StructureMODERATE
What is the bond order in N2+\text{N}_2^+?

[Q53 · 13th May Shift 2 · 2024]

Ions can have a fractional bond order

Removing or adding one electron changes Nb−NaN_b - N_a by 1, so the bond order shifts by 12\tfrac12. N2+\text{N}_2^+ and O2+\text{O}_2^+ are both 2.5, O2−\text{O}_2^- is 1.5 — a half-integer answer is correct, not an arithmetic slip. Do not round it.

MOT bond order can differ from the Lewis picture

The Lewis structure of CO looks like a double bond, but MOT gives bond order 3 (CO is isoelectronic with N2\text{N}_2). Trust the MO count 12(Nb−Na)\tfrac12(N_b - N_a) over a quick Lewis guess.
Drill 3 more on bond order from the mo configuration

Concept 3 of 4

Magnetic behaviour, bond length and stability

Intuition

Once you have the MO configuration, two more properties fall straight out. Any unpaired electron makes the molecule paramagnetic (attracted to a magnet); all-paired means diamagnetic. And a higher bond order pulls the atoms closer, so the bond is shorter and stronger. The famous case is O2: MOT correctly predicts it is paramagnetic, which the Lewis structure cannot.

Definition

Reading properties off the MO configuration:

  • Paramagnetic = one or more unpaired electrons (attracted by a magnetic field); diamagnetic = all electrons paired.
  • O2\text{O}_2 has two unpaired electrons in its π∗2p\pi^*2p orbitals, so it is paramagnetic — a key success of MOT. N2\text{N}_2 and F2\text{F}_2 are fully paired, so diamagnetic.
  • Higher bond order → shorter bond length → greater stability (more energy needed to break it): N2\text{N}_2 (BO 3) has the shortest, strongest bond; Cl2\text{Cl}_2 (BO 1) the longest.
  • Odd-electron molecules like NO (11 valence-shell electrons) carry one unpaired electron → paramagnetic.

Bond order controls length and strength

Bond order↑  ⇒  bond length↓,bond strength↑\text{Bond order} \uparrow \;\Rightarrow\; \text{bond length} \downarrow,\quad \text{bond strength} \uparrow

Worked example

Arrange N2, O2 and Cl2 in decreasing order of bond length.
  1. Find the bond orders: N2=3\text{N}_2 = 3 (triple), O2=2\text{O}_2 = 2 (double), Cl2=1\text{Cl}_2 = 1 (single).
  2. Higher bond order means a shorter bond, so the ordering of bond length is the reverse of bond order.
  3. Decreasing bond length: Cl2>O2>N2\text{Cl}_2 > \text{O}_2 > \text{N}_2.
Answer:Cl2 > O2 > N2 (longest to shortest).
Practice this conceptself-check · 5 quick reps

Try it yourself

State the bond order and magnetic nature of the Li2 molecule.

Practice — Level 1 (5 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    Is O2 paramagnetic or diamagnetic?
  2. 2.
    Number of unpaired electrons in NO?
  3. 3.
    Magnetic nature of N2?
  4. 4.
    Which has the shortest bond: N2, O2 or Cl2?
  5. 5.
    Decreasing bond length order of N2, O2, Cl2?

From the bank · past-year question

Example 3Chemical Bonding and Molecular StructureMODERATE
Identify the bond order and magnetic nature of Li₂ molecule respectively.

[Q85 · 4th May Shift 2 · 2023]

O2 is paramagnetic — the two unpaired electrons

The Lewis structure of O2\text{O}_2 shows all electrons paired, but MOT places two unpaired electrons in the π∗2px\pi^*2p_x and π∗2py\pi^*2p_y orbitals (one each, by Hund's rule). So O2\text{O}_2 is paramagnetic — a classic exam favourite. Never call it diamagnetic.

Higher bond order = shorter bond, not longer

Bond length runs opposite to bond order. N2\text{N}_2 (BO 3) has the shortest, strongest bond; Cl2\text{Cl}_2 (BO 1) the longest, weakest. In a 'decreasing bond length' question the order is the reverse of the bond-order order.
Drill 3 more on magnetic behaviour, bond length and stability

Concept 4 of 4

Bond order and magnetic nature of common species

Intuition

Most PYQs test a handful of standard diatomics and their ions. Memorise this small table — total electrons, bond order and magnetic nature — and you can answer 'which has bond order X', 'which is paramagnetic' or 'order by stability' instantly, without re-deriving each configuration.

Definition

The bank almost always draws from the species below. Two anchors to hold it together:

  • **N2\text{N}_2 and CO (both 14 electrons, isoelectronic) sit at bond order 3** — the most stable.
  • **O2\text{O}_2** is the paramagnetic one (bond order 2, two unpaired electrons); its ions O2+\text{O}_2^+ and O2−\text{O}_2^- shift the bond order up or down by 12\tfrac12.
SpeciesTotal electronsBond orderMagnetic nature
H2\text{H}_221Diamagnetic
Li2\text{Li}_261Diamagnetic
N2\text{N}_2143Diamagnetic
N2+\text{N}_2^+132.5Paramagnetic
One electron removed from a bonding orbital, so bond order drops to 2.5.
O2\text{O}_2162Paramagnetic
Two unpaired electrons in π∗2p\pi^*2p — the classic paramagnetic diatomic.
O2+\text{O}_2^+152.5Paramagnetic
O2−\text{O}_2^-171.5Paramagnetic
F2\text{F}_2181Diamagnetic
CO\text{CO}143Diamagnetic
Isoelectronic with N2\text{N}_2; MOT gives bond order 3, not the Lewis double bond.
NO\text{NO}152.5Paramagnetic
Odd-electron molecule: one unpaired electron in a π∗2p\pi^*2p orbital.
Bond order rises to a maximum of 3 at N2\text{N}_2/CO; paramagnetic species are the ones with an unpaired electron.
Practice this conceptself-check · 5 quick reps

Try it yourself

Among N2, H2, O2 and F2, which molecule has a bond order of exactly 2?

Practice — Level 1 (5 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    Bond order of N2?
  2. 2.
    Which molecule with bond order 2 is also paramagnetic?
  3. 3.
    Bond order and magnetic nature of NO?
  4. 4.
    Bond order of O2+?
  5. 5.
    Which two species are isoelectronic with 14 electrons and bond order 3?

From the bank · past-year question

Example 4Chemical Bonding and Molecular StructureEASY
Which of the following molecule has bond order 2?

[Q82 · 10th May Shift 1 · 2023]

O2 is paramagnetic even though its bond order is a whole number

Bond order being an integer (2) does not make O2\text{O}_2 diamagnetic — magnetic nature depends on *unpaired* electrons, not on the bond order. O2\text{O}_2 has two unpaired π∗2p\pi^*2p electrons, so it is paramagnetic. Read the two properties independently.
Drill 1 more on bond order and magnetic nature of common species

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

  • Bond order from the MO configuration

    Bond order

    Bond order=12(Nb−Na)\text{Bond order} = \tfrac{1}{2}\left(N_b - N_a\right)
  • Magnetic behaviour, bond length and stability

    Bond order controls length and strength

    Bond order↑  ⇒  bond length↓,bond strength↑\text{Bond order} \uparrow \;\Rightarrow\; \text{bond length} \downarrow,\quad \text{bond strength} \uparrow

Reference tables (1)

Bond order and magnetic nature of common species10 rows
SpeciesTotal electronsBond orderMagnetic nature
H2\text{H}_221Diamagnetic
Li2\text{Li}_261Diamagnetic
N2\text{N}_2143Diamagnetic
N2+\text{N}_2^+132.5Paramagnetic
One electron removed from a bonding orbital, so bond order drops to 2.5.
O2\text{O}_2162Paramagnetic
Two unpaired electrons in π∗2p\pi^*2p — the classic paramagnetic diatomic.
O2+\text{O}_2^+152.5Paramagnetic
O2−\text{O}_2^-171.5Paramagnetic
F2\text{F}_2181Diamagnetic
CO\text{CO}143Diamagnetic
Isoelectronic with N2\text{N}_2; MOT gives bond order 3, not the Lewis double bond.
NO\text{NO}152.5Paramagnetic
Odd-electron molecule: one unpaired electron in a π∗2p\pi^*2p orbital.
Bond order rises to a maximum of 3 at N2\text{N}_2/CO; paramagnetic species are the ones with an unpaired electron.

Watch out for (7)

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Chemical Bonding and Molecular StructureMODERATE
What is the number of electrons present in antibonding orbitals of N₂ molecule according to molecular orbital theory?

[Q97 · Shift 1 · 2023]

Example 2Chemical Bonding and Molecular StructureEASY
What is bond order of F2F_2 molecule?

[Q100 · 12th May Shift 2 · 2024]

Example 3Chemical Bonding and Molecular StructureMODERATE
Select the incorrect statement about N2N_{2} molecule.

[Q62 · 20 April Shift II · 2025]

Example 4Chemical Bonding and Molecular StructureMODERATE
Identify the bond order and magnetic nature of Li₂ molecule respectively.

[Q85 · 4th May Shift 2 · 2023]

Example 5Chemical Bonding and Molecular StructureHARD
What is the total number of electrons present in bonding orbitals of O2_2 molecule according to molecular orbital theory?

[Q98 · 16th May Shift 1 · 2023]

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