Using molecular orbital theory, write the electronic configuration of $ ext{O}_2$, calculate its bond order, and explain why $ ext{O}_2$ is paramagnetic — a fact that simple valence bond theory cannot account for.

Question

Using molecular orbital theory, write the electronic configuration of $	ext{O}_2$, calculate its bond order, and explain why $	ext{O}_2$ is paramagnetic — a fact that simple valence bond theory cannot account for.

Accepted Answer

Concept understanding — Molecular Orbital Theory — MO Diagrams, Bond Order & Magnetism

Molecular Orbital Theory — MO Diagrams, Bond Order & Magnetism

The big idea. When atoms bond, their atomic orbitals combine (LCAO) into molecular orbitals that belong to the whole molecule. Each pair of atomic orbitals gives one bonding MO (lower energy, electron density between the nuclei) and one antibonding MO (higher energy, marked with a *). Fill these with the molecule's electrons, and you can read off its bond order, stability and — crucially — its magnetism.

The two orderings

For second-period homonuclear diatomics there are two energy orderings, because s–p mixing pushes the σ2p up:

Li₂ through N₂ (≤ 14 valence e⁻) — s–p mixing order: σ2s, σ2s, (π2px = π2py), σ2pz, (π2px = π2py), σ2pz. Here the π2p bonding MOs sit below σ2pz.
O₂, F₂, Ne₂ — normal order: σ2s, σ2s, σ2pz, (π2px = π2py), (π2px = π2py), σ2pz. Here σ2pz sits below the π2p.

Bond order

Bond order = ½ (N_bonding − N_antibonding).

Fill the MOs (Aufbau, Pauli, Hund) and count. Standard values:

Species	e⁻	Bond order	Unpaired e⁻	Magnetism
H₂	2	1	0	diamagnetic
He₂	4	0	—	does not exist
Li₂	6	1	0	diamagnetic
B₂	10	1	2	paramagnetic
C₂	12	2	0	diamagnetic
N₂	14	3	0	diamagnetic
O₂	16	2	2	paramagnetic
F₂	18	1	0	diamagnetic

Ions follow from adding/removing electrons: O₂⁺ 2.5, O₂⁻ (superoxide) 1.5, O₂²⁻ (peroxide) 1; N₂⁺ and N₂⁻ both 2.5; He₂⁺ 0.5; NO 2.5, NO⁺ 3, CN⁻ 3, CO 3.

Magnetism — MOT's triumph

A species is paramagnetic if it has unpaired electrons, diamagnetic if all are paired. MOT famously predicts that O₂ is paramagnetic — its last two electrons go singly into the degenerate π*2p orbitals (Hund's rule) — which the simple Lewis structure O=O cannot explain. B₂ is also paramagnetic (two electrons singly in π2p).

Bond order, length, energy — and adding/removing an electron

Higher bond order means a shorter bond and a higher dissociation energy. Whether ionisation raises or lowers the bond order depends on which MO the electron leaves or enters:

O₂ → O₂⁺ removes an antibonding π* electron → bond order rises (2 → 2.5).
O₂ → O₂⁻ adds an antibonding π* electron → bond order falls (2 → 1.5). …

Start your 14-day free trial to unlock the full solution →

Oxygen contributes 6 valence electrons per atom, so $\text{O}_2$ has 12 valence electrons to place into the molecular orbitals. For $\text{O}_2$ (heavier than $\text{N}_2$ ), the correct energy ordering places $\sigma_{2p_z}$ below the $\pi_{2p}$ pair: $\sigma_{2s} < \sigma^*_{2s} < \sigma_{2p_z} < (\pi_{2p_x}=\pi_{2p_y}) < (\pi^*_{2p_x}=\pi^*_{2p_y}) < \sigma^*_{2p_z}$ . Filling 12 electrons in order gives: $\sigma_{2s}^2\sigma^{*2}_{2s}\sigma_{2p_z}^2\pi_{2p_x}^2\pi_{2p_y}^2\pi^{*1}_{2p_x}\pi^{*1}_{2p_y}$ — note that, by Hund's rule, the last two electrons go singly into each of the two degenerate $\pi^*_{2p}$ orbitals rather than pairing up in just one of them. Counting bonding electrons ( $\sigma_{2s}=2$ , $\sigma_{2p_z}=2$ , $\pi_{2p_x}=2$ , $\pi_{2p_y}=2$ , total $8$ ) and antibonding electrons ( $\sigma^*_{2s}=2$ , $\pi^*_{2p_x}=1$ , $\pi^*_{2p_y}=1$ , total $4$ ), the bond order is $\tfrac{1}{2}(8-4)=2$ , correctly matching the $\text{O}=\text{O}$ double bond. Crucially, the two singly-occupied $\pi^*_{2p}$ orb …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2025Set ANNUAL1 markMCQ
$Q. Which of the following is not paramagnetic?$
(a) N2+
(b) CO
(c) O2-
(d) NO
›Reveal solutionSolution
CO is the only one of the four species that is NOT paramagnetic — it is diamagnetic.

Using molecular orbital theory:

N2+ (13 electrons, isoelectronic-type with an odd electron count) has 1 unpaired electron — paramagnetic.

CO (14 electrons, isoelectronic with N2) has a fully-paired MO configuration — diamagnetic. …
Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 markMCQ
$Q. The number of unpaired electrons in liquid oxygen molecule is$
(a) 1
(b) 2
(c) 3
(d) 4
›Reveal solutionSolution

O2 has 2 unpaired electrons (in the π2px and π2py antibonding orbitals), explaining its paramagnetism (visible in liquid O2 being attracted to a magnet).

By molecular orbital theory, the electronic configuration of O2 is:

$\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^{*1} \pi_{2p_y}^{*1}$

…

Start your 14-day free trial to see the full solution →
CBSE 2023Set ANNUAL1 markMCQ
$Q. Assertion : Oxygen molecule is Paramagnetic. Reason : It has two unpaired electrons in its bonding molecular orbital.$
(a) Assertion is true but reason is false.
(b) Both assertion and reason are true and reason is the correct explanation of assertion.
(c) Both assertion and reason are false.
(d) Both assertion and reason are true but reason is not the correct explanation of assertion.
›Reveal solutionSolution

Oxygen is genuinely paramagnetic (assertion true), but its two unpaired electrons sit in the antibonding pi*2p molecular orbitals, not the bonding ones -- so the reason, as worded, is false.

By molecular orbital theory, the electronic configuration of O2 (16 electrons total) is:

sigma1s2 sigma1s2 sigma2s2 sigma2s2 sigma2pz2 pi2px2 pi2py2 pi2px1 pi2py1

The last two electrons singly occupy the two degenerate antibonding pi* molecular orbitals (pi2px and pi2py), each with one unpaired electron (following Hund's rule). This gives O2 a net of two unpaired electrons, which is exactly why it is paramagnetic (attracted into a magnetic field) -- a result MO theory correctly predicts, unlike simple Lewis/valence-bond structures which show all electrons paired …

Start your 14-day free trial to see the full solution →
CBSE 2022Set ANNUAL1 markMCQ
$Q. Which one of the following is diamagnetic?$
(a) O2 2- (superoxide/peroxide ion)
(b) O2 + (dioxygenyl cation)
(c) O2 (dioxygen)
(d) None of these
›Reveal solutionSolution

Molecular orbital theory shows O2 and O2+ both have unpaired electrons in their pi* antibonding orbitals (paramagnetic), while O2^2- (peroxide, 18 valence-region electrons) has those pi* orbitals completely filled, leaving no unpaired electrons — so it is diamagnetic.

Building the molecular orbital electron configuration for the O2 family (order: sigma2s, sigma2s, sigma2pz, pi2px = pi2py, pi2px = pi2py, sigma2pz):

O2 (16 electrons): (sigma2s)2 (sigma2s)2 (sigma2pz)2 (pi2px)2(pi2py)2 (pi2px)1(pi2py)1 (sigma2pz)0

→ 2 unpaired electrons in the two pi* orbitals → PARAMAGNETIC.

O2+ (15 electrons, one less than O2): removes one electron from a pi* orbital → (pi2px)1(pi2py)0

→ 1 unpaired electron → PARAMAGNETIC.

…

Start your 14-day free trial to see the full solution →

Q.Using molecular orbital theory, write the electronic configuration of O2\text{O}_2O2​, calculate its bond order, and explain why O2\text{O}_2O2​ is paramagnetic — a fact that simple valence bond theory cannot account for.

🎓Unlock everything free for 14 days

Q.Using molecular orbital theory, write the electronic configuration of $\text{O}_2$ , calculate its bond order, and explain why $\text{O}_2$ is paramagnetic — a fact that simple valence bond theory cannot account for.

Unlock everything free for 14 days