Why O₂ is paramagnetic (MOT prediction): Question bank

Reminders you may lean on (all built in the parent):

Paramagnetic = at least one unpaired electron = attracted by a magnet. Diamagnetic = all electrons paired = weakly pushed away. See Paramagnetism and Diamagnetism.
$B.O. = \frac{N _{b} - N _{a}}{2}$ , where $N_{b}$ = bonding electrons, $N_{a}$ = antibonding electrons.
Spin-only moment $μ = n (n + 2)$ BM, with $n$ = number of unpaired electrons.
For O₂/F₂ (more than 14 electrons) the order is $σ_{2 p_{z}} < π_{2 p} < π_{2 p}^{*} < σ_{2 p_{z}}^{*}$ .

True or false — justify

O₂ has all electrons paired because its Lewis structure is

O = O

False. Lewis pairs everything, but MOT puts the last two electrons singly into the two degenerate

π^{*}

orbitals — so two electrons stay unpaired. The dot picture simply cannot show single occupancy of equal-energy orbitals.

A molecule with an even number of electrons must be diamagnetic.

False. O₂ has 16 (even) electrons yet is paramagnetic. Evenness only guarantees the electrons can pair; whether they do depends on whether the last ones land in a degenerate set, where Hund's Rule forces them apart.

If bond order is 2, the molecule is "saturated" and therefore diamagnetic.

False. Bond order measures net bonding

(N_{b} - N_{a}) /2

; magnetism depends only on unpaired electrons. O₂ has B.O. = 2 and two unpaired electrons at the same time — the two ideas are independent.

Paramagnetism is decided by bond order.

False. It is decided solely by the number of unpaired electrons

n

. Two molecules can share a bond order yet differ in magnetism if one has a singly-occupied orbital and the other does not.

For O₂, the

σ_{2 p_{z}}

orbital lies above the two

π_{2 p}

orbitals, just like in N₂.

False. For O₂ and F₂ (electrons > 14)

s

–

p

mixing is weak, so

σ_{2 p_{z}}

drops below the π's. N₂ (≤ 14 electrons) has the opposite order — see N2 vs O2 MO diagram.

Even though O₂ uses a different orbital order than N₂, its unpaired-electron count is unaffected.

True. The reordering swaps

σ_{2 p_{z}}

and

π_{2 p}

, but both are bonding and both are filled with 2 (or 4) electrons either way. The last two electrons still land in

π^{*}

, so

n = 2

regardless.

Removing an electron from O₂ (making O₂⁺) must lower the bond order.

False. The removed electron comes out of an antibonding

π^{*}

orbital, so

N_{a}

drops and B.O. rises to 2.5. Removing an antibonding electron strengthens the bond.

O₂²⁻ (peroxide) is diamagnetic.

True. With 18 electrons the two extra fill the

π^{*}

set completely

(π^{* 2} π^{* 2})

, pairing every electron, so

n = 0

. See Superoxide and Peroxide ions.

Spot the error

"O₂ config = KK

σ_{2 s}^{2} σ_{2 s}^{* 2} σ_{2 p_{z}}^{2} π_{2 p}^{4} π_{2 p_{x}}^{* 2}

" — find the mistake.

The last two electrons are crammed into one

π^{*}

orbital. Hund's Rule demands one electron in each degenerate

π^{*}

: it must be

π_{2 p_{x}}^{* 1} π_{2 p_{y}}^{* 1}

, giving two unpaired electrons.

"B.O.(O₂) = (10 − 6)/2, so the antibonding count of 6 comes from

σ_{1 s}^{*}, σ_{2 s}^{*}, σ_{2 p_{z}}^{*}

." — spot the slip.

σ_{2 p_{z}}^{*}

is empty in O₂. The 6 antibonding electrons are

σ_{1 s}^{*} (2) + σ_{2 s}^{*} (2) + π^{*} (2)

. The two

π^{*}

electrons are the antibonding ones people forget.

"Since O₂ is paramagnetic, its bond must be weaker than a normal double bond." — what's wrong?

The paramagnetism (from

π^{*}

occupancy) and the bond order (still 2) are separate facts. O₂ genuinely has a double bond's strength; being magnetic doesn't dilute that.

"O₂⁺ has 15 electrons, so it's diamagnetic (odd but they all pair)." — locate the error.

An odd electron count can never fully pair — one electron is always left over. O₂⁺ ends

π^{* 1}

, so

n = 1

and it is paramagnetic.

"μ(O₂) = √(2×2) = 2 BM." — fix it.

The formula is

n (n + 2)

, not

n \cdot n

. With

n = 2

2 (2 + 2) = 8 \approx 2.83

BM.

"VBT predicts O₂ correctly because the double bond has 4 shared electrons." — where's the flaw?

VBT/Lewis get the bond count right but predict diamagnetism, which is wrong. Only MOT reveals the unpaired

π^{*}

electrons — see Valence Bond Theory limitations.

Why questions

Why do we divide by 2 in the bond-order formula?

Because one full covalent bond is made of one bonding pair — two electrons. Dividing the net bonding electrons by 2 converts "electrons" into "number of bonds."

Why does Hund's rule (not Aufbau alone) decide O₂'s magnetism?

Aufbau tells us electrons enter the

π^{*}

level; but two electrons entering two equal-energy orbitals could pair or split. Hund's rule resolves the tie by maximising spin, forcing them apart into separate orbitals.

Why does a single unpaired electron make an atom or molecule magnetic?

A lone unpaired electron has uncancelled spin, and spin acts like a tiny bar magnet. Paired electrons have opposite spins that cancel, leaving no net magnetic moment.

Why does the

σ_{2 p_{z}}

π_{2 p}

order flip between N₂ and O₂?

In lighter diatomics (≤14 e⁻) the

2 s

and

2 p_{z}

orbitals are close in energy and mix, pushing

σ_{2 p_{z}}

up above the π's. In O₂/F₂ the

2 s

–

2 p

gap is large, mixing is weak, and

σ_{2 p_{z}}

stays below the π's.

Why does adding electrons to O₂ (→ O₂⁻ → O₂²⁻) lengthen the O–O bond?

The added electrons enter antibonding

π^{*}

orbitals, raising

N_{a}

and lowering bond order (2 → 1.5 → 1). A weaker net bond is a longer bond: 121 → 128 → 149 pm.

Why is "believe the experiment" the moral of this whole topic?

Liquid O₂ visibly clings to a magnet, ruling out the diamagnetic Lewis prediction. When two models disagree, the one matching the measurement (MOT here) wins.

Edge cases

What is the magnetism and bond order of O₂²⁺? (14 electrons)

With 14 electrons the

π^{*}

orbitals are empty:

n = 0

, so it is diamagnetic, and B.O. =

(10 - 4) /2 = 3

— a triple bond, the strongest of the O₂ family.

What happens to

μ

in the limiting case

n = 0

μ = 0 (0 + 2) = 0

BM. A zero moment is exactly the signature of a diamagnetic species — no net electron spin to respond to the field.

Is the neutral O atom (not O₂) paramagnetic?

Yes. Oxygen's

2 p^{4}

has two unpaired electrons (Hund's rule again), so a lone O atom is paramagnetic — the molecule inherits this "unpaired pair" trait through its

π^{*}

orbitals.

Compare O₂⁻ (superoxide) and O₂ — which has more unpaired electrons?

O₂ does. O₂ has

n = 2

; adding one electron to O₂ pairs one

π^{*}

orbital, leaving O₂⁻ with

n = 1

. Adding electrons to a half-filled degenerate set reduces unpaired count.

If we (wrongly) used the N₂ ordering for O₂, would we still predict paramagnetism?

Yes. Even with

σ_{2 p_{z}}

above the π's, the top-filled level is still the degenerate

π^{*}

pair holding two electrons → two unpaired. The prediction survives the ordering mistake here, though not for every molecule.

Two species both have B.O. = 1: O₂²⁻ and (hypothetically) a molecule ending

π^{* 1} π^{* 1} σ^{* 1}

. Must they share magnetism?

No. Equal bond order does not fix

n

. O₂²⁻ has

n = 0

(diamagnetic) while the other has unpaired electrons — bond order and magnetism are decoupled.

Recall One-line self-test before you close this page

Cover the reveals: can you state, in a single breath, why an even-electron molecule (O₂) is still magnetic? ::: Because its last two electrons occupy two degenerate $π^{*}$ orbitals singly (Hund's rule), leaving two unpaired spins.

Connections

Molecular Orbital Theory — the framework every trap tests
Bond Order — decoupled from magnetism here on purpose
Hund's Rule — resolves the degenerate-orbital tie
Aufbau Principle — the filling order these traps exploit
Paramagnetism and Diamagnetism — the verdict every item lands on
Valence Bond Theory limitations — why the "Lewis says diamagnetic" trap exists
N2 vs O2 MO diagram — the ordering-flip traps
Superoxide and Peroxide ions — O₂⁻, O₂²⁻ edge cases