Why O₂ is paramagnetic (MOT prediction): Worked examples

This page is a drill hall. The parent note built the machinery (16 electrons, the MO order, Hund's rule, bond order). Here we run that machinery over every kind of question the exam can ask — every electron count, every sign of charge, the zero cases, the limiting cases, and a word problem — so no scenario can surprise you.

Before any counting, keep this ladder in front of you — it is the only tool we use, and it is the O₂/F₂ ladder (total electrons $> 14$ , so $σ_{2 p_{z}}$ sits below the two $π$ orbitals):

Figure — Why O₂ is paramagnetic (MOT prediction)

Why this picture first? Every case class differs only in how high up this same ladder we fill. Fix the ladder in your mind and the rest is bookkeeping. The dashed line marks the split between bonding (below, glues atoms together) and antibonding (above, star ∗, pulls atoms apart).

The scenario matrix

Every question this topic can ask falls into one of these cells. The examples that follow each carry a [Cell N] tag so you can see the whole space is covered.

Cell	Case class	Representative species	What is special
1	Neutral, unpaired (the star case)	$O_{2}$	2 unpaired in $π^{*}$ → paramagnetic
2	Add electrons → pairs form	$O_{2}^{2 -}$ (peroxide)	fills $π^{}$ fully → zero* unpaired (degenerate/zero case)
3	Add ONE electron → odd count	$O_{2}^{-}$ (superoxide)	odd electrons → must be paramagnetic
4	Remove electrons → higher bond order	$O_{2}^{+}$ (dioxygenyl)	empties $π^{*}$ → shorter, stronger bond
5	Cross-species trend (limiting behaviour)	$O_{2}^{+} \to O_{2} \to O_{2}^{-} \to O_{2}^{2 -}$	bond order decreases, length increases monotonically
6	The other ladder (sign of s–p mixing)	$N_{2}$ vs $O_{2}$	$\leq 14$ e⁻ flips $σ_{2 p_{z}}$ above $π$ — a trap
7	Heteronuclear, same electron count	$N O$ , $C O$	15 e⁻ (odd) and 14 e⁻ — different magnetism
8	Word / real-world problem	liquid-O₂ magnet demo	connect $μ$ to an observation
9	Exam twist (compute $μ$ backward)	"which ion has $μ = 1.73$ ?"	run the logic in reverse

Two formulas do all the arithmetic. We restate them so no symbol is unearned:

Prerequisites, if any move feels shaky: Aufbau Principle (fill low first), Hund's Rule (spread out in degenerate orbitals before pairing), Bond Order, Paramagnetism and Diamagnetism.

Worked Examples

[Cell 1] The star case — neutral O₂

Worked example Example 1 — Configuration and magnetism of

$O_{2}$ State the MO configuration, bond order, and $μ$ .

Forecast: guess now — how many electrons sit unpaired, and will $μ$ be closer to 0 or closer to 3?

Count electrons. Each O has 8, so $O_{2}$ has $8 + 8 = 16$ . Why this step? Filling starts from a headcount — no count, no filling.
Fill 16 up the ladder. Cores $σ_{1 s}^{2} σ_{1 s}^{* 2}$ (the "KK" part, 4 e⁻), then $σ_{2 s}^{2} σ_{2 s}^{* 2}$ (8), $σ_{2 p_{z}}^{2}$ (10), $π_{2 p_{x}}^{2} π_{2 p_{y}}^{2}$ (14). Two electrons remain. Why this step? Aufbau: lowest available rungs first.
Place the last 2 by Hund's rule. The pair $π_{2 p_{x}}^{*}, π_{2 p_{y}}^{*}$ is degenerate (same height). Hund says: one electron in each, spins parallel, before doubling up. Why this step? Same-energy orbitals are filled singly first because parallel spins lower repulsion energy. This is the whole reason O₂ is magnetic. $KK σ_{2 s}^{2} σ_{2 s}^{* 2} σ_{2 p_{z}}^{2} π_{2 p_{x}}^{2} π_{2 p_{y}}^{2} π_{2 p_{x}}^{* 1} π_{2 p_{y}}^{* 1}$
Read magnetism: $n = 2$ singly-occupied orbitals → paramagnetic, $μ = 2 (2 + 2) = 8 \approx 2.83$ BM.
Bond order: $N_{b} = 2 + 2 + 4 = 8$ (the $2 s$ , $2 p_{z}$ , and two $π$ ; the $1 s$ core cancels since $σ_{1 s}$ and $σ_{1 s}^{*}$ both hold 2). $N_{a} = 2 + 2 = 4$ ( $σ_{2 s}^{*}$ and the two $π^{*}$ ). $B.O. = \frac{8 - 4}{2} = 2$ .

Verify: double bond (B.O. 2) matches Lewis's $O = O$ — so bond strength is not contradicted — yet $n = 2$ reveals the magnet Lewis missed. $μ = 2.83$ BM matches the measured value. ✔

(Counting the $1 s$ core in or out gives the same B.O., because a filled bonding+antibonding pair contributes $+ 2 - 2 = 0$ . In Example 1 the parent note included the core; here we cancelled it. Both give 2.)

[Cell 2] Adding electrons kills the magnet — peroxide

Worked example Example 2 — Peroxide ion

$O_{2}^{2 -}$ (the zero/degenerate case) Predict magnetism and bond order.

Forecast: we add 2 electrons to O₂'s two half-filled $π^{*}$ orbitals. Do they spread out further, or pair up?

Count: $16 + 2 = 18$ electrons. Why? Charge $2 -$ means two extra electrons.
Fill: the two new electrons have nowhere lower to go, so they complete the $π^{*}$ set: $π_{2 p_{x}}^{* 2} π_{2 p_{y}}^{* 2}$ . Why this step? The $π^{*}$ orbitals were the highest occupied; they were half-full, so the incoming pair fills them.
Magnetism: every orbital is now doubly occupied → $n = 0$ → diamagnetic, $μ = 0$ .
Bond order: $N_{b} = 8$ , $N_{a}$ now $= 2 + 4 = 6$ . $B.O. = \frac{8 - 6}{2} = 1$ .

Verify: $n = 0$ ⟹ $μ = 0 = 0$ , consistent with peroxide being diamagnetic. Bond order dropped $2 \to 1$ , so the O–O bond should lengthen (it does: 121 pm → 149 pm). ✔ See Superoxide and Peroxide ions.

[Cell 3] Odd electron count must be magnetic — superoxide

Worked example Example 3 — Superoxide ion

$O_{2}^{-}$ (odd-electron case) Predict magnetism, $μ$ , and bond order.

Forecast: an odd number of electrons cannot all pair up. So before any diagram — is it possible for this to be diamagnetic?

Count: $16 + 1 = 17$ electrons — odd. Why note the parity? An odd count guarantees at least one lone electron → guaranteed paramagnetic.
Fill: the extra electron enters one $π^{*}$ orbital: $π_{2 p_{x}}^{* 2} π_{2 p_{y}}^{* 1}$ . Why this arrangement? One $π^{*}$ pairs up (gets 2), the other keeps its single electron.
Magnetism: $n = 1$ → paramagnetic, $μ = 1 (1 + 2) = 3 \approx 1.73$ BM.
Bond order: $N_{a} = 2 + 3 = 5$ . $B.O. = \frac{8 - 5}{2} = 1.5$ .

Verify: odd electrons ⟹ paramagnetic — consistent. Bond order 1.5 lies between O₂ (2) and O₂²⁻ (1), so bond length 128 pm sits between 121 and 149 pm. ✔

[Cell 4] Removing electrons strengthens the bond — dioxygenyl

Worked example Example 4 — Dioxygenyl cation

$O_{2}^{+}$ Predict magnetism and bond order.

Forecast: remove one electron from O₂. It leaves from the highest occupied orbital — a $π^{*}$ . Does bond order go up or down when we empty an antibonding orbital?

Count: $16 - 1 = 15$ electrons. Why minus one? Charge $+$ means one electron removed.
Which electron leaves? The most energetic — from $π^{*}$ . Configuration ends $π_{2 p_{x}}^{* 1} π_{2 p_{y}}^{* 0}$ (one $π^{*}$ singly filled, the other empty). Why the highest? You always ionise from the top of the filled ladder.
Magnetism: $n = 1$ → paramagnetic, $μ = 3 \approx 1.73$ BM.
Bond order: $N_{a} = 2 + 1 = 3$ . $B.O. = \frac{8 - 3}{2} = 2.5$ .

Verify: removing an antibonding electron raises bond order ( $2 \to 2.5$ ), so O₂⁺ has a shorter, stronger bond than O₂ (112 pm vs 121 pm). ✔ (Same $μ$ as superoxide but very different bond!)

[Cell 5] The full trend — a limiting-behaviour sweep

Worked example Example 5 — Order

$O_{2}^{+}, O_{2}, O_{2}^{-}, O_{2}^{2 -}$ by bond length Arrange by increasing O–O bond length, using bond order.

Forecast: each step adds electrons to antibonding $π^{*}$ . Predict the direction of bond length before computing.

List bond orders (from Examples 1–4): $O_{2}^{+} = 2.5$ , $O_{2} = 2$ , $O_{2}^{-} = 1.5$ , $O_{2}^{2 -} = 1$ . Why? Each added electron sits in an antibonding orbital, subtracting $\frac{1}{2}$ from B.O.
Higher B.O. ⟹ shorter bond. So length increases as B.O. falls. Why this rule? More net bonding pairs pull the nuclei closer.
Order (shortest → longest): $O_{2}^{+} < O_{2} < O_{2}^{-} < O_{2}^{2 -}$ , i.e. $112 < 121 < 128 < 149$ pm.

Verify: bond orders $2.5, 2, 1.5, 1$ form a clean arithmetic drop of $0.5$ each time we add an electron to $π^{*}$ ; lengths increase monotonically. ✔ This is the Bond Order–length correlation in one clean series.

[Cell 6] The OTHER ladder — sign of s–p mixing (N₂)

Worked example Example 6 —

$N_{2}$ on its own ladder (the trap) Configuration, bond order, magnetism — and show why the ladder differs.

Forecast: N₂ has 14 electrons ( $\leq 14$ ). Guess: does it have unpaired electrons?

Count: $7 + 7 = 14$ electrons.
Choose the ladder. For $\leq 14$ electrons, strong $s$ – $p$ mixing pushes $σ_{2 p_{z}}$ above the two $π$ orbitals. So the order near the top is $π_{2 p_{x}} = π_{2 p_{y}} < σ_{2 p_{z}} < π^{*} < σ_{2 p_{z}}^{*}$ . Why a different ladder? When $2 s$ and $2 p$ energies are close (light atoms up to N), they mix and repel, lifting $σ_{2 p_{z}}$ . See N2 vs O2 MO diagram.
Fill 14: $KK σ_{2 s}^{2} σ_{2 s}^{* 2} π_{2 p_{x}}^{2} π_{2 p_{y}}^{2} σ_{2 p_{z}}^{2}$ — everything paired. Why? All bonding levels fill completely; nothing reaches the degenerate $π^{*}$ .
Magnetism: $n = 0$ → diamagnetic, $μ = 0$ .
Bond order: $N_{b} = 2 + 4 + 2 = 8$ , $N_{a} = 2$ . $B.O. = \frac{8 - 2}{2} = 3$ (triple bond).

Verify: N₂ is famously diamagnetic with a triple bond — matches. And crucially, for O₂ the two ladders do not change the count in $π^{*}$ , so O₂ stays paramagnetic either way — but you must get the ladder right for bond-order comparisons. ✔

[Cell 7] Heteronuclear, same-family counting — NO and CO

[Cell 8] Real-world word problem

Worked example Example 8 — The liquid-oxygen magnet demonstration

In a classic demo, liquid O₂ poured between the poles of a strong magnet clings to the field. Estimate the number of unpaired electrons per O₂ implied, and the expected $μ$ .

Forecast: the fact that it sticks (attracted, not repelled) already tells you $n$ is 0 or not — which?

Interpret the observation. Attraction into a field ⟹ paramagnetic ⟹ $n \geq 1$ . Why? Only unpaired spins produce a net magnetic moment that a field pulls on.
Use MOT (Example 1): O₂ has exactly $n = 2$ unpaired electrons.
Predicted moment: $μ = 2 (2 + 2) = 8 \approx 2.83$ BM — a clear, large, non-zero value, fully consistent with the strong clinging seen.

Verify: a diamagnetic liquid (say $N_{2}$ , $n = 0$ , $μ = 0$ ) would be weakly repelled instead — and indeed liquid N₂ does not cling. The demo distinguishes the two exactly as MOT predicts. ✔

[Cell 9] Exam twist — run it backward

Worked example Example 9 — "A diatomic oxygen species has

$μ = 1.73$ BM. Which are candidates?" Identify all O₂-derived species consistent with this moment.

Forecast: $μ = 1.73$ pins down $n$ before you even think about which ion.

Invert the formula. $μ = n (n + 2) = 1.73 \Rightarrow n (n + 2) = 3 \Rightarrow n = 1$ . Why invert? The moment is a fingerprint of $n$ ; solve for $n$ first.
Find O₂ species with exactly 1 unpaired electron. From our cases: $O_{2}^{+}$ (15 e⁻, Ex. 4) and $O_{2}^{-}$ (17 e⁻, Ex. 3) both have $n = 1$ . Why two answers? Both an ionisation and an electron-gain leave a single unpaired $π^{*}$ electron.
Distinguish them if the question adds a clue: $O_{2}^{+}$ has B.O. 2.5 (short bond); $O_{2}^{-}$ has B.O. 1.5 (long bond).

Verify: $1 (1 + 2) = 3 = 1.732 \dots$ ✔ Both O₂⁺ and O₂⁻ satisfy $μ = 1.73$ BM but have opposite bond-strength trends — a favourite trap. ✔

Recall Self-test: cover the answers

Bond orders of $O_{2}^{+}, O_{2}, O_{2}^{-}, O_{2}^{2 -}$ ? ::: 2.5, 2, 1.5, 1 Which of the four is diamagnetic? ::: only $O_{2}^{2 -}$ (peroxide), $n = 0$ $μ$ for $n = 1$ ? ::: $3 \approx 1.73$ BM Why is N₂ diamagnetic but O₂ paramagnetic? ::: N₂ (14 e⁻) fills all bonding levels paired; O₂ (16 e⁻) puts 2 unpaired electrons in degenerate $π^{*}$ Is NO paramagnetic? ::: yes, 15 e⁻ (odd), $n = 1$ , B.O. 2.5

Connections

Molecular Orbital Theory — the ladder every example uses
Bond Order — the $\frac{N _{b} - N _{a}}{2}$ tool
Hund's Rule — why the $π^{*}$ electrons stay unpaired
Aufbau Principle — fill-lowest-first
Paramagnetism and Diamagnetism — the $μ$ classification
Superoxide and Peroxide ions — Cells 2 and 3
N2 vs O2 MO diagram — Cell 6 ladder switch
Valence Bond Theory limitations — why we needed MOT at all

Worked examples — Why O₂ is paramagnetic (MOT prediction)