Why O₂ is paramagnetic (MOT prediction): Visual walkthrough

Before we begin, three plain words we will lean on constantly:

Everything below is one long chase after the question: do any arrows end up unpaired?

Step 1 — Two atoms bring their electrons

WHAT. Each oxygen atom carries 8 electrons. We put two atoms side by side, so the pile we must house is $8 + 8 = 16$ electrons.

WHY. Just like you cannot seat people before you count heads, we cannot fill boxes before we know how many electrons there are. The number 16 is the budget we will spend, one box at a time.

PICTURE. Two atoms on the left and right, each showing its 8 electrons as small dots; between them, the running tally "16 to place."

Step 2 — The two atoms' boxes merge into a shared ladder

WHAT. When the atoms bond, their separate boxes fuse into a new set of shared boxes that belong to the whole molecule. We call these shared boxes molecular orbitals (MOs). Two atomic boxes always make two molecular boxes: one lower (a bonding box, glue that holds atoms together) and one higher (an antibonding box, marked with a star $^{*}$ , which pushes atoms apart).

WHY. This is the one idea of Molecular Orbital Theory: electrons in a molecule do not stay on one atom, they live in boxes spread across both. Lower box = stable = bonding; higher starred box = costs energy = antibonding.

PICTURE. Left and right atomic boxes at middle height; arrows show them merging into a lower bonding box and a higher starred antibonding box.

Step 3 — Stack the boxes by height (the energy ladder)

WHAT. All the molecular boxes for O₂ arrange into one vertical energy ladder, lowest rung at the bottom. Reading up:

$σ_{1 s} < σ_{1 s}^{*} < σ_{2 s} < σ_{2 s}^{*} < σ_{2 p_{z}} < (π_{2 p_{x}} = π_{2 p_{y}}) < (π_{2 p_{x}}^{*} = π_{2 p_{y}}^{*}) < σ_{2 p_{z}}^{*}$

WHY. Electrons, following the Aufbau Principle, always fill the lowest empty box first — like water settling to the bottom. So we need the boxes sorted by height before we pour electrons in.

Two names to earn, right where they appear:

$σ$ (sigma) = a box shaped head-on along the bond axis (a fat sausage between the atoms).
$π$ (pi) = a box shaped sideways, above and below the axis (two lobes like a hamburger bun).
The pairs written with an equals sign, e.g. $π_{2 p_{x}} = π_{2 p_{y}}$ , sit at the exact same height. Same-height boxes are called degenerate. Remember that word — Step 6 lives on it.

PICTURE. The full ladder drawn as labelled rungs from bottom to top, with the two π boxes drawn at one shared height and the two π* boxes at another shared height.

Step 4 — Pour in the first 10 electrons

WHAT. We spend our 16-electron budget from Step 1, two per box, climbing the ladder. The first five rungs swallow 10 electrons:

Rung filled	electrons here	budget spent
$σ_{1 s}$	2	2
$σ_{1 s}^{*}$	2	4
$σ_{2 s}$	2	6
$σ_{2 s}^{*}$	2	8
$σ_{2 p_{z}}$	2	10

WHY. Aufbau again: lowest box first, and each box tops out at 2 (one ↑, one ↓ — opposite spins, so they pair up and cancel, no magnet yet). Notice these ten arrows are all paired.

PICTURE. The bottom five rungs, each holding a neat ↑↓ pair; a side counter reads "10 placed, 6 left."

Step 5 — The next 4 electrons fill both π boxes

WHAT. We spend 4 more electrons on the degenerate bonding pair $π_{2 p_{x}}$ and $π_{2 p_{y}}$ . Two boxes, four electrons → each box gets a full ↑↓ pair. Budget spent: 14.

WHY. These two boxes sit at the same height and are the next rung up, so Aufbau fills them next. Four electrons split perfectly two-and-two, so again everything pairs — still no lone arrow.

PICTURE. The two side-by-side π boxes, each now holding ↑↓; counter reads "14 placed, 2 left."

Recall Where does the tension sit now?

Placed ::: 14 electrons, all paired, zero magnetism so far. Remaining ::: exactly 2 electrons — and the next rung up is the degenerate π* pair.

Step 6 — The last 2 electrons: the moment of truth (Hund's rule)

WHAT. Two electrons remain. The next rung is the degenerate pair $π_{2 p_{x}}^{*}$ and $π_{2 p_{y}}^{*}$ — two equal-height boxes. Hund's Rule says: when boxes are equal in height, electrons spread out one-per-box with parallel spins before any box is forced to double up.

So instead of crowding one box (↑↓, empty), the two electrons sit one in each box, both ↑: $π_{2 p_{x}}^{* 1} π_{2 p_{y}}^{* 1} \Rightarrow ↑ ↑$

WHY. Two electrons in the same box repel harder (same cramped space) — nature avoids that cost when a same-height empty box is available. The result: two unpaired arrows. Recall Step 0's one fact: an unpaired arrow is a tiny magnet. Oxygen is therefore paramagnetic. This single step is the answer the parent note promised.

PICTURE. The two π* boxes side by side, each holding a single up-arrow; a red "unpaired!" tag over each, and the wrong alternative (↑↓ crammed into one box) drawn faded-out and crossed off.

Step 7 — Cross-check with bond order (nothing broke)

WHAT. Count bonding electrons $N_{b}$ (in un-starred boxes) and antibonding electrons $N_{a}$ (in starred boxes), then:

$B.O. = \frac{N _{b} - N _{a}}{2}$

$N_{b} = σ_{1 s} (2) + σ_{2 s} (2) + σ_{2 p_{z}} (2) + π (4) = 10$
$N_{a} = σ_{1 s}^{*} (2) + σ_{2 s}^{*} (2) + π^{*} (2) = 6$

$B.O. = \frac{10 - 6}{2} = 2$

WHY. In $\frac{N _{b} - N _{a}}{2}$ : the == $N_{b} - N_{a}$ == is net glue (bonding boxes minus the anti-glue that undoes them), and the == $\div 2$ == is because one bond = one pair = 2 electrons. See Bond Order. The result 2 matches the old Lewis double bond — so MOT does not fight the known bond strength; it simply adds the two unpaired electrons Lewis could never show.

PICTURE. Two stacked bars — a tall blue "bonding = 10" bar and a shorter orange "antibonding = 6" bar — with the subtraction and the ÷2 annotated to land on 2.

Step 8 — Edge cases: change the electron count, watch magnetism flip

WHAT. Rerun Steps 4–6 with a different budget. Only the top rung (the π* pair) changes.

WHY. This proves the magnetism was decided entirely at Step 6. Same ladder, same O–O frame — just add or remove electrons at the π* top and the answer swings. Details in Superoxide and Peroxide ions.

Species	electrons	π* filling	unpaired $n$	magnetism	B.O.
$O_{2}$	16	$\uparrow\ \big	\ \uparrow$	2	para
$O_{2}^{-}$ (superoxide)	17	$\uparrow\downarrow\ \big	\ \uparrow$	1	para ( $μ = 3 \approx 1.73$ )
$O_{2}^{2 -}$ (peroxide)	18	$\uparrow\downarrow\ \big	\ \uparrow\downarrow$	0	dia

PICTURE. Three little π*-box pictures side by side (16, 17, 18 electrons) showing the arrows filling in, each tagged para/para/dia, with bond order dropping 2 → 1.5 → 1.

The one-picture summary

Everything — budget, ladder, the two lonely arrows, the bond order — compressed into a single labelled MO diagram.

Recall Feynman: the whole walkthrough in plain words

We had 16 electrons to seat (Step 1). Two atoms' boxes merged into a shared ladder of molecular boxes, low bonding ones and high starred anti-boxes (Steps 2–3). We poured electrons bottom-up, two to a box, opposite spins that cancel — the first 14 all paired up, boring, no magnet (Steps 4–5). Then came the last 2 electrons, and the next rung was two boxes at the exact same height. Given the choice of cramming into one box or taking a box each, electrons take a box each and even spin the same way (Hund's rule) — leaving two lonely arrows (Step 6). A lonely spinning electron is a tiny magnet, so oxygen is paramagnetic. We double-checked the bond order is still 2 (Step 7), and saw that adding electrons (superoxide, peroxide) fills those top boxes and flips the magnetism (Step 8). That one same-height rung at the top is where the entire mystery lives.

Connections

Molecular Orbital Theory — the shared-box idea used in every step
Aufbau Principle — lowest box first (Steps 3–5)
Hund's Rule — the parallel-spin rule that makes Step 6 magnetic
Bond Order — the net-glue count of Step 7
Paramagnetism and Diamagnetism — unpaired-arrow = magnet
Valence Bond Theory limitations — why Lewis misses Step 6
N2 vs O2 MO diagram — the ladder-ordering caveat of Step 3
Superoxide and Peroxide ions — the edge cases of Step 8