MO diagrams of H₂, He₂, N₂, O₂, F₂, NO, CO — bond order, magnetism: Worked examples

The scenario matrix

Before any numbers, let us list what can vary when a problem hands you a diatomic species. Each row is a "cell" — a distinct thing that can trip you. Every worked example below is tagged with the cell(s) it covers.

Cell	The variable	What could go wrong	Covered by
A	Neutral, even e⁻, all paired	forget which MO order	Ex 1 (N₂), Ex 5 (F₂)
B	Neutral, even e⁻, degenerate half-fill	miss Hund → wrong magnetism	Ex 2 (O₂)
C	Add electrons (anion)	where does the extra e⁻ go?	Ex 3 (O₂⁻, O₂²⁻)
D	Remove electrons (cation)	which e⁻ leaves — top MO	Ex 4 (O₂⁺, N₂⁺)
E	Odd total electrons	half-integer bond order	Ex 6 (NO), Ex 4 (N₂⁺)
F	Heteronuclear, isoelectronic	can I borrow N₂'s diagram?	Ex 7 (CO, CN⁻)
G	Zero / degenerate input (BO = 0)	"does it exist?"	Ex 8 (He₂, Be₂)
H	Order-choice boundary ( $Z \approx 8$ )	mixed vs unmixed order flip	Ex 9 (C₂ vs "if O₂ used wrong order")
I	Word problem / real world	translate physics → chemistry	Ex 10 (liquid O₂ on a magnet)
J	Exam twist (trend / ranking)	compare BO across a series	Ex 3, Ex 4 wrap-up

The two MO energy orderings we will keep reaching for (from the parent):

How to read the ordering figure

The figure below draws both energy ladders. Energy increases upward (see the vertical arrow). Each horizontal black line is one MO; its height is its energy. The only difference between the two ladders is the height of the red $σ_{2 p}$ line: on the left (mixed) it sits above the two $π_{2 p}$ lines; on the right (unmixed) it drops below them. To fill any molecule, start at the bottom line and climb — put 2 electrons per line, and where two lines are at the same height (degenerate, like the two $π_{2 p}$ ) put one in each before pairing. Watch the red line: its height is the single fact that decides whether $C_{2}$ / $N_{2}$ or $O_{2}$ / $F_{2}$ behaviour applies.

Figure — MO diagrams of H₂, He₂, N₂, O₂, F₂, NO, CO — bond order, magnetism

Worked examples

Ex 1 — N₂ (Cell A: neutral, even, all paired, mixed order)

Worked example Statement

Find the bond order and magnetism of $N_{2}$ (7 electrons per N atom → 14 total).

Forecast: guess before reading — how many bonds, and will it stick to a magnet?

Step 1 — Which order? Nitrogen has atomic number $Z = 7 \leq 7$ , so use the mixed order (in the figure, red $σ_{2 p}$ above the $π_{2 p}$ ). Why this step? The order choice decides everything downstream; picking it first prevents the classic mistake #2 from the parent.

Step 2 — Fill 14 electrons lowest-first. $σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} π_{2 p}^{4} σ_{2 p}^{2}$ Why this step? Aufbau: two per orbital, the $π_{2 p}$ pair of orbitals holds 4 before we reach $σ_{2 p}$ .

Step 3 — Count. $N_{b} = 2 + 2 + 4 + 2 = 10$ (bonding), $N_{a} = 2 + 2 = 4$ (the two starred sets). $BO = \frac{10 - 4}{2} = 3$ Why this step? Net bonds = the definition.

Step 4 — Magnetism. Every filled MO has a pair. No unpaired electrons → diamagnetic.

Verify: $2 + 2 + 2 + 2 + 4 + 2 = 14$ electrons ✔. BO = 3 matches the famously short, strong N≡N triple bond. ✔ Why verify the electron total? A miscount by even one electron changes the filling of the top MO and can flip both BO and magnetism — so re-summing the superscripts is the cheapest way to catch a slip.

Ex 2 — O₂ (Cell B: degenerate half-fill, the Hund case)

Worked example Statement

Bond order and magnetism of $O_{2}$ (8 per O → 16 total).

Forecast: Lewis says O=O all-paired. Trust it? (Spoiler in Ex 10.)

Step 1 — Which order? Oxygen has atomic number $Z = 8$ , so unmixed ( $σ_{2 p}$ below $π_{2 p}$ — the red line drops in the figure). Why this step? Wrong order here still gives BO = 2, but it can flip the magnetism prediction — see Ex 9.

Step 2 — Fill 16. $σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} σ_{2 p}^{2} π_{2 p}^{4} π_{2 p}^{* 2}$ The last 2 electrons enter the two degenerate $π_{2 p}^{*}$ orbitals. Why this step? By Hund's Rule and Pauli Exclusion, equal-energy orbitals each take one electron with parallel spin before any pairs up.

Step 3 — Count. $N_{b} = 10, N_{a} = 6$ . $BO = \frac{10 - 6}{2} = 2$ Why this step? We apply the bond-order definition $BO = \frac{N _{b} - N _{a}}{2}$ : net bonds are bonding electrons minus antibonding electrons, per pair.

Step 4 — Magnetism. Two singly-occupied $π^{*}$ orbitals → 2 unpaired electrons → paramagnetic. See the red half-filled pair in the figure. Why this step? By Paramagnetism and Diamagnetism, any unpaired electron carries a net magnetic moment, so a species with ≥1 unpaired electron is attracted to a magnetic field (paramagnetic); only fully paired species are diamagnetic.

Verify: total $= 2 + 2 + 2 + 2 + 2 + 4 + 2 = 16$ ✔. BO 2 = double bond. Real liquid O₂ is drawn to a magnet — MO wins over Lewis. ✔ Why verify the electron total? If the total came out as 15 or 17 we would know a superscript was wrong; getting exactly 16 confirms the two half-filled $π^{*}$ orbitals are genuinely the last electrons placed, which is precisely what makes O₂ paramagnetic.

Ex 3 — O₂⁻ and O₂²⁻ (Cell C + Cell J: adding electrons, a trend)

Worked example Statement

Superoxide $O_{2}^{-}$ (17 e⁻) and peroxide $O_{2}^{2 -}$ (18 e⁻). Bond orders and magnetism; then rank the series $O_{2}, O_{2}^{-}, O_{2}^{2 -}$ by bond strength.

Forecast: each added electron — does it strengthen or weaken the bond?

Step 1 — Same (unmixed) order, keep adding into $π_{2 p}^{*}$ .

$O_{2}^{-}$ : one extra e⁻ → $π_{2 p}^{* 3}$ (one orbital paired, one single).
$O_{2}^{2 -}$ : two extra → $π_{2 p}^{* 4}$ (both filled). Why this step? The added electrons go into the lowest available MO, which is the antibonding $π^{*}$ still holding vacancies.

Step 2 — Count.

$O_{2}^{-}$ : $N_{b} = 10, N_{a} = 7 \Rightarrow BO = \frac{10 - 7}{2} = 1.5$ .
$O_{2}^{2 -}$ : $N_{b} = 10, N_{a} = 8 \Rightarrow BO = \frac{10 - 8}{2} = 1$ . Why this step? Antibonding electrons subtract — each one erodes bond order by ½.

Step 3 — Magnetism.

$O_{2}^{-}$ : $π^{* 3}$ has one unpaired → paramagnetic.
$O_{2}^{2 -}$ : $π^{* 4}$ all paired → diamagnetic. Why this step? By Paramagnetism and Diamagnetism, a single unpaired electron gives a net magnetic moment (paramagnetic); once every orbital is doubly filled the moments cancel (diamagnetic).

Step 4 — Rank. $BO : O_{2} (2) > O_{2}^{-} (1.5) > O_{2}^{2 -} (1)$ . By Bond Order, Bond Length, Bond Energy correlation, higher BO ⇒ shorter, stronger bond. So bond strength: $O_{2} > O_{2}^{-} > O_{2}^{2 -}$ ; bond length reverses.

Verify: electron totals 17 and 18 ✔; each extra antibonding e⁻ dropped BO by 0.5 as expected. ✔ Why verify the electron total? Charge sign errors are the commonest slip on ions — confirming 16+1 = 17 and 16+2 = 18 guarantees we added electrons (anion) rather than removed them, so the extra e⁻ correctly landed in the antibonding $π^{*}$ .

Ex 4 — O₂⁺ and N₂⁺ (Cell D + Cell E: removing electrons, one odd)

Worked example Statement

Dioxygenyl $O_{2}^{+}$ (15 e⁻) and $N_{2}^{+}$ (13 e⁻). Bond orders and magnetism.

Forecast: removing an electron — from where, and does the bond get stronger or weaker?

Step 1 — $O_{2}^{+}$ : remove the top electron from $O_{2}$ . The highest occupied MOs in $O_{2}$ are the $π_{2 p}^{*}$ (antibonding). Remove one: $\dots σ_{2 p}^{2} π_{2 p}^{4} π_{2 p}^{* 1}$ Why this step? You always ionise the highest-energy occupied electron — here an antibonding one, which helps the bond. $N_{b} = 10, N_{a} = 5 \Rightarrow BO = \frac{10 - 5}{2} = 2.5$ . One unpaired $π^{*}$ e⁻ → paramagnetic.

Step 2 — $N_{2}^{+}$ : remove top electron from $N_{2}$ (mixed order). The highest occupied is $σ_{2 p}$ (bonding). Remove one: $\dots π_{2 p}^{4} σ_{2 p}^{1}$ Why this step? In the mixed order the topmost filled MO is the bonding $σ_{2 p}$ — so removing it weakens the bond, unlike $O_{2}^{+}$ . $N_{b} = 9, N_{a} = 4 \Rightarrow BO = \frac{9 - 4}{2} = 2.5$ . One unpaired $σ_{2 p}$ e⁻ → paramagnetic.

Step 3 — The lesson. $O_{2} \to O_{2}^{+}$ raises BO ( $2 \to 2.5$ , remove antibonding); $N_{2} \to N_{2}^{+}$ lowers BO ( $3 \to 2.5$ , remove bonding). Same "remove one electron", opposite effect — because of which MO sits on top.

Verify: totals 15 and 13 ✔. Both BO = 2.5, both paramagnetic (odd electron guarantees ≥1 unpaired) ✔. Why verify the electron total? For cations you must confirm you subtracted the right number: 16−1 = 15 and 14−1 = 13. An odd total (15, 13) also acts as an instant check that magnetism must be paramagnetic, since an odd number of electrons can never fully pair.

Ex 5 — F₂ (Cell A: neutral, even, unmixed, all paired)

Worked example Statement

Bond order and magnetism of $F_{2}$ (9 per F → 18 total).

Forecast: stronger or weaker than O₂?

Step 1 — Unmixed order (fluorine has atomic number $Z = 9 \geq 8$ ). Fill 18. $\dots σ_{2 p}^{2} π_{2 p}^{4} π_{2 p}^{* 4}$ Why this step? Two more electrons than O₂ complete both $π^{*}$ orbitals.

Step 2 — Count. $N_{b} = 10, N_{a} = 8 \Rightarrow BO = \frac{10 - 8}{2} = 1$ . Why this step? We apply the bond-order definition $BO = \frac{N _{b} - N _{a}}{2}$ : the eight antibonding electrons cancel eight of the ten bonding ones, leaving one net bond.

Step 3 — Magnetism. All MOs full → diamagnetic. Why this step? By Paramagnetism and Diamagnetism, with every orbital doubly filled there are no unpaired electrons, so all magnetic moments cancel — the species is diamagnetic.

Verify: total 18 ✔. BO 1 < O₂'s 2, consistent with $N_{2} (3) > O_{2} (2) > F_{2} (1)$ : piling electrons into antibonding tears the bond down. ✔ Why verify the electron total? Confirming 18 tells us both $π^{*}$ orbitals are completely full (4 electrons), which is exactly why F₂ is diamagnetic; a miscount to 17 would have wrongly left one $π^{*}$ unpaired.

Ex 6 — NO (Cell E: heteronuclear, odd electron)

Worked example Statement

Nitric oxide $N O$ (7 + 8 = 15 e⁻). Bond order, magnetism, and predict $N O^{+}$ .

Forecast: integer or half-integer BO?

Step 1 — Which order? N is on the light side, so NO uses the N₂-type (mixed) order. Why this step? For heteronuclear molecules of light 2nd-period atoms the mixed order is the standard exam convention.

Step 2 — Fill 15. $\dots π_{2 p}^{4} σ_{2 p}^{2} π_{2 p}^{* 1}$ The 15th electron sits alone in a $π_{2 p}^{*}$ . Why this step? Aufbau: electrons fill MOs strictly lowest-energy first, two per orbital, so after the $π_{2 p}^{4}$ and $σ_{2 p}^{2}$ are full the leftover 15th electron must go into the next level up, a $π_{2 p}^{*}$ .

Step 3 — Count. $N_{b} = 10, N_{a} = 5 \Rightarrow BO = \frac{10 - 5}{2} = 2.5$ . One unpaired → paramagnetic.

Step 4 — $N O^{+}$ (14 e⁻). Remove that lone antibonding electron: $N_{a} = 4$ , $BO = \frac{10 - 4}{2} = 3$ , now all paired → diamagnetic, isoelectronic with N₂ and CO. Why this step? Removing an antibonding electron always raises BO — a clean self-check.

Verify: 15 ✔; $N O \to N O^{+}$ gives $2.5 \to 3$ ✔ (bond shortens, matches experiment). ✔ Why verify the electron total? The odd total 15 immediately guarantees NO is paramagnetic (odd electrons can't all pair), and confirming 14 for NO⁺ proves it is isoelectronic with N₂/CO — the fastest sanity check that BO = 3 must follow.

Ex 7 — CO and CN⁻ (Cell F: isoelectronic borrowing)

Worked example Statement

$C O$ (6 + 8 = 14 e⁻) and cyanide $C N^{-}$ (6 + 7 + 1 = 14 e⁻). Bond orders and magnetism.

Forecast: should they match N₂?

Step 1 — Count electrons and compare. Both have 14 electrons, exactly like N₂. By Isoelectronic species (CO, N₂, CN⁻, NO⁺), same electron count on the same MO scaffold ⇒ same diagram. Why this step? Isoelectronic species share bond order and magnetism — no need to refill from scratch.

Step 2 — Borrow N₂'s filling (mixed order). $\dots π_{2 p}^{4} σ_{2 p}^{2}$ $N_{b} = 10, N_{a} = 4 \Rightarrow BO = 3$ , all paired → diamagnetic for both. Why this step? Isoelectronic species have identical MO fillings on the same energy ladder, hence identical bond order and identical magnetism — so we can copy N₂'s result directly rather than refill.

Verify: both totals = 14 ✔; BO = 3 matches CO's very short strong triple bond (and why CO binds haemoglobin so tightly). ✔ Why verify the electron total? Getting 14 for both (6+8 and 6+7+1) is the whole justification for "borrowing" N₂'s diagram — if either total differed we could not claim isoelectronic, and the copied BO = 3 would be invalid.

Ex 8 — He₂ and Be₂ (Cell G: the zero-bond degenerate case)

Worked example Statement

Do $H e_{2}$ (4 e⁻) and $B e_{2}$ (8 e⁻) exist as stable molecules?

Forecast: any net bond at all?

Step 1 — He₂. Fill 4: $σ_{1 s}^{2} σ_{1 s}^{* 2}$ . $N_{b} = 2, N_{a} = 2 \Rightarrow BO = \frac{2 - 2}{2} = 0$ . Why this step? The antibonding pair exactly cancels the bonding pair — a degenerate/zero input where the machine returns "no bond".

Step 2 — Be₂. Fill 8: $σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2}$ . $N_{b} = 4, N_{a} = 4 \Rightarrow BO = 0$ . Why this step? Same story one shell up — every bonding pair is matched by an antibonding pair.

Step 3 — Conclusion. BO = 0 ⇒ no net glue ⇒ neither is a stable molecule. Both diamagnetic (all paired) even so. Why this step? Bond order counts the net number of electron pairs holding the nuclei together; when $N_{b} = N_{a}$ the destabilising antibonding electrons exactly undo the stabilising bonding ones, so there is no net lowering of energy on approach — nothing binds the atoms, and the molecule does not form.

Verify: He₂ $2 + 2 = 4$ ✔, Be₂ $2 + 2 + 2 + 2 = 8$ ✔; both BO = 0 ✔. Why verify the electron total? The count is the proof: seeing that bonding and antibonding electrons are equal in number (2 = 2, and 4 = 4) is exactly what forces BO = 0 — the totals confirm no unmatched bonding pair survives.

Ex 9 — The order-flip trap: C₂, and "what if we used the wrong order for O₂" (Cell H)

Worked example Statement

(a) Magnetism of $C_{2}$ (12 e⁻). (b) Show that using the wrong MO order for a molecule can flip its predicted magnetism. Does the order choice ever change the answer in practice?

Forecast: is C₂ paramagnetic? Can picking the wrong ladder ever change magnetism?

Step 1 — C₂, correct (mixed) order. Carbon has $Z = 6 \leq 7$ , so use the mixed ladder (red $σ_{2 p}$ above $π_{2 p}$ ). Fill 12: $\dots σ_{2 s}^{* 2} π_{2 p}^{4}$ . The two $π_{2 p}$ orbitals fill completely (4 electrons, all paired). $N_{b} = 8, N_{a} = 4 \Rightarrow BO = \frac{8 - 4}{2} = 2$ . Diamagnetic. Why this step? Because $π_{2 p}$ lies below $σ_{2 p}$ in the mixed order, both π orbitals fill and pair — no unpaired electrons.

Step 2 — C₂ with the WRONG (unmixed) order. Then $σ_{2 p}$ fills first ( $σ_{2 p}^{2}$ ), leaving the last 2 electrons to split across the two degenerate $π_{2 p}$ → 2 unpaired → paramagnetic (incorrect!). Same BO = 2, but wrong magnetism. Why this step? This is exactly parent-mistake #2: the order choice can change which orbitals sit at the filling frontier, hence magnetism.

Step 3 — Does it ever matter for O₂? With the correct unmixed order, O₂'s frontier is the degenerate $π^{*}$ → 2 unpaired → paramagnetic (Ex 2). If you wrongly used the mixed order for O₂, the frontier would still be the degenerate $π^{*}$ (it lies below $σ_{2 p}^{*}$ in both ladders), so O₂ reads paramagnetic either way. Why this step? The order choice changes magnetism only when it moves a degenerate orbital pair to the filling frontier: for C₂ the wrong order exposes the degenerate $π_{2 p}$ (creating false unpaired electrons), whereas for O₂ both ladders leave the degenerate $π^{*}$ on top.

Step 4 — The moral. Always match the order to $Z$ : for early elements like C₂ the choice genuinely changes the magnetism answer; for O₂ it happens not to — but never rely on that luck, always pick by $Z$ .

Verify: C₂ total 12 ✔, correct BO = 2, diamagnetic ✔ (experimentally C₂ is diamagnetic — mixed order is right). ✔ Why verify the electron total? Confirming 12 shows the four bonding $π_{2 p}$ electrons are the last placed under the correct order — that full, paired $π_{2 p}^{4}$ is exactly what makes C₂ diamagnetic, so the count directly backs the magnetism claim.

Ex 10 — Real-world: liquid O₂ clings to a magnet (Cell I)

Worked example Statement

In a famous demo, liquid oxygen poured between the poles of a strong magnet bridges the gap and sticks. Which MO fact explains this, and what would liquid N₂ do?

Forecast: which molecule is attracted, and why?

Step 1 — Recall the magnetic rule. A substance is attracted to a magnet (Paramagnetism and Diamagnetism) only if it has unpaired electrons (paramagnetic). Why this step? This links the demo directly to electron pairing, the output of our machine.

Step 2 — Apply to O₂. From Ex 2, O₂ has 2 unpaired electrons in its $π_{2 p}^{*}$ → paramagnetic → attracted → it bridges the poles. Why this step? Each unpaired electron has a net spin magnetic moment; with two of them uncancelled, O₂ behaves like a tiny magnet that is drawn into the strong field, so it physically sticks between the poles.

Step 3 — Apply to N₂. From Ex 1, N₂ is fully paired → diamagnetic → not attracted → liquid N₂ pours straight through. Why this step? With every electron paired the spin moments cancel exactly, leaving no net moment; a diamagnetic substance is in fact weakly repelled by the field, so it will not cling — it just falls through the gap.

Step 4 — Why this matters. Lewis theory draws O=O with all electrons paired and predicts no attraction; only MO theory predicts the sticking. The magnet demo is direct experimental proof of MO theory over Lewis.

Verify: O₂ paramagnetic (2 unpaired) ✔, N₂ diamagnetic (0 unpaired) ✔ — matches every physics classroom's liquid-O₂ demo. ✔ Why verify against the demo? An MO prediction is only trustworthy if it survives experiment: the fact that real liquid O₂ visibly sticks (and liquid N₂ does not) is the physical "plug-back" that confirms our unpaired-electron count for each molecule.

Recall Quick self-test (reveal answers)

Bond order of $O_{2}^{+}$ ? ::: $2.5$ (remove one antibonding $π^{*}$ electron from O₂'s BO of 2). Bond order of superoxide $O_{2}^{-}$ ? ::: $1.5$ , paramagnetic (one unpaired $π^{*}$ electron). Bond order of peroxide $O_{2}^{2 -}$ ? ::: $1$ , diamagnetic (both $π^{*}$ filled). Is $C_{2}$ para- or diamagnetic, and its BO? ::: Diamagnetic, BO = 2 (both $π_{2 p}$ filled and paired). $N_{2}^{+}$ vs $O_{2}^{+}$ bond order? ::: Both $2.5$ — but N₂⁺ lost a bonding electron (BO fell) while O₂⁺ lost an antibonding one (BO rose). Why do CO and CN⁻ share N₂'s bond order? ::: All three have 14 electrons — isoelectronic, same MO filling, BO = 3, diamagnetic. What does the asterisk in $π_{2 p}^{*}$ mean? ::: Antibonding orbital — node between nuclei, higher energy, its electrons weaken the bond.

Connections

Molecular Orbital Theory (LCAO basics)
Hund's Rule and Pauli Exclusion
Paramagnetism and Diamagnetism
Bond Order, Bond Length, Bond Energy correlation
Isoelectronic species (CO, N₂, CN⁻, NO⁺)
s–p mixing and orbital energy ordering
Valence Bond Theory vs MO Theory