Why do two hydrogen atoms form an ordinary stable H₂ molecule while two helium atoms do not form a comparable ordinary covalent He₂ molecule? The simplest molecular orbital model gives a remarkably clear answer. When two 1s atomic orbitals combine, they create a lower-energy bonding σ1s molecular orbital and a higher-energy antibonding σ*1s molecular orbital. H₂ has two electrons, both in the bonding level, giving bond order 1. Hypothetical He₂ has four electrons, filling both bonding and antibonding levels, giving bond order 0 in this simple covalent MO model.
For a JC1–JC2 student seeking an H3 Chemistry tutor in Hougang, this is more than a pleasing comparison of two small molecules. It teaches how to build a molecular orbital diagram from the atomic orbitals, how electrons fill levels and how bonding and antibonding occupancy determine whether the model predicts a net covalent bond. Once that construction is secure, more demanding O₂ and F₂ diagrams become easier to understand.
The 2027 Singapore–Cambridge H3 Chemistry syllabus 9813 explicitly names simple homonuclear diatomics such as H₂, O₂ and F₂ for qualitative molecular orbital diagrams and HOMO/LUMO identification. He₂ and H₂⁺ are included here as supporting comparative examples; they are not separate named examination requirements. The syllabus excludes quantitative LCAO treatment and detailed orbital mixing.
Official SEAB H3 Chemistry 9813 syllabus (2027) · OpenStax molecular orbital theory, H₂ and He₂ · Hougang JC Chemistry learning hub.
Jump to: The one-minute answer · Start with atoms · Build two MOs · Fill H₂ electrons · Fill hypothetical He₂ · H₂⁺ and He₂⁺ enrichment · What bond order does—and does not—mean · Electron pairing and magnetism · HOMO and LUMO · Worked questions · Learning and careers · FAQs.
The One-Minute Molecular Orbital Comparison
- Two H atoms: one 1s electron from each atom, so two electrons in the molecular system.
- Available MOs: the two atomic 1s orbitals form lower-energy σ1s bonding and higher-energy σ*1s antibonding orbitals.
- H₂ configuration: (σ1s)²; no electron occupies σ*1s. Bond order (2−0)/2 = 1.
- Two He atoms: two 1s electrons per atom, so four electrons altogether.
- Hypothetical He₂ configuration: (σ1s)²(σ*1s)². Bond order (2−2)/2 = 0.
- Interpretation: H₂ has a net bonding advantage in the simple MO model. Filling the antibonding level in ordinary neutral He₂ cancels the net covalent bond order.
- Important nuance: a very weak helium dimer can exist under specialised ultracold conditions through van der Waals interactions. Bond order zero in the simple covalent 1s MO model is not a claim that no form of He–He association is ever observable.
Count the participating atomic orbitals. Construct bonding and antibonding levels. Count all electrons. Fill the diagram. Only then calculate bond order.
Step 1: Count the Atomic Electrons Before Combining Anything
A neutral hydrogen atom has one proton and one electron. In its ground state, the electron occupies a 1s atomic orbital: H = 1s¹. Two hydrogen atoms therefore contribute a total of two 1s electrons when the molecular system is constructed.
A neutral helium atom has two protons and two electrons, both in the 1s orbital: He = 1s². Two helium atoms provide a total of four 1s electrons to the corresponding two-orbital construction.
These totals matter because the number of electrons is not determined by the number of printed chemical bonds. Both systems have two atoms and two participating 1s orbitals; their electron counts differ. That difference determines which molecular orbitals are occupied.
Students sometimes try to force helium into a Lewis-style shared pair simply because two nuclei are present. But molecular orbital theory asks whether the occupied molecular states create a net energetic stabilisation. A bonding orbital can be filled while an antibonding orbital is also occupied, reducing or eliminating the bond-order advantage.
Step 2: Build One Bonding and One Antibonding σ Molecular Orbital
In the simplest LCAO picture, two hydrogen-like 1s atomic orbitals combine into two molecular orbital combinations. One has constructive overlap between the nuclei, producing increased electron density in the bonding region and a lower-energy σ1s orbital. The other has opposite relative phase between contributions, a nodal region between the nuclei and a higher-energy σ*1s orbital.
The star symbol in σ* identifies the orbital as antibonding. It does not mean the orbital is forbidden or imaginary. An electron can occupy it if lower levels are already filled or the electronic state requires it. Such occupancy reduces the net bonding stabilisation associated with the interacting atoms.
These are σ-symmetry orbitals because the electron distribution is cylindrically symmetric about the internuclear axis in the simple case. No π orbitals arise from combining only two s orbitals, because s orbitals do not have the side-on directional character of p orbitals that creates π symmetry.
There are exactly two MOs in this chosen two-atomic-orbital basis. That is a general construction rule: combining N independent atomic orbitals in an LCAO model produces N molecular orbitals. It does not create only one orbital merely because the Lewis diagram might show one bond.
An electron capacity check
Each spatial MO can hold at most two electrons of opposite spin under the Pauli exclusion principle. The lower bonding σ1s can therefore hold two electrons; if a third and fourth electron are added within this two-level picture, they cannot simply squeeze into the same bonding orbital. They occupy the available higher-energy σ*1s level.
Step 3: Fill the H₂ Molecular Orbital Diagram
H₂ has two electrons. Begin with the lower-energy σ1s bonding MO and place one electron, then a second of opposite spin, in that orbital. The higher-energy σ*1s antibonding level remains empty. The resulting occupation is (σ1s)².
Now apply the molecular orbital bond-order formula: bond order = (bonding electrons − antibonding electrons)/2. For H₂ there are two bonding electrons and zero antibonding electrons, so (2 − 0)/2 = 1.
The result is consistent with the simple Lewis electron-pair picture H–H. In the MO account, the shared electron density is represented by a molecular state extending over both nuclei rather than by two atomic electrons permanently assigned to separate atoms. Both representations explain a stable H–H interaction at their intended level.
What can we infer about H₂?
- Its ordinary ground-state model has a positive bond order of one.
- Its two electrons are paired in the σ1s bonding MO.
- The simple ground-state diagram is consistent with diamagnetism because there are no unpaired electrons.
- The highest occupied MO is σ1s and the lowest unoccupied MO in the two-level basis is σ*1s.
- The model predicts a net bond between the two H atoms, but not an exact bond dissociation energy from electron counting alone.
An important teaching move is to ask the student to predict what would happen if one electron were removed or added. This invites them to see molecular orbitals as a system whose bonding depends on occupancy—not simply a static diagram to be copied.
Step 4: Fill the Hypothetical Neutral He₂ Diagram
Two helium atoms supply four electrons. Place the first two in σ1s, as before. The remaining two occupy the higher σ*1s antibonding MO with opposite spins. The resulting configuration is (σ1s)²(σ*1s)².
Count bonding electrons: two. Count antibonding electrons: two. The simple bond order is (2 − 2)/2 = 0. The stabilising contribution from the bonding occupancy is offset in the bond-order accounting by antibonding occupancy. Therefore the elementary covalent MO model predicts no ordinary net He–He covalent bond of the H₂ type.
Students occasionally say that the helium pair has bond order two because four electrons occupy ‘two orbitals’. That is wrong because the σ* level is antibonding, not a second bonding orbital. A correct calculation labels every occupied level by its effect before adding or subtracting its electron count.
The real-world qualification: weak helium association
Neutral helium atoms can show weak van der Waals attraction, and an extremely fragile He₂ dimer is possible under specialised very low-temperature conditions. This association is not the strong ordinary covalent bond predicted for H₂ by the simple σ1s calculation. Thus the useful school conclusion is ‘no stable ordinary covalent He₂ molecule from this MO model’, not the categorical claim that two helium atoms can never interact at all.
An H3 Chemistry answer should state the level of approximation. The syllabus asks students to construct and interpret qualitative MO diagrams, not to derive every weak intermolecular potential from first principles. Being accurate about the model’s scope is a sign of strength, not an unnecessary complication.
Step 5: Use H₂⁺ and He₂⁺ as Optional Enrichment
H₂⁺: one electron and bond order one-half
H₂⁺ contains two hydrogen nuclei but only one electron. It occupies the lower-energy σ1s bonding orbital, while σ*1s is empty. Its bond order in the simple MO formula is (1 − 0)/2 = 0.5. The single electron is unpaired, so its ordinary electronic state is paramagnetic.
The important conceptual lesson is that MO theory can represent a bonded species with an odd number of electrons. A simple two-electron Lewis bond picture cannot communicate this situation in the same way. A fractional bond order in a qualitative MO calculation is a useful indicator of net bonding, not literally ‘half a physical stick’ joining the nuclei.
He₂⁺: three electrons and a different route to bond order one-half
He₂⁺ has four protons overall and three electrons in the simple 1s-derived system. Two occupy σ1s; one occupies σ*1s. Its occupation is (σ1s)²(σ*1s)¹, giving bond order (2 − 1)/2 = 0.5. One unpaired electron is present.
This is a useful contrast to neutral He₂. Removing one electron from the antibonding level can create a positive net bond order in the simple model. The chemistry is not that an ion automatically bonds because it is positive; the electron was removed from an orbital whose occupancy opposes bonding.
A four-species summary
- H₂⁺: 1 electron, σ1s¹, bond order 0.5, one unpaired electron.
- H₂: 2 electrons, σ1s², bond order 1, all electrons paired.
- He₂⁺: 3 electrons, σ1s²σ*1s¹, bond order 0.5, one unpaired electron.
- Hypothetical ordinary covalent He₂: 4 electrons, σ1s²σ*1s², bond order 0, all electrons paired.
- Syllabus boundary: H₂ is an explicitly named H3 diatomic example. The ions and helium dimer comparison are supporting enrichment to make the occupancy reasoning transferable.
What Bond Order Does—and Does Not—Tell You
Bond order is an electron-counting measure of net bonding occupancy in the molecular orbital model. A positive value often indicates a net bonding contribution, while an increase in antibonding occupancy tends to weaken that contribution. It is especially powerful for comparing related simple diatomics, including predicting whether adding electrons to an antibonding level may reduce bond strength.
However, bond order alone does not yield an exact measured bond length or dissociation enthalpy. Those depend on the full molecular potential-energy surface, electron interactions, nuclear repulsion and other features not contained in a two-line qualitative diagram. A student who writes ‘bond order one means exactly 436 kJ mol⁻¹ for every single bond’ has confused an illustrative property of one bond with a universal law.
H₂ happens to have a short, strong ordinary covalent bond, while the hypothetical neutral He₂ two-level model yields zero covalent bond order. But applying the same numerical value of bond order to an unrelated molecule would not guarantee the same energy or length. The context of the model must remain visible.
What happens if an electron enters an antibonding MO?
If another electron occupies a previously empty antibonding orbital while the bonding occupancy stays fixed, the formal bond order decreases by one-half in this elementary formula. For example, going from H₂ with two bonding electrons to a hypothetical three-electron occupation σ1s²σ*1s¹ lowers bond order from 1 to 0.5. The direction of the trend is more valuable than pretending the fractional result gives a complete quantitative prediction.
Electron Pairing and Magnetic Behaviour
A molecule with an unpaired electronic spin can be paramagnetic in the ordinary electronic-state picture. If every occupied spatial orbital contains two electrons of opposite spin, the molecule is ordinarily diamagnetic. This distinction can be predicted from the same MO filling diagram used to calculate bond order.
For H₂, two electrons fill σ1s as a paired set, so it is diamagnetic in this simple ground-state treatment. For H₂⁺, one electron remains unpaired, giving paramagnetism. For He₂⁺, one electron occupies σ*1s unpaired, also giving paramagnetism. The hypothetical neutral He₂ configuration contains two pairs, though it lacks a normal stable covalent He–He bond in the model.
Do not confuse bond strength with magnetic response. A system can have paired electrons and zero covalent bond order, or unpaired electrons and a positive fractional bond order. The questions are separate: bond order counts bonding minus antibonding occupancy; magnetism asks whether unpaired spins remain.
For the classic case of a stable paramagnetic diatomic with two unpaired electrons, see the O₂ molecular orbital diagram and paramagnetism. The F₂ comparison is covered separately in fluorine’s bond order and diamagnetism.
HOMO and LUMO in the Simple H₂ Two-Level Diagram
HOMO stands for highest occupied molecular orbital. In neutral H₂, σ1s is occupied by two electrons while σ*1s is empty. Therefore the HOMO is σ1s and the LUMO is σ*1s within the stated 1s-only two-level basis.
In H₂⁺, the single occupied orbital is still σ1s, so it is the HOMO; σ*1s remains unoccupied and is the LUMO in that simple diagram. These labels refer to orbital occupancy rather than being synonyms for ‘bonding’ and ‘antibonding’ in every possible molecule.
For the hypothetical He₂ two-level case, σ*1s is itself filled, so it becomes the highest occupied orbital in the chosen basis. The simplistic two-orbital diagram contains no higher orbital and therefore does not by itself identify a true LUMO for He₂. A student should not call σ*1s the LUMO while also showing it filled.
This is a small but important sign of disciplined modelling. If the basis does not contain an unoccupied orbital above the filled levels, the correct response is to state that the diagram is insufficient to locate a LUMO. Do not invent an empty orbital in order to complete a familiar acronym pair.
Original H3 Chemistry Practice Questions and Solutions
Question 1: Count orbitals
Two 1s atomic orbitals combine in an LCAO description. How many molecular orbitals are formed, and what are they?
Answer: Two: a lower-energy bonding σ1s orbital and a higher-energy antibonding σ*1s orbital.
Question 2: Calculate H₂ bond order
Two electrons occupy σ1s and σ*1s is empty. Find the bond order.
Answer: (2−0)/2 = 1.
Question 3: Calculate He₂ bond order
A hypothetical neutral He₂ diagram has σ1s²σ*1s². Find the bond order.
Answer: (2−2)/2 = 0 in this simple covalent MO model.
Question 4: Explain the electron-count difference
Why does neutral He₂ contain two more electrons than neutral H₂?
Answer: Each helium atom has two 1s electrons whereas each hydrogen atom has one. Two He atoms contribute four electrons; two H atoms contribute two.
Question 5: H₂⁺ bond order
Remove one electron from H₂ to form H₂⁺. Where does the remaining electron go and what bond order follows?
Answer: The one electron occupies σ1s; bond order (1−0)/2 = 0.5.
Question 6: He₂⁺ bond order
A three-electron He₂⁺ species has σ1s²σ*1s¹. Calculate its bond order.
Answer: (2−1)/2 = 0.5.
Question 7: Magnetic behaviour
Which of H₂ and H₂⁺ has an unpaired electron in this model?
Answer: H₂⁺ has one unpaired electron and is paramagnetic; H₂ has a pair in σ1s and is diamagnetic.
Question 8: The wrong conclusion about helium
A student says bond order zero proves two He atoms can never attract each other under any circumstances. Explain the limitation.
Answer: Zero in the elementary 1s MO model excludes an ordinary stable covalent He–He bond of the H₂ type. Very weak van der Waals attraction and ultracold helium dimers are a different physical situation.
Question 9: HOMO and LUMO
Identify H₂’s HOMO and LUMO within the two-1s-orbital basis.
Answer: HOMO σ1s; LUMO σ*1s.
Question 10: Missing LUMO
A student labels σ*1s the LUMO of a filled hypothetical He₂ two-level diagram. What is wrong?
Answer: σ*1s is occupied in that electron configuration. The restricted two-level basis does not include a higher unoccupied orbital from which a proper LUMO can be identified.
Question 11: Bond order and exact energy
If two molecules each have formal bond order one in different qualitative MO models, must their bond energies be equal?
Answer: No. Bond order alone does not calculate exact dissociation energy, which depends on the molecular system and full energy landscape.
Question 12: A counterintuitive ion
What happens to formal bond order when one electron is removed from an antibonding level of neutral hypothetical He₂ to form He₂⁺?
Answer: It rises from 0 to 0.5 in this model because one electron opposing net bonding has been removed.
A Four-Stage Study Sequence for MO Theory
Stage 1: Construct from atomic orbitals
Give the learner two empty 1s atomic orbitals and ask for both possible qualitative molecular combinations. They should explain why one is bonding and the other antibonding, rather than simply drawing one lower and one higher line with no physical reasoning.
Stage 2: Change only electron count
Keep the energy diagram fixed and give the student H₂⁺, H₂, He₂⁺ and hypothetical He₂ in random order. They must count electrons, fill the levels, calculate bond orders and note unpaired spins. Using a single fixed orbital framework highlights how much information comes from occupancy.
Stage 3: Make an observation-based prediction
Ask which systems should have unpaired electrons, which have a net positive covalent MO bond order and which have no such bond under the simple model. The student must defend the answer with electrons in orbitals, not with an element-name stereotype.
Stage 4: Transfer to 2p systems
Only after the 1s construction is secure should the student move to O₂ and F₂. Those have more orbitals, π degeneracy and different occupancy, but the underlying workflow is unchanged: count atomic contributions, fill molecular levels and interpret the result. The learner then sees the growing MO library as one method rather than a set of unrelated drawings.
This sequence also protects H2 revision. There is little benefit to memorising highly specialised H3 diagrams if the student is uncertain about atomic electron configurations, the Pauli principle or the meaning of a covalent bond. The best advanced teaching repairs the earliest weak link.
What I Would Look for in an H3 Chemistry Tutor
A suitable specialist tutor should be able to start with a blank energy axis and ask the learner to construct a bond-order argument. If the student can reproduce H₂ but not explain why hypothetical He₂ differs, they have learned a diagram but not its governing logic. The teacher should diagnose that gap before introducing more difficult diatomics.
I would also value a tutor who is comfortable saying that He₂⁺ and ultracold helium dimers are enrichment examples rather than assigning them as compulsory extra syllabus memorisation. An advanced student needs intellectual depth without an uncontrolled workload that weakens simultaneous H2 Chemistry preparation.
eduKate’s JC2 H3 Chemistry small-group guide outlines its three-student tutorial approach. Families enquiring from Hougang should confirm specialist availability, teaching time, examination-year fit and the current venue near Sixth Avenue MRT. The locality term does not imply a separate Hougang classroom.
Why Molecular Orbital Thinking Matters in Later Science
Molecular orbital theory provides a language for predicting bonding, magnetism, electronic transitions and reactivity. It becomes important in university physical chemistry, inorganic chemistry, materials research, spectroscopy and molecular modelling. The H₃ learner is beginning to use the framework qualitatively; later study may quantify its approximations.
The most valuable general habit is to ask what a model includes. Our two-1s-orbital calculation says something meaningful about ordinary covalent He–He bonding, but it does not describe the full van der Waals potential or every excited state. Reliable scientific judgement requires both a confident calculation and an equally clear statement of its limits.
H₂, He₂ and Molecular Orbital FAQs
Why does H₂ have bond order one?
Its two electrons occupy the bonding σ1s molecular orbital, leaving the antibonding σ*1s orbital empty. The formal bond order is (2−0)/2 = 1.
Why does hypothetical neutral He₂ have bond order zero?
Its four electrons fill both σ1s and σ*1s, giving (2−2)/2 = 0 in the simple covalent molecular orbital model.
Does He₂ literally never exist?
Very weak helium dimers can occur at extremely low temperatures through van der Waals interactions. They are not the ordinary stable covalent He₂ molecule excluded by the elementary 1s MO bond-order argument.
What is the bond order of H₂⁺?
One-half. Its single electron occupies σ1s and the antibonding level remains empty.
Which of H₂ and H₂⁺ is paramagnetic?
H₂⁺ has one unpaired electron and is paramagnetic in the simple ground-state picture. H₂ has paired electrons and is diamagnetic.
What are H₂’s HOMO and LUMO?
In the two-level 1s basis, σ1s is the highest occupied molecular orbital and σ*1s the lowest unoccupied molecular orbital.
Does the H3 syllabus require He₂⁺ and H₂⁺?
The 2027 syllabus 9813 explicitly names H₂ among the simple diatomic MO examples. The ions and helium species here are supporting illustrations to reinforce the electron-filling and bond-order logic, not additional named compulsory syllabus cases.
Can bond order be fractional?
Yes. The simple MO formula can give values such as 0.5 when there are odd electron counts. Such a value describes net bonding occupancy within the model; it is not a physical half-stick between nuclei.
Two lines and four possible electron counts can tell us why a hydrogen molecule has an ordinary covalent bond, why filling an antibonding orbital can remove that advantage, and how an ion can retain partial bonding. Once a student can reconstruct those conclusions from the atoms and electron configurations, molecular orbital theory becomes a tool they can actually use rather than one more picture to memorise.
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