Sharing electrons gives atoms a bond.
One H₂ molecule contains two hydrogen atoms, each initially with one electron. They can stay together by sharing a pair of electrons. What does “sharing” mean here? The atoms do not take turns using the pair. The electrons and both nuclei form a stable arrangement together. By following the atoms as they approach, we can give that sentence something visible to refer to. Water, sugar and DNA also contain covalent bonds; the two-hydrogen example gives us a starting point for understanding larger molecules.
If attraction brings the atoms closer, you may reasonably ask: why do they not keep approaching until they crowd together? That is why we vary their separation below. The potential energy records energy associated with their relative positions. Match each separation to its point on the curve and look for a position lower than its neighbours. This gives the typical stable separation, the bond length. When we examine bonds with different shapes later, we will return to how the electrons are distributed and how that distribution helps hold the atoms together.
01One pair, attracted by two nuclei
Hydrogen has one proton and one electron. Two hydrogen atoms can form H₂ by sharing two electrons. The shared negative charge is attracted by both positive nuclei; the arrows below show the resulting attraction on the nuclei towards the shared region. A single bond line stands for one shared pair.
A qualitative, normalized combination of the two innermost (1s) hydrogen wavefunctions. The cloud shows probability density, not a solid surface. The approach is a guided comparison of electronic distributions, not a simulation of a collision or a calculated reaction rate. Nuclei and electron badges are enlarged for visibility.
02Why does a bond have a particular length?
Keep the same H₂ molecule for now. The nuclei repel one another, as do the electrons; nuclei and electrons attract. Quantum electronic energy also changes strongly at short distances. In atoms with occupied inner shells, Pauli restrictions on overlapping occupied states add an important short-range contribution. All of these contributions determine the total potential-energy curve. The lowest point gives the equilibrium separation between nuclei: the bond length.
Use the three distance buttons in order. At 61 pm, compare the outward force arrows with the high point on the left of the energy curve. At 74 pm, both arrows vanish at the minimum. At 107 pm, the arrows point inwards. Keep the atoms paused while comparing these states; then use “Watch bond vibration” to see how motion carries them across the minimum. Keep three sentences in mind: the lowest point gives the bond length; at that point the slope is zero, so the net force is zero; away from it, the force pulls the atoms back. Below the model, we will work out why the slope gives this force direction. The subsections after that are optional reading; skip them if you like and keep the three sentences.
Watch bond vibration keeps the mechanical energy: the two nuclei oscillate on either side of the potential minimum. Drag and release adds damping to show energy passing to an imagined environment. Motion is greatly slowed and uses a classical teaching potential, not a quantum simulation of H₂.
First connect force with a change in energy
Let r be the distance between the two nuclei, measured outwards from one nucleus towards the other. Positive radial force Fᵣ means a force tending to increase r. Work is energy transferred by a force acting through a displacement. Over a short displacement Δr, during which the force changes very little, the electrical force does work approximately FᵣΔr. Potential energy U is defined so that this work is the decrease in U: energy lost from U can become motion.
This fraction is the change in the graph’s height divided by the change in distance. We can divide while Δr is nonzero. As this nonzero displacement approaches zero, the ratio approaches the local slope, written dU/dr. The minus sign matters: if U rises as r increases, the slope is positive and the force points towards smaller r. If U falls, the force points towards larger r. This explains why the arrows on both sides of a smooth energy minimum point back towards it.
Why two point charges have a 1/r energy term
Coulomb’s law is an experimental starting point: Fᵣ = kq₁q₂/r² for two point charges. Here q₁ and q₂ are signed charges, k is the Coulomb constant, and r > 0. A positive product gives repulsion, while a negative product gives attraction. We now want to add the work as the separation changes; multiplying one force value by the whole distance would be wrong because the force changes along the way.
Divide a change from r to a larger distance R into many short intervals. In one interval from a to b, choose the intermediate distance √(ab), which lies between them. This choice is useful because squaring the distance gives ab. The approximate work on that short interval becomes a difference of two fractions:
The next interval begins at b, so its +1/b cancels this −1/b. Every internal endpoint cancels in the same way. As the largest interval shrinks to zero, a continuous force is nearly constant on each interval and the sum tends to the exact work. Only the first and last endpoints remain:
Choose zero potential energy at infinite separation. As R grows without bound, 1/R tends to zero and U(R) tends to our chosen zero. Therefore U(r) = kq₁q₂/r. For unlike charges this is negative: bringing the pair together lowers energy below the separated reference. A negative value is relative to our chosen zero, not a negative amount of matter.
Read the bond curve before releasing the atoms
A molecule contains several charge interactions and quantum electronic energy. We cannot replace their total by the single point-charge term. The graph we have just used follows a chosen teaching curve with a minimum at r₀ = 74 pm, a representative H₂ equilibrium length. A picometre, pm, is one trillionth of a metre. The curve is a model input, not a quantum calculation of that measured bond length.
For the curve drawn here, write the separation as s = r/r₀. This compares the distance with the equilibrium distance r₀. We use q = exp[−2.8(s−1)] and set the relative energy to U = (1−q)²−1. Here exp(x) means e raised to the power x, where e is the fixed number about 2.718. This positive function grows when x grows; exp(0) = 1. The factor 2.8 sets the width of this example well, and the energy unit makes its depth 1. We can check the minimum without a calculus rule: a square cannot be negative, so U ≥ −1. Equality requires q = 1, hence s = 1 and r = r₀.
For a larger separation, s > 1 gives 0 < q < 1; increasing s makes q smaller, so (1−q)² grows and the energy rises towards zero. The force therefore pulls inwards. For s < 1, q > 1; moving inwards increases q, so (q−1)² grows and the energy rises. The force then pushes outwards. At the smooth minimum the slope is zero, so the net force is zero. That does not make a moving atom stop immediately: it may cross the minimum and then slow down on the other side.
Now use the separation slider. Watch the point move along the energy graph and compare the arrows on the atoms. “Place at equilibrium” sets r = 74 pm. “Watch bond vibration” starts a slow classical motion around the minimum; pause it whenever you want to inspect a position. Releasing a dragged atom includes damping, meaning energy is passed to an imagined surrounding environment.
The electrical starting point
For two point charges, force magnitude grows with charge and falls with separation squared. Opposite signs attract; like signs repel. and are charges, their separation, and the Coulomb constant.
Force is not energy
With zero energy at infinite separation, a two-charge pair has this electrical potential energy. Force tells us how energy changes with position. Molecular energy includes many interactions and quantum motion, so this one formula is not the whole bond.
A minimum is stable
On either side of the minimum, a small displacement produces a restoring force. Real bonds vibrate around equilibrium. At large separation, attraction becomes weak; atoms do not snap back from any arbitrary distance. The displayed H₂ equilibrium distance is approximately 74 pm.
03Build a triple bond: one and two
Our molecule is ethyne, H–C≡C–H. Focus on the two carbons in its centre. An orbital describes the possible distribution of an electron, rather than a track. In this bonding model, each carbon uses inward-pointing sp orbitals for bonding and two perpendicular p orbitals for bonding. We reveal these contributions in order so that you can inspect them.
What do these names mean in the construction? An function has no preferred direction; a function has two signed lobes on opposite sides of its nucleus. In the sp description, one and one combine to give two functions pointing along the carbon chain. Two other directions remain perpendicular to it. The inward-pointing functions overlap along the line joining the nuclei: this is called , pronounced sigma. The remaining functions overlap beside that line: this is called , pronounced pi. “Separate contributions” shows what each atom supplies. Showing just the end-on overlap lets us examine that bond on its own; adding a contribution shows what each remaining pair of functions contributes.
— end-on overlap
Pronounced “sigma”. Density surrounds the line joining the nuclei. Each single bond is a bond in this model. The central C–C pair is shared along the axis.
π — side-on overlap
Pronounced “pi”. Each contributing p function is zero on the plane between its lobes. The two nuclei lie in that same plane, so adding the two functions still gives 0 + 0 = 0 there; squaring gives zero density. The two lobes together describe one orbital, not two separate bonds. Ethyne has two perpendicular orbitals.
Six electrons, not six dots
Each carbon supplies 4 outer, or valence, electrons and each hydrogen supplies 1: . The C–C triple bond uses 3 pairs, so electrons. The two C–H bonds use another , giving in total. Each carbon also has 2 inner 1s electrons, omitted from this display. Bonding rearranges the electrons; it does not create them.
The sequential reveal is a teaching sequence, not a claim that real ethyne forms these bonds one after another in time. The two-lobed outlines show each p orbital’s angular shape; the surrounding soft cloud samples a Gaussian approximation to its probability density. The same outlines continuously stretch and join as the shared density forms; this shape transition is a teaching guide, not a calculated reaction trajectory. This is not a molecular quantum-chemistry calculation; C–H densities and carbon core electrons are omitted. Colour identifies the orbital, not its phase, and no cloud has a hard boundary.
04Why do we draw opposite spins?
An electron has an intrinsic quantum property called spin. Relative to one chosen measurement axis, its two possible spin outcomes are labelled ↑ and ↓; this is a quantum input, not the rotation of a small ball. The Pauli exclusion principle is another starting rule: two electrons cannot occupy the same complete one-electron state. In one spatial orbital, the spatial part is already the same. We can distinguish two electrons by using the two different spin states. A third electron would have to repeat one of those states, which the rule forbids. Thus one orbital holds at most two electrons with opposite spins.
Opposite spin is the occupancy condition for a pair in one orbital; it is not a promise that any two opposite-spin electrons will form a bond. Stable bonding also requires a suitable orbital and a lower total energy. We saw that energy condition in the H₂ curve; opposite spins alone do not replace it.
Check: do the upper and lower π lobes contain a pair each?
No. Together they belong to one spatial orbital, occupied by two electrons in total. Ethyne’s other pair occupies a second orbital in a perpendicular plane.
