Skip to main content
CHEMISTRY / ORBITAL LAB 02

Molecular orbitals: why addition can make a bond

When two atoms approach, an electron can be described by an orbital extending over both nuclei. Two atomic orbitals supply two independent combinations: one bonding, one antibonding. They are alternatives that electrons can occupy, not two stages of a reaction.

01Watch the region between the nuclei

Small dots mark nuclei A and B, enlarged for visibility. The surfaces show equal probability density |ψ|², not electron paths. Start with orbital A, then increase B’s contribution in both combinations. Drag either model to inspect its shape.

2.30

φA and φB are the starting atomic wavefunctions. t controls the contribution of B; N rescales the result so total probability stays 1.

σ · BondingA + B
AB
Drag to rotate · arrow keys when focused
ψ+(t)=N+(t)(ϕA+tϕB)\psi_+(t)=N_+(t)(\phi_A+t\phi_B)

Teal family: constructive overlap. At t = 1, probability is shared across the two centres.

σ* · AntibondingA − B
AB
Drag to rotate · arrow keys when focused
ψ(t)=N(t)(ϕAtϕB)\psi_−(t)=N_−(t)(\phi_A−t\phi_B)

Coral / plum family: destructive overlap. At t = 1, an extra nodal plane lies halfway between the nuclei.

0.00
Building the combination. Both views start as A alone. B grows with the slider, with a plus sign on the left and a minus sign on the right. Only at t = 1 are these the symmetric bonding and antibonding combinations.
Teal = bonding familyCoral = antibonding familyLighter / darker lobes within a family = + / − phase, not charge.

02A negative sign is not a negative probability

Take a line through the 3D model and compare amplitudes. The dashed curves are the two contributions. Teal adds them; coral subtracts them. At the midpoint, equal contributions cancel exactly in the antibonding combination. Squaring happens afterwards.

Section: y = z = 0, along A—B · Curves show amplitudes before normalization
ψx0AB
Sample positionx = 0.00
Normalized |ψ₊|²0.0477
Normalized |ψ₋|²0.0477
0

Why a lower energy?

In the bonding state, the shared density allows attraction to both nuclei. The total quantum energy includes kinetic energy and all electrostatic terms; the density picture is an intuition, not an energy calculation.

Why an antibonding node?

Opposite amplitudes cancel between the nuclei. The additional node changes the wave’s curvature and, in this simple two-orbital model, gives the higher-energy partner. The star * labels that antibonding partner.

Does a minus sign always mean antibonding?

No: it depends on the original phase choices. Here the starting lobes have matching signs in the overlap region. What matters physically is reinforcement or cancellation there; flipping an entire orbital’s sign does not change its density.

03An available orbital is not an occupied orbital

Now use the simplest s–s energy diagram. In this two-orbital model, electrons fill the lower state first; each orbital holds at most two, with opposite spins. Add electrons and compare the stabilizing and destabilizing occupations.

Eσ*σs As BRelative energy only · not to scale
bond order=NbondingNantibonding2\text{bond order}=\frac{N_{\rm bonding}-N_{\rm antibonding}}{2}

These are normalized Gaussian LCAO illustrations. N accounts for orbital overlap; changing the separation does not simulate a collision or calculate a real bond energy. The geometric surfaces are not photographs. The π view illustrates one p pair; a full molecule needs its other orbitals and electron interactions as well.