Bonding and Antibonding Orbitals

Introduction

Two suitable atomic orbitals can combine in two mathematically distinct ways. One combination increases electron density between nuclei and is bonding; the other introduces a node or depleted density there and is antibonding. Filling these states in different amounts determines net stabilisation in a simple molecular-orbital model.

Core explanation

Consider two hydrogen 1s orbitals. Adding their wave amplitudes in phase produces a σ1s bonding MO with enhanced amplitude between the nuclei. Opposite-phase combination produces σ1s with a node between them. The bonding orbital is lower in energy than the separated-atom reference over suitable internuclear distances; the antibonding orbital is higher. The star in σ1s marks antibonding character, not an excited atom or multiplication sign.

Electron density is the square of wavefunction amplitude, so wavefunction sign itself is not a positive or negative electric charge. Opposite signs can cancel amplitude in a region even though electrons remain negatively charged everywhere. This distinction is important: “destructive interference” refers to wavefunction combination, not annihilation of electrons.

The two orbitals formed from two atomic orbitals provide places for electrons according to Pauli's rule. H₂ has two electrons in σ1s and none in σ1s , giving bond order one. A simple hypothetical He₂ configuration has four electrons across σ1s and σ1s , two in each, giving (2 − 2)/2 = 0. Net covalent bond order cancels in that elementary diagram. Very weak van der Waals interactions between He atoms may still exist; bond order zero is not a claim of absolutely no interaction at all.

With p orbitals, orientation determines symmetry. End-on p overlap can generate σ2p and σ2p MOs, while side-on overlap gives π2p and π2p pairs. An antibonding pi orbital carries a star and typically has a nodal relationship that reduces bonding density. The precise order of these energies can vary among second-period diatomics due to s–p mixing; energy ordering must be selected for the molecule being studied.

The bond-order formula is a count of occupied bonding and antibonding orbitals in the chosen MO scheme. It is not an exact universal relationship to bond length or energy across unrelated compounds, though within related species higher bond order often correlates with a shorter, stronger bond.

Step-by-step reasoning

1. Identify atomic orbitals of compatible symmetry and similar energy. 2. Form in-phase and out-of-phase combinations. 3. Label lower bonding and higher antibonding levels. 4. Fill electrons according to the species' count and spin rules. 5. Compare bonding and antibonding occupation for net bond order.

Visual explanation

Draw two 1s circles with plus-sign wave phases combining into one continuous shaded region between nuclei. Below draw plus on one and minus on the other yielding a blank nodal plane in the middle. Label σ1s and σ1s with low and high energy.

Real-world analogy

Two water waves can reinforce or cancel at a location depending on phase. Orbital wavefunctions can likewise combine constructively or destructively. The analogy concerns wave amplitude; electrons do not disappear at the molecule's node.

Real-world example

The difference between H₂ and formal He₂ MO filling explains why H₂ has a normal covalent bond while a stable ordinary He₂ molecule is not predicted by this simple bond-order model. The extra helium electrons occupy antibonding as well as bonding levels.

Why?

Why does an antibonding electron weaken net bonding? Its orbital distribution has reduced stabilising density between nuclei and raises the system's energy relative to putting that electron in a bonding level. It offsets some of the lower-level bonding effect.

Common misconception

“A node is a place with positive charge because the electron wavefunction is zero.” A node is zero wavefunction amplitude for that orbital, not a positive-charge sheet or absence of all electrons from the entire molecule.

Worked example

Fill H₂ and He₂ in the simplest 1s MO diagram. H₂: σ1s², σ1s ⁰, bond order (2 − 0)/2 = 1. He₂: σ1s²σ1s ², bond order (2 − 2)/2 = 0. The two extra electrons of He₂ cancel the net bond-order contribution in the simple model. This calculation does not address transient weak dispersion-bound helium pairs.

Quick check

1. What does the star in π indicate? Answer: An antibonding molecular orbital.

Exam focus

Distinguish wavefunction sign from electric charge. Draw bonding below antibonding for a simple orbital pair, fill with Pauli's rule, and calculate occupation difference rather than counting total electrons only.

Advanced insight

Orbital energies and bond order are approximate quantities in many-electron molecules. Electron correlation and orbital relaxation can alter quantitative results, but the bonding/antibonding symmetry idea remains broadly valuable.

Summary

Compatible atomic orbitals form lower bonding and higher antibonding combinations. Nodes reflect wavefunction interference, and antibonding occupation reduces net bond order. H₂ and He₂ show how different electron counts change the result.

Practice questions

1. What is the bond order for two bonding and two antibonding electrons? Answer: Zero in the simple (Nb − Na)/2 model. 2. Which 1s MO has greater density between H nuclei, σ1s or σ1s ? Answer: σ1s bonding MO. 3. Do electrons vanish at a node of one orbital? Answer: No. That orbital has zero amplitude there; electrons are not annihilated. 4. Why does filling σ1s reduce net H–H bonding? Answer: Antibonding occupation offsets stabilisation from σ1s occupation.