Hydrogen and Helium Diatomics

Introduction

H₂, H₂⁺, He₂ and He₂⁺ fit into the simplest molecular-orbital diagram: one bonding σ1s and one antibonding σ1s orbital. Their different electron counts make a clear laboratory for bond-order bookkeeping and show why adding an electron can either help or weaken bonding.

Core explanation

Two 1s atomic orbitals produce σ1s below σ1s . H₂ has two electrons, both in σ1s with opposite spins. Its configuration is σ1s²; bond order (2 − 0)/2 = 1, with no unpaired electrons. H₂⁺ has one electron in σ1s: order 0.5 and one unpaired electron. Removing a bonding electron from H₂ lowers bond order, which is consistent with weaker net bonding in the simple comparison.

Formal He₂ has four electrons. Fill σ1s² and σ1s ², yielding (2 − 2)/2 = 0, with no unpaired electrons. The simple diagram therefore predicts no ordinary covalent He–He bond in the ground state. Weak dispersion interactions can still produce transient or very weakly associated helium pairs; zero MO bond order is not a claim that two He atoms never influence one another.

He₂⁺ has three electrons: σ1s²σ1s ¹. Bond order (2 − 1)/2 = 0.5, with one unpaired electron in the antibonding orbital. Removing one antibonding electron from formal He₂ increases its calculated bond order from zero to one-half. This example is the mirror image of H₂ → H₂⁺, where removing a bonding electron decreases order.

These species should not be ranked solely by electron count or charge. The identity of the occupied orbital controls the bonding contribution. A half-integer bond order is a model measure of net stabilisation, not a literal half of a classical chemical stick. Ion stability also depends on environment, potential-energy curves and whether the species can be generated and observed under particular conditions.

The 1s model is unusually simple. Larger diatomics have 2s and 2p orbital levels, some degenerate, with different ordering across elements. The same filling principles apply, but one must use the correct diagram. The four light examples establish the logic before those more complex cases.

Step-by-step reasoning

1. Draw σ1s bonding below σ1s antibonding. 2. Count electrons: H₂⁺ 1, H₂ 2, He₂⁺ 3, He₂ 4. 3. Fill lower level first, with opposite spins if paired. 4. Calculate half the bonding-minus-antibonding count. 5. Count unpaired electrons separately from bond order.

Visual explanation

Draw four vertical two-box MO diagrams side by side, adding one electron at each step from H₂⁺ through He₂. Label σ1s and σ1s identically on each, then write orders 0.5, 1, 0.5 and 0 under the appropriate species.

Real-world analogy

Adding a person to a team can improve a task until a new person begins working against it. Filling bonding then antibonding orbitals similarly raises and then offsets net bond order in the simple tally. Electrons are not workers with intentions, so the analogy is only accounting.

Real-world example

H₂ is an ordinary stable gas molecule. H₂⁺ and He₂⁺ can be studied as molecular ions in specialised gas-phase settings. Comparing them demonstrates how a charged ion can retain a positive bond order despite having fewer or more electrons than a neutral counterpart.

Why?

Why is He₂⁺ predicted to bond while formal He₂ is not in this diagram? He₂⁺ has only one antibonding electron against two bonding electrons, leaving a net difference of one. He₂ has two in each and complete cancellation.

Common misconception

“More electrons always produce a stronger bond.” Adding an electron to an antibonding MO lowers bond order. H₂ and He₂ show how the location of electrons matters more than total count alone.

Worked example

Compute He₂⁺. Each He atom contributes two electrons, then the + charge removes one, leaving three. Put ↑↓ in σ1s and ↑ in σ1s . Nb = 2, Na = 1, so bond order = (2 − 1)/2 = 0.5. The σ1s electron is unpaired, making the simple ground-state ion paramagnetic. Both the bond-order and magnetic predictions come from one filled diagram.

Quick check

1. What is the simple MO bond order of formal He₂? Answer: Zero, from two bonding and two antibonding electrons.

Exam focus

Adjust ion electron counts before filling. State which orbital an electron is removed from and count spins explicitly. Qualify He₂ bond order zero as absence of ordinary covalent bonding in the model, not absence of all weak interactions.

Advanced insight

H₂⁺ is a one-electron two-nucleus system with a particularly direct quantum-mechanical bonding calculation. Its existence demonstrates that an ordinary two-electron pair is not strictly required for a positive molecular bond order.

Summary

In the 1s MO model, H₂⁺ and He₂⁺ each have bond order 0.5, H₂ has 1, and formal He₂ has 0. Antibonding occupancy explains why adding electrons beyond H₂ can weaken net bonding.

Practice questions

1. Give H₂'s 1s MO configuration and bond order. Answer: σ1s², bond order 1. 2. Give H₂⁺ bond order and unpaired count. Answer: 0.5 and one unpaired electron. 3. What is He₂⁺'s 1s MO configuration? Answer: σ1s²σ1s ¹. 4. Why does He₂ have zero order in this simple diagram? Answer: Its two antibonding electrons cancel the net bonding contribution of two bonding electrons.