MO Bond-Order Problem Set

Introduction

Molecular-orbital exercises connect three outcomes: bond order, stability trends and magnetism. The safest method is to fill orbitals explicitly, mark paired and unpaired electrons, then calculate. Memorising a single diagram for every second-period diatomic can fail because the relative 2p orbital order differs across the period.

Core explanation

The introductory MO bond order is (N b − N a)/2, where N b and N a are electrons in bonding and antibonding molecular orbitals. Filling a bonding orbital raises the count; filling an antibonding orbital lowers it. A positive bond order indicates net bonding within this model, but it is not a full guarantee of a stable, isolable molecule under every condition. Magnetism is separate arithmetic: any unpaired electrons make a species paramagnetic; all paired electrons make it diamagnetic in the elementary classification.

Hydrogen examples show the method without 2p complications. H₂ has two electrons in σ1s, giving N b = 2, N a = 0 and order 1; its spins pair. H₂⁺ has one σ1s electron, giving order 0.5 and an unpaired electron. He₂ in the simplest configuration fills σ1s and σ 1s with two electrons each, giving order 0. He₂⁺ removes one antibonding electron, giving (2 − 1)/2 = 0.5 and paramagnetism. Do not interpret the simple He₂ result as saying helium atoms can never interact at all; weak dispersion is a separate phenomenon.

For homonuclear second-period diatomics up to nitrogen, the familiar ordering places π2p bonding orbitals below σ2p bonding due to s–p mixing. For oxygen and fluorine, σ2p bonding is below π2p bonding. The core 1s orbitals and paired 2s bonding/antibonding contributions often cancel in bond-order comparisons, but they still occupy electrons. Using only valence MOs can simplify calculations if that convention is made clear.

N₂ has ten valence electrons. Its 2s pair fills σ2s² and σ 2s², cancelling. Six 2p electrons fill two π2p orbitals with four electrons, then σ2p with two under the B₂–N₂ order. The net valence bonding count is 8 and antibonding 2, giving order (8 − 2)/2 = 3. All electrons are paired, so N₂ is diamagnetic. N₂⁺ loses an electron from its highest occupied bonding σ2p orbital under this ordering, reducing bond order to 2.5 and leaving an unpaired electron.

O₂ has twelve valence electrons. After cancelling the 2s bonding/antibonding pair, its 2p electrons fill bonding σ2p² and π2p⁴, then two degenerate π 2p orbitals with one electron each according to Hund's rule. The total bonding count is 8 and antibonding count 4, giving order 2. Its two unpaired π electrons explain paramagnetism. O₂⁺ removes one electron from a π orbital, increasing order to 2.5 while leaving one unpaired electron. O₂⁻ adds one electron to π , lowering order to 1.5 while leaving one unpaired electron.

Ion formation therefore has no one-direction effect on bond order. Removing an antibonding electron strengthens the modeled bond, but removing a bonding electron weakens it. Identify the highest occupied orbital and its bonding character before predicting length or strength. Electron count alone is insufficient without an MO order.

Step-by-step reasoning

1. Count total or valence electrons and state which convention is used. 2. Choose the suitable second-period 2p ordering. 3. Fill orbitals with Aufbau, Pauli and Hund rules. 4. Count bonding and antibonding electrons and calculate order. 5. Count unpaired electrons separately for the magnetic prediction.

Visual explanation

Draw O₂'s two equal-energy π boxes side by side, each with one upward arrow. Beneath, write order 2 and “two unpaired.” For O₂⁺ erase one π arrow: antibonding occupancy falls by one and order rises by 0.5.

Real-world analogy

A bond-order account resembles a team score with helpful and opposing players. Adding a player to the helpful side and adding one to the opposing side have opposite effects; counting only team size misses the result. Spin pairing is an additional property, not the score itself.

Real-world example

Oxygen's paramagnetism is a classic success of the MO account: a liquid oxygen sample responds to a magnetic field. A simple paired-electron Lewis drawing does not display the two unpaired electrons that the MO diagram predicts.

Why?

Why does O₂⁺ have higher modeled bond order than O₂? Ionisation removes a highest-occupied electron from an antibonding π orbital. The antibonding count drops while bonding occupancy stays the same.

Common misconception

“Removing any electron weakens a diatomic bond.” It depends on whether the electron occupied a bonding or antibonding MO. N₂⁺ loses a bonding electron, while O₂⁺ loses an antibonding one in their standard diagrams.

Worked example

For O₂⁻, use O₂'s valence MO pattern and add one electron to a π 2p orbital. Bonding electrons remain 8; antibonding electrons increase from 4 to 5. Bond order becomes (8 − 5)/2 = 1.5. One π orbital is paired and the other singly occupied, so the ion is paramagnetic.

Quick check

1. Is N₂⁺'s bond order higher or lower than N₂'s in the standard MO diagram? Answer: Lower: 2.5 instead of 3, because a bonding electron is removed.

Exam focus

Show the orbital filling, not only the final order. State the 2p ordering convention, distinguish bonding from antibonding electron removal and make an explicit unpaired-electron count for magnetism.

Advanced insight

MO energies can shift with bond length and environment. The textbook diagrams are qualitative models for simple diatomics; quantitative spectroscopy and computation provide more detailed orbital energies and bond descriptions.

Summary

MO occupancy predicts bond order and spin-based magnetism. N₂ has order 3 and paired electrons, while O₂ has order 2 and two unpaired electrons. Ion effects follow the character of the orbital filled or emptied, not charge sign alone.

Practice questions

1. What is the simple MO bond order of H₂⁺? Answer: 0.5. 2. Why is O₂ paramagnetic? Answer: It has two unpaired electrons in degenerate π 2p orbitals. 3. What is the bond order of O₂⁺? Answer: 2.5 in the standard valence MO diagram. 4. Why does He₂ have zero simple MO bond order? Answer: Its two bonding and two antibonding 1s electrons cancel in (N b − N a)/2.