MO Predictions for Diatomic Ions

Introduction

Adding or removing one electron changes a diatomic molecule's MO occupancy. The effect on bonding depends on whether the changed orbital is bonding or antibonding. A charge symbol alone cannot tell the direction of the bond-order change; the occupied diagram can.

Core explanation

For N₂, the highest occupied valence MO in the standard simple diagram is bonding σ2p. Neutral N₂ has bond order three. Removing one electron to make N₂⁺ removes a bonding electron, so Nb falls from eight to seven while Na remains two. Order becomes (7 − 2)/2 = 2.5. The resulting singly occupied σ2p orbital gives one unpaired electron in this introductory model.

For O₂, the highest occupied orbitals are antibonding π2p . Neutral O₂ has bond order two. Removing one antibonding electron to make O₂⁺ reduces Na from four to three while Nb remains eight, so order becomes (8 − 3)/2 = 2.5. Although both N₂ and O₂ lose one electron in these comparisons, their bond orders move in opposite directions because their top electrons occupy different kinds of MO.

Adding one electron to O₂ makes O₂⁻. It goes into a π2p orbital in the simple model, increasing Na to five and reducing order to (8 − 5)/2 = 1.5. Adding a second produces O₂²⁻ with Na six and order one. These ions are often called superoxide and peroxide, respectively, in contexts where their formula units and chemistry are relevant. Their bond lengths tend to follow the bond-order trend, but a precise number depends on environment and measurement.

H₂⁺ and He₂⁺ give analogous simple 1s cases. H₂⁺ is formed by removing a bonding electron from H₂, lowering order from one to one-half. He₂⁺ can be considered by removing an antibonding electron from formal He₂, raising order from zero to one-half. These examples make the orbital identity rule general rather than oxygen-specific.

Magnetism also changes when electron count changes. O₂⁺ has one electron in its π pair and is paramagnetic in the simple model. O₂²⁻ fills the pair completely and is diamagnetic in that diagram. Bond order and spin count must be evaluated separately; one number does not determine the other uniquely.

Step-by-step reasoning

1. Fill the neutral species' correct MO diagram. 2. Identify the highest occupied orbital for removal or lowest available for addition. 3. Determine whether that orbital is bonding or antibonding. 4. Update Nb or Na by one and recalculate bond order. 5. Recount unpaired electrons after the change.

Visual explanation

Draw N₂ and O₂ ladders side by side. For N₂⁺, cross out one electron from bonding σ2p and write “order down 0.5.” For O₂⁺, cross out one from antibonding π and write “order up 0.5.”

Real-world analogy

Removing a support beam weakens a structure, while removing a load that opposed the structure can strengthen it. Electron removal can likewise lower or raise MO bond order depending on the orbital's contribution. The analogy describes the tally, not a mechanical bond.

Real-world example

Superoxide O₂⁻ appears in several chemical and biological contexts. Its extra electron occupies an antibonding orbital in the simple model, giving lower O–O bond order than neutral O₂ and one unpaired electron.

Why?

Why does O₂⁺ have higher calculated order than O₂? Ionisation removes an electron that was offsetting bonding in a π antibonding MO, so the bonding-minus-antibonding difference increases by one.

Common misconception

“Every positive ion has a weaker bond because it lost an electron.” N₂⁺ loses a bonding electron and weakens by this measure, but O₂⁺ loses an antibonding electron and strengthens by this measure.

Worked example

Rank O₂⁺, O₂, O₂⁻ and O₂²⁻ by simple MO bond order. Starting with O₂ at 2, removing one π electron gives O₂⁺ 2.5. Adding one gives O₂⁻ 1.5; adding two gives O₂²⁻ 1. The ranking is O₂⁺ > O₂ > O₂⁻ > O₂²⁻ by calculated bond order. Within these related species, shorter bond length is generally expected toward the higher-order end, but exact lengths require data.

Quick check

1. Does removing an antibonding electron raise or lower simple MO bond order? Answer: It raises order by 0.5 for one electron.

Exam focus

Do not infer bond order from charge sign alone. Mark the changed MO as bonding or antibonding and show the arithmetic. Refill degenerate levels correctly before stating magnetism.

Advanced insight

Ions in solids or solutions can interact strongly with their surroundings, and orbital energies can shift. Gas-phase MO diagrams are clean qualitative baselines; environmental effects may alter measured lengths or magnetic behaviour.

Summary

One-electron changes alter simple MO bond order by one-half, but the direction depends on orbital type. N₂⁺ loses a bonding electron and drops to 2.5; O₂⁺ loses an antibonding electron and rises to 2.5. Spin occupancy requires a separate check.

Practice questions

1. What is N₂⁺ bond order in the standard model? Answer: 2.5. 2. What is O₂⁻ bond order? Answer: 1.5. 3. Which orbital type loses an electron on making O₂⁺ from O₂? Answer: An antibonding π2p orbital. 4. Is O₂²⁻ predicted to have unpaired electrons in the simple ground-state diagram? Answer: No. Its π pair is fully occupied with paired spins.