Magnetism from Unpaired Electrons

Introduction

MO diagrams can predict whether a molecule has unpaired electrons. Ground-state O₂ has two and is paramagnetic, while N₂ has paired valence electrons and is diamagnetic. The magnetic conclusion follows spin occupancy, not merely a Lewis bond line or the numerical bond order.

Core explanation

An unpaired electron carries a magnetic moment associated with its spin. In a paramagnetic species, these moments can align partly with an applied magnetic field, giving attraction. Diamagnetic species have no unpaired electrons in the simple ground-state picture and show a weak opposing response. In classroom classification, “paramagnetic” usually means at least one unpaired electron; “diamagnetic” means all paired, although all matter can have more subtle magnetic contributions.

N₂'s standard valence MO configuration fills σ2s, σ2s , both π2p orbitals and σ2p with pairs. It has bond order three and no unpaired electrons, so diamagnetic. O₂ fills its last two electrons singly into degenerate π2p orbitals under Hund's rule. It has bond order two and two unpaired electrons, so paramagnetic. Lewis N≡N and O=O diagrams help with connectivity but do not reveal this magnetic contrast directly.

B₂ provides another useful test. After filling σ2s²σ2s ², its remaining two valence electrons occupy the two degenerate bonding π2p orbitals separately in the light-diatomic ordering. It is predicted paramagnetic with two unpaired electrons and bond order one in the basic model. Correct orbital ordering and Hund filling are necessary; a wrong σ2p-first diagram would pair them incorrectly.

Ions can change spin count. H₂ has all electrons paired and is diamagnetic, while H₂⁺ has one unpaired electron and is paramagnetic. O₂⁻ has one unpaired π electron in the simple filling; O₂²⁻ has all π electrons paired. These examples demonstrate that a bond order such as one or one-half does not by itself give a unique magnetic classification across different species.

Magnetic measurements can test an electronic model. However, temperature, aggregation and solid-state interactions can affect measured response. The classroom MO diagram predicts isolated ground-state molecular occupancy, and a real sample's magnetic behaviour must be interpreted with its physical state and composition.

Step-by-step reasoning

1. Count electrons for the exact molecule or ion. 2. Select correct MO ordering and fill using Hund and Pauli rules. 3. Count orbitals with single electron arrows. 4. Classify nonzero unpaired count as paramagnetic in the basic model. 5. Calculate bond order separately from the same filled diagram.

Visual explanation

Draw N₂ with paired arrows in all occupied boxes and O₂ with one arrow in each π box. Place “0 unpaired: diamagnetic” beneath N₂ and “2 unpaired: paramagnetic” beneath O₂. A separate bond-order column shows 3 and 2 to emphasise distinct outputs.

Real-world analogy

Unpaired dancers each retain a free hand; paired dancers do not. Counting unpaired electron arrows is similarly separate from counting how many people support a stage. The analogy helps remember the test but does not explain magnetic quantum moments.

Real-world example

Liquid oxygen can be attracted toward a strong magnetic field, a striking demonstration of its paramagnetism. The two unpaired π electrons from the MO model explain this property without requiring an unpaired Lewis dot in the ordinary O=O drawing.

Why?

Why does Hund's rule matter for magnetism? Filling two degenerate orbitals singly leaves unpaired spins. Pairing prematurely would falsely predict a diamagnetic state for O₂ or B₂ despite the correct total electron count.

Common misconception

“A double bond means all electrons are paired.” O₂ has overall bond order two yet two unpaired electrons in antibonding orbitals. Bond order and magnetism are different calculations.

Worked example

Classify O₂⁻ and O₂²⁻ using π occupation. Neutral O₂ has two π electrons, one in each orbital. Adding one for O₂⁻ makes a pair in one π box and a single in the other, leaving one unpaired electron: paramagnetic. Adding another for O₂²⁻ pairs both boxes, leaving zero unpaired electrons: diamagnetic in the simple ground-state diagram. Their orders are 1.5 and 1, respectively, but the spin result required direct box inspection.

Quick check

1. Is ground-state N₂ paramagnetic in the standard simple MO model? Answer: No. All occupied electrons are paired, so it is diamagnetic.

Exam focus

Show singly occupied boxes explicitly and report unpaired count. Do not infer magnetic class from bond order or a Lewis bond line alone. Note the model concerns the specified species and electronic state.

Advanced insight

Paramagnetic susceptibility can have a temperature dependence, and condensed-phase coupling between moments can produce behaviours beyond isolated-molecule classification. MO spin count is a starting prediction, not a complete materials-magnetism theory.

Summary

Magnetic classification follows unpaired electron count in a filled MO diagram. N₂ is paired and diamagnetic; O₂ has two unpaired π electrons and is paramagnetic. Bond order and magnetic response are related to the same occupancy but must be calculated separately.

Practice questions

1. How many unpaired electrons does the basic O₂ diagram show? Answer: Two. 2. Why is B₂ paramagnetic in the standard light-diatomic order? Answer: Its last two valence electrons occupy separate degenerate π2p bonding orbitals. 3. Is H₂⁺ paramagnetic in the simple model? Answer: Yes, it has one unpaired σ1s electron. 4. Does bond order two guarantee diamagnetism? Answer: No. O₂ has order two and is paramagnetic.