Chemical Bonding and Molecular Structure · Lesson 10 of 12

Molecular Orbital Configurations and Predictions

“Fill molecular orbitals to predict bonding, relative bond strength, and magnetic properties of diatomic molecules and ions.”

Learning Objectives

• Write molecular-orbital occupancies using total or valence-electron counts consistently. • Calculate bond order and use the simple MO model to assess covalent bonding. • Identify paramagnetic and diamagnetic species from unpaired electrons. • Compare homonuclear diatomic molecules from hydrogen to neon. • Predict the effects of removing or adding electrons to oxygen molecules.

Turn an energy diagram into a prediction

An MO diagram becomes useful when electrons are placed into it. Count the electrons, select the correct energy order, and fill from low to high while respecting Pauli and Hund rules. The final occupancy tells us how strongly bonding and antibonding influences compete and whether any electrons remain unpaired.

For a neutral homonuclear diatomic molecule, double the electron count of one atom. For an anion add electrons; for a cation subtract them. You may show all electrons or omit a completely filled cancelling core pair, but the bond-order calculation must follow the same convention throughout.

Bond order from bonding and antibonding electrons

An electron in a bonding MO contributes to holding the nuclei together, while an antibonding electron opposes that effect. Subtract the antibonding count from the bonding count and divide by two because an ordinary bond corresponds to a net pair of bonding electrons.

Molecular-orbital bond orderLaTeX

BO = \frac{N _{b} - N _{a}}{2}

N_b is the number of electrons in bonding MOs and N_a is the number in antibonding MOs. Count electrons, not occupied orbital boxes.

A positive bond order predicts a net covalent bonding contribution in this simple model. Zero order means the bonding and antibonding contributions cancel. A negative value would mean the antibonding contribution dominates. Real stability also depends on the full energy balance, so use this criterion as the chapter’s simple MO prediction, not as a complete treatment of every possible weak association.

Within comparable species, a larger bond order normally corresponds to a stronger, shorter bond. Integral values one, two, and three connect with single, double, and triple bonding. Fractional values are also possible because changing one electron changes the formula by one-half.

Paired electrons and magnetic behaviour

A paramagnetic species has one or more unpaired electrons and is attracted by a magnetic field. A diamagnetic species has its occupied orbitals filled in electron pairs and is weakly repelled. The number of electrons alone does not establish magnetism: the way they occupy the orbitals is decisive.

Example — Hydrogen and helium

Problem
Compare the simple MO predictions for H₂ and He₂.

1.H₂ has two electrons. Both fill the lower $σ_{1 s}$ orbital with opposite spins, giving $(σ_{1 s})^{2}$ .
2.Its bond order is $(2 - 0) /2 = 1$ . No unpaired electrons remain, so H₂ is diamagnetic and has a net covalent bond.
3.He₂ would have four electrons, giving $(σ_{1 s})^{2} (σ_{1 s}^{*})^{2}$ . Its bond order is $(2 - 2) /2 = 0$ .
4.The simple MO treatment therefore predicts no ordinary stable covalent He–He molecule. This is the sense of the chapter’s “does not exist” statement; it is not a detailed description of very weak interactions under specialised conditions.

A filled core can cancel out

The four electrons in the filled $σ_{1 s}$ and $σ_{1 s}^{*}$ core orbitals contribute two to each side of the subtraction, so they cancel. The chapter sometimes abbreviates this closed core as KK. Showing or omitting it gives the same bond order if the counting is consistent.

Example — Lithium and beryllium

Problem
Explain why the simple MO model predicts bonding for Li₂ but not ordinary covalent bonding for Be₂.

1.Li₂ has six total electrons. After the four-electron core, two valence electrons occupy $σ_{2 s}$ . The valence bond order is $(2 - 0) /2 = 1$ .
2.All electrons are paired, so Li₂ is diamagnetic. Such molecules are observed in lithium vapour.
3.Be₂ has eight total electrons. Its four valence electrons occupy $(σ_{2 s})^{2} (σ_{2 s}^{*})^{2}$ .
4.The simple bond order is $(2 - 2) /2 = 0$ . The elementary model predicts no net ordinary covalent bond; more refined descriptions of weak bonding are beyond this treatment.

Boron, carbon, and nitrogen require the lighter-molecule order

For B₂, C₂, and N₂, the bonding pi-2p orbitals lie below sigma-2p-z. This matters as soon as electrons start entering the 2p-derived orbitals. Fill the two equal-energy pi orbitals singly before pairing them.

Example — From B₂ to C₂ to N₂

Problem
Predict bond order and magnetism for the three lighter second-period molecules.

1.B₂ has six valence electrons. Four fill the cancelling sigma-2s pair. The next two enter the two bonding pi orbitals singly. Thus $N_{b} = 4$ and $N_{a} = 2$ , giving order 1 and two unpaired electrons: paramagnetic.
2.C₂ has eight valence electrons. The bonding pi pair now contains four electrons. Thus $N_{b} = 6$ and $N_{a} = 2$ , giving order 2 with no unpaired electrons: diamagnetic.
3.N₂ has ten valence electrons. Two more enter the bonding sigma-2p-z orbital. Thus $N_{b} = 8$ and $N_{a} = 2$ , giving order 3 with all electrons paired: diamagnetic.
4.In the elementary C₂ diagram, the sigma-2s contribution cancels its antibonding partner, leaving net bonding from the pi orbitals. This delocalised MO accounting is different from assigning one localised sigma plus one pi bond to an ordinary double-bond drawing.

Oxygen explains why magnetic evidence matters

O₂ has twelve valence electrons and uses the oxygen-type energy order. After filling the lower sigma and bonding pi orbitals, two electrons must enter the degenerate antibonding pi pair. Hund’s rule places one in each before pairing. This leaves two unpaired electrons.

Molecular-orbital occupancy of oxygen— Read upward from sigma-2s. The two singly occupied pi-star orbitals are the key to oxygen’s paramagnetism. The vertical spacings are schematic rather than quantitative energies.

Example — Oxygen bond order

Problem
Calculate O₂ bond order using its valence-only diagram.

1.Bonding electrons: 2 in $σ_{2 s}$ , 2 in $σ_{2 p_{z}}$ , and 4 in the bonding pi pair, so $N_{b} = 8$ .
2.Antibonding electrons: 2 in $σ_{2 s}^{*}$ and 2 in the antibonding pi pair, so $N_{a} = 4$ .
3.The bond order is $(8 - 4) /2 = 2$ . The two unpaired pi-star electrons make O₂ paramagnetic.
4.Including the core would instead give $N_{b} = 10$ and $N_{a} = 6$ , with the same result: $(10 - 6) /2 = 2$ .

A simple paired-electron Lewis drawing of O₂ does not reveal this magnetic behaviour. MO theory’s unpaired-electron prediction agrees with observation and demonstrates why different bonding models are needed. A correct bond-order value alone would not be enough to predict magnetism.

Compare the second-period series

As more electrons are added across the series, bonding orbitals fill first and then antibonding orbitals. Bond order therefore rises to nitrogen and then falls towards neon. The chapter’s comparative diagram pairs this occupancy trend with bond energies, lengths, and magnetic properties.

Species	Valence electrons	Simple MO order	Bond order	Magnetic prediction	Representative energy / length
B₂	6	Pi below sigma	1	Paramagnetic	290 kJ per mole / 159 pm
C₂	8	Pi below sigma	2	Diamagnetic	620 kJ per mole / 131 pm
N₂	10	Pi below sigma	3	Diamagnetic	945 kJ per mole / 110 pm
O₂	12	Sigma below pi	2	Paramagnetic	498 kJ per mole / 121 pm
F₂	14	Sigma below pi	1	Diamagnetic	159 kJ per mole / 143 pm
Ne₂	16	Sigma below pi	0	All electrons paired; no net covalent bond in this model	No ordinary covalent bond in this treatment

The table makes another useful distinction. B₂ and F₂ both have bond order one, but their different atoms and electronic structures give different measured bond energies and lengths. Bond order is a guide within appropriate comparisons, not a universal numerical formula for bond energy.

Species	Valence MO configuration, filled 1s core omitted	Unpaired electrons
B₂	$(σ_{2 s})^{2} (σ_{2 s}^{*})^{2} (π_{2 p_{x}})^{1} (π_{2 p_{y}})^{1}$	2
C₂	$(σ_{2 s})^{2} (σ_{2 s}^{*})^{2} (π_{2 p_{x}})^{2} (π_{2 p_{y}})^{2}$	0
N₂	$(σ_{2 s})^{2} (σ_{2 s}^{*})^{2} (π_{2 p_{x}})^{2} (π_{2 p_{y}})^{2} (σ_{2 p_{z}})^{2}$	0
O₂	$(σ_{2 s})^{2} (σ_{2 s}^{})^{2} (σ_{2 p_{z}})^{2} (π_{2 p_{x}})^{2} (π_{2 p_{y}})^{2} (π_{2 p_{x}}^{})^{1} (π_{2 p_{y}}^{*})^{1}$	2
F₂	$(σ_{2 s})^{2} (σ_{2 s}^{})^{2} (σ_{2 p_{z}})^{2} (π_{2 p_{x}})^{2} (π_{2 p_{y}})^{2} (π_{2 p_{x}}^{})^{2} (π_{2 p_{y}}^{*})^{2}$	0
Ne₂	$(σ_{2 s})^{2} (σ_{2 s}^{})^{2} (σ_{2 p_{z}})^{2} (π_{2 p_{x}})^{2} (π_{2 p_{y}})^{2} (π_{2 p_{x}}^{})^{2} (π_{2 p_{y}}^{})^{2} (σ_{2 p_{z}}^{})^{2}$	0

Read a superscript as the number of electrons in that orbital, not as the bond order. The paired pi orbitals have been written separately so that Hund’s-rule occupancy is visible. The hypothetical Ne₂ configuration fills every bonding and antibonding member, cancelling the net covalent contribution in this elementary model.

Adding or removing oxygen electrons

The highest occupied electrons in O₂ are antibonding pi electrons. Removing one therefore reduces the opposing contribution and increases bond order. Adding one increases that opposing contribution and reduces bond order. The effect of ionisation depends on which orbital loses or gains the electron.

Example — The oxygen family

Problem
Compare O₂⁺, O₂, superoxide, and peroxide.

1.For $O_{2}^{+}$ , remove one antibonding pi electron from O₂. Bond order increases from 2 to 2.5; one unpaired electron remains, so the ion is paramagnetic.
2.For $O_{2}^{-}$ , add one electron to the antibonding pi pair. One orbital becomes paired and the other remains singly occupied. Bond order decreases to 1.5 and the ion is paramagnetic.
3.For $O_{2}^{2 -}$ , add two electrons relative to O₂. Both pi-star orbitals become paired. Bond order is 1 and the ion is diamagnetic.
4.The expected order of decreasing bond strength and increasing bond length in this related family is $O_{2}^{+}$ , $O_{2}$ , $O_{2}^{-}$ , $O_{2}^{2 -}$ .

Adding an electron does not always strengthen a bond

An electron added to a bonding orbital raises bond order, but one added to an antibonding orbital lowers it. Likewise, removing an antibonding electron can strengthen a bond. Identify the affected orbital before reasoning from the charge.

Check your understanding

Use a small occupancy diagram whenever a magnetic prediction is required. Counting the total number of electrons or merely looking at the molecular charge can lead to a plausible but wrong answer.

Quick check

A species has 8 bonding and 5 antibonding electrons. Its bond order is:

Quick check

Which feature makes O₂ paramagnetic in the MO model?

Quick check

Removing one electron from an antibonding orbital changes bond order by:

Quick check

Which oxygen species has the lowest bond order among these choices?

Quick check

Why does the filled 1s core not change bond order?

Practice Problems

Write the MO configurations of H₂, He₂, Li₂, and Be₂ in the simple model and calculate their bond orders.
Use the lighter-molecule energy order to compare B₂, C₂, and N₂, including unpaired-electron counts.
Draw the valence-only O₂ diagram and demonstrate its bond order and magnetic behaviour.
Compare O₂⁺, O₂, O₂⁻, and O₂²⁻ by occupancy, bond order, magnetism, and expected relative bond length.
Explain why F₂ has a lower bond order than O₂ although it has more electrons.
A hypothetical configuration has 7 bonding and 4 antibonding electrons. Calculate its order and explain why magnetic behaviour still requires the occupancy pattern.
Explain why the simple C₂ MO diagram should not be read as an ordinary localised double bond with one independently assigned sigma bond.

Key Takeaways

• Count electrons consistently, choose the energy order, and fill orbitals before predicting properties. • Bond order is half the bonding-minus-antibonding electron count. • Unpaired electrons produce paramagnetism; paired occupancy produces diamagnetism. • O₂ has two unpaired electrons in antibonding pi orbitals and bond order two. • Removing an antibonding electron strengthens the related bond; adding one weakens it.

Previous · Lesson 9

Building Molecular Orbitals

Next · Lesson 11

Hydrogen Bonding and Its Effects

Turn an energy diagram into a prediction

Bond order from bonding and antibonding electrons

Paired electrons and magnetic behaviour

A filled core can cancel out

Boron, carbon, and nitrogen require the lighter-molecule order

Oxygen explains why magnetic evidence matters

Compare the second-period series

Adding or removing oxygen electrons

Check your understanding

Quiz

Practice Problems

Key Takeaways