B2 Bond Order: MO Diagram and Calculation

The bond order of B₂, the diatomic molecule made of two boron atoms, is 1. Its molecular-orbital electron count gives (4 bonding electrons − 2 antibonding electrons) ÷ 2 = 1, indicating a net single-bond order in the standard molecular-orbital model.

Formula

For a diatomic molecule described by molecular orbital (MO) theory, the bond order is:

Bond order = (N_b − N_a) / 2

• N_b is the number of electrons in bonding molecular orbitals.
• N_a is the number of electrons in antibonding molecular orbitals.
• Bond order is a dimensionless quantity; it has no unit.

The calculator asks for the two electron counts and subtracts the antibonding count from the bonding count, then divides by two. For neutral B₂, the occupied orbitals contain four bonding electrons and two antibonding electrons, so the result is 1.

Worked Example

Using the calculator’s default values:

1. Bonding electrons: N_b = 4.
2. Antibonding electrons: N_a = 2.
3. Subtract: 4 − 2 = 2.
4. Divide by two: 2 / 2 = 1.

The calculated bond order is 1, matching the standard MO result for neutral B₂.

Why B₂ Has This Electron Count

A neutral boron atom has atomic number 5, so it has five electrons. Two neutral boron atoms therefore contribute ten electrons to B₂. In introductory MO diagrams, the two lowest-energy electrons occupy the 1s-derived σ orbital, and the next two occupy the 1s-derived σ* orbital. These core orbitals are filled as a bonding–antibonding pair and cancel in the bond-order subtraction.

The remaining six electrons are valence electrons: boron’s configuration is 1s² 2s² 2p¹, so each atom contributes three beyond its 1s core. In the commonly taught ordering for B₂, the valence electrons fill the σ(2s) orbital with two electrons, the σ*(2s) orbital with two, and the two degenerate π(2p) bonding orbitals with one each. That gives:

• Bonding orbitals: two in σ(2s) plus two in π(2p), for four valence bonding electrons.
• Antibonding orbitals: two in σ*(2s), for two valence antibonding electrons.
• The 1s bonding and antibonding core electrons cancel each other’s contribution.

Counting only valence electrons gives (4 − 2) / 2 = 1. Counting all ten electrons gives five bonding and five antibonding electrons, again producing (5 − 5) / 2 = 0 only if one incorrectly assigns the 1s bonding population as five; the correct all-electron bookkeeping is six bonding and four antibonding electrons, yielding (6 − 4) / 2 = 1. The key is to count orbital populations consistently, including the two electrons in each occupied orbital rather than treating orbitals as electrons.

What Bond Order Says—and Does Not Say

A positive bond order means bonding molecular orbitals contain more electrons than antibonding orbitals. In this model, B₂ has a net bond corresponding to bond order 1. A higher bond order often goes with a shorter, stronger bond when comparing related molecules, but bond order is not a direct measurement of bond length or bond energy. Those properties also depend on the atoms, orbital interactions, and the model used.

Bond order can be fractional. For example, if a molecule has three bonding electrons and one antibonding electron, its bond order is (3 − 1) / 2 = 1. A fractional value does not mean a molecule contains a literal fraction of a conventional bond; it reflects the net electron population in the MO description.

Why B₂ Is Paramagnetic

The MO diagram also predicts a notable magnetic property. After the σ(2s) and σ*(2s) orbitals are filled, B₂ has two electrons occupying the two equal-energy π(2p) bonding orbitals separately. Hund’s rule favors placing one electron in each orbital before pairing them, so the molecule has two unpaired electrons and is paramagnetic: it is attracted to a magnetic field.

This is a useful reminder that bond order and magnetism answer different questions. The bond-order calculation uses the total bonding and antibonding populations. Magnetism depends on whether any electrons remain unpaired. A species can have a positive bond order and still be paramagnetic, as B₂ demonstrates.

Common Counting Mistakes

• Using the wrong orbital ordering: For the lighter diatomic molecules, including B₂, the valence σ(2p) and π(2p) levels are commonly ordered with π(2p) below σ(2p). Applying the ordering used for heavier second-period molecules can give the wrong electron arrangement and magnetic prediction.
• Confusing electrons with orbitals: An orbital can hold up to two electrons. Bond order uses electron counts, not the number of occupied orbital boxes.
• Forgetting antibonding electrons: Electrons in σ* or π* orbitals reduce bond order; they must be subtracted.
• Mixing neutral B₂ with an ion: Adding or removing an electron changes the MO population and may change both bond order and magnetism. The neutral molecule’s result should not be transferred automatically to B₂⁺ or B₂⁻.
• Assuming the number is a measured bond strength: Bond order is a useful theoretical descriptor, not a value in energy units.

Frequently Asked Questions

What is the bond order of B₂?
The standard molecular-orbital calculation gives B₂ a bond order of 1: (4 bonding electrons − 2 antibonding electrons) / 2.

Is B₂ a single bond?
Its MO bond order is 1, which is commonly described as a net single-bond order. This is an electron-counting description, not a claim that every detail of the molecule is identical to a typical single bond in a larger molecule.

Why does B₂ have unpaired electrons?
Its two highest occupied valence electrons occupy separate, equal-energy π(2p) bonding orbitals, leaving two unpaired electrons under Hund’s rule.

Is B₂ paramagnetic or diamagnetic?
Neutral B₂ is paramagnetic in the standard MO description because it has two unpaired electrons.

Does bond order have a unit?
No. It is a dimensionless value calculated from electron counts. It is not measured in angstroms, joules, or any other physical unit.

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in electrons | Bonding electrons | - | 4 | 0
in antibonding | Antibonding electrons | - | 2 | 0
out bond_order | Bond order | - | (electrons - antibonding) / 2 | 0