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What is the bond order?

Pick a two-atom molecule of one element, such as O₂ or N₂⁺, or type your own counts of bonding and antibonding electrons. The calculator gives the bond order, the molecular orbital configuration and whether the molecule is paramagnetic.

Your numbers

From
Bond order
Bond order = 2

Bond order = 2 for O₂.

Bond order (number)
2
Bonding electrons
10
Antibonding electrons
6
Stability
Above 0: a stable bond is expected; a higher bond order means a stronger bond.
Molecular orbital configuration
(σ1s)² (σ*1s)² (σ2s)² (σ*2s)² (σ2p)² (π2p)⁴ (π*2p)²
Unpaired electrons
2
Magnetism
Paramagnetic (2 unpaired electrons)
For
O₂

Bond order: Bond order = 2. Bond order = 2 for O₂.

How to calculate

Finds the bond order of a diatomic molecule or ion from H₂ to Ne₂ by filling its molecular orbitals, with its configuration and magnetism, or from counts of bonding and antibonding electrons.

Example with the default inputs (From Diatomic molecule, Molecule O₂ (oxygen), Charge 0): Bond order = 2 for O₂.

Method: Bond order = (electrons in bonding orbitals − electrons in antibonding orbitals) ÷ 2. A diatomic molecule fills σ1s, σ*1s, σ2s, σ*2s, then π2p before σ2p for Li to N (σ2p before π2p for O, F, Ne), then π*2p and σ*2p.

  • Homonuclear diatomic molecules and ions from H₂ to Ne₂ with a charge from −2 to +2, in their ground state; 1 to 20 electrons.
  • The two π2p orbitals and the two π*2p orbitals each fill one electron per orbital before pairing (Hund’s rule).
  • A bond order above 0 means a stable bond is expected; 0 means none.

Machine-readable copies: Markdown, JSON.

Worked examples

Each example is checked against the calculator on every build.

  1. From Diatomic molecule, Molecule O₂ (oxygen), Charge 0 gives Bond order (number) 2, Unpaired electrons 2, Magnetism Paramagnetic (2 unpaired electrons).Source: OpenStax, Chemistry 2e, §8.4 Molecular Orbital Theory (bond order = (bonding − antibonding electrons) ÷ 2), https://openstax.org/books/chemistry-2e/pages/8-4-molecular-orbital-theory, Example 8.6 (O₂): bond order 2, two unpaired electrons, paramagnetic
  2. From Diatomic molecule, Molecule N₂ (nitrogen), Charge 0 gives Bond order (number) 3, Unpaired electrons 0, Magnetism Diamagnetic (all electrons paired).Source: OpenStax, Chemistry 2e, §8.4 Molecular Orbital Theory (bond order = (bonding − antibonding electrons) ÷ 2), https://openstax.org/books/chemistry-2e/pages/8-4-molecular-orbital-theory, Check Your Learning after Example 8.6 (N₂): bond order 3, diamagnetic
  3. From Diatomic molecule, Molecule C₂ (carbon), Charge -2 gives Bond order (number) 3, Bonding electrons 10, Antibonding electrons 4.Source: OpenStax, Chemistry 2e, §8.4 Molecular Orbital Theory (bond order = (bonding − antibonding electrons) ÷ 2), https://openstax.org/books/chemistry-2e/pages/8-4-molecular-orbital-theory, Example 8.7 (C₂²⁻): bond order 3, six more bonding than antibonding electrons
  4. From Diatomic molecule, Molecule Be₂ (beryllium), Charge -2 gives Unpaired electrons 2, Bond order (number) 1, Molecular orbital configuration (σ1s)² (σ*1s)² (σ2s)² (σ*2s)² (π2p)².Source: OpenStax, Chemistry 2e, §8.4 Molecular Orbital Theory (bond order = (bonding − antibonding electrons) ÷ 2), https://openstax.org/books/chemistry-2e/pages/8-4-molecular-orbital-theory, Check Your Learning after Example 8.7 (Be₂²⁻): two unpaired electrons, paramagnetic
  5. From Diatomic molecule, Molecule Be₂ (beryllium), Charge 0 gives Bond order (number) 0, Stability Bond order 0: no stable bond is expected..Source: OpenStax, Chemistry 2e, §8.4 Molecular Orbital Theory (bond order = (bonding − antibonding electrons) ÷ 2), https://openstax.org/books/chemistry-2e/pages/8-4-molecular-orbital-theory, Table 8.3: Be₂ bond order 0, unstable
  6. From Diatomic molecule, Molecule F₂ (fluorine), Charge 0 gives Bond order (number) 1, Unpaired electrons 0.Source: OpenStax, Chemistry 2e, §8.4 Molecular Orbital Theory (bond order = (bonding − antibonding electrons) ÷ 2), https://openstax.org/books/chemistry-2e/pages/8-4-molecular-orbital-theory, Table 8.3: F₂ bond order 1
  7. From Diatomic molecule, Molecule H₂ (hydrogen), Charge 1 gives Bond order (number) 0.5, Bond order Bond order = 0.5, Unpaired electrons 1.Source: OpenStax, Chemistry 2e, §8.4 Molecular Orbital Theory (bond order = (bonding − antibonding electrons) ÷ 2), https://openstax.org/books/chemistry-2e/pages/8-4-molecular-orbital-theory (bond order formula)
  8. From Electron counts, Bonding electrons 8, Antibonding electrons 2 gives Bond order (number) 3, Bond order Bond order = 3.Source: OpenStax, Chemistry 2e, §8.4 Molecular Orbital Theory (bond order = (bonding − antibonding electrons) ÷ 2), https://openstax.org/books/chemistry-2e/pages/8-4-molecular-orbital-theory (bond order formula)

How it works

Formula. Bond order = (number of bonding electrons − number of antibonding electrons) ÷ 2.

Electron counts. Type the electrons in bonding and in antibonding orbitals (whole numbers from 0 to 40). The page refuses more antibonding than bonding electrons, which would give a bond order below 0. The answer names the counts: “8 bonding and 2 antibonding electrons”, or each with its own noun when a count is 1 (“3 bonding electrons and 1 antibonding electron”).

Diatomic molecule. Pick X₂ for one element from hydrogen to neon (atomic number Z from 1 to 10) and a charge q from −2 to +2. The molecule has 2Z − q electrons, which must be from 1 to 20. They fill the molecular orbitals in this order, lowest energy first, each holding 2 electrons (π orbitals come in pairs holding 4):

  • Li to N (Z ≤ 7, and H and He): σ1s, σ*1s, σ2s, σ*2s, π2p, σ2p, π*2p, σ*2p.
  • O, F and Ne (Z ≥ 8): σ1s, σ*1s, σ2s, σ*2s, σ2p, π2p, π*2p, σ*2p.

Orbitals with a star are antibonding; the others are bonding. Within a π pair, electrons go one per orbital before pairing (Hund’s rule), so k electrons in a pair (2 orbitals) leave k unpaired for k ≤ 2 and 4 − k for k > 2; a σ orbital with 1 electron has 1 unpaired.

Outputs. The bond order (whole or a half), the bonding and antibonding counts, the configuration written as (σ1s)² (σ*1s)² …, the unpaired electrons, and the magnetism: paramagnetic with any unpaired electron, diamagnetic with none. A bond order above 0 means a stable bond is expected; 0 means none (OpenStax).

Worked examples by hand

O₂ (OpenStax Example 8.6). 16 electrons, O ordering: (σ1s)² (σ*1s)² (σ2s)² (σ*2s)² (σ2p)² (π2p)⁴ (π*2p)². Bonding 2 + 2 + 2 + 4 = 10, antibonding 2 + 2 + 2 = 6. Bond order (10 − 6) ÷ 2 = 2. The two π*2p electrons are unpaired: paramagnetic.

N₂. 14 electrons, mixed ordering: (σ1s)² (σ*1s)² (σ2s)² (σ*2s)² (π2p)⁴ (σ2p)². Bonding 10, antibonding 4: bond order 3, no unpaired electrons, diamagnetic.

C₂²⁻ (Example 8.7). 2 × 6 + 2 = 14 electrons, the same filling as N₂: bonding 10, antibonding 4, bond order 3.

Be₂²⁻. 2 × 4 + 2 = 10 electrons: (σ1s)² (σ*1s)² (σ2s)² (σ*2s)² (π2p)². Bonding 6, antibonding 4: bond order 1; the two π2p electrons are unpaired (2), so it is paramagnetic.

Be₂. 8 electrons fill σ1s, σ*1s, σ2s and σ*2s: bonding 4, antibonding 4, bond order 0.

F₂. 18 electrons: bonding 10, antibonding 8, bond order 1.

H₂⁺. 1 electron in σ1s: bond order (1 − 0) ÷ 2 = 0.5, 1 unpaired electron.

Counts: 8 bonding and 2 antibonding (the valence electrons of N₂): (8 − 2) ÷ 2 = 3.

Other questions people ask

How do I calculate bond order?

Bond order = (bonding electrons − antibonding electrons) ÷ 2. Fill the molecular orbitals from the lowest energy, count the electrons in bonding orbitals and in antibonding (starred) orbitals, then halve the difference. For N₂, 10 bonding and 4 antibonding electrons give (10 − 4) ÷ 2 = 3.

What does the bond order tell me?

A bond order above 0 means the molecule should be stable, and a higher bond order means a stronger bond: N₂ (3) is stronger than O₂ (2), which is stronger than F₂ (1). A bond order of 0, as for He₂, Be₂ and Ne₂, means no stable bond forms.

Why is O₂ paramagnetic?

Its last two electrons go into the two π*2p orbitals, one in each (Hund’s rule), so O₂ has 2 unpaired electrons and is attracted into a magnetic field. A Lewis structure with a double bond does not show this; the molecular orbital picture does.

Why does the orbital order change between N₂ and O₂?

For Li₂ to N₂ the 2s and 2p orbitals are close in energy and mix (s-p mixing), which raises σ2p above the π2p pair. For O₂, F₂ and Ne₂ the mixing is negligible, so σ2p stays below π2p. The bond order is the same either way; the order changes which orbital a few electrons go into, and so the magnetism of B₂ and C₂.

Can a bond order be a fraction?

Yes. An odd number of electrons gives a half: H₂⁺ has 1 electron in σ1s, so its bond order is (1 − 0) ÷ 2 = 0.5; O₂⁺ has 2.5. A half bond order still means a bond, weaker than the next whole one.

Do the 1s electrons count?

They cancel. For Li₂ to Ne₂ the σ1s and σ*1s orbitals each hold 2 electrons, adding 2 bonding and 2 antibonding electrons, so leaving them out gives the same bond order. The page counts them so the configuration shows every electron.