Q. how to calculate bond order from lewis structure

Definition. Bond order is a measure of how many chemical bonds connect a pair of atoms. From a Lewis structure perspective, bond order is the average number of electron pairs (or bond multiplicity) between two atoms, taking resonance into account when necessary.

Step 1. Draw a correct Lewis structure or all equivalent resonance structures for the molecule or ion.

Detailed actions for Step 1.

1. Count total valence electrons for the molecule or ion. Include extra electrons for negative charge and subtract electrons for positive charge.

2. Connect atoms with single bonds to give a skeleton. Place remaining electrons as lone pairs to complete octets where appropriate. Use double or triple bonds to satisfy the octet rule if needed.

3. If more than one Lewis structure can be drawn by moving pi electrons without changing atom positions, write all equivalent resonance structures. These will be used to compute an average bond order.

Step 2. For a single, nonresonance Lewis structure determine the bond order between two atoms by counting the bond multiplicity directly.

Detailed actions for Step 2.

1. A single bond counts as 1. A double bond counts as 2. A triple bond counts as 3. For a given pair of atoms, the bond order is simply that count when there is only one Lewis structure relevant.

Step 3. For species with resonance, calculate bond order by averaging the bond multiplicities for that bond over all equivalent resonance structures.

Formula for resonance-averaged bond order.

\[ \text{Bond order} \;=\; \frac{\text{sum of bond multiplicities between the two atoms in each resonance form}}{\text{number of resonance forms}}. \]

Detailed actions for Step 3.

1. For each resonance structure, record the multiplicity of the bond of interest (1 for single, 2 for double, 3 for triple).

2. Add those multiplicities across all resonance forms.

3. Divide the sum by the number of resonance forms. The result is the bond order for that bond.

Step 4. Optional alternate method for diatomic molecules using molecular orbital (MO) theory.

Formula using MO electron counts.

\[ \text{Bond order}_{\text{MO}} \;=\; \frac{N_{b} – N_{a}}{2}, \]

where \(N_{b}\) is the number of electrons in bonding molecular orbitals and \(N_{a}\) is the number of electrons in antibonding molecular orbitals. This gives the same numerical bond order as the Lewis/resonance method for typical diatomics when MO occupancy is known.

Worked examples.

Example A. Nitrogen molecule, \( \text{N}_{2} \).

Lewis structure shows a triple bond between the two nitrogen atoms. Bond multiplicity is 3. Therefore

\[ \text{Bond order} \;=\; 3. \]

Example B. Ozone, \( \text{O}_{3} \).

There are two equivalent resonance structures. In each structure one O–O bond is double and the other is single. For either outer O‑central O bond, the multiplicities across the two resonance forms are 2 and 1. Therefore

\[ \text{Bond order} \;=\; \frac{2 + 1}{2} \;=\; 1.5. \]

Example C. Carbonate ion, \( \text{CO}_{3}^{2-} \).

There are three equivalent resonance structures. Each C–O bond is double in one structure and single in the other two. The bond multiplicities for a given C–O bond are 2, 1, and 1. Therefore

\[ \text{Bond order} \;=\; \frac{2 + 1 + 1}{3} \;=\; \frac{4}{3} \approx 1.33. \]

Example D. Benzene, \( \text{C}_{6}\text{H}_{6} \).

There are two main Kekulé resonance forms. Each C–C bond is double in one form and single in the other. Thus each C–C bond has multiplicities 2 and 1 across the two forms. Therefore

\[ \text{Bond order} \;=\; \frac{2 + 1}{2} \;=\; 1.5. \]

Quick reference.

1. If no resonance, bond order equals bond multiplicity: single 1, double 2, triple 3.

2. If resonance, average the multiplicities across equivalent resonance forms using the formula in Step 3.

3. For diatomic species where MO information is available, use the MO formula \( \text{Bond order}_{\text{MO}} = \frac{N_{b} – N_{a}}{2} \) as an alternative check.