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Chemistry · Glossary

What is Bond order?

Definition 14.9 University Chemistry — Year 2 · Chapter 14 — Molecular Orbitals of Diatomic Molecules

The bond order of a diatomic molecule is b=(nb−na)/2b = (n_b - n_a)/2, where nbn_b and nan_a are the numbers of electrons in bonding and antibonding orbitals.

Examples

Example 14.15 (Nitrogen and oxygen ions)

NX2\ce{N2} (I=15.58 eVI = 15.58\,\mathrm{eV}) loses a bonding electron: D0(NX2X+)=9.76+14.53−15.58=8.71 eVD_0(\ce{N2+}) = 9.76 + 14.53 - 15.58 = 8.71\,\mathrm{eV}, weaker than NX2\ce{N2}. OX2\ce{O2} (I=12.07 eVI = 12.07\,\mathrm{eV}) loses a π∗\pi^* electron: D0(OX2X+)=5.12+13.62−12.07=6.66 eVD_0(\ce{O2+}) = 5.12 + 13.62 - 12.07 = 6.66\,\mathrm{eV}, stronger than OX2\ce{O2}, in line with bond orders 2.5 and 2.5 against 3 and 2.

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