OlympIQ
Read before you practise · 5 min

Read MO Diagrams Before Ranking Bond Order or Stability

Bond order = (bonding electrons − antibonding electrons)/2, and every ranking or 'fractional bond order' question is just this formula applied to different electron counts across a diatomic and its ions.

NSEJS · Chemical Bonding and Molecular Structure · 18 linked questions · 100% coverage
Know this first

Chemistry questions first require the material, particles or reaction condition to be identified before a formula or rule is selected.

Observation · What the stem states or showsCondition · The factor that controls the change
Study deeper: NCERT Science · Chemical Bonding and Molecular Structure

Removing or adding an electron changes bond order predictably

Removing an electron from a bonding MO (as in O2 → O2+) increases bond order and shortens/strengthens the bond;

removing an electron from an antibonding MO (as in NO → NO+, since NO's highest electron sits in π*) also increases bond order — the direction of the shift depends on which type of orbital (bonding or antibonding) held the removed electron, not simply on whether an electron was added or removed.

Fractional bond order comes from odd electron counts in antibonding orbitals

O2- has bond order 1.5, O2²⁻ has bond order 1.0 — each extra antibonding electron beyond O2's own 2 subtracts 0.5 from bond order. Any species where the antibonding electron count is odd (relative to a filled/half-filled pattern) gives a fractional bond order; this is a fast way to scan a list of species for the fractional-bond-order ones without drawing every diagram.

Stability tracks bond order, with N2 as the classic exception to memorise

For most diatomics, higher bond order means greater stability (and shorter bond length) — but N2 itself has the HIGHEST bond order (3) of the N2/N2+/N2-/N2²⁻ series and both N2+ and N2- have equal, lower bond order (2.5), making N2 uniquely the most stable of that whole series, unlike O2 where O2+ is actually more stable than neutral O2.

OlympIQ exam lens

  • ✓Bond order = (bonding e- − antibonding e-)/2 — recompute from the actual MO filling, don't guess from formula.
  • ✓Removing an electron from an antibonding orbital increases bond order (e.g. NO→NO+); removing from a bonding orbital decreases it (opposite for some series — check which orbital type is affected).
  • ✓A fractional bond order signals an odd number of antibonding electrons relative to the parent diatomic.
  • ✓N2 is the bond-order/stability maximum of its own ion series; O2, by contrast, has O2+ as MORE stable than neutral O2 — these two series behave oppositely, don't assume the same pattern for both.
OlympIQ trap

Don't assume removing an electron always increases stability the same way across different diatomics — check whether the removed electron came from a bonding or antibonding orbital in THAT specific species' MO diagram; N2 and O2 behave oppositely on ionisation.

Ready to try the pattern?

100% coverage across the 18 linked questions.

Practise 18 linked questions
Sign in to practise — your XP, streak and analysis update as you solve.