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Bond order and magnetic behaviour for JEE: count, fill, check

Distinguish bond order from unpaired-electron count in simple homonuclear diatomic species.

Concept guide · About 4 min to read

TopAll Editorial · Source check: 23 October 2026

Published: 23 October 2026 · Updated: 23 October 2026

Technical review: Codex (automated technical review)

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Original concept diagram for Bond order and magnetic behaviour for JEE: count, fill, check.
Illustration: Bond order and magnetic behaviour for JEE: count, fill, check

What you’ll understand

Distinguish bond order from unpaired-electron count in simple homonuclear diatomic species.

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Quick answer

Use the appropriate molecular-orbital ordering for the species. Fill electrons with Pauli’s principle and Hund’s rule, then calculate bond order b=(Nb−Na)/2b=(N_b-N_a)/2. Determine magnetism from unpaired electrons, not from bond order: an unpaired electron indicates paramagnetism. The simple orbital picture below concerns ground-state homonuclear diatomic species.

Work through the reasoning

Example 1: oxygen and its ions. In O2O_2, the two highest occupied antibonding π∗\pi^* orbitals each hold one electron. Its bond order is 2 and it is paramagnetic. Adding one electron to form O2−O_2^- places it in an antibonding orbital: bond order becomes 1.5, with one unpaired electron. Adding a second to form O22−O_2^{2-} gives bond order 1 and pairs those electrons, giving diamagnetism in this elementary model.

Example 2: remove rather than add. Forming O2+O_2^+ removes an antibonding electron. Bond order rises from 2 to 2.5, while one unpaired electron remains. First identify the orbital that gains or loses an electron; an ion’s charge alone does not tell you whether bond order rises.

See the reasoning

Oxygen ions: bond order and magnetism

The highest two degenerate pi-star orbitals contain 1, 2, 3 and 4 electrons in O2+, O2, O2− and O2²− respectively. Bond orders are 2.5, 2, 1.5 and 1; unpaired counts are 1, 2, 1 and 0.Open full-size diagram ↗
These are occupancies of the highest π* pair, not complete molecular-orbital energy diagrams. In this valence-electron count, the bonding count stays 8; the antibonding counts are 3, 4, 5 and 6 respectively. Any unpaired electron gives paramagnetism in this elementary ground-state model.

Check it without looking: Explain why adding an electron lowers bond order by ½, then inspect whether it pairs an electron. Do not infer magnetism from bond order.

Common mistakes to catch

  • Wrong: a larger bond order must mean diamagnetism. Instead: inspect occupancy separately.
  • Wrong: reuse one energy-order diagram for every diatomic species. Instead: choose the ordering appropriate to the elements.
  • Wrong: pair electrons in degenerate orbitals before singly occupying them. Instead: apply Hund’s rule.

Try it yourself

1. Reveal the answer: removing an antibonding electron changes bond order by how much?

It increases bond order by 1/21/2, since NaN_a falls by one while NbN_b is unchanged. This does not by itself determine the number of unpaired electrons.

2. Reveal the answer: compare O₂⁻ and O₂²⁻.

O2−O_2^- has bond order 1.5 and one unpaired electron; O22−O_2^{2-} has bond order 1 and no unpaired electrons in the elementary ground-state molecular-orbital picture. Count occupancy before naming the magnetic behaviour.

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Sources and method

The NCERT source supports the underlying concepts. All examples, self-checks and their explanations here are original editorial exercises, not past-paper questions, frequency estimates or promises of marks. Technical review: Codex (automated); human subject review is not claimed.

  1. NCERT Chemistry: Chemical Bonding and Molecular Structure

    NCERT online textbook · Sections 4.7–4.8: molecular-orbital theory and homonuclear diatomic molecules · Checked 2026-10-23

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