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Magnetic Properties and the Spin-Only Formula

Magnetic behaviour of transition-metal ions depends mainly on the number of unpaired electrons. Paramagnetic species have unpaired electrons, while diamagnetic species have all electrons paired.

In one line: First find the ion, then count the single d-electrons; the formula measures singles, not charge.

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Understand it

  • Unpaired electrons behave like tiny magnetic centres because each electron has spin.
  • In transition-metal ions, the number of unpaired d-electrons is therefore the main school-level reason for paramagnetism.
  • The spin-only magnetic moment uses only spin contribution and ignores orbital contribution, which is a reasonable approximation for many first transition-series ions in CBSE problems.
  • A species with n = 0 is diamagnetic and has no spin-only magnetic moment, while species with larger n values are more strongly paramagnetic.
  • Correct counting must be done for the ion, not the neutral atom, because cations lose ns electrons before (n-1)d electrons.

Example

Mn2+ has 3d5 configuration with five unpaired electrons. Its spin-only magnetic moment is √(5(5+2)) = √35 = about 5.92 BM.

Rule to remember

Spin-only formula: magnetic moment = √(n(n+2)) BM, where n is the number of unpaired electrons and BM means Bohr magneton.

Paramagnetic: n greater than 0; diamagnetic: n = 0.

See it worked out

Follow the example line by line, then use the same method on new questions.

Worked example

Find the magnetic moment of Fe3+.

  1. 1Fe has atomic number 26; Fe3+ is [Ar] 3d5 after removing two 4s electrons and one 3d electron.
  2. 2Five unpaired electrons give magnetic moment = √(5(7)) = √35 = about 5.92 BM.

Method you can reuse

Find oxidation state, write the ion configuration, remove ns electrons first, fill d-orbitals according to Hund's rule for the expected case, count unpaired electrons, and substitute n in √(n(n+2)) BM.

Visual tip

An orbital box diagram is useful for counting unpaired electrons, especially for d1 to d10 configurations.

Check yourself

Answer the question in your head before opening the answer.

Try it yourself

Calculate the spin-only magnetic moment of a d3 ion.

Think first, then show answer
For d3, there are 3 unpaired electrons. Magnetic moment = √(3(3+2)) = √15 = about 3.87 BM.

Common wrong answer

A common wrong answer is using n as oxidation number instead of number of unpaired electrons, leading to a wrong magnetic moment.

Don't mix this up

Students often count electrons in the neutral atom instead of the ion. For cations of transition metals, ns electrons are removed before (n-1)d electrons.

Exam tip

Always find the oxidation state first, write the d-configuration of the ion, count unpaired electrons, then substitute in the spin-only formula.

Write it in the exam

Each point is worth about one mark. Write point 1 for a 1-mark question and keep adding points as the marks go up.

How CBSE asks it

Questions may ask for magnetic moment, number of unpaired electrons, comparison of magnetic strength, or identification of para- and diamagnetic ions.
  1. 1

    Enough for 1 mark

    Magnetic behaviour of transition-metal ions depends mainly on the number of unpaired electrons. Paramagnetic species have unpaired electrons, while diamagnetic species have all electrons paired.

  2. 2

    Add for 2 marks

    Example: Mn2+ has 3d5 configuration with five unpaired electrons. Its spin-only magnetic moment is √(5(5+2)) = √35 = about 5.92 BM.

  3. 3

    Add for 3 marks

    Unpaired electrons behave like tiny magnetic centres because each electron has spin.

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