ChemistryCBSE Class 12 · The d- and f-Block Elements

Magnetic Properties and Colour of Transition Metal Ions

Q: What is the spin-only magnetic moment formula in Class 12 Chemistry?

\mu = \sqrt{n(n+2)} BM, where n is the number of unpaired electrons in the ion.

Q: Why are transition metal compounds coloured?

Their partly filled d orbitals, split by surrounding ligands, let an electron absorb visible light and move to a higher d orbital. We see the complementary colour of the absorbed light.

Q: Why is Zn²⁺ colourless and diamagnetic?

Zn²⁺ is 3d^{10}: all d orbitals are full, so no d–d transition is possible and there are no unpaired electrons.

Q: Which 3d ions have a magnetic moment of 5.92 BM?

Ions with five unpaired electrons (d^5): Mn^{2+} and Fe^{3+}.

Q: Is the colour of KMnO₄ due to d–d transitions?

No. Manganese in KMnO_4 is +7 and d^0, so there are no d electrons; the colour arises from a different electron-transfer process.

Updated 15 October 2026 · Fact-checked

Transition metal ions with unpaired d electrons are paramagnetic; their spin-only magnetic moment is $\mu = \sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons. Most of these ions are also coloured because a d electron absorbs visible light and jumps to a higher d orbital; ions with $d^0$ or $d^{10}$ configurations are colourless.

Understand Magnetic Properties and Colour of Transition Metal Ions

Magnetic behaviour. Every electron spins and so behaves like a tiny magnet; it also moves in an orbital. When electrons are paired, their spins cancel. A substance with all electrons paired is diamagnetic: it is weakly repelled by a magnetic field. A substance with one or more unpaired electrons is paramagnetic: it is attracted into a field. Ferromagnetism, as in iron, is an extreme form of paramagnetism. Many transition metal ions are paramagnetic because their partly filled d subshell holds unpaired electrons.

Spin-only formula. In the first-row (3d) ions, the orbital contribution is largely quenched, so the magnetic moment comes almost entirely from spin. It is given by $μ = n (n + 2)$ BM (Bohr magnetons), where $n$ is the number of unpaired electrons. One unpaired electron gives 1.73 BM, two 2.83, three 3.87, four 4.90 and five 5.92 BM. The moment rises with $n$ , peaking for $d^{5}$ ions such as $M n^{2 +}$ and $F e^{3 +}$ . The process is reversible: from a measured moment you can find $n$ and identify the ion.

To count unpaired electrons, write the configuration of the ion correctly: remove the $4 s$ electrons first, then $3 d$ . $F e^{2 +}$ is $3 d^{6}$ (four unpaired), $C u^{2 +}$ is $3 d^{9}$ (one), $C r^{3 +}$ is $3 d^{3}$ (three). For free ions in water, fill the five d orbitals singly before pairing (Hund's rule).

Colour. Visible light covers wavelengths from roughly 400 to 700 nm. When a transition metal ion is surrounded by ligands (water molecules in an aqueous solution), its five d orbitals no longer have the same energy: they split into groups. An electron in a lower d orbital can absorb light of just the right energy and jump to a higher d orbital: a d–d transition. The light absorbed is in the visible region, so the colour we see is the complementary colour of the light absorbed. If red light is absorbed, the solution looks blue-green; $[C u (H_{2} O)_{6}]^{2 +}$ looks blue.

The exact colour depends on the size of the energy gap, which depends on the ligand. That is why the same ion can show different colours with different ligands, and why anhydrous $C u S O_{4}$ (no water ligands) is white while $C u S O_{4} \cdot 5 H_{2} O$ is blue.

Colourless ions. If there are no d electrons ( $d^{0}$ : $S c^{3 +}$ , $T i^{4 +}$ ) or the d subshell is full ( $d^{10}$ : $Z n^{2 +}$ , $C u^{+}$ ), no d–d transition is possible, so these ions are colourless. Colours of some aqueous ions (NCERT): $T i^{3 +}$ purple, $V^{3 +}$ green, $C r^{3 +}$ violet, $M n^{2 +}$ pink, $M n^{3 +}$ violet, $F e^{2 +}$ green, $F e^{3 +}$ yellow, $C o^{2 +}$ pink, $N i^{2 +}$ green, $C u^{2 +}$ blue. The intense colours of $K M n O_{4}$ and $K_{2} C r_{2} O_{7}$ are not from d–d transitions (Mn(VII) and Cr(VI) are $d^{0}$ ); they come from a different kind of electron transfer between oxygen and the metal.

Magnetic Properties and Colour of Transition Metal Ions: formulas and key facts

Spin-only magnetic moment: $μ = n (n + 2) BM$; $n$ = number of unpaired electrons; 1 BM = Bohr magneton.
Spin-only values: $n = 1, 2, 3, 4, 5 \Rightarrow μ = 1.73, 2.83, 3.87, 4.90, 5.92 BM$; Learn these to identify ions from data.
Unpaired electrons from d count (free ion): $n = d for d^{1} - d^{5}; n = 10 - d for d^{6} - d^{10}$; Fill singly, then pair (Hund's rule).
Finding n from μ: $n^{2} + 2 n - μ^{2} = 0 \Rightarrow n = - 1 + 1 + μ^{2}$; Round to the nearest whole number.
Condition for colour (d–d): $d^{1} to d^{9} \Rightarrow coloured; d^{0}, d^{10} \Rightarrow colourless$; Colour seen = complementary colour of light absorbed.

How to solve Magnetic Properties and Colour of Transition Metal Ions questions

Questions here are short: calculate a magnetic moment, identify an ion from its moment, arrange ions by moment, or explain colour. Each starts from the configuration of the ion.

1Find the number of 3d electrons in the ion: atomic number − 18 − charge (for 3d ions with charge ≥ 2; $C u^{+}$ is $3 d^{10}$ ).
2Count unpaired electrons: $n = d$ up to $d^{5}$ ; $n = 10 - d$ from $d^{6}$ to $d^{10}$ .
3Apply $μ = n (n + 2)$ BM and give two decimal places.
4To identify an ion from $μ$ , solve for $n$ , then list the 3d ions with that many unpaired electrons and use any other clue (charge, colour).
5For arrangements, compare $n$ directly; equal $n$ gives equal spin-only moments.
6For colour, check whether the ion is $d^{0}$ or $d^{10}$ (colourless) or has partly filled d orbitals (coloured by d–d transitions, with ligands present).
7State 'paramagnetic' or 'diamagnetic' alongside the moment when the question asks about magnetic behaviour.

Magnetic Properties and Colour of Transition Metal Ions in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Paramagnetism depends on unpaired electrons

Suppose a teacher hangs small packets of blue copper sulphate crystals ( $C u S O_{4} \cdot 5 H_{2} O$ ) and white zinc sulphate crystals ( $Z n S O_{4} \cdot 7 H_{2} O$ ) from threads next to a strong magnet. One packet is pulled slightly towards the magnet; the other is not.

Predict which packet is attracted, calculate the spin-only magnetic moment of each metal ion, and link the result to the colours of the crystals.

Try it first, then show the solutionSolution

$C u^{2 +}$ : Cu is $3 d^{10} 4 s^{1}$ ; $C u^{2 +}$ is $3 d^{9}$ , one unpaired electron, $μ = 1 \times 3 = 1.73$ BM: paramagnetic.
$Z n^{2 +}$ : $3 d^{10}$ , no unpaired electrons, $μ = 0$ : diamagnetic.
So the copper sulphate packet is attracted.
The same partly filled $d^{9}$ subshell allows a d–d transition, so copper sulphate is blue; $Z n^{2 +}$ ( $d^{10}$ ) cannot, so zinc sulphate is white.

Answer: Copper sulphate (1.73 BM, paramagnetic, blue); zinc sulphate (0 BM, diamagnetic, white).

Memory hook: An unpaired electron gives an ion both a pull towards a magnet and a colour.

Tricks to remember Magnetic Properties and Colour of Transition Metal Ions

Short memory aids. Say each one aloud once, then use it on a question the same day.

1.73, 2.83, 3.87, 4.90, 5.92
Read as 'one-seventy-three, two-eighty-three, three-eighty-seven, four-ninety, five-ninety-two': the first digit always matches $n$ .
Use it when: Instant identification of $n$ from a given moment.
Empty or full means no colour
$d^{0}$ and $d^{10}$ : nowhere to jump or no room to jump, so no colour (Sc³⁺, Ti⁴⁺, Zn²⁺, Cu⁺).
Use it when: Colourless-ion MCQs.
Five is the peak
Unpaired electrons climb from $d^{1}$ to $d^{5}$ and fall back to $d^{10}$ ; $M n^{2 +}$ and $F e^{3 +}$ top the chart at 5.92 BM.
Use it when: Arranging ions by magnetic moment.

Magnetic Properties and Colour of Transition Metal Ions: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: The colourless ion · 1 mark · MCQBoard level

Which of the following ions is colourless in aqueous solution?
(a) $T i^{3 +}$ (b) $S c^{3 +}$ (c) $C r^{3 +}$ (d) $M n^{2 +}$

Try it first, then show the solutionSolution

$S c^{3 +}$ is $3 d^{0}$ : no d electron, so no d–d transition.

Answer: (b) $S c^{3 +}$ .

Example 2: Spin-only moments · 2 marks · Very short answerBoard level

Calculate the spin-only magnetic moments of $F e^{2 +}$ (Z of Fe = 26) and $C r^{3 +}$ (Z of Cr = 24).

Try it first, then show the solutionSolution

$F e^{2 +}$ : $3 d^{6}$ , 4 unpaired; $μ = 4 \times 6 = 24 = 4.90$ BM (1 mark).
$C r^{3 +}$ : $3 d^{3}$ , 3 unpaired; $μ = 3 \times 5 = 15 = 3.87$ BM (1 mark).

Answer: $F e^{2 +}$ : 4.90 BM; $C r^{3 +}$ : 3.87 BM.

Example 3: Identifying ions from their moments · 3 marks · Short answerHarder board level

(a) A divalent ion of the 3d series has a spin-only magnetic moment of 5.92 BM. Identify it. (b) A trivalent 3d ion has a moment of 3.87 BM. Identify it. Show your reasoning.

Try it first, then show the solutionSolution

(a) $n (n + 2) = 5.92$ gives $n = 5$ , i.e. a $d^{5}$ ion (1 mark).
For $M^{2 +}$ , $d^{5}$ means the atom has $5 + 2 = 7$ electrons beyond argon, i.e. Z = 25: $M n^{2 +}$ (½ mark).
(b) 3.87 BM gives $n = 3$ , a $d^{3}$ ion. For $M^{3 +}$ : $3 + 3 = 6$ electrons beyond argon, Z = 24: $C r^{3 +}$ (1 mark).
Check: $F e^{3 +}$ is also $d^{5}$ but trivalent; $V^{2 +}$ is also $d^{3}$ but divalent (½ mark).

Answer: (a) $M n^{2 +}$ . (b) $C r^{3 +}$ .

Example 4: Ordering by magnetic moment · 3 marks · Short answerHarder board level

Arrange $V^{3 +}$ , $C r^{3 +}$ , $M n^{2 +}$ and $C u^{2 +}$ in increasing order of spin-only magnetic moment, giving the value for each. (Z: V 23, Cr 24, Mn 25, Cu 29)

Try it first, then show the solutionSolution

$V^{3 +}$ : $3 d^{2}$ , 2 unpaired, 2.83 BM. $C r^{3 +}$ : $3 d^{3}$ , 3 unpaired, 3.87 BM (1 mark).
$M n^{2 +}$ : $3 d^{5}$ , 5 unpaired, 5.92 BM. $C u^{2 +}$ : $3 d^{9}$ , 1 unpaired, 1.73 BM (1 mark).
Order: $C u^{2 +} < V^{3 +} < C r^{3 +} < M n^{2 +}$ (1 mark).

Answer: $C u^{2 +}$ (1.73) < $V^{3 +}$ (2.83) < $C r^{3 +}$ (3.87) < $M n^{2 +}$ (5.92 BM).

Example 5: Copper: white, blue and colourless · 3 marks · Short answerHOTS / competency

Explain: (a) $C u S O_{4} \cdot 5 H_{2} O$ is blue but anhydrous $C u S O_{4}$ is white. (b) Copper(I) compounds such as CuCl are white and diamagnetic. (c) A solution of $C u^{2 +}$ looks blue: what colour region of light is it absorbing?

Try it first, then show the solutionSolution

(a) In the hydrate, water molecules act as ligands around $C u^{2 +}$ ( $d^{9}$ ) and split the d orbitals; a d–d transition absorbs visible (red-orange) light, so it looks blue. Without water ligands, the splitting needed for a visible-region absorption is absent, so anhydrous $C u S O_{4}$ is white (1 mark).
(b) $C u^{+}$ is $3 d^{10}$ : completely filled, no d–d transition (colourless) and no unpaired electron (diamagnetic) (1 mark).
(c) The colour seen is complementary to the colour absorbed: blue is seen, so mainly orange-red light is absorbed (1 mark).

Answer: (a) Water ligands make d–d absorption possible. (b) $d^{10}$ : no transition, no unpaired electron. (c) Orange-red light.

Where students lose marks in Magnetic Properties and Colour of Transition Metal Ions

The slips examiners see most often, and what to write instead.

Mistake: Removing 3d electrons before 4s
Wrong working
$F e^{2 +}$ : $3 d^{4} 4 s^{2}$ , 4 unpaired… or $C u^{2 +}$ : $3 d^{8} 4 s^{1}$
Right working
$F e^{2 +}$ : $3 d^{6}$ ; $C u^{2 +}$ : $3 d^{9}$
Do this: Cations lose 4s electrons first.
Usually costs: The whole calculation.
Mistake: Counting d electrons instead of unpaired electrons
Wrong working
$N i^{2 +}$ ( $d^{8}$ ): $μ = 8 \times 10$
Right working
$d^{8}$ has 2 unpaired: $μ = 2 \times 4 = 2.83$ BM
Do this: Above $d^{5}$ , unpaired electrons $= 10 - d$ .
Usually costs: 1 mark.
Mistake: Saying colour is due to 'unpaired electrons'
Mixing up magnetism and colour.
Do this: Colour comes from d–d transitions in partly filled d orbitals split by ligands; magnetism comes from unpaired electrons.
Usually costs: ½ to 1 mark.
Mistake: Explaining $K M n O_{4}$ 's colour by d–d transitions
Do this: Mn(VII) is $d^{0}$ ; its colour is not from d–d transitions.
Usually costs: 1 mark in reasoning questions.

Exam tips

Always write the configuration of the ion first; examiners give a step mark for it.
Quote $μ$ with units (BM) and two decimal places.
For colour answers use the phrase 'd–d transition' and mention 'complementary colour'.
In identification questions, check both the number of unpaired electrons and the charge given.
Remember the colourless set: $S c^{3 +}$ , $T i^{4 +}$ , $C u^{+}$ , $Z n^{2 +}$ .

Practice questions on Magnetic Properties and Colour of Transition Metal Ions

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The spin-only magnetic moment of $T i^{3 +}$ (Z of Ti = 22) is:
(a) 0 BM (b) 1.73 BM (c) 2.83 BM (d) 3.87 BM
Hint
$T i^{3 +}$ is $3 d^{1}$ .
Answer
(b) 1.73 BM.
Q2 · 2 marks · Very short answerBoard level
Why are most transition metal ions coloured? Name one 3d ion that is colourless and give the reason.
Hint
d–d transitions.
Answer
Their partly filled d orbitals, split in energy by ligands, allow an electron to absorb visible light and jump to a higher d level (d–d transition); the complementary colour is seen. $Z n^{2 +}$ ( $3 d^{10}$ ) or $S c^{3 +}$ ( $3 d^{0}$ ) is colourless because no d–d transition is possible.
Q3 · 2 marks · Very short answerHarder board level
Calculate the spin-only magnetic moments of $C o^{2 +}$ (Z = 27) and $N i^{2 +}$ (Z = 28).
Hint
$d^{7}$ and $d^{8}$ .
Answer
$C o^{2 +}$ : $3 d^{7}$ , 3 unpaired, $15 = 3.87$ BM. $N i^{2 +}$ : $3 d^{8}$ , 2 unpaired, $8 = 2.83$ BM.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Salts of $Z n^{2 +}$ are white.
Reason (R): $Z n^{2 +}$ has a completely filled 3d subshell.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
No d–d transitions in $d^{10}$ .
Answer
(a).
Q5 · 2 marks · Very short answerHarder board level
The magnetic moment of a 3d ion is found to be 2.84 BM. How many unpaired electrons does it have? Name one divalent and one trivalent ion that fit.
Hint
$n (n + 2) \approx 8$ .
Answer
$n = 2$ . Divalent: $N i^{2 +}$ ( $d^{8}$ ) (or $T i^{2 +}$ , $d^{2}$ ). Trivalent: $V^{3 +}$ ( $d^{2}$ ).
Q6 · 3 marks · Short answerHarder board level
Which of the following ions are paramagnetic: $S c^{3 +}$ , $T i^{3 +}$ , $C u^{+}$ , $N i^{2 +}$ ? Justify each, and say what 'diamagnetic' means.
Hint
Count unpaired electrons.
Answer
$T i^{3 +}$ ( $d^{1}$ , 1 unpaired) and $N i^{2 +}$ ( $d^{8}$ , 2 unpaired) are paramagnetic. $S c^{3 +}$ ( $d^{0}$ ) and $C u^{+}$ ( $d^{10}$ ) have no unpaired electrons and are diamagnetic. A diamagnetic substance has all electrons paired and is weakly repelled by a magnetic field.
Q7 · 3 marks · Short answerHOTS / competency
A 3d metal ion in aqueous solution is pale green and has a spin-only magnetic moment of 4.90 BM. On standing in air the solution turns yellow-brown. Identify the ion before and after, and give the magnetic moment of the product ion. Why does the change happen readily?
Hint
4 unpaired electrons, green; the product is yellow.
Answer
4.90 BM means 4 unpaired ( $d^{4}$ or $d^{6}$ ); a green ion with $d^{6}$ is $F e^{2 +}$ . Air oxidises it to $F e^{3 +}$ (yellow), $3 d^{5}$ , 5 unpaired, 5.92 BM. The change is easy because $F e^{3 +}$ has the stable half-filled $d^{5}$ configuration.
Q8 · 4 marks · Case-basedHOTS / competency
Read and answer.
Suppose a student prepares two solutions: titanium(III) chloride, which is purple, and titanium(IV) chloride in a non-aqueous solvent, which is colourless. She tests both with a sensitive balance placed near a magnet.
(a) Write the 3d configurations of $T i^{3 +}$ and $T i^{4 +}$ . (1 mark)
(b) Which solution is attracted by the magnet? Calculate its spin-only moment. (1 mark)
(c) Explain why one is coloured and the other is not. (1 mark)
(d) The purple solution absorbs light mainly in the yellow-green region. Why does it look purple? (1 mark)
Hint
Ti is $3 d^{2} 4 s^{2}$ .
Answer
(a) $T i^{3 +}$ : $3 d^{1}$ ; $T i^{4 +}$ : $3 d^{0}$ . (b) $T i C l_{3}$ (Ti³⁺), $μ = 1.73$ BM. (c) $T i^{3 +}$ has one d electron that can undergo a d–d transition; $T i^{4 +}$ has none. (d) The colour seen is the complementary colour of the light absorbed; removing yellow-green leaves a purple appearance.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Spin-only formula?Show answer
$μ = n (n + 2)$ BM.
2.μ for 4 unpaired electrons?Show answer
4.90 BM.
3.Which 3d divalent ion has the highest spin-only moment?Show answer
$M n^{2 +}$ (5.92 BM).
4.Colour of aqueous $N i^{2 +}$ ?Show answer
Green.
5.Why is $C u^{+}$ colourless?Show answer
It is $3 d^{10}$ ; no d–d transition is possible.
6.Paramagnetic or diamagnetic: $S c^{3 +}$ ?Show answer
Diamagnetic.

Magnetic Properties and Colour of Transition Metal Ions: frequently asked questions

What is the spin-only magnetic moment formula in Class 12 Chemistry?

\mu = \sqrt{n(n+2)} BM, where n is the number of unpaired electrons in the ion.

Why are transition metal compounds coloured?

Their partly filled d orbitals, split by surrounding ligands, let an electron absorb visible light and move to a higher d orbital. We see the complementary colour of the absorbed light.

Why is Zn²⁺ colourless and diamagnetic?

Zn²⁺ is 3d^{10}: all d orbitals are full, so no d–d transition is possible and there are no unpaired electrons.

Which 3d ions have a magnetic moment of 5.92 BM?

Ions with five unpaired electrons (d^5): Mn^{2+} and Fe^{3+}.

Is the colour of KMnO₄ due to d–d transitions?

No. Manganese in KMnO_4 is +7 and d^0, so there are no d electrons; the colour arises from a different electron-transfer process.