ChemistryISC Class 12 · d- and f-Block Elements

Magnetic Properties and Colour of Transition Metal Ions (ISC Class 12)

Q: How do you calculate the magnetic moment of a transition metal ion?

Write the d configuration, count unpaired electrons n by Hund's rule, and use \mu = \sqrt{n(n+2)} BM.

Q: Why are transition metal ions coloured?

Their partly filled d orbitals are split by ligands, and d-d transitions absorb visible light; we see the complementary colour.

Q: Why is anhydrous copper sulphate white?

Without water ligands the d orbitals of Cu²⁺ are not split, so no d-d transition occurs in the visible region.

Q: What is the difference between paramagnetic and diamagnetic?

Paramagnetic substances have unpaired electrons and are attracted by a magnetic field; diamagnetic substances have all electrons paired and are weakly repelled.

Q: Why is Cu considered a transition element though Cu atom is 3d¹⁰?

Its common ion Cu²⁺ has an incompletely filled 3d^9 configuration.

Updated 15 October 2026 · Fact-checked

Transition metal ions with unpaired d electrons are paramagnetic; their spin-only magnetic moment is μ = √(n(n+2)) BM, where n is the number of unpaired electrons. Ions with partly filled d orbitals are coloured because d-d transitions absorb visible light. Ions with d⁰ or d¹⁰ configurations are colourless and diamagnetic.

Each extra unpaired electron raises μ = √(n(n+2)); the maximum, 5.92 BM, belongs to d⁵ ions like Mn²⁺ and Fe³⁺.

Illustration drawn for QuestionBuddy.

Understand Magnetic Properties and Colour of Transition Metal Ions

What counts as a transition element. A transition element has an incompletely filled d subshell in its atom or in any of its common ions. Zn, Cd and Hg have $(n - 1) d^{10} n s^{2}$ configurations and keep $d^{10}$ in their common ions (Zn²⁺, Cd²⁺, Hg²⁺), so they are d-block elements but not regarded as transition elements. Cu ( $3 d^{10} 4 s^{1}$ ) is a transition element because Cu²⁺ is $3 d^{9}$ ; similarly Ag shows +2 with an incomplete 4d set. To write the configuration of a first-series ion, remove the 4s electrons first, then 3d: Fe ( $3 d^{6} 4 s^{2}$ ) → Fe²⁺ ( $3 d^{6}$ ) → Fe³⁺ ( $3 d^{5}$ ).

Magnetic behaviour. In a magnetic field, diamagnetic substances are weakly repelled and paramagnetic ones are attracted; strongly attracted substances are ferromagnetic (an extreme form of paramagnetism). Paramagnetism comes from unpaired electrons, each of which behaves as a tiny magnet. For first-series ions the orbital contribution is effectively quenched, so the moment depends only on the number of unpaired electrons $n$ : the spin-only formula is $μ = n (n + 2)$ BM (Bohr magnetons). Values: $n$ = 1, 1.73; 2, 2.83; 3, 3.87; 4, 4.90; 5, 5.92 BM. The moment rises with $n$ , reaching its maximum for $d^{5}$ ions (Mn²⁺, Fe³⁺). Working backwards, a measured moment tells you $n$ and hence the configuration and oxidation state. Observed values for hydrated ions are close to the spin-only values (for Fe²⁺, Co²⁺ and Ni²⁺ they are a little higher).

Unpaired electrons in free ions (Hund's rule). Fill five d orbitals singly before pairing: $d^{1}$ – $d^{5}$ give 1 to 5 unpaired; $d^{6}$ 4; $d^{7}$ 3; $d^{8}$ 2; $d^{9}$ 1; $d^{0}$ and $d^{10}$ none.

Why ions are coloured. In an isolated ion the five d orbitals have equal energy. When ligands (water molecules in aqueous solution) surround the ion, the d orbitals split into groups of slightly different energy. An electron in a lower d orbital can absorb light of a particular frequency, usually in the visible region, and jump to a higher d orbital: a d-d transition. We see the complementary colour of the light absorbed. The frequency absorbed depends on the ligand, so the same ion can have different colours with different ligands, and with no ligands there is no splitting: anhydrous CuSO₄ is white while hydrated CuSO₄·5H₂O is blue.

When there is no colour. A d-d transition needs a partly filled d set. ** $d^{0}$ ions (Sc³⁺, Ti⁴⁺) have no d electron to excite; $d^{10}$ ** ions (Zn²⁺, Cu⁺, Ag⁺, Cd²⁺) have no vacant d orbital to receive one. Both are colourless and, having no unpaired electrons, diamagnetic. Colours of some aquated ions: Ti³⁺ purple, V³⁺ green, Cr³⁺ violet, Mn²⁺ pale pink, Fe²⁺ green, Fe³⁺ yellow, Co²⁺ pink, Ni²⁺ green, Cu²⁺ blue. Intensely coloured species such as $M n O_{4}^{-}$ and $C r_{2} O_{7}^{2 -}$ have $d^{0}$ metals; their colour comes from charge transfer, not d-d transitions.

Copper(I) and copper(II). Cu⁺ ( $3 d^{10}$ ) compounds such as CuCl are white and diamagnetic; Cu²⁺ ( $3 d^{9}$ ) salts are coloured and paramagnetic (one unpaired electron, 1.73 BM). In water Cu⁺ disproportionates, $2 C u^{+} \to C u^{2 +} + C u$ , because the much more negative hydration enthalpy of the smaller, more highly charged Cu²⁺ more than makes up for the second ionisation enthalpy.

Related properties. The same partly filled d orbitals explain complex formation (small, highly charged ions with vacant d orbitals of suitable energy accept ligand lone pairs), catalytic activity (variable oxidation states and surfaces that adsorb reactants) and alloy formation (similar atomic sizes let one metal replace another in the lattice).

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.
Values to remember: $n = 1, 2, 3, 4, 5 \Rightarrow μ = 1.73, 2.83, 3.87, 4.90, 5.92 BM$; Maximum for $d^{5}$ (Mn²⁺, Fe³⁺).
Unpaired electrons in free 3d ions: $d^{0} : 0, d^{1} : 1, d^{2} : 2, d^{3} : 3, d^{4} : 4, d^{5} : 5, d^{6} : 4, d^{7} : 3, d^{8} : 2, d^{9} : 1, d^{10} : 0$; Hund's rule: singly fill all five d orbitals before pairing.
Condition for colour (d-d): $d^{1} to d^{9} \Rightarrow coloured; d^{0}, d^{10} \Rightarrow colourless$; Colour seen is complementary to the light absorbed.
Configuration of a first-series ion: $M^{x +} : 3 d^{(Z - 18 - x)} (4s emptied first)$; For Cr and Cu use the atom configurations $3 d^{5} 4 s^{1}$ and $3 d^{10} 4 s^{1}$ ; the ion rule still gives the right count.

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

Every question here follows the same route: configuration of the ion, count unpaired electrons, then magnetic moment or colour.

1Find the number of d electrons in the ion: Z − 18 − charge for the 3d series (4s electrons are removed first).
2Place the electrons in five d boxes by Hund's rule and count the unpaired ones, $n$ .
3Substitute in $μ = n (n + 2)$ and give the answer in BM to two decimal places.
4If μ is given, square it: $n^{2} + 2 n = μ^{2}$ ; solve for $n$ (or match with 1.73, 2.83, 3.87, 4.90, 5.92), then deduce the configuration and the ion.
5For colour, check for a partly filled d set: $d^{0}$ and $d^{10}$ are colourless; others are coloured by d-d transitions.
6In 'give a reason' answers, write the configuration, the number of unpaired electrons, and the words 'd-d transition' or 'paramagnetic' as appropriate.

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.

Ligands are needed for d-d colour

Suppose a student heats blue copper(II) sulphate crystals, CuSO₄·5H₂O, in a test tube. They turn white. When she lets a few drops of water fall on the cooled white powder, it turns blue again and the tube becomes warm.

Explain both colour changes, and say whether the white powder is paramagnetic. Calculate the spin-only magnetic moment of the copper ion in it.

Try it first, then show the solutionSolution

Heating removes the water ligands; without ligands around Cu²⁺ the d orbitals are not split, so no d-d transition occurs in the visible region: the anhydrous salt is white.
Adding water puts water ligands back around Cu²⁺; the d orbitals split again, a d-d transition absorbs part of visible light and the complementary blue colour returns.
Cu²⁺ is $3 d^{9}$ in both forms, with one unpaired electron, so the white powder is still paramagnetic.
$μ = 1 (1 + 2) = 3 = 1.73$ BM.

Answer: Water ligands cause the d-orbital splitting that gives the blue colour; removing them makes the salt white. Cu²⁺ ( $3 d^{9}$ ) stays paramagnetic, μ = 1.73 BM.

Memory hook: No ligands, no splitting, no d-d colour, but the unpaired electron (and the magnetism) stays.

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.

Root of n times n plus two
1.73, 2.83, 3.87, 4.90, 5.92: each value is a little less than $n + 1$ . Learn the list like a times table.
Use it when: Quick magnetic moment answers and working backwards from μ.
Zero and ten are boring
$d^{0}$ and $d^{10}$ ions have nothing to excite (or nowhere to go): no colour, no magnetism. Sc³⁺, Ti⁴⁺, Zn²⁺, Cu⁺.
Use it when: Spotting colourless, diamagnetic ions.
4s leaves first
When a 3d atom forms an ion, the 4s electrons leave before any 3d electrons: Fe²⁺ is $3 d^{6}$ , not $3 d^{4} 4 s^{2}$ .
Use it when: Writing ion configurations correctly.

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: Which ion has 5.92 BM? · 1 mark · MCQBoard level

Which of the following ions has a spin-only magnetic moment of about 5.92 BM? (Z: Fe 26, Co 27, Cu 29, Zn 30)
(a) Fe³⁺
(b) Co³⁺
(c) Cu²⁺
(d) Zn²⁺

Try it first, then show the solutionSolution

5.92 BM needs 5 unpaired electrons, i.e. a $d^{5}$ ion.
Fe³⁺ is $3 d^{5}$ ; Co³⁺ $3 d^{6}$ (4 unpaired), Cu²⁺ $3 d^{9}$ (1), Zn²⁺ $3 d^{10}$ (0).

Answer: (a) Fe³⁺.

Example 2: Two spin-only moments · 2 marks · Short answerBoard level

Calculate the spin-only magnetic moments of Cu²⁺ (Z = 29) and Co³⁺ (Z = 27) as free ions. Which of them would you expect to be coloured in aqueous solution?

Try it first, then show the solutionSolution

Cu²⁺: $3 d^{9}$ , $n = 1$ , $μ = 3 = 1.73$ BM (½ mark).
Co³⁺: $3 d^{6}$ , $n = 4$ , $μ = 24 = 4.90$ BM (1 mark).
Both have partly filled d orbitals, so both are coloured by d-d transitions (½ mark).

Answer: Cu²⁺ 1.73 BM; Co³⁺ 4.90 BM; both coloured.

Example 3: Identify an ion from its magnetic moment · 3 marks · Short answerHarder board level

A tripositive ion M³⁺ of the first transition series has a spin-only magnetic moment of 3.87 BM. Find the number of unpaired electrons, the d configuration and the identity of M. Is the ion coloured?

Try it first, then show the solutionSolution

$n (n + 2) = 3.8 7^{2} = 15$ , so $n = 3$ (1 mark).
Three unpaired electrons fit $3 d^{3}$ (a $3 d^{7}$ ion also has three, but as M³⁺ that would be the uncommon Ni³⁺), so take $3 d^{3}$ (½ mark).
$3 d^{3}$ for M³⁺ means the atom has $Z = 18 + 3 + 3 = 24$ : chromium, Cr³⁺ (1 mark).
Cr³⁺ has a partly filled d set, so it is coloured (violet in water) (½ mark).

Answer: $n = 3$ ; $3 d^{3}$ ; M = Cr (Cr³⁺), coloured.

Example 4: Copper(I) versus copper(II) · 3 marks · Short answerHarder board level

(a) Explain why copper(II) salts are coloured and paramagnetic while copper(I) compounds are colourless and diamagnetic. (Z of Cu = 29) (b) Why does Cu⁺ not survive in aqueous solution? Write the equation.

Try it first, then show the solutionSolution

(a) Cu²⁺ is $3 d^{9}$ : one unpaired electron (paramagnetic) and a partly filled d set, so d-d transitions give colour (1 mark).
Cu⁺ is $3 d^{10}$ : no unpaired electron (diamagnetic) and no vacant d orbital, so no d-d transition and no colour (1 mark).
(b) Cu⁺ disproportionates: $2 C u^{+} (a q) \to C u^{2 +} (a q) + C u (s)$ , because the much more negative hydration enthalpy of Cu²⁺ outweighs its second ionisation enthalpy (1 mark).

Answer: (a) $3 d^{9}$ versus $3 d^{10}$ . (b) $2 C u^{+} \to C u^{2 +} + C u$ , driven by the high hydration enthalpy of Cu²⁺.

Example 5: Patterns across the 3d series · 5 marks · Long answerHOTS / competency

Consider the aquated ions Sc³⁺, Ti³⁺, V³⁺, Mn²⁺, Fe²⁺, Ni²⁺, Cu²⁺ and Zn²⁺. (Z: Sc 21, Ti 22, V 23, Mn 25, Fe 26, Ni 28, Cu 29, Zn 30)
(a) Which ion has the highest spin-only magnetic moment? Calculate it. [1]
(b) Calculate the spin-only moments of Fe²⁺ and Ni²⁺. [2]
(c) Which ions are colourless? Explain. [1]
(d) An iron salt has a measured moment close to 5.9 BM. What is the oxidation state of iron in it? [1]

Try it first, then show the solutionSolution

(a) Mn²⁺ ( $3 d^{5}$ , 5 unpaired): $μ = 35 = 5.92$ BM (1 mark).
(b) Fe²⁺ $3 d^{6}$ , 4 unpaired: $24 = 4.90$ BM (1 mark). Ni²⁺ $3 d^{8}$ , 2 unpaired: $8 = 2.83$ BM (1 mark).
(c) Sc³⁺ ( $3 d^{0}$ ) and Zn²⁺ ( $3 d^{10}$ ): no d-d transition possible (1 mark).
(d) 5.9 BM means 5 unpaired electrons, $3 d^{5}$ : iron is +3 (Fe³⁺) (1 mark).

Answer: (a) Mn²⁺, 5.92 BM. (b) Fe²⁺ 4.90 BM; Ni²⁺ 2.83 BM. (c) Sc³⁺ and Zn²⁺. (d) +3.

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: Choosing 'incompletely filled' and 'paramagnetic' for colourless ions
ISC examiners reported this in a 2025 fill-in-the-blanks item.
Do this: Colourless ions have empty ( $d^{0}$ ) or completely filled ( $d^{10}$ ) d orbitals and are diamagnetic.
Usually costs: 1 mark.
Mistake: Not mentioning d-d transition
In 2024 and 2025, explanations of coloured Cu²⁺ and Ti³⁺ salts often left out the d-d transition or confused paired and unpaired electrons.
Wrong working
Ti³⁺ is coloured because it has a charge of +3.
Right working
Ti³⁺ is $3 d^{1}$ : the single d electron can undergo a d-d transition by absorbing visible light, so it is coloured; Ti⁴⁺ ( $3 d^{0}$ ) cannot.
Do this: Always write the configuration and the phrase 'd-d transition'.
Usually costs: 1 mark per part.
Mistake: Wrong reason why Zn, Cd and Hg are not transition elements
Examiners noted confusion between partially and completely filled d orbitals (2024).
Do this: State that Zn, Cd and Hg have completely filled $(n - 1) d^{10}$ orbitals in the atom and in their common ions.
Usually costs: 1 mark.
Mistake: Removing 3d electrons before 4s
Students write Fe²⁺ as $3 d^{4} 4 s^{2}$ .
Wrong working
Fe²⁺: $3 d^{4} 4 s^{2}$ , 4 unpaired.
Right working
Fe²⁺: $3 d^{6}$ , 4 unpaired.
Do this: Empty 4s first. The count may agree by chance, but the configuration mark is lost.
Usually costs: 1 mark.

Exam tips

Learn the spin-only values for 1 to 5 unpaired electrons; they save time in MCQs.
Give magnetic moments with the unit BM.
In 'reason' questions on colour, write the configuration of both ions being compared.
Remember the special colourless list: Sc³⁺, Ti⁴⁺ ( $d^{0}$ ) and Zn²⁺, Cu⁺, Cd²⁺, Hg²⁺ ( $d^{10}$ ).
For KMnO₄ and K₂Cr₂O₇, do not use d-d transitions: the metal is $d^{0}$ .

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
Which of the following ions is diamagnetic? (Z: Cu 29, Fe 26, Ni 28)
(a) Cu⁺
(b) Cu²⁺
(c) Fe²⁺
(d) Ni²⁺
Hint
Look for a $d^{10}$ ion.
Answer
(a) Cu⁺ ( $3 d^{10}$ ).
Q2 · 2 marks · Short answerBoard level
Why are most transition metal ions coloured? Why is the colour we see different from the colour of light absorbed?
Hint
d-d transitions and complementary colours.
Answer
Ligands split the d orbitals into sets of slightly different energy; an electron in a partly filled d set absorbs visible light and moves to a higher d orbital (d-d transition). The light that is not absorbed reaches the eye, so we see the complementary colour of the light absorbed.
Q3 · 2 marks · Short answerHarder board level
Calculate the spin-only magnetic moments of Mn³⁺ (Z = 25) and Ti²⁺ (Z = 22).
Hint
Mn³⁺ is $3 d^{4}$ ; Ti²⁺ is $3 d^{2}$ .
Answer
Mn³⁺: 4 unpaired, $μ = 24 = 4.90$ BM. Ti²⁺: 2 unpaired, $μ = 8 = 2.83$ BM.
Q4 · 3 marks · Short answerHarder board level
A dipositive ion of the first transition series has a spin-only magnetic moment of 4.90 BM. Show that it could be either Cr²⁺ or Fe²⁺. Which of the two is the stronger reducing agent, and why? (Z: Cr 24, Fe 26)
Hint
4.90 BM means 4 unpaired electrons: $d^{4}$ or $d^{6}$ .
Answer
$n (n + 2) = 24$ , $n = 4$ . Cr²⁺ is $3 d^{4}$ (4 unpaired) and Fe²⁺ is $3 d^{6}$ (4 unpaired), so both fit. Cr²⁺ is the stronger reductant: Cr²⁺ → Cr³⁺ gives the stable $d^{3}$ configuration in water, while Fe²⁺ → Fe³⁺ goes from $d^{6}$ to $d^{5}$ , a smaller gain in stability in aqueous solution.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Zinc salts are white.
Reason (R): Zn²⁺ has a completely filled 3d subshell, so no d-d transition is possible.
(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
Zn²⁺ is $3 d^{10}$ .
Answer
(a).
Q6 · 3 marks · Short answerHarder board level
Arrange Cr³⁺, Fe²⁺, Cu²⁺ and Mn²⁺ in increasing order of spin-only magnetic moment, showing the number of unpaired electrons in each. (Z: Cr 24, Mn 25, Fe 26, Cu 29)
Hint
Write the d count for each ion.
Answer
Cu²⁺ ( $d^{9}$ , 1, 1.73) < Cr³⁺ ( $d^{3}$ , 3, 3.87) < Fe²⁺ ( $d^{6}$ , 4, 4.90) < Mn²⁺ ( $d^{5}$ , 5, 5.92 BM).
Q7 · 3 marks · Short answerHOTS / competency
A $d^{5}$ ion can have five unpaired electrons (high spin) or one unpaired electron (low spin), depending on its ligands. (a) Calculate the spin-only moment in each case. (b) A complex of Fe³⁺ shows a moment of about 1.7 BM. What does this tell you about the arrangement of its d electrons?
Hint
Strong-field ligands force pairing.
Answer
(a) High spin, $n = 5$ : 5.92 BM; low spin, $n = 1$ : 1.73 BM. (b) 1.7 BM means one unpaired electron: the five d electrons are paired up as far as possible in the lower set of d orbitals, so the ligands must be strong-field ligands that cause a large splitting (low-spin complex).
Q8 · 3 marks · Case-basedHOTS / competency
Read and answer.
Suppose a student is given three solid samples, labelled P, Q and R, each a sulphate of one of Zn²⁺, Cu²⁺ and Ni²⁺. P is white and is repelled slightly by a strong magnet. Q dissolves to give a blue solution, and its moment is about 1.8 BM. R gives a green solution, and its moment is about 2.9 BM.
(i) Identify P, Q and R. [1]
(ii) Calculate the spin-only values for Q and R and compare them with the measured values. [1]
(iii) Why is P both white and diamagnetic? [1]
Hint
Count unpaired electrons: $d^{10}$ , $d^{9}$ , $d^{8}$ .
Answer
(i) P = ZnSO₄, Q = CuSO₄, R = NiSO₄. (ii) Cu²⁺ $3 d^{9}$ : 1.73 BM (measured 1.8); Ni²⁺ $3 d^{8}$ : 2.83 BM (measured 2.9); the measured values are close to spin-only values. (iii) Zn²⁺ is $3 d^{10}$ : no unpaired electrons (diamagnetic) and no vacant d orbital for a d-d transition (colourless).

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.Moment for 3 unpaired electrons?Show answer
3.87 BM.
3.Why is Sc³⁺ colourless?Show answer
It is $3 d^{0}$ : no d electron for a d-d transition.
4.Why are Zn, Cd and Hg not transition elements?Show answer
They have completely filled d orbitals in the atom and in their common ions.
5.Colour of aqueous Ni²⁺ and Co²⁺?Show answer
Ni²⁺ green; Co²⁺ pink.
6.Which first-series ions have the highest spin-only moment?Show answer
$d^{5}$ ions, Mn²⁺ and Fe³⁺ (5.92 BM).

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

How do you calculate the magnetic moment of a transition metal ion?

Write the d configuration, count unpaired electrons n by Hund's rule, and use \mu = \sqrt{n(n+2)} BM.

Why are transition metal ions coloured?

Their partly filled d orbitals are split by ligands, and d-d transitions absorb visible light; we see the complementary colour.

Why is anhydrous copper sulphate white?

Without water ligands the d orbitals of Cu²⁺ are not split, so no d-d transition occurs in the visible region.

What is the difference between paramagnetic and diamagnetic?

Paramagnetic substances have unpaired electrons and are attracted by a magnetic field; diamagnetic substances have all electrons paired and are weakly repelled.

Why is Cu considered a transition element though Cu atom is 3d¹⁰?

Its common ion Cu²⁺ has an incompletely filled 3d^9 configuration.