Generally transition elements and their salts are coloured due to the presence of unpaired electrons in metal

NCERT Exemplar · Q50

Q. Generally transition elements and their salts are coloured due to the presence of unpaired electrons in metal ions. Which of the following compounds are coloured?

(i)

KMnO_4

(ii)

Ce(SO_4)_2

(iii)

TiCl_4

(iv)

Cu_2Cl_2

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Concept understanding — Magnetism and Color

Magnetism and Colour: An Intuitive First Look

You've probably noticed that some materials are magnetic (like iron) and others aren't (like wood). And you've seen that objects have different colours — a rose is red, the sky is blue. At first glance, these two properties seem completely unrelated. But at the deepest level, both magnetism and colour come from the same source: how electrons behave inside atoms.

Let's start with a simple picture.

The Intuition: Electrons as Tiny Magnets and Painters

Imagine an electron orbiting the nucleus of an atom. That moving charge is like a tiny loop of electric current — and any loop of current creates a magnetic field. So every electron is a microscopic magnet. In most materials, these tiny magnets point in random directions and cancel out. But in iron, they align, and the material becomes magnetic.

Now, colour. When light hits an atom, electrons can absorb some of its energy and jump to a higher orbit. The colour we see is the light that wasn't absorbed — the leftover wavelengths. Different atoms absorb different colours because their electrons have different "jump sizes" (energy levels).

So both magnetism and colour are about how electrons move and interact with their environment. One is about the direction of electron spin and orbit (magnetism), the other about the energy of electron jumps (colour).

The Precise Statement

Important

Magnetism and colour are both consequences of the electronic structure of atoms, but they arise from different aspects of electron behaviour:

Magnetism originates from the magnetic moments of electrons — their spin and orbital motion. A material is magnetic when these moments align cooperatively.
Colour originates from the absorption of specific wavelengths of light by electrons, which occurs when the photon energy matches the energy difference between two electron states.

How They Connect (and How They Don't)

The two phenomena are linked because they both depend on the arrangement of electrons in orbitals — the so-called electronic configuration. But they are not the same thing, and one does not cause the other.

Here's a table to make the distinction clear:

Property	Origin	What determines it?	Example
Magnetism	Electron spin and orbital motion	Unpaired electrons, crystal structure	Iron is magnetic because it has 4 unpaired electrons per atom
Colour	Electron transitions between energy levels	Energy gap between orbitals	Copper is reddish because its electrons absorb blue-green light

Note

A material can be magnetic and colourless (like pure iron — it's silvery, not colourful). A material can be brilliantly coloured and non-magnetic (like a ruby). The two properties are independent in most everyday cases.

The Deeper Link: Transition Metals

The most interesting connection appears in transition metals (elements like iron, cobalt, nickel, copper). These atoms have partially filled d orbitals. That partial filling does two things:

It leaves unpaired electrons, which can align to produce magnetism. …

Why this formula?

Magnetism and Color: Why the Key Formulas Hold

This is a fascinating intersection of physics and perception. The core idea is that color is not a property of light itself, but of our brain's interpretation of different wavelengths. Magnetism, in turn, can influence how these wavelengths are produced or absorbed.

Let's break down the key formulas and their why.

1. The Fundamental Link: Energy, Frequency, and Color

The most important formula connecting magnetism and color is the Planck-Einstein relation:

E = h \nu

Where:

$E$ = energy of a photon (light particle)
$h$ = Planck's constant ( $6.626 \times 10^{-34} \text{ J·s}$ )
$\nu$ = frequency of the light

Why does this hold?

Quantum nature of light: Light is not a continuous wave, but comes in discrete packets called photons.
Energy quantization: The energy of a photon is directly proportional to its frequency. Higher frequency means higher energy.
Magnetism's role: When an electron in an atom jumps from a higher energy level to a lower one, it emits a photon. The energy difference ( $\Delta E$ ) between these levels determines the photon's frequency:

\Delta E = h \nu

Color perception: Our eyes detect different frequencies as different colors. For example:
- Red light: $\nu \approx 4.3 \times 10^{14} \text{ Hz}$ (lower energy)
- Blue light: $\nu \approx 6.7 \times 10^{14} \text{ Hz}$ (higher energy)

Key insight: The color you see is determined by the energy gap between electron orbits. Magnetism can alter these energy gaps (via the Zeeman effect, see below).

2. The Zeeman Effect: How Magnetic Fields Split Colors

When a magnetic field is applied to an atom, a single spectral line (one color) splits into multiple lines. This is described by:

\Delta E = \mu_B \cdot B \cdot m_l

Where:

$\Delta E$ = energy shift of the spectral line
$\mu_B$ = Bohr magneton ( $9.274 \times 10^{-24} \text{ J/T}$ )
$B$ = magnetic field strength (in Tesla)
$m_l$ = magnetic quantum number (integer: $-l, ..., +l$ )

Why does this hold?

Electron as a tiny magnet: An electron orbiting a nucleus behaves like a tiny current loop, creating a magnetic dipole moment.
Energy in a magnetic field: This dipole moment interacts with an external magnetic field. The interaction energy depends on the orientation of the electron's orbit relative to the field.
Quantized orientations: The magnetic quantum number $m_l$ tells us which orientation is allowed. Each orientation has a slightly different energy.
Result: A single energy level splits into $2l+1$ sub-levels. Transitions between these sub-levels produce photons with slightly different energies — hence different colors appear.

Example: A sodium lamp emits yellow light. In a strong magnetic field, that yellow line splits into three closely spaced lines (normal Zeeman effect).

3. Faraday Rotation: Magnetic Field Twists Light's Color

When polarized light passes through a material in a magnetic field, its plane of polarization rotates. The rotation angle is:

\theta = V \cdot B \cdot d

Where:

$\theta$ = rotation angle (in radians)
$V$ = Verdet constant (material-specific, depends on wavelength)
$B$ = magnetic field strength
$d$ = path length through the material

Why does this hold?

Circular birefringence: In a magnetic field, the material has different refractive indices for left- and right-circularly polarized light.
Phase difference: These two components travel at different speeds, creating a phase difference.
Recombination: When they recombine, the resulting linear polarization is rotated. …

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Colour in transition compounds arises from d–d transitions (unpaired electrons) or charge transfer (no unpaired electrons needed). Among the given, $KMnO_4$ and $Ce(SO_4)_2$ are coloured (charge transfer), $TiCl_4$ is colourless ( $d^0$ ), and $Cu_2Cl_2$ is white/colourless ( $d^{10}$ , no unpaired electrons). The coloured compounds are (i) and (ii).

The usual rule — “colour comes from unpaired electrons in d-orbitals” — is a good starting point, but it’s incomplete. Many intensely coloured compounds, like $KMnO_4$ , have no unpaired electrons at all. The real story involves two distinct mechanisms: d–d transitions (which need unpaired electrons) and charge transfer transitions (which do not). Let’s see how each compound fits.

$KMnO_4$ — deep purple

Manganese here is in the +7 oxidation state: $[Ar]\,3d^0$ . No d-electrons, so no d–d transition possible. The colour comes from a ligand-to-metal charge transfer (LMCT). An electron from a filled oxygen orbital jumps into an empty d-orbital of Mn(VII). This transition absorbs green-yellow light, leaving purple.

Tip

Charge transfer colours are usually far more intense than d–d colours — a tiny amount of $KMnO_4$ colours a large volume of water visibly.
$Ce(SO_4)_2$ — yellow/orange

Cerium(IV) has the configuration $[Xe]\,4f^0\,5d^0$ — no f or d electrons. Again, no d–d or f–f transition. The colour arises from charge transfer from the sulfate/oxygen ligands to the empty 4f or 5d orbitals of Ce(IV). This is common for Ce(IV) salts; Ce(III) salts are colourless or very pale.
$TiCl_4$ — colourless

Titanium is in the +4 state: $[Ar]\,3d^0$ . No d-electrons. And unlike $KMnO_4$ , the charge transfer band in $TiCl_4$ lies in the ultraviolet region, not the visible. So the compound appears colourless. …

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Method: Unpaired Electron Check (d-orbital Configuration Analysis)

This method determines colour based on the presence of unpaired electrons in the metal ion's d-orbitals. Colour arises from d–d transitions, which require at least one unpaired electron.

Steps

Step 1: Find the oxidation state of the metal

(i) $KMnO_4$ : Mn is in +7 state.
(ii) $Ce(SO_4)_2$ : Ce is in +4 state.
(iii) $TiCl_4$ : Ti is in +4 state.
(iv) $Cu_2Cl_2$ : Cu is in +1 state.

Step 2: Write the electronic configuration of the metal ion

(i) Mn $^{7+}$ : [Ar] $3d^0$ — no unpaired electrons
(ii) Ce $^{4+}$ : [Xe] $4f^0$ — no unpaired electrons (Ce is a f-block element; colour here arises from f–f or charge transfer, but d–d is absent)
(iii) Ti $^{4+}$ : [Ar] $3d^0$ — no unpaired electrons
(iv) Cu $^{+}$ : [Ar] $3d^{10}$ — no unpaired electrons

Step 3: Apply the rule

If $d^0$ or $d^{10}$ → colourless (no d–d transitions possible).

If $d^1$ to $d^9$ (except $d^{10}$ ) → coloured (unpaired electrons present).

Result

Compound	Metal ion	d-configuration	Unpaired e⁻?	Colour?
$KMnO_4$	Mn $^{7+}$	$3d^0$	No	Coloured (purple, via charge transfer)

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✗ Mistake 1: Assuming all transition metal compounds are coloured

Why it happens:

Students memorise “transition elements give coloured compounds” and apply it blindly — without checking the d-electron configuration of the metal ion.

How to avoid:

Always determine the oxidation state of the metal and then its d-electron count. Colour arises only if there are unpaired d-electrons (d¹–d⁹). If the ion is d⁰ or d¹⁰, the compound is colourless (or white).

✗ Mistake 2: Forgetting that $KMnO_4$ gets its colour from charge transfer, not d–d transitions

Why it happens:

Students see Mn in +7 state (d⁰) and think “no unpaired electrons → colourless”. But $KMnO_4$ is deep purple.

How to avoid:

Remember: d⁰ and d¹⁰ ions can still be coloured due to charge transfer (ligand-to-metal or metal-to-ligand). This is a separate mechanism from d–d transitions.

$KMnO_4$ (Mn⁷⁺, d⁰) → purple due to O²⁻ → Mn⁷⁺ charge transfer.
$Ce(SO_4)_2$ (Ce⁴⁺, 4f⁰) → yellow/orange due to charge transfer.

✗ Mistake 3: Thinking $TiCl_4$ is coloured because Ti is a transition metal

Why it happens:

Ti is in group 4, so students assume it behaves like other transition metals.

How to avoid:

Check the oxidation state: In $TiCl_4$ , Ti is +4 → d⁰ configuration. No unpaired electrons → colourless (no d–d transitions). Charge transfer is also absent here because the energy gap is too large for visible light.

✗ Mistake 4: Assuming $Cu_2Cl_2$ is coloured because Cu(I) is a transition metal ion

Why it happens:

Cu is a well-known coloured transition metal (Cu²⁺ is blue). Students forget that Cu⁺ is d¹⁰ → no unpaired electrons.

How to avoid:

Cu⁺ = [Ar] 3d¹⁰ → colourless (white).
Cu²⁺ = [Ar] 3d⁹ → blue (unpaired electron). Always check the oxidation state before concluding.

--- …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2025Set 56/4/11 markMCQ
$Q. For the following question, two statements are given — one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) given below. Assertion (A) : Cuprous salts are diamagnetic. Reason (R) : Cuprous ion has completely filled 3d-orbitals. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true.$
›Reveal solutionSolution
Cuprous salts are diamagnetic because the Cu⁺ ion has a completely filled 3d¹⁰ configuration, leaving no unpaired electrons. Both Assertion and Reason are true, and the Reason correctly explains the Assertion — so the answer is (A).

Why magnetism and electron configuration matter

Magnetic behaviour in transition metal compounds comes down to one thing: unpaired electrons. A substance with any unpaired electrons is paramagnetic (attracted to a magnetic field); one with all electrons paired is diamagnetic (weakly repelled). So to judge whether cuprous salts are diamagnetic, we need to know the electron configuration of the cuprous ion, Cu⁺.

Copper’s atomic number is 29. The neutral atom has the configuration $[\text{Ar}]\,3d^{10}4s^1$ — that one 4s electron is what gives copper its typical +1 and +2 oxidation states. When copper loses one electron to form Cu⁺, it loses that 4s electron first (not a 3d electron, because the 4s orbital is higher in energy once occupied). So Cu⁺ becomes $[\text{Ar}]\,3d^{10}$ .

That’s a completely filled d-subshell. All ten 3d electrons are paired up in five orbitals. No unpaired electrons means diamagnetic.

Now let’s check the Assertion and Reason carefully.

Assertion (A): “Cuprous salts are diamagnetic.”

True. As we just saw, Cu⁺ has no unpaired electrons. Every cuprous salt — CuCl, Cu₂O, Cu₂S, etc. — contains Cu⁺, so the salt as a whole is diamagnetic. (The anion doesn’t contribute unpaired electrons either, so the whole compound is diamagnetic.)

Reason (R): “Cuprous ion has completely filled 3d-orbitals.”

True. The configuration is $3d^{10}$ , which is indeed completely filled.

Is (R) the correct explanation of (A)? …
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CBSE 2022Set ANNUAL1 markMCQ
$Q. Trivalent ion of which of the following lanthanide metals is colourless ?$
(a) La
(b) Ce
(c) Pr
(d) Nd
›Reveal solutionSolution
La³⁺ has an empty 4f subshell, so it has no f-electrons to undergo f–f transitions and is colourless; the other listed ions all carry unpaired f-electrons and are coloured.

Colour in lanthanide ions arises mainly from f–f electronic transitions, which require partially-filled f orbitals. Checking the configurations of $M^{3+}$ :

La³⁺: $[Xe]4f^0$ — no f-electrons → colourless.

Ce³⁺: $[Xe]4f^1$ — coloured (pale).

Pr³⁺: $[Xe]4f^2$ — coloured.

Nd³⁺: $[Xe]4f^3$ — coloured. …
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CBSE 2020Set 56/3/11 mark
$Cu^{2+}$
›Reveal solutionSolution
The colour of transition metal ions arises from d-d electronic transitions, which require partially filled d-orbitals. $Cu^{2+}$ has a $3d^9$ configuration (one unpaired electron) allowing such transitions, while $Zn^{2+}$ has a $3d^{10}$ configuration (completely filled d-subshell) with no vacant d-orbital for electron excitation, making it colourless.

The Concept: Colour and Electronic Structure

Colour in transition metal compounds is a direct consequence of their electronic configuration. When white light falls on a substance, certain wavelengths are absorbed, and the complementary colour is transmitted or reflected. For a transition metal ion in solution, the absorption typically occurs in the visible region due to d-d transitions — electrons jumping from a lower-energy d-orbital to a higher-energy d-orbital within the same subshell.

This is only possible when the d-subshell is partially filled. If the d-orbitals are completely empty ( $d^0$ ) or completely filled ( $d^{10}$ ), no d-d transition can occur because there is either no electron to excite or no vacant orbital to receive it. Such ions appear colourless.

Step-by-Step Reasoning

Electronic configuration of the atoms

Copper (Cu, atomic number 29): $[Ar]\,3d^{10}4s^1$

Zinc (Zn, atomic number 30): $[Ar]\,3d^{10}4s^2$

Formation of the +2 ions

When forming $Cu^{2+}$ , the atom loses the 4s electron and one 3d electron:

$Cu^{2+}: [Ar]\,3d^9$

When forming $Zn^{2+}$ , the atom loses both 4s electrons:

$Zn^{2+}: [Ar]\,3d^{10}$

d-orbital occupancy in aqueous solution

In aqueous solution, water molecules act as ligands and create a crystal field around the metal ion. For octahedral complexes (common for both ions in water), the five d-orbitals split into two sets: the lower-energy $t_{2g}$ set ( $d_{xy}, d_{xz}, d_{yz}$ ) and the higher-energy $e_g$ set ( $d_{z^2}, d_{x^2-y^2}$ ).

$Cu^{2+}$ ( $3d^9$ ): Nine electrons occupy the d-orbitals. The configuration is $(t_{2g})^6(e_g)^3$ , with one unpaired electron in the $e_g$ level. There is a vacant orbital in the $e_g$ set, and an electron from the filled $t_{2g}$ set can be excited into it by absorbing visible light. This d-d transition gives $Cu^{2+}$ its characteristic blue colour in aqueous solution.

$Zn^{2+}$ ( $3d^{10}$ ): All five d-orbitals are completely filled. There is no vacant d-orbital to accept an excited electron. The only possible electronic transitions would involve much higher energy levels (like 4s or 4p), which require ultraviolet light, not visible. Hence, no visible light is absorbed, and the solution appears colourless. …
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Q.Generally transition elements and their salts are coloured due to the presence of unpaired electrons in metal ions. Which of the following compounds are coloured?

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