Predict which of the following will be coloured in aqueous solution? $Ti^{3+}$, $V^{3+}$, $Cu^+$, $Sc^{3+}$, $Mn^{2+}$, $Fe^{3+}$ and $Co^{2+}$. Give reasons for each.

> [!TLDR] > Colour in aqueous solution arises from d–d transitions in ions with partially filled d-orbitals. Among the given ions, $Ti^{3+}$, $V^{3+}$, $Mn^{2+}$, $Fe^{3+}$, and $Co^{2+}$ are coloured; $Cu^+$ and $Sc^{3+}$ are colourless because they have fully filled or empty d-subshells. The key idea is simple: colour in transition metal ions comes from electrons jumping between d-orbitals of slightly different energy. When light hits the ion, it absorbs certain wavelengths to make that jump, and the colour we see is the complementary colour of the absorbed light. But this only works if there is at least one d-electron *and* at least one empty d-orbital — in other words, a partially filled d-subshell. Let’s go through each ion one by one. 1. **$Ti^{3+}$** Titanium in the +3 state has lost three electrons. The ground state configuration of Ti is $[Ar]\,3d^2\,4s^2$. Removing three electrons gives $Ti^{3+}$: $[Ar]\,3d^1$. There is exactly one electron in the d-subshell. In an octahedral aqueous environment, the d-orbitals split into two sets — the lower-energy $t_{2g}$ and higher-energy $e_g$. The single electron occupies a $t_{2g}$ orbital and can absorb visible light to jump to an $e_g$ orbital. This d–d transition gives colour. **Result:** Coloured (violet/purple in solution). 2. **$V^{3+}$** Vanadium atomic number 23: $[Ar]\,3d^3\,4s^2$. Removing three electrons gives $V^{3+}$: $[Ar]\,3d^2$. Two d-electrons, both unpaired, in a partially filled d-shell. d–d transitions are possible. **Result:** Coloured (green in solution). 3. **$Cu^+$** Copper: $[Ar]\,3d^{10}\,4s^1$. Losing one electron gives $Cu^+$: $[Ar]\,3d^{10}$. The d-subshell is completely filled. There is no empty d-orbital for an electron to jump into — all d-orbitals are occupied. Therefore, no d–d transition can occur. **Result:** Colourless. > [!WARNING] > A common mistake is to think $Cu^+$ is coloured because $Cu^{2+}$ is blue. But $Cu^{2+}

Predict which of the following will be coloured in aqueous solution? Ti³⁺, V³⁺, Cu^+, Sc³⁺, Mn²⁺, Fe³⁺ and Co²

Exercises · 4.18

Ti^{3+}

CBSENCERTSubjective· 3mImportance★★★★★

29% · 38/132 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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. …

Start your 14-day free trial to unlock the full solution →

Colour in aqueous solution arises from d–d transitions in ions with partially filled d-orbitals. Among the given ions, $Ti^{3+}$ , $V^{3+}$ , $Mn^{2+}$ , $Fe^{3+}$ , and $Co^{2+}$ are coloured; $Cu^+$ and $Sc^{3+}$ are colourless because they have fully filled or empty d-subshells.

The key idea is simple: colour in transition metal ions comes from electrons jumping between d-orbitals of slightly different energy. When light hits the ion, it absorbs certain wavelengths to make that jump, and the colour we see is the complementary colour of the absorbed light. But this only works if there is at least one d-electron and at least one empty d-orbital — in other words, a partially filled d-subshell.

Let’s go through each ion one by one.

$Ti^{3+}$

Titanium in the +3 state has lost three electrons. The ground state configuration of Ti is $[Ar]\,3d^2\,4s^2$ . Removing three electrons gives $Ti^{3+}$ : $[Ar]\,3d^1$ .

There is exactly one electron in the d-subshell. In an octahedral aqueous environment, the d-orbitals split into two sets — the lower-energy $t_{2g}$ and higher-energy $e_g$ . The single electron occupies a $t_{2g}$ orbital and can absorb visible light to jump to an $e_g$ orbital. This d–d transition gives colour.

Result: Coloured (violet/purple in solution).
$V^{3+}$

Vanadium atomic number 23: $[Ar]\,3d^3\,4s^2$ . Removing three electrons gives $V^{3+}$ : $[Ar]\,3d^2$ .

Two d-electrons, both unpaired, in a partially filled d-shell. d–d transitions are possible.

Result: Coloured (green in solution).
$Cu^+$

Copper: $[Ar]\,3d^{10}\,4s^1$ . Losing one electron gives $Cu^+$ : $[Ar]\,3d^{10}$ .

The d-subshell is completely filled. There is no empty d-orbital for an electron to jump into — all d-orbitals are occupied. Therefore, no d–d transition can occur.

Result: Colourless.

Watch out

A common mistake is to think $Cu^+$ is coloured because $Cu^{2+}$ is blue. But $Cu^{2+}$ has a $3d^9$ configuration (one hole), while $Cu^+$ has a full $3d^{10}$ — no d–d transitions possible.
$Sc^{3+}$

Scandium: $[Ar]\,3d^1\,4s^2$ . Removing three electrons gives $Sc^{3+}$ : $[Ar]$ .

The d-subshell is completely empty. No d-electrons means no d–d transitions.

Result: Colourless.
$Mn^{2+}$

Manganese: $[Ar]\,3d^5\,4s^2$ . Removing two electrons gives $Mn^{2+}$ : $[Ar]\,3d^5$ . …

Start your 14-day free trial to unlock the full solution →

🧠 Core Concept (The Why)

Colour in aqueous transition metal ions arises from d-d transitions — an electron jumps from a lower-energy d-orbital to a higher-energy d-orbital by absorbing visible light.

Two conditions are essential:

The ion must have at least one electron in the d-subshell (i.e., not $d^0$ or $d^{10}$ ).
The d-subshell must be partially filled ( $d^1$ – $d^9$ ) — a completely full $d^{10}$ leaves no possible d-d excitation.

Key rule: Colour is seen only for ions with $d^1$ to $d^9$ configuration.

$d^0$ and $d^{10}$ are colourless (no d-d transition possible).

✗ Common Mistake #1: Forgetting that $d^0$ and $d^{10}$ are colourless

Example: Students often think $Sc^{3+}$ or $Cu^+$ will be coloured.

Why it’s wrong:
- $Sc^{3+}$ : $[Ar] 3d^0$ → no d-electron → colourless.
- $Cu^+$ : $[Ar] 3d^{10}$ → all d-orbitals filled → no vacant d-orbital → colourless.

✓ How to avoid:

Always write the electronic configuration of the ion first. If it’s $d^0$ or $d^{10}$ , stop — it’s colourless.

✗ Common Mistake #2: Confusing $d^1$ with $d^9$ — both are coloured, but for different reasons

Example: $Ti^{3+}$ is $d^1$ — one electron, many vacant orbitals → coloured.

$Cu^{2+}$ is $d^9$ — one vacancy, many electrons → coloured.

✓ How to avoid:

Just check: is the configuration between $d^1$ and $d^9$ ? If yes → coloured. No need to overthink.

✗ Common Mistake #3: Ignoring the effect of high spin vs low spin (for $d^4$ – $d^7$ )

Example: $Fe^{3+}$ is $d^5$ . In weak field (aqueous), it’s high spin — all 5 orbitals singly occupied → no vacant orbital for d-d transition?

Wrong conclusion: Some students say it’s colourless.

✓ Reality:

In high-spin $d^5$ every one of the five d-orbitals is singly occupied — there is no empty d-orbital — yet d-d transitions still occur (an electron is promoted into the higher, singly occupied $e_g$ set, which requires a spin flip). Such transitions are spin-forbidden and therefore very weak. Still, $Fe^{3+}$ solutions are pale yellow/violet and $Mn^{2+}$ pale pink — coloured, just faintly.

✓ How to avoid:

Remember: All $d^1$ – $d^9$ ions are coloured in solution, though intensity varies. Spin-forbidden transitions give pale colours, but they are still coloured.

✗ Common Mistake #4: Thinking $Cu^+$ is coloured because copper compounds are often coloured

Example: $Cu^+$ compounds (like $Cu_2O$ ) are red — so students think $Cu^+$ (aq) is coloured.

✓ Correction:

The colour in $Cu_2O$ comes from charge transfer (ligand-to-metal), not d-d transition. In aqueous solution, $Cu^+$ is $d^{10}$ → colourless.

$Cu^{2+}$ ( $d^9$ ) is the coloured one (blue).

✓ How to avoid: …

Start your 14-day free trial to unlock the full solution →

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)? …
Start your 14-day free trial to see the full solution →
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. …
Start your 14-day free trial to see the full solution →
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. …
Start your 14-day free trial to see the full solution →

←Previous question4.17 — For M²⁺/M and M³⁺/M²⁺ systems the E^° values for some metals are as fo…→Next question4.19 — Compare the stability of +2 oxidation state for the elements of the fi…

Q.Predict which of the following will be coloured in aqueous solution? $Ti^{3+}$ , $V^{3+}$ , $Cu^+$ , $Sc^{3+}$ , $Mn^{2+}$ , $Fe^{3+}$ and $Co^{2+}$ . Give reasons for each.

Concept understanding — Magnetism and Color

Magnetism and Colour: An Intuitive First Look

The Intuition: Electrons as Tiny Magnets and Painters

The Precise Statement

How They Connect (and How They Don't)

The Deeper Link: Transition Metals

Why this formula?

Magnetism and Color: Why the Key Formulas Hold

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

Why does this hold?

2. The Zeeman Effect: How Magnetic Fields Split Colors

Why does this hold?

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

Why does this hold?

Method: Crystal Field Theory (CFT) — Colour Prediction via d-orbital Configuration

Steps to predict colour

Predictions for each ion

1. $\mathbf{Ti^{3+}}$

2. $\mathbf{V^{3+}}$

3. $\mathbf{Cu^+}$

4. $\mathbf{Sc^{3+}}$

5. $\mathbf{Mn^{2+}}$

6. $\mathbf{Fe^{3+}}$

🧠 Core Concept (The Why)

✗ Common Mistake #1: Forgetting that $d^0$ and $d^{10}$ are colourless

✗ Common Mistake #2: Confusing $d^1$ with $d^9$ — both are coloured, but for different reasons

✗ Common Mistake #3: Ignoring the effect of high spin vs low spin (for $d^4$ – $d^7$ )

✗ Common Mistake #4: Thinking $Cu^+$ is coloured because copper compounds are often coloured

Why magnetism and electron configuration matter

The Concept: Colour and Electronic Structure

Step-by-Step Reasoning

Q.Predict which of the following will be coloured in aqueous solution? Ti3+Ti^{3+}Ti3+, V3+V^{3+}V3+, Cu+Cu^+Cu+, Sc3+Sc^{3+}Sc3+, Mn2+Mn^{2+}Mn2+, Fe3+Fe^{3+}Fe3+ and Co2+Co^{2+}Co2+. Give reasons for each.

🎓Unlock everything free for 14 days

Q.Predict which of the following will be coloured in aqueous solution? $Ti^{3+}$ , $V^{3+}$ , $Cu^+$ , $Sc^{3+}$ , $Mn^{2+}$ , $Fe^{3+}$ and $Co^{2+}$ . Give reasons for each.

Unlock everything free for 14 days