Why Transition Metal Ions Are Coloured (d-d Transitions)
Chemistry · D And F Block Elements · NEET
Transition metal ions are coloured because they have partly filled d orbitals. When light hits the ion, an electron jumps from a lower d orbital to a higher d orbital. This jump is called a d-d transition, and it absorbs one colour of light. The ion then shows the leftover (complementary) colour. Memory hook: "Empty or full d = no colour; half-empty d = colour."
A d-d transition: an electron in a lower d orbital absorbs visible light and jumps to a higher d orbital. The ion then shows the complementary (leftover) colour. d0 and d10 ions have no possible jump, so they stay colourless.
Your doubts, answered
What exactly is a d-d transition in simple words?
In a transition metal ion, the five d orbitals are not all at the same energy. When ligands (like water) surround the ion, the d orbitals split into a lower group and a higher group. A d-d transition is when an electron in a lower d orbital absorbs light and jumps to a higher d orbital. The energy gap is small, so the light absorbed falls in the visible range. That is why we see colour.
Why are Sc3+ and Zn2+ colourless but Ti3+ and Cu2+ are coloured?
For colour you need a partly filled d subshell (d1 to d9), so that an electron has an empty higher d orbital to jump into. Sc3+ is d0 (no d electron to jump) and Zn2+ is d10 (d orbitals are completely full, no empty d orbital to receive the electron). So both are colourless. Ti3+ (d1) and Cu2+ (d9) have partly filled d orbitals, so a d-d transition is possible and they are coloured.
What does 'complementary colour' mean here?
White light contains all colours. The ion absorbs one colour (one wavelength) for its d-d transition. The colour you actually see is what is LEFT OVER after that absorption. This leftover colour is called the complementary colour. Example: [Ti(H2O)6]3+ absorbs blue-green light, so it looks violet.
If MnO4- (purple) and CrO4^2- (yellow) are so bright, is that colour from d-d transitions?
No, and this is the most common NEET trap. In MnO4- the Mn is +7 (d0) and in CrO4^2- / Cr2O7^2- the Cr is +6 (d0). A d0 ion has NO d electron, so a d-d transition is impossible. Their strong colour comes from a different cause called charge transfer (ligand-to-metal charge transfer, LMCT), where an electron moves from the oxygen ligand to the metal. These ions are also diamagnetic (no unpaired electrons).
Does the type of ligand change the colour?
Yes. The ligand decides how much the d orbitals split, so it decides how much energy (which wavelength) is absorbed. A stronger-field ligand causes a bigger split, so a higher-energy (shorter wavelength) light is absorbed and the colour changes. That is why the same metal ion can look different colours with different ligands. In NCERT, water is the standard ligand in aqueous solutions.
⚠️ The NEET trap ✗ MnO4- is intensely purple, so its colour must come from a d-d transition of manganese. ✓ In MnO4- manganese is +7, which is d0 (no d electrons). A d-d transition needs at least one d electron, so it is impossible here. The purple colour comes from charge transfer, not d-d transition, and MnO4- is diamagnetic. For a real d-d transition you need a d1 to d9 ion like MnO4^2- (Mn is +6, d1). 🧠 d0 or d10 = NO d-d transition. Bright colour there means charge transfer, not d-d.
Real NEET questions
NEET 2018
Which one of the following ions exhibits d-d transition and paramagnetism as well?
A · MnO4^-
B · Cr2O7^2-
C · CrO4^2-
D · MnO4^2- ✓
Solution: For a real d-d transition AND paramagnetism, the metal must have at least one unpaired d electron (d1 or higher). MnO4^- (Mn +7), Cr2O7^2- and CrO4^2- (Cr +6) are all d0: no d electron, so no d-d transition, and they are diamagnetic. Their colour is due to charge transfer only. MnO4^2- (manganate) has Mn in the +6 state = d1, with one unpaired d electron. So it is paramagnetic and can undergo a genuine d-d transition. Answer: (d) MnO4^2-.
NEET 2021
The incorrect statement among the following is:
A · Lanthanoids are good conductors of heat and electricity.
B · Actinoids are highly reactive metals, especially when finely divided.
C · Actinoid contraction is greater for element to element than lanthanoid contraction.
D · Most of the trivalent lanthanoid ions are colourless in the solid state. ✓
Solution: Statements (a), (b) and (c) are all correct. Statement (d) is wrong because MOST trivalent lanthanoid ions (Ln3+) are actually coloured, in both the solid state and in solution. Their colour comes from f-f transitions of the 4f electrons (a similar idea to d-d transitions, but with f orbitals). Only a few, such as La3+ (f0) and Lu3+ (f14), are colourless. So the incorrect statement is (d).
Solved D And F Block Elements NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Do transition metal ions need unpaired electrons to be coloured?
They need partly filled d orbitals (d1 to d9), which usually means unpaired electrons are present. A d0 ion (like Sc3+) and a d10 ion (like Zn2+) are colourless because no d-d transition is possible.
Why is the light absorbed in the visible region?
The energy gap between the split d orbitals is small. Small energy gaps match the low-energy photons of visible light, so visible light is absorbed and we see colour.
Is d-d transition the same as f-f transition?
They are the same idea but in different orbitals. d-d transitions colour transition metal ions; f-f transitions colour many lanthanoid ions using their 4f electrons.
Why is Cu+ colourless but Cu2+ is blue?
Cu+ is 3d10 (fully filled d), so no d-d transition is possible and it is colourless. Cu2+ is 3d9 (partly filled), so a d-d transition can occur and it appears blue.
