Which of the following statements is incorrect?

The correct answer is Option (3) → Observed color of the complex is the color of the absorbed light/wavelength

In a complex the ligands split the d orbitals of the metal ion. When white light falls on the complex, an electron absorbs light of the energy that matches the splitting and is promoted from a lower to a higher d orbital (a d–d transition). The light that comes out has lost that part of the spectrum, so the colour observed is complementary to the colour absorbed.

For example, \( \ce{[Ti(H2O)6]^3+} \) absorbs blue-green light (about 498 nm) and appears violet.

• Option (1) When white light falls on the complex, some part of it is absorbed — A correct statement about the colour of complexes. The part whose energy equals the crystal field splitting is absorbed for the d–d transition.

• Option (2) Higher the crystal field splitting, lower will be the wavelength absorbed by the complex — A correct statement about the colour of complexes. The energy absorbed equals the splitting, and energy and wavelength are inversely related:

      \( \displaystyle \Delta_o = \frac{hc}{\lambda} \)

  So a larger \( \Delta_o \) means a shorter wavelength absorbed.

• Option (3) Observed color of the complex is the color of the absorbed light/wavelength — Not a correct statement about the colour of complexes. The colour seen is the complementary colour of the light absorbed, not the absorbed colour itself.

• Option (4) In the absence of ligand, compound is colourless — A correct statement about the colour of complexes. Without ligands there is no crystal field splitting and no d–d transition; anhydrous \( \ce{CuSO4} \) is white, while \( \ce{CuSO4.5H2O} \) is blue.

NCERT (2026-27 edition):
Colour in Coordination Compounds

Colour in Coordination Compounds – “… It is important to note that in the absence of ligand, crystal field splitting does not occur and hence the substance is colourless. For example, removal of water from \( \ce{[Ti(H2O)6]Cl3} \) on heating renders it colourless. Similarly, anhydrous \( \ce{CuSO4} \) is white, but \( \ce{CuSO4.5H2O} \) is blue in colour.”