Transition metals form coloured ions or compounds due to partially filled d orbitals. In the presence of solvent molecules (in solutions), ligands (in complexes) or counter ions (in crystals), the d orbitals split into two sets. An electron in a transition metal ion that occupies the set of d orbitals of lower energy can be excited to the set of higher energy by absorbing energy from visible light. Since the energy difference (ΔE) between the two sets of d orbitals is small, only light of the visible region is absorbed during this excitation. The colour of a transition metal ion is due to this d–d excitation (d–d transition) of the electron. During d–d excitation the electron absorbs one colour of visible light, and the ion appears in the complementary colour of the absorbed light. Complementary colours can be identified from the colour wheel shown below (complementary colours lie opposite each other).
The number of electrons undergoing d–d transition and the energy difference between the two sets of orbitals decide the colour.
The hydrated copper(II) ion appears blue because it absorbs light of which colour?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Orange-red
The passage states that a transition metal ion absorbs one colour of visible light in a d–d transition and appears in the complementary colour of the light absorbed, and that complementary colours lie opposite each other on the colour wheel.
The hydrated copper(II) ion (3d9) appears blue. On the wheel, blue lies opposite orange, between yellow and red; of the options only orange-red lies on that side, so the ion absorbs orange-red light and transmits blue.
Why the other options are incorrect
Option (1) Green — Incorrect. Green lies opposite purple; an ion absorbing green would look purple.
Option (2) Blue — Incorrect. The colour seen is the light not absorbed; an ion absorbing blue would look orange.
Option (4) Yellow — Incorrect. On the wheel yellow lies opposite indigo, not blue.