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 cuprous ion (Cu+) is colourless while the cupric ion (Cu2+) is coloured because
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Cu+ has completely filled d orbitals and Cu2+ has incompletely filled d orbitals
The passage states that transition metal ions are coloured because of partially filled d orbitals: an electron in the lower set of split d orbitals absorbs visible light and is excited to the higher set (a d–d transition). An ion with no d electron or a full d subshell has no such transition.
Configuration of Cu: Cu (Z = 29) is [Ar]3d104s1.
\( \ce{Cu^+} \): losing the 4s electron gives 3d10, a completely filled d subshell; no d–d transition is possible, so the ion is colourless.
\( \ce{Cu^2+} \): losing the 4s electron and one 3d electron gives 3d9, an incompletely filled d subshell; a d–d transition absorbs visible light, so the ion is coloured.
Why the other options are incorrect
Option (1) both ions have unpaired electrons in the d orbitals — Incorrect. In 3d10 \( \ce{Cu^+} \) all ten d electrons are paired; only \( \ce{Cu^2+} \) (3d9) has one unpaired electron.
Option (2) both ions have half-filled d orbitals — Incorrect. A half-filled d subshell is d5; the two ions are d10 and d9.
Option (3) Cu+ has incompletely filled d orbitals and Cu2+ has completely filled d orbitals — Incorrect. This is the reverse of the real configurations: \( \ce{Cu^+} \) is 3d10 and \( \ce{Cu^2+} \) is 3d9.