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 oxocation \( \ce{VO^2+} \) gives a coloured aqueous solution. What is the number of 3d electrons on vanadium in \( \ce{VO^2+} \)? (Atomic number of V = 23)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 1
The number of d electrons is found from the oxidation state of the metal: first the charge balance, then the configuration of the ion. When a 3d metal forms a cation, the 4s electrons are lost before the 3d electrons.
Oxidation state of V in VO2+ (O = −2): x + (−2) = +2, so x = +4.
V (Z = 23): [Ar] 3d3 4s2.
V(IV): the two 4s electrons and one 3d electron are removed, leaving [Ar] 3d1.
Number of 3d electrons = 1.
This single d electron can be excited from the lower to the higher set of d orbitals, which is why the ion is coloured.
NCERT (2026-27 edition):
4.4.1 Oxides and Oxoanions of Metals – “…Besides the oxides, the oxocations stabilise VV as VO2+, VIV as VO2+ and TiIV as TiO2+.”