Answer: Sc3+ (d 0 ) and Ti4+ (d 0 ) and Zn 2+ (d 10 ).
- A Sc3+ (d<sup>0</sup>) and Ti4+ (d<sup>0</sup>) and Zn<sup>2+</sup> (d<sup>10</sup>)
- B Fe<sup>3+</sup> (d<sup>5</sup>), which gives pale yellow complexes from a spin-forbidden d-d transition
- C Cu<sup>2+</sup> (d<sup>9</sup>), which gives characteristically blue complexes
- D Ni<sup>2+</sup> (d<sup>8</sup>), which gives characteristically green complexes
Correct answer: A. Sc3+ (d<sup>0</sup>) and Ti4+ (d<sup>0</sup>) and Zn<sup>2+</sup> (d<sup>10</sup>)
Explanation: Colourless complexes form with d<sup>0</sup> (no d electrons for d-d transitions) and d<sup>10</sup> (completely filled, no d-d transitions). Sc3+ and Zn<sup>2+</sup> are classic examples.
Aqueous solutions of transition metal ions show characteristic colours caused by electrons absorbing visible light to jump between split d-orbitals; the exact colour depends on the metal, its oxidation state, and surrounding ligands.
Concept context
Transition metals (d-block) and lanthanides/actinides (f-block). Known for coloured compounds, variable oxidation states, complex formation, and catalytic properties. Frequently tested in JEE and NEET.