Identify which of the following homoleptic metal complex is colorless?

The correct answer is Option (2) → [Zn(H₂O)₆ ]²⁺

The colour of most transition metal complexes comes from d–d transitions. The ligands split the d orbitals into two sets of different energy. An electron in the lower set absorbs visible light of a particular wavelength and moves to the upper set, and the complex shows the complementary colour.

A d–d transition needs a partially filled d subshell (d¹ to d⁹). Ions with d⁰ (empty) or d¹⁰ (completely filled) configurations have no d–d transition and are colourless.

A homoleptic complex has only one kind of ligand; a heteroleptic complex has more than one kind.

In \( \ce{[Zn(H2O)6]^{2+}} \), zinc is \( \ce{Zn^{2+}} \) (water is neutral). Zn (Z = 30) is [Ar]3d¹⁰4s²; losing the two 4s electrons leaves 3d¹⁰, and water is the only ligand. So it is a colourless homoleptic complex.

Why the other options are incorrect

• Option (1) [Ti(H₂O)₆ ]³⁺ — Incorrect. It is homoleptic, but Ti (Z = 22) is [Ar]3d²4s², and \( \ce{Ti^{3+}} \) (two 4s and one 3d electron removed) is 3d¹. It absorbs green-yellow light (about 498 nm) and appears violet.

• Option (3) [ScCl(NH₃)₅]²⁺ — Incorrect. Here x + (−1) + 5(0) = +2, so x = +3; Sc (Z = 21) is [Ar]3d¹4s², and \( \ce{Sc^{3+}} \) has lost all three electrons (3d⁰), so this complex is also colourless, but it has two kinds of ligands (\( \ce{Cl^{-}} \) and \( \ce{NH3} \)), so it is heteroleptic.

• Option (4) [Co(NH₃)₅(H₂O)]³⁺ — Incorrect. It is heteroleptic (\( \ce{NH3} \) and \( \ce{H2O} \)), and Co (Z = 27) is [Ar]3d⁷4s², so \( \ce{Co^{3+}} \) (two 4s and one 3d electron removed) is 3d⁶, so it is coloured.

NCERT (2026-27 edition):
Definitions of Some Important Terms Pertaining to Coordination Compounds – …

Formation of Coloured Ions