Werner was the first to describe the bonding features in coordination compounds. But his theory could not answer basic questions like :
(i) Why only certain elements possess the remarkable property of forming coordination compounds ?
(ii) Why the bonds in coordination compounds have directional properties ?
(iii) Why coordination compounds have characteristic magnetic and optical properties ?
Many approaches have been put forth to explain the nature of bonding in coordination compounds viz. Valence Bond Theory (VBT), Crystal Field Theory (CFT), Ligand Field Theory (LFT) and Molecular Orbital Theory (MOT). We shall focus our attention on elementary treatment of the application of VBT and CFT to coordination compounds.
According to this theory, the metal atom or ion under the influence of ligands can use its (n−1)d, ns np or ns, np, nd orbitals for hybridisation to yield a set of equivalent orbitals of definite geometry such as octahedral, tetrahedral, square planar and so on. These hybridised orbitals are allowed to overlap with ligand orbitals that can donate electron pairs for bonding.
Which of the following pair of ions is colourless according to CFT ?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Ti4+ , Zn2+
According to crystal field theory, the colour of a transition metal ion comes from a d–d transition: light of a particular wavelength is absorbed when an electron jumps from a lower to a higher d level split by the ligands, and the complementary colour is seen. A d–d transition is possible only if the ion has a partly filled d subshell. Ions with d0 (no d electron) or d10 (no vacancy) configurations are colourless.
Ti4+: Ti (Z = 22) is [Ar] 3d24s2; losing four electrons gives 3d0 — colourless.
Zn2+: Zn (Z = 30) is [Ar] 3d104s2; losing the two 4s electrons gives 3d10 — colourless.
So the colourless pair is Ti4+ and Zn2+.
Why the other options are incorrect
Option (2) V2+ , Mn2+ — Incorrect. V (Z = 23) [Ar] 3d34s2 → \( \ce{V^{2+}} \) 3d3 (violet); Mn (Z = 25) [Ar] 3d54s2 → \( \ce{Mn^{2+}} \) 3d5 (pale pink). Both have partly filled d subshells.
Option (3) Sc3+ , Cu2+ — Incorrect. \( \ce{Sc^{3+}} \) (3d0) is colourless, but Cu (Z = 29) [Ar] 3d104s1 → \( \ce{Cu^{2+}} \) 3d9 is blue.
Option (4) Cr3+ , Fe3+ — Incorrect. Cr (Z = 24) [Ar] 3d54s1 → \( \ce{Cr^{3+}} \) 3d3 (violet); Fe (Z = 26) [Ar] 3d64s2 → \( \ce{Fe^{3+}} \) 3d5 (yellow). Both are coloured.
NCERT (2026-27 edition):
Formation of Coloured Ions (rows of Table 4.8: Colours of Some of the First Row (aquated) Transition Metal Ions)
| Configuration | Example | Colour |
|---|---|---|
| 3d0 | Sc3+ | colourless |
| 3d0 | Ti4+ | colourless |
| 3d3 | V2+ | violet |
| 3d3 | Cr3+ | violet |
| 3d5 | Mn2+ | pink |
| 3d5 | Fe3+ | yellow |
| 3d9 | Cu2+ | blue |
| 3d10 | Zn2+ | colourless |