The aqueous solution containing which one of the following ions will be colourless (Atomic number of Sc = 21, Fe = 26, Ti = 22, Mn = 25)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Sc3+
An ion of a transition metal is coloured when it has unpaired d electrons that can undergo d–d transitions by absorbing visible light. An ion with d0 (or d10) has no such transition and is colourless.
Sc3+: Sc (Z = 21) is [Ar] 3d1 4s2; removing three electrons gives [Ar] 3d0, 0 unpaired electrons → colourless.
\( \ce{Fe^{2+}} \): Fe (Z = 26) is [Ar] 3d6 4s2; \( \ce{Fe^{2+}} \) is 3d6 (↑↓ ↑ ↑ ↑ ↑), 4 unpaired → green.
\( \ce{Ti^{3+}} \): Ti (Z = 22) is [Ar] 3d2 4s2; \( \ce{Ti^{3+}} \) is 3d1, 1 unpaired → purple.
\( \ce{Mn^{2+}} \): Mn (Z = 25) is [Ar] 3d5 4s2; \( \ce{Mn^{2+}} \) is 3d5 (↑ ↑ ↑ ↑ ↑), 5 unpaired → pink.
NCERT (2026-27 edition):
Formation of Coloured Ions (Table 4.8: Colours of Some of the First Row (aquated) Transition Metal Ions)
| Configuration | Example | Colour |
|---|---|---|
| 3d0 | Sc3+ | colourless |
| 3d0 | Ti4+ | colourless |
| 3d1 | Ti3+ | purple |
| 3d1 | V4+ | blue |
| 3d2 | V3+ | green |
| 3d3 | V2+ | violet |
| 3d3 | Cr3+ | violet |
| 3d4 | Mn3+ | violet |
| 3d4 | Cr2+ | blue |
| 3d5 | Mn2+ | pink |
| 3d5 | Fe3+ | yellow |
| 3d6 | Fe2+ | green |
| 3d6 3d7 | Co3+ Co2+ | blue pink |
| 3d8 | Ni2+ | green |
| 3d9 | Cu2+ | blue |
| 3d10 | Zn2+ | colourless |
Note: The same/similar question has been asked in different years by NTA as per the details given below:
| 2026 | 2025 | 2024 | 2023 | 2022 |
|---|---|---|---|---|
| – | – | – | 20-06 S3 10-06 S3 | – |