Match the metal ion given in List-I with its property given in List-II
|
List-I Transition metal ion |
List-II Property |
|
(A) \( Mn^{3+} \) |
(I) \( 3d^6 \), green |
|
(B) \( Fe^{2+} \) |
(II) \( 3d^4 \), violet |
|
(C) \( Ti^{4+} \) |
(III) \( 3d^0 \), colorless |
|
(D) \( V^{2+} \) |
(IV) \( 3d^3 \), violet |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(II), (B)-(I), (C)-(III), (D)-(IV)
The colour of a hydrated transition metal ion comes from d-d transitions of its partly filled d subshell. An ion with no d electrons (3d0) or a full set (3d10) has no such transition and is colourless. Each configuration is obtained from the atomic number by removing the 4s electrons first.
(A) \(Mn^{3+}\) — This matches (II) \(3d^4\), violet. Mn (Z = 25) is \( \mathrm{[Ar]\,3d^{5}\,4s^{2}} \); losing two 4s and one 3d electron gives \( \mathrm{3d^{4}} \). The aquated ion is violet.
(B) \(Fe^{2+}\) — This matches (I) \(3d^6\), green. Fe (Z = 26) is \( \mathrm{[Ar]\,3d^{6}\,4s^{2}} \); losing the two 4s electrons gives \( \mathrm{3d^{6}} \). The aquated ion is green.
(C) \(Ti^{4+}\) — This matches (III) \(3d^0\), colorless. Ti (Z = 22) is \( \mathrm{[Ar]\,3d^{2}\,4s^{2}} \); losing all four outer electrons leaves \( \mathrm{3d^{0}} \), so there is no d-d transition.
(D) \(V^{2+}\) — This matches (IV) \(3d^3\), violet. V (Z = 23) is \( \mathrm{[Ar]\,3d^{3}\,4s^{2}} \); losing the two 4s electrons gives \( \mathrm{3d^{3}} \). The aquated ion is violet.
Shortcut: \( \ce{Ti^4+} \) with no d electrons must be colourless, so (C)-(III) leaves Options (1) and (2); \( \ce{Mn^3+} \) is 3d4, (A)-(II), which leaves Option (2).
NCERT (2026-27 edition):
Formation of Coloured Ions
Table 4.8: Colours of Some of the First Row (aquated) Transition Metal Ions (rows)
| Configuration | Example | Colour |
|---|---|---|
| 3d0 | Ti4+ | colourless |
| 3d3 | V2+ | violet |
| 3d4 | Mn3+ | violet |
| 3d6 | Fe2+ | green |