Match List-I with List-II
| List-I | List-II |
| Coordination complex ion | Wavelength of light absorbed (nm) |
| (A) [Co(NH3)6]3+ | (I) 310 |
| (B) [CoCl(NH3)5]2+ | (II) 475 |
| (C) [Co(CN)6]3- | (III) 500 |
| (D) [Co(NH3)5(H2O)]3+ | (IV) 535 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) - (II), (B) - (IV), (C) - (I), (D) - (III)
The colour of these cobalt(III) complexes comes from a d–d transition, and the energy absorbed equals the crystal field splitting Δo. A stronger field ligand gives a larger Δo, so light of higher energy, that is shorter wavelength, is absorbed. The spectrochemical series gives the ligand strength:
\( \displaystyle \ce{Cl^-} < \ce{H2O} < \ce{NH3} < \ce{CN^-} \)
\( \displaystyle \lambda=\frac{hc}{\Delta_o}\quad\Rightarrow\quad \text{larger }\Delta_o,\ \text{shorter }\lambda \)
So the wavelength absorbed decreases in the order \( \ce{[CoCl(NH3)5]^{2+}} \) > \( \ce{[Co(NH3)5(H2O)]^{3+}} \) > \( \ce{[Co(NH3)6]^{3+}} \) > \( \ce{[Co(CN)6]^{3-}} \).
(A) \( \ce{[Co(NH3)6]^{3+}} \) — This matches (II) 475. Six \( \ce{NH3} \) ligands give a fairly large splitting.
(B) \( \ce{[CoCl(NH3)5]^{2+}} \) — This matches (IV) 535. Replacing one \( \ce{NH3} \) by the weak field \( \ce{Cl^{-}} \) gives the smallest splitting, so the longest wavelength is absorbed.
(C) \( \ce{[Co(CN)6]^{3-}} \) — This matches (I) 310. \( \ce{CN^{-}} \) is the strongest field ligand here, so the splitting is largest and the shortest wavelength (ultraviolet) is absorbed.
(D) \( \ce{[Co(NH3)5(H2O)]^{3+}} \) — This matches (III) 500. \( \ce{H2O} \) is a weaker ligand than \( \ce{NH3} \) but stronger than \( \ce{Cl^{-}} \), so the wavelength lies between.
Shortcut: The strongest ligand gives the shortest wavelength, so (C)–(I), which appears in Options (2) and (4). Then (A)–(II) leaves only Option (4).
NCERT (2026-27 edition):
Colour in Coordination Compounds
