Match List I with List II (wavelength of light absorbed in nm): (A) [CoCl(NH3)5]2+ (B) [Co(NH3)6]3+ (C) [Co(CN)6]3- (D)…

🧠 λ Inversely Tracks Δ

$\lambda \propto \frac{1}{\Delta}$

Stronger field → larger Δ → shorter λ absorbed.

🗺️ Rank Each Complex by Field Strength

| Complex | Field assessment | Δ rank | |---|---|---| | (C) $[\mathrm{Co(CN)_6}]^{3-}$ | 6 × strong CN⁻ | largest Δ (smallest λ) | | (B) $[\mathrm{Co(NH_3)_6}]^{3+}$ | 6 × NH₃ (strong) | high Δ | | (A) $[\mathrm{CoCl(NH_3)_5}]^{2+}$ | 5 NH₃ + 1 Cl⁻ (Cl⁻ weak) | lower Δ than B | | (D) $[\mathrm{Cu(H_2O)_4}]^{2+}$ | Cu(II) + weak H₂O | smallest Δ (largest λ) |

Order of λ (small → large): C < B < A < D.

Match to the four wavelengths (310, 475, 535, 600):

• C → 310 (I)
• B → 475 (II)
• A → 535 (III)
• D → 600 (IV)

⚡ Why Cu(II) Sits at Long λ

Cu²⁺ aquo absorbs in the red (≈ 600–800 nm) → transmits the famously pale-blue colour. The reasons: low charge (+2), Jahn-Teller distortion, weak ligand. All three lower Δ.

⚠️ Replacing One Strong Ligand with One Weaker Ligand Shifts λ

[Co(NH₃)₆]³⁺ (475 nm) → [CoCl(NH₃)₅]²⁺ (535 nm): swapping one NH₃ for one Cl⁻ adds ~60 nm. Small change in ligand identity = real shift in λ.

$\boxed{\text{Answer: (2) A-III, B-II, C-I, D-IV}}$