Which of the following complexes will show absorbance of light at the lowest wavelength?

$[Co(CN)_6]^{3-}$

The correct answer is Option (4) → $[Co(CN)_6]^{3-}$

These are octahedral Co(III) (d⁶, low-spin) complexes, where visible color arises from d-d transitions. The energy of the absorbed light equals the crystal field splitting parameter Δ₀.

• Higher Δ₀ → higher energy absorbed → shorter wavelength.
• Lower Δ₀ → lower energy absorbed → longer wavelength.

The spectrochemical series orders ligands by increasing field strength: Cl⁻ < H₂O < NH₃ << CN⁻.

Thus, the order of Δ₀ (and absorbed energy) is: [CoCl(NH₃)₅]²⁺ (weakest, mostly Cl⁻ influence) < [Co(NH₃)₅(H₂O)]³⁺ < [Co(NH₃)₆]³⁺ < [Co(CN)₆]³⁻ (strongest).

Therefore, [Co(CN)₆]³⁻ has the largest Δ₀, absorbs the highest-energy (shortest-wavelength) light (often shifting into UV, appearing pale yellow), while the chloro complex absorbs the longest wavelength.