Which of the following complexes will show absorbance of light at the lowest wavelength?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \ce{[Co(CN)6]^3-} \)
The colour of a complex comes from a d-d transition: an electron absorbs light and moves from the lower set of d orbitals to the higher set. The energy absorbed equals the crystal field splitting \( \Delta_o \), and energy and wavelength are inversely related:
\( \displaystyle \Delta_o = \frac{hc}{\lambda} \)
So the larger \( \Delta_o \), the shorter the wavelength absorbed. \( \Delta_o \) depends on the field strength of the ligands, given by the spectrochemical series:
\( \displaystyle \ce{Cl-} \lt \ce{H2O} \lt \ce{NH3} \lt \text{en} \lt \ce{CN-} \)
All four complexes contain Co(III). \( \ce{CN-} \) is the strongest field ligand, so \( \ce{[Co(CN)6]^3-} \) has the largest \( \Delta_o \) and absorbs light of the lowest wavelength.
Why the other options are incorrect
Option (1) \( \ce{[Co(NH3)6]^3+} \) — Incorrect. \( \ce{NH3} \) is a weaker field ligand than \( \ce{CN-} \), so \( \Delta_o \) is smaller and the wavelength absorbed is longer.
Option (2) \( \ce{[CoCl(NH3)5]^2+} \) — Incorrect. One \( \ce{NH3} \) is replaced by the weak field \( \ce{Cl-} \), which lowers \( \Delta_o \) further; of the four it absorbs at the longest wavelength.
Option (3) \( \ce{[Co(NH3)5(H2O)]^3+} \) — Incorrect. \( \ce{H2O} \) is weaker than \( \ce{NH3} \), so \( \Delta_o \) is a little smaller than in \( \ce{[Co(NH3)6]^3+} \).
NCERT (2026-27 edition):
Crystal Field Theory – “… In general, ligands can be arranged in a series in the order of increasing field strength as given below:
I– < Br– < SCN– < Cl– < S2– < F– < OH– < C2O42– < H2O < NCS– < edta4– < NH3 < en < CN– < CO
Such a series is termed as spectrochemical series. It is an experimentally determined series based on the absorption of light by complexes with different ligands.”