Arrange the following in decreasing order of the wavenumber of absorption in the visible region:
A. [Ni(CN)6]4−
B. [Ni(H2O)6]2+
C. [Ni(NH3)6]2+
D. [Ni(en)3]2+
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → A, D, C, B
In an octahedral complex the d orbitals split into t2g and eg, separated by the crystal field splitting energy Δo. The complex absorbs the visible light whose energy matches this gap, and the energy of light is directly proportional to its wavenumber (1/λ):
\( \displaystyle E=\frac{hc}{\lambda} \)
\( \displaystyle \text{wavenumber}=\frac{1}{\lambda}=\frac{\Delta_o}{hc} \)
So the order of wavenumber is the same as the order of Δo. The metal ion is \( \ce{Ni^{2+}} \) in all four complexes, so the ligand alone decides Δo, through the spectrochemical series. Taking only the four ligands present:
\( \displaystyle \mathrm{CN^- > en > NH_3 > H_2O} \)
Matching the given complexes:
(A) \( \ce{[Ni(CN)6]^{4-}} \): cyanido, the strongest field ligand here
(D) \( \ce{[Ni(en)3]^{2+}} \): ethane-1,2-diamine, a chelating diamine, stronger than ammonia
(C) \( \ce{[Ni(NH3)6]^{2+}} \): ammine
(B) \( \ce{[Ni(H2O)6]^{2+}} \): aqua, the weakest field ligand here
Shortcut: The four complexes differ only in the ligand, so you need just the two ends of the series: \( \ce{CN^{-}} \) is the strongest and \( \ce{H2O} \) the weakest. Only Option (1) starts with A and ends with B.
Note: It seems that there is a mistake in the original NTA paper. Complex A is printed as \( \ce{[Ni(CN)6]^{4-}} \); nickel(II) and cyanide ions actually form the square planar \( \ce{[Ni(CN)4]^{2-}} \), not a six-coordinate cyanido complex.
NCERT (2026-27 edition):
Crystal Field Theory – …
