The increasing order of wavelength of absorption for the complex ions: i) $[Cr(NH_3)_6]^{3+}$, ii) $[CrCl_6]^{3-}$, iii) $[Cr(OH_2)_6]^{3+}$, iv) $[Cr(CN)_6]^{3-}$, is
Crystal Field Theory and Absorption Wavelength
The energy required for an electron to transition to a higher energy level in a transition metal complex is directly related to the Crystal Field Splitting Energy ($\Delta_o$). This energy is inversely proportional to the wavelength of light absorbed ($\lambda$). The relationship is given by the equation:
$ \Delta_o = \frac{hc}{\lambda} $
where \(h\) is Planck's constant and \(c\) is the speed of light. Therefore, a complex with a larger $\Delta_o$ absorbs light of shorter wavelength, and a complex with a smaller $\Delta_o$ absorbs light of longer wavelength.
Spectrochemical Series and Ligand Strength
The strength of a ligand, which determines the magnitude of $\Delta_o$, is given by the spectrochemical series. For the ligands present in the given complexes, the order of increasing ligand strength (and thus increasing $\Delta_o$) is:
- $Cl^-$ (Chloride)
- $H_2O$ (Water)
- $NH_3$ (Ammonia)
- $CN^-$ (Cyanide)
This order corresponds to:
$ [CrCl_6]^{3-} < [Cr(OH_2)_6]^{3+} < [Cr(NH_3)_6]^{3+} < [Cr(CN)_6]^{3-} $
The ligands are listed in increasing order of their ability to split d-orbitals, meaning $\Delta_o$ increases in this sequence.
Determining Wavelength Order
Since wavelength ($\lambda$) is inversely proportional to $\Delta_o$, the increasing order of absorption wavelength will be the reverse of the order of increasing $\Delta_o$. This means the complex with the strongest ligand ($CN^-$) will have the shortest absorption wavelength, and the complex with the weakest ligand ($Cl^-$) will have the longest absorption wavelength.
The complexes are:
- i) $[Cr(NH_3)_6]^{3+}$
- ii) $[CrCl_6]^{3-}$
- iii) $[Cr(OH_2)_6]^{3+}$
- iv) $[Cr(CN)_6]^{3-}$
Based on the ligand strength order ($CN^- > NH_3 > H_2O > Cl^-$), the order of increasing $\Delta_o$ is:
$ \Delta_o([Cr(CN)_6]^{3-}) > \Delta_o([Cr(NH_3)_6]^{3+}) > \Delta_o([Cr(OH_2)_6]^{3+}) > \Delta_o([CrCl_6]^{3-}) $
And the corresponding order of increasing absorption wavelength ($\lambda$) is:
$ \lambda([Cr(CN)_6]^{3-}) < \lambda([Cr(NH_3)_6]^{3+}) < \lambda([Cr(OH_2)_6]^{3+}) < \lambda([CrCl_6]^{3-}) $
Mapping this to the roman numerals:
$ \text{iv} < \text{i} < \text{iii} < \text{ii} $
Final Answer Mapping
The increasing order of wavelength of absorption is: iv) $[Cr(CN)_6]^{3-}$ < i) $[Cr(NH_3)_6]^{3+}$ < iii) $[Cr(OH_2)_6]^{3+}$ < ii) $[CrCl_6]^{3-}$.
This corresponds to option C.
Important Questions from Electronic Spectra
- Consider the given pairs of complex ions. The correct pair(s) in which each complex ion exhibits three spin-allowed $d-d$ transitions (excluding those arising due to distortions) is(are)
- The UV-visible spectrum of $[Ni(en)_3]^{2+}$ (en = ethylenediamine) shows absorbance maxima at $11200 \text{ cm}^{-1}$, $18350 \text{ cm}^{-1}$, and $29000 \text{ cm}^{-1}$.
Absorbance maximum Electronic transition (a) $11200 \text{ cm}^{-1}$ (i) $^3A_{2g} \to ^3T_{1g} (F)$ (b) $18350 \text{ cm}^{-1}$ (ii) $^3A_{2g}\to^3T_{2g}$ (c) $29000 \text{ cm}^{-1}$ (iii) $^3A_{2g}\to^3T_{1g} (P)$
[Given: Atomic number of Ni = 28]
The correct match(es) between absorbance maximum and electronic transition is/are - In aqueous solution of $K_4[Fe(CN)_6]$, the allowed transition(s) is (are)
- The lowest energy d $\rightarrow$ d transition of the complexes follow the order
In the first row high-spin transition metal complexes $[M(H_2O)_6]Cl_2$ with $d^5$ and $d^7$ metal ions, the $d-d$ transitions are
