The standard example is $\ce{CrCl3.6H2O}$, which can have three distinctly different crystalline compounds now known as

$\ce{[Cr(H2O)6]Cl3}$ (violet),

$\ce{[CrCl(H2O)5]Cl .H2O}$ (blue-green), and

$\ce{[CrCl2(H2O)4]Cl.2H2O}$ (dark green).

How does this difference of ligands in octahedral system affect the colour of compound? I think it may be due to the change of splitting energy but if that so then how does it actually work?


1 Answer 1


Your idea appears to be correct.

Initially, the structure of $\ce{CrCl3.6H2O}$ corresponds to the dark green compound $\ce{[CrCl3(H2O)3].3H2O}$. This complex shows hydrate isomerism; i.e. the chlorido ligands are step by step replaced by water molecules:

$$\ce{$\underset{\text{dark green}}{\ce{[CrCl3(H2O)3].3H2O}}$ <=> $\underset{\text{dark green}}{\ce{[CrCl2(H2O)4]Cl.2H2O}}$ <=> $\underset{\text{blue green}}{\ce{[CrCl(H2O)5]Cl2.H2O}}$ <=> $\underset{\text{violet}}{\ce{[Cr(H2O)6]Cl3}}$}$$

Thus, the actual complexes are:

$\ce{[CrCl3(H2O)3]}$ (dark green)
$\ce{[CrCl2(H2O)4]+}$ (dark green)
$\ce{[CrCl(H2O)5]^2+}$ (blue green)
$\ce{[Cr(H2O)6]^3+}$ (violet)

Each of these complexes has octahedral geometry. Cr(III) has a $\mathrm{d^3}$ electronic configuration. In the ground state, three electrons occupy the $\mathrm{t_{2g}}$ orbitals ($\mathrm{d}_{xy}$, $\mathrm{d}_{xz}$, and $\mathrm{d}_{yz}$). The two $\mathrm{e_g}$ orbitals ($\mathrm{d}_{x^2{-}y^2}$ and $\mathrm{d}_{z^2}$) are empty.

Three spin allowed d–d transitions are expected for Cr(III) complexes. The corresponding absorption maxima of $\ce{[Cr(H2O)6]^3+}$ are:

$$\begin{alignat}{2} &\mathrm{^4A_{2g}\to{^4T_{2g}}}\quad&&\tilde\nu_1=17\,400\ \mathrm{cm^{-1}}\\ &\mathrm{^4A_{2g}\to{^4T_{1g}(F)}}\quad&&\tilde\nu_2=24\,500\ \mathrm{cm^{-1}}\\ &\mathrm{^4A_{2g}\to{^4T_{1g}(P)}}\quad&&\tilde\nu_3=38\,600\ \mathrm{cm^{-1}} \end{alignat}$$

The first transition $\mathrm{^4A_{2g}\to{^4T_{2g}}}$ corresponds to promoting one electron from the $\mathrm{t_{2g}}$ level to the $\mathrm{e_g}$ level. Thus, it directly gives the value of $10\ \mathrm{Dq}$ as $17\,400\ \mathrm{cm^{-1}}$.

The energy splitting of the spectral terms depends on the ligand field strength. If the $\ce{H2O}$ ligands are replaced by the low-field ligand $\ce{Cl-}$, the value of $10\ \mathrm{Dq}$ and the wavenumber of the corresponding absorption maximum are lowered:

$$\begin{alignat}{2} &\ce{[Cr(H2O)6]^3+}\quad&&\tilde\nu_1=17\,400\ \mathrm{cm^{-1}}\\ &\ce{[CrCl(H2O)5]^2+}\quad&&\tilde\nu_1=16\,400\ \mathrm{cm^{-1}}\\ &\ce{[CrCl2(H2O)4]+}\quad&&\tilde\nu_1=15\,600\ \mathrm{cm^{-1}}\\ &\vdots \\ &\ce{[CrCl6]^3-}\quad&&\tilde\nu_1=13\,200\ \mathrm{cm^{-1}} \end{alignat}$$

By way of comparison, if the $\ce{H2O}$ ligands are replaced by the strong-field ligand $\ce{CN-}$, the wavenumber of the corresponding absorption maximum is increased ($\tilde\nu_1=26\,700\ \mathrm{cm^{-1}}$ for $\ce{[CrCN6]^3-}$).

The wavenumbers can be converted to the wavelength of the absorbed light:

$$\begin{alignat}{2} &\ce{[Cr(H2O)6]^3+}\quad&&\tilde\nu_1=17\,400\ \mathrm{cm^{-1}}\quad\lambda=575\ \mathrm{nm}\ \text{(yellow)}\\ &\ce{[CrCl(H2O)5]^2+}\quad&&\tilde\nu_1=16\,400\ \mathrm{cm^{-1}}\quad\lambda=610\ \mathrm{nm}\ \text{(orange)}\\ &\ce{[CrCl2(H2O)4]+}\quad&&\tilde\nu_1=15\,600\ \mathrm{cm^{-1}}\quad\lambda=640\ \mathrm{nm}\ \text{(red)}\\ \end{alignat}$$

Since the colour of the absorbed light in this series changes from yellow to orange to red, the corresponding colour of the remaining visible light changes from violet to blue green to green.


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