Using Crystal Field Theory, answer the following question.
Predict, with explanation, the relative magnitude of wavelengths of the observed colours of the following complex ions: [CuBr6]3-, [Cu(H2O)6]3+, [Cu(CO)6]3+.
In each case, state whether the complex would be high-spin or low-spin by drawing their d-orbital electronic configurations.
To predict the relative magnitude of wavelengths of the observed colors of complex ions using Crystal Field Theory (CFT), we analyze the splitting of d-orbitals in the presence of ligands. This splitting determines the energy difference between the orbitals, which in turn affects the absorbed and emitted light wavelengths.
First, let's determine the d-orbital electronic configurations and whether the complexes would be high-spin or low-spin.
[CuBr6]3-:
For [CuBr6]3-, we have copper(III) surrounded by six bromide ligands. Copper (Cu) has an electron configuration of [Ar] 3d^10 4s^1. In the presence of strong-field ligands like bromide (Br-), the t2g and eg orbitals experience significant splitting. This leads to the promotion of one electron from the t2g to the eg orbital, resulting in the d^8 configuration. Since the number of unpaired electrons in the d-orbitals is two (d^8), it is considered a high-spin complex.
[Cu(H2O)6]3+:
For [Cu(H2O)6]3+, we have copper(III) surrounded by six water (H2O) ligands. Copper (Cu) still has an electron configuration of [Ar] 3d^10 4s^1. Water is a weak-field ligand, so the splitting of d-orbitals is small. This causes no promotion of electrons, resulting in the d^9 configuration. Since the number of unpaired electrons in the d-orbitals is one (d^9), it is considered a low-spin complex.
[Cu(CO)6]3+:
For [Cu(CO)6]3+, we have copper(III) surrounded by six carbonyl (CO) ligands. Copper (Cu) retains its electron configuration of [Ar] 3d^10 4s^1. CO is a strong-field ligand, leading to significant splitting of d-orbitals. One electron is promoted from t2g to eg, resulting in the d^8 configuration. Similar to [CuBr6]3-, the number of unpaired electrons in the d-orbitals is two (d^8), making it a high-spin complex.
Now, let's consider the relative magnitude of wavelengths of the observed colors. In CFT, the energy difference between the high-energy (eg) and low-energy (t2g) orbitals determines the color absorbed or emitted. The larger the energy difference, the shorter the wavelength (higher frequency) of absorbed or emitted light.
Since [CuBr6]3+ and [Cu(CO)6]3+ both have the same d-orbital configuration (d^8), they will have a similar energy difference and, therefore, a similar magnitude of wavelengths. [Cu(H2O)6]3+ has a different d-orbital configuration (d^9), resulting in a different energy difference.
In general, [Cu(H2O)6]3+ would have a larger energy difference and, consequently, shorter wavelength colors compared to [CuBr6]3+ and [Cu(CO)6]3+. Therefore, the observed colors would be:
1. [Cu(H2O)6]3+: Shorter wavelength (higher frequency) colors
2. [CuBr6]3- and [Cu(CO)6]3+: Relatively longer wavelength colors compared to [Cu(H2O)6]3+
Remember that specific wavelengths and colors depend on the ligand and metal ion. Experimentally, you would need to consider factors such as the crystal field splitting energy, ligand field strength, and other factors to determine the precise colors.