Problem 99 - Chemistry The Central Science

Question

The value of \(\Delta\) for the \(\left[\mathrm{CrF}_{6}\right]^{3-}\) complex is \(182 \mathrm{~kJ} / \mathrm{mol}\). Calculate the expected wavelength of the absorption corresponding to promotion of an electron from the lowerenergy to the higher-energy \(d\) -orbital set in this complex. Should the complex absorb in the visible range?

Step-by-Step Solution

Answer

The expected wavelength of absorption for the \(\left[\mathrm{CrF}_{6}\right]^{3-}\) complex is \(6.57 \times 10^{-7}\text{ m}\), which falls within the visible range (380 nm to 760 nm). Thus, the complex should absorb in the visible range.

Step 1: Calculate the energy of photon

The energy of the photon is given by the value of \(\Delta = 182 \mathrm{~kJ/mol}\). We'll first convert this energy from kJ to J. Multiply by 1000 to get the energy in J/mol: \[ \Delta = 182 \times 1000 \mathrm{~J/mol} = 182000 \mathrm{~J/mol} \]

Step 2: Convert energy per mole to energy per photon

Next, we need to convert the energy per mole to the energy per photon. To do this, divide the energy in J/mol by Avogadro's number, \(N_A = 6.022 \times 10^{23} \text{ mol}^{-1} \). This will give us the energy in J/photon: \[ \Delta = \frac{182000 \mathrm{~J/mol}}{6.022 \times 10^{23} \text{ mol}^{-1}} = 3.02 \times 10^{-19} \mathrm{~J/photon} \]

Step 3: Use Planck's equation to find the wavelength

Planck's equation relates energy (\(E\)), frequency (\(\nu\)), and wavelength (\(\lambda\)) by: \[ E = h \nu = \frac{hc}{\lambda} \] Here, \(h\) is the Planck's constant (\(h = 6.626 \times 10^{-34} \text{ J} \cdot \text{s}\)) and \(c\) is the speed of light in a vacuum (\(c = 2.998 \times 10^8 \text{ m/s} \)). We need to find the wavelength (\(\lambda\)) corresponding to the energy (\(E = 3.02 \times 10^{-19} \text{ J/photon}\)). Rearranging Planck's equation gives: \[ \lambda = \frac{hc}{E} \] Now, substitute the values of \(E\), \(h\), and \(c\) to find the wavelength: \[ \lambda = \frac{6.626 \times 10^{-34} \text{ J} \cdot \text{s} \times 2.998 \times 10^8 \text{ m/s}}{3.02 \times 10^{-19} \text{ J/photon}} = 6.57 \times 10^{-7} \text{ m} \]

Step 4: Determine if the complex absorbs in the visible range

The visible range of light falls between 380 nm and 760 nm (3.80 x 10^{-7} m and 7.60 x 10^{-7} m). The calculated wavelength for the chromium hexafluoride complex is 6.57 x 10^{-7} m, which falls within the visible range. Therefore, the complex should absorb in the visible range.

Key Concepts

d-orbital splitting

d-orbital splitting is a phenomenon that occurs in transition metal complexes, such as \([\text{CrF}_6]^{3-}\). When ligands, which are molecules that donate electrons, approach these metals, the degeneracy (or equality in energy) of the metal's d-orbitals is broken. This results in what we call splitting. The energy difference between the higher and lower sets of d-orbitals is represented by \(\Delta\). This splitting happens because the electrons in the ligands repel the electrons in the d-orbitals. As a result, some d-orbitals experience more repulsion than others, causing the energy levels to shift. The amount of this shift depends on several factors, including the strength of the ligand field and the geometry of the complex. For transition metals, this d-orbital splitting is important because it determines how the metal ions interact with light, including which wavelengths they might absorb or reflect.

wavelength calculation

In order to calculate the wavelength of light absorbed by a complex, we use the relationship between energy and wavelength described by Planck's equation. The energy \(E\) of the photon absorbed can be found from \(\Delta\). In our example with \([\text{CrF}_6]^{3-}\), \(\Delta\) is given as 182 kJ/mol, which needs to be converted to energy per photon.- Start by converting from kJ/mol to J/mol by multiplying by 1,000.- Then, divide the result by Avogadro's number to find energy per photon.Using Planck's equation \(E = \frac{hc}{\lambda}\), we can then solve for wavelength \(\lambda\):\[ \lambda = \frac{hc}{E} \]Where:

• \(h\) is Planck's constant \(6.626 \times 10^{-34} \text{ J} \cdot \text{s}\)
• \(c\) is the speed of light \(2.998 \times 10^8 \text{ m/s}\)

With the given energy \(E\) of \(3.02 \times 10^{-19} \text{ J/photon}\), we find the wavelength \(\lambda\) of 6.57 \times 10^{-7} \text{ m}. This value helps to determine if the absorbed light falls within the visible spectrum.

visible spectrum absorption

The visible spectrum is the range of wavelengths that can be seen by the human eye, typically between 380 nm and 760 nm. When a substance absorbs a wavelength within this range, it affects the color of the substance as perceived by our eyes. For the \( [\text{CrF}_6]^{3-} \) complex, we calculated a wavelength of 6.57 \times 10^{-7} \text{ m}, which falls within this visible range, specifically around 657 nm. This confirms that the complex does indeed absorb visible light. - Since it absorbs this light, a color complementary to this wavelength will be reflected or transmitted. - The specific color seen is due to the wavelengths that are not absorbed.Thus, understanding visible spectrum absorption is crucial for predicting the color properties of various transition metal complexes.