Problem 100 - Chemistry The Central Science In Si Units

Question

The value of \(\Delta\) for the \(\left[\mathrm{MoI}_{6}\right]^{3-}\) complex is \(198.58 \mathrm{~kJ} / \mathrm{mol}\). Calculate the expected wavelength of the absorption corresponding to promotion of an electron from the lower energy to the higher-energy \(d\) -orbital set in this complex. Should the complex absorb in the visible range?

Step-by-Step Solution

Answer

The complex absorbs at 604 nm, which is in the visible range.

Step 1: Understanding the Problem

We are given the crystal field splitting energy (\(\Delta\)) for the \(\left[\mathrm{MoI}_{6}\right]^{3-}\) complex as \(198.58 \mathrm{~kJ/mol}\). We need to calculate the wavelength corresponding to this energy change and determine if it falls in the visible spectrum.

Step 2: Convert Energy to Joules

Since \(\Delta\) is given in kJ/mol, convert it to joules per electron using the conversion factor from kJ to J. Since 1 kJ = 1000 J and using Avogadro's number, we get:\[ 198.58 \times 10^3 \text{ J/mol} \times \frac{1 \text{ mol}}{6.022 \times 10^{23} \text{ photons/mol}} = 3.297 \times 10^{-19} \text{ J/photon} \]

Step 3: Use the Energy-Wavelength Relationship

Use the equation \(E = \frac{hc}{\lambda}\), where \(h\) is Planck's constant \((6.626 \times 10^{-34} \text{ J s})\), \(c\) is the speed of light \((3.00 \times 10^8 \text{ m/s})\), and \(\lambda\) is the wavelength, to find \(\lambda\).Rearrange the formula to find \(\lambda\):\[ \lambda = \frac{hc}{E} \]

Step 4: Calculate the Wavelength

Substituting the known values into \(\lambda = \frac{hc}{E}\):\[ \lambda = \frac{(6.626 \times 10^{-34} \text{ J s})(3.00 \times 10^8 \text{ m/s})}{3.297 \times 10^{-19} \text{ J/photon}} = 6.04 \times 10^{-7} \text{ m} = 604 \text{ nm} \]

Step 5: Determine the Visible Range

The visible light spectrum ranges from approximately 380 nm (violet) to 750 nm (red). The calculated wavelength of 604 nm falls within this range, specifically in the orange part of the spectrum.

Key Concepts

Wavelength Calculation

When dealing with complexes, it's often important to calculate the wavelength of light that corresponds to the energy used to promote an electron from a lower energy level to a higher one. This is particularly relevant in the context of crystal field theory, where the splitting of the d-orbitals in transition metal complexes can absorb light of specific wavelengths.

To calculate the wavelength from the energy, we use the formula derived from the relationship between energy and wavelength:

• The formula is given by \( E = \frac{hc}{\lambda} \), where:
  • \( E \) is the energy in joules (J)
  • \( h \) is Planck's constant, approximately \( 6.626 \times 10^{-34} \text{ J s} \)
  • \( c \) is the speed of light, about \( 3.00 \times 10^8 \text{ m/s} \)
  • \( \lambda \) is the wavelength in meters (m)

Once we know the energy \( E \), we rearrange the formula to solve for \( \lambda \), the wavelength:\[ \lambda = \frac{hc}{E} \] This equation allows us to translate the energy absorbed or emitted in a transition to the corresponding wavelength of light.

Visible Spectrum

Light that we see is only a small portion of the total electromagnetic spectrum. The visible spectrum ranges from about 380 nanometers (nm) to 750 nm in wavelength, covering all the colors we can see from violet to red. This is crucial for understanding how the \( [\mathrm{MoI}_{6}]^{3-} \) complex interacts with light.

When we calculate the wavelength of absorbed light, like in the case of our complex where the calculated wavelength is 604 nm, we are determining whether this absorption will be visible to our eyes.

• The spectrum can be divided:
  • Short wavelengths (~380 nm) are seen as violet.
  • Longer wavelengths (~750 nm) appear red.
  • The region around 604 nm, as in this complex, corresponds to an orange color within the visible spectrum.

By understanding which wavelengths are absorbed, we get a hint about the color of the complex. If a complex absorbs a particular wavelength strongly, that color might be missing from the visible light reflected by the complex, changing how it presents itself visually.

Electron Promotion

In the context of transition metal complexes and crystal field theory, electron promotion refers to the movement of an electron from a lower energy set of d-orbitals to a higher energy set. This occurs when the complex absorbs a photon with energy equal to the energy difference between these orbitals.

For example, consider the \( [\mathrm{MoI}_{6}]^{3-} \) complex. It absorbs light at a wavelength calculated to be 604 nm, which corresponds to the energy needed to promote an electron in this complex.

• This promotion can cause changes in:
  • Electronic configuration
  • Stability of the complex
  • Its color, as particular wavelengths of light are absorbed

Understanding electron promotion gives insight into the functional properties of the complex and can also explain color changes seen in various experimental conditions. This characteristic absorption and promotion are central to the applications and study of coordination chemistry.