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 lower-energy to the higher-energy \(d\) -orbital set in this complex. Should the complex absorb in the visible range? (You may need to review Sample Exercise 6.3; remember to divide by Avogadro's number.)

Short answer

The expected wavelength of absorption for the \(\left[\mathrm{CrF}_{6}\right]^{3-}\) complex is \(657 \mathrm{~nm}\), which falls within the visible range (400 to 700 nm). Therefore, the complex should absorb in the visible range.

Step-by-step solution

Convert energy to joules per particle

First, let's convert the given value of \(\Delta\) to joules per particle by dividing by Avogadro's number (\(6.022 \times 10^{23} \mathrm{~particles} / \mathrm{mol}\)): \[\frac{182 \mathrm{~kJ}}{\mathrm{mol}} \times \frac{1000 \mathrm{~J}}{1 \mathrm{~kJ}} \times \frac{1 \mathrm{~mol}}{6.022 \times 10^{23} \mathrm{~particles}} = 3.018 \times 10^{-19} \mathrm{~J} / \mathrm{particle} \]

Use the energy-wavelength relationship

Next, we will use the energy-wavelength relationship from the Planck equation: \[E = h\nu = \frac{hc}{\lambda}\] Where \(E\) is the energy, \(h\) is the Planck constant (\(6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\)), \(c\) is the speed of light (\(2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}\)), and \(\lambda\) is the wavelength. We will solve for \(\lambda\): \[\lambda = \frac{hc}{E}\]

Calculate the expected wavelength

Substitute the values of \(h, c,\) and \(E\) that we derived and calculated in Steps 1 and 2 and solve for the wavelength \(\lambda\): \[\lambda = \frac{(6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s})(2.998 \times 10^{8} \mathrm{~m} / \mathrm{s})}{3.018 \times 10^{-19} \mathrm{~J} / \mathrm{particle}} = 657 \mathrm{~nm}\]

Determine if the complex absorbs in the visible range

The visible range of the electromagnetic spectrum is typically between 400 nm and 700 nm. Since the calculated expected wavelength of absorption of the complex \(\left[\mathrm{CrF}_{6}\right]^{3-}\) is 657 nm, we can conclude that the complex should absorb in the visible range.