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?

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: \[ \Delta = 182 \times 1000 \mathrm{~J/mol} = 182000 \mathrm{~J/mol} \]

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: \[ \Delta = \frac{182000 \mathrm{~J/mol}}{6.022 \times 10^{23} \text{ mol}^{-1}} = 3.02 \times 10^{-19} \mathrm{~J/photon} \]

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: \[ \lambda = \frac{6.626 \times 10^{-34} \text{ J} \cdot \text{s} \times 2.998 \times 10^8 \text{ m/s}}{3.02 \times 10^{-19} \text{ J/photon}} = 6.57 \times 10^{-7} \text{ m} \]

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.