(a) If a complex absorbs light at \(610 \mathrm{nm},\) what color would you expect the complex to be? (b) What is the energy in joules of a photon with a wavelength of 610 \(\mathrm{nm} ?\) (c) What is the energy of this absorption in \(\mathrm{kJ} / \mathrm{mol}\) ?

To find the color of the complex, we must first determine which color is complementary to the color it absorbs. Visible light has a wavelength range of about 380 nm to 750 nm. We can use the following approximate ranges for colors: - Violet: 380 - 450 nm - Blue: 450 - 495 nm - Green: 495 - 570 nm - Yellow: 570 - 590 nm - Orange: 590 - 620 nm - Red: 620 - 750 nm Since the complex absorbs light at 610 nm, it falls within the orange color range. This means that the complex will likely appear blue because blue is the complementary color to orange.

To calculate the energy of a photon, we can use Planck's equation: \( E = h\cdot\nu \), where \(E\) is the energy of the photon, \(h\) is Planck's constant (\(6.626\times10^{-34} \, \mathrm{Js}\)), and \(\nu\) is the frequency of light. Note that we are given the wavelength of light (\(\lambda\)), rather than its frequency. We can convert wavelength to frequency using the speed of light (\(c\)) and the equation: \( \nu = \dfrac{c}{\lambda} \). First, we need to convert the wavelength to meters. Since 1 nm = \(1\times10^{-9} \, \mathrm{m}\), 610 nm = 610×\(1\times10^{-9} \, \mathrm{m}\) = \(6.10\times10^{-7} \, \mathrm{m}\). Next, we can find the frequency: \( \nu = \dfrac{c}{\lambda} = \dfrac{3\times10^{8} \, \mathrm{m/s}}{6.10\times10^{-7} \, \mathrm{m}} = 4.918\times10^{14} \, \mathrm{s^{-1}} \). Finally, we can plug the frequency into Planck's equation to find the energy of a photon: \( E = h\cdot\nu = (6.626\times10^{-34} \, \mathrm{Js})(4.918\times10^{14} \, \mathrm{s^{-1}}) = 3.260\times10^{-19} \, \mathrm{J} \).

To find the energy per mole, we need to multiply the energy of a single photon by Avogadro's number (\(6.022\times10^{23}\, \mathrm{mol^{-1}}\)). Then, we need to convert the energy from Joules to kilojoules. \( E_\mathrm{mol} = E \cdot N_A = (3.260\times10^{-19} \, \mathrm{J})(6.022\times10^{23}\, \mathrm{mol^{-1}}) = 1.9630\times10^{5} \, \mathrm{J/mol} \) Now, convert to kJ/mol: \( \dfrac{1.9630\times10^{5} \, \mathrm{J/mol}}{1\times10^{3} \, \mathrm{J/kJ}} = 196.3 \, \mathrm{kJ/mol} \). To summarize our results: (a) The complex is expected to be blue. (b) The energy of a photon with a wavelength of 610 nm is \(3.260\times10^{-19} \, \mathrm{J}\). (c) The energy of this absorption in kJ/mol is 196.3 kJ/mol.