In the \(\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}{ }^{3+}\) ion, the splitting between the \(\mathrm{d}\) levels, \(\Delta_{\mathrm{o}}\) is \(55 \mathrm{kcal} / \mathrm{mol}\). What is the color of this ion? Assume that the color results from a transition between upper and lower d levels.

We are given the energy difference (\(\Delta_\mathrm{o}\)) in kcal/mol. First, we need to convert this to Joules per mole by using the conversion factor 1 kcal/mol = 4184 J/mol. Then, we will use the energy-wavelength relationship (\(E = \frac{hc}{\lambda}\)) to find the wavelength of the absorbed light, where \(E\) is the energy difference, \(h\) is the Planck's constant (6.626 * 10⁻³⁴ J s), \(c\) is the speed of light (2.998 * 10⁸ m/s), and \(\lambda\) is the wavelength. \(\Delta_\mathrm{o} = 55\,\mathrm{kcal/mol}\) Converting to Joules per mole: \(\Delta_\mathrm{o} = (55\,\mathrm{kcal/mol}) * (4184\,\mathrm{J/mol})\) \(\Delta_\mathrm{o} = 230,020\,\mathrm{J/mol}\) Now, we can use the energy-wavelength relationship to find the wavelength, \(\lambda\): \(\lambda = \frac{hc}{\Delta_{\mathrm{o}}}\)

Substituting the given values of \(h\), \(c\), and \(\Delta_{o}\) into the equation, we can calculate the wavelength of the absorbed_light: \(\lambda = \frac{(6.626\mathrm{* 10⁻³⁴\ J\,s) * (2.998\mathrm{* 10⁸\ m/s})}{230,020\, \mathrm{J/mol}}\) Solving for \(\lambda\), we obtain: \(\lambda = 8.609 * 10^{-19}\, \mathrm{m} * \mathrm{mol}\) Knowing that 1 wave number (cm⁻¹) is equal to 100,000,000 meter-wavenumber per meter, we can convert the wavelength into nanometers: \(\lambda = \frac{8.609*10^{-19}\,\mathrm{m}*\mathrm{mol}}{100,000,000\, \mathrm{m^{-1}}}\) \(\lambda = 8.609 * 10^{-27}\,\mathrm{m}^2*\,\mathrm{mol}\) Converting meters to nanometers: \(\lambda = 860.9\, \mathrm{nm}\)

Now that we have the wavelength of the absorbed light, we can determine the color of the ion. Using the visible light spectrum, we know that: - 400-450 nm: Violet - 450-490 nm: Blue - 490-570 nm: Green - 570-590 nm: Yellow - 590-620 nm: Orange - 620-750 nm: Red Since the calculated wavelength is 860.9 nm, which is outside the visible light spectrum. We must check the complementary color to determine the observed color of the ion. Subtracting our calculated wavelength (860.9 nm) from the maximum visible wavelength (750 nm) gives: Complementary wavelength = 750 nm - 860.9 nm = -110.9 nm Since the difference is negative, this indicates the transition is in the infrared region. However, to be complete we can find the complementary color in the visible spectrum to relate the color: Complementary color = 750 nm + 110.9 nm = 860.9 nm (as expected) Thus, the color of the \(\mathrm{Ti(H_{2}O)_{6}^{3+}}\) ion is not visible because it lies in the infrared region.