Question

In the \(\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}{ }^{3+}\right]\) ion, the splitting between the d levels, \(\Delta_{0}\), 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.

Short answer

Answer: The color of the \(\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}{ }^{3+}\right]\) ion is green.

Step-by-step solution

Convert energy to Joules

We are given the energy difference in kcal/mol, but we need to use the SI unit, which is Joules. To convert kcal/mol to Joules/mol, use the conversion factor: 1 kcal = 4.184 kJ. So, our energy difference becomes: \(\Delta_0 = 55 \frac{\text{kcal}}{\text{mol}} \cdot \frac{4.184 \text{kJ}}{1 \text{kcal}} \cdot \frac{1,000 \text{J}}{1 \text{kJ}} = 230,200 \frac{\text{J}}{\text{mol}}\)

Calculate the energy per photon

To find the energy per photon, we use the Avogadro's number (NA) conversion factor: \(E_\text{photon} = \frac{\Delta_0}{N_A} = \frac{230,200 \frac{\text{J}}{\text{mol}}}{6.022 \times 10^{23} \frac{1}{\text{mol}}} = 3.822 \times 10^{-19} \text{J} \)

Determine the wavelength of the light

Now, we will use Planck's equation to calculate the wavelength of light corresponding to this energy difference: \(E = h \cdot c / \lambda\) where: \(\lambda\) = wavelength (in meters) \(E\) = energy per photon (Joules) \(h\) = Planck's constant (\(6.626 \times 10^{-34} \text{J} \cdot \text{s}\)) \(c\) = speed of light (\(2.998 \times 10^8 \frac{\text{m}}{\text{s}}\)) Rearranging for wavelength, we get: \(\lambda = \frac{h \cdot c}{E} = \frac{6.626 \times 10^{-34} \text{J} \cdot \text{s} \cdot 2.998 \times 10^8 \frac{\text{m}}{\text{s}}}{3.822 \times 10^{-19} \text{J}} = 5.18 \times 10^{-7} \text{m}\)

Identify the color

Now that we have calculated the wavelength, we can match it to the visible light spectrum to find the corresponding color. The visible spectrum ranges from about 400 nm (violet) to 700 nm (red). Our calculated wavelength is: \(\lambda = 5.18 \times 10^{-7} \text{m} = 518 \text{nm}\) 518 nm falls within the green portion of the visible light spectrum. Therefore, the color of the \(\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}{ }^{3+}\right]\) ion is green.