Question

\(\mathrm{Ti}\left(\mathrm{NH}_{3}\right)_{6}{ }^{3+}\) has a d-orbital electron transition at \(399 \mathrm{nm}\). Find \(\Delta_{\mathrm{o}}\) at this wavelength.

Short answer

Answer: The crystal field splitting energy, \(\Delta_\mathrm{o}\), for the hexaammine titanium(III) complex ion at the given wavelength of 399 nm is approximately \(4.98 \times 10^{-19}\,\text{J}\).

Step-by-step solution

Identify the known values

We know the following information: - Wavelength of the transition, \(\lambda = 399\,\text{nm}\) - Speed of light, \(c = 3.00 \times 10^8\,\text{m/s}\) - Planck's constant, \(h = 6.626 \times 10^{-34}\,\text{J s}\)

Convert the wavelength to meters

First, let's convert the given wavelength from nanometers to meters, to make sure our units are consistent in our calculations. 1 nm = \(10^{-9}\) m, so \(\lambda = 399\,\text{nm} = 399 \times 10^{-9}\,\text{m}\)

Calculate the energy of the transition

We can calculate the energy of the transition using the following formula: \(E = \dfrac{hc}{\lambda}\) Plugging in the known values: \(E = \dfrac{(6.626 \times 10^{-34}\,\text{J s})(3.00 \times 10^8\,\text{m/s})}{(399 \times 10^{-9}\,\text{m})}\)

Calculate the crystal field splitting energy, \(\Delta_\mathrm{o}\)

The energy of the transition is equal to the crystal field splitting energy. Therefore, \(\Delta_\mathrm{o} = E\) Now, we can evaluate the expression for \(E\) we found in Step 3: \(\Delta_\mathrm{o} = \dfrac{(6.626 \times 10^{-34}\,\text{J s})(3.00 \times 10^8\,\text{m/s})}{(399 \times 10^{-9}\,\text{m})}\)

Calculate the final result

After evaluating the expression for \(\Delta_\mathrm{o}\), we find: \(\Delta_\mathrm{o} \approx 4.98 \times 10^{-19}\,\text{J}\) This is the crystal field splitting energy for the hexaammine titanium(III) complex ion at the given wavelength of 399 nm.