1. Find the electronic configuration and CFSE relation

For $\left[\mathrm{Ti}(\mathrm{H_2O})_6\right]^{3+}$:

• Titanium atomic number $=22$
• $\mathrm{Ti}^{3+} = 3d^1$

In an octahedral field, a $d^1$ ion has configuration: $t_{2g}^1 e_g^0$

So its crystal field stabilization energy is: $\mathrm{CFSE} = -0.4\Delta_o$

Given: $\mathrm{CFSE} = -96.0\ \text{kJ mol}^{-1}$

Thus, $-0.4\Delta_o = -96.0$ $\Delta_o = \frac{96.0}{0.4} = 240\ \text{kJ mol}^{-1}$

2. Relate $\Delta_o$ to absorbed photon energy

For a $d^1$ octahedral complex, maximum absorption corresponds to transition: $t_{2g} \to e_g$ Hence absorbed energy per mole is approximately: $E = \Delta_o = 240\times 10^3\ \text{J mol}^{-1}$

3. Convert molar energy to energy per photon

$E_{\text{photon}} = \frac{240\times 10^3}{6\times 10^{23}}$ $= 4.0\times 10^{-19}\ \text{J}$

4. Use $E = \dfrac{hc}{\lambda}$

$\lambda = \frac{hc}{E}$

Given:

• $h = 6.4\times 10^{-34}\ \text{J s}$
• $c = 3.0\times 10^8\ \text{m s}^{-1}$

So, $\lambda = \frac{(6.4\times 10^{-34})(3.0\times 10^8)}{4.0\times 10^{-19}}$

$\lambda = \frac{19.2\times 10^{-26}}{4.0\times 10^{-19}}$ $\lambda = 4.8\times 10^{-7}\ \text{m}$

5. Convert into nm

$4.8\times 10^{-7}\ \text{m} = 480\ \text{nm}$

Final Answer

$\boxed{480}$