The electronic spectrum of $$\left[Ti(H_2O)_6\right]^{3+}$$ shows a single broad peak with a maximum at 20,300 cm$$^{-1}$$. The crystal field stabilization energy (CFSE) of the complex ion, in kJ mol$$^{-1}$$, is: (1 kJ mol$$^{-1}$$ = 83.7 cm$$^{-1}$$)

We have the octahedral complex $$\left[Ti(H_2O)_6\right]^{3+}$$. The oxidation state of titanium is +3, so the metal ion is $$Ti^{3+}$$.

Titanium in the elemental state has the configuration $$[Ar]\,3d^2\,4s^2$$. Removing three electrons (first the two 4s and then one 3d) gives

$$Ti^{3+} : [Ar]\,3d^1$$

Thus the ion is a $$d^1$$ system placed in an octahedral crystal field provided by six water molecules.

In an octahedral field, the single $$d$$ electron enters the lower-energy $$t_{2g}$$ set. A $$d\!\rightarrow\!d$$ electronic transition therefore corresponds to

$$t_{2g}^1 \;\longrightarrow\; e_g^1$$

The energy gap for this transition is the octahedral crystal-field splitting parameter, usually denoted $$\Delta_0$$ (or $$10Dq$$). The spectrum shows one broad band with its maximum at

$$\tilde{\nu}_{\text{max}} = 20{,}300 \text{ cm}^{-1}$$

Hence,

$$\Delta_0 = 20{,}300 \text{ cm}^{-1}$$

The crystal field stabilization energy (CFSE) for an octahedral $$d^1$$ ion is obtained from the standard CFSE expression. First we state the formula:

For each electron in the $$t_{2g}$$ set, the stabilization is $$-0.4\Delta_0$$, while each electron in the $$e_g$$ set is destabilized by $$+0.6\Delta_0$$. With one electron in $$t_{2g}$$ and none in $$e_g$$ we have

$$\text{CFSE} = 1 \times (-0.4\Delta_0) = -0.4\Delta_0$$

Substituting $$\Delta_0 = 20{,}300 \text{ cm}^{-1}$$:

$$\text{CFSE} = -0.4 \times 20{,}300 \text{ cm}^{-1}$$ $$\text{CFSE} = -8{,}120 \text{ cm}^{-1}$$

The negative sign simply denotes stabilization, so the magnitude of the stabilization energy is

$$|\text{CFSE}| = 8{,}120 \text{ cm}^{-1}$$

We are asked for the value in kJ mol$$^{-1}$$. The conversion factor given is

$$1 \text{ kJ mol}^{-1} = 83.7 \text{ cm}^{-1}$$

So we convert:

$$|\text{CFSE}| = \frac{8{,}120 \text{ cm}^{-1}}{83.7 \text{ cm}^{-1}\,\text{per kJ mol}^{-1}}$$

Now dividing:

$$|\text{CFSE}| = 97 \text{ kJ mol}^{-1}\;(\text{approximately})$$

Therefore, the crystal field stabilization energy of $$\left[Ti(H_2O)_6\right]^{3+}$$ is $$97 \text{ kJ mol}^{-1}$$.

Hence, the correct answer is Option D.