The electronic spectrum of `[Ti(H_(2)O)_(6)]^(3+)` shows a single broad peak with a maximum at `"20,300 cm"^(-1)`.
The crystal field stabillization energy (CFSE) of the complex ion, in `"kJ mol"^(-1)`, is :
`("1 kJ mol"^(-1) =" 83.7 cm"^(-1))`

Options

A. 83.7
B. 242.5
C. 145.5
D. 97

Explanation

To solve the problem of finding the crystal field stabilization energy (CFSE) of the complex ion \([Ti(H_2O)_6]^{3+}\), we can follow these steps:

### Step 1: Determine the oxidation state and electron configuration of Titanium
Titanium in the complex \([Ti(H_2O)_6]^{3+}\) has a +3 oxidation state. The atomic number of Titanium (Ti) is 22, so in its neutral state, it has the electron configuration:
\[ \text{[Ar]} 3d^2 4s^2 \]
In the +3 oxidation state, it loses 3 electrons (two from 4s and one from 3d), resulting in:
\[ \text{Ti}^{3+}: 3d^1 \]

### Step 2: Identify the geometry of the complex
The complex \([Ti(H_2O)_6]^{3+}\) is octahedral because it has six water ligands coordinated to the titanium ion.

### Step 3: Understand the splitting of d-orbitals in an octahedral field
In an octahedral field, the d-orbitals split into two sets:
- \(t_{2g}\) (lower energy)
- \(e_g\) (higher energy)

For \(Ti^{3+}\) with a \(d^1\) configuration, the single electron will occupy one of the \(t_{2g}\) orbitals.

### Step 4: Calculate the CFSE
The CFSE can be calculated using the formula:
\[ \text{CFSE} = (n_{t_{2g}} \times -0.4 \Delta_o) + (n_{e_g} \times 0.6 \Delta_o) \]
Where:
- \(n_{t_{2g}}\) = number of electrons in \(t_{2g}\) orbitals
- \(n_{e_g}\) = number of electrons in \(e_g\) orbitals
- \(\Delta_o\) = octahedral splitting energy

For \([Ti(H_2O)_6]^{3+}\):
- \(n_{t_{2g}} = 1\) (one electron in \(t_{2g}\))
- \(n_{e_g} = 0\) (no electrons in \(e_g\))

Thus, the CFSE becomes:
\[ \text{CFSE} = (1 \times -0.4 \Delta_o) + (0 \times 0.6 \Delta_o) = -0.4 \Delta_o \]

### Step 5: Substitute the value of \(\Delta_o\)
We are given that the maximum peak in the electronic spectrum corresponds to:
\[ \Delta_o = 20,300 \, \text{cm}^{-1} \]

Now substituting this value into the CFSE equation:
\[ \text{CFSE} = -0.4 \times 20,300 \, \text{cm}^{-1} \]
\[ \text{CFSE} = -8120 \, \text{cm}^{-1} \]

### Step 6: Convert CFSE from \(\text{cm}^{-1}\) to \(\text{kJ mol}^{-1}\)
We know from the problem statement that:
\[ 1 \, \text{kJ mol}^{-1} = 83.7 \, \text{cm}^{-1} \]

To convert \(-8120 \, \text{cm}^{-1}\) to \(\text{kJ mol}^{-1}\):
\[ \text{CFSE} = \frac{-8120}{83.7} \, \text{kJ mol}^{-1} \]
Calculating this gives:
\[ \text{CFSE} \approx -97.01 \, \text{kJ mol}^{-1} \]

Since CFSE is typically expressed as a positive value for stabilization energy:
\[ \text{CFSE} \approx 97 \, \text{kJ mol}^{-1} \]

### Final Answer
The crystal field stabilization energy (CFSE) of the complex ion \([Ti(H_2O)_6]^{3+}\) is approximately:
\[ \boxed{97 \, \text{kJ mol}^{-1}} \]

---