For the `[Cr(H_(2)O)_(6)]^(2+)` ion, the mean pairing energy P is found to `23500cm^(-1)`. The magnitude of `Delta_(0)` is `13900cm^(-1)`. Calculate the C.F.S.E. `(cm^(-1))` for this complex ion corresponding to high spin state (x) and low spin state (y). Write your answer as `((y-x)/(100))`.

Explanation

To solve the problem, we need to calculate the Crystal Field Stabilization Energy (C.F.S.E.) for the complex ion \([Cr(H_2O)_6]^{2+}\) in both high spin and low spin states.

### Step 1: Identify the electronic configuration
The chromium ion \([Cr^{2+}]\) has an atomic number of 24. In its +2 oxidation state, it loses two electrons, resulting in a configuration of \([Ar] 3d^4\).

### Step 2: Determine the high spin configuration
For the high spin state of \(d^4\):
- The electrons will occupy the \(t_{2g}\) and \(e_g\) orbitals in a way that maximizes the number of unpaired electrons.
- The configuration will be: \(t_{2g}^3 e_g^1\).

### Step 3: Calculate C.F.S.E. for high spin state (x)
The formula for C.F.S.E. in a high spin state is:
\[ \text{C.F.S.E.} = -0.4 \Delta_0 \times n_{t_{2g}} - 0.6 \Delta_0 \times n_{e_g} \]
Where:
- \(n_{t_{2g}} = 3\) (number of electrons in \(t_{2g}\))
- \(n_{e_g} = 1\) (number of electrons in \(e_g\))

Substituting the values:
\[ \text{C.F.S.E. (high spin)} = -0.4 \times 13900 \, \text{cm}^{-1} \times 3 - 0.6 \times 13900 \, \text{cm}^{-1} \times 1 \]
Calculating:
\[ = -0.4 \times 13900 \times 3 - 0.6 \times 13900 \]
\[ = -16680 \, \text{cm}^{-1} - 8340 \, \text{cm}^{-1} \]
\[ = -25020 \, \text{cm}^{-1} \]

### Step 4: Determine the low spin configuration
For the low spin state of \(d^4\):
- The electrons will pair up in the \(t_{2g}\) orbitals before occupying the \(e_g\) orbitals.
- The configuration will be: \(t_{2g}^4 e_g^0\).

### Step 5: Calculate C.F.S.E. for low spin state (y)
The formula for C.F.S.E. in a low spin state is:
\[ \text{C.F.S.E.} = -0.4 \Delta_0 \times n_{t_{2g}} - 0.6 \Delta_0 \times n_{e_g} + P \]
Where:
- \(n_{t_{2g}} = 4\) (number of electrons in \(t_{2g}\))
- \(n_{e_g} = 0\) (number of electrons in \(e_g\))

Substituting the values:
\[ \text{C.F.S.E. (low spin)} = -0.4 \times 13900 \, \text{cm}^{-1} \times 4 - 0.6 \times 13900 \, \text{cm}^{-1} \times 0 + 23500 \]
Calculating:
\[ = -0.4 \times 13900 \times 4 + 23500 \]
\[ = -22240 \, \text{cm}^{-1} + 23500 \]
\[ = 1260 \, \text{cm}^{-1} \]

### Step 6: Calculate \((y - x)/100\)
Now we have:
- \(x = -25020 \, \text{cm}^{-1}\) (C.F.S.E. for high spin)
- \(y = 1260 \, \text{cm}^{-1}\) (C.F.S.E. for low spin)

Calculating \(y - x\):
\[ y - x = 1260 - (-25020) = 1260 + 25020 = 26280 \, \text{cm}^{-1} \]

Finally, calculate \((y - x)/100\):
\[ \frac{y - x}{100} = \frac{26280}{100} = 262.8 \]

### Final Answer
\[ \frac{y - x}{100} = 262.8 \]