The enthalpy of hydration of `Cr^(+2)` is `-460` kcal `mo1^(-1)` In the absence of `CFSE` the value for `DeltaH=-424` kcal `mo1^(-1)` What is the value of `Delta_(0)` for `[Cr(H_(2)O)_(6))]^(2+)` .

Options

A. `60kcal "mole"^(-1)`
B. `-60kcal "mole"^(-1)`
C. `25.7 kcal "mole"^(-1)`
D. `-25.7kcal "mole"^(-1)`

Explanation

To find the value of \( \Delta_0 \) for the complex \([Cr(H_2O)_6]^{2+}\), we will follow these steps:

### Step 1: Understand the Given Data
We are provided with the following data:
- Enthalpy of hydration of \( Cr^{2+} \) = \( -460 \) kcal/mol
- Enthalpy change in the absence of CFSE = \( -424 \) kcal/mol

### Step 2: Calculate the Crystal Field Stabilization Energy (CFSE)
The CFSE can be calculated using the formula:
\[ \text{CFSE} = \text{Enthalpy of hydration} - \text{Enthalpy change without CFSE} \]
Substituting the values:
\[ \text{CFSE} = -460 \text{ kcal/mol} - (-424 \text{ kcal/mol}) = -460 + 424 = -36 \text{ kcal/mol} \]

### Step 3: Determine the Electron Configuration of \( Cr^{2+} \)
The electron configuration of chromium (\( Cr \)) is:
\[ [Ar] 3d^5 4s^1 \]
For \( Cr^{2+} \), two electrons are removed (typically from the 4s and then 3d):
\[ Cr^{2+} : [Ar] 3d^4 \]

### Step 4: Identify the Splitting of d-Orbitals
In an octahedral field created by six water molecules (a strong field ligand), the d-orbitals split into two sets:
- \( t_{2g} \) (lower energy)
- \( e_g \) (higher energy)

For \( Cr^{2+} \) with 4 d-electrons, the distribution will be:
- \( t_{2g}^3 \) (3 electrons)
- \( e_g^1 \) (1 electron)

### Step 5: Calculate the CFSE Contribution
The CFSE can be calculated using the formula:
\[ \text{CFSE} = (n_{t_{2g}} \times -0.4 \Delta_0) + (n_{e_g} \times 0.6 \Delta_0) \]
Where:
- \( n_{t_{2g}} = 3 \) (number of electrons in \( t_{2g} \))
- \( n_{e_g} = 1 \) (number of electrons in \( e_g \))

Substituting the values:
\[ \text{CFSE} = (3 \times -0.4 \Delta_0) + (1 \times 0.6 \Delta_0) = -1.2 \Delta_0 + 0.6 \Delta_0 = -0.6 \Delta_0 \]

### Step 6: Relate CFSE to the Calculated Value
From Step 2, we found that CFSE = \( -36 \) kcal/mol. Therefore:
\[ -0.6 \Delta_0 = -36 \text{ kcal/mol} \]

### Step 7: Solve for \( \Delta_0 \)
Now, we can solve for \( \Delta_0 \):
\[ \Delta_0 = \frac{-36 \text{ kcal/mol}}{-0.6} = 60 \text{ kcal/mol} \]

### Conclusion
The value of \( \Delta_0 \) for \([Cr(H_2O)_6]^{2+}\) is \( 60 \) kcal/mol.

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