Corystal field splitting energy `(CFSE)` for the complex `[Cr(H_(2)O)_(6)]^(2+)` is when `(DeltaltP)` .

Options

A. ` 1.2Delat_(0)`
B. `-0.6Delta_(0)`
C. `0.6Delta_(0)`
D. `1.2Delta_(0)`

Explanation

To solve the problem of calculating the Crystal Field Splitting Energy (CFSE) for the complex \([Cr(H_2O)_6]^{2+}\), we will follow these steps:

### Step 1: Determine the Oxidation State of Chromium
The complex is \([Cr(H_2O)_6]^{2+}\). Since water (H2O) is a neutral ligand, the oxidation state of chromium can be calculated as follows:

Let the oxidation state of chromium be \(x\).
\[ x + 6(0) = +2 \implies x = +2 \]
Thus, the oxidation state of chromium in this complex is +2.

**Hint:** Always start by determining the oxidation state of the metal in the coordination complex.

### Step 2: Write the Electron Configuration of Chromium
The atomic number of chromium (Cr) is 24. The ground state electron configuration is:
\[ \text{Cr: } [Ar] 4s^1 3d^5 \]
For the +2 oxidation state, two electrons are removed (first from the 4s and then from the 3d):
\[ \text{Cr}^{2+}: [Ar] 3d^4 \]

**Hint:** Remember that when determining the electron configuration for cations, electrons are removed from the outermost shell first.

### Step 3: Distribute Electrons in the Crystal Field
In an octahedral field, the \(d\) orbitals split into two sets: \(t_{2g}\) (lower energy) and \(e_g\) (higher energy). For \(d^4\) configuration:
- The first three electrons will fill the \(t_{2g}\) orbitals.
- The fourth electron will go into the \(e_g\) orbital because the splitting energy (\(\Delta\)) is less than the pairing energy.

The distribution will be:
- \(t_{2g} = 3\) electrons
- \(e_g = 1\) electron

**Hint:** In an octahedral field, fill the lower energy orbitals first before moving to higher energy orbitals, especially when the splitting energy is low.

### Step 4: Calculate the CFSE
The formula for CFSE is given by:
\[ \text{CFSE} = (0.4 \times \text{number of electrons in } t_{2g}) - (0.6 \times \text{number of electrons in } e_g) \]
Substituting the values:
\[ \text{CFSE} = (0.4 \times 3) - (0.6 \times 1) = 1.2 - 0.6 = 0.6 \]

Since CFSE is typically expressed as a negative value (indicating stabilization), we write:
\[ \text{CFSE} = -0.6 \]

**Hint:** Remember the coefficients for \(t_{2g}\) and \(e_g\) when calculating CFSE; they reflect the energy associated with electron placement in these orbitals.

### Conclusion
The Crystal Field Splitting Energy (CFSE) for the complex \([Cr(H_2O)_6]^{2+}\) is \(-0.6\).

**Final Answer:** \(-0.6\)