Crystal field splitting energy `(CFSE)` for the complex `[Cr(NH_(3))_(6)]^(2+)` is (when `Delta gtP`) .

Options

A. `-1.6Delta_(0)+P`
B. `1.6Delta_(0)+P`
C. `-c-24Delta_(0) +P`
D. `2.4Delta_(0)+P`

Explanation

To solve the problem of calculating the Crystal Field Splitting Energy (CFSE) for the complex \([Cr(NH_3)_6]^{2+}\), we will follow these steps:

### Step 1: Determine the oxidation state of chromium
The complex \([Cr(NH_3)_6]^{2+}\) indicates that chromium is in the +2 oxidation state.

**Hint:** The charge of the complex gives the oxidation state of the central metal ion.

### Step 2: Identify the electron configuration of chromium
The atomic number of chromium (Cr) is 24. The electron configuration for neutral chromium is \([Ar] 3d^5 4s^1\). For \(Cr^{2+}\), we remove two electrons, which will be from the 4s and one from the 3d subshell. Thus, the electron configuration for \(Cr^{2+}\) is \(3d^4\).

**Hint:** Remember to remove electrons from the outermost shell first when determining the oxidation state.

### Step 3: Determine the nature of the ligand
Ammonia (NH₃) is a strong field ligand. This means that it causes a larger splitting of the d-orbitals compared to weak field ligands.

**Hint:** Strong field ligands lead to larger splitting energy (Δ) and can cause electron pairing.

### Step 4: Understand the splitting of d-orbitals in octahedral complexes
In an octahedral field, the \(d\) orbitals split into two sets: \(t_{2g}\) (lower energy) and \(e_g\) (higher energy). The \(t_{2g}\) set contains three orbitals, while the \(e_g\) set contains two orbitals.

**Hint:** Visualize the splitting pattern of the d-orbitals in an octahedral field.

### Step 5: Fill the d-orbitals with electrons
For \(Cr^{2+}\) with a \(3d^4\) configuration:
- The first three electrons will fill the \(t_{2g}\) orbitals (one in each orbital).
- The fourth electron will pair up in one of the \(t_{2g}\) orbitals due to the strong field nature of NH₃.

Thus, the distribution will be:
- \(t_{2g} = 4\) electrons (3 unpaired + 1 paired)
- \(e_g = 0\) electrons

**Hint:** Remember that pairing occurs when the splitting energy is greater than the pairing energy.

### Step 6: Calculate the CFSE
The formula for CFSE in an octahedral complex is given by:
\[ CFSE = (-0.4 \Delta_0 \times \text{number of electrons in } t_{2g}) + (-0.6 \Delta_0 \times \text{number of electrons in } e_g) + \text{pairing energy} \]

For our case:
- Number of electrons in \(t_{2g} = 4\)
- Number of electrons in \(e_g = 0\)
- Pairing energy = \(P\)

Thus, the CFSE becomes:
\[ CFSE = (-0.4 \Delta_0 \times 4) + (-0.6 \Delta_0 \times 0) + P = -1.6 \Delta_0 + P \]

**Hint:** Substitute the values into the CFSE formula carefully.

### Final Answer:
The CFSE for the complex \([Cr(NH_3)_6]^{2+}\) is:
\[ CFSE = -1.6 \Delta_0 + P \]

This indicates that the CFSE is influenced by both the crystal field splitting energy and the pairing energy.