Crystal field splitting energy `(CFSE)` for the complex `[Cr(NH_(3))_(6)]^(2+)` is when ` P =125` and `Delta_(0) =250kjmol^(-1)` .

Options

A. `525kJmo1^(-1)`
B. `275kJmo1^(-1)`
C. `-275kJmo1^(-1)`
D. `-525kJm01^(1-)`

Explanation

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

### Step 1: Determine the oxidation state of chromium
The complex \([Cr(NH_3)_6]^{2+}\) has an overall charge of +2. Since ammonia (NH₃) is a neutral ligand, the oxidation state of chromium (Cr) in this complex is +2.

### Step 2: Write the electronic configuration of Cr²⁺
The atomic number of chromium (Cr) is 24. The electronic configuration of neutral chromium is:
\[ Cr: [Ar] 3d^5 4s^1 \]
For Cr²⁺, we remove two electrons (the 4s electron and one 3d electron):
\[ Cr^{2+}: [Ar] 3d^4 \]

### Step 3: Identify the splitting in an octahedral field
In an octahedral field, the 3d orbitals split into two sets: \(t_{2g}\) (lower energy) and \(e_g\) (higher energy). The splitting can be represented as:
- \(t_{2g}\) level: 3 orbitals
- \(e_g\) level: 2 orbitals

### Step 4: Determine the ligand field strength
We are given:
- \(P = 125 \, \text{kJ/mol}\)
- \(\Delta_0 = 250 \, \text{kJ/mol}\)

Since \(P < \Delta_0\), this indicates that the ligands create a strong field, leading to electron pairing.

### Step 5: Fill the electrons in the split d-orbitals
For \(Cr^{2+}\) with 4 electrons in the \(3d\) subshell:
- The first three electrons will fill the \(t_{2g}\) orbitals singly.
- The fourth electron will pair with one of the existing electrons in the \(t_{2g}\) level.

The filling will look like this:
- \(t_{2g}^4\)
- \(e_g^0\)

### Step 6: Calculate the CFSE
The formula for CFSE is:
\[ CFSE = (n_{t_{2g}} \times -0.4 \Delta_0) + (n_{e_g} \times +0.6 \Delta_0) + (n_{paired} \times P) \]
Where:
- \(n_{t_{2g}} = 4\)
- \(n_{e_g} = 0\)
- \(n_{paired} = 1\) (since one electron is paired)

Substituting the values:
\[ CFSE = (4 \times -0.4 \times 250) + (0 \times 0.6 \times 250) + (1 \times 125) \]
\[ CFSE = (-400) + 0 + 125 \]
\[ CFSE = -275 \, \text{kJ/mol} \]

### Final Answer
The CFSE for the complex \([Cr(NH_3)_6]^{2+}\) is \(-275 \, \text{kJ/mol}\).

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