Calculate CFSE for complex `[Co(H_2O)_3F_3] [Given triangle_@lt P]`

Options

A. `-0.8triangle_@ + 2P`
B. `-0.4triangle_@ + P`
C. `-0.8triangle_@`
D. `-0.4triangle_@`

Answer

To calculate the Crystal Field Stabilization Energy (CFSE) for the complex \([Co(H_2O)_3F_3]\), we will follow these steps:

### Step 1: Determine the oxidation state of cobalt
The complex \([Co(H_2O)_3F_3]\) consists of three water molecules (neutral) and three fluoride ions (each with a charge of -1).

Let the oxidation state of cobalt be \(x\):
\[ x + 0 - 3 = 0 \implies x = +3 \]
So, cobalt is in the +3 oxidation state.

### Step 2: Write the electronic configuration of cobalt in +3 state
Cobalt in its elemental form has the electronic configuration of \([Ar] 3d^7 4s^2\). When cobalt is in the +3 oxidation state, it loses two 4s electrons and one 3d electron:
\[ \text{Cobalt (III)}: [Ar] 3d^6 \]

### Step 3: Identify the geometry of the complex
The coordination number of the complex is 6 (3 from \(H_2O\) and 3 from \(F\)), indicating that it has an octahedral geometry.

### Step 4: Determine the splitting of d-orbitals in octahedral field
In an octahedral field, the \(d\) orbitals split into two sets:
- \(t_{2g}\) (lower energy): consists of \(d_{xy}, d_{xz}, d_{yz}\)
- \(e_g\) (higher energy): consists of \(d_{x^2-y^2}, d_{z^2}\)

### Step 5: Fill the electrons in the split d-orbitals
Since we have \(3d^6\) configuration, we will fill the electrons in the \(t_{2g}\) and \(e_g\) orbitals. Given that the pairing energy is greater than the crystal field splitting energy (\(\Delta\)), electrons will occupy the \(t_{2g}\) orbitals before pairing occurs in the \(e_g\) orbitals.

- Fill \(t_{2g}\) first: 6 electrons will fill as follows:
- \(t_{2g}\): 6 electrons (3 pairs)
- \(e_g\): 0 electrons

### Step 6: Calculate CFSE
The formula for CFSE in octahedral complexes is:
\[ \text{CFSE} = \left(-\frac{2}{5} \Delta_0 \times \text{(number of electrons in } t_{2g})\right) + \left(\frac{3}{5} \Delta_0 \times \text{(number of electrons in } e_g)\right) \]

Substituting the values:
- Number of electrons in \(t_{2g} = 6\)
- Number of electrons in \(e_g = 0\)

\[ \text{CFSE} = \left(-\frac{2}{5} \Delta_0 \times 6\right) + \left(\frac{3}{5} \Delta_0 \times 0\right) \]
\[ \text{CFSE} = -\frac{12}{5} \Delta_0 \]

### Final Answer
Thus, the CFSE for the complex \([Co(H_2O)_3F_3]\) is:
\[ \text{CFSE} = -\frac{12}{5} \Delta_0 \]

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