Calculate |C.F.S.E| (mod value) is term of Dq. For complex ion `[MnF_(6)]^(3-)`.

Explanation

To calculate the crystal field stabilization energy (C.F.S.E) for the complex ion \([MnF_6]^{3-}\) in terms of \(Dq\), we will follow these steps:

### Step 1: Determine the oxidation state of manganese in \([MnF_6]^{3-}\)

The oxidation state can be calculated as follows:

Let the oxidation state of manganese be \(x\). The fluorine ion has a charge of \(-1\) and there are 6 fluorine atoms. Therefore, we can set up the equation:

\[ x + 6(-1) = -3 \]

Solving for \(x\):

\[ x - 6 = -3 \implies x = +3 \]

### Step 2: Determine the electron configuration of \(Mn^{3+}\)

The atomic number of manganese (Mn) is 25. The electron configuration of neutral manganese is:

\[ [Ar] 3d^5 4s^2 \]

For \(Mn^{3+}\), we remove 3 electrons (2 from 4s and 1 from 3d):

\[ Mn^{3+} \rightarrow [Ar] 3d^4 \]

### Step 3: Understand the splitting of d-orbitals in an octahedral field

In an octahedral field, the \(3d\) orbitals split into two sets:

- \(t_{2g}\) (lower energy, 3 orbitals)
- \(e_g\) (higher energy, 2 orbitals)

For \(3d^4\), the distribution of electrons in the split d-orbitals will be:

- \(t_{2g}\): 3 electrons (all singly occupied)
- \(e_g\): 1 electron (one electron occupies one of the higher energy orbitals)

### Step 4: Calculate the C.F.S.E

The C.F.S.E can be calculated using the formula:

\[ \text{C.F.S.E} = (n_{t_{2g}} \times -\frac{2}{5}Dq) + (n_{e_g} \times \frac{3}{5}Dq) \]

Where:
- \(n_{t_{2g}} = 3\) (number of electrons in \(t_{2g}\))
- \(n_{e_g} = 1\) (number of electrons in \(e_g\))

Substituting the values:

\[ \text{C.F.S.E} = (3 \times -\frac{2}{5}Dq) + (1 \times \frac{3}{5}Dq) \]

Calculating this gives:

\[ \text{C.F.S.E} = -\frac{6}{5}Dq + \frac{3}{5}Dq = -\frac{3}{5}Dq \]

### Step 5: Express C.F.S.E in terms of \(Dq\)

Since the problem asks for the absolute value of C.F.S.E in terms of \(Dq\):

\[ |\text{C.F.S.E}| = \frac{3}{5}Dq \]

### Step 6: Convert to the required format

To express this in terms of \(10Dq\):

\[ |\text{C.F.S.E}| = \frac{3}{5}Dq = 6Dq \text{ (since } 1Dq = 10Dq) \]

### Final Answer

Thus, the final answer is:

\[ |\text{C.F.S.E}| = 6Dq \]

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