Correct order of spin only magnetic moment of the following complex ions is: (Given At. no..Fe: 26 ,Co:27)

Options

A. `[FeF_(6)]^(3-)gt[Co(C_(2)O_4)_(3)]^(3-)gt[CoF_(6)]^(3-)`
B. `[Co(C_(2)O_(4))_(3)]^(3-)gt[CoF_(6)]^(3-)gt[FeF_(6)]^(3-`
C. `[FeF_(6)]^(3-)gt[CoF_(6)]^(3-)gt[Co(C_(2)O_(4))_(3)]^(3-)`
D. `[CoF_(6)]^(3-)gt[FeF_6]^(3-)gt[Co(C_(2)O_(4))_(3)]^(3)-`

Answer

To determine the correct order of spin-only magnetic moments for the given complex ions, we will follow these steps:

### Step 1: Determine the oxidation states of the central metal ions.

1. **For FeF6^3-**:
- Let the oxidation state of Fe be \( x \).
- The oxidation state of F is -1 and there are 6 fluorine atoms.
- Therefore, the equation is:
\[ x + 6(-1) = -3 \implies x - 6 = -3 \implies x = +3 \]
- So, Fe is in the +3 oxidation state.

2. **For Co(C2O4)3^3-**:
- Let the oxidation state of Co be \( x \).
- The oxidation state of oxalate (C2O4) is -2, and there are 3 oxalate ligands.
- Therefore, the equation is:
\[ x + 3(-2) = -3 \implies x - 6 = -3 \implies x = +3 \]
- So, Co is also in the +3 oxidation state.

3. **For CoF6^3-**:
- Let the oxidation state of Co be \( x \).
- The oxidation state of F is -1 and there are 6 fluorine atoms.
- Therefore, the equation is:
\[ x + 6(-1) = -3 \implies x - 6 = -3 \implies x = +3 \]
- So, Co is again in the +3 oxidation state.

### Step 2: Determine the electronic configurations of the metal ions.

1. **Fe^3+**:
- The electronic configuration of Fe (atomic number 26) is [Ar] 3d^6 4s^2.
- For Fe^3+, we remove 3 electrons (2 from 4s and 1 from 3d):
\[ \text{Fe}^{3+}: 3d^5 \]

2. **Co^3+**:
- The electronic configuration of Co (atomic number 27) is [Ar] 3d^7 4s^2.
- For Co^3+, we remove 3 electrons (2 from 4s and 1 from 3d):
\[ \text{Co}^{3+}: 3d^6 \]

### Step 3: Determine the number of unpaired electrons.

1. **For FeF6^3-**:
- Fluoride (F) is a weak field ligand, leading to a high-spin complex.
- The 3d^5 configuration will have 5 unpaired electrons:
\[ \text{Unpaired electrons} = 5 \]

2. **For Co(C2O4)3^3-**:
- Oxalate (C2O4) is a strong field ligand, leading to a low-spin complex.
- The 3d^6 configuration will have 0 unpaired electrons after pairing:
\[ \text{Unpaired electrons} = 0 \]

3. **For CoF6^3-**:
- Fluoride (F) is a weak field ligand, leading to a high-spin complex.
- The 3d^6 configuration will have 4 unpaired electrons:
\[ \text{Unpaired electrons} = 4 \]

### Step 4: Calculate the spin-only magnetic moment.

The formula for the spin-only magnetic moment (\( \mu_s \)) is given by:
\[ \mu_s = \sqrt{n(n + 2)} \]
where \( n \) is the number of unpaired electrons.

1. **For FeF6^3-**:
\[ \mu_s = \sqrt{5(5 + 2)} = \sqrt{35} \]

2. **For Co(C2O4)3^3-**:
\[ \mu_s = \sqrt{0(0 + 2)} = \sqrt{0} = 0 \]

3. **For CoF6^3-**:
\[ \mu_s = \sqrt{4(4 + 2)} = \sqrt{24} \]

### Step 5: Order the complexes based on their magnetic moments.

- FeF6^3-: \( \mu_s = \sqrt{35} \)
- Co(C2O4)3^3-: \( \mu_s = 0 \)
- CoF6^3-: \( \mu_s = \sqrt{24} \)

### Final Order:
The correct order of spin-only magnetic moments is:
\[ \sqrt{35} > \sqrt{24} > 0 \]

Thus, the correct answer is:
1. FeF6^3- (highest)
2. CoF6^3-
3. Co(C2O4)3^3- (lowest)