The correct order of spin only magnetic moments for the following complex ions is

Options

A. `[Fe(CN)_(6)]^(3-) lt[CoF_(6)]^(3-) lt[MnBr_(4)]^(2-)lt[Mn(CN)_(6)]^(3-)`
B. `[Fe(CN)_(6)]^(3-) lt[Mn(CN)_(6)]^(3-) lt[CoF_(6)]^(3-) lt[MnBr_(4)]^(2-)`
C. `[Mn(CN)_(4)]^(2-) lt[CoF_(6)]^(3-)lt[Fe(CN)_(6)]^(3-) lt[Mn(CN)_(6)]^(3-)`
D. `[CoF_(6)]^(3-) lt[MnBr_(4)]^(2-)lt[Fe(CN)_(6)]^(3-) lt[Mn(CN)_(6)]^(3-)`

Answer

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

### Step 1: Identify the oxidation states and electron configurations
- **Iron (Fe)** in +3 oxidation state:
- Ground state configuration: \( [Ar] 3d^6 4s^2 \)
- In +3 state: \( 3d^5 \)

- **Cobalt (Co)** in +3 oxidation state:
- Ground state configuration: \( [Ar] 3d^7 4s^2 \)
- In +3 state: \( 3d^6 \)

- **Manganese (Mn)** in +2 oxidation state:
- Ground state configuration: \( [Ar] 3d^5 4s^2 \)
- In +2 state: \( 3d^5 \)

- **Another Manganese (Mn)** in +3 oxidation state:
- In +3 state: \( 3d^4 \)

### Step 2: Determine the type of ligand and its effect on spin state
- **Cyanide (CN⁻)** is a strong field ligand, leading to low spin configurations (pairing of electrons).
- **Fluoride (F⁻)** is a weak field ligand, leading to high spin configurations (no pairing).
- **Bromide (Br⁻)** is also a weak field ligand, leading to high spin configurations.

### Step 3: Fill the d-orbitals and determine the number of unpaired electrons
1. **For Fe(CN)₆³⁻ (Fe +3)**:
- Configuration: \( 3d^5 \) (low spin)
- Electrons: Pairing occurs, resulting in 1 unpaired electron.
- Magnetic moment: \( \sqrt{n(n+2)} = \sqrt{1(1+2)} = \sqrt{3} \)

2. **For Co(F)₆³⁻ (Co +3)**:
- Configuration: \( 3d^6 \) (high spin)
- Electrons: 4 unpaired electrons.
- Magnetic moment: \( \sqrt{4(4+2)} = \sqrt{24} \)

3. **For Mn(Br)₄²⁻ (Mn +2)**:
- Configuration: \( 3d^5 \) (high spin)
- Electrons: 5 unpaired electrons.
- Magnetic moment: \( \sqrt{5(5+2)} = \sqrt{35} \)

4. **For Mn(CN)₄³⁻ (Mn +3)**:
- Configuration: \( 3d^4 \) (low spin)
- Electrons: 0 unpaired electrons (all paired).
- Magnetic moment: \( 0 \)

### Step 4: Compare the magnetic moments
- **Fe(CN)₆³⁻**: \( \sqrt{3} \)
- **Co(F)₆³⁻**: \( \sqrt{24} \)
- **Mn(Br)₄²⁻**: \( \sqrt{35} \)
- **Mn(CN)₄³⁻**: \( 0 \)

### Step 5: Order the complexes based on magnetic moments
From highest to lowest magnetic moment:
1. **Mn(Br)₄²⁻**: \( \sqrt{35} \)
2. **Co(F)₆³⁻**: \( \sqrt{24} \)
3. **Fe(CN)₆³⁻**: \( \sqrt{3} \)
4. **Mn(CN)₄³⁻**: \( 0 \)

### Final Answer:
The correct order of spin-only magnetic moments for the given complex ions is:
**Mn(Br)₄²⁻ > Co(F)₆³⁻ > Fe(CN)₆³⁻ > Mn(CN)₄³⁻**