All the metal ions contain `t_(2g)^(6)e_(g)^(0)` configuaration. Which of the following complex will be paramagnetic ?

Options

A. `[FeCl(CN)_(4)(O_(2))]^(4-)`
B. `K_(4)[Fe(CN)_(6)]`
C. `[C(NH_(3))_(6)]Cl_(3)`
D. `[Fe(CN)_(5)(O_(2))]^(-5)`

Answer

To determine which of the given complexes is paramagnetic, we need to analyze the oxidation states of the metal ions and their electron configurations. Here's a step-by-step solution:

### Step 1: Identify the Complexes
The complexes given are:
1. FeCl₄²⁻
2. K₄[Fe(CN)₆]
3. [Co(NH₃)₆]Cl₃
4. [Fe(CN)₅O₂]⁻

### Step 2: Determine the Oxidation State of Iron in Each Complex

1. **FeCl₄²⁻**:
- Let the oxidation state of Fe be x.
- The charge of the complex is -2.
- Chloride (Cl⁻) has a -1 charge, and there are 4 Cl⁻ ions.
- The equation is: x + 4(-1) = -2
- Solving gives: x - 4 = -2 → x = +2

2. **K₄[Fe(CN)₆]**:
- Let the oxidation state of Fe be x.
- The charge of the complex is neutral (0).
- Potassium (K⁺) has a +1 charge, and there are 4 K⁺ ions.
- The equation is: 4(+1) + x + 6(-1) = 0
- Solving gives: 4 + x - 6 = 0 → x = +2

3. **[Co(NH₃)₆]Cl₃**:
- Let the oxidation state of Co be x.
- The charge of the complex is +3 (due to 3 Cl⁻).
- The equation is: x + 6(0) = +3 (NH₃ is neutral)
- Solving gives: x = +3

4. **[Fe(CN)₅O₂]⁻**:
- Let the oxidation state of Fe be x.
- The charge of the complex is -1.
- The equation is: x + 5(-1) + 2(-2) = -1
- Solving gives: x - 5 - 2 = -1 → x = +6

### Step 3: Determine the Electron Configuration

1. **Fe²⁺ (from FeCl₄²⁻)**:
- Configuration: Argon 3d⁶
- Since Cl⁻ is a weak field ligand, there will be unpaired electrons. Thus, this complex is **paramagnetic**.

2. **Fe²⁺ (from K₄[Fe(CN)₆])**:
- Configuration: Argon 3d⁶
- CN⁻ is a strong field ligand and will cause pairing. Thus, this complex is **diamagnetic**.

3. **Co³⁺ (from [Co(NH₃)₆]Cl₃)**:
- Configuration: Argon 3d⁶
- NH₃ is also a strong field ligand and will cause pairing. Thus, this complex is **diamagnetic**.

4. **Fe⁶⁺ (from [Fe(CN)₅O₂]⁻)**:
- Configuration: Argon 3d⁶
- CN⁻ is a strong field ligand and will cause pairing. Thus, this complex is **diamagnetic**.

### Conclusion
The only complex that is paramagnetic is **FeCl₄²⁻**.