Which of the following complexes diamagnetic:

Options

A. `[AuI_(4)]^(Θ)`
B. `[Co(H_(2)O)_(6)]^(3+)`
C. `[CoI_(6)]^(3-)`
D. `[Co(CO)_(4)]^(Θ)`

Explanation

To determine which of the given complexes are diamagnetic, we need to analyze each complex for the presence of unpaired electrons. A complex is considered diamagnetic if it has no unpaired electrons. Let's go through the complexes step by step.

### Step 1: Analyze the first complex - [AuI4]⁻
1. **Determine the oxidation state of Gold (Au)**:
- Let the oxidation state of Au be \( x \).
- The charge of the complex is -1, and each iodide ion (I) has a charge of -1.
- Therefore, the equation is:
\[ x + 4(-1) = -1 \implies x - 4 = -1 \implies x = +3 \]

2. **Electronic configuration of Au in +3 state**:
- The atomic number of Au is 79, with the electronic configuration:
\[ [Xe] 4f^{14} 5d^{10} 6s^1 \]
- In the +3 oxidation state, Au loses 3 electrons:
- 1 from 6s and 2 from 5d, resulting in:
\[ 5d^8 \]

3. **Determine the magnetic property**:
- The 5d orbitals will be filled as follows:
- All 8 electrons will pair up.
- Conclusion: **Diamagnetic** (no unpaired electrons).

### Step 2: Analyze the second complex - [Co(H2O)6]³⁺
1. **Determine the oxidation state of Cobalt (Co)**:
- Let the oxidation state of Co be \( x \).
- Water is neutral, so:
\[ x + 0 = +3 \implies x = +3 \]

2. **Electronic configuration of Co in +3 state**:
- The atomic number of Co is 27, with the electronic configuration:
\[ [Ar] 3d^7 4s^2 \]
- In the +3 oxidation state, Co loses 3 electrons (2 from 4s and 1 from 3d):
\[ 3d^6 \]

3. **Determine the magnetic property**:
- Water is a weak field ligand, leading to pairing:
- The 6 electrons will fill the d orbitals as follows:
- All electrons will pair up.
- Conclusion: **Diamagnetic** (no unpaired electrons).

### Step 3: Analyze the third complex - [CoI6]³⁻
1. **Determine the oxidation state of Cobalt (Co)**:
- Let the oxidation state of Co be \( x \).
- The equation is:
\[ x + 6(-1) = -3 \implies x - 6 = -3 \implies x = +3 \]

2. **Electronic configuration of Co in +3 state**:
- The electronic configuration remains:
\[ 3d^6 \]

3. **Determine the magnetic property**:
- Iodide is a weak field ligand, so pairing does not occur:
- The 6 electrons will fill the d orbitals as follows:
- 4 unpaired electrons remain.
- Conclusion: **Paramagnetic** (presence of unpaired electrons).

### Step 4: Analyze the fourth complex - [Co(CO)4]⁻
1. **Determine the oxidation state of Cobalt (Co)**:
- Let the oxidation state of Co be \( x \).
- The equation is:
\[ x + 0 = -1 \implies x = -1 \]

2. **Electronic configuration of Co in -1 state**:
- The electronic configuration for Co is:
\[ 3d^7 4s^2 \]
- In the -1 oxidation state, Co gains an electron:
\[ 3d^8 4s^2 \]

3. **Determine the magnetic property**:
- CO is a strong field ligand, leading to pairing:
- The 10 electrons will fill the d orbitals as follows:
- All electrons will pair up.
- Conclusion: **Diamagnetic** (no unpaired electrons).

### Final Conclusion:
The complexes that are diamagnetic are:
- Complex 1: [AuI4]⁻
- Complex 2: [Co(H2O)6]³⁺
- Complex 4: [Co(CO)4]⁻

Thus, the diamagnetic complexes are **1, 2, and 4**.