Which of the pair of complex compounds are tettrahedral as well as diamagnetic ? .

Options

A. a.`[CoCl_(4)]^(Θ)` and `[Co(CO)_(4)]^(Θ)`
B. b.`[Ag(SCN)_(4)]^(2-)` and `[NiCl_(4)]^(2-)`
C. c.`[Co(CO)_(4)]^(Θ)` and `[Ni(CN)_(4)]^(4-)`
D. d.`[PdCl_(4)]^(2-)` and `[Ni(CN)_(4)]^(2-)`

Explanation

To determine which pair of complex compounds are tetrahedral and diamagnetic, we will analyze each option step by step.

### Step 1: Analyze Option 1 - CoCl₄²⁻
1. **Identify the oxidation state of cobalt (Co)**:
- Cobalt has an atomic number of 27, with an electron configuration of [Ar] 3d⁷ 4s².
- In CoCl₄²⁻, the oxidation state of Co is +2 (since Cl is -1 and there are four Cl atoms: 4(-1) + x = -2 → x = +2).
- Thus, Co in this complex has a configuration of 3d⁷.

2. **Determine the ligand field strength**:
- Chloride (Cl⁻) is a weak field ligand, which means it does not cause pairing of electrons in the d-orbitals.

3. **Electron configuration in the presence of ligands**:
- For Co²⁺ (3d⁷), the arrangement would be: ↑ ↑ ↑ ↑ ↑ (5 unpaired electrons).
- Since there are unpaired electrons, this complex is paramagnetic, not diamagnetic.

**Conclusion for Option 1**: Not tetrahedral and not diamagnetic.

### Step 2: Analyze Option 2 - AgCl and NiCl₂
1. **AgCl**:
- Silver (Ag) has an atomic number of 47, with an electron configuration of [Kr] 4d¹⁰ 5s¹.
- In AgCl, Ag is in the +1 oxidation state, leading to a configuration of 4d¹⁰ (fully filled).
- A fully filled d-orbital means there are no unpaired electrons, making it diamagnetic.
- However, AgCl is not tetrahedral; it is typically linear.

2. **NiCl₂**:
- Nickel (Ni) has an atomic number of 28, with an electron configuration of [Ar] 3d⁸ 4s².
- In NiCl₂, Ni is in the +2 oxidation state, leading to a configuration of 3d⁸.
- Chloride is a weak field ligand, so there will be unpaired electrons (4 unpaired).
- Therefore, NiCl₂ is also paramagnetic.

**Conclusion for Option 2**: Not tetrahedral and not diamagnetic.

### Step 3: Analyze Option 3 - Co(CO)₄²⁻
1. **Identify the oxidation state of cobalt (Co)**:
- In Co(CO)₄²⁻, Co is in the zero oxidation state since CO is a neutral ligand.
- The electron configuration remains 3d⁷ 4s².

2. **Determine the ligand field strength**:
- Carbon monoxide (CO) is a strong field ligand, which causes pairing of electrons.

3. **Electron configuration in the presence of ligands**:
- For Co in the zero oxidation state, all electrons can pair up: 3d¹⁰.
- This results in no unpaired electrons, making it diamagnetic.

4. **Hybridization**:
- The geometry is tetrahedral due to sp³ hybridization.

**Conclusion for Option 3**: Tetrahedral and diamagnetic.

### Step 4: Analyze Option 4 - Ni(CN)₄²⁻
1. **Identify the oxidation state of nickel (Ni)**:
- In Ni(CN)₄²⁻, Ni is in the +2 oxidation state, leading to a configuration of 3d⁸.

2. **Determine the ligand field strength**:
- Cyanide (CN⁻) is a strong field ligand, which causes pairing of electrons.

3. **Electron configuration in the presence of ligands**:
- For Ni in the +2 oxidation state, the electrons pair up: 3d¹⁰.
- This results in no unpaired electrons, making it diamagnetic.

4. **Hybridization**:
- The geometry is tetrahedral due to sp³ hybridization.

**Conclusion for Option 4**: Tetrahedral and diamagnetic.

### Final Conclusion:
- The complexes that are tetrahedral and diamagnetic are **Co(CO)₄²⁻** and **Ni(CN)₄²⁻**.

### Summary of Correct Options:
- **Correct Answers**: Option 3 (Co(CO)₄²⁻) and Option 4 (Ni(CN)₄²⁻).