Among `[Ni(CO)_(4)], [NiBr_4]^(2-), [Co(NH_3)_(4)Cl_2]Cl, Na_3[CoF_6], BaO_2 and CsO_2`, the total number of diamagnetic compounds is

Options

A. 5
B. 3
C. 4
D. 2

Answer

To determine the total number of diamagnetic compounds among the given species, we will analyze each compound step by step, focusing on the oxidation states of the central metal ions and the nature of the ligands involved.

### Step 1: Analyze `[Ni(CO)_(4)]`
1. **Determine the oxidation state of Ni**:
- CO is a neutral ligand, so the oxidation state of Ni is 0.
2. **Electron configuration of Ni (0)**:
- Ni has the electron configuration of [Ar] 3d^8 4s^2.
3. **Pairing of electrons**:
- CO is a strong field ligand, which causes pairing of electrons in the d-orbitals.
- After pairing, the configuration will be: 3d^10 (all paired).
4. **Unpaired electrons**:
- There are no unpaired electrons.
5. **Conclusion**:
- `[Ni(CO)_(4)]` is diamagnetic.

### Step 2: Analyze `[NiBr_4]^(2-)`
1. **Determine the oxidation state of Ni**:
- Each Br is -1, and there are 4 Br atoms, giving a total charge of -4. The overall charge of the complex is -2.
- Therefore, the oxidation state of Ni is +2.
2. **Electron configuration of Ni (2+)**:
- Ni^2+ has the electron configuration of 3d^8 (4s^0).
3. **Pairing of electrons**:
- Br is a weak field ligand, so there will be no pairing.
- The configuration will be: 3d^8 (2 unpaired electrons).
4. **Unpaired electrons**:
- There are 2 unpaired electrons.
5. **Conclusion**:
- `[NiBr_4]^(2-)` is paramagnetic.

### Step 3: Analyze `[Co(NH_3)_(4)Cl_2]Cl`
1. **Determine the oxidation state of Co**:
- NH3 is neutral and Cl is -1. The overall charge of the complex is +1.
- Therefore, the oxidation state of Co is +3.
2. **Electron configuration of Co (3+)**:
- Co^3+ has the electron configuration of 3d^6 (4s^0).
3. **Pairing of electrons**:
- NH3 is a strong field ligand, which causes pairing.
- The configuration will be: 3d^6 (all paired).
4. **Unpaired electrons**:
- There are no unpaired electrons.
5. **Conclusion**:
- `[Co(NH_3)_(4)Cl_2]Cl` is diamagnetic.

### Step 4: Analyze `Na_3[CoF_6]`
1. **Determine the oxidation state of Co**:
- F is -1, and there are 6 F atoms, giving a total charge of -6. The overall charge of the complex is -3.
- Therefore, the oxidation state of Co is +3.
2. **Electron configuration of Co (3+)**:
- Co^3+ has the electron configuration of 3d^6 (4s^0).
3. **Pairing of electrons**:
- F is a weak field ligand, so there will be no pairing.
- The configuration will be: 3d^6 (4 unpaired electrons).
4. **Unpaired electrons**:
- There are 4 unpaired electrons.
5. **Conclusion**:
- `Na_3[CoF_6]` is paramagnetic.

### Step 5: Analyze `BaO_2`
1. **Determine the oxidation state of Ba**:
- Ba is in the +2 oxidation state.
2. **Electron configuration of O2 (peroxide)**:
- O2 has a total of 16 electrons (8 from each O atom).
3. **Pairing of electrons**:
- The total number of electrons is even, so they will pair up.
4. **Unpaired electrons**:
- There are no unpaired electrons.
5. **Conclusion**:
- `BaO_2` is diamagnetic.

### Step 6: Analyze `CsO_2`
1. **Determine the oxidation state of Cs**:
- Cs is in the +1 oxidation state.
2. **Electron configuration of O2 (superoxide)**:
- O2 has a total of 17 electrons (8 from each O atom + 1 extra electron).
3. **Pairing of electrons**:
- The total number of electrons is odd, so there will be unpaired electrons.
4. **Unpaired electrons**:
- There is 1 unpaired electron.
5. **Conclusion**:
- `CsO_2` is paramagnetic.

### Summary of Results
- **Diamagnetic Compounds**:
- `[Ni(CO)_(4)]` (1st compound)
- `[Co(NH_3)_(4)Cl_2]Cl` (3rd compound)
- `BaO_2` (5th compound)

- **Paramagnetic Compounds**:
- `[NiBr_4]^(2-)` (2nd compound)
- `Na_3[CoF_6]` (4th compound)
- `CsO_2` (6th compound)

### Total Number of Diamagnetic Compounds
- The total number of diamagnetic compounds is **3**.