Among `[Ni(CO)_(4)], [NiCl_(4)]^(2-), [Co (NH_(3))_(4) Cl_(2)] Cl, Na_(3) [CoF_(6)], Na_(2)O_(2)` and `CsO_(2)`, the total number of paramagnetic compounds is

Options

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

Answer

To determine the total number of paramagnetic compounds among the given complexes, we need to analyze each compound for the presence of unpaired electrons. A compound is considered paramagnetic if it has one or more unpaired electrons.

### Step-by-Step Solution:

1. **Identify the Compounds**:
The compounds given are:
- `[Ni(CO)_(4)]`
- `[NiCl_(4)]^(2-)`
- `[Co(NH_(3))_(4)Cl_(2)]Cl`
- `Na_(3)[CoF_(6)]`
- `Na_(2)O_(2)`
- `CsO_(2)`

2. **Analyze Each Compound**:

**a.** **[Ni(CO)_(4)]**:
- Nickel (Ni) has an atomic number of 28, with an electronic configuration of `[Ar] 3d^8 4s^2`.
- In `[Ni(CO)_(4)]`, Ni is in the 0 oxidation state (since CO is neutral).
- CO is a strong field ligand, causing pairing of electrons.
- Result: All electrons are paired → **Diamagnetic**.

**b.** **[NiCl_(4)]^(2-)**:
- Ni is in the +2 oxidation state here (since Cl is -1 and there are 4 Cl).
- The electronic configuration for Ni^2+ is `3d^8`.
- Cl is a weak field ligand, so no pairing occurs.
- Result: There are 2 unpaired electrons → **Paramagnetic**.

**c.** **[Co(NH_(3))_(4)Cl_(2)]Cl**:
- Cobalt (Co) has an atomic number of 27, with an electronic configuration of `[Ar] 3d^7 4s^2`.
- In this complex, Co is in the +3 oxidation state (due to NH3 being neutral and Cl being -1).
- The electronic configuration for Co^3+ is `3d^6`.
- NH3 is a strong field ligand, which causes pairing.
- Result: All electrons are paired → **Diamagnetic**.

**d.** **Na_(3)[CoF_(6)]**:
- Co is in the +3 oxidation state (since F is -1).
- The electronic configuration for Co^3+ is `3d^6`.
- F is a weak field ligand, so no pairing occurs.
- Result: There are 4 unpaired electrons → **Paramagnetic**.

**e.** **Na_(2)O_(2)**:
- The peroxide ion (O2^2-) has a total of 18 electrons.
- Since 18 is an even number, all electrons are paired → **Diamagnetic**.

**f.** **CsO_(2)**:
- The superoxide ion (O2^-) has a total of 17 electrons.
- Since 17 is an odd number, there is 1 unpaired electron → **Paramagnetic**.

3. **Count the Paramagnetic Compounds**:
- From the analysis:
- `[Ni(CO)_(4)]`: Diamagnetic
- `[NiCl_(4)]^(2-)`: Paramagnetic
- `[Co(NH_(3))_(4)Cl_(2)]Cl`: Diamagnetic
- `Na_(3)[CoF_(6)]`: Paramagnetic
- `Na_(2)O_(2)`: Diamagnetic
- `CsO_(2)`: Paramagnetic

The total number of paramagnetic compounds is **3**: `[NiCl_(4)]^(2-)`, `Na_(3)[CoF_(6)]`, and `CsO_(2)`.

### Final Answer:
The total number of paramagnetic compounds is **3**.