The number of paramagnetic species from the following is__________
`[Ni(CN)_4]^(2-),[Ni(CO)_4],[NiCl_4]^(2-)[Fe(CN)_6]^(4-),[Cu(NH_3)_4]^(2+)[Fe(CN)_6]^(3-),[Fe(H_2O)_6]^(2+)`

Answer

To determine the number of paramagnetic species from the given coordination complexes, we will analyze each complex based on the nature of the ligands and the oxidation states of the metal ions involved.

### Step-by-Step Solution:

1. **Identify the complexes and their oxidation states:**
- `[Ni(CN)_4]^{2-}`
- `[Ni(CO)_4]`
- `[NiCl_4]^{2-}`
- `[Fe(CN)_6]^{4-}`
- `[Cu(NH_3)_4]^{2+}`
- `[Fe(CN)_6]^{3-}`
- `[Fe(H_2O)_6]^{2+}`

2. **Analyze each complex for paramagnetism:**
- **[Ni(CN)₄]²⁻:**
- Ni in this complex is in the +2 oxidation state (3d⁸ configuration).
- CN⁻ is a strong field ligand, which causes pairing of electrons.
- **Result:** Diamagnetic (0 unpaired electrons).

- **[Ni(CO)₄]:**
- Ni is in the 0 oxidation state (3d⁹ configuration).
- CO is also a strong field ligand, leading to pairing.
- **Result:** Diamagnetic (0 unpaired electrons).

- **[NiCl₄]²⁻:**
- Ni is in the +2 oxidation state (3d⁸ configuration).
- Cl⁻ is a weak field ligand, which does not cause pairing.
- Distribution: ↑↓ ↑↓ ↑ ↑ (2 unpaired electrons).
- **Result:** Paramagnetic (2 unpaired electrons).

- **[Fe(CN)₆]⁴⁻:**
- Fe is in the +2 oxidation state (3d⁶ configuration).
- CN⁻ is a strong field ligand, leading to pairing.
- **Result:** Diamagnetic (0 unpaired electrons).

- **[Cu(NH₃)₄]²⁺:**
- Cu is in the +2 oxidation state (3d⁹ configuration).
- NH₃ is a weak field ligand, leading to unpaired electrons.
- Distribution: ↑↓ ↑↓ ↑ ↑ ↑ (1 unpaired electron).
- **Result:** Paramagnetic (1 unpaired electron).

- **[Fe(CN)₆]³⁻:**
- Fe is in the +3 oxidation state (3d⁵ configuration).
- CN⁻ is a strong field ligand, but in this case, one electron remains unpaired.
- **Result:** Paramagnetic (1 unpaired electron).

- **[Fe(H₂O)₆]²⁺:**
- Fe is in the +2 oxidation state (3d⁶ configuration).
- H₂O is a weak field ligand, leading to unpaired electrons.
- Distribution: ↑↓ ↑ ↑ ↑ (4 unpaired electrons).
- **Result:** Paramagnetic (4 unpaired electrons).

3. **Count the total number of paramagnetic species:**
- From the analysis, the paramagnetic complexes are:
- [NiCl₄]²⁻ (2 unpaired electrons)
- [Cu(NH₃)₄]²⁺ (1 unpaired electron)
- [Fe(CN)₆]³⁻ (1 unpaired electron)
- [Fe(H₂O)₆]²⁺ (4 unpaired electrons)

- Total paramagnetic species: 4

### Final Answer:
The number of paramagnetic species is **4**.