Amongst the following,the total number of species which are diamagnetic is ………………. .
`K_(4)[Fe(CN)_(6)],K_(3)[Cr(CN)_(6)],K_(3)[Co(CN)_(6)],K_(2)[Ni(CN)_(4)],[Co(NH_(3))_(6)]^(3+),K_(2)[TiF_(6)],[Pt(NH_(3))_(4)]^(2+)`

Answer

To determine the total number of diamagnetic species among the given complexes, we will analyze each complex one by one, focusing on their oxidation states, electronic configurations, and the presence of unpaired electrons.

### Step-by-Step Solution:

1. **K₄[Fe(CN)₆]**:
- **Oxidation State of Fe**: In this complex, CN⁻ is a strong field ligand. The oxidation state of Fe is +2.
- **Electronic Configuration**: Fe (atomic number 26) has the configuration [Ar] 3d⁶ 4s². For Fe²⁺, it becomes 3d⁶.
- **Filling of Electrons**: Since CN⁻ is a strong field ligand, it causes pairing of electrons in the 3d orbitals. Thus, the configuration becomes (↑↓)(↑↓)(↑↓) (↑)(↑).
- **Unpaired Electrons**: 0 unpaired electrons → **Diamagnetic**.

2. **K₃[Cr(CN)₆]**:
- **Oxidation State of Cr**: Here, Cr is in the +3 oxidation state.
- **Electronic Configuration**: Cr (atomic number 24) has the configuration [Ar] 3d⁵ 4s¹. For Cr³⁺, it becomes 3d³.
- **Filling of Electrons**: With CN⁻ as a strong field ligand, the electrons will fill as (↑)(↑)(↑).
- **Unpaired Electrons**: 3 unpaired electrons → **Paramagnetic**.

3. **K₃[Co(CN)₆]**:
- **Oxidation State of Co**: Co is in the +3 oxidation state.
- **Electronic Configuration**: Co (atomic number 27) has the configuration [Ar] 3d⁷ 4s². For Co³⁺, it becomes 3d⁶.
- **Filling of Electrons**: With CN⁻ as a strong field ligand, the configuration becomes (↑↓)(↑↓)(↑)(↑).
- **Unpaired Electrons**: 0 unpaired electrons → **Diamagnetic**.

4. **K₂[Ni(CN)₄]**:
- **Oxidation State of Ni**: Ni is in the +2 oxidation state.
- **Electronic Configuration**: Ni (atomic number 28) has the configuration [Ar] 3d⁸ 4s². For Ni²⁺, it becomes 3d⁸.
- **Filling of Electrons**: With CN⁻ as a strong field ligand, the configuration becomes (↑↓)(↑↓)(↑↓)(↑)(↑).
- **Unpaired Electrons**: 0 unpaired electrons → **Diamagnetic**.

5. **[Co(NH₃)₆]³⁺**:
- **Oxidation State of Co**: Co is in the +3 oxidation state.
- **Electronic Configuration**: For Co³⁺, it becomes 3d⁶.
- **Filling of Electrons**: NH₃ is a strong field ligand, so the configuration becomes (↑↓)(↑↓)(↑)(↑).
- **Unpaired Electrons**: 0 unpaired electrons → **Diamagnetic**.

6. **K₂[TiF₆]**:
- **Oxidation State of Ti**: Ti is in the +4 oxidation state.
- **Electronic Configuration**: For Ti⁴⁺, it becomes 3d⁰.
- **Filling of Electrons**: There are no electrons to fill.
- **Unpaired Electrons**: 0 unpaired electrons → **Diamagnetic**.

7. **[Pt(NH₃)₄]²⁺**:
- **Oxidation State of Pt**: Pt is in the +2 oxidation state.
- **Electronic Configuration**: For Pt²⁺, it has a configuration of 5d⁸.
- **Filling of Electrons**: NH₃ is a neutral ligand and does not cause pairing. The configuration becomes (↑↓)(↑↓)(↑)(↑)(↑).
- **Unpaired Electrons**: 0 unpaired electrons → **Diamagnetic**.

### Summary of Results:
- **Diamagnetic Species**:
- K₄[Fe(CN)₆] → Diamagnetic
- K₃[Co(CN)₆] → Diamagnetic
- K₂[Ni(CN)₄] → Diamagnetic
- [Co(NH₃)₆]³⁺ → Diamagnetic
- K₂[TiF₆] → Diamagnetic
- [Pt(NH₃)₄]²⁺ → Diamagnetic

### Total Number of Diamagnetic Species: **6**