To determine the total number of diamagnetic species among the given compounds, we need to analyze each compound to check if they have unpaired electrons in their electronic configurations. A species is considered diamagnetic if all its electrons are paired.
### Step-by-Step Solution:
1. **Identify the oxidation state of the central atom in each compound:**
- **K₄[Fe(CN)₆]:**
- CN⁻ contributes -6, K⁺ contributes +4.
- Therefore, Fe must be +2.
- **K₃[Cr(CN)₆]:**
- CN⁻ contributes -6, K⁺ contributes +3.
- Therefore, Cr must be +3.
- **K₃[Co(CN)₆]:**
- CN⁻ contributes -6, K⁺ contributes +3.
- Therefore, Co must be +3.
- **K₂[Ni(CN)₄]:**
- CN⁻ contributes -4, K⁺ contributes +2.
- Therefore, Ni must be +2.
- **[Co(NH₃)₆]³⁺:**
- NH₃ is neutral, so Co must be +3.
- **K₂[TiF₆]:**
- F⁻ contributes -6, K⁺ contributes +2.
- Therefore, Ti must be +4.
- **[Pt(NH₃)₄]²⁺:**
- NH₃ is neutral, so Pt must be +2.
2. **Determine the electronic configuration of each central atom:**
- **Fe²⁺ (26 electrons):**
- Configuration: [Ar] 3d⁶ (4 unpaired electrons, paramagnetic)
- **Cr³⁺ (24 electrons):**
- Configuration: [Ar] 3d³ (3 unpaired electrons, paramagnetic)
- **Co³⁺ (27 electrons):**
- Configuration: [Ar] 3d⁶ (4 unpaired electrons, paramagnetic)
- **Ni²⁺ (28 electrons):**
- Configuration: [Ar] 3d⁸ (2 unpaired electrons, paramagnetic)
- **Co³⁺ (from [Co(NH₃)₆]³⁺):**
- Configuration: [Ar] 3d⁶ (4 unpaired electrons, paramagnetic)
- **Ti⁴⁺ (22 electrons):**
- Configuration: [Ar] 3d⁰ (0 unpaired electrons, diamagnetic)
- **Pt²⁺ (78 electrons):**
- Configuration: [Xe] 4f¹⁴ 5d⁸ (2 unpaired electrons, paramagnetic)
3. **Count the number of diamagnetic species:**
- From the analysis, only **K₂[TiF₆]** is diamagnetic (0 unpaired electrons).
- All other species are paramagnetic due to the presence of unpaired electrons.
### Conclusion:
The total number of diamagnetic species among the given compounds is **1** (K₂[TiF₆]).
