Which of the following is the high spin complex?

Options

A. `[Cr(gly)_(3)]`
B. `[CoBr_(2)CI_(2)(SCN)_(2)]^(3-)`
C. `[Fe(CN)_(6)]^(3-)`
D. `Na[PtBrCI(NO_(2))_(2)(NH_(3))_(2)]`

Answer

To determine which of the given complexes is a high spin complex, we need to analyze each complex based on the nature of the ligands and the electronic configuration of the metal ion involved. High spin complexes are typically formed with weak field ligands, which do not cause pairing of electrons in the d-orbitals, allowing for a greater number of unpaired electrons.

### Step-by-Step Solution:

1. **Identify the Metal Ion and its Oxidation State**:
- For each complex, identify the metal ion and determine its oxidation state based on the ligands attached.

2. **Determine the Electronic Configuration**:
- Based on the oxidation state, write the electronic configuration of the metal ion. Transition metals have variable oxidation states, and their configurations will change accordingly.

3. **Identify the Ligands**:
- Classify the ligands as either weak field or strong field. Weak field ligands (like Cl⁻, Br⁻) do not cause electron pairing, while strong field ligands (like CN⁻) do.

4. **Predict the Coordination Number and Geometry**:
- Most transition metal complexes have a coordination number of 6, leading to an octahedral geometry. This will influence the splitting of the d-orbitals.

5. **Draw the Splitting Diagram**:
- For octahedral complexes, the d-orbitals split into two sets: t₂g (lower energy) and e_g (higher energy).

6. **Fill the d-Orbitals**:
- Fill the d-orbitals according to Hund's rule and the Pauli exclusion principle. For high spin complexes, fill each orbital singly before pairing.

7. **Count the Unpaired Electrons**:
- Count the number of unpaired electrons in the d-orbitals. The more unpaired electrons, the higher the spin state.

8. **Conclusion**:
- Compare the number of unpaired electrons in each complex. The complex with the highest number of unpaired electrons is the high spin complex.

### Application to the Given Complexes:

1. **Complex 1: [Cr(Gly)₃]**:
- Metal: Cr (Chromium)
- Oxidation State: +3
- Electronic Configuration: 3d³
- Ligands: Glycine (weak field)
- Unpaired Electrons: 3 (high spin)

2. **Complex 2: [CoBr₂Cl₂(SCN)]**:
- Metal: Co (Cobalt)
- Oxidation State: +3
- Electronic Configuration: 3d⁶
- Ligands: Br⁻, Cl⁻ (weak field)
- Unpaired Electrons: 4 (high spin)

3. **Complex 3: [Fe(CN)₆]**:
- Metal: Fe (Iron)
- Oxidation State: +3
- Electronic Configuration: 3d⁵
- Ligands: CN⁻ (strong field)
- Unpaired Electrons: 1 (low spin)

4. **Complex 4: [Pt(NH₃)₂Br₂Cl₂]**:
- Metal: Pt (Platinum)
- Oxidation State: +3
- Electronic Configuration: 5d⁷
- Ligands: NH₃ (weak field), Br⁻, Cl⁻ (weak field)
- Unpaired Electrons: 1 (low spin)

### Final Answer:
The high spin complex among the given options is **Complex 2: [CoBr₂Cl₂(SCN)]**, which has 4 unpaired electrons.