Find the sum of number of geometrical isomers for following complexes.
`(a) [CoCl_(2)Br_(2)]^(2-) " " (b) [Rh(en)_(3)]^(3+) " " ( c) [Cr(en)_(2)Br_(2)]^(+)`
`(d)[Pt en Cl_(2)] "

Answer

To find the sum of the number of geometrical isomers for the given complexes, we will analyze each complex one by one.

### Step 1: Analyze the complex \([CoCl_2Br_2]^{2-}\)

- **Type of complex**: Octahedral
- **Ligands**: 2 Chloride (Cl) and 2 Bromide (Br)
- **Geometrical isomers**: For octahedral complexes with two pairs of different ligands, we can have cis and trans isomers.
- **Possible isomers**: The two Cl and two Br can be arranged in two ways:
- **Cis isomer**: Both Cl ligands are adjacent.
- **Trans isomer**: Cl ligands are opposite each other.

Thus, the number of geometrical isomers for \([CoCl_2Br_2]^{2-}\) is **2**.

### Step 2: Analyze the complex \([Rh(en)_3]^{3+}\)

- **Type of complex**: Octahedral
- **Ligands**: 3 Ethylenediamine (en) ligands
- **Geometrical isomers**: Since ethylenediamine is a bidentate ligand, all three ligands are identical.
- **Possible isomers**: There are no geometrical isomers possible because all ligands are the same.

Thus, the number of geometrical isomers for \([Rh(en)_3]^{3+}\) is **0**.

### Step 3: Analyze the complex \([Cr(en)_2Br_2]^{+}\)

- **Type of complex**: Octahedral
- **Ligands**: 2 Ethylenediamine (en) and 2 Bromide (Br)
- **Geometrical isomers**: Similar to the first complex, we can have cis and trans arrangements.
- **Possible isomers**:
- **Cis isomer**: Both Br ligands are adjacent.
- **Trans isomer**: Br ligands are opposite each other.

Thus, the number of geometrical isomers for \([Cr(en)_2Br_2]^{+}\) is **2**.

### Step 4: Analyze the complex \([Pt(en)Cl_2]\)

- **Type of complex**: Square planar
- **Ligands**: 1 Ethylenediamine (en) and 2 Chloride (Cl)
- **Geometrical isomers**: For square planar complexes with two different ligands, we can have:
- **Cis isomer**: Both Cl ligands are adjacent.
- **Trans isomer**: Cl ligands are opposite each other.

Thus, the number of geometrical isomers for \([Pt(en)Cl_2]\) is **2**.

### Final Calculation

Now, we sum up the number of geometrical isomers from all complexes:

- \([CoCl_2Br_2]^{2-}\): 2
- \([Rh(en)_3]^{3+}\): 0
- \([Cr(en)_2Br_2]^{+}\): 2
- \([Pt(en)Cl_2]\): 2

**Total = 2 + 0 + 2 + 2 = 6**

### Conclusion

The sum of the number of geometrical isomers for the given complexes is **6**.

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