Consider the following complex compounds:
(i) `[Pt(NH_(3))_(2)(SCN)_(2)]`
(ii) `[Co(NH_(3))_(2)(NO_(2))_(3)]`
(iii) `[Pt(en)Cl_(2)]`
(iv) `[Cr(en)_(2)Br_(2)]^(+)`
(v) `[Rn(en)_(3)]^(3+)`
(vi) `[CoCl_(2)Br_(2)]`
Then calculate sum of total number of geometrical isomers in all above complex compounds.

Answer

To calculate the total number of geometrical isomers for the given coordination compounds, we will analyze each complex one by one.

### Step 1: Analyze the first complex `[Pt(NH₃)₂(SCN)₂]`
- This complex has a coordination number of 4 and is square planar.
- It can have two geometrical isomers:
- **Cis**: Both NH₃ ligands are adjacent.
- **Trans**: Both NH₃ ligands are opposite each other.

**Total geometrical isomers for (i)**: 2

### Step 2: Analyze the second complex `[Co(NH₃)₂(NO₂)₃]`
- This complex has a coordination number of 6 and is octahedral.
- It can also have two geometrical isomers:
- **Cis**: Both NH₃ ligands are adjacent.
- **Trans**: Both NH₃ ligands are opposite each other.

**Total geometrical isomers for (ii)**: 2

### Step 3: Analyze the third complex `[Pt(en)Cl₂]`
- This complex has a coordination number of 4 and is square planar.
- It does not have geometrical isomers because the bidentate ligand (en) occupies two coordination sites in a fixed manner.

**Total geometrical isomers for (iii)**: 0

### Step 4: Analyze the fourth complex `[Cr(en)₂Br₂]⁺`
- This complex has a coordination number of 6 and is octahedral.
- It can have two geometrical isomers:
- **Cis**: Both Br ligands are adjacent.
- **Trans**: Both Br ligands are opposite each other.

**Total geometrical isomers for (iv)**: 2

### Step 5: Analyze the fifth complex `[Rh(en)₃]³+`
- This complex has a coordination number of 6 and is octahedral.
- It does not exhibit geometrical isomerism because all three bidentate ligands (en) are identical and can only occupy the coordination sites in a symmetrical manner.

**Total geometrical isomers for (v)**: 0

### Step 6: Analyze the sixth complex `[CoCl₂Br₂]`
- This complex has a coordination number of 6 and is octahedral.
- It can have three geometrical isomers:
- **Cis**: Two Cl ligands and two Br ligands are adjacent.
- **Trans**: Two Cl ligands are opposite to each other and two Br ligands are opposite to each other.
- **Facial (fac)**: Two Cl ligands and two Br ligands occupy one face of the octahedron.

**Total geometrical isomers for (vi)**: 3

### Step 7: Sum the total number of geometrical isomers
Now, we will sum the total geometrical isomers from all complexes:
- (i) 2
- (ii) 2
- (iii) 0
- (iv) 2
- (v) 0
- (vi) 3

**Total = 2 + 2 + 0 + 2 + 0 + 3 = 9**

### Final Answer:
The total number of geometrical isomers in all the given complex compounds is **9**.