Choose the correct code regarding, possible number of geometrical isomers exhibited by following complexes:
(I) `[CrCl_(2)(NO_(2))_(2)(NH_(3))_(2)]^(-)`
(II) `[Co(NO_(2))_(3)(NH_(3))_(3)]`
(III) `[PtCl(NO_(2))(NH_(3))(py)]`
(IV) `[PtBrCl(en)]`

Options

A. (I)-4,(II)-2,(III)-3,(IV)-1
B. (I)-5,(II)-2,(III)-4,(IV)-0
C. (I)-3,(II)-2,(III)-0,(IV)-1
D. (I)-5,(II)-2,(III)-3,(IV)-0

Explanation

To determine the possible number of geometrical isomers for the given coordination complexes, we will analyze each complex based on its structure and the types of ligands involved.

### Step 1: Analyze the first complex `[CrCl₂(NO₂)₂(NH₃)₂]⁻`
- This complex has the formula MA₂B₂C₂, where:
- M = Cr (Chromium)
- A = Cl (Chloro)
- B = NO₂ (Nitro)
- C = NH₃ (Amine)
- The arrangement allows for geometrical isomers due to the different positions of the ligands.
- **Possible Geometrical Isomers**: 5 (including 1 optical isomer).

### Step 2: Analyze the second complex `[Co(NO₂)₃(NH₃)₃]`
- This complex has the formula MA₃B₃, where:
- M = Co (Cobalt)
- A = NO₂ (Nitro)
- B = NH₃ (Amine)
- This type of complex can exhibit facial (fac) and meridional (mer) isomers.
- **Possible Geometrical Isomers**: 2 (fac and mer).

### Step 3: Analyze the third complex `[PtCl(NO₂)(NH₃)(py)]`
- This complex has the formula MABCD, where:
- M = Pt (Platinum)
- A = Cl (Chloro)
- B = NO₂ (Nitro)
- C = NH₃ (Amine)
- D = py (Pyridine)
- The different arrangements of these ligands allow for geometrical isomers.
- **Possible Geometrical Isomers**: 3.

### Step 4: Analyze the fourth complex `[PtBrCl(en)]`
- This complex has the formula MAA (where A = Br and Cl are the same type of ligand), and en is a bidentate ligand.
- Since the bidentate ligand is symmetrical and the other two ligands are identical, this complex does not exhibit geometrical isomerism.
- **Possible Geometrical Isomers**: 0.

### Summary of Results:
- For complex (I): 5 geometrical isomers
- For complex (II): 2 geometrical isomers
- For complex (III): 3 geometrical isomers
- For complex (IV): 0 geometrical isomers

### Final Answer:
The correct code regarding the possible number of geometrical isomers is: **5, 2, 3, 0**.