The number of geometric isomers that can exist for square planar `[Pt(C1)(py)(NH_(3)) (NH_(2)OH)^(+)]` is (py = pyridine).

Options

A. 3
B. 4
C. 6
D. 2

Answer

To determine the number of geometric isomers for the square planar complex \([Pt(Cl)(py)(NH_3)(NH_2OH)]^+\), we will follow these steps:

### Step 1: Identify the Ligands
The complex has four ligands:
1. Chlorine (Cl)
2. Pyridine (py)
3. Ammonia (NH3)
4. Hydroxylamine (NH2OH)

### Step 2: Understand the Geometry
The geometry of the complex is square planar. In a square planar arrangement, the ligands are positioned at the corners of a square with the central metal ion (Pt) at the center.

### Step 3: Determine Possible Arrangements
In square planar complexes, the geometric isomers arise from the different arrangements of the ligands. The two types of isomers we can have are:
- **Cis Isomers**: where similar ligands are adjacent to each other.
- **Trans Isomers**: where similar ligands are opposite each other.

### Step 4: Analyze Ligand Positions
For our complex, we can consider the positions of the ligands:
- We can have Cl and NH2OH in one pair of positions and NH3 and py in the other pair.
- We can interchange Cl and NH2OH to create different arrangements.

### Step 5: Count the Isomers
1. **Arrangement 1**: Cl and NH2OH are adjacent (cis), NH3 and py are adjacent (cis).
2. **Arrangement 2**: Cl and NH3 are adjacent (cis), NH2OH and py are adjacent (cis).
3. **Arrangement 3**: Cl and py are adjacent (cis), NH3 and NH2OH are adjacent (cis).

Each of these arrangements can yield different geometric isomers based on the positioning of the ligands.

### Conclusion
After analyzing the arrangements, we find that there are a total of **3 geometric isomers** for the complex \([Pt(Cl)(py)(NH_3)(NH_2OH)]^+\).

### Final Answer
The number of geometric isomers that can exist for the complex is **3**.

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