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

Options

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

Answer

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

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

### Step 2: Understand the Geometry
The complex is square planar, which means that the ligands are arranged in a two-dimensional square around the central platinum (Pt) atom.

### Step 3: Fix One Ligand
To find the geometric isomers, we can fix one ligand in a specific position. Let's fix pyridine (py) in one position.

### Step 4: Arrange the Remaining Ligands
With pyridine fixed, we can arrange the remaining three ligands (Cl, NH₃, and NH₂OH) around the platinum atom. The arrangement can lead to different isomers based on the positions of the other ligands.

### Step 5: Identify Possible Arrangements
1. **Isomer 1**: Fix pyridine at the top. Place Cl at the bottom, NH₃ on the left, and NH₂OH on the right.
2. **Isomer 2**: Keep pyridine at the top. Swap NH₃ and NH₂OH, placing NH₂OH on the left and NH₃ on the right.
3. **Isomer 3**: Again, keep pyridine at the top. Now swap NH₃ with Cl, placing Cl on the left and NH₃ on the right.

### Step 6: Count the Isomers
After considering the arrangements, we find that there are three distinct geometric isomers based on the positions of the ligands around the fixed pyridine.

### Conclusion
Thus, the number of geometric isomers that can exist for the square planar complex \([Pt(Cl)(py)(NH_3)(NH_2OH)]^+\) is **3**.

---