How many geometrical isomers are possible for the square planar complex `[Pt(NO_(2))(py)(NH_(3))(NH_(2)OH)]NO_(2)`
(a) `Four
(b) Five
(c ) Eight
(d) Three .

Explanation

To determine the number of geometrical isomers possible for the square planar complex \([Pt(NO_2)(py)(NH_3)(NH_2OH)]NO_2\), we can follow these steps:

### Step 1: Identify the Ligands
The complex contains the following ligands:
- \(NO_2\) (nitrito)
- \(py\) (pyridine)
- \(NH_3\) (ammonia)
- \(NH_2OH\) (hydroxylamine)

All the ligands are different, which is crucial for the formation of geometrical isomers.

**Hint:** Check if all ligands attached to the central metal atom are different.

### Step 2: Determine the Geometry
The complex is square planar. In square planar complexes, geometrical isomers arise due to the different spatial arrangements of the ligands around the central metal atom.

**Hint:** Remember that square planar complexes can have different arrangements leading to isomers.

### Step 3: Analyze Possible Arrangements
For square planar complexes with four different ligands, we can analyze the possible arrangements. We can fix one ligand and permute the others around it.

1. **Fix \(NH_2OH\)**:
- Arrangement 1: \(NH_2OH\) at the top, \(NO_2\) on one side, \(NH_3\) on the other side, and \(py\) at the bottom.
- Arrangement 2: \(NH_2OH\) at the top, \(NH_3\) on one side, \(py\) on the other side, and \(NO_2\) at the bottom.

2. **Fix \(NO_2\)**:
- Arrangement 3: \(NO_2\) at the top, \(NH_2OH\) on one side, \(NH_3\) on the other side, and \(py\) at the bottom.
- Arrangement 4: \(NO_2\) at the top, \(NH_3\) on one side, \(py\) on the other side, and \(NH_2OH\) at the bottom.

3. **Fix \(NH_3\)**:
- Arrangement 5: \(NH_3\) at the top, \(NH_2OH\) on one side, \(NO_2\) on the other side, and \(py\) at the bottom.
- Arrangement 6: \(NH_3\) at the top, \(NO_2\) on one side, \(py\) on the other side, and \(NH_2OH\) at the bottom.

4. **Fix \(py\)**:
- Arrangement 7: \(py\) at the top, \(NH_2OH\) on one side, \(NO_2\) on the other side, and \(NH_3\) at the bottom.
- Arrangement 8: \(py\) at the top, \(NO_2\) on one side, \(NH_3\) on the other side, and \(NH_2OH\) at the bottom.

However, not all of these arrangements will yield unique geometrical isomers due to symmetry.

**Hint:** Consider the symmetry of the complex when determining unique arrangements.

### Step 4: Count Unique Isomers
After analyzing the arrangements, we find that there are three unique geometrical isomers:
1. Isomer 1: \(NH_2OH\) at the top, \(NO_2\) and \(NH_3\) adjacent to each other, with \(py\) at the bottom.
2. Isomer 2: \(NH_2OH\) at the top, \(NH_3\) and \(py\) adjacent to each other, with \(NO_2\) at the bottom.
3. Isomer 3: \(NO_2\) at the top, \(NH_2OH\) and \(NH_3\) adjacent to each other, with \(py\) at the bottom.

Thus, the total number of geometrical isomers for the complex is **3**.

### Final Answer
The answer to the question is (d) Three.

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