The total number of possible isomers of sqaure-planar `[Pt(Cl)(NO_(2))(NO_(3))(SCN)]^(2-)` is :

Options

A. 8
B. 12
C. 16
D. 24

Answer

To find the total number of possible isomers of the square-planar complex \([Pt(Cl)(NO_2)(NO_3)(SCN)]^{2-}\), we will analyze the ligands and their possible arrangements.

### Step-by-Step Solution:

1. **Identify the Ligands**:
- The complex contains four ligands:
- \(Cl\) (unidentate)
- \(NO_2\) (ambidentate, can bind through N or O)
- \(NO_3\) (unidentate)
- \(SCN\) (ambidentate, can bind through S or N)

2. **Determine the Type of Ligands**:
- We have two ambidentate ligands (\(NO_2\) and \(SCN\)) and two unidentate ligands (\(Cl\) and \(NO_3\)).
- Ambidentate ligands can bind through two different atoms, which increases the number of possible isomers.

3. **Arrangement of Ligands**:
- In a square-planar geometry, we can arrange the ligands in different ways. We can fix one ligand and arrange the others around it.
- Let's consider fixing \(Cl\) at one position. The remaining ligands \(NO_2\), \(NO_3\), and \(SCN\) can be arranged in different ways.

4. **Count the Isomers from Ligand Arrangement**:
- For each arrangement of the unidentate ligands, we can have different combinations of the ambidentate ligands.
- For example, if we fix \(Cl\) and arrange \(NO_2\), \(NO_3\), and \(SCN\), we can have:
- \(Cl\) opposite \(NO_3\) with \(NO_2\) and \(SCN\) in the other two positions.
- \(Cl\) opposite \(NO_2\) with \(NO_3\) and \(SCN\) in the other two positions.
- \(Cl\) opposite \(SCN\) with \(NO_2\) and \(NO_3\) in the other two positions.

5. **Consider Ambidentate Ligand Binding**:
- Each ambidentate ligand can bind in two different ways.
- For \(NO_2\), it can bind through either N or O.
- For \(SCN\), it can bind through either S or N.
- This gives us \(2 \times 2 = 4\) combinations for each arrangement of the unidentate ligands.

6. **Calculate Total Isomers**:
- From the arrangements of the unidentate ligands, we have 3 arrangements (as shown in step 4).
- Each arrangement can lead to 4 different isomers due to the binding modes of the ambidentate ligands.
- Therefore, the total number of isomers is \(3 \text{ arrangements} \times 4 \text{ combinations} = 12\).

### Final Answer:
The total number of possible isomers of the square-planar complex \([Pt(Cl)(NO_2)(NO_3)(SCN)]^{2-}\) is **12**.