How many geometrical isomers are possible for `[Pd^(2+)(NH_(2)-CH(CH_(3))-CO_(2)^(-))_(2)]^(0)`

Options

A. 2
B. 3
C. 5
D. 6

Explanation

To determine the number of geometrical isomers for the coordination compound \([Pd^{2+}(NH_2-CH(CH_3)-CO_2^-)_{2}]^{0}\), we can follow these steps:

### Step 1: Identify the Coordination Complex
The given complex is \([Pd^{2+}(NH_2-CH(CH_3)-CO_2^-)_{2}]^{0}\). Here, palladium (Pd) is in the +2 oxidation state and is coordinated to two bidentate ligands.

### Step 2: Analyze the Ligand
The ligand \(NH_2-CH(CH_3)-CO_2^-\) is a bidentate ligand because it has two donor atoms: nitrogen (N) from the amino group and oxygen (O) from the carboxylate group. This means that each ligand can form two bonds with the metal center.

### Step 3: Determine the Possible Geometries
For a coordination complex with a metal center that has a coordination number of 4 (like Pd in this case), the possible geometrical arrangements are typically square planar or tetrahedral. However, for this specific complex, we will focus on the square planar geometry, which is common for d8 metal complexes like Pd(II).

### Step 4: Draw the Structures
1. **Cis Isomers**: In the cis configuration, both donor atoms of the bidentate ligands are on the same side of the metal center. This can lead to two distinct cis isomers based on the arrangement of the substituents on the ligand.

2. **Trans Isomer**: In the trans configuration, the donor atoms of the bidentate ligands are opposite each other. This arrangement will yield one distinct trans isomer.

### Step 5: Count the Isomers
From the analysis:
- **Cis Isomers**: 2 distinct configurations
- **Trans Isomer**: 1 configuration

Thus, the total number of geometrical isomers is:
\[ \text{Total Isomers} = \text{Cis Isomers} + \text{Trans Isomer} = 2 + 1 = 3 \]

### Final Answer
The total number of geometrical isomers possible for the complex \([Pd^{2+}(NH_2-CH(CH_3)-CO_2^-)_{2}]^{0}\) is **3**.

---