Give the ration of geometrical isomers in `[M(A A)_(2)b_(2)]` and optical isomers of `[M(A A)_(3)]` .

Explanation

To solve the problem of finding the ratio of geometrical isomers in the complex \([M(A A)_{2}B_{2}]\) and optical isomers in the complex \([M(A A)_{3}]\), we will follow these steps:

### Step 1: Analyze the complex \([M(A A)_{2}B_{2}]\)

1. **Identify the ligand arrangement**: The complex has two bidentate ligands \(A\) (denoted as \(AA\)) and two monodentate ligands \(B\).
2. **Determine possible geometrical isomers**:
- **Cis form**: Both \(AA\) ligands are adjacent to each other.
- **Trans form**: The \(AA\) ligands are opposite each other.

Thus, there are **two geometrical isomers**: one cis and one trans.

### Step 2: Analyze the complex \([M(A A)_{3}]\)

1. **Identify the ligand arrangement**: The complex has three bidentate ligands \(A\) (denoted as \(AA\)).
2. **Determine possible optical isomers**:
- The arrangement of three bidentate ligands leads to a chiral structure, meaning it can exist in two enantiomeric forms (D and L).
- Since there are no planes of symmetry in this arrangement, it will have **two optical isomers**.

### Step 3: Calculate the ratio of geometrical to optical isomers

- **Geometrical isomers** in \([M(A A)_{2}B_{2}]\) = 2
- **Optical isomers** in \([M(A A)_{3}]\) = 2

The ratio of geometrical isomers to optical isomers is:
\[ \text{Ratio} = \frac{\text{Geometrical Isomers}}{\text{Optical Isomers}} = \frac{2}{2} = 1 \]

### Final Answer
The ratio of geometrical isomers in \([M(A A)_{2}B_{2}]\) to optical isomers in \([M(A A)_{3}]\) is **1**.

---