Trans form of `[M(A A)_(2)a_(2)]^(n+)` complex does not show ________isomerism .

Answer

To solve the question regarding the trans form of the complex \([M(AA)_{2}A_{2}]^{n+}\) and its isomerism properties, we can follow these steps:

### Step 1: Identify the Coordination Number
The coordination number of the complex can be determined by analyzing the ligands:
- The ligand \(AA\) is bidentate, meaning it can bind through two donor atoms.
- The ligand \(A\) is monodentate, meaning it binds through one donor atom.

Thus, the total coordination number is calculated as follows:
- For \(2\) bidentate ligands: \(2 \times 2 = 4\)
- For \(2\) monodentate ligands: \(2 \times 1 = 2\)

Adding these together gives:
\[ \text{Total Coordination Number} = 4 + 2 = 6 \]

### Step 2: Draw the Trans Form
In the trans configuration, the bidentate ligands \(AA\) will be positioned opposite each other. The monodentate ligands \(A\) will occupy the remaining positions.

### Step 3: Analyze Symmetry
Now, we need to analyze the symmetry of the trans form:
- The trans form has a plane of symmetry because the two \(AA\) ligands are opposite each other and the two \(A\) ligands are also opposite each other.

### Step 4: Determine the Type of Isomerism
Since the trans form has a plane of symmetry, it cannot exhibit optical isomerism. Optical isomerism occurs when a compound can exist in two non-superimposable mirror images, which is not possible in this case due to the presence of the plane of symmetry.

### Conclusion
Thus, the trans form of the complex \([M(AA)_{2}A_{2}]^{n+}\) does not show **optical isomerism**.

### Final Answer
The trans form of \([M(AA)_{2}A_{2}]^{n+}\) complex does not show **optical isomerism**.

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