The magnetic property, dipole moment, plane of symmetry, colour and absorption band cann be helpful in structure elucidation of complex compounds:
Q. Complex compound() having even number of space (stereo) isomers is/are: (Where AA-symmetrical bidentate ligand ad a,b,c,d,e-monodentate ligands)

Options

A. `[M(A A)_(2)b_(2)]^(n+-)`
B. `[Ma_(3)b_(3)]n^(+-)`
C. `[Ma_(3)bcd]^(n+-)`
D. `[Ma_(2)bcde]^(n+-)`

Answer

To solve the question regarding the complex compounds having an even number of stereoisomers, we will analyze the given complexes systematically. The complexes are defined by the ligands involved, where "A" represents a symmetrical bidentate ligand and "B", "C", "D", and "E" represent monodentate ligands.

### Step-by-Step Solution:

1. **Identify the Complexes**: We have to evaluate the complexes of the form M(AA)2B2, MA3B3, MA3BCD, and MA2BCD.

2. **Determine the Geometry**:
- For M(AA)2B2, the geometry is octahedral since it has 6 coordination sites (4 from bidentate ligands and 2 from monodentate ligands).
- For MA3B3, the geometry is also octahedral.
- For MA3BCD, the geometry is octahedral.
- For MA2BCD, the geometry is octahedral.

3. **Calculate Total Isomers**:
- **For M(AA)2B2**:
- Stereoisomers: 3 (1 enantiomeric pair + 2 geometrical isomers).
- **For MA3B3**:
- Stereoisomers: 2 (0 enantiomers + 2 geometrical isomers: facial and meridional).
- **For MA3BCD**:
- Stereoisomers: 5 (1 enantiomeric pair + 4 geometrical isomers).
- **For MA2BCD**:
- Stereoisomers: 15 (6 enantiomeric pairs + 9 geometrical isomers).

4. **Identify Even Number of Stereoisomers**:
- M(AA)2B2: 3 (odd)
- MA3B3: 2 (even)
- MA3BCD: 5 (odd)
- MA2BCD: 15 (odd)

5. **Conclusion**: The only complex that has an even number of stereoisomers is **MA3B3**.

### Final Answer:
The complex compound having an even number of stereoisomers is **MA3B3**.