Consider the following reactions of complex compounds A,B and C
(i) `ClCl_(2)Br*5NH_(3) +excess" "Ag^(+)(aq)to1AgCl(s)`
(ii) `underset("Compound B")(CoCl_(2)Br*5NH_(3))+"excess "Ag^(+)(aq) to 2AgCl(s)`
(iii) `underset("Compound C")(CoCl_(2)Br*4NH_3)+"excess "Ag^(+)(aq.) to 1AgCl(s)`
Then according to the given information the correct statement(s) is/are:

Options

A. Compounds(A) and (B) are ionisation isomers
B. Molar conductivity of compounds (A) and (B) are almost same
C. Compounds (A), (B) and (C) do not exhibit geometrical isomerism
D. Order of CFSE values, `Delta_(0)(A) gt Delta_(0)(B) gt Delta_(0)(C)`

Explanation

To solve the problem regarding the complex compounds A, B, and C, we will analyze the reactions and the properties of these compounds step by step.

### Step 1: Identify the Complex Compounds
From the reactions provided:
1. **Compound A**: \( \text{[CoCl}_2\text{Br(NH}_3)_5] \)
2. **Compound B**: \( \text{[CoCl}_2\text{Br(NH}_3)_5] \)
3. **Compound C**: \( \text{[CoCl}_2\text{Br(NH}_3)_4] \)

### Step 2: Analyze the Reactions
1. **For Compound A**:
\[ \text{[CoCl}_2\text{Br(NH}_3)_5] + \text{excess } \text{Ag}^+ \text{(aq)} \rightarrow 1 \text{AgCl}(s) \]
This indicates that one chloride ion is released.

2. **For Compound B**:
\[ \text{[CoCl}_2\text{Br(NH}_3)_5] + \text{excess } \text{Ag}^+ \text{(aq)} \rightarrow 2 \text{AgCl}(s) \]
This indicates that two chloride ions are released.

3. **For Compound C**:
\[ \text{[CoCl}_2\text{Br(NH}_3)_4] + \text{excess } \text{Ag}^+ \text{(aq)} \rightarrow 1 \text{AgCl}(s) \]
This indicates that one chloride ion is released.

### Step 3: Determine Ionization Isomers
- **Compounds A and B**: Both contain the same ligands but differ in the arrangement of bromide and chloride ions. They are ionization isomers because they can produce different ions upon dissociation.
- **Compound C**: It has one less ammonia ligand compared to A and B, which may affect its ionization properties.

### Step 4: Molar Conductivity
- **Molar Conductivity of A and B**: Since both compounds yield the same number of ions upon dissociation (3 ions), their molar conductivities will be similar.
- **Compound C**: It will yield fewer ions compared to A and B, leading to lower molar conductivity.

### Step 5: Geometrical Isomers
- **Compounds A and B**: They do not exhibit geometrical isomerism because they have five identical ligands (ammonia).
- **Compound C**: It can exhibit geometrical isomerism due to the presence of different ligands (bromide and chloride) and fewer identical ligands.

### Step 6: Crystal Field Stabilization Energy (CFSE)
- The CFSE depends on the nature of the ligands. Strong field ligands (like NH3) will stabilize the complex more than weak field ligands (like Br and Cl).
- **Order of CFSE**: A (with 5 NH3) > B (with 5 NH3 and 2 Cl) > C (with 4 NH3 and 1 Cl).

### Conclusion
Based on the analysis, we can summarize the statements:
1. **A and B are ionization isomers**: True
2. **Molar conductivities of A and B are almost the same**: True
3. **Compounds A, B, C do not exhibit geometrical isomers**: False (C can exhibit geometrical isomers)
4. **Order of CFSE is A > B > C**: True

### Final Statements
The correct statements are:
- A and B are ionization isomers.
- Molar conductivities of A and B are almost the same.
- The order of CFSE is A > B > C.