Consider the following reactions of complex compounds A,B and C
(i) `CoCl_(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, we need to analyze the reactions of the complex compounds A, B, and C and derive their structures based on the information provided. Let's break down the steps:

### Step 1: Analyze Complex A
The reaction for complex A is:
\[ \text{CoCl}_2\text{Br} \cdot 5\text{NH}_3 + \text{excess } \text{Ag}^+ \text{(aq)} \to 1 \text{AgCl (s)} \]

- Since only 1 mole of AgCl is formed, it indicates that only 1 chloride ion is outside the coordination sphere. The remaining chloride and bromide ions must be inside the coordination sphere.
- Therefore, the structure of complex A can be formulated as:
\[ \text{[Co(NH}_3\text{)}_5\text{Cl]}^{2+} \text{Br}^- \]

### Step 2: Analyze Complex B
The reaction for complex B is:
\[ \text{CoCl}_2\text{Br} \cdot 5\text{NH}_3 + \text{excess } \text{Ag}^+ \text{(aq)} \to 2 \text{AgCl (s)} \]

- Here, 2 moles of AgCl are formed, indicating that there are 2 chloride ions outside the coordination sphere.
- Thus, the structure of complex B can be formulated as:
\[ \text{[Co(NH}_3\text{)}_5\text{Cl}_2]^{2+} \text{Br}^- \]

### Step 3: Analyze Complex C
The reaction for complex C is:
\[ \text{CoCl}_2\text{Br} \cdot 4\text{NH}_3 + \text{excess } \text{Ag}^+ \text{(aq)} \to 1 \text{AgCl (s)} \]

- Again, since only 1 mole of AgCl is formed, it indicates that there is 1 chloride ion outside the coordination sphere.
- Therefore, the structure of complex C can be formulated as:
\[ \text{[Co(NH}_3\text{)}_4\text{Cl}]\text{Br}^- \]

### Step 4: Identify Relationships Between Compounds
1. **Ionization Isomers**: A and B are ionization isomers because they differ in the arrangement of chloride ions inside and outside the coordination sphere.
2. **Molar Conductivity**: Both A and B will dissociate into three ions in solution (1 complex ion + 2 halides), leading to similar molar conductivities.
3. **Optical Isomerism**: Complex C can exhibit geometrical isomerism due to the presence of different ligands (Cl and Br), while A and B do not show optical isomerism.
4. **Crystal Field Stabilization Energy (CFSE)**: The CFSE will be higher for A than for B due to the stronger ligand (Cl) compared to Br, and C will have the lowest CFSE since it has fewer NH3 ligands.

### Final Summary
Based on the analysis:
- **Correct Statements**:
- A and B are ionization isomers.
- Molar conductivity of A and B is similar.
- A and B do not show optical isomerism, but C can show geometrical isomerism.
- CFSE order: A > B > C.