To solve the problem regarding the hexa-coordinated complex `CoCl3.6H2O`, we need to analyze the statements given in the question based on the information provided about the complex and its behavior in aqueous solution.
### Step-by-Step Solution:
1. **Understanding the Complex**:
- The complex is `CoCl3.6H2O`, which means it contains one cobalt ion (Co), three chloride ions (Cl), and six water molecules (H2O).
- Cobalt typically has a +3 oxidation state in such complexes.
2. **Ionization in Aqueous Solution**:
- The problem states that the complex undergoes 100% ionization in aqueous solution.
- When `CoCl3.6H2O` ionizes, it dissociates into one `Co^3+` ion and three `Cl^-` ions, resulting in a total of 4 particles (1 Co^3+ + 3 Cl^-).
3. **Van't Hoff Factor**:
- The Van't Hoff factor (i) is given as 3, which indicates that the effective number of particles in solution is 3.
- However, we have calculated 4 particles from the dissociation, which suggests that the complex might be behaving differently due to the presence of water molecules or other interactions.
4. **Effective Atomic Number (EAN)**:
- The effective atomic number (EAN) can be calculated using the formula:
\[
\text{EAN} = Z + n - c
\]
where Z = atomic number of Co (27), n = number of ligands (6 from H2O + 3 from Cl = 9), and c = charge on the ion (+3).
- Thus, EAN = 27 + 9 - 3 = 33. However, we need to consider the coordination correctly; the EAN is actually calculated as:
\[
\text{EAN} = 27 + 6 \times 1 + 3 \times (-1) = 27 + 6 - 3 = 30
\]
- This indicates that the EAN is not 36, so this statement is incorrect.
5. **Weight Loss with Anhydrous CaCl2**:
- Anhydrous CaCl2 is a dehydrating agent. When `CoCl3.6H2O` is treated with it, it will lose water molecules, resulting in a decrease in weight. Thus, this statement is correct.
6. **Molar Conductivity**:
- Since the complex undergoes 100% ionization, it should exhibit significant molar conductivity due to the presence of ions in solution. Therefore, the statement about negligible molar conductivity is incorrect.
7. **Formation of AgCl Precipitate**:
- When `CoCl3` is treated with excess AgNO3, it will react with the chloride ions to form AgCl precipitate. Since there are three Cl^- ions, it will produce three moles of AgCl. Therefore, the statement about producing 2 moles of AgCl is incorrect.
### Summary of Correct Statements:
- The correct statements regarding the complex `CoCl3.6H2O` are:
- The complex loses some weight when treated with anhydrous CaCl2 (Correct).
- The aqueous solution of the complex has negligible molar conductivity (Incorrect).
- 1 mole of the complex gives 2 moles of AgCl precipitate with excess AgNO3 (Incorrect).
Thus, the correct options are:
- Option 2: The complex loses weight when treated with anhydrous CaCl2.
