Select the correct statements : `CaCl_(2)` doped with NaCl solid solution results in

Options

A. substitutional cation vacancy
B. Frenkel's defect
C. Schottky defect
D. decrease of density

Answer

To solve the question regarding the solid solution of CaCl₂ doped with NaCl, we will analyze the situation step by step.

### Step 1: Understand the composition of CaCl₂ and NaCl
CaCl₂ consists of calcium ions (Ca²⁺) and chloride ions (Cl⁻). The formula indicates that for every calcium ion, there are two chloride ions. NaCl consists of sodium ions (Na⁺) and chloride ions (Cl⁻), with one sodium ion for each chloride ion.

### Step 2: Analyze the doping process
When NaCl is doped into CaCl₂, Na⁺ ions will replace Ca²⁺ ions in the crystal lattice. Given that Ca²⁺ has a +2 charge and Na⁺ has a +1 charge, one Ca²⁺ ion can be replaced by two Na⁺ ions to maintain charge neutrality.

### Step 3: Determine the effect on the crystal lattice
The replacement of one Ca²⁺ ion by two Na⁺ ions creates a vacancy in the lattice because the charge balance is disrupted. This results in a deficiency of cations in the crystal structure.

### Step 4: Identify the type of defect
The loss of one cation (Ca²⁺) leads to a decrease in density, which indicates a defect in the crystal structure. This situation is characteristic of a Schottky defect, where both cations and anions are missing from the lattice. However, since only cations (Ca²⁺) are being replaced by Na⁺ ions, it does not fully satisfy the criteria for a Schottky defect.

### Step 5: Conclusion
The correct statements regarding the solid solution of CaCl₂ doped with NaCl are:
- The density of the solid solution decreases due to the loss of cations.
- The defect created is not a Frankel defect, as that involves smaller cations occupying interstitial sites, which is not the case here.

### Final Answer
The correct options are:
- A: The density decreases due to the loss of cations.
- D: The defect is not a Frankel defect.