Select the corrent statement `(s)`
(a) The `C.N.` of cation occupying a tetrahedral hole is `4`
(b) The `C.N.` of cation occupying an octabedral hole is `6`
(c ) In schottky defects, density of the lattice decreases

Options

A. I, II
B. II, III
C. I, II, III
D. I, III

Answer

To solve the question, we need to analyze each statement regarding the coordination number of cations in tetrahedral and octahedral holes, as well as the effect of Schottky defects on the density of the lattice.

### Step-by-Step Solution:

1. **Understanding Coordination Number**:
- The coordination number (C.N.) is defined as the number of nearest neighboring atoms or ions surrounding a central atom or ion in a crystal lattice.

2. **Analyzing Statement (a)**:
- **Statement (a)**: "The C.N. of cation occupying a tetrahedral hole is 4."
- In a tetrahedral void, there are four atoms surrounding the void: three at the base and one at the apex. Therefore, the cation in a tetrahedral hole is in contact with four neighboring atoms.
- **Conclusion**: This statement is **true**.

3. **Analyzing Statement (b)**:
- **Statement (b)**: "The C.N. of cation occupying an octahedral hole is 6."
- In an octahedral void, there are six atoms surrounding the void: three on the bottom and three on the top. Thus, the cation in an octahedral hole is in contact with six neighboring atoms.
- **Conclusion**: This statement is **true**.

4. **Analyzing Statement (c)**:
- **Statement (c)**: "In Schottky defects, density of the lattice decreases."
- A Schottky defect occurs when equal numbers of cations and anions are missing from the crystal lattice. This results in a decrease in the total number of ions present in the lattice, which in turn leads to a decrease in the density of the lattice.
- **Conclusion**: This statement is **true**.

5. **Final Conclusion**:
- Since all three statements (a), (b), and (c) are true, the correct answer is that all statements are correct.

### Final Answer:
All statements (a), (b), and (c) are correct.

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