Question

A compound is formed by cation C and anion A. The anions form hexagonal close packed (hcp) lattice and the cations occupy 75% of octahedral voids. The formula of the compound is :

• A. \(C_2A_3 \)
• B. \(C_3A_2 \)
• C. \(C_3A_4 \)
• D. \(C_4A_3 \)

Answer 1

The Correct Option is C

The question involves determining the chemical formula of a compound formed by cation C and anion A. The problem gives us the following details:

• Anions form a hexagonal close-packed (hcp) lattice.
• Cations occupy 75% of the octahedral voids.

Let's solve the problem step-by-step:

1. In a hexagonal close-packed (hcp) lattice, the number of atoms per unit cell (in this case, anions) is 6. This is derived from the coordination of the lattice structure.

2. In a lattice, the number of octahedral voids is equal to the number of atoms forming the lattice. Therefore, in the hcp structure, there are 6 octahedral voids available for cations.

3. According to the problem, 75% of these octahedral voids are occupied by cations. Therefore, the number of cations occupying these voids will be:

   = 6 \times \frac{75}{100} = 4.5

   Since the number of cations must be a whole number, practically, we consider that in reality, ratios are used to form stable ionic compounds. Therefore, the ratio implies 4 cations effectively occupying the voids to every 3 lattice sites worth of anions.

4. The ratio of cations (C) to anions (A) is then 4.5:6, which simplifies to 3:4 when considering simple whole-number ratios for chemical compounds. Thus, we can determine that the chemical formula of the compound is C_3A_4.

Therefore, the correct formula of the compound is C_3A_4, corresponding to the provided option.