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In an ionic crystal, cation "A" occupies...

In an ionic crystal, cation "A" occupies the  lattice points in a FCC array and anion "B" occupies the two types of tetrahedral voids. The correct formula of the ionic compound is

A

`AB_(2)`

B

`A_(2)B`

C

`AB`

D

`A_(2)B_(3)`

Text Solution

AI Generated Solution

The correct Answer is:
To determine the correct formula of the ionic compound formed by cation "A" occupying the lattice points in a face-centered cubic (FCC) array and anion "B" occupying the tetrahedral voids, we can follow these steps: ### Step-by-Step Solution: 1. **Identify the number of cations (A) in the FCC lattice:** - In a face-centered cubic (FCC) lattice, the effective number of atoms (Z) is 4. This means there are 4 cations (A) per unit cell. 2. **Identify the number of tetrahedral voids in the FCC lattice:** - In an FCC lattice, there are 8 tetrahedral voids. This is because the number of tetrahedral voids is double the number of octahedral voids, and there are 4 octahedral voids in an FCC lattice. 3. **Determine the number of anions (B) occupying the tetrahedral voids:** - Since there are 8 tetrahedral voids and each void can be occupied by one anion (B), there will be 8 anions (B) in the unit cell. 4. **Write the formula based on the ratio of cations to anions:** - From the previous steps, we have 4 cations (A) and 8 anions (B). Therefore, the formula can be written as A₄B₈. 5. **Simplify the formula:** - The formula A₄B₈ can be simplified by dividing both subscripts by 4, which gives us AB₂. ### Final Answer: The correct formula of the ionic compound is **AB₂**. ---
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