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In an ionic solid, B^(x-) ions constitut...

In an ionic solid, `B^(x-)` ions constitute a ccp lattice, if `A^(y+)` ions occupy 25%, of the tetrahedral voids, the possible ions in the solid are

A

`A^(+)` and `B^(-)`

B

`A^(2+)` and `B^(4-)`

C

`A^(2+)` and `B^(-)`

D

`A^(4+)` and `B^(-)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze the given information about the ionic solid and determine the possible ions present in it. Let's break down the solution step by step: ### Step-by-Step Solution: 1. **Understanding the Lattice Structure:** - The question states that `B^(x-)` ions form a cubic close-packed (ccp) lattice. In a ccp lattice, the number of atoms (or ions) per unit cell is 4. 2. **Identifying Tetrahedral Voids:** - In a ccp lattice, there are 8 tetrahedral voids per unit cell. Therefore, for `x` number of `B^(x-)` ions, the number of tetrahedral voids is given by: \[ \text{Number of tetrahedral voids} = 2 \times \text{Number of B ions} = 2 \times 4 = 8 \] 3. **Calculating the Number of A Ions:** - It is given that `A^(y+)` ions occupy 25% of the tetrahedral voids. Since there are 8 tetrahedral voids, the number of tetrahedral voids occupied by `A^(y+)` ions is: \[ \text{Number of A ions} = 25\% \text{ of } 8 = \frac{25}{100} \times 8 = 2 \] 4. **Determining the Ratio of Ions:** - We have 2 `A^(y+)` ions and 4 `B^(x-)` ions in the unit cell. Therefore, the formula for the compound can be expressed as: \[ \text{Formula} = A_2B_4 \] 5. **Simplifying the Formula:** - To simplify the formula, we can divide both the subscripts by 2: \[ \text{Simplified Formula} = AB_2 \] 6. **Identifying the Charges of the Ions:** - Since `A` is a cation and `B` is an anion, we need to assign the charges. Assuming `A` has a charge of +2 (i.e., `A^(2+)`) and `B` has a charge of -1 (i.e., `B^(-)`), the overall charge balance in the compound is: \[ 2 \times (+2) + 4 \times (-1) = 0 \] - This indicates that the charges are balanced. 7. **Final Conclusion:** - The possible ions in the solid are `A^(2+)` and `B^(-)`. ### Final Answer: The possible ions in the solid are `A^(2+)` and `B^(-)`.
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