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In a crystal of an ionic compound, the i...

In a crystal of an ionic compound, the ions B form the close packed lattice and the ions A occupy all the tetrahedral voids. The formula of the compound is

A

`AB_(2)`

B

`A_(2)B`

C

AB

D

`AB_(3)`

Text Solution

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The correct Answer is:
To determine the formula of the ionic compound based on the given information, we can follow these steps: ### Step-by-Step Solution: 1. **Identify the Lattice Structure**: - The question states that ions B form a close-packed lattice. In a close-packed structure, specifically a cubic close-packed (CCP) lattice, there are 4 ions per unit cell. 2. **Determine the Number of Tetrahedral Voids**: - In a cubic close-packed lattice, there are 8 tetrahedral voids. This is derived from the arrangement of the ions in the lattice. Each unit cell contains 8 tetrahedral voids. 3. **Assign the Ions to the Lattice and Voids**: - According to the question, ions A occupy all the tetrahedral voids. Since there are 8 tetrahedral voids and each void can accommodate one ion A, we have 8 ions of A. 4. **Combine the Ions to Form the Compound**: - Since there are 8 ions of A and 4 ions of B in the unit cell, we can express the formula of the compound as A₈B₄. 5. **Simplify the Formula**: - The formula A₈B₄ can be simplified by dividing both subscripts by 4, resulting in A₂B. ### Final Answer: The formula of the compound is A₂B.
To determine the formula of the ionic compound based on the given information, we can follow these steps: ### Step-by-Step Solution: 1. **Identify the Lattice Structure**: - The question states that ions B form a close-packed lattice. In a close-packed structure, specifically a cubic close-packed (CCP) lattice, there are 4 ions per unit cell. 2. **Determine the Number of Tetrahedral Voids**: ...
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