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In a hypothetical ionic crystal, B is ar...

In a hypothetical ionic crystal, B is arranged in cubic close packing and A occupies all octahedral voids and alternative tetrahedral voids. The correct formula of the compound 

A

`AB`

B

`A_(2)B_(3)`

C

`AB_(2)`

D

`A_(2)B`

Text Solution

AI Generated Solution

The correct Answer is:
To determine the correct formula of the compound formed by ions A and B in a hypothetical ionic crystal, we can follow these steps: ### Step 1: Determine the number of B atoms in the cubic close packing (CCP) In a cubic close packing (CCP) arrangement, which is also known as face-centered cubic (FCC), there are 4 atoms per unit cell. - **Calculation**: - Number of B atoms = 4 ### Step 2: Determine the number of A atoms in the unit cell A occupies all octahedral voids and alternating tetrahedral voids. 1. **Octahedral voids**: In a CCP unit cell, there are 4 octahedral voids. 2. **Tetrahedral voids**: There are 8 tetrahedral voids in a CCP unit cell. Since A occupies alternating tetrahedral voids, it occupies half of them. - **Calculation**: - Number of A atoms from octahedral voids = 4 (since all octahedral voids are occupied) - Number of A atoms from tetrahedral voids = 8 / 2 = 4 (since only half of the tetrahedral voids are occupied) - Total number of A atoms = 4 (from octahedral) + 4 (from tetrahedral) = 8 ### Step 3: Write the empirical formula Now that we have the number of A and B atoms, we can write the empirical formula of the compound. - **Formula**: - A: 8 - B: 4 - Therefore, the formula is A₈B₄. ### Step 4: Simplify the formula To express the formula in its simplest form, we can divide both the subscripts by their greatest common divisor, which is 4. - **Simplification**: - A₈B₄ → A₂B (dividing both subscripts by 4) ### Final Answer The correct formula of the compound is A₂B. ---
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