An ionic compound is made up of A & B only. lons of A occupy all the corners and alternate face centers while that of B occupy body center and edge centers. If B contains -1 charge then charge on atom A will be

Options

A. +1
B. +2
C. +3
D. +4

Answer

To solve the problem, we need to determine the charge on atom A given the arrangement of ions in the ionic compound and the charge on ion B. Let's go through the solution step by step.

### Step 1: Identify the positions of ions A and B in the unit cell.
- Ions of A occupy all the corners and alternate face centers of the cubic unit cell.
- Ions of B occupy the body center and edge centers of the cubic unit cell.

### Step 2: Calculate the number of ions A per unit cell.
- There are 8 corners in a cubic unit cell, and each corner ion is shared by 8 unit cells. Therefore, the contribution of A from the corners is:
\[ \text{Contribution from corners} = 8 \times \frac{1}{8} = 1 \]
- There are 2 alternate face centers, and each face center ion is shared by 2 unit cells. Therefore, the contribution of A from the face centers is:
\[ \text{Contribution from face centers} = 2 \times \frac{1}{2} = 1 \]
- Total number of A ions per unit cell:
\[ \text{Total A} = 1 + 1 = 2 \]

### Step 3: Calculate the number of ions B per unit cell.
- There is 1 ion of B at the body center, which is not shared with any other unit cell:
\[ \text{Contribution from body center} = 1 \]
- There are 12 edges in a cubic unit cell, and each edge center ion is shared by 4 unit cells. Therefore, the contribution of B from the edge centers is:
\[ \text{Contribution from edge centers} = 12 \times \frac{1}{4} = 3 \]
- Total number of B ions per unit cell:
\[ \text{Total B} = 1 + 3 = 4 \]

### Step 4: Determine the empirical formula of the compound.
- The ratio of A to B is:
\[ \text{Ratio of A to B} = 2 : 4 = 1 : 2 \]
- Therefore, the empirical formula of the compound is \( AB_2 \).

### Step 5: Determine the charges on A and B.
- Given that the charge on ion B is -1, and the empirical formula is \( AB_2 \), we can set up the charge balance:
\[ \text{Charge on A} + 2 \times (\text{Charge on B}) = 0 \]
\[ \text{Charge on A} + 2 \times (-1) = 0 \]
\[ \text{Charge on A} - 2 = 0 \]
\[ \text{Charge on A} = +2 \]

### Conclusion:
The charge on atom A is +2.

### Final Answer:
The charge on atom A is +2 (Option 2).

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