In a face centered cubic lattice atoms A are at the corner points and atoms B at the face centered points. If atom B is missing from one of the face centered points, the formula of the ionic compound is:

In a face-centered cubic (FCC) lattice, atoms A are located at the corners and atoms B are at the face centers. Normally, a standard FCC unit cell has:

For corner atoms (A): There are 8 corners, and each corner atom is shared by 8 adjacent unit cells. So, the contribution of each corner atom to one unit cell is $$\frac{1}{8}$$. Therefore, the total number of A atoms per unit cell is $$8 \times \frac{1}{8} = 1$$.

For face-centered atoms (B): There are 6 faces, and each face-centered atom is shared by 2 unit cells. So, the contribution of each face-centered atom to one unit cell is $$\frac{1}{2}$$. Therefore, the total number of B atoms per unit cell in a perfect FCC lattice is $$6 \times \frac{1}{2} = 3$$.

However, the problem states that one atom B is missing from one of the face-centered points. This means that instead of 6 face-centered atoms, there are only 5 present (since one is missing).

Now, recalculate the contribution of B atoms per unit cell: With 5 face-centered atoms, each contributing $$\frac{1}{2}$$ to the unit cell, the total number of B atoms per unit cell is $$5 \times \frac{1}{2} = \frac{5}{2}$$.

The number of A atoms per unit cell remains unchanged: $$1$$.

So, per unit cell, we have:

• A atoms: $$1$$
• B atoms: $$\frac{5}{2}$$

To find the simplest whole-number ratio of A to B, multiply both by 2 to eliminate the fraction:

• A atoms: $$1 \times 2 = 2$$
• B atoms: $$\frac{5}{2} \times 2 = 5$$

Thus, the formula of the compound is A$$_2$$B$$_5$$.

Comparing with the options:

• A. A$$_2$$B$$_3$$
• B. AB$$_2$$
• C. A$$_5$$B$$_2$$
• D. A$$_2$$B$$_5$$

Hence, the correct answer is Option D.