In a face-centred cubic (fcc) lattice, atom A occupies all corner lattice positions and atom B occupies all face-centred lattice positions. If exactly one B atom is missing from one of the face centres, what is the simplest formula of the resulting crystalline solid?

Introduction / Context:
This question tests your understanding of solid state chemistry, specifically how to determine the empirical formula of an ionic or atomic solid from occupancy of lattice points in a face centred cubic (fcc) crystal. In such questions, it is essential to understand how many atoms from corners, edges, faces and body centres actually belong to a single unit cell. Here, atom A sits at all corners, atom B sits at all face centres, and one B atom is missing from one of the face centres. You are asked to convert these occupancy details into a simplest whole number formula of the type A x B y for the solid.

Given Data / Assumptions:

• The lattice is face centred cubic (fcc).
• Atom A occupies all eight corner lattice positions of the cubic unit cell.
• Atom B occupies all six face centred positions in the ideal case.
• Exactly one face centre is vacant, so there are B atoms only on five faces.
• We assume a perfect infinite crystal in which each unit cell is identical apart from this systematic missing B per cell.

Concept / Approach:
In crystallography, each lattice point can be shared by neighbouring unit cells. Corner atoms are shared by eight cells, so each corner contributes only one eighth to a given cell. Face centred atoms are shared by two neighbouring cells, so each face atom contributes one half to one unit cell. The basic strategy is to calculate the effective number of A and B atoms per unit cell using these fractions, then simplify the ratio to the smallest whole numbers. This ratio gives the empirical formula A x B y of the crystalline solid.

Step-by-Step Solution:
Step 1: Count the number of A atoms at corners. There are 8 corners, each shared by 8 cells, so effective A atoms per cell = 8 * (1/8) = 1. Step 2: In a normal fcc cell, there are 6 faces and thus 6 face centred atoms. Each face centred atom is shared between 2 unit cells, so normally B atoms per cell = 6 * (1/2) = 3. Step 3: The question states that one face centred B is missing. So there are B atoms only at 5 faces instead of 6. Step 4: Effective number of B atoms per cell = 5 * (1/2) = 2.5. Step 5: Now the ratio of A:B in the crystal is 1 : 2.5. Step 6: To convert 1 : 2.5 into whole numbers, multiply both by 2 to get 2 : 5. Step 7: Hence the empirical formula of the solid is A2B5.

Verification / Alternative check:
You can verify the result by reversing the process. Start with A2B5 per formula unit. For every 2 A atoms and 5 B atoms, normalising to one A atom gives 1 A and 2.5 B atoms. This matches the effective atoms per unit cell that we calculated directly from the lattice positions. Because no simpler whole number ratio exists than 2:5, A2B5 must be the correct empirical formula. This confirms that the approach of counting fractional contributions from corners and faces is consistent.

Why Other Options Are Wrong:
Option A2B3 would correspond to a B count of 1.5 per A, which does not match the calculated 2.5 B per A. A3B2 implies B is less abundant than A, contradicting the fact that there are more B atoms than A atoms in the unit cell. A3B4 gives a ratio B:A of 4:3 (about 1.33), again not matching the required 2.5:1. Therefore these alternative formulas do not reflect the actual occupancy of lattice points in the given fcc structure.

Common Pitfalls:
Many students forget that corner and face atoms are shared between multiple unit cells and incorrectly count them as one full atom per position. Another common error is to ignore the missing face centred B atom and assume 6 B atoms at faces, which would lead to a formula A1B3 instead of A2B5. Some also forget to convert fractional numbers into the simplest whole number ratio, leaving answers like A1B2.5. Always compute effective atoms per cell using sharing fractions, then simplify to whole numbers to get the final formula.

Final Answer:
The correct empirical formula of the crystalline solid is A2B5.