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 :

Question

What are the constituents of Natalite?

Answer 1

To solve this problem, we need to understand the distribution of atoms in a face-centered cubic (FCC) lattice and how missing atoms affect the stoichiometry of the compound.

Step 1: Understand the FCC lattice structure.

In a face-centered cubic (FCC) lattice:

There are 8 atoms at the corners of the cube.
There is 1 atom at each of the 6 face centers of the cube.

In this question:

Atoms $ A $ are located at the corner points.
Atoms $ B $ are located at the face-centered points.
One atom of $ B $ is missing from one of the face-centered points.

Step 2: Calculate the net number of equivalent atoms.

Corner atoms ($ A $) contribution: Each corner is shared by 8 cubes, so the contribution per cube is $ \frac{1}{8} $ of each corner atom. Thus, the total contribution of 8 corner atoms is $ 8 \times \frac{1}{8} = 1 $.
Face-center atoms ($ B $): Each face-centered atom is shared by 2 cubes, so the contribution per cube is $ \frac{1}{2} $. With 6 face-centered atoms, the total contribution is $ 6 \times \frac{1}{2} = 3 $.
Since one of the $ B $ atoms is missing from a face, we subtract 0.5 from the faces: The available face-face atoms become $ 3 - 0.5 = 2.5 $.

Step 3: Determine the empirical formula.

From the calculations:

Number of $ A $ atoms: 1
Number of $ B $ atoms: 2.5

We need to express this in whole numbers. Multiplying both atom counts by 2 to eliminate the fraction:

Number of $ A $ atoms becomes $ 1 \times 2 = 2 $
Number of $ B $ atoms becomes $ 2.5 \times 2 = 5 $

Thus, the empirical formula is A_2B_5.

Conclusion: Therefore, considering the positions and the missing atom in the FCC lattice, the correct formula of the compound is A_2B_5.

Answer 2