In a solid AB, A atoms are in ccp arrangement and B atoms occupy all the octahedral sites. If two atoms from the opposite faces are removed, then the resultant stoichiometry of the compound is $$A_xB_y$$. The value of $$x$$ is ______ [nearest integer]

We need to find the value of x in the stoichiometry $$A_xB_y$$ after removing two A atoms from opposite faces of a solid AB with ccp arrangement.

In a ccp (FCC) unit cell, A atoms are arranged in cubic close packing:

Corner A atoms: 8 corners × $$\frac{1}{8}$$ = 1

Face-center A atoms: 6 faces × $$\frac{1}{2}$$ = 3

Total A atoms per unit cell = 1 + 3 = 4

B atoms occupy all octahedral sites:

Body center: 1 × 1 = 1

Edge centers: 12 × $$\frac{1}{4}$$ = 3

Total B atoms per unit cell = 1 + 3 = 4

Initial formula: $$A_4B_4$$ = AB (confirmed).

Two A atoms are removed from opposite faces. Each face-center atom contributes $$\frac{1}{2}$$ to the unit cell.

Removing 2 face-center atoms reduces A by: $$2 \times \frac{1}{2} = 1$$

New A atoms = 4 - 1 = 3

B atoms remain = 4 (unchanged)

The formula becomes $$A_3B_4$$.

So $$x = 3$$.

Therefore, the value of x is 3.