The empirical formula for a compound with a cubic close packed arrangement of anions and with cations occupying all the octahedral sites is $A_x B$. The value of $x$ is ________. (Integer answer)
Show Hint
Standard Lattice-Void Ratios:
- Octahedral voids = $N$ (number of atoms in the unit cell).
- Tetrahedral voids = $2N$.
In ccp/fcc, $N=4$; in hcp, $N=6$.
In a cubic close packed (ccp) arrangement, the unit cell contains four anions. In such a structure, all anions form a face-centered cubic (FCC) lattice. For a standard FCC lattice, each unit cell contains:
8 anions at the corners, each shared by 8 unit cells, resulting in 1 anion contribution (8 × 1/8 = 1).
6 anions on the faces, each shared by 2 unit cells, resulting in 3 anion contributions (6 × 1/2 = 3).
Thus, the total number of anions per unit cell is 1 + 3 = 4.
Now, considering the octahedral sites:
In a ccp structure, the number of octahedral sites is equivalent to the number of anions, i.e., 4. This means there are 4 octahedral sites in the unit cell.
All these octahedral sites are occupied by cations.
Consequently, the ratio of cations to anions is 4:4 or simply 1:1. Therefore, for every anion (B), there is one cation (A).
Thus, the empirical formula of the compound becomes A1B, and therefore the value of x is 1.
Verifying the value of x against the given range (1,1), it clearly fits within the expected range.
