Question

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)

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, $N=6$.

Answer 1

Correct Answer: 1

Step 1: Understanding the Concept:
In a crystal lattice, the number of atoms per unit cell and the number of specific interstitial sites (tetrahedral and octahedral) follow a fixed ratio. For a cubic close-packed (ccp/fcc) lattice, if there are $N$ atoms forming the lattice, there are $N$ octahedral sites and $2N$ tetrahedral sites.
Step 2: Detailed Explanation:
1. Let the number of anions ($B$) forming the ccp arrangement be $Z$. For ccp (fcc), $Z = 4$.
2. The number of octahedral sites in a ccp lattice is equal to the number of lattice atoms ($Z$).
\[ \text{Number of octahedral sites} = 4 \]
3. Cations ($A$) occupy all the octahedral sites:
\[ \text{Number of cations } (A) = 4 \]
4. The ratio of $A : B = 4 : 4 = 1 : 1$.
5. The empirical formula is $AB$.
6. Comparing $AB$ with the given $A_x B$, we get $x = 1$.
Step 3: Final Answer:
The value of $x$ is 1.