The empirical formula for a compound with a cubic close packed arrangement of anions and with cations occupying all the octahedral sites in AₓB. The value of x is _____________. (Integer answer)

Solution

Chemistry ma’am · 4 steps

In ccp, octahedral voids match the packed atoms one for one, so filling them all gives a 1 : 1 ratio.

1. Count the anions. A cubic close packed (fcc) unit cell has 8 × 1/8 + 6 × 1/2 = 4 anions.
2. Count octahedral voids. The octahedral voids sit at the body centre and at the 12 edge centres. That gives 1 + 12 × 1/4 = 4 voids per unit cell, one for every packed anion.
3. Fill all of them. Cations occupy all 4 octahedral voids. So the cell holds 4 cations (A) and 4 anions (B), a ratio of 4 : 4 = 1 : 1.
4. So. The empirical formula is AB, which matches the NaCl structure, and x = 1.

The common slip: 2. Using the tetrahedral void count, which is 2 per packed atom, gives A₂B and x = 2. Octahedral voids are only 1 per packed atom.

The idea to keep. For N close-packed atoms there are N octahedral voids and 2N tetrahedral voids. Multiply by the fraction filled to get the formula.

Answer from the NTA final answer key. Solution by PrepWiser.