Question

$\text{KF}$ has CCP structure. How many ${{\text{F}}^{-}}$ ions and octahedral voids are here in this unit cell respectively?
A. 4 and 4
B. 4 and 8
C. 8 and 4
D. 6 and 6

Answer 1

Hint: CCP or cubic close packing are those lattice structures in which the atoms are present at 8 corners and on the centre of each face. Octahedral voids are formed when two triangle voids of second and first join then the interstitial void formed is known as octahedral void.

Complete answer:
-In $\text{KF}$, the anions i.e. Fluorine ion is present at the corner and face.
-So, firstly the total number of atoms which will contribute to the structure will be calculated.
-As we know that a cube has 8 corners and 6 faces so we will multiply it with 1/8 and 1/2 because each atom will contribute 1/8th part of it in the corner and 1/2th part in the face centre.
-So, the number of atoms will be:

At corner: $\text{8 }\times \text{ }\dfrac{1}{8}\text{ = 1}$
At face: $\text{6 }\times \text{ }\dfrac{1}{2}\text{ = 3}$
-So, a total of 4 atoms will contribute to the structure of potassium fluoride.
-Now, we will calculate the octahedral voids.
-Each charge should be balanced i.e. if there is one cation then one anion should be present for it.
-So, as we have calculated that are 4 atoms then there should be $\text{4}{{\text{F}}^{-}}\ \text{and 4}{{\text{K}}^{+}}$ are present.
-Now, the octahedral void will also be 4 because there is only 1 octahedral void per atom in cubic close packing and hexagonal close packing.

Note:
There are two tetrahedral voids per atom in CCP and HCP structure. Students should not confuse between the octahedral and tetrahedral void. The tetrahedral void is formed between the two layers when the atom of the second layer rests on the hollow part of the first layer.