Question

In chromium chloride($CrC{l_3}$) ,$C{l^ - }$ ions have cubic close packed arrangement and $C{r^{3 + }}$ ions are present in the octahedral holes .The fraction of total number of holes occupied is
(A) $\dfrac{1}{3}$
(B)$\dfrac{1}{6}$
(C) $\dfrac{1}{9}$
(D)$\dfrac{1}{{12}}$

Answer 1

Hint:The cubic close packed arrangement is also known as the face centred cubic close packed arrangement . If n is the number of atoms per unit cell then the number of octahedral voids is n and the number of tetrahedral voids is $2n$.

Complete step-by-step solution:We know that The cubic close packed arrangement is also known as the face centered cubic close packed arrangement . If n is the number of atoms per unit cell then the number of tetrahedral voids is n and the number of octahedral voids is $2n$.In cubic close packed arrangement the atoms are present at the corners and the face centres of the unit cell. For a cubic close packed unit cell the octahedral holes are present at the edge centres and the body centres. Also the tetrahedral holes are present at the body diagonal of the face centred cubic unit cell.
Now in the cubic close packed arrangement each chloride ion would have one octahedral void associated with it.
So number of octahedral voids with three chloride ions $ = 3$
Number of tetrahedral voids with three chloride ions= $3 \times 2 = 6$
Then total number of voids with three chloride ions $ = 9$
Number of octahedral voids occupied by $C{r^{3 + }}$$ = 1$
Fraction of octahedral voids occupied $ = \dfrac{1}{3}$
Fraction of total number of voids occupied$ = \dfrac{1}{9}$

Hence the correct answer is option (C).

Note: The octahedral holes are formed by six atoms and look like an octahedron. The number of atoms of fcc lattice is four . So the number of octahedral voids in fcc lattice is four.