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Question: In chromium chloride \(\left( {CrC{l_3}} \right),C{l^ - }\) ions have cubic close packed arrangement...

In chromium chloride (CrCl3),Cl−\left( {CrC{l_3}} \right),C{l^ - } ions have cubic close packed arrangement and Cr3+C{r^{3 + }} ions are present in the octahedral holes. The fraction of the total number of holes occupied is.
A) 13\dfrac{1}{3}
B) 16\dfrac{1}{6}
C) 19\dfrac{1}{9}
D) 112\dfrac{1}{12}

Explanation

Solution

We know that a unit is the smallest representation of a whole crystal. During a unit, an atom's coordination number is the number of atoms it's touching. The coordination number of face-centered cubic (FCC) is twelve and we know that cubic close packing is also similar to that of FCC.

Complete step by step answer:
In cubic close packing the second layer is placed over the spheres of the upper layer that are exactly above those of the primary layer. In other words, spheres of both the layers are perfectly aligned, horizontally and vertically. The three-dimensional close-packed structure is often generated by placing layers one over the opposite. The spheres of the second layer are placed in the empty space formed by the primary layer. Since the spheres of the two layers are aligned differently, if the primary layer is termed as ‘A; the second layer is often termed as ‘B’. The tetrahedral void is made wherever a sphere of the second layer is above the empty space formed by the primary layer whereas at other places, the triangular voids formed within the second layer are above the triangular voids within the first layer, such the triangular shapes of those don't overlap. Such voids are known octahedral voids and are surrounded by six spheres.
We know that in a cubic close packed arrangement each chloride ion would have one octahedral void and two tetrahedral voids related to it.
The number of octahedral voids with three chloride ions =3 = 3
The number of tetrahedral voids with three chloride ions =3×2 =6 = 3 \times 2{\text{ }} = 6
The total number of voids with three chloride ions =9 = 9
The number of octahedral voids occupied by Cr(III)=1Cr\left( {III} \right) = 1
The fraction of octahedral voids occupied 13\dfrac{1}{3}
The fraction of total number of voids occupied 19\dfrac{1}{9}

So, the correct answer is Option C.

Note: One can easily calculate the amount of those two sorts of voids. Let us take the amount of close packed spheres as N, then:
The number of octahedral voids generated =N = N
The number of tetrahedral voids generated =2N = 2N.