
Percentage of free space in cubic close packed structure and in body centered packed structure are respectively ------------.
A \[{\text{30% }}\] and ${\text{26% }}$
B ${\text{26% }}$ and ${\text{32% }}$
C ${\text{48% }}$ and ${\text{48% }}$
D ${\text{32% }}$ and ${\text{26% }}$
Answer
570.6k+ views
Hint: In this question, we will use the concept of packing fraction. By knowing the number of atoms in a unit cell and their volume we can calculate the ratio of volume spherical atoms to that of volume of unit cell which leads us to the free space in the given arrangements.
Complete step by step answer:
As we learnt in this chapter that the packing percentage in a cubic close packed structure is equal to ${\text{74% }}$. So the percentage of free space in a cubic close packed structure will be equal to \[{\text{100% - 74 % }}\] further which is equal to ${\text{26% }}$.
But the packing percentage in a body centered structure is ${\text{68% }}$. So here the percentage of free space body centered packed structure will be equal to \[{\text{100 % - 68 % }}\] = $32\% $.
Therefore, the free spaces are ${\text{26% }}$ and ${\text{32% }}$ respectively.
And we must note that point that, in a unit cell, coordination number is that number in which the number of atoms it is touching. Moreover, the (hcp) which is a hexagonal closed packet has a coordination number of $12$ and it contains $6$ atoms per unit cell. Talking about the (bcp) body centered cubic which has a coordination number of $8$ and contains $2$ atoms per unit cell.
So, the correct answer is Option B .
Additional Information:
Let us consider three layers, so here the gaps in between the first layer are covered by the second layer. But the third layer is also offset relative to the interspaces gaps of the first layer and the third layer does not show any alignment with the first layer. And now we can say that this configuration is referred to as “ABC “and this is known as the cubic closest packing (CCP).
Note:
The CCP arrangement has a total of $4$ spheres per unit cell and HCP has $8$ spheres per unit cell. So however, both the configurations have a coordination number of $12$.
Complete step by step answer:
As we learnt in this chapter that the packing percentage in a cubic close packed structure is equal to ${\text{74% }}$. So the percentage of free space in a cubic close packed structure will be equal to \[{\text{100% - 74 % }}\] further which is equal to ${\text{26% }}$.
But the packing percentage in a body centered structure is ${\text{68% }}$. So here the percentage of free space body centered packed structure will be equal to \[{\text{100 % - 68 % }}\] = $32\% $.
Therefore, the free spaces are ${\text{26% }}$ and ${\text{32% }}$ respectively.
And we must note that point that, in a unit cell, coordination number is that number in which the number of atoms it is touching. Moreover, the (hcp) which is a hexagonal closed packet has a coordination number of $12$ and it contains $6$ atoms per unit cell. Talking about the (bcp) body centered cubic which has a coordination number of $8$ and contains $2$ atoms per unit cell.
So, the correct answer is Option B .
Additional Information:
Let us consider three layers, so here the gaps in between the first layer are covered by the second layer. But the third layer is also offset relative to the interspaces gaps of the first layer and the third layer does not show any alignment with the first layer. And now we can say that this configuration is referred to as “ABC “and this is known as the cubic closest packing (CCP).
Note:
The CCP arrangement has a total of $4$ spheres per unit cell and HCP has $8$ spheres per unit cell. So however, both the configurations have a coordination number of $12$.
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