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In HCP or CCP constituent particles occu...

In HCP or CCP constituent particles occupy 74% of the available space. The remaining space (26%) in between the spheres remains unoccupied and is called interstitial voids or holes. Considering the close packing arrangement, each sphere in the second layer rests on the hollow space of the first layer, touching each other. The void created is called tetrahedral void. If R is the radius of the spheres in the close packed arrangement then,
R (radius of tetrahedral void) = 0.225 R
In a close packing arrangement, the interstitial void formed by the combination of two triangular voids of the first and second layer is called octahedral coid. Thus, double triangular void is surrounded by six spheres. The centre of these spheres on joining, forms octahedron. If R is the radius of the sphere. in a close packed arrangement then,
R (radius of octahedral void = 0.414 R).
If the anions (A) form hexagonal close packing and cations (C ) occupy only 2/3rd octahedral voids in it, then the general formula of the compound is

A

AB

B

`A_(3)B_(2)`

C

`A_(2)B`

D

`AB_(3)`

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To solve the problem, we need to determine the general formula of the compound formed when anions (A) occupy a hexagonal close packing (HCP) structure and cations (C) occupy only 2/3 of the octahedral voids in that structure. Let's break down the solution step by step. ### Step-by-Step Solution: 1. **Understanding the HCP Structure**: - In a hexagonal close packing (HCP) arrangement, the coordination number is 12, and the number of spheres (or particles) per unit cell is 6. This means that the effective number of anions (A) in the unit cell is 6. 2. **Identifying the Octahedral Voids**: - In HCP, the number of octahedral voids is equal to the number of atoms in the unit cell. Therefore, there are also 6 octahedral voids in the HCP structure. 3. **Calculating the Number of Cations (C)**: - According to the problem, cations (C) occupy only 2/3 of the octahedral voids. - Since there are 6 octahedral voids, the number of cations (C) that will occupy these voids is: \[ \text{Number of C} = \frac{2}{3} \times 6 = 4 \] 4. **Writing the General Formula**: - Now that we know the number of anions (A) and cations (C) in the unit cell, we can write the general formula. - The number of A is 6 and the number of C is 4, so the formula can be represented as: \[ \text{General formula} = A_6C_4 \] 5. **Simplifying the Formula**: - To simplify the formula \( A_6C_4 \), we can divide both subscripts by 2: \[ A_6C_4 \rightarrow A_3C_2 \] ### Final Answer: The general formula of the compound is \( A_3C_2 \).
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In HCP or CCP constituent particles occupy 74% of the available space. The remaining space (26%) in between the spheres remains unoccupied and is called interstitial voids or holes. Considering the close packing arrangement, each sphere in the second layer rests on the hollow space of the first layer, touching each other. The void created is called tetrahedral void. If R is the radius of the spheres in the close packed arrangement then, R (radius of tetrahedral void) = 0.225 R In a close packing arrangement, the interstitial void formed by the combination of two triangular voids of the first and second layer is called octahedral coid. Thus, double triangular void is surrounded by six spheres. The centre of these spheres on joining, forms octahedron. If R is the radius of the sphere. in a close packed arrangement then, R (radius of octahedral void = 0.414 R). In Schottky defect

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In HCP or CCP constituent particles occupy 74% of the available space. The remaining space (26%) in between the spheres remains unoccupied and is called interstitial voids or holes. Considering the close packing arrangement, each sphere in the second layer rests on the hollow space of the first layer, touching each other. The void created is called tetrahedral void. If R is the radius of the spheres in the close packed arrangement then, R (radius of tetrahedral void) = 0.225 R In a close packing arrangement, the interstitial void formed by the combination of two triangular voids of the first and second layer is called octahedral coid. Thus, double triangular void is surrounded by six spheres. The centre of these spheres on joining, forms octahedron. If R is the radius of the sphere. in a close packed arrangement then, R (radius of octahedral void = 0.414 R). In the figure given below, the site marked as S is a

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