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
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
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
A
cations are missing from the lattice sites and occupy the interstitial sites
B
equal number of cations and anions are missing
C
anions are missing and electrons are present in their place
D
equal number of extra cations and electrons are present in the interstitial sites
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To solve the question regarding Schottky defect, we will analyze the information provided step by step.
### Step-by-Step Solution:
1. **Understanding Schottky Defect**:
- A Schottky defect is a type of stoichiometric defect in ionic crystals where equal numbers of cations and anions are missing from their lattice sites. This results in vacancies in the crystal structure.
2. **Characteristics of Schottky Defect**:
- It occurs in ionic compounds where the cations and anions are of similar size.
- The absence of these ions leads to a decrease in the density of the crystal since there are fewer particles in the lattice.
3. **Analyzing the Options**:
- **Option 1**: Cations are missing from the lattice site and occupy the interstitial site.
- This is incorrect because in a Schottky defect, both cations and anions are missing, not just cations.
- **Option 2**: Equal number of cations and anions are missing.
- This is correct as it aligns with the definition of Schottky defect.
- **Option 3**: Anions are missing and electrons are present in their place.
- This is incorrect because the defect does not involve electrons replacing anions.
- **Option 4**: Equal number of extra cations and electrons are present in the interstitial site.
- This is also incorrect as it does not describe the Schottky defect.
4. **Conclusion**:
- The correct answer is **Option 2**: Equal number of cations and anions are missing.
### Final Answer:
The correct option regarding the Schottky defect is **Option 2**: Equal number of cations and anions are missing.
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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). Mark the false statement :
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 spinel structure, oxide ions are cubic close packed whereas 1/8th of tetrahedral voids are occupied by A^(2+) cations and 1/2 of octahedral voids are occupied by B^(3+) cations. The general formula of the compound having spinel structure is
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
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
If the radius of the spheres in the close packing is R and the radius pf pctahedral voids is r , then r = 0.414R .
If R is the radius of the octahedral voids and r is the radius of the atom in close packing, then r//R is equal to
In a close packed array of N spheres, the number of tetrahedral holes are
Correct the following statement by changing the underlined part of the sentence. If R is the radius of the spheres forming closest packing arrangement, then radius r of the octahedral void will be ul("0.225 R") .
Correct the following statement by changing the underlined part of the sentence. If R is the radius of spheres forming closest packing arrangement, then radius r of the tetrahedral void will be ul(0.732R) .
How many octahedral & tetrahedral voids are per sphere in a unit cell ?
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