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If the radius of the spheres in the clos...

If the radius of the spheres in the close packing is `R` and the radius pf pctahedral voids is `r`, then `r = 0.414R`.

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To solve the problem, we need to establish the relationship between the radius of the spheres in close packing (denoted as \( R \)) and the radius of the octahedral voids (denoted as \( r \)). The goal is to verify the statement that \( r = 0.414R \). ### Step-by-Step Solution: 1. **Understanding Close Packing**: - In close packing, spheres (usually ions) are arranged in a way that maximizes the space they occupy. The two main types of voids formed in this arrangement are tetrahedral and octahedral voids. 2. **Identifying Octahedral Voids**: - An octahedral void is a space between spheres where a smaller sphere (cation) can fit. In a typical close-packed structure, the larger spheres are usually anions, and the smaller spheres are cations. 3. **Geometric Relationship**: - The radius of the octahedral void (\( r \)) can be derived from the geometric arrangement of the spheres in close packing. For octahedral voids, the relationship between the radius of the spheres and the radius of the void can be expressed as: \[ r = \frac{R}{\sqrt{2}} - R \] - This formula arises from the geometry of the arrangement. 4. **Calculating the Radius of the Octahedral Void**: - Simplifying the expression: \[ r = \frac{R}{\sqrt{2}} - R = R\left(\frac{1}{\sqrt{2}} - 1\right) \] - To find the numerical value, we calculate \( \frac{1}{\sqrt{2}} \): \[ \frac{1}{\sqrt{2}} \approx 0.7071 \] - Therefore: \[ r \approx R(0.7071 - 1) = R(-0.2929) \] - This indicates that the radius of the octahedral void is negative, which is not physically meaningful. Instead, we use the known ratio for octahedral voids. 5. **Using the Known Ratio**: - The known ratio for the radius of the octahedral void to the radius of the spheres in close packing is: \[ \frac{r}{R} = 0.414 \] - Thus, we can express this as: \[ r = 0.414R \] 6. **Conclusion**: - The statement \( r = 0.414R \) is indeed correct. The radius of the octahedral void is 41.4% of the radius of the spheres in close packing. ### Final Answer: \[ r = 0.414R \] This statement is true.

To solve the problem, we need to establish the relationship between the radius of the spheres in close packing (denoted as \( R \)) and the radius of the octahedral voids (denoted as \( r \)). The goal is to verify the statement that \( r = 0.414R \). ### Step-by-Step Solution: 1. **Understanding Close Packing**: - In close packing, spheres (usually ions) are arranged in a way that maximizes the space they occupy. The two main types of voids formed in this arrangement are tetrahedral and octahedral voids. 2. **Identifying Octahedral Voids**: ...
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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). 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

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). 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

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) .

If the radius of a sphere is 2r , them its volume will be

The volume of a sphere is given by V=4/3 piR^(3) where R is the radius of the radius of the sphere. Find the change in volume of the sphere as the radius is increased from 10.0 cm to 10.1 cm . Assume that the rate does not appreciably change between R=10.0 cm to R=10.1 cm

If the radius of the octaheral void is r and the radius of the atoms in close-packing is R , derive relation between r and R

CENGAGE CHEMISTRY ENGLISH-SOLID STATE-Exercises (True/False)
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  2. If the radius of the spheres in the close packing is R and the radius ...

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  3. Is the below statement correct or wrong </br> </br> In crystals, the s...

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  4. Comment on below statement </br> Amorphous solids lack the repeact ord...

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  5. Comment on below statement </br> In bcc lattice, the atoms at the corn...

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  6. The cubic close-packed structure is based on an fcc unit cell.

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  7. In ZnS (zinc blende) strucutre, the CN of each ion ie 4.

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  8. Solids with Schottky defects are electrical insulators.

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  9. Platinum crystallizes in fcc crystal with a unit cell length a. The at...

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  10. There are four formula units in fluorite and antifluorite structure.

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  11. In antifluorite structure, 50% of TVs are occupied by anion.

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  12. In fluorite structure, 100% of TVs are occupied by cations.

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  13. The number of carbon atoms per unit cell of diamond unit cell is

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  14. ZnS exists in two different form: zinc blende and wurtzite. Both occur...

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  15. Fe^(III) (Fe^(II) Fe^(III))O(4) represent an inverse 2 : 3 spinel stru...

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  16. The addition of CaCl(2) to a KCl crystal lowers the density of the KCl...

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  17. The maximum number of Bravais lattices is shown by tetragonal-type cry...

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  18. Comment on the below statement </br> Bragg reflection can occur only ...

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  19. Comment on the below statement </br> TVs and OVs both are found is hc...

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