The radius of an octahedral void is 'r' and the radius of an atom is 'R' when these are in close packing. Derive the relation between the two.

If the radius of a tetrahedral void is 'r' and radius of atom in close packing is 'R', derive the relation between r and R.

If radius of an octahedral void is r and atomic radius of atoms assuming cubical close pacting is R. Then the relation between r and R is

(a) If the radius of octahedral void is 'r' and the radius of the atoms in close packing is 'R' what is the relation between 'r' and 'R' ? (b) A metal crystallises in body centred cubic structure. If 'a' is the edge length of its unit cell, 'r' is the radius of the sphere, what is the relation between 'r' and 'a' ?

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

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

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

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

If the radius of the octahedral void is r and radius of the atom in closet packed structure is R then

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