The tetrahedral voids present in 0.2 of compound forming hexagonal closed packing crystal structure are`Xxx10^23`.Then X is..........(given `N_A=6xx10^23`)

Total no. of voids in 0.5 mole of a compound forming hexagonal closed packed structure are :

How many octahedral voids are present in 1 mole of a compound having cubic close packed structure ?

The tetrahedral voids are present in 0.5 mole of hcp crystal structure.

Which of the following is not true about the voids formed in 3 dimensional hexagonal close packed structure?

A metal forms hexagonal close-packed structure? How many voids are present in 0.5 mol of it?

Only those atoms which four covalent bonds produce a repetitive three dimesional structure using only covalent bonds., e.g. , diamond structure.The latter is based on a FCC lattice where lattice points are occupied by carbon atoms.Every atom in this structure is surrounded tetrahedrally by four others.Germanium, silicon and grey tin also crystallize in the same way as diamond. (Given : N_A=6xx10^(23) sin 54^@44'=0.8164 ) If the edge length is 3.60Å , density of diamond crystal is

The number of F^- ions in 2.1 g AlF_3 is (N_A=6xx10^23)

A compound forms hexagonal close-packed structure. What is the total number of voids in 0.5 mol of it? How many of these are tetrahedral voids?

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

How many atoms are there in 5 moles of silver (N_(A)=6xx10^(23))