Home
Class 12
CHEMISTRY
A compound forms hexagonal close-packed...

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?

Text Solution

Verified by Experts

No. of atoms in 0.5 mole of the compound `=0.5xxN_(0)=0.5xx6.022xx10^(23)=3.011xx10^(23)`
No. of octahedral voids = No. of atoms `=3.011xx10^(23)`
No. of tetrahedral voids `=2xx3.011xx10^(23)=6.022xx10^(23)`
Total no. of voids = `(3.011+6.022)xx10^(23)=9.033xx10^(23)`
Promotional Banner

Similar Questions

Explore conceptually related problems

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

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

In a crystal of an ionic compound, the ions B form the close packed lattice and the ions A occupy all the tetrahedral voids. The formula of the compound is

A compound is formed by cation C and anion A. The anions form hexagonal close packed (hcp) lattice and the cations occupy 75% of octahedral voids. The formula of the compound is

Show that in a cubic close packed structure eight tetrahedral voids are present per unit cell .

Show that in a cubic close packed structure, eight tetrahedral voids are present per unit cell.

The total number of tetrahedral voids in the face centred unit cell is

Element 'B' forms ccp structure and 'A' occupies half of the octahedral voids, while oxygen atoms occupy all the tetrahedral voids. The structure of bimetallic oxide is:

Total number of T-voids and O-voids in HCP and CCP structures are different . (True/False)

In chromium (III) chloride CrCl_(3) chloride ions have cubic close packed arrangement and Cr (III) ions present in the octahedral voids. What fraction of the octahedral void is occupied ? What fraction of the total number of voids is occupied?