Advertisements
Advertisements
प्रश्न
What is the formula of a compound in which the element Y forms hcp lattice and atoms of X occupy 2/3rd of tetrahedral voids?
उत्तर
Suppose the number of atoms Y in hcp lattice = n
As the number of tetrahedral voids is double the number of atoms in close packing, the number of tetrahedral voids = 2n
As atoms X occupy 2/3rd of the tetrahedral voids, the number of atoms X in the lattice
`2/3xx2n=(4n)/3`
`:. "Ratio of X:Y"=(4n)/3:n=4/3:1=4:3`
Hence, the formula of the compound is X4Y3.
संबंधित प्रश्न
Total number of voids in 0.5 mole of a compound which forms hexagonal close packed structure is ____________.
What is the coordination number of sodium in Na2O?
Which set of following characteristics for ZnS crystal is correct?
In which of the following crystals alternate tetrahedral voids are occupied?
If Germanium crystallises in the same way as diamond, then which of the following statement is not correct?
In the cubic close packing, the unit cell has ______.
In which of the following arrangements octahedral voids are formed?
(i) hcp
(ii) bcc
(iii) simple cubic
(iv) fcc
In a compound, nitrogen atoms (N) make cubic close packed lattice and metal atoms (M) occupy one-third of the tetrahedral voids present. Determine the formula of the compound formed by M and N?
Show that in a cubic close packed structure, eight tetrahedral voids are present per unit cell.
A solid compound XY has Nacl structure. If the radium of cation (X+) is 100 pm, the radium of anion (r–) will be:-