Advertisements
Advertisements
प्रश्न
The number of tetrahedral voids per unit cell in NaCl crystal is:
(i) 4
(ii) 8
(iii) twice the number of octahedral voids.
(iv) four times the number of octahedral voids.
उत्तर
(ii) 8
(iii) twice the number of octahedral voids.
Explanation:
NaCl has fee arrangement of CF ions. Thus,
Number of CF ions in packing per unit cell = 4
Number of tetrahedral voids = 2 × No. of particles present in close packing = 2 × 4 = 8
Number of tetrahedral voids = 2 × No. of octahedral voids
APPEARS IN
संबंधित प्रश्न
If the radius of the octachedral void is r and radius of the atoms in close packing is R, derive relation between r and R.
The unit cell of a substance has cations A+ at the corners of the unit cell and the anions B− at the center. The simplest formula of the substance is ____________.
A compound is formed by two elements A and B. The element B forms cubic close packing and atoms of A occupy `1/3"rd"` of tetrahedral voids. The formula of this compound is ____________.
The Ca2+ and F– are located in CaF2 crystal, respectively at face centered cubic lattice points and in ____________.
Which set of following characteristics for ZnS crystal is correct?
In which of the following crystals alternate tetrahedral voids are occupied?
In the hexagonal close-packed structure of a metallic lattice, the number of nearest neighbours of a metallic atom is ____________.
In the cubic close packing, the unit cell has ______.
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.
