Answer: (c)

Answer:(a)

Answer: (a)

Number of Y atoms per unit cell in ccp lattice (N) = 4 Tetrahedral voids = 32N = 2 × 4 = 8 number of tetrahedral voids occupied by X = (2/3)2/3 of the tetrahedral void = (2/3) × 8 = 16/3 Hence the formula of the compound will be X4Y3 Answer: (b)

the hexagonal called hcp ,The hcp packing has an ABABAB... two-layer sequence, where A and B represent the location of the layers. In other word, the third layer is exactly above the first layer in hcp Answer : (d)

Na+ and Cl- ions both have six coordination number Answer: (c)

Amorphous soids are isotropic Answer:(b)

NaCl has face centred cubic close packinh in which lattice points are occupied by Cl- ions where as Na+ ions occupy all octahedral holes. Here Na+ and Cl- both have a coordination number of 6 (i) Effective Number of A- = [8 × (1/8)] + (6 × (1/2)] = 4 the face-centred atoms along one of the axis are removed Thus effective number of A- from edge and body centre = [ 8 × 91/8) ] + [ 4 × (1/2)] = 3 (ii) Effective number of B- = [ 12 × (1/4) ] + 1 = 4 ∴ formula is A3B4 Answer: (d)

In Frenkel defect, ions get only displaced from their original position to interstitial site Answer: (c)

For B.C.C. structure number atom per unit cell = Z=2 Atomic weight M = 56 Volume V = a3= (2.961×10-8) 3 Avogadro's number N = 6.0323×10-23 Now formula for density d is- Answer: (a)

Frenkel defects occur in compounds having low coordination number and large difference in size of cations and anions Answer:(a)

Solid O2 is molecular solids involving weak van der Waal's force Answer : (b)

In ZnS, Zn is present in tetrahedral voids, thus its coordination nummber is 4 Answer: (a)

Atom 'A' are at corners Number of atoms of A = 8 × (1/8) = 1 Atoms B occupy face-centres Number of atoms of B = 6 × (1/2) = 3 Simplest formula of solute = AB3 Answer : (b)

Answer : (d)

Answer : (c)

Answer : (a)

ABCABCABC... arrangement gives fcc Number of atoms in packing = Z Number of tetrahedral voids = 2Z Number of octahedral voids = Z Answer : (b)

Answer : (d)

CsCl has bcc structure; Cs+ at body-centre is surrounded by eight Cl- ions at centre Answer : (c)

In hcp arrangement ( ABCABCABC ..) the coordination number is 12 Answer : (d)

Ca2+ ions are present in ccp arrangement and F- ions occupy all tetrahedral voids Answer : (a)

In Schottky defect, equal number of cation and anion vacancies are found; the density of solid decreases Answer : (a)

Number of O2- ions in packing = n Number of tetrahedral voids = 2n Number of octahedral voids = n Number A2+ ions = (1/4) ×(2n) = n/2 Number B+ = n A2+:B2+:O2- = (n/2):n:n = 1:2:2 Simplest formula of solid = AB2O2 Answer : (d)

Answer : (b)

Sodium chloride has a face centred cubic (fcc) structure. Na+ ions occupy the corners and face centres of a cube.In the rock salt structure Na+ and Cl– ions are alternatively arranged such that Cl– ion is surrounded by six Na+ ions and similarly Na+ ion is surrounded by six chloride ions. In other words the ions are capable of achieving a co-ordination number of 6. Now it is given 14 chloride ions. If we draw the structure it will according to image below Here Na+ is present at octahedral voids and Cl- is present at FCC The Cl– ions are present at the body and edge centres.One unit cell of sodium chloride consists of 14 Na+ ions and 13 Cl- ions or vice versa. Unit cells demonstrate alignment and relative position of atoms within a crystal. It do not provide stoichiometric information. The unit cell model is not necessarily balance atoms or charges Answer : (a)

A superconductor is a material that can conduct electricity or transport electrons from one atom to another with no resistance. In a superconductor, the resistance drops abruptly to zero when the material is cooled below its critical temperature. Answer : (c)

For face-centred unit cell Answer : (c)

given that KCl crystallizes in the same type of lattice as does NaCl Sodium chloride (NaCl) is an ionic compound of AB type. Two ions (cation and anions) are in ratio of 1:1. Thus side consists of one positive and one negative ion or side = Y+ + r- Now Dividing equation (2) by (1) we get Answer : (a)

For sodium chloride, Na+Cl-, there are 6 anions around each cation (coordination number Na = 6); because of the 1:1 stoichiometry there must also be 6 Na cations around each Cl anion. Answer : (c)