a. 2, 4 b. 2, 2 c. 4, 2 d. 4, 4 Ans. c (3)
a. 4 x 10 25 b. 3x 1025 c. 2x 1025 d. 1x 1025 Ans. a (6)
a. nλ = 2 θ sin θ b. nλ = 2d sin θ c. 2nλ = dsin θ d. n (θ /2) = d/2 θ Ans. b (12)
6.023 x 1023 mol -1 CsBr. the density of CsBr is a. 8.25 g/ cm3 b. 4.25 g/cm3 c. 42.5g /cm3 d. 0.425g / cm3 Ans. b (41)
a. 20/3 (πr 3 ) b. 24/ 3 (πr 3 ) c. 12/3 (πr 3 ) d. 16/3 (πr 3 ) Ans. d (11)
a. π/6 b. π 3√2 c. π /4√2 d. π/ 4 Ans. a (5)
a. 1/8 and 1/2 b. 1/2 and 1/4 c. 1/4 and 1/2 d. 1/4 and 1/8 Ans. a (10)
a. 65% b. 45% c. 90% d. 75% Ans. c (8)
Na+ and Cl – ions remaining in the unit cell are a. 4 and 4 b. 3 and 3 c. 1 and 1 d. 4 and 3 Ans. b (16)
a. 39.27 % b. 68.02% c. 74.05 % d. 78. 54 % Ans. d (5)
a. C 4 A 3 b. C2 A3 c. C 3 A 2 d. C 3 A 4 Ans. d (5)
a. 0.047 a b. 0.027 a c. 0.134 a d. 0.067 a Ans. d (11)
900º C It transforms to fcc structure. the ratio of density of iron at room temperature to that at 900 º C (assuming molar mass and atomic radii of iron remains constant with temperature ) is a. √3 /√2 b. …
a. 6. 022 x 1023 b. 3. 011 x 1023 c. 9. 033 x 1023 d. 4. 516 x 1023 Ans. c (5)
a. The number of carbon atoms in an unit cell of diamond is 4 b. The number of bravais latties in which a crystal can be categorized is 14 c. The fraction of the total volume occupied by the atoms …
Na+ ion is 0.95 Å and that of Cl– ion is 1.81 Å hence in the close packed lattice of Cl – ions Na+ ions prefer to occupy a. Tetrahedral site b. Octahedral site c. Cubic site d. Trigonal site …
a. 285 pm b. 398 pm c. 144 pm d. 618 pm Ans. c (10)
a. 4/3 π r 3 b. 8/3 π r 3 c. 16/3 π r 3 d. 64π r 3 / 3√3 Ans. c (8)
a. 1/2 b. 1/4 c. 1 d. 1/8 Ans. b (31)
a. r cs + + r cl- = 3a b. r cs + + r cl- = 3a/ 2 c. r cs + + r cl- = √3/2 a d. r cs + + r cl- …
a. 75 pm b. 300 pm c. 240 pm d. 152 Ans. d (4)
a. 288 pm b. 408 pm c. 144 pm d. 204 pm Ans. a (15)
a. 300 pm b. 335 pm c. 250 pm d. 204 pm Ans. b (10)
a. 108 pm b. 127 pm c. 157 pm d. 181 pm Ans. b (9)
a. 1. 86 Å b. 3.22 Å c. 5.72 Å d. 0.93 Å Ans. a (3)
Cs + and Br – ions is a. 1. 86 Å b. 3.72 Å c. 4. 3 Å d. 7. 44 Å Ans. b (7)
a. 1 b. 3 c. 2 d. 4 Ans. a Explanation : Number of octahedral voids in ccp, is equal to effective number of atoms. In ccp, effective number of atoms are 4 so, 4 octahedra voids. So, 1 octahedral …
a. 1 : 1 b. 1 : 2 c. 1 : 3 d. 2 : 1 Ans. b (8)
a. ABC ABA b. ABC ABC c. ABABA d. ABBAB Ans. b Explanation : As, a = b not = c and alpha = beta not = gamma (9)
a. Z b. 2 Z c. Z/2 d. Z/4 Ans. b Explanation : Number of tetrahedral voids in the unit cell = 2 x number of atoms = 2 Z (9)
a. Octahedral close packing b. Hexagonal close packing c. Tetragonal close packing d. Cubic close packing Ans. d Explanation : It represents ccp arrangement (8)
a. 26 % b. 48 % c. 23 % d. 32 % Ans. d Explanation : Packing efficiency in bcc lattice = 68 % vacant space in bcc lattice = 100- 68 = 32 % (7)
Cs + is a. Equal to that of Cl – , that is 6 b. Equal to that of Cl – , that is 8 c. Not Equal to that of Cl – , that is 6 d. Not Equal to that of Cl – …
a. Cl– ions are in fccarrangement b. Na+ ions has coordination number 4 c. Cl– ions has coordination numnber 6 d. Each unit cell contains 4NaCl molecules Ans. b (17)