Tiêu đề Chemistry ÔÇó Final Step-B ÔÇó SOL.pdf ✅

Vidyamandir Classes VMC | Final Step | Part-B 1 Class XII | Chemistry Solutions to Final Step - B | Chemistry The Solid State 1.(B) Crystalline solid’s physical properties depends on the directions of measurement, so they are not isotropic. 2.(B) Quartz glass (SiO2) is super cooled liquid, so it is amorphous. 3.(D) In anti-ferromagnetic substances magnets are aligned in opposite direction. 4.(A) Since glass is amorphous, its refractive index is same in all direction 5.(A) I2 is non-polar molecule, so in solid state I2 molecules are held by London dispersion forces. 6.(AC) Graphite and diamond both are network solid 7.(C) 8.(C) Lattice point is occupied by one atom, ion or molecule. 9.(B) 10.(B) For p-type semiconductor, that are formed by doping IIIA group element with IV group element in neutral form. 11.(B) In face centered unit cell, 8 tetrahedral void exist on the body diagonals. 12.(AB) Schottky as well as Frenkel defects are observed in AgBr crystal. 13.(B) hcp and ccp both are closest packing, so, they are most efficient. 14.(C) The packing efficiency of BCC is 68% , so empty space is 32%. 15.(ABC) 16.(AD) In NaCl and ZnS, both cation and anion has same coordination number 17.(C) For 2D packing coordination number is 4. 18.(D) Doping creates holes or extra electrons which are helpful for electrical conductivity. 19.(B) Electron rich impurity leaves extra electrons which creates n-type semi-conductor 20.(ACD) 21.(ABC) 22.(A) In ferromagnetic solids, all magnets are aligned in same direction (direction of external magnetic field) 23.(A) In Frenkel defect, specie slips into void, so, it is also called dislocation defect. 24.(D) In ccp, 8 tetrahedral voids are inside unit cell, they are not shared by other unit cell. 25.(A) In fcc, spheres are in contact with each other at surface diagonal. Thus edge length   4r a 2 2 r 2   In bcc, spheres are in contact with each other at body diagonal. Thu edge length 4r (a) 3  In simple cubic, spheres are in contact at edge thus a = 2r 26.(C) Cs Cl 2r 2r a 3      Cs Cl a 3 r r 2     27.(B) Let the fraction of 3 M  is x x 3 (0.98 x) 2 2       3x 1.96 2x 2     x 0.04  So % = 0.04 100 4.08% 0.98  
Vidyamandir Classes VMC | Final Step | Part-B 2 Class XII | Chemistry 28.(D) For bcc, 3 a 4r  3 351 r 151.98 152 pm 4     29.(D) For regular fcc with A at center and B at face centred point formula  AB3 losting one face point decreases efficiently 1/2of B. Thus formula  1 2 5 3 2 AB A B   30.(A) The mass of one unit cell 23 58.5 4 gm 6.02 10    Thus number of unit cells = 23 6.02 10 4 58.5   = 21 2.58 10 cells  31.(D) a r r 254 2      r 254 110 144 pm     32.(B) Packing efficiency of fcc and bcc are 74% and 68% respectively, so empty spaces are 26% and 32%. 33.(C) Number of atoms of Na(bcc type) = 2 and Number of atoms of Mg (fcc type) = 4 34.(A) Forumula: X Y X Y X Y 8 2 3   4 4 3 4 3   35.(D) The effective no of atoms in fcc are 4, so, volume 4 16 3 3 r 4 r 3 3      36.(B) Empirical formula = AB3 Since corner contributes 1/8th part of ion and surface center contributes 1/2 part of ion to the unit cell. 37.(B) 38.(C) Packing efficiency of bcc lattice  68% Hence, empty space  32% 39.(D) The distance between the body centered atom and one corner atom is 3a 2 i.e. half of the body diagonal. 40.(D) Diamond is like ZnS (Zinc blende). Carbon forming ccp (fcc) and also occupying half of tetrahedral voids. Total number of carbon atoms per unit cell (Comers) (Free (Tetrahedral centrered) void) 1 1 8 6 4 8 8 2       41.(A) Number of octahedral voids is same as number of atoms. 42.(D) Number of atoms in ccp 2 4 O    Number of tetrahedral     2 N 2 4 Number of 2 A  ions 1 8 2 4    Number of octahedral volts = Number of B  ions   N 4 Ratio, 2 2 O : A : B 4 : 2 2 : 1 : 2      Formula of oxide  AB O2 2 43.(C) Packing fraction for a cubic unit cell is given by 3 3 4 z r 3 f a    where a  edge length, r  radius of cation and anion. 44.(C) For Simple cubic : r a / 2  ; For Body centered : r a 3 / 4  r = radius of sphere
Vidyamandir Classes VMC | Final Step | Part-B 3 Class XII | Chemistry For Face centered : a r 2 2   Ratio of radii of the three will be a a 3 a : : 2 4 2 2 45.(B) F-centres are the sites where anions are missing and instead electron are present. They are responsible for colours. 46.(7) For ZnS structure, eff (Z of ZnS 4)  Number of B 4  Θ /Unit cell (corner + face centre) Number of A 4  Θ /Unit cell (in alternate TVs) Number of B Θ ion removed = 4 (Two from each face centre)  1 2 (per face centre share) = 2 Number of BrΘ ion left    4 2 2 /unit cell Number of 2 Z  ions entering in place of B 1  Θ [To maintain electrical neutrality, 2 2B 1 Z   Θ ] Formula A B Z 4 2 1   x y c 4 2 1 7       47.(2) Consider one face of unit cell as shown below: Number of atoms on one face 1 1 4(corners) (Per corner share) 1(face center) (face center share) 8 2     1 1 1/ per face 2 2    Given number of atoms on all faces 30  6 10 Given number of atoms on one face 1 30 30 6 10 10 atoms 6     Number of unit cells at one face of crystal 30 6 10 30 10 6    So, number of unit cells at the edge of crystal 30 15   10 10 Now, edge length of unit cell 4 50 nm 2   Edge length of cubical crystal 4 15 50 10 nm 2    So, area of face of crystal 2 15 4 nm 2 50 10          16 2 30 25 10 10 2     34 2  2 10 nm 18 34 2 2 10 m     16 2  2 10 m  16 2 16 2 A 10 m 2 10 m    A 2  48.(2) Void volume = 0.22/unit volume of unit cell 0.11A 0.22 A 2    49.(2) Figures A and D 50.(8) Diameter of 8 Cs 2 2.6 5.2Å 5.2 10 cm       Number of atoms in 2.50 cm row 8 2.50 5.2 10   8 7     0.48 10 4.8 10 Cs atoms
Vidyamandir Classes VMC | Final Step | Part-B 4 Class XII | Chemistry  Moles of Cs atoms 7 16 23 4.8 10 0.8 10 6 10        17 17 x 10 8 10 x 8        51.(4) a. Number of X atoms 1 8 1/ unt cell 8    Number of Y atoms  1/ unit cell Number of O atoms 1 12 3 / unt cell 4    Formula is: XYO X Y O 3 a b c  b. Number of O atoms missing from two edge centers per unit cell 1 1 2 / unt cell 4 2    Number of O atoms left 1 3 2.5 / unit cell 2    Formula is XYO X Y O X Y O 2.5 2 2 5 x y z    The value of (x y z) (a b c) (2 2 5) (1 1 3) 4             52.(7) Let radius of hollow sphere B r   Edge length (a) 4r / 3  Volume of unit cell 3 2   a (4r / 3) Volume of B unoccupied by A (having radius  r / 2 ) in unit cell 3 4 4 r 3 2 r 3 3 2                    Volume of B unoccupied by A  2 4 7r 2 in unit cell 7 3 3 8 volume of unit cell 64 4r 3             7 3 7 3 A 64 64      A 7  53.(6) Statements (a), (b), and (c) are correct. So total score     1 2 3 6 Statement (a, b) are factual. Statement (c): r 0.35 0.368 r 0.95    The radius ratio lies in the range of 0.225 0.414  , which corresponds to TV and the CN of TV = 4 54.(13) Single unit cell consists of three cubes Number of 2 Ba  ions = 2 per unit cell Number of 4 Y  ions = 1 per unit cell Number of 2 Cu  ions = 8 in each cube at corners  8 3 (in three cubes) 1 8  (per corner share) = 3/unit cell Number of 2 o  ions = 10 (at edge center of cube 1) +8 (at edge center of cube 2) +10 (at edge center of cube 3)