test venue: lajpat bhawan, madhya marg, sector …...a face centered atom is neighbour of corner...

Dr. Sangeeta Khanna Ph.D 1 CHEMISTRY COACHING CIRCLE D:\Important Data\2018\+2\Physical\Test\Grand Test - 1\Obj- Grand Test -1 (6.5.2018).doc

Test Venue:

Lajpat Bhawan, Madhya Marg, Sector 15-B, Chandigarh

Dr. Sangeeta Khanna Ph.D


READ INSTRUCTIONS CAREFULLY

INSTRUCTIONS 1. The test is of 2 hours duration.

2. The maximum marks are 270.

3. This test consists of 64 questions.

SECTION – A (Single Correct Choice Type) Negative Marking [-1]

This Section contains 36 multiple choice questions. Each question has four choices A), B), C) and D) out of which ONLY ONE is correct. Marks: 36 × 4 = 144

1. Which of the statement is true regarding electric properties of solids?

a. Conductivity)metals << Conductivity)insulator < Conductivity)semiconductor b. Depending upon temperature copper can behave as insulator or conductor c. I2(s) cannot conduct electricity d. n-type semiconductor will have conductivity less than pure semiconductor (intrinsic) C

2. A piece of copper and another of Ge are cooled from room temperature to 80 K. The resistance of:

a. each of them increases b. Cu increases and that of Ge decreases c. Cu decreases and that of Ge increases d. each of them decreases C

Sol. Copper is a metal its conductance decreases with temperature and resistance increases but Ge is semiconductor and its conductance increases with temperature and resistance decreases.

3. A ferromagnetic substance becomes a permanent magnet when it is placed in a magnetic field because :

a. domains get randomly oriented. b. all the domains get aligned opposite to the direction of magnetic field. c. all the domains get aligned in the direction of magnetic field d. some of the domains get aligned either in same direction or in opposite direction. C

4. Which of the following unit cells will have at least two lengths same and at least two angles same?

a. Monoclinic b. Rhombohedral c. Orthorhombic d. Triclinic B

5. Figure shows a cube of unit cell of CCP arrangement with face centered atoms marked 1, 2, 3. Which of the following is true?

a. Atom 3 is twice as far from 1 as from 2 b. Atom 2 is equidistant from atoms 1 and 3 c. Atom 2 is nearer to 1 than to 3 d. All atoms lie on a right angled triangle B

Sol. A face centered atom is neighbour of corner atom & Adjacent face centred atom. 6. You are given 6 identical balls. What is the maximum number of square voids and triangular voids (in

separate arrangements) that can be created?

a. 2, 4 b. 4, 2 c. 4, 3 d. 3, 4

3

2

1



A Sol.

7. The spinel structure (AB2O4) consists of an fcc array of O2– ions in which the :

a. A cation occupies one-eighth of the tetrahedral holes and B cation occupies one-half of

octahedral holes

b. A cation occupies one-fourth of the tetrehdral holes and the B cations the octahedral holes

c. A cation occupies one-eighth of the octahedral hole and the B cations the tetrahedral holes

d. A cation occupies one-fourth of the octahedral holes and B cations the tetrahedral holes

A

8. In a crystalline solid, anions C are arranged in cubic close packing. Cation A occupies 50 % of the

tetrahedral voids and cation B occupies 50% of octahedral voids. What is the formula of solid ?

a. ABC2 b. A2BC2 c. A2BC3 d. A2B2C3

B

Sol. Let no. of anions, C = 100. We know that :

In a close packed structure of N spheres, there are 2N tetrahedral sites and N octahedral sites. So,

No. of octahedral voids = 100; No. of tetrahedral voids = 200

But 50% of both the voids are occupied. Hence, Tetrahedral voids contain A cations = 100

Octahedral voids contain B cations = 50. Hence,

A : B : C is

100 : 50 : 100 i.e., 50

100:

50

50:

50

100 i.e.,2:1:2

Formula of the compound = A2B1C2 or A2BC2

9. A compound of formula PQ2R, have lattice made up of R atom. P goes in tetrahedral void & Q is

equally distributed in octahedral & tetrahedral void. What percentage of tetrahedral void is occupied.

a. 50% b. 25% c. 100% d. 75% C

Sol. Out of two tetrahedral void, one is occupied by P & other by Q. 10. Calculate the radius of sodium atom if sodium metal crystallises in b.c.c. lattice with the cell edge,

a = 4.29 Å. a. 4.29 Å b. 1.86 Å c. 2.83 Å d. 1.30 Å B

Sol. The body diagonal (Fig.2) in b.c.c. lattice = 4r and 4r = .3 a

and r = radius of atom, a = edge length.

r = Å86.14

Å29.4732.1

4

a.3

Ans.

11. Calculate the radius ratio, r+/r- and the co-ordination number of Li+ and F- ions in LiF crystal structure

from the following data. Lir = 60 pm; F

r = 136 pm.

a. 0.225 ; Co. No. = 4 b. 0.441 ; Co. No. = 6 c. 0.731 ; Co. No. = 8 d. 0.531 ; Co. No. = 6 B

Square void (2)

Triangular void (4)

4r



Sol. Radius of Li+ = 60 pm Radius of F- = 136 pm

Hence, Radius ratio,

F

LI

r

r=

136

60= 0.441

This value (0.441) falls within the range, 0.414 and 0.732. Hence, the coordination number of Li+ & F- = 6 each.

12. A certain oxide of metal M crystallizes in such a way that O2– ions adopts HCP arrangement following ABAB ….. pattern. The metal ions however occupy two third of the octahedral voids. The formula of the compound is

a. M2O3 b. M3O c. M8/3O3 d. MO2

A Sol. The number of oxide ions in hexagonal unit cell = 6 Octahedral voids = 6

Octahedral voids occupied by M = 6 × 3

2 = 4

The formula is M4O6 or M2O3

13. In FCC unit cell if on one of the face, unoccupied space between atom is 10 pm. What will be the edge length.

a. 33.3 pm b. 66.6 pm c. 28.4 pm d. 333 pm A

Sol. r22

a2

r = 22

a

a = 2r + 10 Substitute value of r

a = 1022

a2

a = 0.70a + 10 0.3a = 10

a = 3.0

10 =

3

100 = 33.3 pm

14. Greater the value of r+/r–

a. the lower will be the coordination number b. the higher the value of coordination number c. the higher will be the number of cations d. the lower will be the number of anions B

15. In a simple cubic crystal, each atom is shared by

a. 2 unit cells b. 4 unit cells c. 6 unit cells d. 8 unit cells D

16. In a cubic type unit cell, atoms of X are at the corners as well as at the centre of a cube. Atoms of Y are at one half faces of the cube. Drive the formula of the compound.

a. X4Y3 b. X2Y3 c. X2Y5 d. X3Y4 A

Sol. No. of atoms of X = 8 corners 8

1 atom per unit cell = 1

No. of atoms of X at centre = 1. Hence, total no. of atoms of X = 1+1=2

No. of atoms of Y = 2

1(6 faces)

2

1 atom per unit cell =

2

3

Formula of compound = X2

2

Y3 or X4 Y3

10 cm = a – 2r

4r = 2 a



17. If three elements P, Q and R crystallize in a cubic lattice with P atoms at the corners, Q atoms at the cube centre and R atoms at the centre of edges of the cube, then write the formula of the compound.

a. P2Q2R3 b. PQR c. PQR2 d. PQR3 D

Sol. No. of P atoms per unit cell = 8 corners 8

1 atom per unit cell = 1

No. of Q atoms per unit cell = 1

No. of R atoms per unit cell = 12 × 4

1 = 3

Formula of compound = PQR3 18. In face centred cubic arrangement of A and B atoms, whose A atoms are at the corners of the unit

cell and B atoms at the face centres and A atoms are missing from two corners in each unit cell. What is the formula of the compound.

a. AB4 b. AB3 c. A2B d. AB A

Sol. No. of A atoms per unit cell

= 6(8 – 2 = 6) corners 8

1 atom per unit cell =

4

3

8

6

No. of B atoms per unit cell = 6 faces 2

1atom per unit cell = 3

Hence, the formula of the compound = A3/4B3 or A3B12 i.e., AB4 Ans. 19. A non stoichiometric compound Cu1.8S is formed due to incorporation of Cu2+ ions in the lattice of

cuprous sulphide. What percentage of Cu2+ ion in the total copper content is present in the compound?

a. 88.88 b. 11.11 c. 99.8 d. 89.8 B

Sol. Let x Cu2+ and (1.8 – x)Cu+ ions are present in the compound Cu1.8S. Compound is electrically neutral.

+2x + (1.8 – x) = 2 2x + 1.8 – x = 2

x = 0.2

% Cu2+ = 8.1

2.0 × 100 = 11.11

20. Radii of Cs+ and Cl– ions are 1.69 Å and 1.81 Å; what is the edge length of CsCl unit cell?

a. 4.50 Å b. 4.04 Å c. 3.50 Å d. 3.80 Å B

Sol. ClCs

rr2

3a

= 1.69 + 1.81 a = 4.04 Å 21. Dipole-dipole interaction energy between stationary polar molecules (as in solids) is proportional to

....A... and between the rotating polar molecules is proportional to .....B.... . Here, A and B refer to

a. 36 r

1B ,

r

1A b.

63 r

1B ,

r

1A c.

43 r

1B ,

r

1A d.

34 r

1B ,

r

1A

B Sol. Dipole-dipole interaction energy between stationery polar molecules (as in solids) is proportional to

3r

1 and that between rotating polar molecules is proportional to

6r

1, where r is the distance between

the polar molecules. Besides dipole-dipole interaction, polar molecules can interact by London forces also.



22. An ionic solid A+B– crystallizes as a fcc structure of NaCl. If the edge length of cell is 508 pm and radius of anion is 144 pm, the radius of cation is:

a. 110 pm b. 364 pm c. 220 pm d. 288 pm A

Sol. r+ + r– = 2

afor fcc

23. A molecule AB2 (mol. Wt. = 166.4) occupies orthorhombic lattice with a = 5 Å, b = 8 Å and c = 4 Å. If density of AB2 is 5.2 g cm–3, the number of molecules present in one unit cell is:

a. 2 b. 3 c. 4 d. 5 B

Sol. Volume of unit cell = a × b × c = 5 × 10–8 × 8 × 10-8 = 1.6 × 10-22 cm3 Mass of unit cell = 1.6 × 10-22 × 5.2 = 8.32 × 10-22 g

Number of molecules in one unit cell = 4.166

1061032.8 2322 = 3

24. The volume of atoms present in a face-centred cubic unit cell of a metal (r is atomic radius) is:

a. 3r3

20 b. 3r

3

24 c. 3r

3

12 d. 3r

3

16

D

Sol. Volume of atom in a cell = nr3

4 3 (n = 4 for fcc)

= 33 r3

164r

3

4

25. In a CCP lattice of A and B, A atoms are present at the corners while B atoms are at face centres. Then the formula of the compound would be if one of the A atoms from a corner is replaced by C atoms (also monovalent)? a. A7B24C8 b. A7B24C c. A24BC d. AB24C B

Sol. AB3

If one A is missing A = 8

7

C = 8

1

3CBA87

26. Which of the following expression is correct for packing fraction of NaCl if the ions along one face are diagonally removed?

a. 3

33

)rr(8

r3

16r

3

13

b. 3

33

)rr(8

r3

4r

3

13

c. 3

33

)rr(8

r3

16r

3

16

d. 3

33

)rr(8

r3

13r

3

4

A

Sol. Effective number of remaining Cl– = 4

13

2

1

8

124

Effective number of Na+ = 4

Packing fraction = volume Total

volume Occupied

= 3

33

)rr(8

r3

44r

3

4

4

13



27. If the radii of A+ and B– in the crystalline solid AB are 96 pm and 200 pm respectively. Then expected structure of void will be:

a. trigonal b. octahedral c. hexagonal d. cubic B

Sol. 48.0200

96

r

r

28. Fe0.95O can be due to presence of iron in +2 and +3 oxidation number. The iron present in +3 oxidation state will be a. 15% b. 13.5% c. 10.5% d. 8.85% C

Sol. Fe+3 = 0.1

10095.0

1.0 =

95

1000

29. If AgCl is doped with 10-4 mol% of CdCl2, then the concentration of cation vacancies per mol will be a. 6.02 × 1016 b. 6.02 × 1017 c. 6.02 × 1018 d. 6.02 × 1014 B

Sol. 17234

10023.61002.6100

10

(No. of vacancy is same as Number of Cd+2 ion)

30. Which of the following have four Bravais lattices? a. Tetragonal b. Orthorhombic c. Triclinic d. Monoclinic B

31. Which of the following is not a covalent solid? a. silicon b. graphite c. SiC d. Rhombic sulphur D

Sol. It is a molecular solid. 32. I. Crystalline solids have definite heat of fusion whereas amorphous solids lack it.

II. Crystalline solids have definite geometrical shape whereas amorphous solids have no definite shape.

III. Crystalline solids are anisotropic whereas amorphous solids are isotropic. IV. Crystalline solid do not have sharp melting point whereas amorphous solids have sharp melting

point. The correct statement(s) is/are

a. I, II, IV b. I, II, III c. II, III, IV d. I, II, III, IV B

33. Match the following Columns

Column – I (Solid) Column – II (Example)

(a) Non-polar molecular (1) CCl4

(b) Metallic (2) ZnS

(c) Network (3) SiO2

(d) Ionic (4) Mg

a. A (1); B (2); C (3); D (4) b. A (1); B (4); C (3); D (2)

c. A (3); B (4); C (1); D (2) d. A (3); B (2); C (1); D (4) B

34. The theoretical density of ZnS is d g/cm3. If the crystal has 4% Frenkel defect, then the actual density of ZnS should be

a. d g/cm3 b. 0.04d g/cm3 c. 0.96d g/cm3 d. 1.04d g/cm3 A

Sol. No change in density in Frenkel defect.



35. A ferromagnetic substance becomes a permanent magnet when it is placed in a magnetic field because

a. all the domains get oriented in the direction of magnetic field b. all the domains get oriented in the direction opposite to the direction of magnetic field c. domains get oriented randomly d. domains are not affected by magnetic field A

36. A metallic element exists as cubic lattice. Each edge of the unit cell is 4.0 Å. The density of the metal is 6.25 g/cm3. How many unit cells will be present in 100 g of the metal?

a. 1.0 ×1022 b. 2.5 × 1023 c. 5.0 × 1023 d. 2.0 × 1023 B

Sol. 23

303105.2

10)400(25.6

100

cell unit one of Mass

Masscell unit of .No

SECTION – B (Assertion & Reason Type) Negative Marking [-1] This Section contains 7 multiple choice questions. Each question has four choices A), B), C) and

D) out of which ONLY ONE is correct. Marks = 7 × 4 = 28

(A) Statement–1 and Statement -2 are true; Statement–2 is the correct explanation of Statement -1. (B) Statement–1 & Statement -2 are true but Statement–2 is not a correct explanation for Statement-1. (C) Statement–1 is true and Statement–2 is false. (D) Statement–1 is false and Statement–2 is true.

1. Statement-1: In ccp arrangement, a tetrahedral void is surrounded by four spheres whereas an

octahedral void is surrounded by six spheres. Statement-2: Size of tetrahedral void is smaller than that of octahedral void. a. (a) b. (b) c. (c) d. (d) B 2. Statement-1: For atomic crystalline solids, the packing efficiency lies in the sequence:

Face centred cubic > body centre cubic > simple cubic unit cell.

Statement-2: Packing efficiency = 100a

r3

4Z

3

3

Z = Number of atoms per unit cell r = Radius of atom a = Edge length of unit cell

a. (a) b. (b) c. (c) d. (d) B

3. Statement – 1: When potassium chloride is heated in the vapour of potassium then its colour becomes light blue-violet.

Statement – 2: Anionic vacancy having trapped electrons is called F-centre. F-centre is responsible for colour of ionic solid.

a. (a) b. (b) c. (c) d. (d) A 4. Statement – 1: On heating a ferromagnetic solid behave as paramagnetic solid. Statement – 2: Heating increases enthalpy of solid a. (a) b. (b) c. (c) d. (d) B



5. Statement – 1: The stability of a crystal gets reflected in its melting point. Statement – 2: The stability of a crystal depends upon the strength of the inter particle force. The melting

point of a solid depends on the strength of the attractive force acting between the constituent particles. a. (a) b. (b) c. (c) d. (d) A 6. Statement – 1: HCl, SO2 are the examples of polar molecular solids. Statement – 2: These are good conductors of electricity. a. (a) b. (b) c. (c) d. (d) C Sol. HCl, SO2 are polar molecular solids. These are non-conductors of electricity 7. Statement – 1: Increasing temperature increases the density of point defects.

Statement – 2: The process of formation of point defects in solids is endothermic and has S > 0 a. (a) b. (b) c. (c) d. (d) A

SECTION – C (PARAGRAPH TYPE) Negative Marking [-1]

This Section contains 2 paragraphs. Each of these questions has four choices A), B), C) and D) out of which ONLY ONE is correct. 7 × 4 = 28 Marks

Comprehension – 1

Answer the following on the basis of given diagram: 1. The shaded plane represents diagonal plane of symmetry. How many such planes are possible in the

cubic system?

a. 1 b. 2 c. 4 d. 6 D

2. The line joining the points A and C is called:

a. face diagonal and three fold axis of symmetry b. body diagonal and three fold axis of symmetry c. four fold axis of symmetry d. only two fold axis of symmetry B

3. How many body diagonal like AC are possible in the cubic system?

a. 2 b. 4 c. 6 d. 8 B

B

A

D

C



Paragraph – 2 The mineral perovskite has a structure that is the prototype of many ABX3 solids; particularly oxides. The perovskite structure is cubic with A atoms surrounded by 12X atoms and the B atoms surrounded by 6X atoms. The sum of charges on the cations A and B must be +6; it can be achieved in several ways (A2+ B4+ and A3+ B3+ among them).

The perovskite structure is closely related to the materials that show interesting electrical properties, such as piezoelectricity, ferroelectricity, ferroelectricity and high temperature superconductivity. Perovskite is the name of a mineral containing calcium, oxygen and titanium having following unit cell structure:

4. Molecular formula of perovskite is:

a. CaTiO3 b. CaTiO c. CaTiO2 d. CaTiO4 A

5. Total number of atoms in the unit cell of perovskite is:

a. 2 b. 3 c. 4 d. 5 D

Sol. Number of Ca2+ ions = 8 × 18

1

Number of Ti4+ = 1

Number of O2– ions = 6 × 2

1 = 3

Total Number of atoms = 5

6. Oxidation state and co-ordination number of titanium in CaTiO3 (Perovskite) is:

a. + 2, 6 b. + 3, 8 c. + 4, 6 d. + 4, 8 C

Sol. CaTiO3 +2 + x – 6 = 0 x = + 4 7. What is coordination number of oxygen in pervoskite structure? a. 4 b. 6 c. 8 d. 12 B

Oxygen

Titanium Calcium



SECTION – D (More than One Correct Choice Type) No Negative Marking

This Section contains 6 multiple choice questions. Each question has four choices A), B), C) and

D) out of which MORE THAN ONE is correct. Marks: 6 × 5 = 30

1. The correct statement(s) regarding defects in solids is (are) :

a. Frenkel defect is usually favoured by a very small difference in the sizes of cation and anion b. Frenkel defect is a dislocation defect c. Trapping of an electron in the lattice leads to the formation of F-center. d. Schottky defects have no effect on the physical properties of solids. B,C

Sol. Frenkel defect exist when cation is much smaller than anion. 2. Which of the following statement is correct for two dimensional packing

a. Square packing is more efficient than hexagonal packing b. In hexagonal packing atoms of 2nd row are placed in depression of 1st row. c. In square packing, atoms are aligned horizontally & vertically d. Hexagonal packing is more efficient than square packing B,C,D

3. Which of the following is not true for metallic solid. a. Strength of metallic bond is directly proportional to size

b. They are good conductor of electricity c. Conductance increases with increase in temperature d. Localised metal ions are embedded in sea of electrons. A,C

4. Which of the following is correct for three dimensional packing?

a. If atoms of third layer are placed over octahedral void of 2nd layer, it will be ABCABC ….. type packing

b. If atoms of 3rd layer form tetrahedral void with 2nd layer, it will be ABAB …. packing c. Coordination number of ABAB….. packing is more than ABCABC packing d. Unit cell of ABCABC packing is Body centred cubic A,B

5. Which of the following statements are incorrect with respect to crystalline defects?

a. n-type semiconductor always increases density of crystal while p-type semiconductor decreases density.

b. Dislocation defect does not change the density of defects. c. Increase in temperature increases density of the defects. d. Ionic substance crystallizing as CsCl structure will have greater tendency for Schottky defect as

compared to Frenkel defect. A,B

6. A cubical crystal is such that the corners atoms are occupied by P atoms, face centers are occupied by Q atoms, alternate tetrahedral voids are occupied by R atoms and edge centers are occupied by S atoms. Based on this info identify the correct option.

a. If there are no defects in the crystal, then the formula should be PQ3R4S3. b. If all the atoms lying on one of the body diagonal are missing, then formula should be PQ4R8/3S4 c. If all the atoms lying on a line passing connecting edge centers of a face are missing, then

formula should be P2Q5R8S5 d. If all the atoms lying on a line passing through body center and connecting opposite face centers

are missing, then formula will be PQ2R4S3.



A,C,D Sol. P = 1; Q = 3; R = 4; S = 3 (a) PQ3R4S3

P = 8

6; Q = 3; R = 3; S = 3 (b) PQ4R8/3S4

P = 1; Q = 2; R = 4; S = 3 (d) PQ2R4S3

SECTION – E (Matrix Type) No Negative Marking

This Section contains 2 questions. Each question has four choices (A, B, C and D) given in

Column I and five statements (p, q, r, s and t) in Column II. Any given statement in Column I can have correct matching with one or more statement(s) given in Column II. Marks = 2 × 8 = 16

1. Match the crystal system/unit cells mentioned in Column I with their characteristic features

mentioned in Column II. (One or more than one Match)

Column – I Column – II

A. simple cubic and face-centred cubic p. have these cell parameters

a = b = c and a = =

B. cubic and rhombohedral q. are two crystal systems of different unit cell

C. cubic and tetragonal r. have only two crystallographic angles of 90°

D. hexagonal and monoclinic s. belong to crystal system of same unit cell

Sol. A p, s; B p, q; C q; D q, r 2. Match Column – I with Column – II. One or More than One Answer

Column – I Column – II

(A) The defect that lowers the density (p) Schottky defect

(B) The defect in which e– are present in voids, left vacant by the anion.

(q) Presence of F-centre

(C) Extrinsic semi-conductor mainly shows conductance due to

(r) Presence of impurity of higher valency in a covalent solid

(D) Insulator turn semiconductor (s) Electronic imperfection

Sol. A p, q; B q; C r; D p, q, r, s

SECTION – E (Integer Type) No Negative Marking This Section contains 6 questions. The answer to each question is a single digit integer ranging

from 0 to 9. Marks: 6 × 4 = 24

1. What is the co-ordination number of Rb+ in RbBr unit cell? The ionic radii of Rb+ and Br – ions being

148 and 195 respectively. Sol. 8

The ratio r+/r– = 195

148 = 0.759. Since this value is more than 0.732 it implies that Rb+ ion occupies a

cubical hole in the crystal lattice. In other words, the co-ordination number of Rb+ is 8.

2. If edge length is 2 cm. What will be the distance between next neighbour of FCC unit cell. Sol. 2 (next neighbours are corner atoms) 3. How many moles of total tetrahedral & octahedral voids will be found in a ccp arrangement having

3 moles of sphere in lattice. Sol. 9 mole 4. If No. of cation vacancy in AgCl is 4 moles, due to impurity of AuCl3. What will be the no. of moles of

AuCl3. Sol. 2 mol; No. of vacancy is double to the impurity of Au+3



5. The total number of ions per unit cell in sphalerite (zinc blend) structure if there is one Schottky pair defect per unit cell,

Sol. 6; one pair of schottky defect will have two ions missing Sphalerite structure have 4 Zn+2 & 4S-2 ion i.e. 8 ions. In A pair of schottky defect two ions will be

missing so total ions left = 6 6. How many of the following statements are correct for defects in solids

(a) Schottky defect is a vacancy defect

(b) Solids with smaller cation have tendency of schottky defects

(c) Zinc oxide become non-stoichiometric on heating

(d) In ferrous oxide crystal, Number of cation vacancy is half, of the number of ferric ion as three ferrous ions are replaced by two ferric ion.

(e) Transition metal salts having metal in lower oxidation state have tendency to have non-stoichiometric defect

(f) F-centres are cation vacancies having e–.

(g) Boron doped silicon crystal is a p-type semiconductor

(h) Density of crystal remains same both in schottky and Frenkel defect.

(i) Compounds with schottky defect donot obey law of constant composition Sol. 5

a,c,d,e,g

test venue: lajpat bhawan, madhya marg, sector …...a face centered atom is neighbour of corner...

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