1 SOLID STATE

EXERCISE - 4 : PREVIOUS YEAR JEE ADVANCED QUESTION

1. In a solid ‘AB’ having the NaCl structure, ‘A’ atoms occupy known as octahedral and tetrahedral voids. If r1 is the
radius of void and r2 is the radius of atom creating these
the corners of the cubic unit cell. If all the face-centred
voids then
atoms along one of the axes are removed, then the resultant
stoichiometry of the solid is (2001)  r1 
  =0.414
 r2  octa
(a) AB2 (b) A2B
and
(c) A4B3 (d) A3B4  r1 
  =0.225
Ans. (d)  r2  tetra
Sol. In NaCl, Na+ occupies body centre and edge centres while The above radius ratio values indicate that octahedral void
Cl– occupies corners and face centres, giving four Na+ has larger radius, hence for maximum diameter of atom to
and four Cl– per unit cell. In the present case A represent be present in interstitial space :
Cl– and B represents Na+. Two face centres lies on one r1 = 0.414 r2
axis.
Also in fcc, 4r2  2a
1
Þ No. of A removed = 2  = 1 Þ Diameter required (2r1) = (2r2)×0.414
2
Number of B not is removed because it is not present on a 400×0.414
= ×0.414 = =117 pm
face centres. 2 2
Þ A remaining = 4 - 1 = 3, B remaining = 4
Formula = A3B4
4. Which of the following fcc structure contains cations in
2. The crsytal AB (rock salt structure) has molecular weight alternate tetrahedral voids ? (2005)
6.023 y u. where, y is an arbitrary number in u. If the mini-
mum distance between cation and anion is y1/3 nm and the (a) NaCl (b) ZnS
observed density is 20 kg/m3. Find the (a) density in
kg/m3 and (b) type of defect (2004) (c) Na2O (d) CaF2
Ans. (b)
Ans. (a) 5Kg/m3, (b) metal excess defect
Sol. In ZnS, S2- (sulphide ions) are present at fcc positions
Sol. (i) In rock salt like crystal AB, there are four AB units per
giving four sulphide ions per unit cell. To comply with 1:1
unit cell.
stoichoimetry, four Zn2+ ions must be present in four
4×6.023 y alternate tetrahedral voids out of eight tetrahedral voids
Therefore, density (d) is = present. In NaCl, Na+ ions are present in all its tetrahedral
6.023×1023 ×8y×10-27
voids giving the desired 2:1 stoichiometry. In CaF2, Ca2+
[ a = 2y1/3 nm = 2y1/3  10-9m]
ions occupies fcc positions and all the tetrahedral voids
= 5  103 g/m3 = 5 kg/m3 are occupied by fluoride ions.
(ii) Since, observed density is greater than expected,
theoretical density, there must be some excess metal 5. The edge length of unit cell of a metal having molecular
occupying interstitial spaces. This type of defect is known weight 75 g/mol is 5 Å which crystallizes in cubic lattice. If
as metal excess defect the density is 2 g/cc then find the radius of metal atom.
(NA = 6 × 1023). Give the answer in pm. (2006)
3. An element crystallizes in fcc lattice having edge length Ans. 217 pm
400 pm. Calculate the maximum diameter of atom which can Sol. From the given information, the number of atoms per unit
be placed in interstitial site without distorting the structure. cell and therefore, type of unit cell can be known as
(2005)
Ans. 117 pm
Sol. In a cubic crystal system, there are two types of voids
 SOLID STATE 2
of 90 . 0
ZM
=
NA .a3
Comprehension (Ques. 10 to 12)

 NA a3 2  6  1023  (5  10-8 cm)3 In hexagonal systems of crystals, a frequently encountered

Z= = = 2(bcc)
M 75 arragenement of atoms is described as a hexagonal prism.
Here the top and bottom of the cell are regular hexagons
 In bcc, 4r  3a and three atoms are sandwiched in between them. A space-
filling model of this structure, called hexagonal close-
packed (HCP), is constituted of a sphere on a flat surface
3 3 surrounded in the same plane by six identical spheres as
 r= a= ×5×10-10 m
4 4 closely as possible. Three spheres are then placed over
the first layer so that they touch each other and represent
the second layer. Each one of these three spheres touches
=2.17×10-10 m= 217 pm
three spheres of the bottom layer. Finally, the second layer
is covered with a third layer that is identical to the bottom
layer that is identical to the bottom layer in relative position.
Match the Column
Assume radius of every sphere to be ‘r’. (2008)
6. Match the crystal system/unit cells mentioned in Column-
7. The number of atoms on this HCP unit cell is :
I with their characteristic features mentioned in Column-II.
Indicate your answer by darkening the appropriate bubbles (a) 4 (b) 6
of the 4 × 4 matrix given in the ORS. (2007) (c) 12 (d) 17
Column I Column II Ans. (b)

(A) Simple cubic and (p) have these cell

face-centred cubic parameters a = b = c and

Sol.
  

(B) Cubic and (q) are two crystal

rhombohedral systems
A hcp unit cell
(C) Cubic and tetragonal r) have only two 1
Contribution of atoms from corner 
6
crystallography
1
angles of 90º Contribution from face centre 
2
(D) Hexagonal and (s) belong to same 1 1
Total number of atoms per unit cell = 12  + 2  + 3 = 6
6 2
monoclinic crystal system
Ans. A - p, s ; B - p, q ; C - q ; D - q, r 8. The volume of this HCP unit cell is :
Sol. A. Simple cubic and face centred cubic both have cell
parameters a = b = c and  =  =  = 900 . Also both of (a) 24 2 r 3 (b) 16 2r3
them belongs to same, cubic, crystal system.
B. The cubic and rhombohedral crystal system belongs 64r 3
to different crystal system. (c) 12 2 r 3 (d)
3 3
C. Cubic and tetragonal are two different types of crystal
systems having different cell parameters. Ans. (a)
D. Hexagonal and monoclinic are two different crystal
system and both have two of their crystallographic angles Sol. In close packed arrangement, side of the base = 2r
 3 SOLID STATE
Þ RS = r No of Mg ions = 8n
2+

Molecule of mineral is neutral.

RS 3 So,
In triangle PRS,cos 300 = = 4(-2) + 4m(+3) + 8n(+2) = 0
PR 2
12m + 16n = 8
3m + 4n = 2
2 2 m = 1/2 , n = 1/8
 PR= RS= r
3 3
11. The Correct statement(s) for cubic close packed (ccp) three
2 2 2 dimensional structure is (are) (2016)
In right angle triangle PQR : PQ= QR -PR =2 r
3
(a) The number of the nearest neighbours of an atom
present in the topmost layer is 12
2
 Height of hexagon = 2PQ = 4 r
3 (b) The efficiency of atom packing is 74%

(c) The number of octahedral and tetrahedral voids per

3 2 atom are 1 and 2, respectively
 Volume = Area of base×Height =6 (2r)2 ×4 r
4 3
(d) The unit cell edge length is 2 2 times the radius of
3
=24 2 r . the atom
9. The empty space in this HCP unit cell is : Ans. b,c,d
(a) 74% (b) 47.6% Sol. a) Top most layers nearest neighbours = 9
b) Efficiency = 74%
(c) 32% (d) 26% c) No of O-voids = 1 × atom per unit cell
No of T-voids = 2 × no of unit cell atoms
Ans. (d)
d) a  2 2 r
Volume occupied by atoms
Sol. Packing fraction=
Volume of unit cell 12. A crystalline solid of a pure substance has a face-centred
4 1 cubic structure with a cell edge of 400 pm. If the density of
=6× π r 3 × =0.74 the substance in the crystal is 8 g cm–3, then the number of
3 24 2 r 3
atoms present in 256 g of the crystal is N × 1024. The value
Þ Fraction of empty space = 1 - 0.74 = 0.26 = 26% of N is (2017)

Ans. 2
10. If the unit cell of a mineral has cubic close packed (ccp)
array of oxygen atoms with m fraction of octahedral holes zm
Sol. 8
occupied by aluminium ions and n fraction of tetrahedral Na  a 3
holes occupied by magnesium ions, m and n, respectively,
4 m
are (2015) 8
 
3
6.022  10  400  10 10 cm
23

1 1 1
(a) , (b) 1, M = 76.8 g mol–1
2 8 4 76.8 g mol–1 = 6 × 1023 atoms
1 1 1 1 6  1023
(c) , (d) , 1 g mol–1 = atom
2 2 4 8 76.8
Ans (a) 6  1023  256
Sol. In CCP arrangement, 256 g mol 1   2 1024
No of O atoms = 4 76.8
No of O-voids = 4
No of T-voids = 8 13. Consider an ionic solid MX with NaCl structure. Construct
No of Al3+ ions = 4m a new structure (Z) whose unit cell is constructed from the
 SOLID STATE 4
unit cell of MX following the sequential instructions given (b) The cation M and anion X have different coordination
below. Neglect the charge balance.
geometries.
(i) Remove all the anions (X) except the central one (c) The ratio of M-X bond length to the cubic unit cell
(ii) Replace all the face centered cations (M) by anions (X) edge length is 0.866.
(d) The ratio of the ionic radii of cation M to anion X is
(iii) Remove all the corner cations (M)
0.414.
(iv) Replace the central anion (X) with cation (M)
Ans. (a,c)

 number of anions  1
Sol. (a) Z M  2  1
The value of   in Z is .......... . 2
 number of cations 
1
ZX  4  1
(2018) 4
Ans. 3  Empirical formula is MX
Sol. MX has NaCl structure (b) Coordinate numbers of both M and X is 8.
M ® CCP (c) Bond length of M – X bond
X ® occupy O-voids
a
i) No of anions left = 1  AB  3 .  0.866a
ii) No of anions added = 3 2
No of cations left = 1
iii) No of cations left = 0
iv) No of cations added = 1
No of anions added = 3

Final no. of cations :- 1

Final no. of anions :- 3
3 (d) radius ratio for coodination no. 8 is in
ratio  3
1 between of 0.732-1.00

14. The cubic unit cell structure of a compound containing 15. For the given close packed structure of a salt
made of cation X and anion Y shown below (ions of
cation M and anion X is shown below. When compared only one face are shown for clarity), the packing
to the anion, the cation has smaller ionic radius. Choose fraction is approximately (packing fraction
the correct statement(s). packing efficiency
 ).
(2020/Paper 1) 100

(a) The empirical formula of the compound is MX. (a) 0.74 (b) 0.63
(c) 0.52 (d) 0.48
Ans. (b)
 5 SOLID STATE
Sol.
3 8 3
rz  ry ; ry  rx ; M z  M y and MZ = 3MX,
2 3 2
then the correct statement (s) is (are)
[Given : Mx, My, and Mz are molar masses of metals x, y,
and z, respectively. rx, ry, and rz are atomic radii of metals x,
y, and z, respectively.] (2023)
(a) Packing efficiency of unit cell of x > Packing efficiency
of unit cell of y > Packing efficiency of unit cell of z
(b) Ly > Lz
(c) Lx > Ly
(d) Density of x > Density of y
Packing fraction
Ans. (a, b, d)
Volume occupied by anion + Volume occupied by cation
 Sol.
Volume of unit cell
Here anion = 1 effective and cations = 3
effective
r
Also,
r
 2 1 
Packing fraction

4 4  3
 3   r   3 3   r    1
 3

= 
 2r  
3      3
 6
   
2 1
3

8 3 3 8
  Now, ry  rx & rz  ry   rx  rz  4rx
3 2 2 3
Packing fraction
 1 22
 
6 7

1 3  2 1   16  227 1.2132  0.635
3
So, L x  2 2rx , L y 
4
3

8
3
rx , L z  8rx

16. Atom X occupies the fcc lattice sites as well as alternate

tetrahedral voids of the same lattice. The packing efficiency 32
L x  2 2rx , L y  rx , L z  8rx
(in %) of the resultant solid is closest to (2022) 3

(a) 25 (b) 35 So Ly > Lz > Lx

(c) 55 (d) 75
4M x 2  M y
Ans. (a) Density L3 , L3
x y
Sol. Z=4+4

4 3 4 3M y
 r 8  8 Now, 3M x  or M x  2  M y
3 2
PF  3  3   35%
 8r 
3
28  8  8 16
  3 3 3
 3  32 
density  x  4M x L y 4M x  3 
3

17. Atoms of metals x, y, and z form face-centred cubic (fcc)    

density  y  2M y L3x 4M x 2 2 3
unit cell of edge length Lx, body-centred cubic (bcc) unit  
cell of edge length Ly, and simple cubic unit cell of edge
length Lz, respectively. If Hence d(x) > d(y)