Solved Objective Problems
Ex.1. Select right statement(s)
(A) 8 Cs ions occupy the second nearest neighbour locations of a Cs ion.
(B) Each sphere is surrounded by six voids in two dimensional hexagonal close packed layer .
(C) If the radius of cations and anions are 0.3 Å and 0.4 Å then coordinate number of cation of the crystal is 6.
(D) In AgCl, the silver ion is displaced from its lattice position to an interstitial position such a defect is called a frenkel defect.
(A) 6 Cs ion second nearest neighbour
(C) r+\r- = 0.75 ( BCC) 8 : cordination no.
Ex.2. Which of the following statement is/are incorrect in the rock-salt structure if an ionic compounds?
(A) coordination number of cation is four where as that of anion is six.
(B) coordination number of cation is six where as that of anion is four.
(C) coordination number of each cation and anion is four.
(D) coordination number of each cation and anion is six.
Coordination no. of cation = 6
coordination no of anion = 6
Ex.3. Which of the following statements are correct?
(A) The coordination number of each type of ion in CsCl is 8.
(B) A metal that crystallizes in BCC structure has a coordination number 12.
(C) A unit cell of an ionic crystal shares some of its ions with other unit cells.
(D) The length of the unit cell in NaCl is 552 pm.
Sol. r + r- = a/2
⇒ (95 + 181)= a/2
⇒ a = 276 × 2
⇒ a = 552 pm
Ex.4. A cubic solid is made up of two elements A and B. Atoms B are at the corners of the cube and A at the body center. What is the formula of compound?
Sol. A = 1 (Body center)
formula = AB
Ex.5. A compound alloy of gold and copper crystallizes in a cubic lattice in which gold occupy that lattice point at corners of the cube and copper atom occupy the centres of each of the cube faces. What is the formula of this compound?
Formula = AuCu3
Ex.6. A cubic solid is made by atoms A forming close pack arrangement, B occupying one. Fourth of tetrahedral void and C occupying half of the octahedral voids. What is the formula of compound?
A = = 4
B = no. of void (tetrahedral) = 2
C = Total no. of octahedral voids = 2
no. of octahedral voids (effective)= 1 12 × 1/4 = 4
C = 4/2 = 2
formula = A4B2C2
Ex.7. What is the percent by mass of titanium in rutile, a mineral that contain Titanium and oxygen, if structure can be described as a closet packed array of oxide ions, with titanium in one half of the octahedral holes. What is the oxidation number of titanium ?
Sol. O2- = × 6 =4
Ti (At mass) = 47.88
Ox. State = 4
Ex.8. Spinel is a important class of oxides consisting of two types of metal ions with the oxide ions arranged in CCP pattern. The normal spinel has one-eight of the tetrahedral holes occupied by one type of metal ion and one half of the octahedral hole occupied by another type of metal ion. Such a spinel is formed by Zn2+ , Al3+ and O2-, with Zn2+ in the tetrahedral holes. Give the formulae of spinel.
Sol. Zn2+ = 8/8 = 1
Al3+ = 4/2 = 2
O2- = CCP (4)
O2- = CCP arrangment (1 3) = 4
Ex.9. Iron occurs as bcc as well as fcc unit cell. If the effective radius of an atom of iron is 124 pm. Compute the density of iron in both these structures.
In bcc, z=2
, r =
⇒ a = in pm
r = 7.92 × 106 g/cm3
In FCC , Z = 4,
= 8.6 g/cm3
Ex.10. A closed packed structure of uniform spheres has the edge length of 534 pm. Calculate the radius of sphere, if it exist in
(A) simple cubic lattice (B) BCC lattice (C) FCC lattice
(A) Simple cubic lattice
2r = a ⇒ r = = = 267 pm
(B) 4r = ⇒ r = = 231.23 pm
(C) 4r = ⇒ r = = 188.79 pm
Ex.11. Calculate the density of diamond from the fact that it has face centered cubic structure with two atoms per lattice point and unit cell edge length of 3.569 Å.
Sol. In simple FCC, Z = 4
but here from questions two atoms per lattice point
⇒ z = 8
= 3.5 × 106 g/m3 = 3.5 g/cm3
Ex.12. An element crystallizes into a structure which may be described by a cubic type of unit cell having one atom on each corner of the cube and two atoms on one of its body diagonals. If the volume of this unit cell is 24 × 10-24 cm3 and density of element is 7.2 g cm-3, calculate the number of atoms present in 200 g of element.
Sol. z = 3
⇒ M = 34.68
no. of atoms in 200 gram = × 6.022 × 1023
= 3.47 × 1024 atoms
Ex.13. Silver has an atomic radius of 144 pm and the density of silver is 10.6 g cm-3. To which type of cubic crystal, silver belongs?
Sol. 10.6 × 106 g/m3
= 74 % ⇒ F.C.C.
Ex.14. AgCl has the same structure as that of NaCl. The edge length of unit cell of AgCl is found to be 555 pm and the density of AgCl is 5.561 g cm-3. Find the percentage of sites that are unoccupied.
If 100% sites occupied =5.575 but density is 5.561
% sites occupied =
% sites unoccupied = 100 - (99.75) %
Ex.15. Xenon crystallizes in the face-centred cubic lattice and the edge of the unit cell is 620 pm. What is the nearest neighbour distance and what is the radius of xenon atom ?
Sol. Nearest neighbour distance = 2r (in FCC)
⇒ = 4r
⇒ 2r = =
2r = 438.47
r = 219.23
Ex.16. The two ions A and B- have radii 88 and 200 pm respectively. In the closed packed crystal of compound AB, predict the co-ordination number of A.
(r+\r-) = 0.44
It could be square planer or octahedral void but for closed packed crystal it is octahedral void
Its coordination no. = 6
Ex.17. CsCl has the bcc arrangement and its unit cell edge length is 400 pm. Calculate the interionic distance in CsCl.
Sol. Interionic distane
r + r- =
= = 346.4
Ex.18. Gold crystallizes in a face centered cubic lattice. If the length of the edge of the unit cell is 407 pm, calculate the density of gold as well as its atomic radius assuming it to be spherical. Atomic mass of gold = 197 amu.
Sol. FCC : Z = 4
= 19.41 × 106 g/m3 = 19.41 g/cm3
Ex.19. The density of KBr is 2.75 g cm-3. The length of the edge of the unit cell is 654 pm. Show that KBr has face centered cubic structure.
(N = 6.023 × 1023 mol-1, At. mass : K = 39, Br = 80)
= 7.18 × 10-23
z = (654 × 10-10)3 / (7.18 × 10-23) = 4
Ex.20. An element crystallizes in a structure having FCC unit cell of an edge 200 pm. Calculate the density, if 200 g of this element contains 24 × 1023 atoms.
Sol. 24 × 1023 atom contain by 200 g
6 × 1023 atom contains by
= 50 g
M » 50 g/mole
r = 41.67 g/cm3
Ex.21. The effective radius of the iron atom is 1.42 Å. If has FCC structure. Calculate its density
(Fe = 56 amu)
Sol. r =
⇒ a = 4.016 Å
= 0.574 × 107 g/m3
= 5.74 × 106 g/m3
r = 5.74 g/cm3
Ex.22. A crystal of lead (II) sulphide has NaCl structure. In this crystal the shortest distance between Pb2 ion and S2- ion is 297 pm. What is the length of the edge of the unit cell in lead sulphide ? Also calculate the unit cell volume.
Sol. NaCl type FCC
r r- = = 297
⇒ a = 297 × 2 = 594 pm
= 594 × 10-10 cm
= 5.94 × 10-8 cm
a3 = volume % = 2.096 × 10-22 cm3
Ex.23. If the length of the body diagonal for CsCl which crystallizes into a cubic structure with Cl- ions at the corners and Cs ions at the centre of the unit cells is 7 Å and the radius of the Cs ion is 1.69 Å, what is the radii of Cl- ion ?
r r- = =
1.69 r- = 3.5
r- = 3.5 - 1.69 = 1.81 Å
Ex.24. In a cubic closed packed structure of mixed oxides the lattice is made up of oxide ions, one eighth of tetrahedral voids are occupied by divalent ions (A2 ) while one half of the octahedral voids occupied trivalent ions (B3 ). What is the formula of the oxide ?
Sol. Total tetrahedral voids = 8
A2 occupied tetrahedral void
Total octahedral void = 4
B3 Occupy = = 2
A2 = 1, B3 = 2, O2- = 4
formula = AB2O4
Ex.25. A solid A and B- has NaCl type closed packed structure. If the anion has a radius of 250 pm, what should be the ideal radius of the cation ? Can a cation C having a radius of 180 pm be slipped into the tetrahedral site of the crystal of A B-? Give reasons for your answer.
Sol. for ideal NaCl crystal
⇒ r- =
r = 103.55 pm
tetrahedral sites for tetrahedral voids
0.225 < < 0.414
no a cation can't slipped into tetrahedral site of crystal
Ex.26. If the radius of Mg2 ion, Cs ion, O2- ion, S2- ion and Cl- ion are 0.65 Å, 1.69 Å, 1.40 Å, 1.84 Å, and 1.81 Å respectively. Calculate the co-ordination numbers of the cations in the crystals of MgS, MgO and CsCl.
Sol. for MgS
coordination no = 4
MgO ⇒ = = 0.464
coordination no. = 6
BCC coordination no. = 8
Ex.27. In a cubic crystal of CsCl (density = 3.97 gm/cm3) the eight corners are occupied by Cl- ions with Cs ions at the centre. Calculate the distance between the neighbouring Cs and Cl- ions.
Sol. 3.94 =
a3 = 7.05 × 10-23 cm3
a = 4.13 × 10-8 cm
r r- = = 3.577 × 10-8
= 3.577 × 10-10 cm
= 3.577 Å
Ex.28. Calculate the value of Avogadro's number from the following data:
Density of NaCl = 2.165 cm-3
Distance between Na and Cl- in NaCl = 281 pm.
Sol. d =
NA = 6 × 1023
Ex.29. KCl crystallizes in the same type of lattice as does NaCl. Given that and Calculate:
(a) The ratio of the sides of unit cell for KCl to that for NaCl and
(b) The ratio of densities of NaCl to that for KCl
(a) = 0.5 = 0.7, = ....(i)
1 = 1 0.5, = 1.5 ....(ii)
¸ = , = , 1 = 1, = ....(iii)
From eq. (i)
= = = 1.143
(b) dNaCl = , dkCl = ,
= . = . (1.143)3 = 1.172
Ex.30. An element A (Atomic weight = 100) having bcc structure has unit cell edge length 400 pm. Calculate the density of A and number of unit cells and number of atoms in 10 gm of A.
Sol. a = 400 pm = 4 × 10-8 cm
z = 2
= = = 5.188 gm/cc
In 100 gm ⇒ 6.023 × 1023 atoms
In 10 gm ⇒ 6.023 × 1022 atoms
Ex.31. The composition of a sample of wustite is Fe0.93 O1.0. What percentage of iron is present in the form of Fe(III) ?
Sol. Fe : O = 93 : 100
Let Fe (III) = x
then Fe (II) = (93 - x)
Balancing charge, We have
x × 3 (93 - x) × 2 = 100 × 2
x = 200 - 186 = 14
% Fe (III) = × 100 = 15.05 %
Ex.32. BaTiO3 crystallizes in the perovskite structure. This structure may be described as a cubic lattice with barium ions occupying the corner of the unit cell, oxide ions occupying the face-centers and titanium ion occupying the center of the unit cell.
(a) If titanium is described as occupying holes in BaO lattice, what type of holes does it occupy?
(b) What fraction of this type hole does it occupy ?
(a) Ti is present at the body centre i.e. it occupies octahedral hole.
(b) No. of octahedral in c.c.p. = 4
Fractional of octahedral hole occupied =
Ex.33. Rbl crystallizes in bcc structure in which each Rb is surrounded by eight iodide ions each of radius 2.17 Å. Find the length of one side of RbI unit cell.
Sol. = 0.732 Q = 0.732 × 2.17 = 1.59 Å
a = 2  a = [1.59 2.17] = 4.34 Å
Ex.34. Find the size of largest sphere that will fit in octahedral void in an ideal FCC crystal as a function of atomic radius 'r'. The insertion of this sphere into void does not distort the FCC lattice. Calculate the packing fraction of FCC lattice when all the octahedral voids are filled by this sphere.
Sol. If Radius of sphere is r
Radius of largest sphere that will fit in the octahedral void is 0.414 r
2[r 0.414 r] = a
2.828 r = a
Volume of spheres = 4 × pr3 4 × p(0.414r)3 = pr3 [1 0.071]
Volume of unit cell = a3 = (2r)3 = 16r3 , f = = = 0.793
Ex.35. NaH crystallizes in the same structure as that of NaCl. The edge length of the cubic unit cell of NaH is 4.88 Å.
(a) Calculate the ionic radius of H-, provided the ionic radius of Na is 0.95 Å.
(b) Calculate the density of NaH.
Sol. a = 4.88 Å = 4.88 × 10-8 cm
(a) = , = 2.44 - 0.95 = 1.49 Å
(b) d = =
= 1.37 gm/cm3
Ex.36. Metallic gold crystallizes in fcc lattice. The length of the cubic unit cell is a = 4.07 Å
(a) What is the closest distance between gold atoms.
(b) How many "nearest neighbours" does each gold atom have at the distance calculated in (a).
(c) What is the density of gold ?
(d) Prove that the packing fraction of gold is 0.74.
Sol. a = 4.07 Å
(a) 4r = a
r = = = 1.44 Å
Closest distance between An atoms = 2.88 Å
(b) C.N. = 12
(c) d = =
= 19.4 gm/cc
Ex.37. Graphite in an example of-
(A) Ionic solid
(B) Covalent Solid
(C) Vander waal's Crystal
(D) Metallic crystal
Graphite is a covalent solid having sp2 hybridised carbon atoms.
Ex.38. Which is amorphous solid -
Amorphous solids neither have ordered arrangement (i.e. no definite shape) nor have sharp melting point like crystals, but when heated, they become pliable until they assume the properties usually related to liquid. It is therefore they are regarded as super cooled liquids.
Ex.39. Xenon crystallizes in face centre cubic lattice and the edge of the unit cell is 620 PM, then the radius of Xenon atom is-
(A) 219.20 PM
(B) 438.5 PM
(C) 265.5 PM
(D) 536.94 PM
For fcc lattice ;
where a = 620 PM
On solving r = 219.20 PM.
Ex.40. The edge length of cube is 400 PM. Its body diagonal would be-
(A) 500 PM
(B) 693 PM
(C) 600 PM
(D) 566 PM
Since in body centre cubic, the body diagonal
= = PM = 692.82 PM or say 693 PM
Ex.41. What is the simplest formula of a solid whose cubic unit cell has the atom A at each corner, the atom B at each face centre and a C atom at the body centre-
An atom at the corner of a cube is shared among 8 unit cells. As there are 8 corners in a cube, number of corner atom (A) per unit cell = 8 × = 1.
A face-centered atom in a cube is shared by two unit cells. As there are 6 faces in a cube, number of face-centered atoms (B) per unit cell = 6 × = 3.
An atom in the body of the cube is not shared by other cells.
Number of atoms (C) at the body centre per unit cell = 1.
Hence, the formula of the solid is AB3C.
Ex.42. A compound alloy of gold and copper crystallizes in a cube lattice in which the gold atoms occupy the corners of a cube and the copper atoms occupy the centres of each of the cube faces.The formula of this compound is-
One-eighth of each corner atom (Au) and one-half of each face-centered atom (Cu) are contained with in the unit cell of the compound.
Thus, the number of Au atoms per unit cell = 8 × = 1 and the number of Cu atoms per unit cell = 6 × = 3. The formula of the compound is AuCu3.
Ex.43. Select the correct statement (s)-
(a) The C.N. of cation occupying a tetrahedral hole is 4.
(b) The C.N. of cation occupying a octahedral hole is 6.
(c) In schottky defects, density of the lattice decreases.
(A) a, b
(B) b, c
(C) a, b, c
(D) a, c
Since tetrahedral holes are surrounded by 4 nearest neighbours. So, the C.N. of cation occupying tetrahedral hole is 4. Since octahedral hole is surrounded by six nearest neighbours. So, C.N. of cation occupying octahedral is 6. In schottky a pair of anion and cation leaves the lattice. So, density of lattice decreases.
Ex.44. Lithium borohydride (LiBH4), crystallises in an orthorhombic system with 4 molecules per unit cell. The unit cell dimensions are : a = 6.81 Å, b = 4.43 Å , c = 717 Å. If the molar mass of LiBH4 is 21.76 g mol-1 . The density of the crystal is-
(A) .668 g cm-3
(B) .585 g cm-3
(C) 1.23 g cm-3
We know that,
r = ; where V = a × b × c
= 0.668 g cm- 3
Ex.45. The unit cell of a metallic element of atomic mass 108 and density 10.5 g/cm3 is a cube with edge length of 409 PM. The structure of the crystal lattice is -
(D) None of these
Here, M = 108, NA = 6.023 x 1023
Put on these values and solving we get-
a = 409 PM = 4.09 x 10-8 cm,
r = 10.5 g/cm3
n = 4 = number of atoms per unit cell
So, The structure of the crystal lattice is fcc.
Ex.46 Among the following types of voids, which one is the largest void-
(A) Triangular system
(B) Tetragonal system
(C) Monoclinic system
The vacant spaces between the spheres in closed packed structure is called void. The voids are of two types, tetrahedral voids and octahedral voids. Also radius of tetrahedral voids and octahedral voids are rvoid = 0.225 × rsphere and rvoid = 0.411 × rsphere respectively. Thus, octahedral void is larger than tetrahedral void.
Ex.47 Close packing is maximum in the crystal which is-
(A) Simple cube
(D) None of these
The close packing in the crystal is 0.52, 0.68 and 0.74 for simple cubic, bcc, and fcc respectively.
i.e the close packing is maximum is fcc.
Ex.48 Bragg's equation is-
(A) nl = 2q sin q
(B) nl = 2d sinq
(C) 2nl = d sin q
(D) l = (2d/n) sin q
Bragg's equation is nl = 2d sin q.
Ex.49 Copper metal has a face-centred cubic structure with the unit cell length equal to 0.361 nm. Picturing copper ions in contact along the face diagonal, The apparent radius of a copper ion is-
For a face-centred cube, we have,
radius = = nm = 0.128.
Ex.50 The rank of a cubic unit cell is 4. The type of cell as-
(A) Body centred
(B) Face centred
The number of atoms present in sc, fcc and bcc unit cell are 1, 4, 2 respectively.
Ex.51 At room temperature, sodium crystallises in a body centred cubic cell with a = 4.24 Å. The theoretical density of sodium is - ( Atomic mass of sodium = 23.0 g mol-1)
(A) 2.05 g cm-3
(B) 3.45 g cm-3
(C) 1.00 g cm-3
(D) 3.55 g cm-3
The value of Z for a bcc unit cell is 2.
Volume V = (4.24 A)3
r = =
= 1.00 g / cm3