Unit 2
SOLID STATE
1. Classify following solids.
1 P4O10 Molecular solid 2 Graphite Covalent solid 3 Brass Metallic solid 4 Ammonium Phosphate Ionic solid 5 SiC Covalent solid 6 Rb Metallic solid 7 I2 Molecular solid 8 LiBr Ionic solid 9 P4 Molecular solid 10 Si Covalent solid 11 Plastic Amorphous solid
2.
a) What is meant by co-ordination number?
It is the number of nearest neighbours of each atom or ion in a solid structure.
b) What is the co-ordination number of atoms?
In cubic close packed structure.
In a body centred cubic structure.
Ans:- a) 6 b) 8
3. How can you determine the atomic mass of an unknown metal if you know it’s density and the
dimension of the unit cell? Explain.
Ans:- Using the equation, ρ=(zM)/(a3 x NA)
ρ=density, z=effective no. of atoms, M= gram atomic mass, a=edge length, NA=avogadro number.
Derivation :
Density of a unit cell(ρ) = Mass of a unit cell/volumeof unit cell
= Mass of all atoms present in a unit cell/ a3
= Mass of one atom of the element x No of atoms in a unit cell/a3
= M x z / NAa3
4. Stability of a crystal is reflected in the magnitude of it’s melting point. Comment. Collect
melting point of water, ethyl alchohol, diethyl ether and methane. What can you say about the
intermolecular forces between the molecules.
If a crystal has high stability, it’s melting point will be high.
Water 273 Ethyl alchohol 159 Diethy ether Methane
5. Distingush between
Hexagonal close packing and cubic close packing.
Close packed structure of solid can be built in different stages. First layer is made by minimum waste of space. Second layer is made in such a way that spheres are kept in the depressions of first layer. At this time two types of voids are formed – octahedral void and tetrahedral void.
If third layer is arranged covering tetrahedral void, packing is called hexagonal close packing.(AB, AB, AB…pattern)
If third layer is arranged covering octahedral void, packing is called cubic close packing.(ABC, ABC,…pattern)
(2) Crystal lattice and unit cell.
A regular three dimensional arrangement of particles in crystal is called crystal lattice. Unit cell
It is the smallest unit in a crystal lattice which repeats itself throughout to get the whole crystal structure.
(3) Tetrahedral void and octahedral void.
Tetrahedral void
Void formed in between four particles is called tetrahedral void.
Octahedral Void
Void formed in between 6 particles.
6. How many lattice points are there in
FCC- 4 points [ (8 x1/8) ) (6 x ½) = 4]
Face centred tetragonal-4 points [ (8 x1/8) ) (6 x ½) = 4]
Body centred cubic – 2 points [(8x1/8) + 1= 2]
7. Explain:-
a) The basis of similarities and differences between metallic and ionic crystals. Metallic and ionic crystals, have similarities because both are crystals, ie, regular
arrangement is there throughout the lattice.
Differences are there in following ways .
Metallic crystal is made up of metal atoms, Ionic crystals are made up of ions of different elements.
Metallic crystals are conductors ionic crystals are insulators.
Metallic bond is there in metals. Ionic or electrostatic force of attraction is there in ionic crystals.
c) Can a cube consisting of Na+ and Cl- at alternate corners serve as a unit cell for NaCl
lattice?
Ans: No. This cube on repetitions along its cell edges by a distance equal to cell edgelength(a) every time will not produce NaCl lattice.
NaCl lattice is formed by interpenetrating Face centred sodium ion cube
and face centred chloride ion cube half the way.
d) Ionic solids are hard and brittle.
Hardness is because of very strong electrostatic force of attraction between
oppositely charged ion.
Ionic solids are brittle because crystallization takes place quickly from
solution and also because of strong ionic bond. (More over crystalline solids can be easil cleaved along certain planes)
8) Calculate the efficiency of packing in case of metal crystal for :-
SC
Let cell edgelength be a and radius of an atom be r
a = 2r
Vol. of cube = a3 = 8r3
z = 1; Volume of one sphere = (4/3) (r3
Efficiency of packing= vol. of all spheres in one cube x 100/ Vol. of cube
= [(4/3) (r3 x 100 ]/ 8r3
= 52.3%
BCC
Body diagonal = 4r
(4r)2=a2+((2a)2=a2+2a2=3a2
16r2=3a2
a=4r/(3
Vol. of cube =a3 = 64r3/3(3
z=2; Vol. of two spheres = (4/3) (r3 x 2
Efficiency of packing= vol. of two spheres x 100/ Vol. of cube
=[2x(4/3)(r3 x 100] / [64r3/3(3]
= 68%
FCC
Face diagonal = (2a = 4r (or) a = 4r/(2
Vol. of cube = a3 = 64r3/2(2,
z = 4 ; Vol. of 4 spheres = 4 x (4/3) (r3
Efficiency of packing = [(16/3)(r3 x 100] / 64r3/2(2
= 74%
9) Silver crystallizes in FCC lattice. If edge length of unit cell is 4.077 x 10-8 cm and density is
10.5 g/cm3 , calculate atomic mass of silver.
ρ= (zM) /(a3 x NA)
10.5 = (4 x M) / [(4.077 x 10-8)3 x 6.023x1023]
M = 107.09 g/mol
10) A cubic solid is made of two elements P & Q. Atoms Q are at the corners of the cube and P
are at the body center. What is the formula of the compound. What are the Co-ordination numbers of P & Q.
No of Q atoms per unit cell = 8 x 1/8 = 1
No of P atoms per unit cell = 1
Formula of compound =PQ
Co-ordination no. of P =8
Co-ordination no. of Q =8
11) Niobium crystallizes in body centred cubic structure.. If Density is 8.55 g/cm3. Calculate atomic radius of niobium (At. Mass of Niobium=93).
ρ=(zM)/(a3NA)
8.55=(2 x 93) / [a3 x 6.023 x 1023]
a3=2x93/(8.55x6.023x1023)
a=[3.6x10-23 ]1/3 = 3.302 x 10-8 cm = 3.302 x 10-10 m
For BCC unit cell (3a = 4r : r = (3a/4 = 1.732 x 3.3 x 10-10 m / 4 = 14.29 nm
12) If the radius of octahedral void is ‘r’ and the radius of atoms in close packing is R, derive a
relation between r & R.
Sphere in void
AB2 = AD2 + BD2
(2r + 2R)2 = (2R)2 + (2R)2 = 8R2
2r + 2R = 2(2R
r + R = (2R
( r/R) + 1 = (2
= 1.414
r/R = 0.414
13) Copper crystallizes in face lattice with edge length 3.61 x 10-8 cm. Show that the calculated
density is in agreement with its measured value of 8.92 g/cm3.
a = 3.61x10-8 cm, z = 4,
p = (4x63.5) / [(3.61x10-8)3 x 6.023x1023 ]
= 8.963 g/cm3
14. Formula mass of NaCl is 58.45 g/mol and density of it’s pure form is 2.167 g/cm3.
Average distance between adjascent sodium and chloride ions in the crystal is 2.814x10-8 cm.
Calculate Avogadro constant.
M =5 8.45 g/mol
ρ = 2.167 g/cm3
a = 2 x distance between Na+ and Cl- = 2 x 2.814x10-8 cm = 5.628 x10-8 cm
ρ = (zM)/(a3 x NA)
NA = 4 x 58.45 / [(5.628x10-8)3 x (2.167)] = 6.0523x1023
15. Analysis shows that nickel oxide has formula Ni0.98 O1.00. What fractions of Nickel exist as Ni2+ and Ni3+ ion?
Let no of Ni2+ be ‘x’ and the no. of Ni3+ be ‘y’.
x + y = 0.98( (1)
The total positive charge per formula unit = 2+, since charge of one oxide ion is 2-. Therefore
2x + 3y = 2( (2)
Multiply (1) with 2 => 2x + 2y = 1.96( (3)
From (3) and (2) y = 0.04 & x = 0.94
No. of Ni3+ = 0.04, No of Ni2+ = 0.94
% of Ni3+ = 0.04x100/0.98 = 4.08%
% of Ni2+ = 95.92%
16. The compound CuCl has ZnS structure. Its density is 3.4 g/cm3 . What is the length of the edge of the unit cell?
ρ = (zM)/(a3 x NA) ie, a = [(4x99)/(3.4x6.023x1023)]1/3 = (19.34 x 10-23)1/3
= 5.75 x 10-8 cm
17. If the radius of the Bromide ion is 0.182 nm, how large a cation can fit in each of the tetrahedral hole?.
For tetrahedral void, 0.225 ≤ (r+/ r-) < 0.414
When r+ is maximum possible, the radius ratio(r+/ r-) should have maximum limit.
(ie) r+ = 0.414 x 0.182 = 0.0753 nm,
A cation having radius upto 0.0753 nm can fit into tetrahedral hole.
18. The first order diffraction of X-rays from a certain set of crystal planes occurs at an angle of 11.80 from the planes occurs at an planes are 0.281 nm apart, what is the wavelength of X-rays? n( = 2d sin(, ( = 11.80
( = 2 x 0.281nm x Sin11.80 = 2 x 0.281nm x 0.2045 = 0.115 nm
19. What is a semi conductor ? describe two main types of semiconductors and contrast their conduction mechanism.
Solids with conductivity intermediate between that of insulators and conductors are called semiconductors. There are 2 types.
P-Type semi conductor N-type semiconductor 1. produced by doping group 13 elements with Si or Ge Produced by doping group 15 elements wih Si or Ge Holes are responsible for conductivity Electrons are responsible for conductivity.
20. Non Stochiometric cuprous oxide Cu2O can be prepared in the laboratory. If this oxide copper to oxygen ratio is slightly less than 2:1. Can you account for the fact that this substance is a P-type semiconductor?
Ans : If ratio is less than 2:1, there will be cuprous ion in addition to cupric ion. During formation of cuprous ion, only one electron is given to ‘O’. Here oxygen does not get sufficient electron in its’ valence shell. This lack of electron leads to formation of holes and substance becomes p-type semiconductor.
21. Ferric oxide Crystallises in a hexagonal close packed array of oxides with two out of every three octahedral holes occupied by ferric ions. Derive the formula of ferric oxide. 3 Oxide ions are required to make 3 octahedral voids
No. of Ferric ions for each oxide ion = 2/3 Chemical formula= Fe2/3 O ( Fe2O3
22. Classify each of the following as p-type and n-type semiconductors.
i) Ge doped with In : Ans : p-type ii) B Doped with Si Ans : n-type
23. Thallium chloride, TlCl crystallizes in either a simple cubic lattice or a face centred cubic lattice of Cl- ion with Tl+ ion in the holes. If the density of the solid is 9 gcm-3 and edge of the unit cell is 3.85 x 10-8 cm, what is the unit cell geometry? p=zM/a3NA ,
9 = (z x 240)/[(3.85x10-8)3x6.023x1023]
z = 9 x 3.85 x 3.85 x 3.85 x 10-24 x 6.023x1023 / 240
z =1.289 ( 1
So TlCl is simple cubic.
24. Gold (atomic radis =0.144 nm). Crystallises in face centred Unit cell. What is the length of a side of a cell?
For FCC, a = 2(2r = 2 x 1.414 x 0.144nm = 0.4072 nm
25. In terms of band theory, what is the difference (1) between a conductor and an insulator Ans: Energy gap between valence band and conduction band is high in insulator. In conductors, there is no energy gap between these bands.
(2) Conductors & Semi conductors In semiconductors there is energy gap between valence band and conduction band, but not so high as in insulators. In conductors there is no energy gap.
26. Schottky defect Vacancy of one ion followed by the vacancy of oppositly charged ion to maintain electrical neutrality in ionic crystals. In NaCl Schottky defect is found. Frenkel defect An ion leaves from it’s normal site and occupies an interstitial void eg. In AgBr, it is found.. Interstitials Atoms which occupy the normally vacant interstitial positions in a crystal are called interstitials. Eg. AgBr F-Centres Electrons trapped in anionic vacancies are called F-centres. Eg. F-centres are created when KCl crystal is heated in the atmosphere of potassium vapour.
27. The ions of NaF and MgO all have the same no. of electrons, and the intermolecular distances are about the same (235pm & 215pm). Why then are the melting points of NaF and MgO so different (9920 C& 2640C).
NaF contains Na+ and F- ions MgO contains Mg2+ and O2- ions
Na+ is monopositive and F- is mononegative. Mg2+ is dipositive and O2- is dinegative.
(Mg 2+ has smaller size and greater charge). Interionic attraction in MgO, are stronger than that of NaCl, since MgO contains dipositive and dinegative ions.
28. (a) Explain radius ratio. Ans: Radius ratio is (r+/r-) where r+ is cation radius and r- is anion radius.
Radius ratio is r+ /r- Co-ordination no. 0.732 and more 8 0.414- 0.732 6 0.225-0.414 4 b) Predict the form of the crystal structure in each case.
Li+=74 Rb+=140
Na+=102 Cs+=170
K+=138 Br-=195
r+ /r- for Li+Br- = 74/195 = 0.379 Li+ is in tetrahedral void of Br-
r+ /r- for Na+Br- = 102/195 = 0.523 Na+ is in octahedral void of Br-
r+ /r- for K+Br- = 138/195 = 0.708 K+ is in octahedral void of Br-
r+ /r- for Rb+Br- = 140/195 = 0.718 Rb+ is in octahedral void of Br-
r+ /r- for Cs+Br- = 170/195 = 0.872 Cs+ is in cubic void of Br-
29. Aluminium crystalises in a cubic close packed structure. Its’ metallic radius is 125 pm.
(a) What is the length of the side of the unit cell.
Ans For CCP ie FCC unit cell: 4r = (2a,
a = 4 x 125pm / 1.414 = 354pm
(b) How many unit cells are there in 1cm3 of aluminium?
Volume of one unit cell of Al = a3,
(354x10-10 cm)3 = 354x354x354x10-30 cm3
No. of unit cells in 1cm3=1/(354x354x354x10-30) = 2.25 x1022
30. If Nacl is doped with 10-3 mol%SrCl2 , what is the concentration of cation vacancies.
Let us take 1 mol NaCl.
Amount of Sr 2+ doped = 10-3 mol% = 10-3 mol / 100 = 10-5 mol
No of Sr 2+ doped = 10-5 mol x 6.022 x 1023 mol-1 = 6.022 x 1018
Each Sr 2+ doped, creates one cation vacancy.
Thus concentration of cation vacancies in one mol NaCl = 6.022 x 1018 mol-1
31. KF has NaCl structure. What is the distance between K+ and F- in KF. If density is 2.48 g/cm3.
ρ = zM/a3NA ,
a =[ (4x58)/(2.48x6.02x1023)]1/3 = 5.374 x 10-8cm
Distance b/w K+ & F- = ½ a = ½ (5.374 x 10-8 cm) = 2.687x10-8 cm = 269 pm
32. Calculate the value of Avogadro constant from the following data. Density of NaCl = 2.165 g/cm3. Distance b/w Na+ & Cl- is 281 pm.
ρ = zM/a3NA , NA= (4 x 58.5)/ [2.165x(562x10-10)3] = 6.023x1023 mol-1
33. Explain the following with suitable examples. a) Ferromagnetism: A spontaneous alignment of magnetic moments in same direction gives rise to ferromagnetism. Eg. CrO2 . Ferromagnetic substances are strongly altered by magnet.
b) Paramagnetism: Paramagnetic substances are weakly attracted by magnet. Eg. O2 c) Ferrimagnetism: When magnetic moments are aligned in opposite direction in unequal numbers it leads to Ferrimagnetism. d) Piezoelectric effect: When certain crystals (in which the net dipole moment is zero) are mechanically stressed, electricity is produced.
e) Antifluorite structure: Crystals in which positive and negative ions are found in the reversed positions of fluorite structure. Eg. Na2O.
f) 12-16 & 13.15 compounds
12-16 compounds : Compounds found b/w group 12 & 16. eg. ZnS, CdS, HgTe 13-15 compounds : Compounds found b/w group 13 & 15, eg InSb, AlP, GaAs
34. Define the term amorphous. Give an example of amorphous solid.
Morphous means crystalline. Amorphous means non-crystalline. Eg. glass, amorphous silicon.
35. what makes a glass different from a solid such as quartz? Under what condition could quartz be converted to glass?
Glass is amorphous. Quartz is crystalline.
Quartz can be converted to glass by melting followed by cooling very rapidly.
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