SOLID STATE STUDY MATERIAL

L-1

THE SOLID STATE

·         POINTS TO BE REMEMBERED:-

·         The state of matter whose Melting Point is above room temp is solid state.

·         Solids have definite shape and volume, having high density and constituent particles are held strongly.

·         Crystalline solids have regular arrangement of constituent particles throughout, melting point is sharp, Anisotropic (Some Physical Properties Like Refractive Index, Electrical Conductance may vary in different directions) in nature and give clear cut cleavage.    

·         Amorphous solids have no regular arrangement, no sharp Melting Point isotropic (Some Physical Properties Like Refractive Index, Electrical Conductance may not vary in different directions) in nature they do not exhibit cleavage property.

·         Amorphous silica is used in photovoltaic cells.

·         Space lattice is a regular three dimensional arrangement of constituent particles in the crystalline solid.

·         Smallest repeating unit in a space lattice is called unit cell.

·         There are 4 types of unit cells, 7 crystal systems and 14 Bravais lattices.

·           Types of unit cell	No. of atoms per unit cell
i. Simple cubic unit cell	8	X 1/8=1
ii. FCC (Face centered cubic)	8	X 1/8 + 6 X 1/2= 4
iii. BCC (Body centered cubic)	8	X 1/8 + 1 X 1 = 2

·         Hexagonal close packing and cubic close packing have equal efficiency i.e 74% and coordination no. is 12.

·         Coordination no.: The number of nearest neighbour points surrounding a particular point is called coordination no (point may be atom, ions & molecules).           

·         Packing efficiency = (volume occupied by total spheres     ) x 100

·                                                 Volume of unit cell

·         For simple cubic unit cell the packing efficiency = 1x4/3 πr3/8xr3x100 = 52.4%

·         The Packing efficiency in BCC = 2x4/3 πr³/64x33/2 r³ x100 = 68%

·         The Packing efficiency in FCC = 4x4/3 πr³/16x21/2 r³ x100 = 74%

·         Packing efficiency in simple cubic unit cell is 52.4%, bcc arrangement in 68% and fcc is 74%.

·         Unoccupied spaces in solids are called interstitial voids or interstitial sites.

·         Two important interstitial voids are (i). Tetrahedral void and (ii). Octahedral void.

·         Radius ratio is the ratio of radius of void to the radius of sphere.

·         For tetrahedral void radius ratio =0.225  For octahedral void radius ratio=0.414

·         No. of tetrahedral void=2 X N (N=No. of closed packed particles)

·         No. of octahedral void=N

·         Formula of a compound depends upon arrangement of constituent particles in the unit cell.

·         Density of unit cell

·         D   =

·         D=density, M=Molar mass, a=side of unit cell (edge length) , NA=6.022 X 10²³

·         The relationship between edge length and radius of atom and interatomic or interionic distance for different types of unit cell is as given below

a.	Simple cubic unit cell                         a=2r
b.		FCC	a=4r/√2
c.	BCC                                             a=4r/√3

·         Interatomic distance=2r

·         Interionic distance=Rc + Ra (Rc=Radius of cation, Ra=Radius of anion)

·         Imperfection is the irregularity in the arrangement of constituent particles.

·         Point defect or Atomic defect - It is the deviation from ideal arrangement of constituent atom. Point defects are two types (a) Vacancy defect (b) Interstitial defect

·         Vacancy defect lowers the density.

·         Interstitial defect increases the density of crystal.

·         Point defects in the ionic crystal may be classified as:

a.       Stoichiometric defect (Ratio of cation and anion is same).

b.       Non Stoichiometric defect (disturb the ratio).

c.       Impurity defects (due to presence of some other ions at the lattice sites)

·         Schottky defect arises due to missing of equal number of cations and anions from lattice sites in the crystalline solid and it lowers the density of crystal e.g. Alkali halides NaCl,KCl,CsCl etc.

·         Frenkel defect is the combination of vacancy and interstitial defects. Cations leave their actual lattice sites and occupy the interstitial space in the solid. In this defect density remains same e.g. AgCl,ZnS,AgI etc.

·         AgBr is the compound which shows both Schottky Defect and Frenkel Defect.

·         Non stoichiometric defect

1. Metal excess defect due to anion vacancy.
2. Metal excess due to presence of extra cation.
3. Metal deficiency due to absence of cation.

·         F-Center - In metal excess defect, trapping of electrons in the anion vacancies which act as color center. E.g. NaCl gives yellow color in excess of Na⁺ ions, excess of Li makes LiCl crystal pink and excess of K makes KCl crystal violet.

SHORT ANSWER QUESTION (1)

Q1. What do you mean by paramagnetic substance?

Ans.: - Substances which are attracted by external magnetic field are called paramagnetic substances. The paramagnetic property is due to the presence of unpaired electrons in atoms or ions e.g. Cu^2+, Fe³⁺ etc.

Q2. Which substance exhibit schottky and Frenkel both defects.

Ans: - AgBr

Q3. Name a salt which is added to AgCl so as to produce cation vacancies.

Ans.:-CdCl₂

Q4. Why Frenkel defects not found in pure Alkali metal halide.

Ans: - Due to larger size of Alkali metal ions.

Q5. What is the use of amorphous silica?

Ans.:- It is used in Photovoltaic cells.

Q6. Analysis shows that a metal oxide has the empirical formula M0.98 O. Calculate the percentage of M²⁺ and M³⁺ ions in the crystal.

Ans: - Let the M²⁺ ion in the crystal be x and M³⁺ = 0.98-x                         

Since total charge on the compound must be zero

   2x+3(0.98-x)-2=0 X=0.88

%of M²⁺ = (0.88/0.96) X 100=91.67

%of M³⁺ =100-91.67=8.33

Q7. What is the co-ordination no. of cation in Antifluorite structure?

Ans: - 4

Q8. What is the Coordination Number of cation and anion in Caesium Chloride.

Ans: 8 and 8

Q9. What is F centre?

Ans.:- F-Center - In metal excess defect, trapping of electrons in the anion vacancies which act as colour center. e.g. NaCl gives yellow color in excess of Na⁺ions,excess of Li makes LiCl crystal pink and excess of K makes KCl crystal violet.

Q10. What makes Alkali metal halides sometimes coloured, which are otherwise   colourless?                            

 Ans.:- It is due to F-Center.

      Q11. What happens when a ferromagnetic substance is heated to high temperature?

Ans: Ferromagnetic substance changes to paramagnetic substance due to randomization of domains (spins) on heating.

Very Short Answers (1 marks) :

1. How does amorphous silica differ from quartz?

-In amorphous silica, SiO₄ tetrahedral are randomly joined to each other whereas in quartz they are linked in a regular manner.

2. Which point defect lowers the density of a crystal?

-Schottky defect.

3. Why is glass called supper cooled liquids?

-It has tendency to flow like liquid.

4. Some of the very old glass objects appear slightly milky instead of being transparent. Why?

-Due to conversion amorphous glass into crystalline.

5. What is anisotropy?

-Physical properties show different values when measured along different direction in crystalline solids.

6.  What is the coordination number of atoms?

a)      in fcc structure               b) in bcc structure

a) 12                                  b) 8

7.

	How many lattice points are there in unit cell of
a)		fcc	b) bcc	c)simple cubic
Ans: a 14			b) 9	c) 8

8. What are the co-ordination numbers of octahedral voids and tetrahedral voids?

-6 and 4 respectively.

9. Why common salt is sometimes yellow instead of being pure white?

-Due to the presence of electrons in some lattice sites in place of anions these sites act as F-centers. These electrons when excited impart color to the crystal.

10. A compound is formed by two elements X and Y. The element Y forms ccp and atoms of X occupy octahedral voids. What is formula of the compound?

-No. of Y atoms be N                         No. of octahedral voids N

No. of X atoms be =N                        Formula XY

HOTS Very Short Answers:

1. Define F centers.
2. What type of stoichiometric defect is shown by
1. ZnS
2. AgBr
3. What are the differences between frenkel and schottky defect?
4. Explain the following with suitable examples

o        Ferromagnetism

o        Paramagnetism

o        Ferrimagnetism

5. In terms of band theory what is the difference between

o        Conductor and Insulator

·         Conductor and Semi-conducto

6.      Solid A is a very hard electrical insulator in solid as well as in   molten state and melts at extremely high temperature. What type of solid is it?

      7. What is meant by “doping” in a semiconductor?

Short Answers (2 Marks): HOTS

1. Explain how electrical neutrality is maintained in compounds showing Frenkel and Schottky defect.

-In compound showing Frenkel defect, ions just get displaced within the lattice. While in compounds showing Schottky defect, equal number of anions and Cations are missing from the lattice. Thus, electrical neutrality is maintained in both cases.

2. Calculate the number of atoms in a cubic unit cell having one atom on each corner and two atoms on each body diagonal.

8 corner X (1/8) atom per unit cell = 1 atom

-There are four body diagonals in a cubic unit cell and each has two body centre atoms. So 4 X 2=8 atoms therefore total number of atoms per unit cell =1+8=9

3.      Gold crystallizes in an FCC unit cell. What is the length of a side of the cell (r=0.144 nm) r=0.144 nm

a=2 X √2r

=2 X 1.414 X 0.144 nm

=0.407 nm

4. Classify each of the following as either p-type or n-type semi-conductor.

a)  Ge doped with In

b)  B doped with Si

Ge is group 14 elements and In is group 13 element. Therefore, an electron deficit hole is created. Thus semi-conductor is p-type.

(a)  Since B is group 13 element and Si group 14 element, there will be a free electron, thus it is n-type semi-conductor.

5.      In terms of band theory what is the difference between a conductor, an insulator and a semi-conductor?

-The energy gap between the valence band and conduction band in an insulator is very large while in a conductor, the energy gap is very small or there is overlapping between valence band and conduction band. In semi-conductor there is small energy gap between the valence band and conduction band.

6.  CaCl₂ will introduce Schottky defect if added to AgCl crystal. Explain

Two Ag⁺ ions will be replaced by one Ca²⁺ ions to maintain electrical neutrality. Thus a hole is created at the lattice site for every Ca²⁺ ion introduced.

7.      The electrical conductivity of a metal decreases with rise in temperature while that of a semi-conductor increases. Explain.

In metals with increase of temperature, the kernels start vibrating and thus offer resistance to the flow of electrons. Hence conductivity decreases. In case of semi-conductors, with increase of temperature, more electrons can shift from valence band to conduction band. Hence conductivity increases.

8.      What type of substances would make better permanent magnets, ferromagnetic or ferrimagnetic, why?

Ferromagnetic substances make better permanent magnets. This is because the metal ions of a ferromagnetic substance are grouped into small regions called domains. Each domain acts as tiny magnet and get oriented in the direction of magnetic field in which it is placed. This persists even in the absence of magnetic field.

9.      In a crystalline solid, the atoms A and B are arranged as follows:-

a.       Atoms A are arranged in ccp array.

b.      Atoms B occupy all the octahedral voids and half of the tetrahedral voids. What is the formula of the compound?

Let no. of atoms of A be N No. of octahedral voids = N

No. of tetrahedral voids= 2N

i)                    There will be one atom of B in the octahedral void

ii)                  There will be one atom of B in the tetrahedral void ((1/2) X 2N)

Therefore, total 2 atoms of B for each atom of A

Therefore formula of the compound =AB₂

10.  In compound atoms of element Y forms ccp lattice and those of element X occupy 2/3^rd of tetrahedral voids. What is the formula of the compound?

No. of Y atoms per unit cell in ccp lattice=4

No. of tetrahedral voids= 2 X 4=8

No. of tetrahedral voids occupied by X= (2/3) X 8=16/3

Therefore formula of the compound =X_16/3 Y₄

=X16 Y12

=X₄ Y₃

HOTS Short Answer:

1.  How many lattice points are there in one unit cell of the following lattices?

o        F.C.C.

o        B.C.C.

o        S.C.C. (Simple Cubic Cell)

2. A cubic solid is made of two elements X and Y. Atom Y are at the corners of the cube and X at the body centers. What is the formula of the compound?

3. Silver forms ccp lattice and X –ray studies of its crystal show that the edge length of its unit cell is 408.6 pm. Calculate the density of silver (Atomic mass= 107.9 u).

4. A cubic solid is made up of two elements P and Q. Atoms of the Q are present at the corners of the cube and atoms of P at the body centre. What is the formula of the compound? What are the co-ordination number of P and Q.

5. What happens when:-

o        CsCl crystal is heated

o Pressure is applied on NaCl crystal.

Short Answers (3 marks):

1.      The density of chromium is 7.2g cm^-3. If the unit cell is a cubic with length of 289pm, determine the type of unit cell (Atomic mass of Cr=52 u and N_A = 6.022 X 10²³ atoms mol^-1).

Z=? , a= 289 pm =289×10^-10 cm, M=52g mol^-1, d=7.2g cm^-3

2. An element crystallizes in FCC structure; 200 g of this element has 4.12 X 10²⁴ atoms. If the density of A is 7.2 g cm^-3, calculate the edge length of unit cell.
3. Niobium crystallizes in bcc structure. If its density is 8.55 g cm^-3, calculate atomic radius of

Niobium .[ At. Mass of Niobium = 92.9u, N_A = 6.022 X 10²³ atoms mol^-1 ].

4. If radius of octahedral void is r and radius of atom in close packing is R, derive the relationship between r and R.
5. Non stoichiometric cuprous oxide can be prepared in the laboratory. In this oxide, copper to oxygen ratio is slightly less than 2:1 can you account for the fact that the substance is a p-type semiconductor?
6. The unit cell of an element of atomic mass 50 u has edge length 290pm. Calculate its density the element has bcc structure (N_A = 6.022 X 10²³ atoms mol^-1).
7. Calculate the density of silver which crystallizes in face centered cubic form. The distance between nearest metal atoms is 287pm (Ag= 107.87g mol^-1, N_A= 6.022 X 10²³).
8. What is the distance between Na⁺ and Cl^- ions in NaCl crystal if its density 2.165 g cm^-3.NaCl crystallizes in FCC lattice?
9. Analysis shows that Nickel oxide has Ni _0.98 O _1.00 what fractions of nickel exist as Ni²⁺ ions and Ni³⁺ ions?
10. Find the type of lattice for cube having edge length of 400pm, atomic mass = 60 and density =6.25 g/cc.

11.  The density of copper metal is 8.95 g cm^–3. If the radius of copper atom

Be127.8 pm, is the copper unit cell simple cubic, body-centred cubic or face-centred cubic? (Given: atomic mass of Cu = 63.54 g mol^–1 and

NA = 6.02 ×10²³ mol^–1)

12.  What is a semi-conductor? Name the two main types of semi-conductors and explain their conduction mechanism.

HOTS Short Answer:

1. Aluminium crystallizes in cubic closed pack structure. Its metallic radius is 125 pm

oWhat is the length of the side of the unit cell?

oHow many unit cell are there in 100 cm³of Aluminium.

2. Classify the following as either p-type or n-type semiconductors.

Ge doped with In

B doped with Si

3.      Zinc oxide is white but it turns yellow on heating. Explain.

4.      What type of semiconductor is obtained when silicon is doped with arsenic?

5.      What type of non-stoichiometric point defect is responsible for the pink colour of LiCl?

6.      Write a distinguishing feature between a metallic solid and an ionic solid.

Long Answer (3 Marks):

1. In a face centered cubic lattice Edge length of lattice cell is 2A⁰. The density of metal is 2.4 g cm^-3. How many units cell are present in 200g of metal.

2.      A metal crystallizes as face centered cubic lattice with edge length of 450pm. Molar mass of metal is 50g mol^-1. Calculate the density of metal.

3. A compound forms hexagonal close packed structure. What is the total number of voids in 0.5 mol of it? How many of these are tetrahedral voids?
4. Copper Crystallizes into FCC lattice with edge length 3.61 X 10^-8 cm. Show that calculated density is in agreement with measured value of 8.92 g/cc.
5. Niobium crystallizes in bcc structure with density 8.55 g/cc, Calculate atomic radius using atomic mass i.e. 93 u.

HOTS Long Answer:

1. The  compound CuCl has Fcc structure like ZnS, its density is 3.4 g cm^-3. What is the length of the edge of unit cell?

Hint: d=((Z X M) /(a³ X N_A)

a³=(4 X 99) / (3.4 X 6.022 X 10²³)

a³=193.4 X 10-24 cm³

a=5.78 X 10^-8cm

2.      If NaCl is doped with 10^-3 mol% SrCl₂. What is the concentration of cation vacancies?

3.      The edge length of the unit cell of metal having molecular mass 75 g/mol is 1 A⁰ which crystallizes into cubic lattice. If the density is 2g/cm³ then find the radius of metal atom (N_A = 6.022 X 10²³)

4.      Iron has a body centred cubic unit cell with a cell dimension of 286.65 pm.    The   density of iron is 7.874 g cm–3. Use this information to calculate    Avogadro’s   number. (Atomic mass of Fe = 55.84 g mol–1)

VALUE BASED QUESTIONS

Q.1. Sudanshu made a model of the unit cell of diamond. It resembled theunit cell of ZnS. If the unit cell of ZnS has 4 units of ZnS per unit cell. Ithas the same packing efficiency as ZnS. But diamond is the hardestknown substance.

a. What is the number of atoms of carbon per unit cell of diamond?

b. Why?

c. What is the value that Sudanshu can derive from these facts?

Ans: a) The number of atoms of Carbon per unit cell is 8 in diamond.

b)The C—C bond is very strong in diamond (due to small size of Carbon)

Unlike the Zn—S bond in ZnS.

c) Though from the same background ie with the same structure theproperty can be different, thus, with a little effort, we can do something differently and bring about major changes.

Q.2 Amorphous solid have number of applications, for example glass is used in buildings, rubber in tyre, shoe soles etc. Plastic in making bucket, toys etc. Amorphous silica in photovoltaic cell etc. On the other hand common salt, sugar, urea etc are crystalline solid which are equally useful in our everyday life. Now answer the following questions?

a) Rubber and synthetic polymers are all amorphous material .They are used in making shoe soles, car mat etc. which one do you think is better and why? How do you compare PVC shoes with leather shoes?

b) Our country is facing problem of shortage of electricity. Use of photovoltaic cell is an excellent solution. How can they contribute to meet the shortage?

c) Is Polyethene used in making polyethene bags a crystalline material or an amorphous material? Why polyethene bag should be banned?

Ans: a)PVC is better than rubber because the former can be given different beautiful colours, can be easily moulded to desired shape, has greater durability as it is more resistant to abrasion. However it is sticky in hot weather. Leather shoes are the best because they can breadth and keep the feet cool, can be easily repaired and are the enviorment friendly because leather is biodegradable.

b) Photovoltaic cell have limited power production. hence they can notbe used on a very large scale.However they can make significant contribution towards street lighting, home lighting, solar geyseretc .

c)Polyethene is polymer and is an amorphous solid. It’s used need to be banned because it is not biodegradable . It may therefore pileup as a waste and swallowed by animal causing death or choke the severe.

*****************