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SOLID STATE
KEY CONCEPTS
As we know that matter exists in different physical states under different conditions of temperature and pressure. For example solid state, liquid gases plasma and BEC etc. Now we will study about different aspects of solid state. Introduction:
- The state of matter whose M.P is above room temp is solid state. Solids have definite shape and volume, having high density and constituent particles are held strongly.
- Based on arrangement of particles types of solid : 1: Crystalline 2:Amorphous
- Crystalline solids have regular arrangement of constituent particles throughout, melting point is sharp, Anisotropic in nature and give clear cut cleavage.
- Amorphous solids have no regular arrangement, no sharp M.P, isotropic in nature they do not exhibit cleavage property.
- Amorphous silica is used in photovoltaic cells.(Applications of amorphous solid)
- Space lattice is the regular 3D, arrangement of constituent particles in the crystalline solid. It shows how the constituents particles(atoms, molecules etc.) are arranged.
- 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 81/8= ii. FCC (Face centered cubic) 81/8+61/2= iii. BCC (Body centered cubic) 81/8+1*1=
- Hexagonal close packing and cubic close packing have equal efficiency i.e 74%
- Packing efficiency =volume occupied by spheres (Particles)/volume of unit cell *
- For simple cubic unit cell the p.f.=1*4/3 πr^3 /8r 3 *100 =52.
- The packing efficiency in fcc =4*4/3 πr 3 /162 1/2^ r 3 *100 =
- The packing efficiency in bcc =2*4/3 πr^3 /643 3/2^ r 3 *100 =
- The packing efficiency in hcp =
- Packing efficiency in bcc arrangement in 68% and simple cubic unit cell is 52.4%
- 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. a. For tetrahedral void radius ratio=0. For octahedral void radius ratio=0.
- No. of tetrahedral void=2*N (N=No. of particles)
- No. of octahedral void=N
- Formula of a compound depends upon arrangement of constituent of particles.
- Density of unit cell D=Z*M/a^3 NA D=density, M=Molar mass, a=side of unit cell, NA=6.02210 23
- The relationship between edge length and radius of atom and interatomic or interionic distance for different types of unit is different as given below a. Simple cubic unit cell a=2R b. F C C a=4R/ c. B C C a=4R/
- Interatomic distance=2R
- Interionic distance=Rc+Ra (Rc=Radius of cation, Ra=Radius of anion)
- Imperfection is the ir‐regularty 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 and
Q6. Analysis shows that a metal oxide has the empirical formula Mo.98 0 1.00. Calculate the percentage of M2+^ and M 3+^ ions in the crystal. Ans: ‐ Let the M2+^ ion in the crystal be x and M3+^ =0.98‐x Since total change on the compound must be zero 2x+3(0.098‐x)‐z= X=0. %of M2+^ 0.88/0.96*100=91. % of M3 +^ =100‐91.91.67=8. Q7. What is the co‐ordination no. of cation in Antifluorite structure? Ans: ‐ 4 Q8. What is the Co.No. of cation and anion in Caesium Chloride. Ans: 8 and 8 Q9. What is F centre? Ans:‐ It is the anion vacancy which contains unpaired electron in non‐stoichiometric compound containing excess of metal ion. Q10. What makes Alkali metal halides sometimes coloured, which are otherwise colourless? Very Short Answers(1 marks) :
1. How does amorphous silica differ from quartz? In amorphous silica, SiO 4 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 glass is 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 crystallization. 5. What is anisotropy? Physical properties show different values when measured along different 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 one cell of ‐ a) fcc b) bcc c) simple cubic 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 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
4. Classify each of the following as either a p‐type or n‐type semi‐conductor. a) Ge doped with In b) B doped with Si (a) Ge is group 14 elements and In is group 13 element. Therefore, an electron deficit hole is created. Thus semi‐conductor is p‐type. (b) Since b group 13 element and Si is group 14 elements, 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. 6. CaCl 2 will introduce Scotty defect if added to AgCl crystal. Explain Two Ag+ ions will be replaced by one Ca 2+^ ions to maintain electrical neutrality. Thus a hole is created at the lattice site for every Ca 2+^ 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 ferromagnetic, 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*2N) Therefore, total 2 atoms of b for each atom of A Therefore formula of the compound =AB 2
10. In compound atoms of element Y forms ccp lattice and those of element X occupy 2/3rd^ of tetrahedral voids. What is the formula of the compound? No. of Y atoms per unit cell in ccp lattice= No. of tetrahedral voids= 24= No. of tetrahedral voids occupied by X= 2/38=16/ Therefore formula of the compound =X16/3 Y 4 =X 16 Y (^12) =X 4 Y 3 HOTS Short Answer: 1. How many lattice points are there in one unit cell of the following lattices? o FCC o BCC o SCC 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 wt= 107.9u). 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):
- 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 = 6.02210 23 atoms mol‐1).* d= Z * M
**3. Zinc oxide is white but it turns yellow on heating. Explain. Long Answer(5 Marks):
- It is face centered cubic lattice A metal has cubic lattice. Edge length of lattice cell is 2A^0. The density of metal is 2.4g cm‐3.**^ **How many units cell are present in 200g of metal.
- A metal crystallizes as face centered cubic lattice with edge length of 450pm. Molar mass of metal is 50g mol‐^1. The density of mental is?
- A compound forms hexagonal close packed structure. What is the total number of voids in 0. mol of it? How many of these are tetrahedral voids?
- Copper Crystallizes into FCC lattice with edge length 3.6110‐*^8 **cm. Show that calculated density is in agreement with measured value of 8.92g/cc.
- Niobium crystallizes in bcc structure with density 8.55g/cc, Calculate atomic radius using atomic mass i.e. 93u. HOTS Long Answer:
- The compound CuCl has Fu structure like ZnS, its density is 3.4g cm‐^3. What is the length of the edge of unit cell? Hint: d=Z X M /a3 X NA a**^3 =4X99 / 3.4 X 6.022 X 10 2. a^3 =193.4 X 10 ‐ 24 cm 3 a=5.78 X 10 ‐^8 **cm
- If NaCl is dropped with 10 ‐**^3 **mol% SrCl 2. What is the concentration of cation valancies?
- If the radius of the octahedral void is r and the radius of the atom in the close packing is R. derive relationship between r and R.
- The edge length of the unit cell of mental having molecular weight 75g/mol is A*^0 which crystallizes into cubic lattice. If the density is 2g/cm^3 then find the radius of metal atom (N (^) A = 6.02210^23 **)
- The density of K Br. Is 2.75 gm cm ‐^3. the length of edge of the unit cell is 654 pm. Predict the type of cubic lattice to which unit cell of KBr belongs. NA=6.02310*^23 ; at mass of K=39: Br. = 80 Ans. Calculate value of z= 4 so it has fcc lattice
6. CsCl has bcc arrangement and its unit cell edge lenth is 400 pm. calculate the interionic **distance of CsCl. Ans. 34604 pm
- The radius of an Iron atom is 1.42 A 0. It has rock salt structure. Calculate density of unit cell.** Ans. 5.74 g cm‐^3 8. What is the distance between na+^ and Cl‐^ in a NaCl crystal if its density is 2.165 gcm‐^3 NaCl **crystalline in the fcc lattice. Ans.281PM
- Copper crystalline with fcc unit cell. If the radius of copper atom is 127.8 pm. Calculate the** density of copper metal. At. Mass of Cu=63.55u NA= 6.0210^23 Ans.a=2√ 2 .r , a3=4.72310‐^23 , d=8.95 5.74 g cm‐^3
It changes with changes temperature.
- Parts per million (ppm) concentration of very dilute solution is expressed in ppm. Ppm = Vapor pressure – It is defined as the pressure exerted by the vapour of liquid over the liquid over the liquid in equilibrium with liquid at particular temperature vapour pressure of liquid depends upon nature of liquid and temperature. Roult’s Law – 1. For the solution containing non‐volatile solute the vapor pressure of the solution is directly proportional to the mole fraction of solvent at particular temperature PA X (^) A P (^) A = P (^0) A .X (^) A 2. For the solution consisting of two miscible and volatile liquids the partial vapor pressure of each component is directly proportional to its own mole fraction in the solution at particular temperature. P (^) A =P (^0) A. X (^) A, P (^) B =P (^0) B .X (^) B And total vapor pressure is equal to sum of partial pressure. P (^) total = P (^) A + P (^) B Ideal solution – The solution which obeys Roult’s law under all conditions of temperature and concentration and during the preparation of which there is no change in enthalpy and volume on mixing the component. Conditions – P (^) A = P (^0) A X (^) A, P (^) B = P (^0) B .X (^) B Mix =^ 0,^ mix =^0 This is only possible if A‐B interaction is same as A‐A and B‐B interaction nearly ideal solution are –
- Benzene and Toluene
- Chlorobenzene and Bromobenzene Very dilute solutions exhibit ideal behavior to greater extent. Non‐ideal solution – (a) P (^) A ≠ P (^0) A .X (^) A (b) P (^) B ≠P (^0) B .X (^) B (b) (^) mix ≠ 0 (d) (^) mix ≠ 0 For non‐ideal solution the A‐B interaction is different from A‐A and B‐B interactions i. For solution showing positive deviation P (^) A > P (^0) A , P (^) B > P (^0) B. X (^) B Mix =^ positive,^ mix =positive^ (A‐B^ interaction^ is^ weaker^ than^ A‐A^ and^ B‐B^ ) E.g. alcohol and water
ii. For the solution showing negative deviation PA < P (^0) A. X (^) A , P (^) B <P (^0) B .X (^) B Mix =^ negative,^ mix =^ negative’ A‐B interaction is stronger than A‐A and B‐B interactions E.g. Chloroform, acetone, HCl and water What is Azeotrope? – The mixture of liquids at particular composition which has constant boiling point which behaves like a pure liquid and cannot be separated by simple distillation. Azeotropes are of two types: (a) minimum boiling Azeotrope (mixture which shows +ve deviations ) ex. alcohol and water (b) maximum boiling Azeotrope (which shows –ve deviations) ex. acetone and chloroform Colligative Properties ‐ Properties of ideal solution which depends upon no. of particles of solute but independent of the nature of particle are called colligative property Relative lowering in vapour pressure: (P oA – PA )/ P oA = X B Determination of molar mass of solute MB =( W (^) A × M (^) A × P oA )/W (^) A ×(P o^ A –P A ) Elevator in Boiling Point ∆TB = Kb. m Where ∆T (^) B = T’B‐ ToB Kb = molal elevator constant M = molality MB =(Kb×1000×W (^) B )/∆TB ×W A Depression in Freezing Point: ∆Tf = kf. m Where ∆Tf – T’ (^) f ; m = molality
Q1‐ What do you mean by Henry’s Law? The Henry’s Law constant for oxygen dissolved in water is 4.34×10^4 atm at 25 o^ C. If the partial pressure of oxygen in air is 0.2 atm, under atmospheric pressure conditions. Calculate the concentration in moles per Litre of dissolved oxygen in water in equilibrium with water air at 25 o^ C. Ans: Partial pressure of the gas is directly proportional to its mole fraction in solution at particular temperature. P (^) A α X (^) A ; KH = Henry’s Law of constant P (^) A = KH ×A K (^) H = 4.34×10 4 atm PO 2 = 0.2 atm Xo 2 = PO 2 / K (^) H =0.2 / 4.34×10 4 = 4.6×10‐^6 If we assume 1L solution = 1L water n water = 1000/18 = 55. XO 2 = nO 2 /(nO 2 + n H 2 O ) ~ = nO 2 /nH 2 O nO 2 = 4.6 X 10 ‐^6 X 55.5 = 2.55 X 10 ‐^4 mol M = 2.55 X 10 ‐^4 M Q.2. What is Vant Hoff factor? Ans. It is the ratio of normal molecular mass to observed molecular mass. H is denoted as ‘i’ i = normal m.m / observed m.m = no. of particles after association or dissociation / no. of particles before Q.3. What is the Vant Hoff factor in K 4 [Fe(CN) 6 ] and BaCl 2? Ans 5 and 3 Q.4. Why the molecular mass becomes abnormal? Ans. Due to association or dissociation of solute in given solvent. Q.5. Define molarity, how it is related with normality? Ans. N = M x Basicity or acidity.
Q.6. How molarity is related with percentage and density of solution? Ans. M = P x d x10/M.M Q.7. What role does the molecular interaction play in the solution of alcohol and water? Ans. Positive deviation from ideal behavior. Q.8. What is Vant Hoff factor , how is it related with a. degree of dissociation b. degree of association Ans. a. α=i – 1/n‐ 1 b. α = i ‐ 1 / 1/n ‐ 1 Q.9. Why NaCl is used to clear snow from roads? Ans. It lowers f.p of water Q10. why the boiling point of solution is higher than oure liquid Ans. Due to lowering in v.p HOTS Q 1. Out of 1M and 1m aqueous solution which is more concentrated Ans. 1M as density of water is 1gm/Ml Q2. Henry law constant for two gases are 21.5 and 49.5 atm ,which gas is more soluble. Ans. KH is inversely proportional to solubility. Q.3. Define azeotrope , give an example of maximum boiling azeotrope. Q.4. Calculate the volume of 75% of H 2 SO 4 by weight (d=1.8 gm/ml) required to prepare 1L of 0.2M solution Hint: M 1 = P x d x 10 / M 1 V 1 = M 2 V (^2) 14.5ml Q.5. Why water cannot be completely separated from aqueous solution of ethyl alcohol? Ans. Due to formation of Azeotrope at (95.4%)
Q.10. what do you mean by hypertonic solution, what happens when RBC is kept in 0.91% solution of sodium chloride? Q 11. (a). define the following terms.
- Mole fraction
- Ideal solutions (b)15 g of an unknown molecular material is dissolved in 450 g of water. The resulting solution frrezez at ‐0.34 0 c. what is the molar mass of material? Kf for water= 1.86 K Kg mol‐^1. Ans. 182.35 glmol Q 12.(a) explain the following :
- Henry’s law about dissolution of a gas in a liquid.
- Boling point elevation constant for a solvent (b)a solution of glycerol (C 3 h 80 3 ) in water was prepared by dissolving some glycerol in in 500 g of water. The solution has a boiling point of 100.42 0 c. what mass of glycerol was dissolved to make this solution? Kb for water = 0.512 k Kg mol‐^1 (hint: atb = bwb Mb*Wa Ans. 37.73 gm Q 13. 2 g of benzoic acid (c 6 h 5 cooh) dissolved in 25 g of benzene shows a depression in freezing point equal to 1.62 K. KF for benzene is 4.9 K Kg mol‐^1. What is the percentage association of acid if it forms dimer in solution. Ans. 99.2% Q14. Osmotic pressure of a 0.0103 molar solution of an electrolite is found to be 0.70 atm at 27 0 c. calculate Vant Hoff factor.( R=0.082 L atom mol‐^1 K‐^1 ) Ans. 2.
UNIT-
ELECTROCHEMISTRY CONCEPTS
Electrochemistry may be defined as the branch of chemistry which deals
with the quantitative study of interrelation ship between chemical energy and
electrical energy and inter-conversion of one form into another.relationships
between electrical energy taking place in redox reactions.
A cell is of two types:-
I. Galvanic Cell
II. Electrolytic cell.
In Galvanic cell the chemical energy of a spontaneous redox reaction is
converted into electrical work.
In Electrolytic cell electrical energy is used to carry out a non-spontaneous redox
reaction.
The Standard Electrode Potential for any electrode dipped in an appropriate
solution is defined with respect to standard electrode potential of hydrogen
electrode taken as zero. The standard potential of the cell can be obtained by
taking the difference of the standard potentials of cathode and anode.
E
0
cell = E
0
cathode-E
0
anode
The standard potential of the cells are related of standard Gibbs energy.
∆r G=-nFE
0 cell
The standard potential of the cells is related to equilibrium constant.
∆r G= -RTink
Concentration dependence of the potentials of the electrodes and the cells are
given by Nernst equation.
aA+bB ne
-
→ cC + dD
Nernst equation can be written as
Ecell = E
0
cell-RT ln[C]
c
[D]
d
nF [A]
a
[B]
b
The conductivity, K of an electrolytic solution depends on the concentration of the
electrolyte, nature of solvent and temperature.
Molar Conductivity, ∆m, is defined by K/C where C is the concentration in Mol L
∆M= k*
m
