CHAPTER 1 SOLID STATE 1 mk ‘Q’
Q 1. ‘Crystalline Solids are anisotropic in nature’. What does this Statement mean?
A 1. The statement means that some of the physical properties like electrical resistance or refractive index of Crystalline Solids show different values when measured along different directions in the same crystal.
Q 2. Why does the presence of excess of lithium make LiCl crystals pink?
A 2. The presence of excess of lithium makes LiCl crystals pink due to e-s trapped in
anionic vacancies (F centers). These electrons absorb some energy of the white light giving pink colour to LiCl crystal.6
Q3. What is meant by anti-ferromagnetism? What type of substances exhibit anti ferromagnetism?
A3. Substance like MnO, MnO2 in which magnetic domains are oppositely oriented
and cancel out each other’s magnetic moment exhibit anti ferromagnetism. Magnetic. Alignment of magnetic moments in antiferromagnetic substance:-
OR
Q3 What type of substance would make better magnets, ferromagnetic or ferromagnetic?
A3. Ferromagnetic substance would make better magnets because when ferromagnetic substance is placed in magnetic field all the domains get oriented in the direction of the magnetic field and a strong magnetic effect is produced eg : Co, Ni Q4. In a compound nitrogen atoms (N) make ccp and metal atoms (M) occupy one third of the tetrahedral voids present. Determine the formula of the compound formed by M & N?
A4. Let the no of nitrogen atoms (N) be x No of tetrahedral voids = 2x
Ratio of M:N = 2/3x : x
Therefore the formula of the compound is M2 N3
Q 5. Classify each of the following as being either a p type or n type semi conductor?
(i) Ge doped with In. (ii) B doped with Si.
A 5. (i) P type semiconductor because when group 14 element is doped with group 13 element, an electron deficit hole is created.
(ii) n type semiconductor because when group 13 element is doped wih group 14 element , free electrons will become available.
Q 6. Write a distinguishing feature between a metallic solid and an ionic solid? A 6.
Ionic Solids Metallic Solids
In solid state ionic solids are electrical insulator :- ions are not free to move. Eg NaCl, CuSO4 etc
Metallic Solids are good electrical conductors in solid state because of the presence of free electrons , Eg :- Copper, Iron etc
2 mk
Q 1. Analysis shows that nickel oxide has the formula Ni 0.98 0 1.00 . What fractions of nickel exist as Ni2+ and Ni3+ ions?
A1. The formula of the oxide is Ni 0.98 0 1.00
Let the number of O2- ions be 100.
Then number of nickel ions = 98 Let the number of Ni2+ be x
Then number of Ni3+ = 98 - x
Since total charge on cations = total charge on anions. X x (2) + (98-x) x (3) = 100 x 2
2x + 294 – 3x = 200 x = 94
% of Ni2+ = 94 x 100 = 96%
% of Ni3+ = 100 -96 = 4%
Q 2. A compound forms hcp structure. What is the total number of voids in 0.5 mol of it? How many of these are tetrahedral voids?
A2. No of atoms = 0.5 mol = 0.5 x 6.022 x 1023
= 3.011 x 1023
No of octahedral voids = no of atoms in hcp structure = 3.011 x 1023
No of tetrahedral voids = 2 x no of atoms in hcp structure = 2 x 3.011 x 1023
= 6.022 x 1023
Total no of voids = 6.022 x 10 23 + 3.011 x 10 23
= 9.033 x 10 23
Q3. Examine the given defective
crystal:-A+ B - A + B - A +
B - B - A + B
-A+ B - A + A +
B - A+ B - A + B
-Answer the following questions
(i) What type of stoichiometric defect is shown by the crystal? (1/2) (ii) How is the density of crystal affected by this defect? (1/2) (iii) What type of ionic substance show such defect? (1)
A 3. (i) Schotty defect is shown by the crystals, since equal number of cations and anions are missing from the crystal lattice.
(ii) Due to this defect, the density of the crystal decreases.
(iii) This defect is shown by those ionic substance in which cations and anions are of almost similar size eg :- NaCl, KCl etc.
Q 4. If NaCl crystals are doped with 2 x 10 -3 mol percent of SrCl2 , calculate the cation vacancies per mole?
A 4. Doping of NaCl with 2 x 10 -3 mol percent of SrCl
2 means 100 moles of NaCl is
doped with 2 x 10 -3 mole of SrCl
Each Sr2+ will occupy the place of Na+ and displace one Na+ from crystal lattice to
create cation vacancies.
Cation vacancies = Number of Sr 2+ ion added.
= 2 x 10 -5 mol = 2 x 10 -5 x 6.022 x 10 23. = 12.046 x 10 18 mol -1
Q 5. Calculate the packing efficiency of a metal crystal for a simple cubic lattice?
A 5. Packing efficiency in simple cubic lattice
rr
a
Volume of one atom x 100 Volume of cubic unit cell (a3)
Since a= 2r for simple cubic Vol of one atom =4/3 π r3
= 4/3 π r3 x 100
(2 r) 3
= 4 x 3.14 x r3 x 100
3 x 8 x r3
= 52.36% ≈ 52.4%
Q 6 (a) In reference to a crystal structure, explain the meaning of coordination number?
(b) What is the number of atoms in a unit cell of (i) Face centered cubic structure?
(ii) Body centered cubic structure?
A 6 (a) The number of nearest neighbours of any constituted particle in the crystal lattice is called its coordination number.
(b) Number of atoms in a unit cell of
(i) Face centered cubic cell structure is 4. (ii) Body centered cubic structure is 2.
3 mk
Q1. In terms of band theory. What is the difference? (a) between a conductor and an insulator. (b) between a conductor and a semi conductor.
Empty conduction band
Filled (valence band)
Partially Over lappling band filled band
Conductors _____
A 1 (a) In conductors, energy band is partially filled or it overlaps with a higher energy unoccupied conduction band. Due to this electrons can flow easily under an applied electric field and show conductivity
In insulators, the gap between filled valence band and the next higher unoccupied band is large, hence electrons cannot jump to it and substance has very small conductivity and behaves as an insulator.
Empty band (conduction band)
Filled band (Valence band)
(b) In semiconductors the gap between valence band and conduction band is small. Therefore some electrons may jump from valence band to conduction band and show some conductivity eg ‘Si’ and ‘Ge’
Q2 ) Aluminum crystallizes in a cubic close packed structure. Its metallic radius is 125 pm.
(a) What is the length of the side of the unit cell?
(b) How many unit cells are there in 1 cm3 of aluminum? A 2) (a) In a cubic close packed structure
:-4 r =
√
2aGiven r = 125 x 10 -12 m
a = 4 r
√
2 or 2√
2 r(c) a3 = (354)3 x (10-12) 3
= 44.21 x 10-30 m 3
No of unit cells in 1 cm3 = Total Volume = 10-6
Vol of one unit cell 44.21 x 10-30
= 2.261x 10 22 unit cells.
Q3 How will you distinguish between the following pairs of terms :-(a) Hexagonal close packing and Cubic close packing? (b) Crystal lattice and unit cell?
(c) Tetrahedral void and Octahedral void?
Hexagonal close packing Cubic close packing
AB AB—type packing is called hexagonal close packing. i.e spheres of the third layer are exactly aligned with those of the first layer . eg : ‘Mg’ and ‘Zn’
ABC ABC—type packing is called cubic close packing. i.e spheres of the third layer are not aligned with those of 1st or 2nd.
When 4th layer is placed its spheres are
aligned with those of 1st layer. eg :- ‘Cu’ and
‘Ag’
Q4
Account for the following
:-(i) Table salt, NaCl sometimes appears yellow in colour. (ii) FeO(s) is not formed in stoichiometric composition.
(iii) Some of the very old glass objects appear slightly milky instead of being transparent.
A4. (i) Yellow colour in NaCl is due to metal excess defect due to which unpaired electrons occupy anionic vacancies. These sites are called F centers. These electrons absorb energy from the visible region and transmits yellow colour.
Tetrahedral void Octahedral void
A tetrahedral void is surrounded by 4 spheres which lie at the vertices of a regular tetrahedron.
An octahedral void is surrounded by six spheres and formed by a combination of 2 triangular voids of the 1st and 2nd layer
There are 2 tetrahedral voids per atom in a crystal
There is one octahedral void per atom in a crystal
Crystal lattice Unit cell
The three dimensional arrangement of constituent particles in the space which represents how the constituent particles (atoms, ions or molecules) are arranged in a crystal
The smallest portion of crystal lattice which when repeated in different directions, generates the entire lattice.
(ii) In the crystal of FeO, some of the Fe2+ cations are replaced by Fe3+ ions.
Three Fe2+ ions are replaced by two Fe3+ ions to make up for the loss of positive
charge. Thus there would be less amount of metal as compared to stoichiometric properties.
(iii) Very old glass objects become slightly milky, because of heating during the day & cooling at nights i.e annealing. Over a number of years, glass acquires some crystalline character.
Q5. The density of copper is 8.95 gcm-3. It has a face centered cubic structure. What is the radius of copper atom? (Atomic mass Cu = 63.5 g mol -1 . NA = 6.022 x 1023 mol -1) ?
A5. Mass per unit cell = Atomic mass of Cu x 4 NA
= ______63.5__x 4______ 6.022 x 1023
= 4.22 x 10 -22 g
Volume of unit cell = __Mass__ Density = 4.22 x 10 -22 8.95 = 4.7 x 10-23 cm3 Edge = (Volume )1/3 = (4.7 x10 -23)1/3 = 3.61 x 10-8 cm or 361pm For fcc, r = ___a___ 2
√
2 r = 2 x 1.41361 r = 128 pmCHAPTER 2 SOLUTIONS
1 mk
Q 1. Define an ideal solution and write one of its characteristics?
A1 The solutions which obey Raoult’s law over the entire range of concentration are known as ideal solutions. i.e PA = PoA x A and PB = PoB x B.
For ideal solutions H mix = 0 and V mix = 0. eg ;- Solution of n – hexane and n- heptane
Q 2. What is meant by reverse osmosis ?
A2 The process in which the solvent flows from the solution into the pure solvent through the semi permeable membrane when a pressure higher than the osmotic pressure is applied on the solution is called reverse osmosis.
Application :- The technique is used in the desalination of sea water. Q3. Define mole fraction.
A3 The mole fraction of a component is the ratio of the number of moles of that component to the total number of moles of all the components present in the solution.
For a binary solution consisting of 2 components A and B if nA is the number of
moles of A and nB is the number of moles of B then. x
A = __nA___ x B = _ nB___
nA + nB nA + nB
Q 4. Define the term azeotrope?
A4 The constant boiling mixture which distill out unchanged in their composition are called azeotropes.
Eg :- A mixture of ethanol and water containing 95.4% of ethanol forms an azeotrope with boiling point 351.15 K.
Q5. Explain boiling point elevation constant for a solvent / Define ebullioscopic constant?
A5 Since Tb = Kb.m where m is molality
When m = 1
Kb = Tb
Therefore Ebullioscopic constant is defined as the elevation in boiling point of a solution when 1 mole of a solute is dissolved in 1 Kg of solvent .
Q6 What are isotonic solutions ?
A6 The solutions of equimolar concentrations having same osmotic pressure at given temperature are called isotonic solutions.
eg :- A 0.9% solution of pure NaCl is isotonic with human red blood cells.
Q7 Explain why aquatic species are more comfortable in cold water rather than in warm water?
A7 Aquatic species need dissolved oxygen for breathing. As solubility of gases decrease with increase of temperature, less oxygen is available in summers in lakes. Hence they feel more comfortable in winter when the solubility is higher.
OR
Q7 Why do gases nearly always tend to be less soluble in liquid as the temperature is raised?
A7 Dissolution of gas in liquid is an exothermic process. Gas + Solvent Solution + heat.
Therefore as per Lechatlier’s principle if the temperature is increased, equilibrium shifts backward i.e the solubility decreases.
2 mk
Q1 How is the vapour pressure of a solvent affected when a non volatile solute is dissolved in it?
A1 When a non volatile solute is added to a solvent, its vapour pressure decreases because some of the surface sites are occupied by solute molecules. Thus less space is available for the solvent molecules to vaporize.
Q2. The depression in freezing point of water observed for the same molar concentrations of acetic acid, trichloro acetic acid and trifluoroacetic acid increases in the order as stated above. Explain?
A2 As depression in f.pt ( Tf) is dependent on degree of dissociation (α) and
fluorine exerts the highest – I effect. So trifluoro acetic acid is the strongest acid and ionizes to a greater extent while acetic acid ionises to the minimum extent. Thus greater the number of ions produced, greater is the depression in freezing point.
Q3. Differentiate between molarity and molality for a solution. How does a change in temperature influence their values?
Or
Q3. State the main advantage of molality over molarity as the unit of concentration?
Molarity Molality
It is defined as the number of moles of solute dissolved in one litre of the solution
It is defined as the number of moles of solute dissolved in 1 Kg of the solvent Mathematically
Molarity (M) =No of moles of solute x 1000 Vol of Solution (ml)
Mathematically
Molality (M) =No of moles of solute x 1000 Mass of Solvent (g)
It decreases with increase in temperature (as V α T)
It does not change with change in temperature
Since molality does not change with a change in temperature therefore it is a better method to express the concentration of a solution.
Q4. What is meant by colligative property. List any four factors on which colligative properties of a solution depend?
A4. The properties of solutions which depend upon the number of solute particles and not upon the nature of the solute are known as colligative properties eg :- Osmotic pressure.
Factors
:-(i) Number of particles of solute. (ii) Concentration of solution. (iii) Temperature.
(iv) Association or dissociation of solute.
Q5. An aqueous solution of sodium chloride freezes below 273K. Explain the lowering in freezing point of water with the help of a suitable diagram?
A5. Freezing point of a substance is the temperature
at which solid and liquid phases of a substance coexist i.e they have the same vapour pressure. As the vapour ……… pressure of the solution is less than that of pure solvent
For solution its vapour pressure will become equal to that of a solid solvent only at a lower temperature.
Tf Tfo
Temperature 3 mk
Q1. A sample of drinking water was found to be severely contaminated with chloroform CHCl3 , supposed to be carcinogen. The level of contamination was 15ppm (by mass)?
(i) Express this in percent by mass.
(ii) Determine the molality of CHCl3 in water sample. A1. 15 ppm means 15 parts in 106 parts by mass in the solution.
i.e 106 parts by mass of solution = 15 parts of solute.
or 100 parts (Mass %) = 15 x 100 106 = 15 x 10 -4 Mass of solvent = (106 -15) g liquid solvent solution solid solvent Vapour pressure
≈ 106 g
Therefore Molality = nB where nB = no of moles of solute
wA
wA = Mass of the solvent
Molality = 15/119.5 X 1000 106
= 1.25 x 10-4m
Q2 State Raoult’s law for a solution containing volatile components. How does Raoult’s law becomes as special case of Henry’s law?
A2 For a solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction?
i.e PA α x A or PA = P0A x A
and PB α x B or PB = P0B x B
Where PA and PB are the partial pressure of A and B, P0A , P0B are the vapour
pressure of pure component and x A and x B are their mole fractions.
If gas is the solute and liquid is the solvent then according to Henry’s law PA = KH x A
Thus Raoult’s laws and Henry’s law become identical except that their proportionality constants are different.
Q3 Some ethylene glycol is added to your car’s cooling system along with 5kg of water. If the freezing point of water-glycol solution is -150c, what is the boiling point of the solution?
Kb = 0.52 K Kgmol-1 and Kf = 1.86 K Kg mol-1
A3 Tf = 15oc Kf = 1.86 k/m
Molality = Tf
Kf
Tb = Kb x m
= 0.52 x 8.06 = 4.190c
Boiling point of pure water = 100oc
Therefore Tb = Tb + Tbo
Tb = 100 + 4.19
= 104.19oC
Q4 Assuming complete dissociation. Calculate the expected freezing point of a solution prepared by dissolving 6 g of Glauber’s salt, Na2SO4.10 H2O in 0.1 Kg mol-1 of water. Kf of water = 1.86 K Kg mol -1.
A 4 Kf = 1.86 K Kgmol-1
WB = 6g WA = 0.1 Kg mol-1
MB = 322 g mol-1
Since there is a complete dissociation Therefore i = 3 Tf = i K f WB MB x WA = 3 x 1.86 x 6 322 x 0.1 = 1.04o
Q5. Calculate the mass of a nonvolatile solute (molar mass 40g mol-1) which should be dissolved in 114 g octane to reduce its vapour its pressure to 80%? A5. Ps= 80% of Po = 0.80 Po No of moles of solute = W 40 No of moles of solvent = 114 = 1 (octane) 114 therefore P o – Ps = X B Po P o – 0.8 Po = W /40 Po W/40 + 1 or 0.2 W +1 = W 40 40 0.8 W = 0.2 40 or W = 10g
Q6 19.5g of CH2FCOOH is dissolved in 500g of water. The depression in freezing point observed is 1oc. Calculate the Van’t Hoff factor and dissociation constant of fluoroacetic acid .Kf for water is1.86 K Kg mol-1.
A6 WB = 19.5g WA = 500g Kf= 1.86K Kgmol-1 (ATf) obs = 1o. MB (obs) = 1000 Kf WB WA ∆ Tf = 1000 X1.86 X 19.5 = 72.54 g mol-1 500 X 1 MB (cal) = 14 + 19 + 45 = 78 g mol-1 L = M(cal) = _78__ = 1.0753 MB (Obs) 72.54
Calculation of dissociation const (K) Ka = _Cα 2 1-α CH2 FCOOH CH2FCOO -+ H+ Initial C MOL L-1 O O C(1-α) Cα Cα i = C (1+α) = 1 + α or α = i - 1 C = 1.0753 – 1 = 0.0753 Taking volume of solution as 500 ml
C =19.5 x _1_ x 1000 = 0.5 M 78 500
Ka = Cα 2 = 0.5 x (0.0753)2
1-α 1-0.0753 Ka = 3.07 x 10-3
Q7. Non ideal solutions exhibit either positive or negative deviations from Raoult’s law. What are these deviations and why are they caused? Explain with one example for each type?
A7 For non ideal solutions, vapour pressure is either higher or lower than that predicted by Raoult’s law. If it is higher the solution exhibits positive deviation and if it is lower it exhibits negative deviation from Raoult’s law.
Positive Deviation Negative Deviation
When solute – solvent interactions are weaker than solute – solute or solvent solvent interactions, vapour pressure increases which result in positive deviation.
When solute solvent interactions are stronger than solute – solute or solvent – solvent interactions, vapour pressure decreases which result in negative deviation.
Eg :- Ethanol +Acetone.
for a solution showing +ve deviation PA > PoA XA and PB > PoB XB
eg :- Chloroform + Acetone.
for a solution showing –ve deviation PA < PoA XA PB < P0B XB
H mix = +ve H mix = - ve
5 mk
Q1. (i) What is Van’t Hoff factor? What types of values can it have if in forming the solution the solute molecules undergo
(a) Dissociation? (b) Association?
(ii) How many ml of 0.1M HCl solution are required to react completely with 1g of a mixture of Na2CO3 containing equimolar amounts of both ? (Molar
Mass Na2CO3 = 106g and NaHC03 = 84g)
A1. (i) Van’t Hoff factor (i) is defined as the ratio of the experimental value of colligative property to the calculated value of colligative property .
i = Observed Colligative property Calculated Colligative property
Also, i = Total number of moles of particles after dissociation/ association Number of moles of particles before association/ dissociation
Therefore , For (a) dissociation i > 1 And (b) Association i < 1
(ii) Let x g of Na2CO3 be present in the mixture
So amt of Na HCO3 = 1 –x
Moles of Na2CO3 in x g = x_
106 Moles of NaHCO3 in (1-x) = _1-x_
84
As mixture contains equimolar amounts of the two. x_ = 1-x_ 106 84 106 – 106 x = 84x X = _106__ = 0.558g 190 Moles of Na2 CO3 = __0.558__ = 0.00526
106
Moles of NaHCO3 = 1- 0.558 = 0.442 = 0.00526
84 84 To calculate the moles of HCl required. Na2CO3 + 2 HCl 2 NaCl + H2O + CO2
NaHCO3 + HCl NaCl + H2O + CO2
1 mole of Na2CO3 requires 2 moles of HCl
Therefore 0.00526 mole of Na2 CO3 requires 2 x 0.00526 moles or
0.0152 moles of HCl.
1 mole of NaHCO3 requires 1 mole of HCl
Therefore 0.00526 mole of NaHCO3 requires 0.00526 moles of HCl.
Total moles of HCl required = 0.01052 + 0.00526 = 0.01578 moles To calculate volume of 0.1M HCl
1.1 mole of 0.1M HCl are present m 1000ml HCl Therefore 0.01578 mole of 0.1M HCl will be present in
1000 x 0.01578 0.1
= 157.8 ml
Q2. (i) The molecular masses of polymers are determined by osmotic pressure method and not by measuring other colligative properties. Give two reasons?
(ii) At 300k, 36g of glucose C6H12O6 present per litre in its solution has a pressure pf 4.98 bar. If the osmotic pressure of another glucose solution is 1.52 bar at the same temperature, calculate the concentration of the other solution?
A2. (i) The osmotic pressure method has the advantage over other methods because
:-(a) It uses molarities instead of molalities and it can be measured at room temperature.
(ii)
π
1 = C1 RT, π2 = C2 RT Therefore _π1 = C1π
2 C2 or 4.98 = 36/180 1.52 C2 or C2 = 0.061 moll-1 = 0.061 x 180 gl-1 = 10.98 gl-1Q 3 (i) Define the terms osmosis and osmotic pressure?
(ii) An aqueous solution containing 12.48g of barium chloride in 1.0 kg of water boils at 373.0832 k. Calculate the degree of dissociation of barium chloride?
(Given Kb for H2O = 0.52 Km-1 molar mass of Ba Cl2 = 208.34 gmol-1) A3. (i) The net spontaneous flow of the solvent molecules from the solvent to the solution or from a less concentrated solution to a more concentrated solution through a semipermeable membrane is called osmosis.
Osmotic pressure :- The minimum excess pressure that has to be applied on the solution to prevent the entry of the solvent into the solution through the semipermeable membrane is called osmotic pressure.
(iii) Given W2 = 12.48g , W1 = 1 Kg = 1000g
(iv)
Tb (solution) = 373.0832 K
Kb for H2O = 0.52 Km-1 and M2 = (Ba Cl2) = 208.34
Tb = Tb - Tbo = 373.0832 – 373 = 0.0832 K M2 (Observed) = Kb x W2 x 1000 Tb x W1 = 0.52 x 12.48 x 1000 0.0832 x 1000 = 78 g mol-1 i = M2 (calculated) = 208.34 = 2.67
M2 (observed) 78
For BaCl2 m = 3 as it gives 3 ions on dissociation
α = i-1 = 2.67 -1 = 1.67 = 0.835 m-1 3-1 2
= 83.5%
Q4. (a) What type of deviation is shown by a mixture of ethanol and acetone? Give reason.
(b) What do you expect to happen when RBC’s are placed in (i) 1% NaCl solution (ii) 0.5 % NaCl solution
(c) Calculate the molarity of 68% (w/w) solution of nitric acid , if the density of the solution is 1.504 g ml-1
A4. (a) Ethanol and acetone shows positive deviation because on mixing the two the forces of attraction decreases and the vapour pressure increases.
(b) Since RBC’s are isotonic with 0.9% NaCl solution therefore (i) In 1% solution of NaCl they will shrink due to plasmolysis (ii) In 0.5% solution of NaCl they will swell or may even burst.
(c) Molarity = % x density x 10 Molar mass
Mass % = 68 , d = 1.504 Molar mass of HNO3 = 63 gmol-1
Therefore Molarity = 68 x 1.504 x 10 63
= 16.23 M
Q5. (a) State Henry’s law and mention two of its important applications
(b) The partial pressure of ethane over a saturated solution containing 6.56 x 10-2g of ethane is 1bar. If the solution were to contain 5 x 10-2 g of ethane, then what will be the partial pressure of the gas.
A5. (a) Henry’s law states that at a constant temperature , the solubility x of a gas in a liquid is directly proportional to the pressure of the gas
P = KH x
Applications
:-- To increase thje solubility of CO2 in soda water, the bottle is sealed under high
pressure
- There is a low concentration of oxygen in the blood and tissues of the people living at high altitudes due to which they feel weak and are unable to think clearly (anoxia).
- (b) M = KH x P
For the 1st case 6.56 x 10-2 = K
H x 1 bar or KH = 6.56 x 10-2 g bar -1 In the 2nd case 5 x 10 -2 = (6.56 x 10-2) x P or P = 5 x 10 -2 6.56 x 10 -2 P = 0.762 bar CHAPTER 5
SURFACE CHEMISTRY 01 MARK
Q1). What is meant by shape selective catalysis?
A1) The catalytic reaction that depends upon the pore structure of the catalyst and the size of the reactant and product molecules is called shape selective catalysis.
eg:- zeolites (eg ZSM 5) converts alcohol directly into gasoline. Q2) What are the dispersed phase and dispersion medium in milk? A2) Dispersed phase : Fat
Dispersion medium : water Q3) What is Collodion?
A3) Collodion is 4% solution of nitro cellulose in a mixture of alcohol and ether. Q4) Define Kraft temperature?
A4) The formation of micelles from the ionic surfactant can take place only above a certain temperature which is called Kraft temperature.
Q5) Why is adsorption always exothermic ? Or
What is the sign of H & S when a gas is adsorbed by adsorbent ? A5) When a gas is adsorbed in the surface of solid, its entropy decreased i.e.
S =-ve. From Gibbs Helmholtz equation : G = H – T S for the process to be spontaneous , G must be negative, which is possible only
when H = -ve . Hence adsorption is always exothermic.
Q6) Why is it essential to wash the precipitate with water before estimating it quantitatively?
A6) Some amount of the electrolytes mixed to form the precipitate remains adsorbed on the surface of the particles of the precipitate. Hence it is essential to wash it with water before it quantitatively.
Q7) What happens when persistent dialysis of colloidal solution is carried out?
A7) The stability of a colloidal sol is due to the presence of a small amount of the electrolyte. On persistent dialysis, the electrolyte is completely removed, so the colloidal sol becomes unstable and gets coagulated.
Q8) What causes Brownian movement in a colloidal solution?
A8) Brownian movement ie zigzag movement of the colloidal particles is due to hitting of these particles by the molecules of the dispersion medium with different forces from different directions.
Q 9) Why it is important to have clean surface in surface studies? A 9) It facilitates the adsorption of species on the adsorbent. Q 10) What happens when gelatin is mixed with gold sol?
A 10) Gold sol is a lyophobic sol. On addition of gelatin, the sol is stabilized.. Q 11) Gelatin which is a peptide is added in ice – creams. What can be it s role?
A 11) Ice –creams are emulsion which get stabilized by emulsifying agents like gelatin. Q 12) What is the role of activated charcoal in gas mask used in coal mine?
A 12) Activated charcoal adsorbs poisonous gases present in coal mine. Q 13) In what way is a sol different from a gel?
A 13) Sols are colloidal solutions of solid dispersed in liquid while gels are colloidal solutions of liquid dispersed in solid.
Q 14) Of NH3 and N2, which gas will be adsorbed more readily on the surface of
charcoal and why?
A 14) NH3 is adsorbed more readily as it is more easily liquefiable compared to N2,
Moreover, NH3 molecule has greater molecular size.
Q 15) How does it become possible to cause artificial rain by spraying silver iodide on the clouds?
A 15) Clouds are colloidal in nature and carry a charge. On spraying silver iodide which is an electrolyte, the charge on the colloidal particles is neutralized. Clouds coagulate to form rain.
Q 16) What is the role of diffusion in heterogeneous catalysis?
A 16) The gas molecules diffuse onto the surface of the catalyst and get adsorbed. After the chemical change, the products formed diffuse away from the surface of the catalyst setting the surface free for other reactant molecules to adsorb on the surface and give the product.
0 2 MARKS
Q1) Write the differences between physisorption and chemisorption with respect to the following
(i) Specificity (ii) Temperature dependence (iii) Reversibility (iv) Enthalpy change
Criteria Physisorption Chemisorption
Specificity It is not specific in nature It is highly specific in nature Temperature
dependence
It decreases with increase in temperature. Thus, low temperature is favorable for physisorption
It increases with increase in temperature. Thus, highly temperature is favorable for chemisorption
Reversibility Reversible in nature Irreversible in nature Enthalpy
change
Low enthalpy of adsorption High enthalpy of adsorption
Q2) Distinguish between homogeneous and heterogeneous catalysis.
A2) When reactants and the catalysts are in the same phase (i.e. liquid or gas) the catalysis is known as homogeneous catalysis.
) ( ) ( 2 3 3 ) ( 2 g O g O NO g
When reactants and the catalysts are in the different phase (i.e. liquid or gas) the catalysis is known as heterogeneous catalysis.
4NH3(g) + 5O2(g) ) ( 6 ) ( 4 2 ) (s NO g H O g Pt
Q3) What is meant by coagulation of a colloidal solution? Describe briefly any three methods by which coagulation of lyophobic sols can be carried out?
A3) The process of aggregating together the colloidal particles is called coagulation of the sol. It is also known as precipitation. Following are the three methods by which coagulation of lyophobic sols can be carried out.
(i)
Electrophoresis. In this process, the colloidal particles move towards oppositely charged electrodes and get discharged resulting in coagulation.
(ii)
Mixing of two oppositely charges sols. When equal proportions of positively charges sols are mixed, they neutralize each other resulting in coagulation.
(iii)
Dialysis. By this method, electrolytes present in sol are removed completely and colloid becomes unstable resulting in coagulation colloid is a heterogeneous system, e.g gold sol, sulphur sol, soap, etc.
Q4) What are emulsions ? Discuss the role of an emulsifier in forming emulsion. A4) Emulsions are one of the types of colloidal system, in which both the dispersed phase and dispersion medium are liquids, e.g milk.
Role of emulsifier Emulsifying agents are added to emulsions to stabilize them. The emulsifying agent forms an interfacial film between suspended particles and the medium. For oil in water emulsions, the principal emulsifying agents are gums, proteins, natural and synthetic soaps.
Q 5) How are the following colloidal solutions prepared? (i) Sulphur in water
(ii) Gold in water
A5) (i) Sulphur sol is prepared by the oxidation of H2S with SO2.
SO2 + 2H2S oxidation 3S + 2H2O
(Sol)
(ii) Gold sol is prepared by Bredig’s arc process or by the reduction of AuCl3 with
HCHO.
2AuCl3 + 3HCHO + 3H20 Reduction 2Au + 3HCOOH + 6HCI
(Sol)
Q6) What are enzymes? Write in brief the mechanism of enzyme catalysis.
A6) Enzymes are biochemical catalysts whih are globular proteins and form macromolecular colloidal solution in water.
The mechanism of enzyme catalysis may be explained on the basis of lock and key theory. Acc to the theory , Enzymes are highly specific due to the presence of active sites on their surface. The shape of active site of any given enzyme is such that only a specific substrate can fit in to it just as one key can fit into a particular lock enzyme catalyzed reactions takes place in two steps as follows
:-Step I :- Formation of Enzyme – substrate complex
E + S ES (fast and reversible) Enzyme Substrate Enzyme substrate complex
Step II :- Dissociation of enzyme substrate complex to form the products ES [EP] E + P (slow and rate determining)
Enzyme Enzyme Enzyme Product Substrate Product (Regenerated)
Complex association
Q7) What do you mean by activity and selectivity of catalysts?
A7) Activity : Ability of a catalyst to accelerate chemical reactions is known as its activity . For example, Pt catalyses the combination of H2 and O2 to form water. It has
been found that for hydrogenation reactions, the catalytic activity increases from Group 5 to Group 11 metals with maximum activity being shown by Group 7-9 elements of the periodic table.
Selectivity : The ability of a catalyst to direct a reaction to yield a particular product is called its selectivity. Combination of CO and H2 yields different products with
different catalysts as given below
:-CO (g) + 3H2 (g) Ni CH4 (g) + H2O (g)
CO (g) + H2 (g) Cu HCHO (g)
CO (g) + H2 (g) Cu I ZnO – Cr2O3 CH3OH (g).
Q8) Describe some features of catalysis by zeolites?
A8) (i) Zeolites are hydrated alumino-silicates which have a three- dimensional network structure containing water molecules in their pores.
(ii) On heating, water of hydration present in the pores is lost and the pores become vacant to carry out catalysis.
(iii) The size of the pores varies from 260 to 740 pm. Thus, only those molecules can be adsorbed in these pores and catalyzed whose size fits these pores. Hence, they act as molecular sieves or shape selective catalysts.
An important catalyst used in petroleum industry is ZSM -5 (Zeolite sieve of molecular porosity – 5). It converts alcohols into petrol by dehydrating them to form a mixture of hydrocarbons.
Alcohols ZSM -5 Hydrocarbons Dehydration
03 MARKS
Q1) Account for the following
:-(a) Medicines are more effective in the colloidal form/ colloidal gold is used for intramuscular injection.
(b) Sky appears blue in colour. (c) Alum is added to purify water. OR
Bleeding stops on rubbing moist alum on the cut surface.
A1) (a) Medicines are more effective in the colloidal form because they have a large surface area and are easily assimilated in this form.
(b) There are dust particles in the atmosphere. These dust particles are of colloidal size and scatter the light. Blue light coming from the sun is scattered and the sky appears blue.
(c) Alum coagulates colloidal impurities present in water. or
Alum brings about and coagulation of blood and stops further bleeding. Q2) Explain clearly how the phenomenon of adsorption finds application in
(i) Production of vacuum in a vessel (ii) Heterogeneous catalysis
A 2) (i) Production of high vacuum. The traces of air can be adsorbed by charcoal from a vessel evacuated by a vacuum pump to give a very high vacuum.
(ii) Heterogeneous catalysis. Adsorption of reactants on the solid surface of the catalyst increases the rate of reaction.
(iii) Froth floatation process. A low grade sulphine ore is concentrated by this method using pine oil and frothing agent. The mineral particles become wet by oils while the gangue particles by water.
Q3) What is an adsorption isotherm? Describe Freundlich adsorption isotherm. A 3) Adsorption isotherm. It is the variation in the amount of gas adsorbed by the adsorbent with pressure at constant temperature.
Freundlich’s adsorption isotherm. It is an empirical relationship between the quantity of gas adsorbed by unit mass of solid adsorbent and pressure at a particular temperature. ) 1 ( 1 kp n m x n …. (i) When, kp m x n1, or m x p
Where x is the mass of gas adsorbed on mass m of the adsorbent at pressure p.k and n are constants which depend on the nature of the adsorbent and the gas at a particular temperature.
Taking log in Eq. (i), gives
p n k m x log 1 log log
The validity of Freundlich isotherm can be verified by plotting m
x
log
on Y-axis and log p on X-axis. If, it comes to be a straight line, the Freundlich isotherm is valid.
Q4) Explain the terms with suitable examples
:-(i) Alcosol (ii) Aerosol (iii) Hydrosol
A4) (i) Alcosol :- It is a colloidal dispersion having alcohol as the dispersion medium. For eg :- Collodion, which is a colloidal sol, of cellulose nitrate in ethyl alcohol.
(ii) Aerosol :- It is a colloidal dispersion of a liquid in a gas eg:- fog. (iii) Hydrosol :- IT is a colloidal dispersion of a solid in water as the dispersion medium.eg :- starch sol.
Q5) Explain the following observations
:-(a) Cotrell’s smoke precipitator is fitted at the mouth of the Chimneys use is factories.
(b) Physical adsorption is multilayered, while chemisorption is monolayered. (c) A white precipitate of silver halide becomes coloured in presence of the dye eosin.
A5) (a) The charged collodial particles of carbon after coming in contact with oppositely charges electrode in Cottrell precipitator lose their charge and settle down at the bottom.
(b) In physical adsorption, there are Weak Vander Waals forces. Therefore it forms multilayers. In Chemisorption, adsorbate is attached by chemical bond. There is a strong force of attraction. Therefore, only one layer is obtained.
(c) Eosin is adsorbed on the surface of silver halide precipitate making it look coloured.
CHAPTER – 6
GENERAL PRINCIPLES OF EXTRACTION
01 MARK
Q1. Name the chief ores of aluminium and zinc.
A1. Chief ores of – Aluminium – Bauxite. A l0x (OH)
3- 2x [Where O<X<1]
Zinc :- zinc blende (ZnS)
Q2. What is the function of collectors in the froth floatation process for the concentration of ores?
or
What is the role of collectors in froth floatation process?
A2. Collectors such as pine oils, fatty acids, xanthates etc, enhance non – wettability of the mineral particles.
Q3. What is the role of flux in metallurgical processes?
A3. Flux is a substance that chemically combines with gangue (earthy impurities) which may still be present in the roasted or the calcined ore to form an easily fusible material called the slag.
Q4. What types of ores can be concentrated by magnetic separation method?
A4. Those ores which are magnetic in nature and associated impurities are non-magnetic in nature or vice-versa, are concentrated by non-magnetic separation method. Q5. Differentiate between a mineral and an ore?
A5. Naturally occurring substances from which metal may or may not be extracted profitably is known as mineral. The minerals from which metal can be extracted chiefly, easily and profitably are known as ores.
Q6. Why is it that only sulphide ores are concentrated by froth floatation process? or
Why is the froth floatation method selected for the concentration of sulphide ores?
A6. Only sulphide ores are concentrated by froth floatation method because sulphide ores are wetted preferentially by pine oil and impurities (gangue) are wetted by water. Q7. Why are sulphide ores converted to oxide before reduction?
A7. It is more easy to reduce an oxide ore than a sulphide ore. This is why sulphide ores are roasted and first converted to its oxide form.
Q8. Although carbon and hydrogen are better reducing agents, but they are not used to reduce metallic oxides are at high temperature. Why?
A8. This is because carbon and hydrogen react with metals to form carbides and hydrides respectively at high temperature.
Q9. What is the role of CO2 in the extractive metallurgy of aluminium from its ore?
or
What role is played by CO2 in the getting pure alumina ( Al2 O3 ) in the
extraction of aluminium?
A9. The aluminate in solution is neutralized by passing CO2 and hydrate Al2
O3 is precipitated
2 Na [Al (OH)4] (aq) + CO2 (g) Al2O3 . xH2O(s) + 2Na HCO3 (aq)
Q10. Why is electrolytic reduction preferred over chemical reduction for the isolation of certain metals?
A10. More reactive metals, i.e. alkali metals and alkaline earth metals are usually extracted by electrolysis of their fused salts because in electrolytic reduction, by applying external potential from outside source, reduction can be brought easily. Further, more reactive metals have more affinity with oxygen than that of carbon. That’s why electrolytic reduction is preferred over chemical reduction.
02 MARKS
A1. Copper is extracted by hydrometallurgy from low grade ores. It is leached out using acid or bacteria. The solution containing 2+Cu¿¿ is treated with scrap iron or H2
.
2+¿
Cu¿ (aq) + H2 .(aq) Cu(s) + 2 +H¿¿ (aq)
2+¿
Cu¿ (aq) + Fe(s) Cu(s ) + 2+¿
Fe¿ (aq)
Q2. How do we separate two sulphide ores by Froth floatation method? Explain with an example?
Or
State the role of depresent in froth floatation process?
A2. Two sulphide ores can be separated by adjusting properties of oil to water or by using depressants which prevent one type of sulphide ore particles from forming the froth with air bubbles.
Eg :-In case of an ore containing Zns and Pbs, the depresent NaCN is used. It forms complex with ZnS and thus prevents it from forming a froth while Pbs forms the froth and hence can be separated from ZnS.
4NaCN + ZnS Na2 [Zn CN)
4] + Na2 S
sodiumtetracyanidozincate(II)complex Q3. State Write the reactions involved in the following process :
(i) Leaching of bauxite ore to prepare pure alumina.
(ii) Recovery of gold after gold ore has been leached with NaCN solution.
A3. Leaching of alumina from bauxite ore Bauxite usually contains Si O2 , iron
oxide and titanium oxide (Ti O2 ) as impurities. Powered ore is digested with conc
NaOH solution at 473-523Kand 35-36 bar pressure. Al2 O3 and SiO2 are
dissolved in the solution while impurities do not.
Al2 O3 (s) + 2NaOH (aq) + 3 H2 O(l) 473−523 K35−36 ¯¿
Filtrate is neutralized by passing CO2 gas and hydrated Al2 O3 is
precipitated.Some fresh alumina is also added to solution to induce precipitation. 2 Na [Al (OH)4] (aq) + 2CO2 (g) Al2O3 . xH2O(s) + 2Na HCO3 (aq).
ppt
Precipitate is filtered, washed, dried and heated to give back pure Al2 O3 .
Al2O3 . s xH2O¿ ) 1470K Al2O3 (s)+¿ . xH2O ( g) Alumina Recovery of gold 4Au(s)+ 8 CN−¿¿
(aq) +2 H2 O (aq) + O2 (g) 4[Au(CN)2]2-(aq) + 4 OH−¿¿ (aq)
2[Au(CN)2]- (aq) + Zn (s) [Zn(CN)42- ] (aq) +2Au(s)
Soluble gold complex
In this reaction, zinc acts as a reducing agent. Q4. Define the following terms :
(i) Roasting (ii) Calcination
A4. (i) Roasting. The process of heating of metal ore below its melting point in the presence of air is called roasting.
(ii) Calcination. The process of heating of metal ore in the absence of air is called calcination.
Q5. Why is an external emf of more than 2.2V required for the extraction of CL2
from brine?
A5. For the reaction 2 −Cl¿¿
(aq) + 2 H2 O (l) 2 OH−¿¿ (aq) + H2 (g) +
Value of Go is + 422kJ. Using the equation Go = -n FEo , the value of
Eo comes out to be -2.2V. Therefore an external emf of more than 2.2V is
required for the extraction of Cl2 from brine.
Q6. (a) What is the composition of copper matte?
(b) What is the function of SiO2 in the metallurgy of copper?
A6. (a) Copper matte contains Cu2 s and FeS.
(b) Silica
SiO
¿ ¿
¿ ) is added in the reverberatory furnace during the extraction of
Cu to remove impurities of iron oxide (FeO) present in the ore. Silica here acts as flux and reacts with iron oxide gangue to remove it as slag, iron silicate.
FeO + SiO2 Fe SiO3
Iron oxide Silica Iron silica slag Q7. Give reason for the following
:-(i) Alumina is dissolved in cryolite instead of being electrolysed directly. (ii) Zinc oxide can be reduced to the metal by heating with carbon but
not Cr2O3 .
A7. (i) Melting point of alumina is very high and it is a bad conductor of electricity. Cryolite is added to alumina to lower its melting point and to make it conductor of electricity.
(ii) Reduction of zinc oxide is done by using coke ZnO +C 1673K Zn + CO Chromium oxide is reduced through alumina (thermic reduction process}.
This is because, the free energy change for the formation of Cr2O3 is
more negative than that of ZnO.
Q8. State the basis of refining a substance by chromatographic method. Under what circumstances is this method specially useful?
A8. Chromatographic Method. This method is based on the principle that different components of a mixture are differently adsorbed on an adsorbent. Adsorbed components are removed by using suitable solvent (eluent).
This method is very useful for the purification of the elements which are available in minute quantities and the impurities are not different in chemical properties from the element to be purified.
Q9. What chemical principle is involved in choosing a reducing agent for getting the metal from its oxide ore? Consider the metal oxides, Al2O3 and Fe2O3 , and justify
the choice of reducing agent in each case?
A9. Thermodynamic factor helps us in choosing a suitable reducing agent for the reduction of a particular metal oxide to metal. From Ellingham diagram, it is evident that metals which have more negative f Go of their oxide can reduce those metal oxide for which f G
o
is less negative. Since, the free energy change for the formation of
Fe2O3 is less negative than CO, CO is a good reducing agent for it.
Fe2O3 + 3CO 823K 2Fe + 3 CO2
, G = -ve [because f G ( Fe2O3 ) is less negative then f G (CO)]
The free energy change for the formation of Al2O3 is highly negative (~ –1000
to –1100), thus, no reducing agent is suitable for its reduction. It is thus,reduced by electrolysis of its oxide.
Q10. How is wrought iron different from steel?
A10. Pig iron contains 4% C and trace elements like, S, P,Si and Mn as impurity. It is heated strongly (in excess of oxygen) in Bessemer converter. By this, volatile impurities like C,P,S are removed as CO, P2O5 and SO2 repectively. Si amd Mn are also
is called wrought iron. When, it is heated with 0.5% C, it gives steel. Thus, steel is less pure3 than wrought iron.
03 MARKS
Q1. Explain the basic principles of following metallurgical operatins : (i) Zone refining
(ii) Vapour phase refining (iii) Electrolytic refining
A1. (i) Zone refining It is based on the fact that the impurities are more soluble in the melt than in the solid state of the metal. When one end of the impure metal rod is heated with the help of mobile heater, the molten zone moves forward along with impurities.
In this way impurities are concentrated at the other end of the rod. This end is cut off. The process is repeated several times to obtain ultrapure metals for producing semiconductors. Ge,Si,Ga and In are purified by this method.
Metal rod Induction –coil heaters moving as
shown
Molten zone
Zone refining process
(ii) Vapour phase refining In this method, the metal is converted into its volatile compound and is collected elsewhere. It is then decomposed to give pure metal. So the two requirements are
(a) the metal should form a volatile compound with an available reagent.
(b) The volatile compound should be easily decomposable, for the recovery of metal.
(iii) Electrolytic refining In this method, the impure metals is made anode and a strip of pure form of same metal is made cathode. Aqueous solution of salt of same metal is taken as electrolyte. On passing electric current, metal ions from the electrolyte are deposited at the cathode in the form of pure metal while an equivalent amount of metal dissolves from the anode and goes into the electrolyte solution as metal ions, i.e. pure metal is dissolved from anode and deposited at cathode through electrolyte in equivalent amount.
Q2. Write down the reactions which occur in upper, middle and lower zones in the blast furnace during the extraction of iron from iron ore?
A2. Reduction of iron oxide in blast furnace (i) Lower zone of the blast furnace
C + O2 CO2 + Heat
C + CO2 2CO
Coke is burnt to give temperature upto 2200K at lower part of the blast furnace.
(ii) Middle zone of the blast furnace
CO and heat move up in the furnace. The temperature range on the middle zone of the blast furnace is 900 – 1500 K.
FeO + CO Fe + CO2
Limestone is also decomposed to CaO which removes silicate impurity of the one as slag.
Ca CO3 1100K CaO + CO2
C + CO2 2CO
CaO + SiO2 Ca SiO3
(iii) Upper zone of the blast furnace
Temperature range in this zone is 500-800K. Following reactions take place in this zone.
3 Fe2O3 + CO 2 Fe3O4 + CO2
Fe3O4 + 4CO 3Fe + 4 CO2
Fe2O3 + CO 2FeO + CO2
The iron obtained from blast furnace is known as pig iron. It contains about 4% carbon.
Q3. (a) The reaction : Cr2O3 + 2Al Al2O3 + 2Cr ( Δ Go =
-421kJ) is thermodynamically feasible as is apparent from the Gibbs energy value. Why does it not take place at room temperature?
(b) The value of Δf G
o
for formation of Cr2O3 is -540kJ mol−1 and
that of Al2O3 is - 827 kJ mol−1 . Is the reduction Cr2O3 possible with AI?
A3. (a) In the given redox reaction, all the reactants and the products are solids at room temperature, therefore, there does not exist any equilibrium between the reactants and the products and hence the reaction does not occur at room temperature. The interpretation of Go is based on K ( Go = -RT in K) where K is the equilibrium constant. Where there is no equilibrium in the solid state and the value of K becomes insignificant. However, at high temperature, when chromium melts, the value of T S increases. As a result, the value of r G
o
becomes more –ve and hence the reaction proceeds rapidly.
4 3
Al (s) +
O2 (g) 2 3 Al2O3 (s) ; f G o AI, Al2O3 = -827kJ mol−1 (i) 4 3Cr (s) +
O2 (g) 2 3 Cr2O3 (s): f G o = -540 kJ mol−1 (ii)Subtracting equation (ii) from equation (i), we get 4 3
Al (s) +
2 3 Cr2O3 (s) 2 3 Al2O3 (s) + 4 3Cr (
s) ;
f Go = -287 kJ mol−1 As f G oof the combined redox reaction is –ve, therefore , reduction of
Cr2O3 by AI is possible.
Q4. Write chemical reactions taking place in the extraction of zinc from zinc blende. A4. The various steps involved in the extraction of zinc from zinc blende are :
(a) Concentration. The ore is crushed and then concentrated by froth floatation process.
(b) Roasting. The concentrated ore is roasted in presence of excess of air at about 1200 K when zinc oxide (ZnO) is formed and SO2 is evolved.
2ZnS + 3 O2 1200K 2ZnO + 2S O2
Zinc blende Zinc oxide
(c) Reduction Zinc oxide obtained in step (b) is mixed with powdered coke and heated to 1673 K in a fire clay retort when it is reduced to zinc metal.
At 1673 K, zinc metal being volatile (b.p 1180K), distils over and is condensed. (d) Electrolytic refining The metal obtained as above is impure. It is purified by electrolytic method. Impure zinc is made the anode while cathode consists of a sheet of pure zinc. The electrolyte consists of Zn SO4 solution
acidified with dil. H2SO4 . On passing electric current, pure Zn is deposited on
the cathode.
Q5. (a) Describe the principle involved in each of the following process :-(i) Mond process for refining of nickel.
(ii) Liquation method
(b) What is the role of graphite rod in the electrometallurgy of aluminium? A5. (a) (i) Mond process for refining of nickel. (vapour phase refining) is based upon the principle that nickel is heated in the presence of carbon monoxide to form nickel tetra carbonyl, which is a volatile complex.
Ni+ 4CO 330-350 K Ni(CO) 4
Nickel tetra carbonyl
Then, the obtained nickel tetra carbonyl decomposed by subjecting it to a higher temperature (450 – 470K) to obtain pure nickel metal.
Ni(CO)4 Ni + 4CO
Nickel tetra Nickel carbonyl
(ii) Liquation method In this method a low melting metal like tin can be made to flow on a sloping surface. In this way, it is separated from higher melting impurities.
(b) In the electrometallurgy of alumina, a fused mixture of alumina, cryolite and fluorspar (CaF2) is electrolysed using graphite as anode and graphite lined as cathode. During electrolysis, Al is liberated at the cathode while CO and CO2 are liberated at the anode. The relevant equations are as under :
At cathode : ) ( 3 3 e Al l Al
At anode : C (s) + 2−O¿¿ (melt) CO(g) + −2 e¿¿
C (s) + 2−2 O¿¿ (melt) CO2 (g) + 4 e−¿¿
Oxygen evolved at the anode burns, the anode material (graphite).
If instead of graphite, some other metal is used as the anode, then O2
liberated will oxidise the metal of the electrode. Since graphite is much cheaper than any metal, therefore, graphite is used as the anode.
CHAPTER – 7
P-BLOCK ELEMENTS 01 MARK
1) Why does PCl3 fume in moisture?
Ans: PCl3 hydrolyses in the presence of moisture giving fumes of HCl.
PCl3 + 3H2O - H3PO3 + 3HCl
2) PH3 forms bubbles when passed slowly in water but NH3 dissolves. Explain why?
Ans:- Due to high electronegativity & small size of N, NH3 forms H- bonds with water and
hence it is water soluble. On the other hand, due to its lower electronegativity and its bigger size. PH3 does not form H-bonds with H2O. As a result, it does not dissolve in H2O and hence escapes
as bubbles.
3) Solid Phosphorus pentachoride behaves as an ionic compound. Explain why?
Ans In the solid state PCl5 exists as an ionic solid [PCl4]+ [PCl6]- in which the cation [PCl4]+
tetrahedral and the anion [PCl6]- is octahedral.
4) Sulphur in vapour state exhibits paramagnetic behavior. Give reason.
Ans In vapour state, sulphur partly exists as S2 molecule which has two unpaired electrons in
the antibonding π-orbitals like O2 and hence, exhibits paramagnetism.
5) In solution of H2SO4 in water, the second dissociation constant, Ka 2 is less than the first
dissociation constant, Ka 1. Explain.
Ans Ka2 is less than Ka1 because the negatively charged HSO4- ion has much less tendency to
donate a proton to H2O as compared to neutral H2SO4.
6) Draw the structure of SO2 molecule. Comment on the nature of two S-O bonds formed in
it. Are the two S-O bonds in this molecule equal? Ans Structure of SO2 molecule is given below:
In SO
2,
sulphur is sp2 hybridised. The molecule of SO
canonical forms. In SO2, both the S-O bonds are covalent and have equal strength due to
resonance.
7) Write the balanced chemical equation for the following reaction : Excess of SO2 reacts with sodium hydroxide solution.
Ans Chemical equation for sodium hydrogen sulphite. 2NaOH + SO2 - Na2SO3 + H2O
Na2SO3 + H2O + SO2 - 2NaHSO3 Sodium hydrogen sulphite
8) Name two poisonous gases which can be prepared from chlorine gas.
Ans Phosgene (COCl2), tear gas (CCl3NO2), mustard gas (ClCH2CH2SCH2CH2Cl) all are
obtained from chlorine gas.. 9) Write two uses of ClO2
Ans i) ClO2 is a powerful oxidizing agent & chlorinating agent
ii) It is an excellent bleaching agent.
10) What inspired N Bartlett for carrying out reaction between Xe and PtF6?
Ans Bartlett found that the first ionization enthalpy of molecular oxygen is almost similar with that of xenon. Thus, after preparing red coloured compound O2+ [PtF6]- , he got inspired for
carrying out reaction between Xe and PtF6 and made efforts to prepare Xe+ [PtF6]- by mixing Xe
and PtF6.
02 MARKS
1) Arrange the following NH3, PH3, A3H3, SbH3 BiH3
in order of –
a) Increasing basic strength b) Increasing reducing character.
Ans a) On moving down the group, size of the elements increases. As the size increases, tendency to attract a proton decreases and thus, the basic character decreases. Hence the
increasing order of basic strength is BiH3 < Sbh3 < AsH3 < Ph3 < NH3
b) The stability of hydrides decreases from NH3 to BiH3 which can be observed from
their bond dissociation enthalpy. Thus the reducing character of hydrides increases in the order – NH3 < PH3 < AsH3 < SbH3 < BiH3
2) Explain the following :
ii) BiCl3 is more stable than BiCl5.
Ans i) NO2 is an odd molecule due to the presence of odd number of valence electrons.
Hence, it readily dimerises to give more stable N2O4 molecule with even number of electrons.
ii) Due to inert pair effect, +3 oxidation state of Bi is more stable than its +5 oxidation state. Thus, BiCl3 is more stable than BiCl5.
3) Draw the structures of white phosphorus and red phosphorus. Which one of these two types of phosphorus is more reactive and why?
Ans Structure of white phosphorus is given below :
Structure of red phosphorus is given below:
White phosphorus is more reactive because of its discrete tetrahedral structure and angular strain.
4) Write the conditions to maximize the yield of H2SO4 by contact process.
Ans The key step in the production of H2SO4 is the oxidation of SO2 to SO3.
SO2(g) + O2(g) 2SO3 (9) ∆
ƒ
Ho = - 196.6 KJmol-1
The reaction is exothermic & reversible and the forward reaction proceeds with decrease in volume. Therefore in accordance with Le chatliers; principle, to maximize the yield of SO3 &
hence of H2SO4, a low temperature (720 K), a high pressure (2 bar) & V2O5 is used as a catalyst.
a) H2S is more acidic than H2O.
b) Sulphur has a greater tendency for catenation than oxygen.
Ans a) Due to decrease in (E-H) bond dissociation enthalpy down the group, acidic character increases. Thus, H2S is more acidic than H2O.
b) Bond energy of S-S bond (213 kJ mol-1) is greater than O-O bond (138 kJ mol-1).
Due to small size of oxygen atom, there is greater lp-bp repulsion in O-O, resulting in weakening of O-O bond more than in S-S bond. Therefore, the tendency of catenation in oxygen is lower than sulphur.
6) Draw the structure of each of the following. a) H2S2O8 b) H2SO4
Ans a) Structure of H2S2O8 is given below:
b) The structure of H2SO4
7) Complete the following chemical equations:-a) NaOH (hot & Conc.) + Cl2 →
b) Cl2 + F2 (excess) →
Ans a) 3Cl2 + 6NaOH → 5NaCl + NaClO3 + 3H2O
(hot & conc)
b) Cl2 + 3F2 (excess) → 2ClF3
8) a) Which neutral molecule would be isoelectronic with ClO-?
b) How are interhalogen compounds formulated and how are they prepared? Ans a) ClO- has 26 electrons. A neutral molecule with 26 electrons is OF
2.
b) Interhalogen compounds are formulated as
| XX XX3| | 5 XX , etc.
The interhalogen compounds can be prepared by direct combination or by the action of halogen on lower interhalogen compounds. The product formed depends upon some specific conditions, eg: Cl2 + F2 437K 2ClF (Equal volume)
Cl2 + 3F2
573K
2CIF3
(Excess)
9) a) How are the following compounds prepared from XeF6?
i) XeOF4
ii) XeO3
Ans i) XeF6 + H2O → XeOF4 + 2HF
ii) XeF6 + 3H2O → XeO3 + 6HF
9) b) XeF2 is linear molecule without a bent. Explain.
Ans XeF2 is linear molecule. According to VSEPR theory, the three lone pairs will occupy the
equalorial positions and two bond pairs will occupy axial positions to minimize lp-lp and lp-bp repulsions.
10) Explain Why –
a) Noble gases have comparatively large atomic sizes.
b) Noble gases form compounds with fluorine & oxygen only.
Ans a) It is due to the reason that noble gases have only vander waals radii while others have covalent radii & by definition, Vander waals radii are larger than covalent radii.
b) Fluorine and oxygen are the most electronegative elements and hence, are very reactive. Therefore, they form compounds with noble gases particularly with xenon.
03 MARKS:
1) What happens when –
a) White phosphorus is heated with conc. NaOH solution in an inert gas atmosphere?
b) Orthophosphorous acid is heated? c) PCl5 is heated.
Ans a) When white phosphorus is heated with Conc. NaOH solution in an inert gas atmosphere, phosphine gas is produced.
b) On heating, H3PO3 disproportionate into H3PO4 and phosphine.
4H3PO3 → 3H3PO4 + PH3
c) When PCl5 is heated, it decomposes into PCl3 and Cl2.
PCl5 → PCl3 + Cl2
2) Complete the following chemical reaction – i) I2 + conc.HNO3 →
ii) HgCl2 + PH3 →
iii) AgCl(s) + NH3 (aq) →
Ans i) I2 + 10HNO3 → 2HIO3 + 10NO2 +4H2O
ii) 3HgCl2 + 2PH3 → Hg3P2 + 6HCl
