1.

METALLURGY
1. What are the difference between minerals and ores?
Minerals Ores
A naturally occurring substance obtained by Minerals that contain a high percentage of metal,
mining which contains the metal in free state or from which it can be extracted conveniently and
in the form of compounds like oxides, sulphides economically are called ores.
etc… is called a mineral.
All minerals are not ores . All ores are minerals.
Contains less percentage of metal Contains high percentage of metal
China clay and Bauxite are minerals of Al. Bauxite is the ore of Al.

2. What are the various steps involved in extraction of pure metals from their ores?
 Concentration of the ore.
 Extraction of crude metal.
 Refining of crude metal.
3. What is the role of Limestone in the extraction of Iron from its oxide Fe2 O3?
 A limestone (CaO) is used as a basic flux, in the extraction of iron from its oxide Fe 2O3.
 The silica gangue present in the ore is acidic in nature, the limestone combines with it to form
Calcium silicate (slag)
CaO(s)+ SiO2(s) → CaSiO3(s)
Flux Gangue Slag

4. Which type of ores can be concentrated by froth floatation method? Give two examples for such ores.
 Sulphide ores are concentrated by froth floatation method.
 Eg : galena (PbS), zinc blende (ZnS)
5. Describe a method for refining nickel.
Nickel is refined by Mond process.
 Impure nickel is heated with carbon monoxide at 350K to form a highly volatile nickel tetracarbonyl.
Ni(s) + 4CO(g) [Ni(CO)4](g)
 When Nickel tetra carbonyl is heated at 460K, it decomposes to give pure nickel.
[Ni(CO)4](g) Ni(s) + 4CO(g)
6. Explain zone refining process with an example.
 This method is based on Fractional Crystallisation
 When an impure metal is melted and allowed to solidify, the impurities will prefer to be in the molten
region.
 The impure metal is taken in the form of a rod
  When the metal rod is heated with a heater, the metal melts.
 The heater is slowly moved from one end to the other end.
 The impurity dissolves in the molten zone. The pure metal will crystallize.
 When the heater moves the molten zone also moves.
 This process is repeated again and again to get the pure metal.
 This process is carried out in an inert gas atmosphere to prevent the oxidation of metals.
 E.g. Germanium, Silicon and Gallium which are used as semiconductor are refined by this process.
7.Using the Ellingham diagram,
(A) Predict the conditions under which
(i) Aluminium might be expected to reduce magnesia.
(ii) Magnesium could reduce alumina.
(B) it is possible to reduce Fe2O3 by coke at a temperature around 1200K
(A) (i) Ellingham diagram for the formation of Al 2O3 and MgO intersects around 1600K. Above this
Temperature Al line lies below the Mg line. Hence we can use Al to reduce Magnesia above 1600K.
(ii) In Ellingham diagram below 1600K Mg line lies below Al line. Hence below 1600K Mg can
reduce alumina.
(B) In Ellingham diagram above 1000K carbon line lies below the iron line. Hence it is possible to reduce
Fe2O3 by coke at a temperature around 1200K.
8. Give the uses of zinc.
 Metallic zinc is used in galvanizing of iron and steel to protect them from rusting and corrosion.
 Used to produce die-castings in the automobile, electrical and hardware industries.
 Zinc oxide is used in the manufacture of paints, rubber, cosmetics, pharmaceuticals, plastics, inks,
batteries, textiles and electrical equipment.
 Zinc sulphide is used in making luminous paints, fluorescent lights and x-ray screens.
9. Explain the electrometallurgy of aluminium.
 Aluminium is extracted by Hall – Herold process
 Cathode :- Iron tank coated with Carbon
Anode: - Carbon rod
 Electrolyte :- Calcium Chloride + Alumina + Cryolite
 Temperature :- 1270K
 Pure Aluminium is deposited at the cathode
 Ionization of aluminia: Al2O3 → 2Al3+ + 3O2-
 Reaction At cathode :- 2Al3+(aq) + 6e- → 2Al(l)
Reaction At anode :- 6O2- → 3O2 + 12e-
  Reaction in Carbon rod :-
C(s) + O2-(aq) → CO + 2 e-
C(s) + 2O2-(aq) → CO2 + 4 e-
 During the above reaction carbon is slowly consumed.
 The pure aluminium is formed at the cathode and settles at the bottom.
 The net electrolysis reaction can be written as follows
4Al3+(melt) + 6O2–(melt) + 3C(s) 4Al(l) + 3CO2(g)
10.Explain the following terms with suitable examples.
(i) Gangue (ii) Slag
(i) Gangue
 The nonmetallic impurities, rocky materials and siliceous matter which are associated with the ore
Which are collectively known as gangue.
Eg : Aluminosilicate is the gangue present in gold ore.
(ii) Slag
 Slag is fusible product, which is formed by the reaction of gangue with a flux.
Eg : CaO + SiO2 CaSiO3
(flux) (gangue) (slag)
11. Give the basic requirement for vapour phase refining.
The two requirements are
 The impure metal should form a volatile compound with suitable reagent.
 The volatile compound should be easily decomposed to give the pure metal.
12. Describe the role of the following in the process mentioned.
(i) Silica in the extraction of copper.
(ii) Cryolite in the extraction of aluminium.
(iii) Iodine in the refining of Zirconium.
(iv) Sodium cyanide in froth floatation.
(i) Silica in the extraction of copper.
In the extraction of copper silica acts as an acidic flux to combine gangue FeO to form slag FeSiO3.
FeO + SiO2 FeSiO3
(ii) Cryolite in the extraction of aluminium.
As Al2O3 is a poor conductor, cryolite improve the electrical conductivity. Cryolite lowers the melting point
of the electrolyte.
(iii) Iodine in the refining of Zirconium.
Iodine is the suitable reagent in the purification of Zr.
Zr(s) (impure) + 2I2 → ZrI4 ( vapour )
 ZrI4 ( vapour ) → Zr(s) (pure) + 2I2
(iv). Sodium cyanide in froth floatation.
In froth floatation process, sodium cyanide is used as a depressing agent. It selectively prevents other metal
sulphides from coming to the froth.
Eg : ZnS is present as impurity in galena (PbS)
13. Explain the principle of electrolytic refining with an example.
 Crude metal is refined by electrolysis. It is carried out in an electrolytic cell.
Cathode: Thin strips of pure metal.
Anode: Impure metal to be refined.
Electrolyte: Aqueous salt solution of the metal with dil. Acid.
 As current is passed, the metal of interest dissolved from the anode and pass into electrolytic solution.
 At the same time same amount of metal ions from solution will be deposited at the cathode.
 Less electro positive impurities in the anode settle down as anode mud.
 Example : Electro refining of silver.
Cathode: pure silver
Anode: impure silver rods.
Electrolyte: Acidified aqueous solution of silver nitrate.
 When current passed, the following reaction takes place.
 Reaction at anode : Ag(s) → Ag+ + e–
 Reaction at cathode : Ag+ + e– → Ag(s)
14. The selection of reducing agent depends on the thermodynamic factor: Explain with an example.
 A suitable reducing agent is selected based on the thermodynamic considerations.
 For a spontaneous reaction, the change in free energy should be negative.
 Thermodynamically, the reduction of metal oxide with a given reducing agent can occur if the free
energy change for the coupled reaction is negative.
 Hence, the reducing agent is selected in such a way that it provides a large negative ΔG value for the
coupled reaction.
 Eg : Reduction of ZnO by carbon monoxide.
ZnO + CO Zn + CO2 ΔG° = +ve (non spontaneous)
 Reduction of ZnO by coke
ZnO + C Zn + CO ΔG° = –ve (spontaneous)
 So, coke is better reducing agent for the reduction of ZnO into Zn.
15. Give the limitations of Ellingham diagram.
 Ellingham diagram is constructed based only on thermodynamic considerations.
 It gives information about the thermodynamic feasibility of a reaction. It does not tell anything about
 the rate of the reaction.
 It does not give any idea about the possibility of other reactions that might be taking place.
 The interpretation of ΔG is based on the assumption that the reactants are in equilibrium with the
products which is not always true.
16. Write a short note on electrochemical principles of metallurgy.

 The reduction of oxides of active metals such as sodium, potassium etc., by carbon is
thermodynamically not feasible.
 Such metals are extracted from their ores by using electrochemical methods. In this technique,
the metal salts are taken in a fused form or in solution form.
 The metal ion present in can be reduced by treating it with some suitable reducing agent or by
electrolysis.
 Gibbs free energy change for the electrolysis process is given by the following expression,
ΔG° = -nFE°
n = number of electrons, F = Faraday, E° = Electrode potential of the redox couple
 If E0 is positive then the ΔG is negative and the reduction is spontaneous and hence a redox
reaction is planned in such a way that the e.m.f of the net redox reaction is positive.

ADDITIONAL QUESTIONS:-
1. Explain the Gravity separation method.
 High specific gravity ore separated from less specific gangue.
 The crushed ore treated with rapidly flowing current of water.
 The lighter gangue particles are washed away by running water.
 Oxide ores and native ores are concentrated by this method.
 Ex. Tin stone, Haematite, Gold ore..
2. Explain the Froth flotation process.
 Principle: The metallic ore particles preferentially wetted by oil while impurities preferentially
wetted by water.
 Example : Sulphide ores like galena (PbS), zinc blende (ZnS)
 Frothing agent : Pine oil
 Collector : Sodium ethyl xanthate
 Depressing agent : Sodiun cyanide
 The crushed ore is mixed with water and pine oil
 Froth is produced by blowing air through this mixture.
 The collector molecule attached to ore particles and makes them water repellent.
 The ore particles rise to the surface along with froth and collected separately.
  Impurities settle at the bottom of the container.
 When sulphide ore of a metal contains other metal sulphide as a impurities depressing agents are
used to selectively prevent other metal sulphides from coming to the froth.
 For Eg. Impurity ZnS present in PbS sodium cyanide (NaCN) is added to depress the
flotation property of ZnS by forming a layer of zinc complex Na 2[Zn(CN)4]
3. Explain the magnetic separation method.

 It is based on the difference in the magnetic properties of the ore and the impurities.
 This method is applicable for ferromagnetic ores.
 Ex. Tin stone ore can be separated from wolframite impurity which is magnetic.
 The crushed ore is added onto an electromagnetic separator. It consisting of a moving belt over two
rollers, one is magnet.
 The magnetic part of the ore attracted towards the magnet and falls near to the magnetic region.
 Non magnetic part of the ore falls away from the magnetic region.

4. Explain acid leaching with example.

 This method is applicable for sulphide ores such as ZnS, PbS etc., it can be done by treating them with
hot aq. H2SO4.
 In this process insoluble sulphide is converted into soluble sulphate and sulphur.

2ZnS + 2H2SO4 + O2 2ZnSO4 + 2S + 2H2O

5. Define Roasting.

 Concentrated ore is heated with excess of oxygen in a furnace below the melting point of the metal.
 This method is applicable for sulphide ores.
 Ex. 2ZnS + 3O2 2ZnO + 2SO2

6. Define calcination.

 Concentrated ore is heated in the absence of oxygen is called calcination.

 This method is applicable for carbonate ores and hydrated ores.
 Ex. CaCO3 CaO + CO2
Fe2O3.3H2O Fe2O3 + 3 H2O

7. Explain Auto reduction reaction.

 Simple roasting of some ores give the crude metal even in the absence of a reducing agent.

HgS + O2 Hg + SO2

8. Explain Liquation

 This method is used to remove high melting point impurities from low melting point metals.
 The impure metal is heated in the absence of air in a sloping furnace.
 Pure metal melts and flows down and collected separately.
 The impurities remain on the slope.
 Eg. Tin, Lead, Mercury, Bismuth..
9. What is cementation?

 Gold is reduced to its elemental state when leached solution is treated with Zinc and the process is
called cementation.
Zn(s) +2[Au(CN)2]-(aq) [Zn(CN)4]2-(aq) + 2Au(s)

10. Explain extraction of copper from its ore.

 Copper extracted from copper pyrites ore.

 Important ore: copper pyrites (CuFeS2)
 Concentration: gravity separation
 Roasting: The concentrated ore is heated in a reverberatory furnace after mixing with silica, an
acidic flux.

2CuFeS2(s) + O2(g) 2FeS(l) + Cu2S(l) + SO2(g)

2FeS(l) + 3O2(g) 2FeO (l) + 2SO2 (g)

FeO (l) + SiO2 (s) FeSiO3 (s)

 The remaining metal sulphides Cu2S and FeS are mutually soluble and form a copper matte.
 It separated from slag and transferred into converting furnace. During the conversion FeS oxidized to
FeO and removed as slag.
 The remaining copper sulphide is further oxidised to its oxide which is subsequently converted to
metallic copper

2Cu2S (l.s) + O2 (g) 2Cu2O (l.s) + 2SO2 (g)

2Cu2O (l) + Cu2S (l) 6Cu (l) + SO2 (g)

 The metallic copper is solidified and it has blistered appearance due to evolution of SO 2 gas formed
in this process. This copper is called blistered copper.

11. Write note Aluminothermic process.

 Metallic oxides such as Cr2O3 can be reduced by an aluminothermic process.

 In this process, the metal oxide is mixed with aluminium powder and placed in a fire clay crucible.
 To initiate the reduction process, ignition mixture magnesium and barium peroxide is used.
BaO2 + Mg BaO + MgO
 During the above reaction a large amount of heat is evolved which facilitates the reduction of
Cr2O3 by aluminium powder.
Cr2O3 + 2Al Al2O3 + 2Cr

12. Define smelting.

 In this method, a flux and a reducing agent such as carbon, carbon monoxide (or) aluminium is added
to the concentrated ore.
 The mixture is melted by heating at an elevated temperature above the melting point of the metal in a
smelting furnace.
 Fe2O3(s) + 3CO (g) 2Fe (s) + 3CO2(g)

13. What is Ellingham diagram?

 The graphical representation of variation of the standard Gibbs free energy of reaction for the formation
of various metal oxides with temperature is called Ellingham diagram.
14. Write observation of Ellingham diagram.
 For most of the metal oxide formation, the slope is positive. Due to Oxygen gas is consumed during the
formation of metal oxides which results in the decrease in randomness.
 The graph for the formation of carbon monoxide is a straight line with negative slope. In this case ΔS is
positive as 2 moles of CO gas is formed by the consumption of one mole of oxygen gas.
 As the temperature increases, generally ΔG value for the formation of the metal oxide become less
negative and becomes zero at a particular temperature.
 Below this temperature, ΔG is negative and the oxide is stable and above this temperature ΔG is positive.
 There is a sudden change in the slope at a particular temperature for some metal oxides like MgO, HgO.
This is due to the phase transition (melting or evaporation).
15. Write note on distillation.
 This method is applicable for low boiling volatile metals.
 In this method the impure metal is heated to evaporate and the vapours are condensed to get pure metals.
 Ex. Zinc, Mercury...
.
 2. P-BLOCK ELEMENTS - I
1. Write a short note on anamolous properties of the first element of p-block.
The following factors are responsible for this anomalous behaviour.
 Small size of the first member
 High ionisation enthalpy and high electronegativity
 Absence of d orbitals in their valance shell
2. Describe briefly allotropism in p- block elements with specific reference to carbon.
 Some elements exist in more than one crystalline or molecular forms in the same physical state.
 This phenomenon is called allotropism. The different forms of an element are called allotropes.
 Allotropes of carbon : Diamond, Graphite, Graphene, Fullerenes, Carbon nanotubes
3. Give the uses of Borax.
 Borax is used for the identification of coloured metal ions
 Used in the manufacture optical and borosilicate glass, enamels and glazes for pottery
 It is also used as a flux in metallurgy and also acts as a good preservative.
4. What is catenation? Describe briefly the catenation property of carbon.
 Catenation is an ability of an element to form chain of atoms. The following conditions are necessary for
catenation.
 The valency of element is greater than or equal to two
 Element should have an ability to bond with itself
 Self bond must be as strong as its bond with other elements
 Kinetic inertness of catenated compound towards other molecules.
 Carbon possesses all the above properties and forms a wide range of compounds with itself and with
other elements such as H, O, N, S and halogens.
5. Write a note on Fisher tropsch synthesis.
 Pressure : less than 50 atm
 Catalyst : metal catalysts
 Temperature: 500 – 700 K yields saturated and unsaturated hydrocarbons.
nCO + (2n+1)H2 CnH(2n+2) + nH2O
nCO + 2nH2 CnH2n + nH2O
6. Give the structure of CO and CO2
Structure of CO :-

 It has a linear structure. The C-O bond distance is 1.128Å

Structure of CO2 :

 Carbon dioxide has a linear structure with equal bond distance for the both C-O bonds.
 In this molecule there is two C-O sigma bond. In addition there is 3c-4e bond covering all the three atoms
7. Give the uses of silicones.
 Used for low temperature lubrication and high temperature oil baths etc...
 They are used for making water proofing clothes
 They are used as insulating material in electrical motor and other appliances
 Mixed with paints to make them resistance towards high temperature, sun light, dampness etc..,
8. Describe the structure of Diborane.
 In diborane two BH2 units are linked by two bridged hydrogens. It has eight B-H bonds.
 It has only 12 valance electrons and are not sufficient to form normal covalent bond.
 The four terminal B-H bonds are normal covalent bonds (2c-2e bond).
 The remaining four electrons have to be used by two bridged B – H – B bonds (3c – 2e bond).
 In diborane, the boron is sp3 hybridised.
Formation of 2c-2e bond:
 Two of the half filled hybridised orbitals of each boron overlap with the 1s orbitals of two
Hydrogens to form four terminal 2c-2e bonds.
Formation of 3c-2e bond:
 B–H–B bond formed by overlapping the half filled hybridised orbital of one boron, the empty hybridised
orbital of the other boron and the half filled 1s orbital of hydrogen.

9. Write a short note on hydroboration.

 Diborane adds on to alkenes and alkynes in ether solvent at room temperature. This reaction is called
hydroboration.
 It is highly used in synthetic organic chemistry, especially for anti Markovnikov addition.
B2H6 + 6RCH=CHR → 2(RCH2-CHR)3B
10. Give one example for each of the following
Icosagens - Ex : Boron
Tetragen - Ex : Carbon
Pnictogen - Ex : Nitrogen
Chalcogen - Ex : Oxygen
11. Write a note on metallic nature of p -block elements.
 The tendency of an element to form a cation by loosing electrons is known as electropositive or metallic
character.
 This character depends on the ionisation energy. Generally on moving down the group ionisation energy
decreases metallic character increases.
 In p-block, the elements present in lower left part are metals while the elements in the upper right part
are non metals.
 All the elements of group 17 and 18 are non metals.
12. Complete the following reactions
a) B(OH) 3 + NH3 ?
B(OH) 3+ NH3 → BN + 3H2O
Boron nitride
b) Na2B4O7 + H2SO4 + H2O →?
Na2B4O7 + H2SO4 + 5H2O → 4H3BO3 + Na2SO4
Boric acid
c) B2H6 + 2NaOH + 2H2O →?
B2H6 + 2NaOH + 2H2O → 2NaBO2+ 6H2
Sodium metaborate
d) B2H6 + CH3OH →?
B2H6 + 6CH3OH → 2B(OCH3)3+ 6H2
Tri methyl borate
e) BF3 + 9H2O →?
12BF3+ 9H2O → 3H3BO3+ 9H++ 9[BF4]–
Boric acid
f) HCOOH + H2SO4→?
HCOOH + H2SO4 → CO + H2O + H2SO4
g) SiCl4 + NH3→?
2SiCl4 + NH3 → Cl3Si-NH-SiCl3 + 2HCl
Chlorosilazanes
h) SiCl4 + C2H5OH → ?
 SiCl4 + 4C2H5OH → Si(OC2H5)4+ 4HCl
i) B + NaOH →?
2B +6NaOH → 2Na3BO3+ 3H2
Sodium borate
j) H2B4O7 → ?
H2B4O7 → 2B2O3+ H2O
13. How will you identify borate radical?
con. H2SO4
B(OH)3+ 3C2H5OH B(OC2H5)3+ 3H2O
Tri ethyl borate

 Tri ethyl borate burns with green edged flame.

14. Write a note on Zeolites.
 Zeolites are three-dimensional crystalline solids containing aluminium, silicon, and oxygen in their
regular 3-D framework.
 They are hydrated sodium alumino silicates with general formula Na 2O.(Al2O3).x(SiO2).yH2O
(x=2 to 10; y=2 to 6).
 It have porous structure in which the mono valent sodium ions and water molecules are loosely held.
 Water molecules freely move in and out these porous.
 Zeolites are similar to clay minerals but they differ in their crystalline structure.
 Zeolites have a three dimensional crystalline structure looks like a honeycomb consisting of a network of
interconnected tunnels and cages.
 It acts as a molecular sieve for the removal of permanent hardness of water.
15.How will you convert boric acid to boron nitride?
Fusion of urea with B(OH)3, in an atmosphere of ammonia at 800 - 1200 K gives boron nitride.
B(OH)3 + NH3 → BN + 3H2O
Boron nitride
16. A hydride of 2nd period alkali metal (A) on reaction with compound of Boron (B) to give a reducing
agent (C). Identify A , B and C.
4 LiH + BF3 → LiBH4 + 3LiF
(A) (B) (C)

Compound Name Formula

A Lithium hydride LiH
B Boron trifluoride BF3
C Lithium Borohydride LiBH4
17. A double salt which contains fourth period alkali metal (A) on heatingat 500K gives (B). Aqueous
solution of (B) gives white precipitate with BaCl2 and gives a red colour compound with alizarin.
Identify A and B.
K2SO4 .Al2(SO4)3 .24H2O K2SO4. Al2 (SO4)3+ 24H2O
(A) (B)

Compound Name Formula

A Potsh alum K2SO4 .Al2(SO4)3 .24H2O

B Burnt alum K2SO4. Al2(SO4)3

18. CO is a reducing agent. Justify with an example.

Carbon monoxide acts as a strong reducing agent and can reduce many metal oxides to metals.
3CO + Fe2O3 → 2Fe + 3CO2

ADDITIONAL QUESTIONS
1. Write the uses of boron.
 10
B5 is used as moderator in nuclear reactors.
 Amorphous boron is used as a rocket fuel igniter.
 Boron is essential for the cell walls of plants.
2. Write the uses of boric acid.
 Boric acid is used in the manufacture of pottery glazes, enamels and pigments.
 It is used as an antiseptic and as an eye lotion.
 It is also used as a food preservative.
3. What is the action of heat on boric acid?

H3BO4 373 K HBO2 + H2O

4HBO2 413 K H2B4O7 + H2O

Red hot
H2B4O7 2B2O3 + H2O

4. What is inorganic benzene? How is prepared?

 Borazole or borazine is called inorganic benzene.
-153 K
3B2H6 + 6NH3 3(B2H6.2NH3)
High temp
3(B2H6.2NH3) clossed vessel 2B3N3H6 + 12H2.
5. Difference between graphite and diamond.
Graphite Diamond
Soft in nature Very hard
Conduct electricity Do not conduct electricity
Sp2 hybridization Sp3 hybridization
Hexagonal sheet shape (flat) Tetrahedral arrangement
Weak vanderwaals force Covalent bond
Used as Lubricant Used for Cutting glasses

6. Why silicones are water repellent?

All silicones are water repellent. This property arises due to the presence of organic side groups that
surrounds the silicon which makes the molecule looks like an alkane.
7. How will you prepare aluminium chloride by McAfee process?
 Aluminium chloride is obtained by heating a mixture of alumina and coke in a current of chlorine.
2Al2O3 + 3C + 6Cl2 4AlCl3 + 3CO2
8. How is potash alum prepared?
 Alum stone is treated with excess of sulphuric acid, the Al(OH) 3 is converted to aluminium sulphate.
 A calculated quantity of potassium sulphate is added.
 Solution is crystallized to get potash alum. It is purified by recrystallisation.
K2SO4.Al2 (SO4)3.4Al (OH)3 + 6H2SO4 K2SO4 + 3Al2(SO4)3 + 12H2O
K2SO4 + Al2(SO4)3 + 24H2O K2SO4.Al2 (SO4)3.24H2O
9. What is burnt alum?
 When potash alum heated at 475 K it loses water of hydration and swells up.
 The swollen mass is known as burnt alum.
K2SO4.Al2 (SO4)3.24H2O K2SO4.Al2 (SO4)3 + 24H2O
10. Give the uses of potash alum.
 Used for purification of water
 Used for water proofing and textiles
 Used in dyeing, paper and leather tanning industries.
11. Write Oxo process.
 Ethene is mixed with carbon monoxide and hydrogen gas to produce propanal.
CO + H2 + CH2 = CH2 CH3-CH2-CHO
12. From Aluminium to Thalium only a marginal difference in ionization enthalphy why?
 Presence of inner d and f- electrons which has poor shielding effect.
 High effective nuclear charge
13. Write water gas equilibrium reaction.
 The equilibrium involved in the reaction between carbon dioxide and hydrogen is called water gas
Equilibrium.
CO2 + H2 CO + H2O
14. What are silicates?
 The minerals which contains silicon and oxygen in tetrahedral units [SiO4]4- linked together in different
patterns are called silicates.
Types of silicates:
 Ortho silicates, Pyro silicates, Cyclic silicates
 Chain silicates, Double chain silicates, Sheet silicates, 3D silicates
15. Write note on Fullerenes.
 They are newly synthesised allotropes of carbon.
 These allotropes are discrete molecules such as C32, C50, C60, C70, C76 etc…
 The C60 molecules have a soccer ball structure and are called buckminster fullerene or buckyballs.
 It has a ring structure and consists of 20 six membered rings and 12 five membered rings.
 Each carbon atom is sp2 hybridised.
16. How is Borax prepared?
 Borax is a sodium salt of tetraboric acid.
 It is obtained from colemanite ore by boiling its solution with sodium carbonate.

2Ca2B6O11 + 3Na2CO3 + H2O 3Na2B4O7 + 3CaCO3 + Ca(OH)2
 6. SOLID STATE
1. Define unit cell.
 A basic repeating structural unit of a crystalline solid is called a unit cell.
2. Give any three characteristics of ionic crystals.
 Ionic solids have high melting points.
 Ionic solids do not conduct electricity, because the ions are fixed in their lattice positions.
 They do conduct electricity in molten state (or) when dissolved in water because, the ions are free.
 They are hard and brittle.
3. Differentiate crystalline solids and amorphous solids.

crystalline solids Amorphous solids

Long range orderly arrangement of Short range, random arrangement of
constituents. constituents.
Definite shape Irregular shape
They are anisotropic in nature They are isotropic nature
They are true solids They are pseudo solids
Definite heat of fusion Heat of fusion is not definite.
They have sharp melting point Do not have sharp melting point.
Examples: NaCl, diamond etc... Examples: Rubber, glass etc...

4. Classify the following solids

a. P4 b. Brass c. diamond d. NaCl e. Iodine

Solid Type
a. P4 Molecular solid
b. Brass Metallic solid
c. Diamond Covalent solid
d. NaCl Ionic solid
e. Iodine Molecular solid

5. Explain briefly seven types of unit cell.

 There are seven primitive crystal systems. They differ in the arrangement of their crystallographic axes
and angles.
 S. No Name of the unit cell Edge length Angles

1. Cubic a=b=c       90

2. Tetragonal a = b c       90

3. Orthorhombic a b c       90

4. Rhombohedral a=b=c       90

5. Hexagonal a = b c     90   120

6. Monoclinic a b c     90   90

7. Triclinic a  b c       90

6. Distinguish between hexagonal close packing and cubic close packing.

Hexagonal close packing Cubic close packing
This is 'aba' pattern of arrangement This is 'abc' pattern of arrangement
Tetrahedral voids of the second layer Octahedral voids of the second layer are
are exactly covered by the sphere of the partially covered by sphere of the third layer.
first layer
6 spheres per unit cell 4 spheres per unit cell

7. Distinguish tetrahedral and octahedral voids.

Tetrahedral voids Octahedral voids
It is formed when a sphere of second layer It is formed when a sphere of second
placed above the void of the first layer. layer partially covers the void in the
first layer.
Coordination number is 4 Coordination number is 6
No. Of tetrahedral voids = 2n No. Of octahedral voids = n
A void surrounded by 4 spheres A void surrounded by 6 spheres

8. What are point defects?

 Point defect is the deviations from ideal arrangement that occurs at point or an atom in a crystalline
substance.
 If imperfections in solids occur due to missing of atoms, displaced atoms or extra atoms. This type
imperfections are called point defects. Ex. NaCl.
9. Explain Schottky defect.
 Schottky defect arises due to the missing of equal number of cations
and anions from the crystal lattice.
 This effect does not change the stoichiometry of the crystal.
 It is occurs in ionic crystal in which cation and anion are almost
similar in size. Example: NaCl.
 Presence of large number of schottky defects in a crystal lowers its density.

10. Write short note on metal excess and metal deficiency defect with an example.
Metal Excess defect:-
 Metal excess defect arises due to the presence of more number
of metal ions as compared to anions. Ex. NaCl, KCl.
 The electrical neutrality of the crystal can be maintained by the
presence of extra cation and electron present in interstitial position.
 The anionic vacancies, which are occupied by unpaired electrons, are called F - centers.

Metal deficiency defect:-

 Metal deficiency defect arises due to the presence of less number
of cations than the anions.
 This defect is observed in a crystal in which, the cations have variable
Oxidation states.
 For example, in FeO crystal, some of the Fe2+ ions are missing from the crystal lattice.
 To maintain the electrical neutrality, twice the number of other Fe 2+ ions in the crystal is oxidized to
Fe3+ ions. In such cases, overall number of Fe2+ and Fe3+ ions is less than the O2– ions.
11. Calculate the number of atoms in a fcc unit cell.
 Number of atoms in a fcc unit cell
Nc Nf 8 6
= + = +
8 4 8 2

=1+3
= 4.
12. Explain AAAA and ABABA and ABCABC type of three dimensional packing with the help of neat
diagram
AAAA type of 3D arrangement:-
 It is occurs in simple cubic arrangement.
 It can be obtained by repeating AAA type two dimensional arrangements in three dimensions.
  Spheres in one layer sitting directly on the top of those in the
previous layers so that all layers are identical.
 Each sphere is in contact with 6 neighbouring spheres.
 Hence coordination number is 6

ABABA type of 3D arrangement:-

 It occurs in bcc arrangement.
 The spheres in the first layer (a) are slightly separated and the
second layer (b) is formed by arranging the spheres in the
depressions between the spheres in layer (a).
 The third layer is the repetition of the first layer.
 This ABABA pattern is repeated throughout the crystal.
 Each sphere is in contact with 8 neighbouring spheres.
 Hence coordination number is 8.

ABCABC type of 3D arrangement:-

 It occurs in fcc arrangement. The first layer (A) is formed by arranging the spheres as in the
case of two dimensional ABAB arrangement.
 The second layer (B) is formed by placing the spheres in the depressions of the first layer.
 A tetrahedral void and octahedral voids are formed in the first layer. The third layer is different
from other two layers.
 Hence it is designated (C).Coordination number is 12

13. Why ionic crystals are hard and brittle?

 Ionic crystals are hard due to strong electrostatic force of attraction between cations and anions.
 They are brittle because ionic bonds are non directional. (Repulsive force)
14. Calculate the percentage efficiency of packing in case of body centered cubic crystal.
Volume of the cube with edge length a is = a3
In ∆ABC,
AC2 = AB2 + BC2
AC2 = a2 + a2
AC2 = 2a2
In ∆ACG,
AG2 = AC2 + CG2
AG2 = 2a2 + a2
AG = 2a 2  a 2
AG = 3a 2
AG = 3 a
3a = 4r
3
r= a
4
Volume of the sphere with radius ‘r’
4
=  r3
3
3
4  3 
=   a 
3  4 

3 3
= a
16

For bcc n=2

 3  3
Total volume occupied by all the spheres in BCC unit cell = 2    a 3  =  a3
 16  8
Total volume occupied by spheres in bcc unit cell
Packing efficiency =  100
Volume of the unit cell

3
 a3  3 
Packing efficiency =
8 100 =    100
a3  8 
= 3 x12.5
= 1.732 x 3.14 x 12.5
= 68%
15. What is the 2D coordination number of a molecule in square close packed layer?
 The two dimensional coordination number of a molecule in square close packed layer is 4.
 In this arrangement each spheres is in contact with four of its neighbours.
16. What is meant by the term “coordination number”? What is the coordination number of atoms in a
bcc structure?
 The number of nearest neighbours that surrounding a particular particle in a crystal is called
coordination number.
 The coordination number of atoms in bcc structure is 8.
18. Aluminium crystallizes in a cubic close packed structure. Its metallic radius is 125pm. calculate the
edge length of unit cell.
 Cubic close packing is based on the fcc unit cell. For fcc

a 2 a
r= (or) r =
4 2 2
a=r2 2
= 125  2 1.414
a = 353.5 pm
19. If NaCl is doped with 10-2 mol percentage of strontium chloride, what is the concentration
of cation vacancy?
 For every Sr2+ ion introduced in NaCl crystal, two Na+ ions are removed to maintain electrical
neutrality. One lattice site occupied by Sr 2+ ion and the other remains vacant.
 One cation of Sr2+ would create one cation vacancy in NaCl.
 Number of cation vacancy created = number of Sr 2+ ions added
 Mole percentage of SrCl2 = 10-2
10-2
Mole of SrCl2 = = 10-4 mol
100
No. of ions = n × NA
= 10-4 × 6.023 × 1023
= 6.023 × 1019 Sr2+ ions
 Number of cation vacancies = 6.023 × 1019
20. KF crystallizes in fcc structure like sodium chloride. Calculate the distance between K + and F− in
KF. (Given: density of KF is 2 48 g cm−3)
Given,
n = 4 (FCC structure)
ρ= 2.48 gcm-3
M = 58 g mol-1
 NA = 6.023 ×1023
a =?
Solution,
nM
=
a3NA

nM 4  58
a3 = =
NA 2.48  6.023 1023

a3 = 0.1553 10-21cm3
a = 0.5375 10-7cm (or) 5.375 10-8cm
1
Distance between K+ and F- =  edge length
2
1
= a
2

= 1  5.375 10-8
2
= 2.6875 10-8 cm

= 2.6875 10-10 cm
= 268.75 10-12 cm

= 268.75 pm
21. An atom crystallizes in fcc crystal lattice and has a density of 10 gcm−3 with unit cell edge length
of 100pm. calculate the number of atoms present in 1 g of crystal.
Given,
ρ= 10 gcm3
n = 4 (FCC structure)
a = 100 pm = 100 ×10-12 m = 100 ×10-10 cm = 1.0 ×10-8 cm
W = 1g; M =?
Solution,
nM
=
a3NA

a3NA 10 (1.010-8)3 6.023 1023

M=
n = = 1.51 g mol-1
4
Molar mass = 1.51 g mol-1
W 1
Number of moles(n) =   0.6623 moles
M 1.51
  Number of atoms in ‘n’ moles = n × N A
 Number of atoms in 0.6623 moles = 0.6623 × 6.023 × 1023
= 3.989 × 1023 atoms
22. Atoms X and Y form bcc crystalline structure. Atom X is present at the corners of the cube and Y is
at the centre of the cube. What is the formula of the compound?
Nc 8
 Number of corner atoms (X) = = =1
8 8

Nf 1
 Number of body centre atoms (Y) = = =1
1 1

 Formula of the compound = XY.

23. Sodium metal crystallizes in bcc structure with the edge length of the unit cell 4. 3 x 108 cm.
Calculate the radius of sodium atom.
 Edge length of the unit cell (a) = 4.3 × 10-8 cm
 Radius of sodium atom (r) =?
3
For BCC structure r = a
4
1.732  4.3  108
=
4
7.4476 108
=
4
= 1.86 108 cm
= 1.86A°

24. Write a on Frenkel defect.

 Frenkel defect arises due to the dislocation of ions from its crystal lattice.
 The ion which is missing from the lattice point occupies an interstitial position.
 This defect occurs in ionic solids in which cation and anion differ in size. This defect does not affect
the density of the crystal.
 For example AgBr, small Ag+ ion leaves its normal site and occupies an interstitial position.
ADDITIONAL QUESTIONS:-
1. What are covalent solids? What are the characteristics of covalent crystals?
 In covalent solids the constituents are atoms.
 They are held by covalent bonds in three dimensional networks.
 Example: Diamond, Silica
 They are hard. They have high melting point.
 Poor thermal and electrical conductors.
2. Write note on molecular solids.
 The constituents are neutral molecules.
 They are held together by weak van der waals forces.
 They are soft, not conduct electricity.
 Example: Glucose, Naphthalene...
3. Write note on metallic solids.
 In metallic solids, the lattice points are occupied by positive metal ions and a cloud of electrons
pervades the space.
 They are hard, and have high melting point.
 They are good conductor of heat and electricity.
 Example: Cu, Fe, Cu-Zn.
4. What is Bragg’s equation?
nλ = 2dsinɵ
Where,
n – Order of diffraction
λ – Wavelength of X-ray
d – Inter planar distance
ɵ – Angle of diffraction
5. Write a short note on impurity defect?
 A general method of introducing defects in ionic solids is by adding impurity ions.
 For example, addition of CdCl2 to AgCl yields solid solutions where the divalent cation Cd2+ occupies
the position of Ag+.
 In order to maintain the electrical neutrality, proportional number of Ag+ ions leaves the lattice.
 This produces a cation vacancy in the lattice.
6. Why ZnO turns yellow on heating?
 On heating ZnO loses oxygen atom and forms a free Zn2+ ion.
 The Zn2+ ion and electrons occupy the interstitial position.
7. What is meant by Isotropy and Anisotropy?
Isotropy Anisotropy
Uniformity in all directions Not uniformity in all the directions
Identical values of physical properties in all Different values of physical properties in
direction different direction
Ex. Glass Ex. NaCl.

8. What is packing efficiency?

 The percentage of total volume occupied by constituent spheres gives the packing efficiency of an
arrangement.
Total volume occupied by spheres in unit cell
Packing efficiency =  100
Volume of the unit cell

9. Calculate the packing efficiency of packing in the case of simple cubic crystal.
 Let us consider the cube with edge length ‘a’.
 Volume of the cube = a × a × a = a3
 Radius of the sphere is (r)
a
Here, a = 2r; r =
2
4
 Volume of the sphere with radius (r) =  r 3
3
3

=    = 
4 a 4 a3
3 2 3 8

 a3
=
6

 For simple cubic n = 1

 a3
 Total volume occupied by spheres in simple cubic = 1×
6

Total volume occupied by spheres in unit cell

Packing efficiency =  100
Volume of the unit cell
 a3
6 100
=  100 =
a3 6

= 52.3%
 Empty space = 47.7 %