KVS REGIONAL OFFICE, RAIPUR STUDY MODULE FOR CHEMISTRY CLASS - XII (MINIMUM LEARNING CONTENTS)

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1 KVS REGIONAL OFFICE, RAIPUR STUDY MODULE FOR CHEMISTRY CLASS - XII (MINIMUM LEARNING CONTENTS) UNIT -1 SOLID STATE Marks=4 Terminology and Definitions:- Solid : The substances having definite shape, Mass and volume are called solid. Characteristic properties of solid : [1] They have definite mass, volume and shape. [2] The inter molecular distances are short. [3] The inter molecular forces are strong. [4] Their constituent particles have fixed positions and can only oscillate about their mean positions. [5] They are incompressible and rigid. Types of Solids : On the basis of the arrangement of the constituents particles they are of 2 types :Crystalline Solid 2. Amorphous solids 1. Crystalline solid [1] In a crystalline solid the particles (atoms, molecules or ions) are arranged in a regular and repetitive three dimensional arrangements [2] These solids have sharp melting point. [3] These solids are Anisotropic. Note : Anisotropic substances/solids are the substances have different values of physical properties (such as electrical conductivity, refractive index, thermal expansion etc.) in different directions. (Phenomenon of showing different values of all physical properties in different directions is known as Anisotropism.) [4] These solids can undergo a clean cleavage. [5] These solids are generally incompressible. Examples: All the metallic elements like iron, copper and silver; Non metallic elements like sulphur, phosphorus and iodine and Compounds like sodium chloride, zinc sulphide and naphthalene 2. Amorphous solid [1] In amorphous solid the particles (atoms, molecules or ions) are arranged in an irregular and non-repetitive three dimensional arrangements. [2] Rapidly solidified liquids are amorphous substances, e.g. Glass, rubber etc. [3] These solids are generally Isotropic. Note : Inotropic substances/solids are the substances have the same values of physical properties (such as electrical conductivity, refractive index, thermal expansion etc.) in all directions. (Phenomenon of showing same values of all physical properties in all direction is known as Isotropism.) [4] These solids on cleavage form smaller pieces with non-planar faces. [5] These solids do not have sharp melting point and boiling point i.e. they melt gradually over a temperature range. [6] These solids are compressible. Que: What makes a glass different from a solid such as Quartz? Under what conditions quartz could be converted into glass?

2 Ans: In glass, amorphous silica ( SiO 2 ) is present. SiO 4 tetrahedral have an irregular arrangement. In quartz, crystalline silica ( SiO 2 ) is present. SiO 4 tetrahedral have a regular arrangement. When quartz (SiO 2 ) is melted and the melt is cooled very rapidly, quartz converted into glass. Que. Why glass is considered as a super cooled liquid? Ans : Amorphous solids have a tendency to flow. Since glass is an amorphous solid, so it is called super cooled liquid or pseudo solid. Que. Why some glass objects from ancient civilizations are found to become milky in appearance? Ans : Glass becomes crystalline at some temperature. For which glass objects from ancient civilizations become milky in appearance because of some crystallization. # Fluid Liquids and gases are called fluids because of their ability to flow. The fluidity is due to the fact that the molecules are free to move about. Que: Why glass pans fixed to windows or doors of old building are invariably found to be slightly thicker at the bottom than at the top? Ans : Due to fluidity property, the glasses flows down very slowly and make the bottom portion slightly thicker. Que: Why solids are rigid in nature? Ans: The constituent particles in solid have fixed position and can only oscillate about their mean position, for which solids are rigid. Que : Name the factors which determine the stability of a substance. Ans: [1] Intermolecular forces tend to keep the molecules or constituent particles closer. [2] Thermal energy tends to keep them apart by making them move faster. # Crystalline Solids can be classified in to the following categories on the basis of the intermolecular forces acting between the constituent particles : Types of solid [1] Molecular solids (a) Non-polar (b) Polar (c)hydrogen bonded Constituent particles Attractive forces Examples Physical nature Molecules Dispersion or London forces Ar, CCl 4, H 2, I 2, CO 2 Soft Dipole-dipole interactions Hydrogen bonding HCl, SO 2 H 2 O ( ice) [2] Ionic solids Ions coulombic or electrostatic NaCl, MgO,ZnS, CaF2 Hard but brittle Soft Hard Electrical conductivity Insulator Insulator Insulator Insulator in solid state but conductors in molten state and in aqueous solutions Melting point Very low Low Low High [3] Metallic solids Positive ions in a sea of delocalised electrons Metallic bonding Fe, Cu, Ag, Mg Hard but malleable and ductile Conductors in solid state as well as in molten state Fairly high [4] Covalent or network solid Atoms Covalent bonding SiO 2 ( quartz), SiC, Diamond, AlN Hard Insulator Very high Graphite Soft Conductor #Lattice point : This is a constituent particles (i.e. spheres, ions, atoms, molecules) of a crystal. # Space lattice or crystal lattice A regular three dimensional arrangement of points in space is called space lattice or crystal lattice. The points represent the constituent particles of the crystal. # Unit cell An unit cell is the smallest portion of the crystal lattice. When it is moved repeatedly a distance equal to its own dimension along each direction, a three dimensional crystal lattice is generated. # Types of unit cell : The unit cells are of 4 types : -

3 1. Simple/Primitive Unit cell : It has lattice points (i.e. spheres, ions, atoms, molecules) at all the corner of a cube. (Therefore, it has total 8 points) 2. Body Centred Unit cell : It has one lattice point at the centre of the cube along with all the points at the corner of a cube. (Therefore, it has total 9 points) 3. Face Centred Unit cell : It has lattice points at the centre of each face along with all the points at the corner of a cube. (Therefore, it has total 14 points) Rank of unit cell ( Z ) The number of particles as points in a unit cell is known as rank. To calculate rank, these points should be noted:- - A point at each corner of unit cell is counted as 1 - The points on an edge are counted as The points at each face are counted as - The points within the unit cell is conted as The rank of unit cell of :- [1] Primitive / simple cubic = 1 8 X 8 = 1 1 [2] Body-centred cubic = X = [3] Face-centred cubic = 8 X X = 4 [4] In hexagonal system,each corner atom shared by 6 hexagonal unit cell.,each face shared by 2 unit cell and inside thebody there are three atoms 1 1 Z= ( X 12 corners) + ( X 2 faces ) + ( 3 inside the body ) = # Packing in metallic crystal:- The identical solid spheres can be packed in a number of ways. [1] In the first layer the spheres are arranged in a hexagonal manner in which each sphere is in contact with six other spheres. [2] In the second layer the spheres will fit into the depression of the first layer. [3] For the third layer, there are two possibilities:- (a) The spheres can be placed in the depression of the second layer i.e. the third layer is directly above the first and the fourth layer is directly above the second. This leads to the arrangement ABABAB This type of arrangement is known as hexagonal closed packed (HCP) structure. e.g. Zinc, magnesium crystallizes in this type of structure. [b] Alternatively, the sphere can be placed in the depressions, of the second layer; do not lie directly above the atoms of the first layer i.e. the spheres in fourth layer lie exactly above the first, fifth above the second, sixth above the third and so on. This leads to the arrangement ABCABCABC. This type of arrangement is cubic closed packed or face centered cubic arrangement. e.g. Cu,Ag.Au crystallizes in this type of structure. # Writing formula of a compound based on the position of its constituent atoms/ions :

4 # Co-ordination number It is the number of atoms or spheres that surrounds the single sphere / atom in a crystal. C.N of tetragonal arrangement = 3 C.N of tetrahedral arrangement = 4 C.N of octahedral arrangement = 6 C.N of body centered cubic arrangement = 8 Any close ( tight ) packing having C.N = 12 i.e. hcp and ccp i.e. fcc having C.N = 12 # Void / Hole/ Interstices The space which is left in between the closest pack arrangement is called void. In close packing two types of voids are created. Relationship between edge length (a) and radius of the sphere ( r ) in unit cell : For. FCC, a = 2 2 r For BCC, a = 3 r For simple cubic, a = 2 r... 4 (Note : Sphere = Atom, molecule, ion or any constituent.) # Relationship between d & a and d & r S. No Type of unit cell d (nearest neighbor distance) and a(edge length) 1 Simple d = a r = a/2 2 BCC d = a r = a FCC d = a r = a a (Edge length) and r (atomic radius) # EFFICIENCY OF PACKING : This is the space occupied by the sphere in a particular unit cell. It is generally shown in % 1. The % of packing fraction in Simple Cubic = 52.36% 2. The % of packing fraction in Body Centred Cubic = 68% 3. The % of packing fraction in Face Centred Cubic = 74% # CRYSTAL DEFECT/Crystal IMPERFECTION/IRREGULARITY (The following defects are found in crystals) Crystal defect Crystal defect Internal irregularities of crystals is known as crystal defect. Line defect Point defect Stoichiometric defect Non-Stoichiometric defect Vacancy defect Interstitial defect Metal excess defect Meta-deficiency defect Schottky defect Frenkel defect Colour centre or F-centre or Anion vacancy Impurity defect Cation vacancy # Single crystals are formed when the process of crystallization occurs at extremely slow rate. # Point defects are the irregularities or deviations from ideal arrangement around a point or an atom in a crystalline substance. # Line defects are the irregularities or deviations from ideal arrangement in entire rows of lattice points.

5 # Point defects do not disturb the stoichiometric of the solid is known as stoichiometric defect or intrinsic defect or thermodynamic defect. # Vacancy defect When some of the lattice sides are vacant, the crystal is said to have vacancy defect. This results the decrease in density of the substance. This defect develops when a substance is heated. # Interstitial defect- When some constituent particles occupy an interstitial site, the crystal is said to have interstitial defect. This defect increases the density of the substance. # Schottky defect [1] It occurs when a pair of ions of opposite charge are missing from the ideal lattice. [2] The presence of a large number of Schottky defect in a crystal lowers its density. [3] This defect occurs if cation and anion having similar size with high coordination number. [4] It is found in NaCl (there is one schottky defect for ions. One c.c of sodium chloride contains ) ions. Therefore, one cubic centimetre (c.c.) of NaCl possesses 10 6 Schottky pair of ions. # Frenkel defect [1] When an ion leaves its position in the lattice and occupy interstitial site leaving a gap in the crystal i.e. it creates a vacancy defect and interstitial defect. [2] This defect will occur if size of cation is smaller than anion, with low coordination number. [3] Frenkel defects are not found in pure alkali halide. Due to larger size of cations, ions can not accommodate in interstitial site. [4] Frenkel defect are found silver halide (AgX), because silver ions are smaller in size and can get into the interstitial site. [5] The Frenkel defect does not change the density of the solid. [6] In AgBr, AgCl both Schottky and Frenkel defects are found. # Differences between Schottky and Frenkel defects : ( These defects are classified under stoichiometric defects i.e. ionic ratio of compound remains same.) Schottky defect Schottky defects Ions leave the lattice structure Electrical neutrality is maintained Density of crystal decreases Found in compounds with higher coordination no. Ionic size should be same e.g. NaCl, CsCl, AgBr, AgCl etc. Frenkel defect Frenkel defects Ions slip into voids in the lattice structure Electrical neutrality is maintained Density of crystal remains same Found in compounds with lower coordination no. Ionic size difference is more, Frenkel defect is found silver halide (AgX), because silver ions are smaller in size and can get into the interstitial site. e.g. AgBr, AgCl etc. N.B.- In AgBr, AgCl both Schottky and Frenkel defects are found. # Metal excess defect : [a] F-Centers ( here F means Farbe (i.e. colour ) : When there is an excess of metal ions in non-stoichiometric compounds, the crystal lattice has vacant anion site. The anion sites occupied by electrons are called F-centre. The F-centers are associated with the colour of the compounds. Excess of K in KCl makes the crystalviolet. Excess of Li in LiCl makes the crystal pink.

6 Solid containing F-centre are paramagnetic, because the electrons occupying the F-centers are unpaired. When the crystal having F-centers are exposed to light, they become photoconductor. # F- centres & its importance F centres : F-centres are sites in an imperfect crystal where electrons are trapped in the anion vacancies. F-centre Importance of F centres: F-centres are responsible for imparting colour to the crystals [b] Metal excess defect due to presence of extra cation at the interstitial site Zinc oxide is white in colour at room temperature on heating, it loses oxygen and turns yellow. Now there is excess of zinc in the crystal and its formula becomes Zn 1+ x O the excess zinc ions move to interstitial site and the electrons to neighboring interstitial site. # Metal deficiency defect FeO, mostly found with a composition of Fe 0.95 O i.e. range from Fe0.93O Fe0.96O. In crystals of FeO, some Fe 2+ ions are missing and the loss of positive charge is made up by the presence of required number of Fe 3+ ions. # Impurity defect This defect arises due to cation vacancies, in some ionic compounds presence of impurity produce some defects they are called impurity defects. e.g. 1. Presence of Cd ++ ion as impurity in AgCl. 2. Presence of Sr ++ ion as impurity in NaCl. # Doping : The process of adding impurities to a crystal so as to change its electrical properties is called Doping. e.g. Mixing of SrCl2 in NaCl solid. Note : 1. Doping increases the conductivity of crystal. 2. Doping is kind of impurity defect. For example, if we mix Strontium chloride ( SrCl2 ) with Sodium chloride, some strontium ( Sr 2+ ) ions occupy the lattice sites of sodium ions ( Na + ) and equal number of sodium ( Na + ) sites remain vacant. Such vacancies in the crystal increase the electrical conductivity because certain ions from the neighboring sites can move into these vacant holes. In this defect the number of positive ions are less as compared to negative ions. Crystals with such defects also act as semiconductor. Since the conductivity is due to holes, these are known as p-type semiconductors. Cation Vacancy Fig. Doping of SrCl 2 in NaCl

7 Que: If NaCl is doped with 10-3 mol % of SrCl2. What is the concentration of cation vacancies? Ans : The addition of SrCl2 to NaCl produces cation vacancies equal in number to that of Sr 2+ ions. No. of moles of SrCl 2 added to one mol of NaCl = 10-3 / 100 = 10-5 mol. No. of holes created in one mole of NaCl = 10-5 X X = x # Semiconductor These are solids whose conductivity lies in between those of conductors and insulators. The conductivity of semiconductors increases with increase of temperature. # Intrinsic semiconductor An insulator capable of conducting electric current at higher temperature or when irradiated with electromagnetic radiations, are known as intrinsic semiconductor. This happens because certain covalent bonds are broken and the released electrons are in a position to conduct electric current. e.g. Silicon, Germanium. # Extrinsic semiconductor These are formed when impurities of certain elements are added (doped) to insulator. # n-type semiconductors It is obtained by doping group 14 elements with group 15 elements. Suppose Si (having 4 valence electrons) is doped with P (with 5 valence electrons), out of 5 valence electrons of P, only 4 valence electrons are involved in bond formation. The 5 th electron is not bound any where and can be easily promoted to the conduction band. The conduction is thus mainly caused by the movement of electrons. # p-type semiconductors It is obtained by doping group 14 elements with group 13 elements. Suppose Si (having 4 valence electrons is doped with Ga (which has 3 valence electrons), 3 valence electrons are involved in bond formation with neighboring Si atom. A vacancy is left which can be filled by the transfer of a valence electron from a neighboring Si atom. The movement of electron into the vacancy leaves behind a hole which carries positive charge. Another electron from a neighboring Si atom can move into the hole leaving behind another hole. It appears as if the hole has moved through the lattice. The movement of positively charged hole is responsible for the conduction of charge. Que. Non-stoichiometric cuprous oxide, Cu2O can be prepared in laboratory. In this oxide, copper to oxygen ratio is slightly less than 2:1. Can you account for the fact that this substance is a p-type semiconductor? Ans : Since Cu 2O is non-stoichiometric oxide, it contains Cu in two oxidation states, +1 and +2. Cu 2+ provides an excess of positive charge. As a result an electron from a neighboring Cu + is transferred to Cu 2+. The transfer of electron leaves behind a hole, which carries an extra positive charge and a negative hole is created. It appears that the positive hole moves through the lattice, hence it appears as P-type semiconductor. # and compounds Combination of elements of Gr 13 and Gr 15 or Gr 12 and Gr 16 produce compounds which stimulate average valence of four as in Ge or Si compounds > ZnS, CdS, CdSe, HgTe compounds > InSn, AlP, GaAs

8 511 Kj/mol Kj/mol # The Band theory to explain electrical properties : The band model of metal is based on molecular orbital theory. When a large no. of orbital overlap in metal, it results a continuous energy level produced by a large number of molecular orbital is called Energy Band Coduction band 2p 0 2s 2 2p band ( conduction band ) 2s- band ( valence band ) [ conductor ] 1s 2 1s- band Valence band [ Semi conductor ] Conductivity of Insulator ohm -1 m -1 [ Insulator ] Semiconductor ohm -1 m -1 Conductor ohm -1 m -1 The lowest unoccupied energy band is known as conduction band The highest occupied energy band is known as valence band The energy difference between the top of valence band and the bottom of the conduction band is known as energy gap. or forbidden zone. # Electrical properties of solids- There are three kinds of electrical properties (i.e. solids may be classified into three categories on the basis of their electrical properties:) 1. Conductors 2. Insulators 3. Semiconductors Conductors- Substances which allow electricity to pass through them easily eg- Metals Insulators- Substances which do not allow electricity to pass through them easily. eg- Rubber, wood, glass Semiconductors- Substances which allow electricity to pass through them partially. eg- Doped Silicon Conductors: The solids having no gap (there is no forbidden zone) between conduction band and filled band. Since energy gap is not there therefore electrons can easily/freely move from one band to another band. Insulators: There is large gap (large forbidden zone) between conduction band and filled band. Electrons can not move easily/freely Semiconductors : There is small gap (Small forbidden zone) between conduction band and filled band. Electrons can move/jump when energy is supplied. # Magnetic properties : [1] Diamagnetic Property/Substance/ Solids : Diamagnetic substances are the substances which are weakly repelled by a magnetic field. The electrons in diamagnetic substances are all paired. They do not contain unpaired electrons. e.g. TiO 2, NaCl, C 6H 6,N 2, Zn

9 [2] Paramagnetic Property/Substance/ Solids Paramagnetic substances are those which are attracted by a magnetic field but they lose their magnetism in the absence of magnetic field. These substances have permanent magnetic dipole, due to presence of atoms, molecules or ions containing unpaired electrons. e.g. Cu 2+, Fe 3+, O 2, NO, CuO, etc. Substances containing unpaired electrons are further classified as: (a) Ferromagnetic Property/Substance/ Solids > Ferromagnetic substances are those substances which are strongly attracted by a magnetic field and can be made into permanent magnets. These substances show magnetism even in the absence of a magnetic field. The large magnetism in these substances is due to the spontaneous alignment of magnetic moment,i.e Unpaired electron in the same direction. These substances lose ferromagnetism and become paramagnetic on certain temperature on heating. (This particular Temp. is called Curie Temperature) Example: Iron, cobalt, nickel, gadolinium and CrO 2 CrO 2 is used to make magnetic tapes for audio recording. (b) Anti-ferromagnetic Property/Substance/ Solids > Antiferromagnetic substances are those substances in which equal number of magnetic moments are aligned in opposite directions so as to give zero net moment. Example: MnO, MnO 2 and Mn 2O 3 (c) Ferrimagnetic substances Ferrimagnetism is observed when the magnetic moments of the domains in the substance are aligned in parallel and anti-parallel directions in unequal numbers They are weakly attracted by magnetic field as compared to ferromagnetic substances. eg : Fe 3O 4 (magnetite) and ferrites like MgFe 2O 4 and ZnFe 2O 4 These substances lose ferrimagnetism and become paramagnetic on certain temperature(called Curie Temperature). on heating.(this particular Temp. is called Curie Temperature) A Few Numericals : Que: Gold (At.radius= 0.144nm) crystallizes in afcc unit cell. What is the length of the side of the unit cell? Ans : Atomic radius= nm So, length of side a = 2 2 r = 2 2 x = nm Que: Aluminum crystallizes in a cubic closed packed structure. Its metallic radius is 125 pm. [1] What is the length of the side of the unit cell? [2] How many unit cells are there in one c.c of aluminium? Ans : [1] For FCC, a = 2 2 r = 2 2 x 125 = 354 pm, So the edge length of the unit cell = 354 x cm [2] Volume of unit cell = a 3 = ( 354 x ) 3 cm 3 Therefore, number of unit cells in 1 cc of aluminum = 1/ ( 354 x ) 3 cm 3 = x Unit cells Que: Ferric oxide crystallizes in a hexagonal closed pack array of oxide ions with two out of every three octahedral holes occupied by ferric ions. Deduce the formula of the ferric oxide. Ans : No. of atoms in one unit cell of hcp structure = 6 No. of oxide ions per unit cell = 6 No. of octahedral voids = 6 Since ferric ions occupy only two out of every three octahedral voids, therefore, no. of octahedral voids occupied by ferric ions = ( 2/3 ) x 6 = 4 Stoichiometric ratio of Fe 3+ and O 2- is 4 : 6 = 2 : 3 Hence the formula of ferric oxide is Fe 2O 3 Que : Analysis shows that nickel oxide has the formula Ni 0.98 O What fraction of nickel exist as Ni 2+ and Ni 3+ ions? Ans : Let amount of Ni 3+ be x mol. Then amount of Ni 2+ is (0.98 x) Total oxidation number of Ni in the compound is 3x + 2 (0.98 x)

10 Oxidation number of oxygen is -2 Since the sum of the oxidation number of all the constituents in a compound is zero. => 3x + 2 (0.98 x) 2 = 0 => 3x x 2 = 0 => x = 0.04 Hence % of Ni 3+ = ( 0.04 / 0.98 ) x 100 = 4.08 % % of Ni 2+ = = % Que : Relationship between density (d) and the dimension of unit cells. Ans : Let the edge length of unit cell be a Therefore volume of unit cell = a 3 Let no. of atoms in unit cell = Z Gm. Atomic mass = M Therefore mass of one atom = M / N A Where N A = Avogadro s number i.e x Mass of Z atoms = ZM N A Density of unit cell ( d ) = Mass of unit cell Volume of unit cell = ZM N A a 3 d = ZM a 3 N A Que : Silver crystallizes in FCC lattice. If edge length of the cell is 4.05 x 10 8 cm and density is 10.5 gm/cm 3. Calculate atomic mass of silver. Ans : Given data- Edge length ( a ) = 4.07 X 10 8 cm Density ( d ) = 10.5 gm/ cm 3 Since silver crystallizes in fcc lattice, so rank, i.e. the no. of silver atoms per unit cell ( z ) = 4 N A = X ZM So d = a 3 M = N A d a 3 N A z M = 10.5 X ( 4.07 X 10-8 ) 3 X X = gm / mol So atomic mass of silver = amu A quick Revision : TERMS Amorphous and Crystalline Solids Molecular solids EXPLANATION Amorphous- short range order, Irregular shape eg-glass Crystalline Solids- long range order, regular shape eg : NaCl Ar, CCl 4, H 2 O (ice) Covalent or Network solid SiO 2 diamond No of atoms per unit cell (z ) Simple cubic -1, BCC- 2, FCC 4, End-Centred- 2 Coordination Number FCC- 6:6 BCC- 8:8 Calculation of number of voids Let the number of close packed spheres be N, then: The number of octahedral voids generated = N The number of tetrahedral voids generated = 2N Relation between r and a Simple Cubic a = 2r, BCC 4r = a 3 FCC 4r = a 2 Packing Efficiency Simple Cubic 52.4%, BCC 68%, FCC 74%

11 Calculations Involving Unit Cell Dimensions M=molar mass (g/mol) cm, N A = a = edge length in Frenkel Defect: Schottky Defect Metal excess defect due to anionic vacancies (F-centres ) Doping This is a defect in which Cation is dislocated to an interstitial site. It does cause change in the density of the solid. Frenkel defect is shown by ionic substance in which there is a large difference in the size of ions, for example, ZnS, AgCl, AgBr and AgI due to small size of Zn 2+ and Ag + ions. It is a vacancy defect. In this defect the number of missing cations and anions are equal. This defect causes decreases in density of the crystal. For example, NaCl, KCl, CsCl and AgBr. When NaCl heated in an atmosphere of Na vapour, the Na atoms deposited on the surface of the solid. The Cl ions diffuse to the surface of the crystal and combine with Na atoms to give NaCl. The released electrons diffuse into the crystal and occupy anionic sites The anionic sites occupied by unpaired electrons are called F - centres They impart yellow colour to the crystals of NaCl. Similarly, excess of lithium makes LiCl crystals pink and excess of potassium makes KCl crystals violet (or lilac). The conductivity of intrinsic semiconductors is increased by adding an appropriate amount of suitable impurity. This process is called doping n / p -type semiconductors n- type : Si + As or Sb or Bi p-type: Si + B or Ga or In or Tl compounds compounds compounds: InSb, AlP and GaAs compounds :ZnS, CdS, CdSe and HgTe Paramagnetic substances Weakly attracted by a magnetic field. Examples: O 2, Cu 2+, Fe 3+, Cr 3+ Diamagnetic substances Weakly repelled by a magnetic field. Example: H 2 O, NaCl and C 6 H 6 Ferromagnetism: A few substances like iron, cobalt, nickel, gadolinium and CrO 2 are attracted very strongly by a magnetic field. Antiferromagnetism Ferrimagnetism: Domains oppositely oriented and cancel out magnetic moment e.g. MnO. domains aligned in parallel and anti-parallel directions in unequal numbers. Example: Fe 3 O 4 (magnetite)

12 (important questions with answer) Q.1) What are the stoichiometric defects found in ionic crystals? (Ans) The point defects in which the ratio of cations to anions remains the same as shown by the molecular formula of the compound are known as stoichiometric defects. Q.2) Write two examples of amorphous solids. (Ans) Rubber and quartz glass are examples of amorphous solids. (Q.3) Name two metals which have cubic close packed structure. (Ans) Silver (Ag) and copper (Cu) have cubic close packed structure. (Q.4) Name the type of solids which are malleable, ductile and electrical conductors. (Ans) Metallic solids are malleable, ductile and electrical conductors. (Q.5) Sometimes, crystals of common salt (NaCl) are yellow instead of being pure white. Why? (Ans) Sometimes, crystals of common salt (NaCl)are yellow instead of being pure white due to the presence of electrons in some lattice sites in place of anions. These sites act as F-centers and impart yellow colour to crystal of common salt. (Q.6) What are voids? (Ans) Voids are the empty spaces present between atoms or ions, when they are packed within the crystal. (Q.7) How many atoms are present in the unit cell of bcc and fcc? (Ans) Number of atoms in unit cell of bcc is2and in fcc it is 4. (Q.8) Fe 3 O 4 is ferrimagnetic at room temperature. What happens to its magnetic properties when it is heated to 850 K? (Ans) When Fe 3 O 4 is heated to 850 K it loses ferrimagnetism and becomes paramagnetic. (Q.9) Name the point defect which lowers the density of a crystal. (Ans) Schottky defect lowers the density of a crystal. (Q.10) Write the unit in which the magnitude of magnetic moment is measured. (Ans) The magnetic moment is measured in Bohr magneton ( B). (Q.11) What do you mean by F-center? (Ans) The anion sites which are occupied by unpaired electron are called F-centres. Q.(12) Name the type of solids which have long range orders. (Ans) crystalline solids have long range order. (Q.13) Name one solid which has both Frenkel and Schottky defects? (Ans) Silver bromide (AgBr) has both Schottky and Frenkel defects. (Q.14) Why is the window glass of the old building thick at the bottom? (Ans) Glass is a pseudo solid, that is, it is a supercooled liquid of high viscosity. It flows down very slowly and makes the bottom portion of window glass of old building slightly thicker. (Q.15) What causes the conduction of electricity by semiconductors? (Ans) Electrons and holes produced by defects cause the conduction of electricity by semiconductors. (Q.16) Name the salt which can be added to AgCl so as to produce cationic vacancies. (Ans) SrCl 2 orcdcl 2 is added to AgCl to produce cationic vacancies. (Q.17) What is a diode? (Ans) Diode is a combination of n- type and p- type semiconductors. It is used as a rectifier. (Q.18) Name a transition metal oxide which has appearance and conductivity like that of Cu? (Ans) Rhenium oxide (ReO 3 ) (Q.19) What is the co-ordination number of an atom present in octahedral void? (Ans) The co-ordination number of atom present in octahedral void is 6. (Q.20) Write one property which is caused due to the presence of F-center in a solid. (Ans) The colours and paramagnetic behaviour of the solid is due to the presence of F-center in a solid. (Q.21) Draw body centred and face centred cubic unit cells. (2 Marks)

13 (Ans) (Q.22) Define: (i) Crystal lattice (ii) Packing efficiency (2 Marks) (Ans) (i)crystal lattice is the well-defined regular arrangement of atoms, molecules or ions in three-dimensional space. (ii) Packing efficiency is the percentage of the total space filled by the particles. (Q.23) Explain the following with suitable example: (i) Molecular solids (ii) Metallic solids (2 Marks) (Ans) (i) In molecular solids, the constituents particles are molecules which are held together byvan der Waals forces of attraction. e.g. I 2. (ii)metallic solids consist of positive ions surrounded by and held together by a sea of free electrons. These free electrons are mobile and are responsible for high thermal and electrical conductivity of metallic solids. e.g. Ag, Cu. (Q.24) What is rank? Find rank of face centered cubic unit cell. (2 Marks) (Ans) Rank is the number of atoms per unit cell of a crystal. In f c c Contribution of atoms present at the corners= Contribution of atoms present at faces = Rank = = 4. (Q.25) What is doping? How does n-type and p-type semiconductors formed? (2 Marks) (Ans) The process of introducing atoms of other elements as impurity into an insulator to make it semiconductor is called doping. Doping of silicon or germanium with electron rich impurities like P, As, Sb results in formation of n-type semiconductors whereas p- type semiconductors are formed by adding elements of group 13 like B, Al,Ga. (Q.26) Explain the following with one example each: (i)ferrimagnetism (ii)antiferromagnetic substances (2 Marks) (Ans) (i) When magnetic moments are aligned in parallel and antiparallel directions in unequal numbers it results in net moment. It is called ferrimagnetism. These substances are weakly attracted by magnetic field as compared to ferromagnetic substances. e.g.: Magnetite (Fe 3 O 4 ) (ii) Antiferromagnetic substances are expected to posses paramagnetism or ferromagnetism but actually they possess zero net magnetic moment. It is due to the presence of equal number of domains in opposite direction. e.g.: MnO. (Q.27) Write two differences between isotropy and anisotropy. (2 Marks) (Ans) Isotropy Anisotropy (i)these substances show identical values of These substances show different values of physical properties in all directions. physical properties in different directions. (ii) Amorphous solids show isotropy. Crystalline solids show anisotropy. (Q.28) Potassium metal crystallizes in bcc. The edge length of unit cell is 4.3 A 0. Find the radius of potassium atom. (2 Marks) (Ans)

14 (Q.29) A solid is made of two elements A and B. Atoms of element A occupy all the tetrahedral sites while atoms of element B are in ccp arrangement. From this data find the formula of the compound. (2 Marks) (Ans) There are 2 tetrahedral sites per atom of B because atoms of element B have ccp arrangement. There are 2 atoms of element A for each atom of element B because all tetrahedral sites are occupied by atoms of element A. Therefore, the formula of the compound is A 2 B. (Q.30) A solid A + B has NaCl closed packed structure. The radius of anion is 245 pm. Find radius of anion. (Ans) The co-ordination number of A + B = 6 ( It has NaCl type structure.) (2 Marks) For this, (Q.31) The edge length of unit cell of NaCl crystal lattice is 5.6A 0. The density of NaCl is 2.2g cm 3. Find the number of formula units of NaCl per unit cell. (2 Marks) (Ans) (Q.32) Find the number of NaCl molecules in a unit cell of its crystal. (Ans) (2 Marks) No. of NaCl molecules in a unit cell of NaCl= 4 (Q.33) A compound is made of two atoms X and Y. Atom X is arranged in ccp and atom Y occupies tetrahedral site. Find the formula of compound. (2 Marks) (Ans) No. of atoms of X = 8 No. of atoms of Y =(8/8 ) + (6/2) = 4 Ratio of X : Y is 2: 1 Formula of compound is X 2 Y. (Q.34) The unit cell of metallic silver is face-centred cubic. What is the mass of a silver unit cell? (Ans) (2Marks) (Q.35) A metal crystallizes as face centered cubic lattice with edge length of 450pm. Molar mass of metal is 50g mol 1. Find the density of metal. (2 Marks) (Ans) (Q.36) A solid has bcc structure. Distance of closest approach between two atoms is 1.73A 0. Find edge length of cell. (2 Marks) (Ans) In bcc, distance of closest approach =[( 3)/2] edge length Or Edge length = [1.73/ 3] = 200pm 2A

15 (Q.37) Classify the following solids as metallic; molecular, amorphous, covalent or ionic. (i) SO 2 (ii) Diamond (iii) I 2 (iv) MgO (iv) Ag (v) Quartz (vi) Ar (3 Marks) (Ans) Metallic solid - Ag Covalent solid - Quartz Molecular solids - I 2, Ar, SO 2 Ionic solids - MgO (Q.38) (i) What are voids? (ii) How a tetrahedral void is different from octahedral void? (iii) Draw structure of tetrahedral and octahedral void. (3 Marks) (Ans) (i) Atoms and ions are spherical in shape. A crystal is formed by close packing of atoms or ions. Since, spheres touch each other only at points, some empty space is left between them. This space is called void or hole. (ii)a tetrahedral void is surrounded by four spheres(atoms), which lie at vertices of regular tetrahedron whereas an octahedral void is surrounded by six spheres(atoms). (iii) (Q.) The density of an atom is 7.2g cm 3. It has bcc structure. The edge length is 288 pm. How many atoms of element does 208g of element has? (3 Marks) (Ans) (Q.39) Find the type of lattice for cube having edge length of 400 pm, atomic wt. = 60 and density = 6.25 g/cc. (3 Marks) (Ans) Let the no. of atoms in a unit cell = x = Mass of one unit cell = Volume of unit cell = (edge length) 3 = = 64 x cm 3 Density = or, Mass = = = 4 The unit cell has 4 atoms It is face centered cubic lattice. (Q.40) A mineral contains Ca, O and Ti. In its unit cell oxygen atoms are present at face centres, calcium atoms at corners and titanium atoms at centre of cube. Find the oxidation number of titanium in the mineral. (3 Marks) (Ans) No. of Ca atoms = No. of O atoms =

16 No. of Ti atoms = Formula of mineral is CaTiO 3 Let oxidation number of Ti = x In CaTiO x + (-2 3) = 0 x = +4 Oxidation state of titanium is + 4 in this mineral. (Q.41) A metal oxide crystallizes in a hexagonal close packed array of oxide ions with two out of every three octahedral holes occupied by metal. Find formula of metal oxide. (3 Marks) (Ans) (Q.42) A metal has cubic lattice. Edge length of lattice cell is 2A 0. The density of metal is 2.4g cm 3.How many units cell are present in 200g of metal. (3 Marks) (Ans) (Q.43) The density of NaCl crystal is 2.155g cm 3 and distance between Na + and Cl is 280 pm. Find value of Avogadro s number. (3 Marks) (Ans) NaCl has fcc structure. In fcc, a (edge length) = = pm = 560 pm For fcc, Z = 4, Na = Avogadro number =? (Q.44) In a face centered cubic lattice atoms of A occupy corner of cell and that of B occupy face centers. One of the A atoms is missing from one corner of a unit cell. Find the simplest formula of compound. (3 Marks) (Ans)

17 UNIT-2 SOLUTIONS Marks : 5 Solid Solutions Temp. Vs Conc. Henry s law. Application of Henry s law. Temp and Solubility of gas Raoult s law for volatile liquids Ideal Solutions Gas in solid Solution of hydrogen in palladium Liquid in solid Amalgam of mercury with sodium Mass %, ppm, mole fraction and molality are independent of temperature, whereas molarity depends on temperature. This is because volume depends on temperature. At a constant temperature, the solubility of a gas in a liquid is directly proportional to the pressure of the gas. p = K H. x K H = Henry s law constant ( greater the K H value means lower the solubility.) 1.To increase the solubility of CO2 in soft drinks, the bottle is sealed under high pressure. 2. To avoid bends, the tanks used by scuba divers are filled with air diluted with helium Solubility of gas increases with decrease of temperature. It is due to this reason that aquatic species are more comfortable in cold waters rather than in warm waters. The partial vapour pressure of each component in the solution is directly proportional to its mole fraction. p 1 α x 1, p 1 = p 1 0 x 1 The solutions which obey Raoult s law over the entire range of concentration are known as ideal solutions. ( For ideal solution ΔmixH = 0, ΔmixV = 0) Example : Solution of n-hexane and n-heptane, Non-ideal Solutions Positive deviation : A-B interactions are weaker than those between A-A or B-B, Example - Mixtures of ethanol and acetone Negative deviations : Forces between A-A and B-B are weaker than those between A-B Example- mixture of phenol + aniline. a mixture of chloroform +acetone Azeotropes Colligative properties Osmosis Osmotic pressure Isotonic solutions Hypertonic Hypotonic Reverse Osmosis Mixtures have same composition in liquid and vapour phase and boil at a constant temp. minimum boiling azeotrope (positive deviation) eg- 95% aq ethanol maximum boiling azeotrope (negative deviation) eg- 68% aq nitric acid Depend on the number of solute particles not upon their nature. Molecules of Solvent flows through the semi permeable membrane from pure solvent to the solution. The extra pressure applied on the solution that just stops the flow of solvent is called osmotic pressure of the solution Two solutions having same osmotic pressure Higher osmotic pressure than a particular soln. Lower osmotic pressure than a particular soln. The direction of osmosis can be reversed if a pressure larger than the osmotic

18 pressure is applied to the solution side. That is, now the pure solvent flows out of the solution Application : Desalination of sea water van t Hoff factor (i) Ratio of normal molar mass to experimentally determined molar mass or as the ratio of observed colligative property to the calculated colligative property. (Q1) Define binary solution? Questions for Practice : (Ans)Binary solution is a solution containing only one solute dissolved in a solvent. (Q2) What is molarity? (Ans) The number of moles of solute dissolved in one litre or 1dm 3 of solution is known as molarity. (Q3) What do you understand by saying that molality of a solution is 0.2? (Ans) This means that 0.2 mol of the solute is dissolved in 1Kg of the solvent. (Q4) Why is the vapour pressure of a liquid remains constant at constant temperature? (Ans) At equilibrium, the rate of evaporation = rate of condensation. Hence the vapour pressure of a liquid is constant at constant temperature. (Q5) Define Azeotropes? (Ans) Constant boiling mixtures are called Azeotropes. (Q6) Which substance is usually added into water in the car radiator to act as antifreeze? (Ans) Ethylene glycol is usually added into water in the car radiator to act as antifreeze. (Q7) Which liquids form ideal solution? (Ans) Liquids having similar structure and polarities. (Q8) Which property of solution depend only upon the number of moles of solute dissolved and not on the nature of the solute? (Ans) Colligative properties. (Q9) Write one example each of solid in gas and liquid in gas solution? (Ans) Solid in gas e.g. Camphor in nitrogen gas. Liquid in gas e.g. Chloroform mixed with N 2 gas (Q10) What is molal elevation constant or ebullioscopic constant? (Ans) The elevation in boiling point which takes place when molality of the solution is unity, is known as ebullioscopic or molal elevation constant. (Q11) Define van t Hoff factor. (Ans) The ratio of the observed colligative property to the theoretical value is called van t Hoff factor. (Q11 A) Two liquids A and B boil at c and c respectively. Which of them has higher vapour pressure at 70 0 c? (Ans)Lower the boiling point, more volatile it is.so liquid A will have higher vapour pressure at 70 0 c. (Q12) What happens when blood cells are placed in pure water? (Ans) Water molecules move into blood cells through the cell walls. So, blood cells swell and may even burst. (Q13) What is the effect of temperature on the molality of a solution? (Ans) No effect. (Q14) Write Henry s law. (Ans) The solubility of a gas in a liquid is directly proportional to the partial pressure of the gas at a given temperature. (Q15) What is the relation between normality and molarity of a given solution. (Ans) Normality = 2 x molarity. (Q16) What is an antifreeze? (Ans) An antifreeze is a substance which is added to water to lower its freezing point. e.g. Ethylene glycol (Q17) Why cutting onions taken from the fridge is more comfortable than cutting onions lying at room temperature? (Ans) The vapour pressure is low at lower temperature. So, less vapours of tear producing chemicals are produced (Q18) What will be the van t Hoff factor for O.1 M ideal solution? (Ans) Van't Hoff factor = 1, because ideal solution does not undergo dissociation or association. (Q19) Name the substances which are used by deep sea divers to neutralize

19 the toxic effects of nitrogen dissolved in the blood. (Ans) They use mixture of helium and oxygen to neutralize the harmful effects of nitrogen. (Q20) Why is Anoxia disease very common at higher altitudes? (Ans) Anoxia is very common at higher altitudes because of lower partial pressure of oxygen at higher altitudes. (Q21) What is the optimum concentration of fluoride ions for cleaning of tooth? (Ans) The optimum concentration of fluoride ions for the cleaning of tooth is 1.5 ppm. [If it is more than 1.5 ppm it can be poisonous and if less than 1.5 ppm it is in effective.] (Q22) What happens to blood cells when they are placed in water containing less than 0.99% (mass/volume) salt? (Ans) The blood cells collapse due to loss of water by osmosis when placed in water containing less than 0.9% (mass/volume) salt. (Q23) Why is molality generally preferred over molarity as unit for expressing concentration of solutions? (Ans) Molality is preferred over molarity as unit for expressing concentration of solutions because it involves mass term which is not affected by change in temperature while molarity involves volume term which changes with temperature. (Q24) Calculate the van't Hoff factor of K 4 [Fe(CN) 6 ]. (Ans) K 4 [Fe(CN) 6 ] 4 K + + [Fe(CN) 6 ] 4- Since one molecule of K 4 [Fe(CN) 6 ] dissociates to produce 5 ions, the value of van't Hofff factor is 5. (Q25) Name the law which explains the relationship between solubility of a gas in a liquid and pressure above the liquid surface. Also write the name of the variable which is kept constant for this law. (Ans) This is Henry s law and temperature is kept constant for this law. [According to Henry's law the partial pressure of gas in vapour phase is proportional to the mole fraction of gas in the solution at constant temperature.] (Q26) How does Henry constant vary with the solubility of gas in the solution at a given pressure? (Ans) Higher the value of K h (Henry constant), lower is the solubility of gas in solution. (Q27) What care should be taken while preparing intravenous injection? (Ans) They should be prepared in aquatic medium and salt concentrations should be same as blood plasma levels. (Q28) Why osmotic pressure is better technique for determination of molar mass of biomolecules? (Ans) Osmotic pressure is better technique for the determination of molar mass of biomolecules because they are generally not stable at higher temperatures and have poor solubility. Therefore, other techniques like elevation in boiling point and depression in freezing point cannot be used to determine molecular masses of biomolecules. (Q29) What will be the molecular mass of acetic acid when it is dissolved in benzene? (Ans) Normal molecular mass of acetic acid (CH 3 COOH) is 60. It dimerises when dissolved in benzene. Therefore, molecular mass of acetic acid will be 120 when it is dissolved in benzene. (Q30) How does increase in temperature affects the solubility in endothermic and exothermic dissolution of substances? (Ans) The solubility in endothermic dissolution increases while in exothermic it decreases. (Q31) What will be the shape of graph which is obtained by plotting partial pressure of a gas against mole fraction of gas in solution? (Ans) It is a straight line since partial pressure of a gas and mole fraction of gas are directly proportional to each other. (Q32) Aquatic animals are more comfortable in cold water than in warm water. Explain? (Ans) This is because Kh (Henry constant) values for both N 2 and O 2 increase with increase in temperature indicating that the solubility of gases increases with decrease in temperature. (Q33) Deep sea divers are advised not to come to surface immediately from deep waters. Why? (Ans) Deep sea divers are advised not to come to surface immidiately from deep waters because sudden change in outside pressure can be fatal for divers because N 2 will bubble out of blood vessels causing severe pain and can be dangerous. (Q34) What are colligative properties? Name them. (2 Marks) (Ans) The properties of solute which depend upon the number of particles present in definite amount of solvent but not on the chemical nature of solute are called colligative properties. They are-

20 (i) Relative lowering of vapour pressure (ii) Elevation in boiling point (iii) Depression in freezing point (iv) Osmotic pressure (Q35) Which is better method for expressing concentration of solution Molarity or Molality? (2 Marks) (Ans) Volume changes with temperature whereas mass does not change with temperature. So molality, which does not have volume term in it is better method for expressing concentration. (Q36) Define- (i) Mole fraction (ii) Molality (2 Marks) (Ans) ( i) The mole fraction of a particular component in a solution 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. (ii) Molality of a solution is defined as a number of moles of solute dissolved per 1000g of solvent. (Q37) The osmotic pressure of human blood is 7.65 atm at 37 C. For injecting glucose solution it is necessary the glucose solution has same osmotic pressure as of human blood. Find the molarity of glucose solution having same osmotic pressure as of human blood. (2 Marks) (Ans) = Or 7.65 = = = Molarity = 0.30 (Q38) Vapour pressure of two liquid A & B are 120 and 180mm Hg at a given temp. If 2 mole of A and 3 mole of B are mixed to form an ideal soln, calculate the vapour pressure of solution at same temperature. (2 Marks) (Ans) Total moles = = 5 P solution = = = = 156mm. (Q39) Density of 1 M soln of glucose 1.18g/cm 3. K f for H 2O is 1.86 Km 1. Find freezing point of solution. (Ans) Mass of solution =volume x density =1000 x 1.18 =1180g Mass of water = =1000g = = = = = 1.86 C = 1 m (Q40) An aqueous solution freezes at C. K f = 1.86, K b = Find elevation in boiling point. (2 Marks) (2 Marks) (Ans) Or m = = = 0 (0.186) = C = = (Q41) Vapour pressure of dilute solution of glucose is 750 mm of Hg at 373K. Calculate the mole fraction of solute. (2 Marks) (Ans) 373K = 100 C Vapour pressure of pure water = 760 mm Hg. P = 750 mm Hg (given)

21 Or (Q42) Ethylene glycol solution having molality 0.5 is used as coolant in a car. Calculate the freezing point of solution (given K f =1.86 C/mole) (2 Marks) (Ans) (Q43) Calculate the molality of solution prepared by dissolving 18g of glucose 500g of water. (Ans) Mol. Wt. of glucose =(12 x 6) + (1 x 12) + (16 x 6) (2 Marks) = 180 Molality = = (Q44) Find the vant Hoff factor for Al 2 (SO 4 ) 3 (Ans) (2 Marks) Total ions produced = = 5 (Q45) On a hill station pure water boils at C. The K b of water is C Kg mol 1. Calculate the boiling point of 0.69m solution of urea. (2 Marks) (Ans) Boiling point of solution = Boiling point of water + = = C (Q46) A soln of ethanol in water is 1.6 molal. How many gms of ethanol is present in 500g of solution. (2 Marks) Ans) Mass of ethanol =Molality x Molecular weight = 1.6 x 46 = 73.6 g Total of mass of solution = = g g of solution contain 7.6g of ethanol. Mass of ethanol in 500g of solution = (73.6 / )x 500 (Q47) List two conditions that ideal solutions must satisfy. (2 Marks) (Ans) 1. H mixing and V mixing of ideal solutions should be zero. 2.They should obey Raoult s law over the entire range of concentrations. (Q48) Explain ideal and non-ideal solutions with respect to intermolecular interactions in a binary solution of A and B. (2 Marks) (Ans) For the given binary solution of A and B, it would be ideal if A-B interactions are equal to A-A and B-B interactions and it would be non-ideal if they are different to each other. The deviation from ideal behaviour will be positive if A-B interactions are weaker as compared to A-A and B-B. The deviation will be negative if A-B interactions are stronger as compared to A-A and B-B. (Q49) (i) What are minimum boiling and maximum boiling azeotropes? (ii) Can azeotropes be separated by fractional distillation? (2 Marks)

22 (Ans) (i) Minimum boiling azeotropes are the non-ideal solutions showing positive deviation while maximum boiling azeotropes are those which show negative deviation. Because of positive deviation their vapour pressures are comparatively higher and so they boil at lower temperatures while in case of negative deviation, the vapour pressures are lesser and so higher temperature are required for boiling them. (ii) No, azeotropes can t be separated by fractional distillation. (Q50) (i) When a non-volatile solute is added to solvent,there is increase in boiling point of solution. Explain. (ii) Define ebullioscopic constant and give its units. (2 Marks) (Ans) (i) When a non-volatile solute is added to a volatile solvent the vapour pressure of pure solvent decreases because a part of the surface is occupied by non-volatile solute which can t volatilise. As a result, the vapour pressure of solution decreases and hence, the solution requires a comparatively higher temperature to boil causing an elevation of boiling point. (ii) Ebullioscopic constant is defined as the elevation in boiling point of a solution of a non-volatile solute when its molality is unity. Its units are K Kg mol -1 (Q51) One molal solution of a given solvent is always less concentrated than one molar solution. Explain. (2 Marks) (Ans) In one molar solution one gram mole of solute is dissolved in one litre of solution while in case of one molal solution same one gram mole of solute is dissolved in 1000 gm of solvent only, which on considering normal density parameters of water, can t be lesser in amount than solvent part present in one litre solution. Therefore,more amount of solvent is present in one molal solution than in one molar solution. (Q52) State Raoult s law. Prove that it is a special case of Henry law? (2 Marks) (Ans) Raoult s law states that partial pressure of a volatile component of a solution is directly proportional to its mole fraction. It is a special case of Henry law because it becomes the same when Kh (Henry constant) is equal to pressure of pure solvent. (Q53) How did Van t Hoff explain the abnormal molecular masses of electrolytes like KCl in water and non-electrolytes like benzoic acid in benzene. (2 Marks) (Ans) The molecular mass of KCl in aqueous medium has been observed to be almost half than expected and it has been explained as dissociation of KCl into K + ions and Cl - ions when actual no. of particles become double and so become the colligative properties but since molecular mass is always inversely proportional to colligative property it becomes almost half. In case of benzoic acid in benzene, association of molecules take place when they dimerise and their no. becomes almost half and so molecular mass doubles as a result. (Q54) Calculate molality of an aerated drink having 2.5 gm of carbonic acid dissolved in 150 gm of water. (2 Marks) (Ans) Molality= moles of solute / Mass of solvent in Kg = [(2.5 / 62.)/ (150 / 1000)] = (Q55) When 3.49 gm of a non-volatile solute was dissolved in 125 gm of benzene, its boiling raised by 1.23K. Calculate the molecular mass of non-volatile solute. (K b for benzene is 2.53 K kg Mol -1 ) (2 Marks) (Ans) (Q56) When 5.29 g of a non-volatile solute was dissolved in 400 g of water, its freezing point decreased by 1.79 K. Calculate the molecular wt of solute if K f for water is 1.86 KkgMol -1 (2 Marks) (Ans) (Q57) Calculate the mass of ethanol which is present in 500g of 1.6m solution of ethanol in water. (2 Marks)

23 (Ans) (Q58) One litre seawater is found to contain g of dissolved oxygen. Calculated the ppm of dissolved oxygen in seawater. (Density of seawater = 1.03 g/cc). (2 Marks) (Ans) (Q59) When a non-volatile solute is added to pure water its vapour pressure decreases by 4 mm Hg. Calculate molality of solution. (Vapour pressure of pure water is 40mm Hg) (2 Marks) (Ans) (Q60) What are azeotropes? Give an example of maximum boiling azeotrope. (2 Marks) (Ans) Azeotropes are binary mixtures having the same composition in liquid and vapour phase and boil at a constant temperature. Mixture of nitric acid and water is an example of maximum boiling azeotrope. (Q61) A decimolar solution of NaCl exerts OP( ) of 4.6 atm at 300K. Find the degree of dissolution. (3 Marks) (Ans) NaCl Na + + Cl - Initial moles Moles at Equilibrium Total moles at equilibrium = = = 86.8% (Q62) An aqueous solution of a non-volatile and non-electrolyte substance boils at C. Calculate osmotic pressure of this solution at 27 C. K b (for water) per 1000g = (3 Marks) (Ans) = = = 0.5 C

24 = Molality of solution = 1 Solvent is water. density of solution = 1 Volume of solution = volume of solvent = 1000/1 = 1000ml = 1 L = nrt = (Q63) The vapour pressure of benzene at certain temperature is 640mm Hg. To of benzene, non-volatile and non-electrolyte solid-weighing 2.175g was added. The vapour pressure of solution was 600mm of Hg. Find the mass of the solute? (3 Marks) (Ans) or = = or m = = (Q64) The vapour pressure of water at 296K is 19.8 mm of Hg, 0.1 mol of glucose dissolved in 178.2g of water. Calculate the vapour pressure of resultant solution. (3 Marks) (Ans) n glucose = 0.1 (given) n H 2 O = x (water) = (Q65) A solution is prepared by dissolving 30g of non-volatile non-electrolyte solute in 90g water. The vapour pressure of solution was 2.8 K Pa at 298K. When 18g of water was further added to it, the vapour pressure became 2.9 k Pa at 298K. Calculate molar mass of solute. (3 Marks) (Ans) (When A is for H 2 O) n = moles of solute 2.8 = (1) 2.9 = (2) Dividing eq. (1) and (2), we get = n = Mass of solute = 30 and molecular mass = 23 (Q66) Vapour pressure of pure water is 40mm. If a non-volatile solute is added to it, vapour pressure falls by 4 mm. Calculate molality of solution. (3 Marks) (Ans) The solution has 0.1 moles of solute in 0.9 moles of water. Mass of water = Molality of solution = = 6.17m (Q67) Conc. H 2SO 4 has a density 1.9g/ml and is.99% H 2SO 4 by weight. Find molarity of solution. (3 Marks) (Ans) Mass of 1000 ml of H 2 SO 4 = density volume = = 1900 g Mass of H 2 SO 4 present in 1900 g (1L) of H 2 SO 4

25 = = 1881 g Mole of H 2 SO 4 present in 1L = Molarity = = = M (Q68) A solution contains 25% water, 25% ethanol and 50% acetic acid by mass. Find mole fraction of each of the component. (3 Marks) (Ans) Moles of water = n 1 = Moles of = n 2 = Moles of = n 3 = Total moles in solution = Mole fraction of water = Mole fraction of = Mole fraction of = (Q69) Which will have more osmotic pressure and why? Solution prepared by dissolving 6g/L of CH 3COOH or Solution prepared by dissolving 7.45g/L of KCl (3 Marks) (Ans) Moles of CH 3COOH = Moles of KCl = Molar concentration of both the solutions is same. KCl ionizes into K + and Cl where as CH 3 COOH does not ionize: Osmotic pressure is colligative property. Its value depend on number of particles. Since, KCl produces more ions so, osmotic pressure of KCl will be more than that of CH 3 COOH. (Q70) A solution is prepared by mixing 50ml of chloroform and 50ml of acetone. What will be the resulting volume of solution? 100 ml or >100 ml or <100 ml. (3 Marks) (Ans) Intermolecular H-Bonding When chloroform and acetone are mixed, they form intermolecular hydrogen bonds. The hydrogen bonds are strong forces of attraction. As a result, volume of the solution will be less than 100ml. (Q71) One litre sample of seawater is found to contain 5.8 x 10-3 g of dissolved oxygen. Calculate the concentration of dissolved oxygen in seawater in ppm. (Density of seawater = 1.03 g/cc). (3 Marks) (Ans) Mass = Volume x density = = 1030g Mass of oxygen present in 10 6 (one million) gof seawater = = (Q72) A solution is prepared by mixing 50ml of chloroform and 50ml of acetone. (i) The volume of the resulting solution will be 100 ml or less than 100ml or more than 100 ml?

26 (ii) What happens to the net vapour pressure of resultant solution- it increases or decreases or remains same? Support your answer with suitable explanation. (3 Marks) (Ans) (i) When chloroform and acetone are mixed they exhibit negative deviation from ideal behavior because intermolecular hydrogen bonds are formed between molecules of acetone and chloroform. Since the hydrogen bonds are comparatively stronger forces of attraction, the volume of the solution decreases. Therefore, the volume of the resulting solution will be less than 100ml. (ii) The net vapour pressure of the solution decreases because newly formed hydrogen bonds are stronger forces of attraction than the original forces of attraction existing in separate solutions. (Q73) Find molarity of a conc. H 2 SO 4 sample having ρ 1.9 g/ml & is 99% pure by weight.(3 Marks) (Ans) Mass of 1000 ml of H 2 SO 4 = Density Volume = 1.9 g/ml 1000 ml = 1900 g Mass of H 2 SO 4 present in 1900 g (1Litre) of H 2 SO 4 = = 1881 g Molarity of sulphuric acid = = = (Q74) What are colligative properties? Why are they also called democratic properties? Is osmotic pressure a colligative property? Prove it. (3 Marks) (Ans) These are the properties of solution that depend on the number of solute particles in the solution not at all on the nature of solute particle. Above mentioned is also the reason as why are they sometime called as democratic properties. Yes, osmotic pressure is a colligative property because osmotic pressure of a solution is proportional to its molarity at a given temperature. (Q75) Why semi permeable membrane is so important in the phenomenon of osmosis? What are isotonic, hypo tonic and hyper tonic solutions? Does osmosis take place in all three types of solutions? (3 Marks) (Ans) The semi permeable membrane is very important in the phenomenon of osmosis because they only permit the movement of solvent molecules through them. Two solutions having similar osmotic pressure at a given temperature are called isotonic solutions. If the given solution has less osmotic pressure it is called hypo tonic and it is hyper tonic if its osmotic pressure is higher than the solution on the other side of semi permeable membrane. Osmosis takes place only in hypo tonic and hypertonic solutions. (Q76) List three general applications of phenomenon of osmosis? (3 Marks) (Ans) 1. Shrinking and swelling of blood cells when put into aqueous NaCl having concentration more or less than 0.9%. 2. Water movement from soil into plant body. 3. Preservation of meat and fruits by salting against bacterial infection. (Q77) Why is reverse osmosis of great practical utility to meet potable water requirements these days? Can it make sea water drinkable? Do we use ordinary semi permeable membranes for this? (3 Marks) (Ans) Reverse osmosis is of great practical utility since it can result in absolute purification of any water sample because semi permeable membrane allows only solvent molecules to pass through. It can make sea water drinkable though ordinary semi permeable membrane may not be able to sustain high pressure observed in desalination of sea water i.e. reverse osmosis of sea water. Semi permeable membrane of cellulose acetate is generally used for this. (Q78) The following solutions which have higher osmotic pressure is 1. Solution having 6gL -1 of CH 3 COOH 2. Solution having 7.45gL -1 of KCl Explain. (3 Marks) (Ans)

27 (Q79) Calculate the boiling point of a solution of urea prepared by dissolving g of urea in 250 g of water on a hill station where pure water boils at C. (K b of water is C Kg mol 1 ) (3 Marks) (Ans) (Q80) Calculate the mole fraction of each of the component in a solution containing 25% water, 25% ethanol and 50% acetic acid by mass. (3 Marks) (Ans) (Q81) A solution has been prepared by dissolving 30g of non-volatile solute in 90g of water. The vapour pressure of solution is 2.8 K Pa at 298K. When 18g of water was added to it, the vapour pressure rose to 2.9 kpa at same temperature. Calculate molar mass of solute. (3 Marks) (Ans) According to Raoult s law the vapour pressure of a component of a solution is equal to the product of its vapour pressure in pure state and its mole fraction So (When A is for H 2 O) n = moles of solute Dividing equation (1) by (2), we get = n = Molecular Mass of solute = 30 x = (Q82) Calculate van t Hoff factor of the following solutions 1. Al 2 (SO 4 ) 3 is dissolved in water 2. MgCl 2 is dissolved in water 3. Benzoic acid is dissolved in benzene 4. Glucose is dissolved in water gm of KCl is dissolved in 1 Litre of water (5 Marks) (Ans) 1.When Al 2 (SO 4 ) 3 is dissolved in water, its one molecule dissociates into 2 Al 3+ ions and 3 SO 4 2- ions, so its van t Hoff factor for this is When MgCl 2 is dissolved in water, its one molecule dissociates into 1 Mg 2+ ions and 2 Cl - ions, so its van t Hoff factoris 3. 3.When benzoic acid is dissolved in benzene, its two molecule associate to form a dimer. Therefore, its van't Hoff factor is 1/2.

28 4. Glucose does not undergo association or dissociation, hence van t 'Hoff factor for glucose is 1. 5.When KCl is dissolved in water, its every molecule dissociates in two ions. So its van t Hoff factor for this is 2. (Q83) The osmotic pressure of human blood is 7.65 atm. Find the molarity of glucose solution that can be injected into human blood. If adding some more amount of glucose into it disturbs its molarity and now it boils at C, find its new molality at 27 0 C. What will be the effect on osmotic pressure of solution? K b (for water) per 1000g = 0.50 (5 Marks) (Ans) As we know for injecting glucose solution into human blood it is necessary that glucose solution has same osmotic pressure as of blood. So osmotic pressure of solution should be 7.65 atm. = 7.65 = (Human body temperature is 37 0 C) = = Molarity = 0.30 Now since elevation in boiling point has taken place with the addition of more glucose, it can be calculated as: = = = 0.5 C = Molality of solution = 1 The osmotic pressure of solution will increase because no of solute particles will increase. (Q84) Calculate the vapour pressure of a solution prepared by mixing gm of component A (Mol. Mass 134) and gm of component B (Mol. Mass 169) at 298 K. Vapour pressures of pure components A and B are 421mm Hg and 562mm Hg respectively. Also calculate mole fraction of A and B in vapour phase. (5 Marks) (Ans) As shown above component B is more volatile than component A and vapour phase is richer in component B (Q85) 3.9 gm of a protein was dissolved in 63 gm of benzene when its freezing point was observed to go down by 3.26 K. Kf for benzene is 4.9 K Kg Mol-1. Calculate the percentage association of protein if it forms dimers in the solution?( molar mass of protein is 925) (5 Marks) (Ans) Wt of protein= 3.9 gm Kf = 4.9 K Kg Mol-1 Wt of benzene= 63 gm Depression in freezing point= 3.26 K Thus experimental molecular mass of protein in benzene is Now suppose the degree of association of protein be X and 1-X mol of protein will be left unassociated and

29 corresponding X/2 as associated at equilibrium. Therefore total no. of moles of particles at equilibrium will be 1-X+X/2 = 1-X/2 So total no moles at equilibrium equals von t hoff factor (i) But as we know i = Theoretical molar mass/ Experimental molar mass Therefore, degree of association of protein in benzene is only 1.2% (Q86) A solution of 10 gm NaCl in 1000 gm water freezes at 0.604K. Calculate the degree of dissociation of NaCl. (K f for water is 1.86Kkgmol -1 ) (5 Marks) (Ans)

30 UNIT -3 ELECTROCHEMISTRY Marks =5 1. Differences between electrochemical reaction and electrolysis. Electrochemical reaction Chemical reaction produce electricity It is spontaneous, G is -ve Electrolysis. Electricity causes chemical reaction to take place It is non-spontaneous, G is +ve 2. Galvanic cell, its representation- Representation of Electrochemical Cell : Anode / anodic electrolyte(m) ǁ Cathodic electrolyte(m) / cathode Example : Zn/ Zn ++ (0.1M) ǁ Cu ++ (0.12M)/Cu Functions of Salt Bridge 1. Salt Bridge maintains electrical neutrality in solutions. 2. It completes circuit internally Alphabet rule for an electrochemical cell- According to this rule we write that term first which come first alphabetically as show below- L.H.S. ANODE OXIDATION NEGATIVE Note :Remember LOAN (Left, Oxidation, Anode, Negative charge)as an abbreviation for anodic properties of Electrochemical Cells R.H.S. CATHODE REDUCTION POSITIVE

31 3. SPECIFIC CONDUCTANCE & MOLAR CONDUCTANCE, And their calculation : Specific conductance- The conductance of 1 cm 3 / 1m 3 of the solution of an electrolyte. It is represented by ƙ (kappa) Molar Conductance- It the conductance due to all the ion produced from 1 mole of an electrolyte in V cm 3 of the solution. It is represented by λ c m at conc. C and λ m at infinite dilution. Relationship between them-

32 4. Effect of dilution on K and λ m λ m for strong electrolyte increases constantly on dilution. λ m can be calculated on extra plotting the curve. λ m for weak electrolyte steeply (sharply) on dilution. λ m can not be calculated on extra plotting the curve. It is determined using Kohlrausch s law. Factors affecting electrolytic conductance of an electrolyte :- 5. Kohlrausch law, its application- Kohlrausch Law- Each ion makes a definite contribution to the total molar conductivity of an electrolyte at infinite dilution irrespective of the nature of other ion of the electrolyte. Application of Kohlrausch Law - It helps us to calculate λ m of weak electrolytes (if the λ m of strong electrolytes or ions is known.) Numericals related to Kohlrausch Law :

33 6. NERNST equation (to show the effect of conc. of electrolyte and temperature on E and E cell ), its different forms, writing Nernst equation for an electrochemical reaction i) Nernst equation for electrode potential- ii) Nernst equation for electrode potential- For an electrochemical reaction at equilibrium E cell = 0 then, From E cell we can calculate G using following formula 7. Application of electrochemical series- i) To compare the relative oxidizing and reducing powers- A chemical species with higher value of E will be stronger oxidizing agent and a chemical species with more ve E will be stronger reducing agent ii) To compare the relative activities of metals-a metal with more ve E value will be more reactive iii) To calculate the E cell of given cell- E cell = E cathode - E anode or E cell = E RHS - E LHS N.B. - E for SHE is taken 0 (Zero) iv) To predict whether a metal will displace hydrogen from its compound or not- All the chemical species which have ve E value will displace hydrogen from its compounds M M n+ + ne - (electron lost is gained by H + ion to undergo reduction and H 2 is produced) 2 H + + 2e - H 2 v) To predict spontaneity of a redox reaction- For a cell E cell / E cell is +ve the cell reaction will be spontaneous

34 8. Commercial cells- These are of 4 types 1) Dry cell (primary cell) It is a Non rechargeable cell. Used in Torch, Wall clock etc. The following reaction takes place at Anode and Cathode in the Dry cell : The above reaction takes place at anode i.e. Zn plate The above reaction takes place at cathode i.e. Carbon rod 2) Mercury cell/button Cell It is used in hearing aids and watches etc. The following reaction takes place at Anode and Cathode in the Mercury cell/button Cell : The above reaction takes place at anode i.e. Zn container The above reaction takes place at cathode i.e. Carbon rod 3) Lead storage battery It is a Rechargeable battery. Used in automobiles, Rechargeable batteries used in house hold appliances etc. 4) Fuel cell : This cell is based on the combustion of fuel like Hydrogen by oxygen: Advantages of Fuel Cells : 1. As works as long as we supply Hydrogen & oxygen gas 2. It causes no pollution 3. Its efficiency is 70% 4. Its By product (Water) is also useful. The following reaction takes place at Anode and Cathode in the fuel cell.

35 3. Predicting products of electrolysis- The nature of the product obtained at cathode or anode depends upon E value of given chemical species. The chemical species which has most ve E will form product at cathode and The chemical species which has least ve/most +ve E will form product at anode. Inert electrodes do not take part in chemical reactions Faraday s 1 st Laws Faraday s 2nd Laws The amount of chemical reaction which occurs at any electrode during electrolysis is proportional to the quantity of electricity passed through the electrolyte (solution or melt). The amounts of different substances liberated by the same quantity of electricity passing through the electrolytic solution are proportional to their chemical equivalent weights (Atomic Mass of Metal Number of electrons required to reduce the cation). Product of Electrolysis NaCl (molten) Cathode : Na + (l) + e Na(s) Anode : Cl ½Cl 2 +e NaCl (aq) Cathode : H 2 O (l ) + e ½H 2 (g) + OH Anode : Cl ½Cl 2 +e H 2 SO 4 (dil) Cathode : H + + e - ½ H 2 Anode: 2H 2 O(l ) O 2 (g) + 4H + (aq) + 4e H 2 SO 4 (conc) Cathode : H + + e - ½ H 2 Anode: 2SO 2 4 (aq) S 2 O 2 8 (aq) + 2e AgNO 3 (aq)-ag electrodes Cathode : Ag + (aq) + e - Ag(s) Anode: Ag(s) Ag + (aq) + e - AgNO 3 (aq)- Pt electrodes Cathode : Ag + (aq) + e - Ag(s) Anode: 2H 2 O(l) O 2 (g)+4h + (aq)+4e CuCl(aq)- Pt electrodes Cathode : Cu + (aq) + e - Cu(s) Anode: 2H 2 O(l) O 2 (g)+4h + (aq)+4e Standard Hydrogen Electrode- Definition of STANDARD HYDROGEN ELECTRODE (S.H.E.) : Electrode is used to determine E value of a given half cell by connecting this electrode to the other half cell. The E value of other Half cell is determined according to the following steps : 1.We will constitute one cell (SHE + Given Half Cell). 2. The EMF of the above cell is determined which is also called E cell. 3. Knowing the EMF/ E cell the E of given half cell is determined as- E cell = E cathode - E anode 4. E value of S.H.E. is taken Zero. (Note:S.H.E. may work as anodic or cathodic half cell) Corrosion- It is a process of eating away of metals on their surfaces, it is an unwanted process as it results in loss of mass of metals. In this process metal surface reacts with atmospheric oxygen to form a layer of oxide. It is an electrochemical reaction

36 The following reaction takes place during rusting : Anodic Reaction Cathodic Reaction Some Important questions : Q1. How much electricity in terms of Coulomb is required to reduce 1 mol of Cr 2 O 2-7 to Cr 3+. Ans:- 2Cr 2 O Cr +3, 2Cr +6 +6e Cr 3+ Therefore the coulomb of electricity required =6F, =6x96500 C= C Q2. What is Fuel Cell? Ans:- Fuel cell is a device which produce the energy during the combustion of fuels like Hydrogen, Methane, Methanol. Q3. A solution of CuSO 4 is electrolysed using a current of 1.5 amperes for 10 minutes. What mass of Cu is deposited at the cathode? (Atomic mass of Cu=63.7) Ans:- The reaction is Cu +2 (aq.) + 2e Cu (s) The mass of copper deposited=e Cu X I X t = 63.7 x 1.5 x10 x 60/ 2x96500C =0.297 g C Over all Reaction Q4. Calculate the equilibrium constant for the reaction Cu (s) + 2Ag+ Cu Ag(s) E o Cu 2+ /Cu = +0.34V, E o Ag + /Ag = V. Q5. Write the Nernst equation and emf of the following cells at 298K: Sn/Sn 2+ (0.050M)//H + (0.020M)/H2(g)/Pt(s) E o Sn 2+ /Sn= V Q6. Calculate the standard free energy change for the following reaction at250c, Au(s) +Ca +2 (1 M) Au 3+ (1M) + Ca (s),the electrode potential values are Ca 2+ /Ca = 2.87V,Au 3+ / Au = +1.50V. Predict whether the reaction will be spontaneous or not at 250C. Q7. How do you account for conductivity of strong and weak electrolyte with concentration? Plot the graphs also. Q8. State Kohlrausch law. Calculate Limiting molar conductivity of NaCl, HCl and NaAc (Sodium Acetate) are 126.4, &91 SCm 2 mol -1.Calculate Limiting molar conductivity of HAc (Acetic Acid). Ans:- According to this law, Molar conductivity of an electrolyte, at infinite dilution can be expressed as the sum of contributions from its individual ions. It the molar conductivity of the cation is denoted by Λ o + and that of the anions by Λ o - then the law of independent migration of ions is Λ o m=v + Λ o + + v - Λ o -. Q9. Resistance of conductivity cell filled with 0.1molL -1 KCl solution is 100 ohm. If the resistance of the same cell when filled with 0.02molL -1 KCl solution is 520 ohm. Calculate the conductivity & molar conductivity of 0.02molL -1 KCl solution. The conductivity of 0.1 moll -1 solution of KCl is 1.29Sm -1. Q10. A Copper silver is set up. The copper ion concentration is 0.10M. The concentration of silver is not known. The cell potential measured 0.422V.Determine the concentration of silver ion in the cell. E o (Ag + /Ag) = +0.80V, E o (Cu 2+ /Cu)= +0.34V. Q11. A voltaic cell is set up at 25 0 C with the following half cells: Al(s)/Al 3+ (0.001M) and Ni 2+ (0.50)/Ni(s),Write the equation for the cell reaction that occurs when the cell generates an electric current and determine the cell potential (given E o Ni 2+ /Ni = -0.25V, E o Al(s)/Al 3+ =-1.66V)

37 Q12. Write the reaction involved in the following cells: (a) Fuel Cell (b) Lead Storage Battery. Q13. Three electrolytic cells A,B,C containing solutions ZnSO 4,AgNO 3,and CuSO 4 respectively are connected in series.a Steady current of 1.5 amperes was respectively are connected in series.a steady current of 1.5 amperes was passed through them until 1.45g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited? Q14. Conductivity of M acetic acid is X 10-6 S cm -1. Calculate its molar conductivity. If Λ 0 for acetic acid is S cm 2 mol -1. What is its dissociation constant? Q15. (a) Two half cell reactions of an electrochemical cell are given below: MnO H+ + 5 e H 2 O E o =+1.51V Sn 2+ Sn e- E o = +0.15V. Construct the redox reaction from the two half cell reaction and predict if the reaction favours formation of reactants or product shown in the reaction (b). How much electricity in terms of Faraday is required to produce (i) 20g of Ca from molten CaCl 2 (ii) 40g of Al from molten AlCl 3 VALUE BASED QUESTIONS : Q.1 In news paper Yogesh read that iron pipes of sewerage system of the town will be protected by cathodic protection method (sacrificial method). His friend Juned said in this method something will replace iron to protect it from rusting. Explain the above method for Yogesh and also write the value associated with. Q.2 Sanjay went to buy dry cell for wall clock. His father instructed him to check date of manufacturing before buying the cells. Sanjay was wondering if buying new cells with seal packed what to do with manufacturing date. If Sanjay is write? Give your opinion and the value associated with. UNIT 5 CHEMICAL KINETICS Marks = 5 The branch of chemistry which deals with the study of reaction rates and their mechanisms is called kinetics. Rate of a chemical reaction Change of molar concentration per unit time rate = Change of molar concentration / time R P Rate of reaction = - [R]/ t = + [P]/ t +ve sign for decreasing in concentration -ve sign for increasing in concentration Average rate = decrease in concentration of R / time taken = -[R]/ t = increase in concentration of P / time taken = + [P]/ t Instantaneous rate. Rate of change of concentration of any one of reactant or product at a particular moment of time R P t 0 - [R]/ t = -d[r]/dt t 0 [P]/ t = -d[p]/dt Calculation of reaction rate 3H 2 + N 2 2NH 3 Rate = - 1/3 d[h 2 ]/dt = -d[n 2 ]/dt = ½ d[nh 3 ]/dt 2HI H 2 + I 2 Rate = -1/2 d[hi]/dt = d[i 2 ]/dt Rate law = it is the mathematical expression in which reaction rate is given in terms of molar concentration of reactions.

38 aa + bb cc + dd Rate α [A] x [B] Y Rate = k [A] x [B] Y Where K = rate constant X and Y may or may not equal to a and b. The value of x and y are determined from experiment. Order of reaction the sum of powers raise to the concentration terms in the rate law equation which is determined experimentally. Rate = k[a] 0 zero order Rate = k[a] 1 first order Rate = k[a] 2 second order Zero order: - The rate of reaction is proportion to the zero power of the concentration of the reactant. R P Rate law for zero order reaction = Rate = K [R] 0 Derivation for zero order Rate = K[R] 0 Rate = -d[r] 0 /dt -d[r]/dt=k[r] 0 =K d[r]=-k dt On integration, d[r]= - K dt [R] = -Kt+C ( C = integration constant) At t=0 [R]=[R] o [R] o = 0 + C ie C=[R] 0 [R] = - Kt +[R] 0 Kt =[R] 0 [R] K = 1/t x {[R] 0 [R]} [R] 0 = initial concentration [R] =final concentration * First Order Reaction Rate of reaction depend on one conc. term i.e. Rate of reaction is proportion to first power of concentration of the reactant. R P, its rate law will be Rate =K [R] 1 Derivation of integrated rate equation for first order reaction- Rate =K [R] 1 Rate = -d[r]/dt =K[R] 1 d[r]/[r]= -K dt On integrating both sides d[r]/[r]= -K dt Taking log both side [R 0 ] = Initial conc. of reactant [R] = Final conc. of reactant ln [R] = -Kt +C, When t=0,[r] =[R 0 ], So ln [R 0 ] = C ln[r]=-kt+ ln[r 0 ] ln[r]/[r 0 ] = -kt or kt = ln[r 0 ]/[R] kt =2.303 log [R 0 ]/[R]

39 k= 1/t x log [R 0 ]/[R] At t=⅟ 2 [R]= [R 0 ]/2 For zero order Half-life:- It is a time at which the concentration becomes half of its initial concentration. K=[R 0 ] {1/2[R 0 ]}/t 1/2 K=[R 0 ]/2k For first order K= 2.303/t x log[r] 0 /[R] K =2.303/t 1/2 x log 2[R 0 ]/[R 0 ] K =2.303/t 1/2 x K= 0.693/ t 1/2 *Pseudo order of reaction- Those reactions which appear to be of high order but actually follow lower Order kinetic Example: hydrolysis of ethyl acetate, decomposition of ammonia, decomposition of HI 2NH 3 N 2 + 3H 2 ; rate law = Rate =K [NH 3 ] 0 HI H 2 + I 2 ; rate law = Rate =K [HI] 0 1 st order reaction N 2 O 5 2NO 2 + 1/2 O 2 H 2 O 2 H 2 O + ½ O 2 2 nd Order reaction - H 2 +I 2 2HI units of rate constants Zero order: mol L -1 s -1 First order: s -1 Second order: mol -1 L s 1 Elementary reaction: The reaction taking place in one step is called elementary reaction. Complex reaction when a sequence of elementary reaction gives us the product, the reaction are called complex reaction. Molecularity the number of reacting molecules taking part in an elementary reaction. Unimolecular when one molecule involved. Bimolecular when two molecule involved. Ex of Bimolecular: H 2 +I 2 2HI Rate determining step: In complex reaction the overall rate of the reaction is controlled by the slowest step in a reaction called rate determining step. Rate of reaction become almost doubled for every 10 o c rise in temp. Arrhenius equation- K= A e -Ea/RT Where, A =frequency factor, Ea =activation energy R =gas constant (8.314 j/k mol) log K =log A E a /2.303RT logk 2 /K 1 =E a /2.303R x(1/t 1-1/T 2 ) Activation energy the excess of energy which must be supplied to the reactant to undergo chemical reaction. It is the energy required to form intermolecular called activation complex. Lower the value of activation energy faster will be rate of a reaction. Collision frequency (Z) : The number of collision that takes place per second per volume of reaction Effective collision The collision among the reacting species which result in the product are called effective collision.

40 Factors affecting the rate of reaction 1. Concentration- The rate of chemical reaction at a given temperature Depends on concentration of one or more reactant and products. Rate =K [A] x [B] y. 2. Temperature most of reaction are accelerated by increase in temperature For a chemical reaction with rise in temperature by 10 o C, the rate constant is double. K= A e -Ea/RT 3. Catalyst A catalyst is a substance which alters the rate of a reaction without itself undergoing any chemical change. Catalyst participates in a chemical change by forming temporary bond the reactant and give complex intermediate. 4. Nature of reaction: if depends on chemical bond. 5. Physical states of reactant: it depends on state. i.e. - solid liquid and gas.

41 Intext Question 1. R P. Concentration of reactant from 0.03M to 0.02M in 25 minutes. Calculate the average rate. ANS: Average rate = - / t= / 25(60) = -0.01M/1500s. 2. 2A Product, concentration of A from 0.5 to 0.4 in 10 minutes. Rate=? ANS: Average rate = = x( )/10min = 0.1/20min=0.005M/min. 3. A+B = Products. Rate Law = r k[a] 1/2 [B] 2, order =?, ANS: Order=1/2+2= X Y. If concentration of X is increase to three times. Rate of formation of Y=? ANS: r = k[x] 2, r =k[3x] 2 r /r =k(3x) 2 /kx 2 = 9. Therefore rate of conc. of Y increases to 9 times. 5. Rate constant = 1.15 x 10-3 /s. How long 5g of this to reduce to 3g? ANS: t= log(r 0 /R) R 0 =5g, R =3g, t = log(5/3) =4445s 6. To decompose SO 2 Cl 2 to half is 60 minutes, given: 1 st order reaction. Calculate k. ANS: k= = =1.925 x 10-4 /sec. 7. Rate of reaction doubles with every 10 0 rise in temp.. If temp. increases from 295K to 305K. E a =? ANS: log(k 2 /k 1 )= (1/T 1 1/T 2 ) Log2 = (1/295 1/305) 0.301= x E a =51855 J/mol. 8. 2HI H 2 + I 2, E a =209.5 KJ at 581K.Calculate fraction of molecules of reactant having energy equal to or greater than E a. ANS: x =n/n = e -Ea/RT ln x =, log x = log x = = x = Antilog( )= 1.471x Examples 1. Rate =k[a] 1/2 [B] 3/2 order =1/2 + 3/2=2

42 2. k =2.3 x 10-5 Lmol/s It is 2 nd order reaction st order reaction, N 2 O 5 2NO 2 + O 2,[R 0 ]=1.24 X 10-2 After 60 minutes [R] =0.2x 10-2,Rate constant =? log(r 1 /R 2 ) = k (t 2 t 1 )/2.303 k = 2.303/(t 2 -t 1 ) X log (R 1 /R 2 ) = xlog(1.24 x 10-2 /0.2x 10-2 ) = log( 6.2) /min k = /min 4 1 st order, k = 5.5 x /s,t 1/2 =? ANS: t 1/2 = t 1/2 =0.693/5.5 x =1.26 x s 5. Show that in 1 st order reaction,time for 99.9 % is 10 times to half-life of the reaction. k= log (R 0 /R) = log t=6.909/k, t 1/2 =0.693k t/t 1/2 = x =10 6. Rate constant at 500K =0.02/s at 700K =0.07 /s. Calculate E a and A. log(k 2 /k 1 )= (T 2 -T 1 /T 1 T 2 ) log(0.07/0.02)= ( /700x500) = E a x x 10-4 /19.15 E a = J k=ae -Ea/RT, 0.02= Ae /8.314(500) A= C 2 H 5 I C 2 H 4 + HI, 1 st order reaction, T=600K,k =1.6 x 10-5 E a =209KJ/mol, at 700 k=? logk 2 logk 1 = (1/T 2-1/T 1 ) logk 2 =logk 1 + (1/T 2-1/T 1 ) =log(1.6x10-5 ) + (1/600-1/700) logk 2 = k 2 =6.36 x 10-3

43 Important questions for Practice : Q1. Define Pseudo order reaction? Q2. The decomposition reaction of ammonia gas on platinum surface has a rate constant = 2.3 x 10-5 L mol -1 s -1. What is the order of the reaction? Q3 Mention the factors that affect the rate of a chemical reaction. Q4 From the rate expression for the following reactions determine their order of reaction and dimensions of the rate constants. a) H2O2 (aq) + 3 I - (aq) + 2H + 2H2O (l) + 3I -1 Rate = k [H2O] [I - ] b) CH 3 CHO (g) CH 4(g) + CO(g) Rate = k [CH3 CHO] 3/2 Q5.. A reaction is first order in A and second order in B. i) Write differential rate equation. ii) How is the rate affected when concentration of B is tripled? iii) How is the rate affected when the concentration of both A and B is doubled? Q6. The decomposition of NH3 on platinum surface is zero order reaction. What are the rates of production of N2 and H2 if k = 2.5 x 10-4 mol L -1 S -1? Q7. Derive the Integrated rate equation for first order reaction. Also find half life period and plot the graph associated to it. Q8. For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction. Q9. A first order reaction has a rate constant min -1.If we begin with 0.10 M concentration of the reactant, what concentration of the reactant will be left after 3 hours. Q10. The half-life for radioactive decay of 14 C is 5730 years. An archaeological Artifact containing wood had only 80% of the 14 C found in a living tree. Estimate the age of the sample Q11. What is the effect of temperature on the rate constant of a reaction? How Can this temperature effect on rate constant be represented quantitatively? Q12. The rate of a reaction quadruples when the temperature changes from 293K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature. Q13. A first order reaction takes 40 min for 30% decomposition. Calculate t1/2.. Q14 (a) Distinguish between order of reaction & Molecularity. (b) For a decomposition reaction the values of rate constant k at two different temperatures are given below:k1 =2.15 x 10-8 L mol -1 s -1 at 650K, k2 =2.39 x 10-7 L mol -1 s -1 at 700K Calculate the value of Activation Energy for this reaction. Q15. (i) Write short notes on the following: (a) Activation energy of a reaction (b) Elementary step in a reaction (c)rate of a reaction (ii) The following result has been obtained during the kinetic studies of the reaction2a + B C+D Experiment [A] mol L-1 [B] mol L-1 Intial rate mol L -1 min -1 I x10-3 II X10-2 III X10-1 IV X10-2 Determine the rate law and rate constant for the reaction.

44 Q16 From the concentrations of C 4 H 9 Cl (butyl chloride) at different times given below, calculate the average rate of the reaction:c 4 H 9 Cl + H 2 O C 4 H 9 OH + HCl during different intervals of time.t/s [C 4 H 9 Cl]/mol L Definition of : UNIT - 5 SURFACE CHEMISTRY Marks = 4 i) Adsorption- A surface phenomenon in which molecules of a substance (adsorbate) are retained on the surface of another substance (adsorbent). It is an exothermic reaction ii) desorption- It is an opposite reaction of adsorption. In this process molecules of adsorbate leave the surface of adsorbent. It is an endothermic reaction iii) sorption- When adsorption and absorption both take place simultaneously the it is known as sorption. i) adsorption isotherm- It a graph plotted between extent of adsorption and pressure of the gas at constant temperature. ii) homogeneous catalysis- The catalysis process in which all the reactants and catalyst are in same phase.

45 iii) Heterogeneous catalysis- The catalysis process in which the reactants and catalyst are not in the same phase. iv) Sol- A colloidal solution in which dispersed phase is solid and dispersion medium is liquid. v) Emulsion- A colloidal solution in which dispersed phase is liquid and dispersion medium is also liquid. vi) Micelle- A colloidal solution which is formed by the association of large no. of species supplied by dissociation of molecules at certain higher concentration. Soap in water at or above CMC CMC- The minimum concentration at or above which formation of micelle takes place. vii) viii) Kraft temperature- The formation of micelle takes place only above a particular temperature known as Kraft temperature ix) Tyndall effect-it may be defined as the scattering of light by the colloidal particles in a colloidal solution kept in a dark room. In this effect the path of the light becomes visible. x) Zeta potential- The potential difference between the fixed layer and diffused layer of opposite charges is called the zeta potential. xiv) Electrophoresis : Electrophoresis- It is migration of colloidal particles towards one of the electrode under the influence of an external electric field.

46 2. Positive and negatively charged colloidal solutions Oil in water and water in oil emulsions- Oil in water In this type of emulsion dispersed phase is oil and medium is water e.g. milk, vanishing cream Water in oil In this type of emulsion dispersed phase is water and medium is oil e.g. butter, cream 3. Applications of adsorption

47 Applications of colloids- 4. Difference- Physical and chemical adsorption- Lyophilic and lyophobic colloids Lyophilic colloids In this type of colloid there is greater affinity between d. phase and d d.mdm These colloids are reversible These are formed easily These are quite stable Lyophobic colloids In this type of colloid there is poor affinity between d. phase and d d.mdm These colloids are irreversible These are not formed easily These are quite unstable

48 Multimolecular and Macromolecular Colloids - Multimolecular These colloids are formed by combination of large no. of molecules of a substance. Exa.- Gold sol, sulphur sol Macromolecular- When molecules of dispersed phase are big enough to form colloidal dimension then they form macromolecular colloids. These colloids are quite stable Exa.- Starch, synthetic rubber Steps involved in the process of Heterogeneous catalysis- Examples of Enzymatic reactions- Examples of Industrial Catalysis - Formation of Micelle- 1. In water soap molecules behave like normal electrolyte at lower conc. 2. The dissociated part of soap molecules get attached with dirt particles (Fig. b) 3. All the soap molecules get attached to dirt particles through their tail( hydrophobic end) keeping head (Hydrophilic end) outward to form micelle

49 Chemical methods of preparation of colloids- 5. Origin of charge on colloidal particles- Preferential adsorption theory- The colloidal particles acquire +ve or ve charge by the adsorption of +ve or ve ion present in the medium. The colloidal particles adsorbs that ion which is common with its composition Some Important Questions : (Q.) What is meant by critical temperature of gas? (Ans) Critical temperature is the minimum temperature above which a gas cannot be liquefied howsoever high the pressure may be applied. (Q.) Give the expression for Fruendlich adsorption isother (Ans) x/m= kp 1/n (Q.) What do x and m represent in the expression x/m=kp 1/n (Ans) m is the mass of the adsorbent and x is the number of moles of the adsorbate when the dynamic equilibrium has been achieved between the free gas and the adsorbed gas. (Q.) Why is heterogeneous catalysis also known as surface catalysis

50 (Ans) In heterogeneous catalysis the reaction always starts at the surface of the catalyst. So, it is also known as surface catalysis. (Q.) What is a hydrosol? (Ans) A colloid in which the dispersion medium is water is known as hydrosol. (Q.) Define peptization? (Ans) Peptization is a process of converting a precipitate into colloidal particlesby adding suitable electrolyte. (Q.) Define Brownian movement? (Ans) Brownian movement can be defined as continuous zig- zag movement of the colloidal particles in a colloidal sol. (Q.) Why is Brownian movement important? (Ans) Brownian movement opposes the force of gravity and does not allow the colloidal particles to settle down, thus making the colloidal solution stable. (Q.) Differentiate between adsorption and absorption. (Ans) Adsorption Absorption a)it occurs only at surface a) it is a bulk phenomena b)concentration on the surface b) concentration is same is more than in the bulk through out the material (Q.) What is the effect of temperature on adsorption? (Ans) Adsorption processes, being exothermic, decreases with increase in temperature. (Q.) When a finely powdered active carbon is stirred into a solution of a dye, the intensity of color in solution decreases. Why? (Ans) The intensity of color in the solution decreases because of gas adsorbedon the surface of carbon. (Q.) Why do finely divided substances have larger adsorption power? (Ans) Finely divided substances have large surface area for adsorption and hence have larger adsorption power. (Q.) What are zeolites? (Ans) Zeolites are aluminosilicates i.e. three dimensional network silicates in which some silicon atoms are replaced by aluminium atoms. (Q.) Why are zeolites called shape selective catalysts? (Ans) Zeolites are called shape selective catalysts because their catalytic action depends upon the size and shape of the reactant and the product molecules as well as on their own pores and cavities. (Q.) A small amount of silica gel and that of anhydrous CaCl 2 are placed separately in two corners of a vessel containing water vapours. What phenomena will occur in the two corners? (Ans) Adsorption would occur where silica gel is kept in the vessel where as absorption will occur in the corner where CaCl 2 is placed. (Q.) Name the substance catalysed by Zymase. (Ans) Glucose-- Zymase ->ethyl alcohol. (Q.) How can colloidal solution of ferric hydroxide be prepared by peptization? (Ans) A colloidal sol. of ferric hydroxide can be prepared by adding small quantity of ferric chloride solution to freshly prepared precipitate of ferric hydroxide. (Q.) What is the cause of Brownian movement? (Ans) Brownian movement is caused by the striking of the colloidal particles with the molecules of dispersion medium due to their kinetic energy. (Q.) Define Tyndall effect? (Ans) It is defined as the scattering of light by the colloidal particles present in a colloidal solution. (Q.) What happens to a gold sol. when gelatin is added to it? (Ans) Gold sol. which is lyophobic starts behaving like lyophilic sol. (Q.) Write down the relation between pressure of the gas and the amount of it adsorbed? (Ans) x/m = K P 1/n (Q.) Which adsorption may be a multilayered formation phenomenon? (Ans) Physisorption (Q.) Which is irreversible and why? Physisorption or chemisorption. (Ans) chemisorption. Because of the formation of chemical bond. (Q.) Name the promoter used in Haber s process? (Ans) Molybdenum.

51 (Q.) What is emulsion? What are their different types? (2 Marks) (Ans) An emulsion is the colloidal dispersion in which both the dispersed phase and the dispersion mediums are liquids. They can be of two types:- i) Emulsion of oil in water. ii) Emulsion of water in oil. (Q.) How are micelles formed in soap solution? (2 Marks) (Ans) Soap is sodium salt of fatty acids (RCOONa) which when dissolved in water dissociates to give RCOO - and Na +. The RCOO - consists of polar group COO - which is hydrophilic and stays at the surface and the non polar group R which being hydrophobic stays away from the surface. At high concentrations RCOO - ions are pulled into the solution to form spherical aggregates with R pointing to the centre COO - part remaining outward. This aggregate is known as ionic micelle. (Q.)How can lyophobic sols be prepared by mechanical disintegration? (2 Marks) (Ans) The coarse suspension of the substance is introduced in the colloid mill that consists of two metal discs close together rotating at a high speed in the opposite directions. Here the suspension particles are broken to the colloidal size. (Q.) Differentiate between chemisorption and physisorption? (2 Marks) (Ans) Physisorption: a)the forces operating are weak vander Waal s forces b)the heat of adsorption is low KJ Mol -1 c)does not require any activation energy d)forms multimoleculer layer Chemisorption: a)forces acting are similar to those of chemical bonds b) The heat of adsorption is high KJ Mol -1 c) Requires activation energy d) Forms unimolecular layer (Q.) Describe the mechanism of peptization? (2 Marks) (Ans) When electrolyte is added to the freshly precipitated substance, the particles of the precipitate preferentially absorb one particular type of ions of the electrolyte and get dispersed due to electrostatic repulsions giving particles of colloidal size and hence cause peptization. (Q.) Give any two reasons for the origin of electrical charge on the colloidal particles. (2 Marks) (Ans) The two reasons are: i) Due to electron capture by sol particles during electro dispersion of metals, due to preferential adsorption of ions from solution ii) Dissociation of colloidal sols. (Q.) How is the electrical charge of the colloidal particles responsible for the stability of colloidal sols? (2 Marks) (Ans) The electrical charges of the particles prevent them from coming together due to electrostatic repulsion. All the dispersed particles in a colloidal solution carry the same charge while the dispersion medium has equal and opposite charge. (Q.) What is demulsification? Name two demulsifiers. (2 Marks) (Ans) The process of separation of the constituent liquids of an emulsion is called demulsification. Demulsification can be done by centrifugation or boiling. (Q.) Why lyophilic colloids are called reversible sols while lyophobic sols are called irreversible sols? (3 Marks) (Ans) In the lyophilic colloids if the dispersed medium is separated from the dispersion medium the sol can be made again by simply remixing with the dispersion medium. So they are called reversible sols. In lyophobic sols if small amount of electrolyte is added, the sols are readily precipitated and do not give back the colloid by simple addition of the dispersion medium. So they are called irreversible sols. (Q.) Describe the preparation of the following colloidal solution. (a) Gold sol (b) Sulphur sol (3 Marks) (Ans) (a) Preparation of Gold sol :- By the reduction of very dilute solution of silver salts with a suitable reducing agent 2AuCl 3 + 3SnCl > 2Au + 3SnCl 4 Gold sol

52 (b) Preparation of Sulphur sol :- By the oxidation of H 2 S in the presence of suitable oxidizing agent like nitric acid, bromine water, etc. H 2 S + Br > S + 2HBr H 2 S + 2HNO > 2H 2 O + 2NO 2 + S (Q.) What are macromolecular and multimolecular colloids? How are they different from associated colloids? (3 Marks) (Ans) Macromolecular colloids:- i)they are molecules of large size. ii)they have lyophobic property. Multimolecular colloids:- i) They are formed by the aggregation of large number of atomsor molecules which have diameter less than 1nm. ii) They have lyophilic property. Associated colloids:- i) They are formed by the aggregation of large number of ions in concentrated solution ii) They contain both lyophilic and lyophobic groups (Q.) What are lyophilic and lyophobic solutions? Give examples for each. (3 Marks) (Ans) Lyophilic solutions are those that can be prepared by directly mixing the dispersed phase with dispersion medium. For example starch dissolved in water. Lyophobic solutions are those that can not be prepared directly but some special methods are used to prepare them. For example metal sulphides when mixed with a dispersion medium directly do not result in any colloid. (Q.) "Action of soap is due to emulsification and micelle formation". Comment. (3 Marks) (Ans) Yes, action of soap is due to emulsification and micelle formation. Soaps are sodium salt of higher fatty acids like sodium stearate, C 17 H 35 COO - Na + The anionic head of stearate ion (-COO - ) is hydrophobic in nature and has great affinity for water, while the hydrocarbon part (C 17 H 35 - ) is hydrophilic in nature and great affinity for oil, grease etc. When soap is used in water, the anions (C 17 H 35 COO - ) form micelle and due emulsification encapsulate oil or grease inside. These micelle are removed by rinsing with water; while free dirt (from oil or grease) either settle down or are washed away by water. Thus the main function of a soap is to entrap oil or grease with the micelles through emulsification, thereby freeing dirt from grease and oil. (Q.) Why the sun looks red at the time of setting? Explain on the basis of colloidal properties. (3 Marks) (Ans) At the time of setting, the sun is at the horizon. The light emitted by the sun has to travel a longer distance through the atmosphere. As a result, blue part of the light is scattered away by the dust particles in the atmosphere. Hence, the red part is visible. (Q.) Explain the reason for these: (3 Marks) (a) Sky looks blue in colour.

53 (b) Delta is formed at the meeting place of river and sea water. (c) Blood coagulate on treatment of alum. (Ans) (a) Sky looks blue in colour because colloidal particles suspended in environment scatter the light and blue light is scattered maximum. (b) The charged colloidal particles of river water neutralized by ions present in sea water so coagulation take place. (c) The charged colloidal particles present in blood are neutralized by ions of alum. VALUE BASED QUESTIONS : Q.1 Use of gas mask is recommended in a crowded place, a place of fire so that poisonous gases are adsorbed. Does it adsorb gases like N 2 and O 2 equally? Explain your answer citing a suitable value behind this. Q.2 Milk and cold dream are examples of colloidal system. Which type of colloid they are? Can they be diluted by adding water? Based upon your observation/ answer differentiate between them. To protect untimely curdling of milk how will you utilize the properties of colloidal solution? Extract the value associated with it. UNIT 6 GENERAL PRINCIPLE OF EXTRACTION OF ELEMENTS 1. Differences between- Ore and minerals- Ores The minerals form which elements are extracted economically and beneficially. Theses contains higher % of metal compound Marks : 3 Minerals The combined form of elements found naturally. These contains lower % of metal compound Gangue and slag- Gangue The impurity part of an ore. Slag A compound form when gangue combines with flux 2. Definition with example a. Depressants- The chemical compound which is used to prevent one kind of sulphide ore to go with froth in the presence of another sulphide ore. Exa.- NaCN b. Collectors- These substances increase the non-wettability of the ore particles. Exa.- Pine oil, fatty acids c. Leaching- When conc. Of ore is done using certain chemicals then this process of concentration is called leaching. Exa.- Bauxite, Ag and Au ore are concentrated by this method. d) Pyro-metallurgy- The process of thermal reduction of an oxide by varying the temperature for a reducing agent is called pyro-metallurgy. e) Copper matter- It is a combination of Cu 2 S and FeS obtained on roasting of copper pyrites in reverberatory furnace.

54 f) Blister copper- The solidified copper obtained after reduction of copper matte has blistered appearance. This is called blister copper. 3. Name and formula of some important ores- 4. Thermodynamic principle of metallurgy- Thermodynamic principle of metallurgy is based on the measurement of G of formation of oxide of reducing agent form metal oxide. If G of formation of oxide of reducing agent is more ve than G of formation of metal oxide than the reducing agent can reduce given metal oxide into metal spontaneously. -ve sign of overall reaction shows its spontaneity. The possibility of reduction of a metal oxide can also be predicted from Ellingham diagram. A metal oxide can be reduced by that metal for which G of formation vs T plot is lower.

55 Examples- 1. Mg can reduce Al 2 O 3 2. Al can redue ZnO 3. C can reduce ZnO at higher Temp. 4. CO can reduce FeO at lower temp. 5. At point A reaction is at equilibrium 6. Bend in a plot shows change in Phase 5. Principles of different Refining Processa) Distillation- This purification method is based on the principle that metal to be purified (Zn, Hg) has lower boiling point that the impurity present in that. b) Liquation- This purification method is based on the principle that metal to be purified (Sn, Pb) has lower melting point that the impurity present in that. c) Electrolytic refining- This purification method is based on the principle that impure metal undergo oxidation and its ion undergo reduction from solution based on their E value. d) Zone refining- This purification method is based on the principle that impurities are more soluble in molten state than solid state of the metal. e) Vapur phase refining- This purification method is based on the principle that metal a) should form a volatile compound b) the volatile compound should be easily decomposable A Few Questions for practice: 1. Give the name and composition of ore chosen for extraction of aluminium. Ans: The ore chosen for the extraction of aluminium is bauxite and its composition is Al 2 O 3.xH 2 O. 2. What is leaching? Ans: Leaching is the process of extracting a substance from a solid by dissolving it in a liquid. In metallurgy leaching is used for the ores that are soluble in a suitable solvent. 3. Why cryolite & fluorospar added to alumina during electrolytic reduction? Ans: Cryolite and fluorospar are added to alumina during electrolytic reduction to reduce the melting point of alumina and to increase its conductivity 4. Reduction with C for Cu 2 O can be done at lower temp. than ZnO. Why? Ans: In the Ellingham diagram the curve for Cu 2 O lies higher than ZnO i.e. for the reduction of Cu 2 O with C the negative value of Gibbs energy can be reached at a lower temperature than ZnO. 5. Although thermodynamically feasible in practice magnesium metal is not used for the reduction of alumina. Why? Ans: Magnesium can reduce alumina at the temperature above the intersection point of the curves for Al 2 O 3 and MgO in the Gibbs Energy vs T plot (Ellingham diagram). But the temperature at which this is feasible is too high to be achieved economically and is also technologically difficult. So this reduction is not done. 6. What is the significance of leaching in extraction of aluminium? Ans: In the extraction of aluminium leaching is used for the concentration of ore by removing the impurities i.e. silica, iron oxides and titatinium oxides. 7. Define Metallurgy. Ans: Metallurgy is the process of extraction of metals from their ores that includes various steps. 8. Why is hydraulic washing a type of gravity separation?

56 Ans: The process of hydraulic washing is based on the differences in gravity of the ore and the gangue particles and so is known as gravity separation. 9. What is the use of van Arkel method? Ans: Van Arkel method is used for removal of impurities like oxygen and nitrogen from the metals like zirconium and titanium. 10. How is distillation used for metal refining? Ans: Distillation is used for the metals with boiling point lower then the impurities. So the metals can be evaporated and separately obtained as distillate. 11. Why do the anodes used in the electrolytic cell for the reduction of alumina need to be replaced regularly? Ans: The oxygen liberated at the anode during the reduction of alumina, reacts with the carbon of the anode to form CO and CO burns away the anode and hence the anodes need to be replaced. 12. What is the role of depressant in froth floatation process? Ans: In froth floatation process the depressant selectively prevents one of the ores from coming to the froth in a mixture of two ores hence enabling the separation of the other one with the froth. 13. State the role of silica in the metallurgy of copper. Ans: Silica in the metallurgy of copper helps in removal of iron oxide as iron silicate (slag). 14. What is the role of graphite rods in the electrometallurgy of aluminium? Ans: In the electrometallurgy of aluminium graphite rods act as anodes in the electrolytic cell of reduction and are the site for release of oxygen 15. Give an example when an element is extracted by oxidation. Ans: Extraction of chlorine from brine is based on oxidation. 16. What will happen if aqueous solution of NaCl is subjected to electrolysis? Ans: If aqueous Solution of NaCl is subjected to electrolysis, Cl 2 will be obtained with NaOH and H 2 gas as the side products. 17. What is refining of metals? Ans: Refining of metal is the process of purification of a metal extracted from its ore. 18. What is vapour phase refining? Ans: Vapour phase refining is the method of metal refining by changing the metal into volatile compound that can be collected separately leaving behind the impurities and can be decomposed to give the pure metal. 19. Give the principle underlying the process used for refining of gallium. Ans: The process used for the refining of gallium is zone refining and the principle underlying it is that the impurities are more soluble in the melt than in solid state of the metal. 20. State the principle on which the chromatographic methods of metal refining are based? Ans: Chromatographic methods of metal refining are based on the principle that different components of a mixture are differently adsorbed on an adsorbent 21. Which is the purest form of iron and what are its uses? Ans: The purest form of iron is wrought iron & is used in making anchors, wires, bolts etc. 22. What are minerals and how are they different from ores? (2 Marks) Ans: Minerals are the naturally occurring chemical substances in the earth s crust obtained by mining. Its different from ores, as ores are the minerals that are used for the extraction of metals profitably. 23. Name one ore each for iron & copper & give their chemical compositions. (2 Marks) Ans: The ore of iron is hematite- Fe 2 O 3 and the ore for copper is copper pyrites- CuFeS What is the purpose of adding collectors and froth stabilisers during froth floatation? Give an example for each. (2 Marks) Ans: During froth floatation process collectors like pine oil and fatty acids are added to enhance non wettability of the mineral particles and the froth stabilisers like cresol and aniline stabilise the froth. 25. How can the ores ZnS and PbS be separated from a mixture using froth floatation process? (2 Marks) Ans: During the froth floatation process a depressant like NaCN is added to the tank. The depressant selectively prevents ZnS from coming to the froth but allows PbS to come to the froth and hence helps the separation of PbS with the froth. 26. Give the equations involved in the concentration of bauxite ore. (2 Marks) Ans: i) Al 2 O 3(s) + 2NaOH + 3H 2 O 2Na[Al(OH) 4 ] (aq) ii) 2Na[Al(OH) 4 ] (aq) + CO 2(g) Al 2 O 3.xH 2 O (s) + 2NaHCO 3

57 iii) Al 2 O 3.xH 2 O (s) Al 2 O 3(s) + xh 2 O 27. Give one reaction each for roasting and Calcination. (2 Marks) Ans: Calcination: ZnCO 3 ZnO(s) + CO 2 Roasting: 2ZnS + 3O 2 2ZnO + 2SO Why is coke preferred over CO for reducing FeO? (2 Marks) Ans: According to Ellingham diagram the point of intersection of the curves of C, CO and Fe, FeO lies at temperature lower than that of the point of intersection of CO,CO 2 and Fe, FeO curves. This means the reduction of FeO will occur at much lower temperature with C than with CO. So C is preferred to CO for reduction. 29. How is cast iron different from pig iron? (2 Marks) Ans: Pig iron has 4% carbon and can be easily cast into verity of shapes. Whereas cast iron has lower carbon content and is extremely hard and brittle. 30. Give the reactions that occur after the copper matte have been fed into silica lined converter. (2 Marks) Ans: 2FeS+3O 2 2FeO + 3O 2 FeO + SiO 2 FeSiO 3 2Cu 2 S + 3O 2 2Cu 2 O + 2SO 2 2Cu 2 O + Cu 2 S 6Cu + SO Give the reactions taking place at the anode and the cathode during the electrolytic reduction of alumina. (2 Marks) Ans: Anode: C(s) + O 2- (melt) CO(g) + 2e - C(s) + 2O 2- (melt) CO 2 (g) + 4e - Cathode: Al 3+ (melt) + 3e - Al(l) 32. Explain the process of magnetic separation for concentration of ores. (3 Marks) Ans: In magnetic separation ore is carried over a conveyer belt which passes over a magnetic roller. If either the ore or the gangue is capable of being attracted by the magnetic field then it will collect near the roller and the particles showing non magnetic behaviour will be collected away from the roller. 33. Differentiate between roasting and Calcination. (3 Marks) Ans: Calcination: i) it involves heating of the ore in the absence of air ii) it is generally used for carbonate ores Calcination: ZnCO 3 ZnO(s) + CO 2 Roasting: i) it involves the heating of the ore in the presence of air ii) it is generally used for sulphide ores Roasting: 2ZnS + 3O 2 2ZnO + 2SO Give the reactions involved in the reduction of iron oxide to give iron in a blast furnace. (3 Marks) Ans: The reactions are as follows: C + O 2 CO 2 CaCO 3 CaO + CO 2 CO 2 + C CO 3Fe 2 O 3 + CO Fe 3 O 4 +CO 2 Fe 3 O 4 + CO 3FeO + CO 2 FeO +CO Fe + CO 2 FeO + C Fe + CO CaO + SiO 2 CaSiO How is copper extracted from low grade ores and scraps? (3 Marks) Ans: For extraction of copper from low grade ores and scraps the ore is first leached out using acid or bacteria. The solution containing Cu 2+ is treated with scarp iron or H 2 and Cu is obtained. Cu 2+ (aq) + H 2 (g) Cu(s) + 2H + (aq) Cu 2+ + Fe Cu(s) + Fe How is gold extracted from its ore? (3 Marks) Ans: Extraction of gold involves leaching the metal with CN- giving metal complex. 4Au + 8CN - (aq) + 2H 2 O + O 2 (g) 4[Au(CN) 2 ] - (aq) + 4OH - (aq) the metal is later recovered by displacement method with zinc acting as reducing agent. 2[ Au(CN) 2 ] - (aq) + Zn(s) 2Au(s) + [Zn(CN) 4 ] - (aq)

58 37. Describe the method used for refining copper metal. (3 Marks) Ans: Copper metal is refined by using electrolytic method with impure copper metal as anode and the pure copper metal strip as cathode. The electrolyte is acidified copper sulphate solution. Copper dissolves from the anode into the electrolyte and get reduced and deposited on the cathode as pure metal. Anode: Cu Cu e - Cathode: Cu e - Cu Impurities deposit as anode mud. 38. How is nickel refined? (3 Marks) Ans: Nickel is refined by Mond s process which is based upon vapour phase refining. In this process nickel is heated in stream of carbon monoxide giving a volatile complex, leaving the impurities behind. The complex is further subjected to higher temperature so that it gets decomposed to giving pure metal K Ni + 4CO > Ni(CO) K Ni(CO) > Ni + 4CO 39. Describe briefly column chromatography. (3 Marks) Ans: Column chromatography is the method of chromatographic refining of metals available in minute quantities and the impurities are not chemically very much different from the element. In this process the column of Al 2 O 3 is prepared in glass tube that forms the stationary phase and the solution of the components to be separated are taken as solution that forms the mobile phase. The components would separate out based on their different solubilities in the mobile phase and the stationary phase. 40. What criterion is followed for the selection of the stationary phase in chromatography? (3 Marks) Ans: Stationary phase is the immobile and immiscible phase in chromatographic method. Stationary phase is such chosen that the components to be separated present in the mobile phase have different solubilities in the mobile phase and the stationary phase. 41. How is zinc extracted from zinc blende? (3 Marks) Ans: Zinc blende is ZnS. For the extraction of zinc from zinc blende, the ore is first concentrated by the method of froth floatation. The concentrated ore is then roasted by heating the ore in the presence of oxygen to give ZnO releasing SO 2. The ZnS is further reduced using coke at temperature of 673k giving zinc metal. 2ZnS + 3O 2 2ZnO + 2SO 2 ZnO + C Zn + CO UNIT - 7 p- BLOCK ELEMENTS Marks= 8 P block elements are placed in groups 13 to 18 of the periodic table. Electronic configuration is ns 2 np 1-6 Group 15 elements- group 15 includes nitrogen, phosphorous, arsenic, antimony and bismuth.down the group metallic character increases nitrogen and phosphorous are non metals arsenic and antimony are metalloids and bismuth is a typical metal. Electronic configuration is ns 2 np 3 both are half filled orbitals so they are extra stable. ionization enthalpy decreases down the group due to gradual increase in atomic size.(ionization enthalpy is the energy requires to loose electron from its outer most shell to form positive ion) The electronegativity decreases down the group with increasing atomic size. (electronegativity is the tendency to attract electron towards itself) Metallic character electron towards itself). Metallic character increases down the group because ionization enthalpy decreases down the group.the common oxidation states are +3,-3and +5,-5 oxidation states decreases down the group and +3 state increases due to inert pair effect.inert pair effect is the reluctance of s electron in chemical bonding. Disproportionation reaction takes place in nitric acid

59 3HNO 2 HNO 3 + H NO Nitrogen is restricted to a maximum covalent of 4 since only for (one s and three p ) orbitals are available for bonding Nitrogen shows anomalous behaviour because of small size, high enthalpy and non availability of d orbitals. Nitrogen forms pπ-pπ multiple bond. N-N bond is weaker than the single p-p bond because of high inter electronic repulsions of non bonding electrons and owing to the small bond length.for this reason also catenation tendency is weaker in nitrogen.nitrogen cannot form d( )-p( ) bond due to absence of d orbital. The stability of hydrides decreases from NH 3 to BIH 3.Also size decrease of hydrides in the group. Reducing character of hydrides also decreases as size of element increases down the group and basicity decreases down the group nitrogen does not form pentahalides due to absence of d orbital In the laboratory dinitrogen is prepared NH 4 Cl(aq)+ NaNO 2 (aq) N 2 + 2H 2 O+NaCl Other methods (NH 4 ) 2 Cr 2 O 7 heat > N 2 +4H 2 O+Cr 2 O 3 Ammonium dichromate pure nitrogen can be obtained by the thermal decomposition of sodium or barium acid Ba(N 3 ) 2 -> Ba+ 3N 2 Dinitrogen is inert at room temperature because of high bond enthalpy. Ammonia is a colourless with pungent odour. It is sp 3 hybridisation, trigonal pyramid structure.it has sp 3 hybridisation, trigonal pyramidal structure.it has three bond pairs and one lone pair of electrons.nh 3 is a Lewis base due to presence of a lone pair of electrons.it form complexes with metal ions because it donates the electron pair and forms linkage with metal ions. Nitric acid is a colourless liquid.concentrated nitric acid is a strong oxidizing agent and attacks most metals except noble metals such as gold and platinum. Some metals ex. (r) Al do not dissolve in conc. Nitric acid because of the formation of a passive film of oxide on the surface.conc. nitric acid also oxidises non metals and their compounds.ex. iodine is oxidised to iodic acid. Phosphorous have allotropic forms.but important are white red,black white phosphorous is waxy solid poisonous,insoluble in water but soluble in carbon disulphide goes in dark.it is ion stable,more of tetrahedral by molecules. Red phosphorous is iron grey lusture. It is odourless, non-poisonous and insoluble in water and carbon disulphide. Red phosphorous is much less reactive than white phosphorous. It does not glow in the dark.red phosphorous is less reactive than white phosphorous because red phosphorous has polymeric structure in which p 4 units are bonded together with strong single covalent bonds. In white phosphorous p 4 units are held together with vander walls force. Phosphine is a colourless gas with rotten fish smell and is highly poisonous, It is slightly soluble in water.it is weak basic. Phosphorous forms two types of halides one is PX 3 and other is PX 5. Pcl 3 is a colourless oily liquid and hydrolysis in the presence of moisture.it has pyramidal shape,where phosphorous is sp 3 hybridised.pcl 5 is a yellowish white powder and in moist air, In gas and liquid phases, it has a trigonal bipyramidal structure. The three equatorial p-d bonds are equivalent while the two axial bonds are longer than equatorial bonds. This is due to the axial bond pairs suffer more repulsion as compared to equatorial bond pairs. In the solid state it exists as an ionic solid [PCl 4 ] + [PCl 6 ] - in which [PCl 4 ] + is tetrahedral and anion is [PCl 6 ] - octahedral. Phosphorus forms a number of oxoacids. For table 7.5 and figure 7.4 are very important. Ortho phosphorus acid on heating disproportionates to give ortho phosphoric acid and phosphine 4H 3 PO 3 3H 3 PO 4 +PH 3 Hypophosphorus acid is a good reducing agent as it contains two P-H bonds and reduces. 4H 3 PO 3 is dibasic as it contains two P-OH bonds. H 3 PO 4 is tribasic as it contains 3P-OH bonds.

60 Group 16 elements Group 16 elements are oxygen sulphur selenium tellurium and polonium. 16 group also called chalcogens. The elements of group 16 have 6 electrons in the outer most cell and configuration is ns 2 np 4.The elements of this group have lower ionization enthalpy compared to 15 group due to extra stability of half filled of 15 groups element. Oxygen has less negative electron gain enthalpy than sulphur because of the compact nature of oxygen atom. Metallic character increases from oxygen to polonium due to electronegativity decreases in a group. The elements shows -2,+2,+4 and+6 oxidation state. The oxygen shows only negative oxidation state due to high electronegativity. But in OF 2 its oxidation state is +2. In this group +6 oxidation state decreases down the group and stability of +4 oxidation state due to inert pair effect. Oxygen shows anomalous behaviour due to small size, high electronegativity and absence of d orbital. Dimeric halides under go Disproportionation reaction, example 2Se 2 Cl 2 Se Cl 4 + 3Se. An oxide that combines with water to give an acid is termed as acidic oxide, example SO 2, CO 2. Non metal oxides are acidic. The oxides which give a base with water are known as basic oxides, example CaO. Metallic oxides are basic. Some metallic oxides exhibits a dual behaviour. They show both acidic and basic property. Such oxides are called amphoteric oxides There are some oxides which are neither acidic nor basic. Such oxides are known as neutral oxides. Example CO,NO, N 2 O. Ozone is an allotropic form of oxygen. Since the formation of ozone from oxygen is an endothermic process, it is necessary to use a silent electrical discharge in its preparation to prevent its decomposition. Sulphur form a number of allotropes. Yellow rhombic (alpha sulphur) and monoclinic (beta sulphur) are important. Sulphur dioxide is a colourless gas with pungent smell and highly soluble in water. When moist sulphur dioxide behaves as a reducing agent it converts iron (III) ions to iron (II). Sulphur forms a number of oxoacids. But sulphuric acid is important. It is manufactured by the contact process which involves 3 steps- 1-burning of sulphur in air to form SO 2 S+O2 SO Conversion of SO 2 to SO 3 by the reaction with oxygen in the presence of catalyst (V ) 2SO 2 + O 2 V2O5 2SO 3 3-Absorption of SO 3 in H 2 SO 4 to give Oleum(H 2 S 2 O 7 ) SO 3 + H 2 SO 4 H 2 S Dilution of Oleium with water gives H 2 SO 4 of the desired concentration. Sulphuric acid is a colourless, dense, oily liquid. The concentrated acid must be added slowly into water with constant stirring. The larger value of Ka, means that H 2 SO 4 is highly dissociated into H + and HSO4-.Greater the value of dissociation constant, the stronger is the acid. The acid forms 2 series of salt normal sulphate example sodium sulphate and acid sulphate example sodium hydrogen sulphate. Group 17 elements (Halogen) It contains fluorine, chlorine, bromine, Iodine and astatine. These are called halogens means salt producers. Astatine is a radioactive element. Electronic configuration is ns 2 np 5.High ionization enthalpy because there is less tendency to loose electon.ionisation enthalpy decreases down the group as atomic size decreases. Halogens have maximum negative electron gain enthalpy in the period due to having 1 electron less than noble gas configuration. The negative electron gain enthalpy of fluorine is less than that of chlorine due to small size of fluorine atom and strong inter electronic repulsion 2 p orbitals of fluorine. Fluorine and chlorine are gases, bromine is a liquid and Iodine is solid. Halogens are coloured due to absorption of radiation in visible region, excite the outer electron to higher energy level. By absorbing different radiations display different colours. F 2 is yellow Cl 2 is greenish yellow, Br 2 is red and I 2 is

61 violet colour. All the halogens show -1 oxidation state. Fluorine atom has no d orbtals, so it can not expand and shows only -1 oxidation state. But Cl,Br and I show +1,+3,+5 and +7 oxidation state. All the halogens are highly reactive. They react with metals and non metals to form halide. The reactivity of the halogens decreases down the group. Halogens are strong oxidizing agent because they accept 1 electron easily to attend noble gas configuration. F 2 is the strongest oxidizing agent.(oxidising agent is one which oxidises other but itself reduced.)fluorine shows anomalous behavior due high ionization enthalpy, highest electronegativity and absence of d-orbital, small size, low F-F bond dissociation energy. Most of the reactions of fluorine are exothermic due to small and strong bond formed with other element Hydrogen fluoride is a liquid with high boiling point due to strong hydrogen bonding. F 2 O does not exist but OF 2 is stable due to high electronegativity. Chlorine is greenish yellow gas with pungent and suffocating odor. It is colorless and pungent smelling gas. It is easily liquefied to a colorless liquid and freezes to a white crystalline solid. High value of dissociation constant (K a ) indicates that it is a strong acid in water fluorine forms only one Oxo acid, due to high electronegativity and small size. When two different halogens react with each other, interhalogen compounds are formed, as XX,XX 3, XX 5 and XX 7 where X is a halogen of larger size and X of smaller size and X is more electropositive than X. Inter halogen compounds can be prepared by the direct combination or by the action of halogen on lower interhalogen compounds. Group 18 elements Group 18 consists of six elements helium, neon, argon, krypton, xenon and radon. All these are gases and chemically unreactive. They form very few compounds because of this they are called noble gases. General electronic configuration is ns 2 np 6 except helium. Gases exhibit very high ionization enthalpy due to stable configuration. It decreases down the group with the increase in atomic size. Noble gases have large positive values of electron gain enthalpy because of stable configuration and have no tendency to accept the electron. They have very low melting and lowest boiling point because of interatomic interaction. Neil Bartlett prepared a red compound formula of O + 2 PtF - 6. Since first ionisation enthalpy of Oxygen is almost identical with that of xenon. So he prepared red colour compound of Xe + 2 PtF - 6 by mixing PtF 6 and xenon. After his discovery a number of xenon compounds with most electronegative elements like fluorine and oxygen have been synthesised. USES (i) Nitrogen: Manufacturing of Ammonia, refrigerant industrial chemicals. (ii) Ammonia: Nitrogenous fertilizer, manufacturing of some inorganic compound. Liquid ammonia is used as refrigerant. (iii) Nitric Acid: Manufacturing of ammonium nitrate for fertilizer, preparation of nitroglycerin trinitrotoluene, other uses are pickling of stainless steel, etching of metals and as an oxidise in rocket fuels. (iv) Phosphine: Home s signals, smoke screens. (v) Oxygen: Oxyacetylene welding, manufacturing of many metals. Oxygen cylinder, combination of fuel eg hydrazine in liquid oxygen. (vi) Ozone: Used as a germicide, disinfectant for sterilising water. Bleaching oils, flour, ivory, starch, manufacturing of potassium permanganate. (vii) Sulphur dioxide: Refining petroleum and sugar bleaching wool and silk, anti-chlor, disinfectant and preservative. Liquid SO 2 is used as a solvent to dissolve a number of organic and inorganic chemicals. (viii) Sulphuric Acid: It is very important industrial chemical. Manufacture of compounds and fertilizer petroleum refining, manufacture of pigments, paints and dyes, detergent industry, in metallurgy cleaning fo metals before enameling, electroplating and galvansing, storage batteries, manufacturing of nitrocellulose product, lab reagent. (ix) Interhalogen Compound are useful fluorinating agents. ClF 3 and BrF 3 are used for the production of UF 6 in the enrichment of 235 U.

62 (x) (xi) (xii) Helium : It is used in filling balloons due to non-inflammable and light gas, in gas cooled nuclear reactor. Liquid Helium used as cryogenic agent, produced superconducting magnets for NMR spectrometer, Magnetic resonance (MRI) for clinical diagnosis, modern diving apparatus. Neon: Neon is used in discharge table and fluorescent bulbs. Neon bulbs are used in botanical gardens and in green houses. Argon: Argon is used in inert atmosphere in high temperature, metallurgical process, filling electric bulbs, for handling substance that are air sensitive. Xenon and Krypton are used in light bulb designed for special purpose. (THE P BLOCK ELEMENTS CHEMICAL REACTION) NH 4 Cl + NaNO 3 N H 2 O + NaCl (NH 4 ) 2 Cr 2 O 7 N 2 + 4H 2 O + Cr 2 O 3 Ammonium dichromate Ba(N 3 ) Ba + 3 N 2 Barium azide 6Li + N 2 2Li 3 N (Lithium nitride) 3 Mg + N 2 Mg 3 N 2 (Magnesium nitride) 2NaN 3 2Na + 3N 2 N H 2 2 NH 3 2 NH 4 Cl + Ca ( OH) 2 2NH 3 + 2H 2 O + CaCl 2 (NH 4 ) 2 SO Na OH 2 NH 3 + 2H 2 O + Na 2 SO 4 NH 3 + H 2 O NH OH - Zn SO 4 + 2NH 4 OH Zn(OH) 2 + (NH 4 ) 2 SO 4 FeCl 3 + NH 4 OH Fe 2 O 3 XH 2 O + NH 4 Cl (BROWN PPT) Cu NH 3 aq [Cu(NH 3 ) 4 ] 2+ aq Blue Ag + aq + Cl - deep blue solution Ag Cl WHITE PPT AgCl +2NH 3 aq [Ag(NH 3 ) 2 ]Cl Colourless NH 4 NO 3 N 2 O + 2H 2 O Nitrousoxide 2Pb(NO 3 ) 2 4NO 2 + 2PbO NaNO 3 + H 2 SO 4 NaHSO 4 + HNO 3 4NH 3 + 5O 2 4NO + 6H 2 O 3NO 2 + H 2 O 2HNO 3 + NO (Nitric oxide) 3 Cu + 8 HNO 3 (Dil.) 3Cu(NO 3 ) 2 + 2NO 2 + 2H 2 O Cu + 4HNO 3 (Conc.) Cu(NO 3 ) 2 + 2NO 2 +2H 2 O 4Zn + 10HNO 3 (Dil.) 4Zn(NO 3 ) 2 +5H 2 O + N 2 O Zn + 4HNO 3 (Conc.) Zn(NO 3 ) 2 + 2H 2 O + 2NO 2 I HNO 3 (Conc.) 2HIO 3 + 4H 2 O + 10NO 2 (IODICACID) C + 4HNO 3 (Conc.) CO 2 + 2H 2 O + 4NO 2 S HNO 3 (Conc.) 8H 2 SO H 2 O + 48NO 2 P HNO 3 (Conc.) 4H 3 PO 4 + 4H 2 O + 20NO 2 P 4 + 3NaOH +3H 2 0 PH 3 + 3NaH 2 PO 2 (SODIUM HYPOPHOSPHITE) P 4 + 5O 2 P Ca 3 P 2 + 6H 2 O 3Ca(OH) 2 + 2PH 3 Ca 3 P 2 + 6HCl 3CACl 2 + 2PH 3 PH 4 I + KOH KI + H 2 O + PH 3 3 CuSO 4 + 2PH 3 Cu 3 P 2 + 3H 2 SO 4 3HgCl 2 + 2PH 3 Hg 3 P 2 + 6HCl MERCURIC PHOSPHIDE PH 3 + HBr PH 4 Br P 4 + 6Cl 2 4PCl 3 P 4 + 8SOCl 2 4PCl 3 +4 SO 2 + 2S 2 Cl 2

63 PCl H 2 O H 3 PO 3 +3HCl (FUMES ) PCl 5 + H 2 O POCl 3 + 2HCl POCl 3 + 3H 2 O H 3 PO 4 + 3HCl 2Ag + PCl 5 2AgCl + PCl 3 Sn +2 PCl 5 SnCl 4 + 2PCl 3 2KClO 3 2 KCl +3 O 2 2Ag 2 O 4Ag(S) +O 2 (G) 2Pb 3 O 4 (S) 6PbO(S) + O 2 (G) 2HgO 2Hg (L) + O 2 (G) 2ZnS +3O 2 2ZnO + 2 S0 2 Al 2 O 3 + 6HCl + 9H 2 O 2[Al(H 2 O) 6 ] C - Al 2 O 3 + 6NaOH +3H 2 0 2Na 3 [Al(OH) 6 ] PbS+ 4O 3 PbSO 4 + 4O 2 4FeS O 2 2Fe 2 O 3 +8SO 2 SO 3 +H 2 O H 2 SO 4 CaF 2 + H 2 SO 4 CaSO 4 + HF 2NaOH + SO 2 Na 2 SO 3 + H 2 O 2Fe 3+ + SO 2 + 2H 2 O 2Fe 2+ + SO H + NaCl + H 2 SO 4 HCl +Na 2 SO 4 C 12 H 22 O C + 11H 2 O Cu + 2H 2 SO 4 (CONC) CuSO 4 +SO 2 + 2H 2 O 3S +2H 2 SO 4 (CONC) 3SO 2 + 2H 2 O C + 2H 2 SO 4 (CONC) CO 2 + 2H 2 O +2SO 2 MnO 2 + 4HCl MnCl 2 + Cl 2 + 2H 2 O H 2 + Cl 2 2HCl 4NaCl + MnO 2 + 4H 2 SO 4 MnCl 2 + 4NaHSO 4 +2H 2 O+Cl 2 2kMnO HCl 2KCl + 2MnCl 2 + 8H 2 O + 5Cl 2 4HCl +O 2 2Cl 2 +2H 2 O 2F 2 + 2H 2 O 4H + + 4F - + O 2 2NaOH + Cl 2 NaCl + NaOCl + H 2 O 2Cl 2 + 2H 2 O HCl + HOCl 2Ca(OH) 2 + 2Cl 2 Ca(OCl) 2 + CaCl 2 + 2H 2 O 2FeSO 4 +Cl 2 + H 2 SO 4 Fe 2 (SO 4 ) 3 + 2HCl NaCl +H 2 SO 4 NaHSO 4 + HCl 2NaCl +2H 2 SO 4 + MnO 2 MnSO 4 + Na 2 SO 4 +2H 2 O + Cl 2 NaHSO 4 + NaCl Na 2 SO 4 + HCl Cl 2 + 2NaI 2NaCl + I 2 Na 2 CO 3 + 2HCl 2NaCl + H 2 O + CO 2 NaHCO 3 + HCl NaCl + H 2 O + CO 2 Na 2 SO 3 + 2HCl 2NaCl + H 2 O + SO 2 Fe + 2HCl Fe + H 2 U + 3ClF 3 UF 6 + 3ClF Xe + F 2 XeF 2 (XENON IN EXCESS) Xe + 2F 2 XeF 4 (1: 5)

64 Xe + 3F 2 XeF 6 (1:20) XeF 4 +O 2 F 2 XeF 6 + O 2 2XF H 2 O 2Xe +4HF + O 2 4XeF H 2 O 4 Xe + 2XeO 3 +24HF +3O 2 XeF 6 +3H 2 O XeO 3 +6HF XeF 6 +H 2 O XeOF 4 + 2HF (PARTIAL HYDROLYSIS) XeF 6 +2H 2 O XeO 3 F 2 + 4HF(PARTIAL HYDROLYSIS) CaCO 3 + H 2 SO 4 CaSO 4 + H 2 O + CO 2 DISPROPORTIONATION REACTION Disproportionation reaction, is a specific type of redox reaction, in which a particular element is simultaneously reduced and oxidized to form two different products. ( it is also known as dismutation Reaction, while the reverse of this reaction is comproportionation reaction) 3HNO 2 HNO 3 + H 2 O +2NO 4H 3 PO 3 3H 3 PO 4 +PH 3 2SeCl 2 SeCl 4 + 3Se

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66 NCERT Intext Questions 1 Why are pentahalides more covalent than trihalides? Ans- higher the positive oxidation state more will be polarising power and more covalent character. 2 Why is BiH 3 the strongest reducing agent amongst all the hydrides of Group 15 elements? Ans-The size of element increases in a group then bond length increases and become weaker. From N H bond can break easily and gives hydrogen.

67 3 Why is N 2 less reactive at room temperature? Ans- Due to presence of a triple bond, bond dissociation energy is higher. 4 Mention the conditions required to maximise the yield of ammonia. Ans- i. Low temp, but optimum temp is 700K. ii. high pressure(200 atm) 5 How does ammonia react with a solution of Cu 2+? Ans -Cu +2 (aq)+nh 4 OH(aq) [Cu(NH 3 ) 4 ] +2 + H 2 O Blue colour complex 6 What is the covalence of nitrogen in N 2 O 5? Ans- Covalancy is 4. 7 Bond angle in PH 4 + is higher than that in PH 3. Why? Ans- In PH 3 and PH 4 + P undergoes sp 3 hybridisation. In PH 4 + there is no lone pair electron but in PH 3 there is one lone pair of electron. due to repulsion, angle decreases in PH 3. 8 What is the basicity of H 3 PO 4? Ans. Basicity is three as it has three P OH. 9 What happens when H 3 PO 3 is heated? Ans. H 3 PO 3 undergoes disproportionation reaction H 3 PO 3 PH 3 + H 3 PO 4 Here P 3+ is oxidised to P 5+ and again it reduced to P 3-, so it is disproportionation reaction. 10 Why is H 2 O a liquid and H 2 S a gas? Ans. In H 2 O, hydrogen bond is formed but in H 2 S no hydrogen bond exists. Because electronegativity of S is less than oxygen. 11 Which of the following does not react with oxygen directly? Zn, Ti, Pt, Fe. Ans. Platinum (Pt) is a noble element, so it does not react with oxygen. 12 Why does O 3 act as a powerful oxidising agent? Ans. Because ozone on heating gives nascent oxygen. 13 How is the presence of SO 2 detected? Ans. SO 2 is a pungent smell gas. It can be detected by passing through acidified potassium dichromate solution which turns green due to Cr 2 +6 changes to Cr Write the conditions to maximise the yield of H 2 SO 4 by Contact process. Ans 24. (a) low temp (b) high pressure (c) V 2 O 5 as catalyst 15 Why is K a2 << K a1 for H 2 SO 4 in water? Ans 25. Ka 2 is less than K a1 because H 2 SO 4-1 has a tendency to donate a proton to water. 16 Sea is the greatest source of some halogens. Comment. Ans. Sea water contain chloride bromide and iodide of sodium, potasium magnesium and calcium 17 Give the reason for bleaching action of Cl 2. Ans. Chloride liberates nascent oxygen in presence of moisture, which bleaches the coloured substances present in organic matter. 18 Name two poisonous gases which can be prepared from chlorine gas. Ans. (a) phosgene Cocl 2 (b) mustard gas 19 Why is ICl more reactive than I 2?Ans. ICl is more reactive than I 2 because I-Cl bond is polar and weaker than I-I bond which is non-polar and stronger.

68 NCERT EXAMPLES Example 1 Though nitrogen exhibits +5 oxidation state, it does not form pentahalide. Give reason. Ans - Due to absence of d orbital it does not form pentahalide. But due to involvement of s and p orbit it exhibits +5 oxidation state. Example 2 PH 3 has lower boiling point than NH 3. Why? Ans- Because of hydrogen bond which is present in NH 3. Example 3 Why does NH 3 act as a Lewis base? Ans- Nitrogen atom in NH 3 has one lone pair of electron for donation (Lewis base is one which donate one lone pair electron) Example 4 Why does NO 2 dimerise? Ans-NO 2 contain odd number of valence electron. On dimirisation it gives a stable N 2 O 4 molecule. Example 5 In what way can it be proved that PH 3 is basic in nature? Ans- PH 3 react with acid like HI to give PH 4 I Example 6 Why does PCl 3 fume in moisture? Ans- PCl 3 hydrogen in the presence of moisture giving fumes of HCl PCl 3 + 3H 2 O H 3 PO 3 + 3HCl Example 7 Are all the five bonds in PCl 5 molecule equivalent? Justify your answer. Ans- PCl 5 has a trigonal bipyramidal structure and three equatorial P-Cl bonds equivalent, while two axial bonds are different and longer. Example 8 How do you account for the reducing behaviour of H 3 PO 2 on the basis of its structure? Ans- H 3 PO 2 contains two hydrogen atoms and bonded directly to P atom. Example 9 Elements of Group 16 generally show lower value of first ionisation enthalpy compared to the corresponding periods of group 15. Why? Ans- Due to extra stable of half filled p orbital configuration present in 15 group element. Example 10 H 2 S is less acidic than H 2 Te. Why? Ans-Bond association decreases down the group as atomic size increases. increases, So acidic character Example 11 Which form of sulphur shows paramagnetic behaviour? Ans- Vapour state Sulphur because it has to unpaired electrons. Example 12 Halogens have maximum negative electron gain enthalpy in the respective periods of the periodic table. Why? Ans- Due to size increases in period and high effective nuclear charge. So accept one electron to acquire noble gas electronic configuration. Example 13 Although electron gain enthalpy of fluorine is less negative as compared to chlorine, fluorine is a stronger oxidising agent than chlorine. Why? Ans- I) due to low enthalpy of dissociation of f-f bond. Ii) high hydration enthalpy of F -

69 Example 14 Fluorine exhibits only 1 oxidation state whereas other halogens exhibit + 1, + 3, + 5 and + 7 oxidation states also. Explain. Ans- Due to high electronegative of fluorine it can not exhibit any positive oxidation state. Other halogens have d orbits so show +1,+3,+5 and +7 oxidation state. Example 15 When HCl reacts with finely powdered iron, it forms ferrous chloride and not ferric chloride. Why? Ans- Libration of hydrogen prevents the formation ferric chloride Fe + HCl FeCl 2 + H 2 Example 16 Noble gases have very low boiling points. Why? Ans - Noble gasses are mono atomic. They have no inter atomic forces, they have van der Wall s forces. So liquefied at very low temp. So they have low boiling points. Textbook Exercise Questions & Answer 1. Why does the reactivity of nitrogen differ from phosphorus? Ans- Nitrogen exists as a diatomic molecule by triple bond. It is gas at room temperature.multiple bond is not possible in phosphorus due to its large size.it exists as P 4 molecule. Nitrogen is less negative than phosphorus due to high bond dissociation energy. 2. The HNH angle value is higher than HPH, HAsH and HSbH angles. Why? [Hint: Can be explained on the basis of sp3 hybridisation in NH 3 and only s p bonding between hydrogen and other elements of the group]. Ans-The central atom is sp 3 hybridised. Electronegativity decreases down the group. The forces of repulsion decreases down the group so angle decreases. 3. Why does R 3 P = O exist but R 3 N = O does not (R = alkyl group)? Ans- Due to absence of d orbitals in nitrogen. It can not extend its covalency to five but in phosphorus vacant d-orbits exists. So exists. 4. Explain why NH 3 is basic while BiH 3 is only feebly basic. Ans- Nitrogen is smaller than bismuth. So electron density is concentrated on nitrogen is maximum. It can donate electron easily. 5. Nitrogen exists as diatomic molecule and phosphorus as P 4. Why? Ans-Due to small size, two nitrogen atoms can be joined by three covalent bonds in order to complete octet. The size of phosphorus is large so it has less tendency to form three bonds. So, p atom complete its octet by sharing electrons with three other p atoms. So, phosphorus exists as p 4 molecule. 6. Write main differences between the properties of white phosphorus and red phosphorus. Ans- White phosphorus Red phosphorus 1. white but turns yellow on exposure dark red 2. waxy solid brittle powder 3. less stable at ordinary temp. more stable at ordinary temp. 4. very active less reactive 7. Why does nitrogen show catenation properties less than phosphorus? Ans- Catenation means self linkage property of nitrogen is less than that of phosphorus because the N- N bond is weaker than P-P bond.

70 8. Can PCl 5 act as an oxidising as well as a reducing agent? Justify. Ans-Phosphorus can show maximum oxidation state. Ex PCl 5. It can not increases its oxidation state beyond 5. So, it can not act as reducing agent. But it can act as an oxidizing agent by decreasing its oxidation state from +5 to +3 Ex. Ag + PCl 5 2AgCl + PCl 3 Oxidizing agent - is one which oxidize other, but it self is reduced. Reducing agent is one which reduce other, but itself is oxidized. 9. Why is dioxygen a gas but sulphur a solid? Ans-Due to small size and high electronegativity oxygen atom forms Pл-Pл double bond due to weak van der Walls force, oxygen is gas. But sulphur does not form stable Pл-Pл bond and do not exists as S 2. It is linked by single eight atoms per molecule S 8 and strongly held together so it exists as solid. 10. Which aerosols deplete ozone? Ans -Chloroflurocarbon is oxides of nitrogen and sulphur. 11. How is SO 2 an air pollutant? Ans-SO 2 dissolves in rain water and produces acid rain. The acid rain contains sulphuric acid SO 2 + H 2 O + O H 2 SO Why are halogens strong oxidising agents? Ans-Halogens have strong tendency to accept electrons due to high electronegativity. 13. Explain why fluorine forms only one oxoacid, HOF. Ans-Due to small size,high electronegativity and absence of d-orbital fluorine shows only single oxidation state (-1).So it forms only one oxoacid HOF. 14. Explain why inspite of nearly the same electronegativity, oxygen forms hydrogen bonding while chlorine does not. Ans-Oxygen has smaller size than chlorine,which favours hydrogen bonding. 15. Write two uses of ClO 2. Ans-(1)It acts as bleaching agent for paper pulp in industry and textile industry. (2)It acts as germicide for disinfecting water. 17. What inspired N. Bartlett for carrying out reaction between Xe and PtF 6? Ans- N. Bartlett first prepared an ionic compound O 2 + [PtF 6 ] - from oxygen and PtF 6.Ionisation enthalpy of xenon gas is close to that oxygen.that is inspired to carry out the reaction between Xe & PtF What are the oxidation states of phosphorus in the following: (i) H 3 PO 3 (ii) PCl 3 (iii) Ca 3 P 2 (iv) Na 3 PO 4 (v) POF 3? Ans- H 3 PO 3 +3; PCl 3 +3; Ca 3 P 2-3 Na 3 PO 4 +5; PoF With what neutral molecule is ClO - isoelectronic? Is that molecule a Lewis base? Ans- ClO - is isoelectronic with ClF is Lewis base as Cl atom can accept a pair of electrons. It is due to presence of vacant d-orbital in chlorine. 20. Arrange the following in the order of property indicated for each set: (i) F 2, Cl 2, Br 2, I 2 - increasing bond dissociation enthalpy. (ii) HF, HCl, HBr, HI - increasing acid strength. (iii) NH 3, PH 3, AsH 3, SbH 3, BiH 3 increasing base strength Ans- (i) I 2 < F 2 < Br 2 < Cl 2 Increasing bond dissociation enthalpy. (ii) HF < HCl < HBr < HI Increasing acid strength (iii) BiH 3 < SbH 3 < AsH 3 < PH 3 < NH 3 Increasing base strength 21. Which one of the following does not exist? (i) XeOF 4 (ii) NeF 2 (iii) XeF 2 (iv) XeF 6 Ans- NeF 2 does not exist 22. Why do noble gases have comparatively large atomic sizes? Ans- In case of noble gases the atomic radii corresponds to Vander Waals radii, which are always large. Vander Waals radii arise simply due to Vander Waal s force.

71 UNIT 8 THE d- and f- BLOCK ELEMENTS Marks =5 Electronic configuration of transition elements is (n-1) 1-10 ns 1-2. The name of transition metal refers to d block. The metal of inner transition metal refers to f block. d block contain 3-12 group elements where d block are progressively filled, where as f block contain elements in which the 4f and 5f orbitals are progressively filled. Three series of transition metals are present. 3d series9sc-zn), 4d series(y cd), 5d series(la Hg omitting Ce Lu) The name transition is given because of there position between s and p block. the configuration Cr is 3d 5 4s and Cu is 3d 10 4s 1. Zn, Cd and Hg are not regarded as transition elements because orbitals are completely filled. Properties Transition metals are quite hard, high melting point, high enthalpies of atomization, high ionization because of strong metallic bond. Covalent bond and unpaired electron. Transition element show vari9able oxidation states due to the participation of ns and (n-i) d electrons in bonding. Highest oxidation state shown by transition element is eight. M = n (n+2) M is the magnetic moment n is the number of unpaired electron. Transition elements and their compounds are paramagnetic in nature because they contain unpaired electrons. Transition metals from a large number of complex compound due to smaller size of the metal ions, high ionic charges and availability of vacant d orbitals. Transition metal form coloured compounds due to d-d transition. The transition metals and some of their compounds act as catalyst due to variables oxidation state, and large surface area. Interstitial compounds are those which are those which are formed when small atoms like H, C or N are trapped inside the critical lattice of metals. Transition metal compounds form interstitial compound because some voids are present in the site. The transition metals form a number of alloys because of similar atomic size. Ex Brass (Copper, Zinc) Bronze (Copper- Tin). Oxides and Oxo anions As the oxidation number of a metal increases, ionic character decreases so Mn 2 O 7 is acidic V 2 O 3 Basic, V 2 O 5 atmospheric, CrO is basic Cr 2 O 3 is atmospheric. MnO 2 is atmospheric preparation K 2 cr 2 O 7 (Potassium dichromate) 4 FeCr 2 O Na 2 CO 3 + 7O 2 8Na 2 CrO 4 + 2Fe 2 O 3 + 8CO 2 Chromite are 2 Na 2 CrO 4 + 2H + Na 2 Cr 2 O 7 + 2Na + + H 2 O Na 2 CrO 7 +2 KCl K 2 Cr 2 O 7 + 2NaCl *on increasing ph -2 Cr 2 O 7 + OH - -2 CrO 4 + H 2 O Orange yellow On decreasing ph -2 CrO 4 + H + Yellow Cr 2 O H 2 O orange Preparation of Potassium Permanganate (KMnO 4 ) 2MnO 2 + 4KOH + O 2 2K 2 MnO H 2 O

72 Black 3MnO H + green 2MnO MnO H 2 O Commercial preparation MnO 2 fused with -2 MnO 4 electrolytic -1 MnO 4 permagnet ion Alkali oxidation (pink colour) Manganate and permanganate are tetrahedral. Chromate is tetrahedral. On heating KMnO 4 K 2 MnO4 MnO 2 + O 2 Pink green USES KMnO 4 - oxidation, bleaching of coal, cotton, silk, analytical chemistry, disinfectant for ater K 2 Cr 2 O 7 oxidant,chrome tanning in leather industry f block elements consists of two series- 1. Lanthanoids 14 elements Cerium to Lutetium are similar to lanthanium. 2. Actinoids 14 elements Thorium to lawrencium are similar to actinium Misch metal 95% lanthanoid metals and 5%iron traces of S, C, Ca, Al. The different oxidation states exhibited by lanthanoids are +2, +3, +4. The actinoids exhibits a large number of oxidation states than the corresponding lanthanoid because of large size. Electronic configuration of lanthanoid is 4f¹ ¹⁴ 5dº ¹ 6s². SIMILARITY BETWEEN LANTHANOIDS AND ACTINOIDS Both shows +3 oxidation states. f orbitals are progressively filled. Same number of unpaired electrons. Electropositive and highly active. Contraction. DIFFERENCE BETWEEN LANTHANOIDS AND ACTINOIDS LANTHANOIDS ACTINOIDS 1. Compounds are less basic. 1. Compounds are less basic. 2. Binding energy of 4f is higher. 2. Binding energy of 5f is lower. 3. Most ions are colourless. 3. Ions are coloured. 4. Not radioactive. 4. Radioactive. 5. Tendency to form complexes is less. 5. Tendency is more. LANTHANOIS CONTRACTION The steady decrease in atomic size of lanthanoid elements with increase atomic number is called lanthanoid contraction. CAUSE Nuclear charge increases from left to right, to be compensated by shielding effect by 4f electrons. F electrons orbitals have poor shielding effect. So, there is steady decrease in atomic size. CONSEQUENCE Separation of lanthanoids difficult as they have similar properties. Similarity in the atomic size of the elements of second and third transition series in the same group. Ex - Zr - Hf Variation in basic strength of hydroxide. The basic strength decreases from La (OH) 3 to Lu(OH) 3. Because of size M+³ ion decreases.

73 A few Ionic Euations related to the d- Block Elements

74 Some questions and Answers for practice : 1. Silver atom has completely filled d orbitals (4d10) in its ground state. How can you say that it is a transition element? ANS: Because silver can exhibit +2 oxidation state, which has incompletely filled d orbital. 2. Why do the transition elements exhibit higher enthalpies of atomisation?

75 ANS: Because Sc (21) has incompletely filled d orbital so it takes part in the formation of metallic bond. But zinc has completely filled d orbital so zinc has weakest bond and lowest enthalpy of atomization. 3. Name a transition element which does not exhibit variable oxidation states. ANS: Mn exhibits large number of oxidation states because s and d orbitals takes in bond formation. 4. The E (M2+/M) value for copper is positive (+0.34V). What is possibly the reason for this? (Hint: consider its high Δ a H and low Δ hyd H ) ANS: both fluorine and oxygen have small size,high electronegativity. 5. Why is the EV value for the Mn3+/Mn2+ couple much more positive than that for Cr3+/Cr2+ or Fe3+/Fe2+? Explain ANS: Fe+² is strongest reducing agent because it will oxidize to Fe+³ which is half filled d orbital, more stable. 6. Calculate the magnetic moment of a divalent ion in aqueous solution. If its atomic number is 25. ANS: M(27) M+²=[Ar] 3d⁷ 4sº Magnetic moment, M = n (n+2) 3x5 = 15 = 3.8BM 7. What is meant by disproportionation of an oxidation state? Give an example. ANS: Because hydration energy of Cu+² ion is high which can overcome 2 nd ionization energy. So, Cu+² is more stable. 8. Name a member of the lanthanoids series which is well known to exhibit +4 oxidation state. ANS: Due to poor shielding by 5f electrons in actinoids as compared to 4f electrons in the lanthanoids. NCERT-EXAMPLES Example 1- Why do the transition elements exhibit higher enthalpies of atomisation? ANS: due to large number of unpaired electrons, strong bonding atoms results in higher enthalpies of atomization. Example 2- Name a transition element which does not exhibit variable oxidation states. ANS: Scandium does not exhibit variable oxidation state. Example 3- Why is Cr 2+ reducing and Mn 3+ oxidising when both have d 4 configuration. ANS: Cr+²is reducing as it changes from d⁴ to d³ is half filled in t2g Mn+² to Mn+³ which half filled d⁵ has extra stability. Example 4- Calculate the magnetic moment of a divalent ion in aqueous solution if its atomic number is 25. ANS: With atomic number 25 Mn+²has d⁵ cofiguration-50 M = 5(5+2) = 5.92BM Example 5- Name a member of the lanthanoids series which is well known to exhibit +4 oxidation state. ANS: Cerium ( Z = 58 )

76 A few more questions & answers for practice 1. Why are Mn 2+ compounds more stable than Fe 2+ towards oxidation to their +3 state? Ans Mn+²s 3d⁵, half filled which is stable, while Fe*² is 3d⁶ not stable. 2. To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples. Ans The presence of half filled or completely filled orbitals. 3. Name the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number. Ans Mno4 ¹ oxidation state +7 equal to group number Cro4 ² oxidation state +6 equal to group number. 4. In what way is the electronic configuration of the transition elements different from that of the non transition elements? Ans In transition elements d orbitals are progressively filled where as in Non transition elements outer most s or p orbitals are progressively filled. 5. What are the different oxidation states exhibited by the lanthanoids? Ans common stable oxidation state of lanthanoid is +3, others are +2 and Predict which of the following will be coloured in aqueous solution? Ti 3+, V 3+, Cu +, Sc 3+, Mn 2+, Fe 3+ and Co 2+. Give reasons for each. Ans - Ti+³, V+³, Mn+², Fe+³, Co+² as they contain unpaired electrons 7. Compare the chemistry of actinoids with that of the lanthanoids with special reference to: (i) electronic configuration (iii) oxidation state (ii) atomic and ionic sizes and (iv) chemical reactivity Ans 1) Electronic configuration In lanthanoids 4f orbitals are progressively filled whereas in actinoids 5f orbitals are progressively filled 2) Oxidation state lanthanoids show+3state some shows +2 and +4. But actinoids show +3, +4, +5, +6, +7 oxidation state. +3 and +4 are Common. 3) Atomic and ionic size decreases from left to right. atomic and ionic sizes decreases from left to right in both. But decreases in more in case of actinoids 4) Chemical reactivity actinoids are more reactive than lanthanoids. 8. Compare the chemistry of actinoids with that of the lanthanoids with special reference to: (i) electronic configuration (iii) oxidation state (ii) atomic and ionic sizes and (iv) chemical reactivity 9. How would you account for the following? (i) Of the d4 species, Cr 2+ is strongly reducing while manganese(iii) is strongly oxidising. (ii) Cobalt(II) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidised. (iii) The d1 configuration is very unstable in ions Ans- (1). Cr +2 is reducing agent, it change to Cr 3+ by loosing electron, Cr 3+ is more stable due to half filled. Mn 3+ is oxidising agent, reduced to Mn 2+ which is half filled(d 5 ) and very stable. (2). cobalt(2) is oxidised to Co(3) because Co(3) is more stable than Co(2). (3). because after loosing electron it become more stable

77 10. What is meant by disproportionation? Give two examples of disproportionation reaction in aqueous solution. Ans. In disproportionation reaction, an element under goes oxidation and reduction both eg. 2Cu + Cu + Cu Which metal in the first series of transition metals exhibits +1 oxidation state most frequently and why? Ans. Cu exhibits +1 oxidation state, by loosing one electron, the cation ion acquires a stable configuration of d orbital. 12. Give examples and suggest reasons for the following features of the transition metal chemistry: (i) The lowest oxide of transition metal is basic; the highest is amphoteric/acidic. (ii) A transition metal exhibits highest oxidation state in oxides and fluorides. (iii) The highest oxidation state is exhibited in oxo-anions of a metal. Ans- (i) In the lowest oxidation state, ionic bond is formed, less number of electrons are involved. Oxides donate electrons and behave like a base. In the highest oxidation state, covalent bond is formed. In high oxidation state more electrons are involved in bonding. Oxide can gain electron, behave like Lewis acid. (ii) Because oxygen and fluorine are strong oxidising, highly electronegative element. (iii)due to high electro negativity of oxygen. 13. What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses. Ans- Alloy are homogeneous mixture of two or more metals. Misch metal is an alloy, which contains 45% lanthanoid metals, iron 5%, traces of S, C, Ca, Al Use- used to produce bullets, shell and lighter flint.3% misch metal to magnesium used in making jet engine parts. 14. What are inner transition elements? Decide which of the following atomic numbers are the atomic numbers of the inner transition elements : 29, 59,74, 95, 102, 104. Ans- Lanthanoids and Actinoids are called inner transition elements because inner f orbitals are progressively filled.z=58 to71 are lanthanoids.z=90 to103 are actinoids,so atomic no. 59,95and102 belong to inner transition elements. Text book ex. Answer 30-Lawrencium (Lr=103) [Rn] 5f 14 6d 1 7s 2, oxidation state=+3 Text book ex. Answer 36- Te 2+ 3d 2,Cr 3+ 3d 3,Fe 2+ 3d 6,Co 2+ 3d 7,Cu 2+ 3d 9,V 2+ 3d 8, Mn 2+ 3d 5,Fe 3+ 3d 5,Ni 2+ 3d The chemistry of the actinoid elements is not so smooth as that of thelanthanoids. Justify this statement by giving some examples from theoxidation state of these elements. Ans. Lanthanoids exhibits +2,+3,+4 oxidation state out of these +3 is most common lanthanoids element shows +3 oxidation state. 16. Which is the last element in the series of the actinoids? Write the electronic configuration of this element. Comment on the possible oxidation state of this element. Ans. Lr, Z=103, is the best actinoids. It electronic configuration is [Rn] 86 5f 14 6d 1 7s 2.The possible oxidation state is +3.

78 UNIT - 9 CO-ORDINATION COMPOUNDS Marks= 3 Double Salt It gives test of all the ions present in the composition of the salt Coordination Compound It does not gives test of all the ions present in the composition of the salt IMPORTANT TERMS a) Coordination entity- A coordination entity is a compound having a central metal ion/atom bonded with fixed number of ions or molecules and exist as a single entity. It is written within [metal ion/atom and ligands] b) Central atom/ion- In a coordination entity the metal ion/atom which is bonded to a fixed number of ion/molecules (Ligands) is called central metal ion/atom c) Ligands- The ion or molecules bonded with central ion/atom with coordinate bond are called ligands. They are of different typesmonodentate or unidentate ligands- These type of ligand are bonded with metal ion through one site(bond). Examples are H 2 O, NH 3, CN - i) Ambidentate ligands- They are unidentate ligands but can donate electron through two different atoms but at a time only through one atom. Exa- CN - can donate electron through C as well as Through N atom, NO 2 - can donate electron through O as well as Through N atom, ii) Didentate ligands- These type of ligand are bonded with metal ion through two site/ form two coordinate bond. Examples are (COO - - ) 2, Ethylenediamine iii) Polydentate ligands- These ligands can form more than two coordinate bonds with metal ion/atom. Example- EDTA iv) Chelating ligands- Didentate or polydentate ligands which form a closed ring structure around metal ion are called chelating ligands. Example- oxalate ion. Ethylene diamine etc. d) Coordination number- Coordination number of a metal ion is equal to the number of coordinate bonds ligands forms with metal ion. Coordination number of a metal ion is = the no. of momodentate ligands = 2 no. of didentate ligands = 3 no. of tridentate ligands e) Coordination sphere- In a coordination compoud [ ] known as coordination sphere as species within [ ] are bonded through coordinate bonds f) Oxidation number of metal ion- It is equal to the charge on metal ion in the complex or coordination compounds g) Chelate compound- The coordination formed by chelating ligands with metal ion is called chelate compound. Chelate compounds are more stable than normal coordination compounds.

79 Description of a Coordination Compound a C s +z q [M +x (L -y ) ] d +/-B A P b -V Symbol C a s Description of symbols Simple cation (may be Na +, K + etc) ( will give its test) Ionisation sphere (species within this sphere supply ion) Number of simple cation +z Charge on simple cation q M Coordination sphere (Species within [ ] exist as single entity and do not supply ions) Metal ion/atom (Any transition metal with a fixed oxidation state) +x Charge on metal ion (It is equal to the oxidation state of TM) L Ligand ( Any ionic or molecular species which can donate electron pair to the TM) -y Charge on ligand d No. of ligand + or -B Charge on complex (It is sum of charge on metal ion and charges on ligands) b A p Ionisation sphere (species within this sphere will supply ion) Simple anion (Will give its test) No. of simple anion -v Charge on simple anion

80 Naming of coordination compounds- 1. Name of cation (simple or complex) is written first. 2. Name of anion (simple or complex) is written at last. 3. Name of coordination entity is written as single name. 4. Within coordination sphere ligands name (with suitable prefix and ending) is written first. When more than one type of ligands are there that their name are written in the order of alphabates 5. Name of metal ion is follow ligands name with their oxidation no. in Roman. Examples- ISOMERISMS IN COORDINATIONCOMPOUNDS 1. GEOMETRICAL ISOMERISM- compounds having same molecular formula, same structural formula but different spatial arrangement of atoms/ions. In cis form both the Cl atoms are in the adjacent position In trans isomer both the Cl atoms are in opposite position Geometrical isomers (cis and trans) of Pt [NH 3 ) 2 Cl 2 ]

81 In cis form both the Cl atoms are in the adjacent position In trans isomer both the Cl atoms are in opposite position In cis form both the Cl atoms are in the adjacent position In trans isomer both the Cl atoms are in opposite position In cis form both the NH 3 atoms are in the adjacent position In trans isomer both the NH 3 atoms are in opposite position 2. OPTICAL ISOMERISM- compounds having same molecular formula, same structural formula but different spatial arrangement of atoms/ions and rotate plane of polarized light towards left or right hand side. It is known as optical activity. Chirality is an essential requirement for showing optical activity. Both the structures are non super imposable mirror images of each other Both the structures are non super imposable mirror images of each other

82 GEOMETRY OF COORDINATION COMPOUNDS- BONDING IN COORDINATION COMPOUNDS- (VBT) In the presence of strong NH 3 ligands 3d electrons pair up. Low spin compound is formed. Magnetic moment of such compound is low They are not able to show d-d transition so they are colourless In the presence of weak F ligands 3d electrons do not pair up. High spin compound is formed. Magnetic moment of such compound is high They are able to show d-d transition so they are coloured In the presence of weak Cl ligands 3d electrons do not pair up. High spin compound is formed. Magnetic moment of such compound is high They are able to show d-d transition so they are coloured

83 In the presence of strong CN ligands 3d electrons pair up. Low spin compound is formed. Magnetic moment of such compound is low They are not able to show d-d transition so they are colourless CRYSTAL FIELD THEORY & CRYSTAL FIELD SPLITING ENERGY- According to crystal field theory bond formed between ligands and metal ion is ionic. Ligands are treated as point ve charged and metal ions are treated as point +ve charged. In order to form ionic bond ligands approach towards metal ion in this process ligands show repulsive interaction with d orbitals of transition metal. Five d orbitals of TM split into two sets. e g and tg. The energy difference between two sets of orbitals is termed as CRYSTAL FIELD SPLITING ENERGY. Its value depends upon the field strength of ligands. Stronger ligands like CN -, NH 3 CO cause greater splitting and weaker ligands like F, Cl, H 2 O cause poor splitting When splitting is more electrons can not jump from lower orbitals to higher energy orbitals so they pair up in lower set of orbitals and show poor magnetic moment and colourless. This happens in the presence of ligands like CN -, NH 3 CO. In case of weal ligands like F, Cl, H 2 O opposite scene is observed

84 When splitting is more electrons can not jump from lower orbitals to higher energy orbitals so they pair up in lower set of orbitals and show poor magnetic moment and colourless. This happens in the presence of ligands like CN -, NH 3 CO. In case of weal ligands like F, Cl, H 2 O opposite scene is observed BONDING IN METAL CARBONYLS Metal carbonyls are special class of coordination compounds. In metal carbonyls metal is found in zero oxidation state. There is a special kind of bonding i.e. pπ-dπ bond by back donation of electrons from metal atom to CO group In metal carbonyl electron pair donated from filled orbital of CO to empty orbitals of transition metal. In addition to this electron pair is donated from filled d orbital of transition metal to empty antibonding orbitals of CO

85 APPLICATIONS OF COORDINATION COMPOUNDS- Some questions & Answers for practice : 1. Write the formula for the following: - (3) (i) Tetrahydroxozincate (II)ion (ii) Hexaammineplatinum (IV) ion (iii) Hexaamminecobalt (III) sulphate Ans: (i) [Zn(OH) 4 ] 2- (ii) [Pt(NH 3 ) 6 ] 4+ (iii)[co(nh3) 6 ] 2 (SO 4 ) 3 2. Write the IUPAC name of the following: - (2) (i)[pt(nh 3 ) 2 Cl(NH 2 CH 3 )]Cl (ii) [Co(NH 3 ) 4 Cl(NO 2 )]Cl Ans (i) Diamminechloridomethyleamineplatinum (ii) chloride (ii) Tetraaminechloridonitrito-N-cobalt (iii) chloride 3. What is meant by ambidentate ligands? Give two examples. (1) Ans : Ligand which can ligate through two different atoms e.g. CN -, SCN - 4. Explain on the basis of VBT, the experimental findings that [Ni(CN) 4 ] 2- ion with a square-planar structure is diamagnetic and the [NiCl 4 ] 2- ion with tetrahedral geometry is paramagnetic. (2)

86 Ans: In [Ni(CN) 4 ] 2- Ni 2+ electronic configuration- 3d 8 Hybridization - dsp 2 Unpaired electron = 0, therefore it is diamagnetic In [NiCl 4 ] 2- Ni 2+ Electronic configuration 3d 8 Hybridization SP 3 Unpaired electron = 2, therefore it is paramagnetic 5. Aqueous copper sulphate solution (blue in colour) gives( a) a green precipitate with aqueous potassium fluoride, and(b) a bright green solution with aqueous potassium chloride. Explain. (2) Ans: (a) [Cu(H 2 O) 4 ] F - [CuF 4 ] H 2 O Blue Green ppt (b) [Cu(H 2 O) 4 ] Cl - [CuCl 4 ] H 2 O Blue Bright green solution 6. Give a chemical test to distinguish between [Co(NH 3 ) 5 Br]SO 4 and [Co(NH 3 ) 5 SO 4 ]Br. What kind of isomerism do they exhibit? (2) Ans : (i) [Co(NH 3 ) 5 Br]SO 4 + Ba 2+ BaSO 4 White ppt [Co(NH3)5Br]SO 4 + Ag+ No ppt (ii) [Co(NH 3 ) 5 SO 4 ]Br. + Ba2+ No ppt [Co(NH 3 ) 5 SO 4 ]Br. + Ag+ AgBr Yellow ppt 7. Name the metal present in (i) Chlorophyll (ii) Haemoglobin (iii) Vitamin B 12 (iv) Cis-platin (2) Ans: (i) Mg (ii) Fe (iii) Co (iv) Pt 8. What is the coordination number of central metal ion in (i) [Fe(C 2 O 4 ) 3 ] 3- (ii) [Co(en) 2 Cl 2 ] +. (1) Ans: (i) 6 (ii) 6 9. Why are cyclic complexes more stable than open one? (1) Ans : Cyclic complexes have more stability because of reduced strain in five or six member rings. e.g. Chlorophyll (cyclic) is more stable than [Co(NH 3 ) 5 Br]SO CuSO 4 on mixing with NH 3 (1:4) does not give test for Cu 2+ ions but gives test for SO 4 2- ions. Why? (2) Ans: It is because when NH 3 coordinates to Cu 2 ions it forms the complex [Cu(NH 3 ) 4 ]SO 4. Copper ions are present in coordination sphere, therefore, they are non- ionisable whereas SO 4 2- ions are counter ions which are ionisable.

87 UNIT-10 HALOALKANES AND HALOARENES Marks = 4

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91 Q :1 Identify A, B in the following:- 2 Dry ether Br + Mg A H 2 O B Ans: A= MgBr B = Q : 2 Give the IUPAC names of the following:- 2 (a) Cl-CH 2 C C-CH 2 -Br (b) CH 3 CH 2 CH(CH 3 )CH(C 2 H 5 )Cl Ans : A) 1-Bromo,4-Chloro But-2- yne B) 3-Chloro-4-Methyl-hexane Q: 3 Explain why Aryl halides are extremely less reactive towards nucleophlic substitution reactions. Ans : Due to resonance in aryl halide in C-X bond will acquire double bond character so difficult to substitute halogen. Q: 4 Haloalkanes are only very slightly soluble in water explain. 1 Ans: Unable to form Hydrogen bond with water. Q : 5 Write short notes on:- (a) Wurtz reaction (b) Swarts reaction 2 Dry Ether Ans :A) R-X + 2Na +R-X R-R + 2Na X B) R-Br + AgF RF + AgBr Q : 6 Arrange the following in increasing order of the property indicated 1 Bromomethane, Bromoform, Chloromethane, Dibromomethane (Boiling point) Ans : Chloromethane < Bromomethane < Dibromomethane < Bromoform. Q : 7 What do you mean by asymmetric carbon? Give one example. 1

92 Ans: If all the four atom or group of atoms attach to a carbon are different then carbon is called asymmetric carbon. e.g. CH 3 CHBrCl. Q : 8 Explain why H 2 SO 4 is not used during the reaction of alcohol with KI. 1 Ans: Because HI produced get oxidize to Iodine in presence of Sulphuric acid. Q : 9 Explain why Reaction of CH 3 Br with KCN yields CH 3 CN while with AgCN yields CH 3 NC? 1 Ans: Because CN - is an ambidentate nucleophile. Q : 10 Write short notes on:- (a) Sandmeyer s reaction (b) Finkelstein reaction 2 CuCl /HCl Ans : (a) C 6 H 5 N 2 Cl C 6 H 5 Cl Acetone (b) C 6 H 5 Br + NaI C 6 H 5 I + Na Br Q11. Convert : -propane-1-ol to 2-iodopropane. 1 Conc. H 2 SO 4 Morkownikove s Rule Ans : CH 3 CH 2 CH 2- OH CH 3 CH=CH 2 + HI CH 3 CH I CH 3 443K Q12. Arrange the following in increasing order of the property indicated 2 (a) 1-Bromobutane, 2-Bromobutane, 2-Bromo-2-methylpropane (boiling point) Ans : 2-Bromo-2-methylpropane <2-Bromobutane < 1-Bromobutane, (b) CH 2 Cl and Cl (reactivity towards S N2 reaction) Ans : CH 2 Cl UNIT 10 ALCOHOL, PHENOL & ETHERS Marks=4 NOMENCLATURE

93 PREVIOUS YEAR QUESTION ON NOMENCLATURES 1. Draw the structure of Hex-1-en-3ol compounds (Delhi 2012)? 2. Draw the structural formula of 2-Methylpropan-2-ol (delhi2012)

94 3. Write the name of the following compound:- (HEX1-EN-3-OL) 2, 5-DIMETHYL HEXANE-1, 3-DI-OL 2,5DIMETHYL PHENOL 2-METHYL-1-METHOXYBUTANE NAME REACTION 1. KOLBE REACTION- (PHENOL TO SALICYLIC ACID) 2. REIMER TIMEN REACTION- (PHENOL TO SALICYLALDEHYDE) 3. WILLIAMSON SYNTHESIS- (ALKYL HALIDE TO ETHER) ALKYL HALIDE SODIUM ALKOXIDE ETHER

95 1. HYDRATION OF ETHENE H 2 C=CH 2 + H 2 O MECHANISM H + CH 3 CH 2 OH 2. DEHYDRATION OF ETHANOL- H + CH 3 CH 2 OH CH 2 =CH 2

96 3. DEHYDTATION OF ETHANOL ( IN EXCESS OF ALCOHOL) H+ CH 3 CH 2 OH CH 3 CH 2 -OCH 2 CH 3 PREVIOUS YEAR QUESTIONS 1. Explain the mechanism of acid catalyzed hydration of alkenes to form corresponding alcohol. ( All India 2012 ) 2. Write the mechanism of hydration of ethane to ethanol. (Foreign 2009, 2010,2011 C ) REASON BASED QUESTIONS Reason based question based on the following facts ACIDIC CHARACTER

97 Phenol is more acidic than aliphatic alcohol Due to resonance peroxide ion is more stable than phenol Due to the higher electronegativity of SP2 hybridised carbon of phenol to which OH is attached, electron density decreases on oxygen &increases the polarity of O-H bond. Effect on acidic character of phenol due to presence of EWG and ERG group The acidic strength of alcohol depends on pka value: the higher the value of pka lower is acidic strength. The pka is inversely proportional to Ka.

98 PREVIOUS YEAR QUESTIONS 1.Explain the following behaviorsa. Alcohols are more soluble than water than hydrocarbons of comparable molecular mass (HINT:- due to hydrogen bonding)? b.ortho nitro phenol is more acidic than Orthomethoxy phenol?[hint:-higher the stability of phenoxide ion more will be the acidic character,no2 group (EWG) increases the stability of phenoxide ion where as methoxy group(erg) destabilizes the phenoxide ion](all INDIA 2012 ). 2.Give reason for the following:- A. The boiling point ethanol is higher than that of methanol? B. Phenol is stronger acid than alcohol (2009,2011 C) CHEMICAL TEST FOR DISTINGUISH PAIRS OF COMPOUNDS 1) LUCAS REAGENT TEST-The pri.,sec.,tert. Alcohols can be distinguished by LUCAS reagent test (ZnCl2 + conc.hcl) a) Organic compound +LUCAS reagent --1.If turbidity appears immediately the given organic compound is tertiary alcohol (t-alcohol or 3 Alcohol) b) If turbidity appears after 5 min. the given organic compound is secondary alcohol (sec. alcohol or 2 Alcohol) c) If turbidity not appears at room temp.-primary alcohol. 2) FERRIC CHLORIDE TEST (only for phenol)- Phenol + neutral Ferric chloride violet color appears. Preparation of alcohol 1. FROM GRIGNARD REAGENT - IMPORTANT REACTION WITH REAGENTS (BASED ON PREPARATION & PROPERTIES)

99 2. FROM CARBONYL COMPOUNDS:- 1. By reduction of aldehyde and ketones - 2.By reduction of carboxylic acid and ester :- \ Preparation of phenol- 1. From chloro benzene 2. From benzene sulphonic acid:- 3. From benzene diazonium chloride:- 4. From cumene

100 PROPERTIES OF ALCOHOL 1. ESTERIFICATION- alcohols and phenol react with carboxylic acid, acid chloride and acid anhydride to form esters. ESTERIFICATION METHOD USED FOR THE CONVERSION OF PHENOL TO ASPIRIN 1.Phenol convt. to salicylic acid.by KOLBE REACTION 2. salicylic acid convt. to aspirin by esterification 2. OXIDATION Oxidation of alcohol involves formation of carbon oxygen double bond (carbonyl grp.) And a mixture of carboxylic acids containing lesser number of carbon atoms is formed.

101 DEHYDROGENATION When the vapours of a pri. Or sec. alcohol are passed over a heated copper at 573K, dehydrogenation takes place and an aldehyde or a ketone is formed while tert. Alcohols undergo dehydration. REACTION OF PHENOL

102 Q : 1 Write the names of reagents and equations for the preparation of 2-Methoxy-2-methylpropane. 1 Ans: CH 3 Br + (CH 3 ) 3 CO - Na > CH 3 -O-C(CH 3 ) 3 + NaBr Q : 2 Give reasons: 2 a. Ethanol has higher boiling point in comparison to methoxymethane. Ans : Due to presence of intermolecular hydrogenbond in ethanol. b. Phenols are more acidic than alcohols. Ans : Due to R effect of phenoxide ion produced after the loss of proton from phenol, phenoxide ion is more stable than alkoxide ion Q : 3 (a) o-nitrophenol is steam volatile while p-nitrophenol not. Ans : Due to presence of intramolecular hydrogen bond (b)cleavage of phenyl alkyl ether with HI always gives phenol and alkyl iodide Ans : Due to resonance C-O bond of phenyl will acquire double bond character hence difficult to break Q : 4 How are the following conversions carried out: - (2) a. Methyl magnesium bromide to 2-methylpropane-2-ol CH 3 COCH 3 H + /H 2 O Ans : CH 3 MgBr (CH 3 ) 3 C-OMgBr (CH 3 ) 3 C-OH Dry Ether b. Propene to propan-2-ol HBr,MarkwnkvRule aq.koh, Ans : CH 3 CH=CH CH 3 CHBrCH CH 3 CHOHCH 3 c. Propan-2-one to 2-methyl-2-propanol CH3MgBr H + /H 2O Ans : CH 3 COCH (CH 3 ) 3 C-OMgBr (CH 3 ) 3 C-OH Dry Ether Q : 5 Write IUPAC name- (2) Ans: i) 2-methoxybutane ii) 4-Chloro,3-methyl butan-1-ol. i. CH 3 -O-CH 2 -CH-CH 3 ii. CH 3 -CH-CH-CH 2 -CH 2 OH CH 3 Cl CH 3 Q : 6 Arrange the following as property indicated: - (2) (i) pentan-1-ol, pentanal, ethoxyethane (increasing order of boiling point) Ans : n-butane <ethoxyethan < pentanal < pentan-1-ol (ii) pentan-1-ol, phenol, 4-methylphenol, 3-nitrophenol ( increasing order of acid strength) Ans : pentan-1-ol < 4-methylphenol < phenol < 3-nitrophenol. Q : 7 Give simple chemical tests to distinguish between the following pairs of compounds: (i) Phenol and Benzoic acid. (ii) Propan-1-ol and Propan-2-ol. (2) Ans: (i) C 6 H 5 OH + NaHCO No effervescence

103 C 6 H 5 COOH + NaHCO C 6 H 5 COONa + H 2 O + CO 2 (effervescence ) Lucas Reasgent (ii) CH 3 -CH 2 CH 2 OH No Turbidity CH 3 -CHOH CH Turbidity produced with in five minutes 2 Q:8 Write the mechanism of dehydration of ethanol. Ans : Conc.H 2 SO 4 CH 3 CH 2 OH CH 2 =CH 2 + H 2 O Mechanism : STEP1 : Formation of a protonated alcohol CH 3 CH 2 OH + H CH 3 CH 2 -O + H H STEP: 2 Formation of a carbocation CH 3 CH 2 -O + H CH 3 CH H 2 O H STEP : 3 Loss of proton : CH 3 CH CH 2 =CH 2 + H + Q : 9 A compound A with molecular formula C 4 H 10 O is a unreactive towards sodium metal. It does not add Bromine water and does not react with NaHSO 3 solution.on refluxing A with excess of HI gives B which react with aq. NaOH to form C. C can be converted into B by reacting with P and I 2. C on heating with aqueous alkali to form E which form F on heating with conc. H 2 SO 4. F decolourises bromine water. Identify A to F and write the reactions involved. Ans : A is not alcohol therefore it does not react with Sodium metal. A is not aldehyde and ketone as it does not react with NaHSO 3 A is not unsaturated hydrocarbon as it does not add Br 2 (aq). It is likely to be ether. Ans : CH 3 CH 2 OC 2 H 5 + 2HI C 2 H 5 I + H 2 O A excess B (C 4 H 10 O) 2C 2 H 5 I + NaOH (aq) C 2 H 5 OH + Na I B C C 2 H 5 OH C P / I C 2 H 5 I B Cu CH 3 CH 2 OH CH 3 CHO C 573 K D OH - CH 3 CHO CH 3 CHOHCH 2 CHO 3. D (3-Hydroxybutanal)

104 UNIT-12 ALDEHYDES, KETONES AND CARBOXYLIC ACIDS Marks = 6

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108 1. Arrange the following compounds in increasing order of their property as indicated: (2) (i) HCOOH, CH 3 COOH, CH 2 (F)COOH, CH 2 (Cl)COOH (acid strength) (ii) Acetaldehyde, Acetone, Formaldehyde, Ethylmethyl ketone (reactivity towards HCN) Ans: (i) CH 3 COOH< HCOOH< CH 2 (Cl)COOH < CH 2 (F)COOH 2. Give reasons (2) Carboxylic acids are stronger acids than phenols. Ans : Carboxylic acids are stronger acids than phenols because the carboxylate ion is stabilised more by the resonace. 3. Give simple chemical test to distinguish between (2) (i) Propanal and propanone Ans: CH 3 CH 2 CHO + Ag 2 O CH 3 CH 2 COOH + Ag (silver mirror) Propanal (ii) CH 3 COCH 3 + Ag 2 O no reaction Propanone Phenol and benzoic acid Ans: i. Phenol does not give sodium carbonate test whereas benzoic acid give this test. ii. Phenol give azo dye test, benzoic acid does not. 4. Identify A, B and C (2) KCN SnCl 2 -HCl Zn-Hg/ HCl CH 3 CH 2 Br A B C Ans: A= CH 3 CH 2 CN, B= CH 3 CH 2 CHO, C= CH 3 CH 2 CH 3 5..Name the following as IUPAC system (2) (i) CH 3 CH( CH 3 )CH 2 CH 2 CHO Ans: 4- methylpentanal (ii) (CH 3 ) 3 C CH2 COOH 3-methylbutanoic acid 6. Give reasons for the following: (2) (a). Ethanal is more reactive towards nucleophilic addition reactions than propanone. Ans The formula of Ethanal is (CH 3 CHO) and that of Propanone is CH 3 COCH 3 It is very clear that, in propanone the presence of two alkyl groups hindere the approach of nucleophile to carbonyl carbon. (b).ethanoic acid is a stronger acid than ethanol. Ans Due to presence of resonance in ethanoate ion.

109 7. An organic compound A (molecular formula C 8 H 16 O 2 ) was hydrolysed with dilute sulphuric (5) acid to give a carboxylic acid (B) and an alcohol (C). Oxidation of (C) with chromic acid produced (B). (C) on dehydration gives but-1-ene. Identify A, B and C. also write equations for the reactions involved. Ans: A= CH 3 -CH 2 -CH 2 -COO-CH 2 -CH 2 -CH 2 -CH 3 B= CH 3 -CH 2 -CH 2 COOH C= CH 3 -CH 2 -CH 2 CH 2 OH Equation H2SO4 CH 3 -CH 2 -CH 2 -COO-CH 2 -CH 2 -CH 2 -CH 3 CH 3 CH 2 CH 2 COOH (B) + CH 3 CH 2 CH 2 CH 2 OH (C) Dehydration CH 3 CH 2 CH 2 CH 2 OH CH 3 CH 2 CH=CH 2 Chromic acid CH 3 CH 2 CH 2 CH 2 OH CH 3 CH 2 CH 2 COOH 8. Complete the following reactions by identifying A, B and C. (2) (i) Pd/BaSO4 A + H 2 (g) (CH 3 ) 2 CH-CHO Pd/BaSO4 Ans: (CH 3 ) 2 CH-CO-Cl + H 2 (g) (CH 3 ) 2 CH-CHO (ii) CH 3 CH 2 CH=CH 2 + B CH 3 CH 2 CHO +CH 2 -O ZnO/H2O Ans: CH 3 CH 2 CH=CH 2 + O 3 CH 3 CH 2 CHO +CH 2 -O 9. Suggest a reason for the large difference in the boiling points of butanol and butanal, although they have the same solubility in water. (2) Ans: The b.p. of butanol is higher than butanal because butanol has strong intermolecular hydrogen bonding while butanal has weak dipole-dipole attraction. However, both of them form H- bonding with water and hance are soluble. 10. Name two methods which are used to convert >C=O group into >CH2 group. (2) Ans: Clemmensen reduction and Wolff-Kishner reduction UNIT 13 NITROGEN CONTAINING ORGANIC COMPOUNDS Marks = 4 Nomeclature of these Compound :

110 5.COUPLING REACTION (the reaction between benzenediazoniamchloride with phenol or aniline to form dyes (coloured) compound) N 2 Cl + H- -OH -N=N OH BDC PHENOL dyes (coloured) compound PREVIOUS YEARS QUESTION 1. Describe the following giving the relevant chemical equation in each case. (i) Carbylamine reaction (ii) Hofmann s bromamide reaction Hints: See the name reaction. 2. Illustrate the following with an example of reaction in each case. [Delhi 2011C] (i) Sandmeyer s reaction (ii) Coupling reaction Hints: (i) application of BDC (ii) See the name reaction. [All India 2012; Delhi 2012] 3. Giving an example for each describe the following reactions. [Delhi 2009, 2008C; Foreign 2008] (i) Hofmann s bromamide reaction (ii) Coupling reaction Hints: See the name reaction.

111 REASON BASED QUESTIONS Reason based question based on the following facts: (i) Gabriel phthalimide synthesis is not suitable for preparation of aromatic primary amine because aryl halides do not under go nucliophilic substitution with the anion formed by phthalimide. (ii) Basic character of amines: (a) Ammonia is lewis base due to presence of lone pair of electron. Amines are derivatives of ammonia. (b) Basic character depends on pka value. Larger the value of pka or smaller the value of Ka weaker is base and vice-versa. (c) Basic character of Primary secondary amine-- Condition No. 1: 3 o amine > 2 o amine > 1 o amine (In vapour phase). This is due to +I effect of alkyl group Condition No. 2: In aqueous solution = 1 o amine > 2 o amine > 3 o amine This is due to solvation effect. The greater the size of ion lesser will be solvation and the less stabilized is the ion. Condition No. 3: In general the basic characters of amines in aqueous medium i.e. inductive effect, solvation effect and steric hindrance the sequence are as follows- Condition No. 4: In case aryl amine (aniline). Aniline are less basic than ammonia because the NH 2 group is attached directly to the benzene ring, hence the unshared electron pair on nitrogen is less available for protonation due to the resonance. Condition No. 5: Effect of the substituent on basic strength of aniline. (a) The presence of ERG group increases the basic strength of aniline whereas (b) The presence of EWG group decreases the basic strength of aniline (iii) Nitration of aniline gives ortho para and meta derivatives-- In strong acidic medium aniline is protonated to form anilinium ion which is meta directing, hence besides ortho and para deravatives significant amount of metaderivative also formed. (iv) Aniline does not undergo friedel craft reaction-- Due to salt formation with aluminium chloride ( lewis acid) hence nitrogen of aniline acquire positive charge act as a strong deactivating group for further reaction.

112 PREVIOUS YEARS QUESTION 1. Assign reason for: [All India 2009C] (i) Amines are less acidic than alcohols of comparable molecular masses. (ii) Aliphatic amines are stronger bases than aromatic amines. 2. Write chemical equation for the following conversion: [Delhi 2012] (i) Nitrobenzene to benzoic acid (ii) Benzyl chloride to 2-phenylethanamine (iii) Aniline to benzyl alcohol CHEMICAL TEST FOR DISTINGUISH PAIRS OF COMPOUNDS 1. Carbyl amine reaction- To distinguish Pri., Sec., Aliphatic or Aromatic amines with others amines 2. HINSBERG S REAGENT TEST (BENZENE SULPHONYL CHLORIDE): Hinsberg s reagent test distinguished primary, secondary and tertiary amine. Primary amine Secondary amine Tertiary amine Reaction with Hinsberg s reagent Reaction with Hinsberg s reagent Reaction with Hinsberg s reagent Product soluble in alkali Product insoluble in alkali No reaction 2. Aniline can be distinguished from other primary amines by coupling reaction. IMPORTANT REACTION WITH REAGENTS (BASED ON PREPARATION & PROPERTIES)

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116 PREVIOUS YEARS QUESTION 1. Convert: (i) Nitrobenzene to phenol (ii) Aniline to chlorobenzene [Delhi 2011C] 2. How will you convert? (i) Aniline to benzonitrile? (ii) ethanamine to ethanoic acid? [Delhi 2011C] 3. How will you convert? (i) Nitrobenzene to aniline? (ii) Aniline to iodobenzene? [Delhi 2011] UNIT-14 BIOMOLECULES Marks =4 KEY POINTS EXPLANATIONS Monosaccharides Cannot be hydrolyzed further.eg- glucose, fructose, ribose Disaccharides Sucrose (α-d- glucose + β-d-fructose), Maltose(α-D- glucose + α-d- glucose) Lactose(β-D-galactose + β-d-glucose ) Polysaccharides Starch (two components Amylose and Amylopectin) polymer of α-d- glucose Amylose Water soluble, 15-20% of starch., unbranched chain, C1 C4 glycosidic linkage. Amylopectin Water insoluble, 80-85% of starch., branched chain polymer, C1 C4 & C1 C6 glycosidic linkage Cellulose Straight chain polysaccharide of β -D-glucose units/ joined by C1-C4glycosidic linkage (β-link), not digestible by human / constituent of cell wall of plant cells Glycogen Highly branched polymer of α-d- glucose.found in liver, muscles and brain. reducing sugars Aldehydic/ ketonic groups free so reduce Fehling s/ Tollens solution and. Egmaltose and lactose Non reducing sugars Aldehydic/ ketonic groups are bonded so cannot reduce Fehling s solution and Tollens reagent. Eg- Sucrose Anomers. The two cyclic hemiacetal forms of glucose differ only in the configuration of the hydroxyl group at C-1, called anomeric carbon Such isomers, i.e., α form and β - form, are called anomers. Invert sugar Sucrose is dextrorotatory but after hydrolysis gives dextrorotatory glucose and laevorotatory fructose. Since the laevorotation of fructose ( 92.4 ) is more than dextrorotation of glucose ( ), the mixture is laevorotatory. Thus, hydrolysis of sucrose brings about a change in the sign of rotation, from dextro (+) to laevo ( ) and the product is named as invert sugar Glycosidic linkage Linkage between two mono saccharide Importance of Carbohydrates -Major portion of our food. / used as storage molecules as starch in plants and glycogen in animals/. -Cell wall of bacteria and plants is made up of cellulose./wood and cloth are cellulose / -Provide raw materials for many important industries like textiles, paper, lacquers and breweries. essential amino acids -Which cannot be synthesised in the body and must be obtained through diet, eg- Valine, Leucine Nonessential amino acids -Which can be synthesised in the body, eg - Glycine, Alanine zwitter ion. In aqueous solution, amino acids exist as a dipolar ion known as zwitter ion.

117 peptide linkage structure of proteins: structure of proteins: Peptide linkage is an amide formed between COOH group and NH2 group of two successive amino acids in peptide chain. sequence of amino acids that is said to be the primary structure of protein Secondary structure of protein refers to the shape in which a long polypeptide chain can exist. They are found to exist in two types of structures viz. α -helix and β -pleated sheet structure. 3 structure of Further folding of the secondary structure. It gives rise to two major molecular proteins: shapes viz. fibrous and globular. Fibrous proteins Polypeptide chains run parallel, held together by hydrogen and disulphide bonds, fibre like structure. Water insoluble.eg- are keratin (in hair, wool, silk) and myosin (present in muscles). Globular proteins Chains of polypeptides coil around to give a spherical shape. Water soluble. Eg- Forces which stabilise 2 & 3 Denaturation of Proteins Fat soluble vitamin Water soluble vitamin Vitamins sources- Deficiency diseases Insulin and albumins Hydrogen bonds, disulphide linkages, van der Waals and electrostatic forces of attraction. When a protein is subjected to physical change like change in temperature or chemical change like change in ph, the hydrogen bonds are disturbed. Due to this, globules unfold and helix gets uncoiled and protein loses its biological activity. This is called denaturation of protein. (During denaturation 2 and 3 structures are destroyed but 1º structure remains intact.) eg- The coagulation of egg white on boiling, curdling of milk These are vitamins A, D, E and K. They are stored in liver and adipose (fat storing) tissues B & C. these vitamins must be supplied regularly in diet because they are readily excreted in urine Vit- A (Fish liver oil, carrots)- Night blindness / Vitamin B1 (Yeast, milk,)- Beriberi Vit-B2 (Milk, eggwhite)- Cheilosis / Vit- B6 (Yeast, milk,)- Convulsions / Vit- B12 (Meat, fish,)- anaemia Vit C(Citrus fruits)- Scurvy, / Vit D(Exposure to sunlight, fish and egg yolk)- Rickets, osteomalacia Vit E(wheat oil, sunflower oil)- fragility of RBCs / Vit K(leafy vegetables)- Increased blood clotting time DNA pentose sugar (D-2-deoxyribose) + phosphoric acid + nitrogenious bases ( A, G, C, T ) RNA pentose sugar (ribose) + phosphoric acid + nitrogenious bases Nucleoside / tides Phosphodiester link Functions of Nucleic Acids (A, G, C, U ) Nucleoside sugar + base Linkage between two nucleotides in polynucleotides Nucleotides sugar + base + phosphate DNA reserve genetic information, maintain the identity of different species e is capable of self duplication during cell division, synthesizes protein in the cell.

118 A FEW QUESTIONS for Practice 1 Mark Questions (Q.) Define the term bio molecules? (Ans) Bio molecules may be defined as the complex lifeless chemical substances which form the basis of life, i.e., they not only build up living systems (creatures) but are also responsible for their growth, maintenance and their ability to reproduce. (Q.) Define the term photosynthesis? Give its general chemical equation? (Ans) Photosynthesis may be defined as a chemical process through which plants make their own food by the reaction of carbon dioxide and water in the presence of sunlight with the help of plant chlorophyll. x CO 2 + y H 2 O ----->C x (H 2 O) y + x O 2 (Q.) Define Monosaccharides. (Ans) These are the simplest carbohydrates which cannot be hydrolysed to smaller molecules. Their general formula is (CH 2 O) n, where n = 3-7. (Q.) Define the term Oligosaccharides? (Ans) Those carbohydrates which give 2 to 10 molecules of monosaccharides in hydrolysis. (Q.) Define Disaccharides. (Ans) Carbohydrates which on hydrolysis give two molecules of the same or different monosaccharides are called disaccharides. e.g., C 12 H 22 O 11 + H 2 O C 6 H 12 O 6 + C 6 H 12 O 6 sucrose glucose Fructoes (Q.) What is difference between Reducing and non-reducing sugars or carbohydrates? (Ans) All those carbohydrates which contain aldehydic and ketonic group in the hemiacetal or hemiketal form and reduce Tollen s reagent or Fehling s solution are called reducing carbohydrates while others which do not reduce these reagents are called non-reducing reagents. (Q.) Explain the term mutarotation? (Ans) Mutarotation is the change in the specific rotation of an optically active compound with time, to an equilibrium mixture. (Q.) Define glycosidic linkage? (Ans) The two monosaccharide units are joined together through an ethereal or oxide linkage formed by the loss of a molecule of H 2 O. Such a linkage between two monosaccharide units through oxygen atoms is called glycosidic linkage. (Q.) Give a chemical equation for obtaining Maltose? (Ans) Maltose is obtained by partial hydrolysis of starch by the enzyme diastase present in malt i.e., sprouted barley seeds. 2(C 6 H 10 O 5 ) n + n H 2 O n C 6 H 12 O 6 (Q.) What are the main sources of vitamins? (Ans) The main sources of vitamins are milk, butter, cheese, fruits, green vegetables, meat, fish, eggs, etc. (Q.) Give two methods for the preparation of glucose? (Ans) The methods for the preparation of glucose are: (i) From sucrose (Cane Sugar). When sucrose is hydrolysed by boiling with dil. HCl or H 2 SO 4 in alcoholic solution, an equimolar mixture of glucose or fructose is obtained.

119 C 12 H 22 O 11 + H 2 O C 6 H 12 O 6 + C 6 H 12 O 6 (ii) From starch. Commercially glucose is obtained by hydrolysis of starch by boiling it with dil. H 2 SO 4 at 393 K under pressure. (C 6 H 10 O 5 )n + n H 2 O n C 6 H 12 O 6 (Q.) Define Carbohydrates? Give their basic classification depending upon their behaviour towards hydrolysis. (Ans) Carbohydrates are defined as optically active polyhydroxy aldehydes or polyhydroxy ketone substances which give these on hydrolysis. These are broadly classified as: (i) Monosaccharides. (ii) Oligosaccharides. (iii) Polysaccharides. (Q.) What is Milk sugar? Give its characteristics. (Ans) Lactose occurs in milk so, it is called milk sugar. Lactose on hydrolysis with dilute acids yields an equimolar mixture of D-glucose and D-galactose. It is a reducing sugar since it forms an osazone. It undergoes mutarotation and also reduces Tollen s or Fehling s solution. (Q.) Define the term vitamins? State its importance. (Ans) Vitamins may be defined as group of bio-molecules (other than fats, carbohydrates and proteins) which are required in small amounts for normal metabolic processes and for the life, growth and health of human beings and animal organisms.vitamins neither supply energy nor help in building tissues of the cells. They play an important role in keeping good health of human beings and animals. Their deficiency causes serious disturbances and diseases in the body. (Q.) What do you understand by denaturation of proteins? (Ans) When a protein in its native form, is subjected to physical change like in temperature or chemical change like change in ph, the hydrogen bonds are disturbed. Due to this, globules unfold and helix get uncoiled and protein loses its biological activity. This is called denaturation of protein. (Q.) Give the D and L configurations of Glyceraldehyde? (Ans) (Q.) Give the chemical structure of sucrose & explain why sucrose is non reducing sugar. (Ans)

120 The two monosaccharide are held together by a glycosidic linkage between C1 of -glucose and C2 of -fructose. Since the reducing groups of glucose and fructose are involved in glycosidic bond formation, sucrose is a non-reducing sugar. (Q.) Give a broad classification of vitamins? (Ans) Vitamins are complex organic molecules.they can be broadly classified as: (i) Water soluble vitamins: These include vitamin B-complex and vitamin C. (ii) Fat soluble vitamins: These are oily substances that are not readily soluble in water. However, they are soluble in fat. These include vitamins A,D,E and K. Nucleic acids are bipolar (i.e. polymers present in the living system). They are also called polynucleotides since the repeating structural unit of nucleic acids is a nucleotide. General structure of a Nucleotide can be given as: (Q.) Write a short note on cellulose and give its chemical structure. (Ans) Cellulose occurs exclusively in plants and it is the most abundant organic substance in plant kingdom. It is a predominant constituent of cell wall of plant cells. Cellulose is a straight chain polysaccharide composed only of -D-glucose units which are joined by glycosidic linkage between C1 of one glucose unit and C4 of t he next glucose unit. (Q.) Give a short note on Zwitter ion? (Ans) Amino acids are usually colourless, crystalline solids. These are water soluble, high melting solids and behave like salts rather than simple amines or carboxylic acids. This behaviour is due to the presence of both acidic (carboxylic group) and basic (amino group) groups in the same molecule. In aqueous solution, the carboxyl group can lose a proton and amino group can accept a proton, giving rise to a dipolar ion known as zwitter ion. (Q.) How are peptides formed? Show the formation of peptide bond with diagram. (Ans) Peptides are amides formed by the condensation of amino group of one -amino acid with the carboxyl group of another molecule of the same or different -amino acid with the elimination of

121 (D) Based on Molecular Forces (C) Based On Mode of Polymerization: (B) Based on Structure (A) Based on Source awater molecule. They are classified as di-, tri-, tetra-, etc. eg. UNIT 15 POLYMERS Marks= 3 CLASSIFICATION OF POLYMERS- (i) Natural (ii) Synthetic Found in plants and animals, e.g. Proteins, cellulose, natural rubber, silk, wool. Man-made e.g. Nylon, polyester, neoprene, Bakelite, Teflon, PVC, polystyrene (i) Linear Polymer (ii) Branched Polymers (iii) Cross Linked Polymers (i) Addition Polymers This consist of long and straight chain repeating units e.g. Polythene (HDPE), PVC, nylon, polyester. This contain linear chains having some branches e.g. amylopectin, glycogen etc. Strong covalent bonds are present between various linear polymer chains. E.g. Bakelite, urea-formaldehyde polymer, melamine, polymer etc. These are formed by the repeated addition of monomer molecules possessing multiple bonds, e.g., polythene, polypropene, polystyrene, PMMA (polymethylmethacrylate) (ii) Condensation Polymers These are formed by the repeated condensation reaction of different bifunctional or trifunctional monomers, with the elimination of small molecules like water, HCL, NH 3, alcohol etc. eg. Bakelite, nylon, polyester, urea-formaldehyde resin. (i) Elastomers (ii) Fibers Forces of interaction between polymer chains are weakest, e.g. natural rubber, neoprene, and vulcanized rubber. Strong hydrogen bonds are present between the polymer chains. They have high tensile strength e.g., nylon, polyester, silk, wool, orlon, rayon etc.

122 (E) Based On Growth Of Polymerization (iii) Thermoplastics (i) Addition Polymers or Chain Growth Polymers (ii) Condensation Polymers or Step Growth Polymers They are linear / slightly branched chains molecules capable of repeated softening on heating and hardening on cooling e.g., polythene, PVC, polystrene, polypropene. They follow mostly free radical mechanism. Because they are formed in gradual steps.

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127 PREVIOUS YEARS QUESTIONS 1. Define the terms: (i) Homopolymerization [2012] (ii) Copolymerization [2010] (iii) Addition polymer [2007, 09,11C] (iv) Condensation polymer [2007, 09,11C] 2. Write the monomers of: (i) Neoprene [2011] (ii) PMMA [2006C, 2010] (iii) Buna-N [2006] (iv) Nylon-6, polypropene [2009C, 2006, 2012] 3. Write the difference between: (i) Elastomer & fibre (ii) Homopolymer & copolymer [2008C] [2008C,2010] (Q.) Give the structure of natural rubber? (Ans) A Few question on Polymers (Q.) Define the term polymerization? (Ans) The repeating structural units formed from some simple and reactive molecules (monomers) are linked together by covalent bonds. This process of formation of polymers from their respective monomers is called polymerization. (Q.) Define Elastomers? (Ans) Elastomers are rubber-like solids with elastic properties. In elastomers, the polymer chain is heldtogether by weak intermolecular forces which allow the polymer to be stretched. e.g. buna-s, buna-n neoprene. (Q.) What are fibers? (Ans) Fibers are the thread forming solids which possess high tensile strength and high modulus. These polymers possess strong intermolecular forces.thus leading to close packing of chains and imparting crystalline nature. e.g., polyamides( nylon 6,6) and polyesters( terylene), etc. (Q.) Define Thermoplastics polymers? (Ans) Thermoplastic polymers are the linear or slightly branched chain molecules capable of repeatedly softening on heating and hardening on cooling. Some examples of thermoplastics are polythene, polystyrene, polyvinyl, etc (Q.) Define Thermosetting plastics? (Ans) Thermosetting polymers are cross linked or heavily branched molecules which on heating undergo extensive cross linking in moulds and again become infusible. Some examples of thermosetting plastics are bakelite, urea-formaldehyde resins, etc. (Q.) Give the method of preparation of polyacrylonitrile? (2 Marks) (Ans) The addition polymerization of acrylonitrile in presence of a peroxide catalyst leads to the formation of polyacrylonitrile. It is used as a substitute for wool in making fibers as orlon or acrilan. (Q.) Define copolymerization? Give chemical reaction showing formation of copolymer.(2 Marks) (Ans) Copolymerisation is a polymerization reaction in which a mixture of more than one monomeric species is allowed to polymerise and form a copolymer. For example, a mixture of a 1,3 butadiene and styrene form a copolymer.

128 (Q.) Describe the method for the preparation of neoprene? (2 Marks) (Ans) Neoprene or poly chloroprene is formed by the free radical polymerization of chloroprene. (Q.) Differentiate between Addition and Condensation polymers? (2 Marks) (Ans) S.No Addition Polymers Condensation polymers 1. They are formed by the repeated addition of molecules possessing double or triple They are formed by repeated condensation reaction between two different bi-functional and tri-functional monomeric units. bonds. 2. E.g. polythene E.g. nylon 66 (Q.) Differentiate between Homopolymers and Copolymers with example. (2 Marks) (Ans) Homopolymers: The addition polymers formed by the repeated addition of monomer molecules possessing double or triple bonds, are known as homopolymers. E.g., nch 2 = CH >---(CH 2 CH 2 )--- Ethene polythene Copolymers: The polymers formed by addition polymerization of two different monomers are termed as copolymers. E.g., nch 2 = CH CH = CH 2 + nc 6 H 5 CH=CH >---[CH 2 CH = CH CH 2 CH 2 CH(C 6 H 5 )] n --- (Q.) Give the differences between homopolymers and copolymers? (2 Marks) (Ans) S.No. Homopolymers Copolymers 1 The addition polymers formed by the polymerisation of a single monomeric unit are called Homopolymers. The polymers formed by the addition polymerisation of two different monomers are termed as copolymers. 2 E.g. Polythene. E.g. Buna-S (Q.) What is synthetic rubber? (2 Marks) (Ans) Synthetic rubber is any vulcanisable rubber like polymer, which is capable of getting stretched twice its length. However, it returns to its original shape and size as soon as the external stretching force is released. (Q.) Give the method of preparation of Teflon and its uses. (3 Marks) (Ans) Polytetrafluoroethene (Teflon) is manufactured by heating tetrafluroethene with a free radical or persulphate catalyst at high temperature. It is chemically inert in nature. It is used for making oil seals and gaskets and also used for non-stick surface coated utensils. Tetraflouroethane Teflon (Q.) What are the different types of structural polymers? Give examples? (3 Marks) (Ans) There are three different types of structural polymers: (1) Linear polymers: These polymers consist of long and straight chains. e.g. polythene, polyvinyl chloride, etc. (2) Branched chain polymers: These polymers contain linear chains having some branches. e.g. low density polythene. (3) Cross linked polymers (network polymers): These are usually formed from bi-functional and tri-functional monomers and contain strong covalent bonds between various linear polymer chains. e.g. bakelite, melamine etc. (Q.) Give the method for preparing Bakelite? (3 Marks) (Ans) Bakelite is manufactured from Phenol-formaldehyde polymers. It is obtained by the condensation reaction of phenol with formaldehyde in thepresence of an acid or base catalyst.

129 The initial product formed is a linear chain called Novolac used in paints. Novolac on heating with formaldehyde undergoes cross linking to form infusible solid mass called bakelite. (Q.) Give the method of preparation and uses of nylon 6? (Ans) Nylon 6 is obtained by heating caprolactum with water at high temperature. (3 Marks) Nylon 6 is used for the manufacture of tyre cords, and fabrics and ropes. (Q.) How are condensation polymers formed? Explain giving one example. (3 Marks) (Ans) The condensation polymers are formed by repeated condensation reaction between two different bi-functional or tri-functional monomeric units. During this process, a small molecule of water, alcohol, hydrogen chloride, etc is eliminated. E.g. terylene, nylon 6,6, nylon 6, etc. Preparation of nylon 6,6: nh 2 N(CH 2 ) 6 NH 2 + nhooc(ch 2 ) 4 COOH ----->---(NH(CH 2 ) 6 NHCO(CH 2 ) 4 CO) n n H 2 O (Q.) What are natural and synthetic polymers? Give 2 examples of each. (3 Marks) (Ans) Natural polymers: Polymers which are found in plants and animals are called natural polymers. e.g. proteins, cellulose starch etc. Synthetic polymers: Man made polymers are called synthetic polymers. They consists of a number of smaller molecules to form large molecules. e.g. nylon 6,6 and Buna-S. (Q.) Define the term polyesters? How is it manufactured? (3 Marks) (Ans) Polyesters are the condensation products of dicarboxylic acids and diols. e.g. Dacron or Terylene. It is manufactured by heating a mixture of ethylene glycol and terephthalic acid at 420 to 460 K in the presence of zinc acetate-antimony trioxide catalyst. (Q.) Explainvulcanisation of rubber? (3 Marks) (Ans) Natural rubber becomes soft at high temperatures and brittle at low temperatures. It has a high water absorption capacity. It is soluble in non-polar solvents and is non-resistant to attack by oxidizing agents. To improve upon these physical properties, the process of vulcanization is carried out. It consists of heating a mixture of raw rubber with sulphur with a temperature range between 373 K to 415 K. On vulcanisation, sulphur forms cross links at reactive sites of double bonds and thus the rubber gets stiffened. (Q.) Give three examples of biodegradable polymers? (3 Marks) (Ans) (1) phbv (2) Nylon-2 (3) nylon-6

130 UNIT 16 CHEMISTRY IN EVERY DAY LIFE Marks=3 1 DRUGS Drugs are chemical of low molecular masses, which interact with macro molecular targets and produce a biological response. 2 CHEMOTHERAPY The use of chemicals to cure illness/ailments is called chemotherapy. Drugs for a particular type of 1. On the basis of pharmacological effects problem. e.g. analgesics for pain relieving. Action on drug on a 3 2. On the basis of Drug Action particular biological CLASSIFICATIONS process OF DRUGS Drugs having 3. On the Basis Chemical Action similar structure. E.g. sulpha drugs Drugs interacting 4. On the Basis of Molecular targets with biomolecules as lipids and protiens (a) Enzymes have active sites which (i) Catalytic Action of Enzymes hold the substrate molecule. It can be attracted by reacting molecules. (b) Substrate is bonded to active sites through hydrogen bonds, ionic bonds, vander waal or dipole - dipole interactions 4 ENZYMES AS DRUG TARGETS (ii) Drug Enzyme Interactions (a) Drug complete with natural substrate for their attachments on the active sites of enzymes. They are called competitive inhibitors. (b) Some drugs binds to a different site of the enzyme called allosteric sites which changes the shape of active sites.

131 RECEPTORS AS DRUG TARGETS Receptors are proteins that are crucial to body s communication process. In the body message between two neurons and that between neurons to muscles is communicated through certain chemicals. These chemicals known as chemical messengers are received at the binding sites of receptor proteins. To accommodate a messenger s shapes of receptors sites are changed. There are a large number of receptors in the body that interact with different chemical messengers hence one medicine is not suitable for different type of diseases. 5 ANTAGONISTS 6 AGONISTS 7 ANTACIDS 8 ANTI HISTAMINES 9 TRANQULIZERS 10 ANALGESICS 11 ANTIMICROBIALS The drugs that bind to the receptor site and inhibit its natural functions. Drugs mimic the natural messenger by switching on the receptor. These are compounds which neutralize excess acid of stomach The drugs which interfere with the natural action of histamines and prevent the allergic reaction. The class of chemical compounds used for treatment of stress, mild or even severe mental diseases. They reduce pain without causing impairment of consciousness, mental confusion or some other disturbance of the nervous system. e.g. Aluminium hydroxide, Magnesium hydroxide. e.g. ranitidine, tegament, avil. e.g. luminal, seconal, equanil, iproniagid. e.g. aspirin: Use antipyretic as well as analgesic. It prevents platelets coagulation, because of its anti-blood clotting action aspirin finds use in prevention of heart attacks. saridon, phenacetin. They tend to prevent / destroy or inhibit the pathogenic actions of microbes as bacteria, virus, fungi etc. They are classified as (i) Antibiotics: Those are chemical substances which are produced by microorganism and used to kill the pathogenic micro organism. e.g. penicillin, offloxacin (a) Narrow spectrum Antibiotics (b) Broad spectrum Antibiotics These are affective mainly against gram positive or gram negative bacteria. e.g. penicillin, streptomycin They kill or inhibit a wide range of micro-organism. E.g. chloramphenicol, tetracycline.

132 12 ANTIFERTILITY DRUGS 13 ARTIFICIAL SWEETNING AGENTS 14 FOOD PRESERVATIVES 15 CLEANSING AGENTS 16 SYNTHETIC DETERGENTS 17 BIO DEGREDABLE DETERGENTS 18 NON BIO DEGREDABLE DETERGENTS: (ii) Antiseptics or Disinfectant: These are which either kill / inhibit the growth of micro - organism antiseptic applied to the living tissues such as wounds, cuts, ulcers etc. These are the chemical substances used to control pregnancy. They are also called contraceptives or birth control pills. These are the chemical compounds which give sweetning effects to the foods without adding calorie. They are good for the diabatic patients. They prevent spoilage of food to microbial growth. 1. SOAPS: They are sodium or potassium salts of long chain fatty acids. They are obtained by the saponification reaction, when fatty acids are heated with aqueous sodium hydroxide. They do not work well in hard water. They are cleaning agents having properties of soaps, but actually contain no soap. They can used in both soft and hard water. They are: The detergents which are linear or less branched and can be attacked by microorganism are bio degradable. The detergents which are highly branched and can not be decomposed by microorganism are called non-bio degradable detergents. It creates water pollution. e.g. furacine, chloroxylenol & terpinol (Dettol). Disinfectant is applied to inanimate objects such as floors, drainage system. e.g. 0.2% solution of phenol is an antiseptic while 1% solution is disinfectant. e.g. Mifepristone, Norethindrone. e.g. aspartame, saccharin, alitame, sucrolose. e.g. salt, sugar and sodium benzoate, BHT, BHA They are biodegradable. (i) Anionic Detergents: They are sodium salts of sulphonated long chain alcohols or hydro carbons. E.g. Sodium lauryl sulphonate. Its anionic part is responsible for cleansing action (ii) Cationic Detergents: They are quarternary ammonium salts of amines with acetates, chlorides or bromides. They are expensive and used to limited extent. E.g. cetyltrimethylammoniumbromide. Its cationic part is responsible for cleansing action (iii) Non-ionic Detergents: They do not contain any ions. Some liquid dish washing detergents which are of non-ionic type E.g. Sodium-4 - (1-dodecyl) benzene / sulphonate. N.B.: SOLVE ALL THE EXERCISE AS WELL AS INTEXT QUESTIONS.

133 IMPORTANT TOOLS / NOTES FOR ORGANIC CHEMISTRY (CLASS XII) KEY FOR CONVERSIONS IN ORGANIC CHEMISTRY S.No Reagent Group Out Group In Remark 1 KMnO 4 / H+ -CH 2 OH -COOH Strong Oxidation (2 0 alcohol gives ketone) 2 LiAlH 4 -COOH -CH 2 OH Strong Reduction (ketone gives 2 0 alcohol) 3 Cu / 573 K or -CH 2 OH -CHO Dehydrogenation CrO 3 4 PCl 5 or SOCl 2 -OH -Cl 5 Cl 2 / Δ or Cl 2 / -H -Cl Free radical substitution UV 6 Aq NaOH / KOH -X -OH Nucleophilic substitution 7 KCN -X -CN Steping Up of a compound 8 AgCN -X -NC 9 Alcoholic KOH -HX = Dehydrohalogenation (Stzf) 10 Mg / dry ether Mg R-X R-MgX 11 HBr >=< H, Br Markovnikov 12 H 2 / Pd-BaSO 4 -COCl -CHO Rosenmund Reduction 13 Zn-Hg / HCl >C=O -CH 2 - Clemmension Reduction 14 NH 3 / Δ -COOH -CONH 2 -COOH + NH 3 -COONH 4 15 Br 2 / NaOH or NaOBr -CONH 2 -NH 2 Step Down ( Hoffmann) 16 HNO 2 or -NH 2 -OH HONO NaNO 2 /HCl 17 CHCl 3 / alc KOH -NH 2 -NC Carbyl amine 18 P 2 O 5 -CONH 2 -CN Dehydration 19 H 3 O + -CN -COOH Hydrolysis 20 OH - -CN -CONH 2 21 LiAlH 4 -CN -CH 2 NH 2 Reduction 22 Red P / Cl 2 α-h of acid -Cl HVZ Reaction KEY FOR CONVERSIONS IN BENZENE RING : 23 Fe / X 2 /dark -H -X Halogination 24 CH 3 Cl / AlCl 3 (anhyd) -H -CH 3 Friedel Craft alkylation 25 CH 3 COCl / -H -COCH 3 Friedel Craft acylation AlCl 3 (anhyd) 26 Conc.HNO 3 /con.h 2 SO 4 -H -NO 2 Nitration 27 Conc H 2 SO 4 -H -SO 3 H Sulphonation 28 KMnO 4 / H+ -R -COOH Oxidation 29 CrO 2 Cl 2 / H+ -CH 3 -CHO Mild oxidation(etard Reaction) 30 Sn / HCl or Fe/HCl -NO 2 -NH 2 Reduction 31 NaOH / 623K / 300 -Cl -OH atm 32 Zn dust / Δ -OH -H 33 NaNO 2 / dil HCl / NH 2 -N + 2 Cl - Diazo reaction 278 K 34 CuCl / HCl or -N + 2 Cl - -Cl Sanmeyer or Gattermann Cu/HCl 35 CuBr / HBr or -N + 2 Cl - -Br Sanmeyer or Gattermann

134 Cu/HBr 36 CuCN / KCN -N + 2 Cl - -CN Sanmeyer 37 KI -N + 2 Cl - -I 38 HBF 4 / Δ -N + 2 Cl - -F 39 H 3 PO 2 or -N + 2 Cl - -H CH 3 CH 2 OH 40 H 2 O / 283 K -N + 2 Cl - -OH 41 HBF 4 / NaNO 2, Cu / Δ -N + 2 Cl - -NO 2 42 C 6 H 5 -OH -N + 2 Cl - -N=N-C 6 H 5 -OH Coupling ( p-hydroxy) 43 C 6 H 5 -NH 2 -N + 2 Cl - -N=N-C 6 H 5 -NH 2 Coupling ( p-amino) Reactions of Grignard Reagent Grignard reagent + Any one below + H 2 O Product H 2 O or ROH or RNH 2 R-H H-CHO R-CH 2 -OH (1 0 alc) R-CHO R-CH(OH)-R (2 0 alc) R-MgX R-CO-R R 2 C(OH)-R (3 0 alc) CO 2 R-COOH R-CN R-CO-R HCOOR Aldehyde RCOOR Ketone NB: i) During reaction generally changes take place in the functional group only so see the functional group very carefully. ii) Remember structural formula of all the common organic compounds ( with their IUPAC and common names) iii) Wurtz Reaction and Aldol Condensation are not included in the table although they are Very important for conversions so study them. iv) By taking examples practice all the above cases (from 1 to 43 and Grignard) v) Practice only from NCERT book. vi) Start practicing NOW! How to use the table? See below. Example: See no 7 in the table Directional Properties of groups in benzene ring for electrophilic substitution Ortho-para directing group: -R, -OH, -NH 2, -X, -OR, -NHR, -NR 2, -NHCOCH 3, -CH 2 Cl, -SH, - Ph Meta-directing group: -NO 2, -CHO, -COOH, COOR, -CN, -SO 3 H, -COCH 3, -CCl 3, - NH 3 +

135 NAME REACTIONS (ORGANIC CHEMISTRY) 1. Finkelstein - CH 3 Br + NaI CH 3 -I + NaBr 2. Swarts - CH 3 Br + AgF CH 3 F + AgBr CH 3 3. Friedel- Crafts- Alkylation + C H 3 Cl Anhydrous AlCl 3 COCH 3 4. Friedel- Crafts- Acylation CH 3 COCl Anhydrous AlCl 3 5. Wurtz - H 3 C Cl + Cl CH 3 2Na H 3 C CH 3 + Na Cl Cl Cl 6. Fittig + 2Na Dry ether + Na Cl Cl 7. Wurtz-Fittig + Cl CH 3 2Na Dry ether CH 3 + Na Cl OH ONa OH 8. Kolbe s reaction Na OH i) CO 2 ii) H + COOH

136 9. Reimer- Tiemann OH CH 3 Cl + Na OH ONa CHO H + OH CHO 10. Williamson s CH 3 -Br + CH 3 -ONa CH 3 -O- CH 3 + NaBr 11. Stephen H 3 C CN+ SnCl 2 + HCl H 3 C CH NH H 3 O + C H 3 CHO CH 3 CHO 12. Etard CrO 2 Cl 2 H 3 O + CHO 13. Gatterman Koch CO / HCl Anhydrous AlCl Rosenmund reduction C H 3 C Cl O H 2 Pd / BaSO 4 O C H 3 C H 15.Clemmensen reduction C H 3 O C CH 3 Zn - Hg Conc. HCl H 3 C CH 2 CH Wolff-Kishner reduction C H 3 O C CH 3 i) NH 2 - NH 2 ii) KOH / Ethylene glycol / H 3 C CH 2 CH Tollens test R-CHO + 2 [Ag(NH 3 ) 2 ] OH - R-COO - + 2Ag + 2H 2 O + 4 NH 3 Silver mirror 18. Fehling s test 19.Iodoform Reaction C H 3 R-CHO + 2 Cu OH - R-COO - + Cu 2 O + 3H 2 O O I 2 / NaOH C CHI 3 + CH 3 COONa CH 3 OR, NaOI Yellow ppt.

137 20. Aldol Condensation 2 H 3 C CHO dil NaOH OH H 3 C CH CH 2 CHO H 3 C CH CH CHO 21. Cannizzaro HCHO + HCHO Conc. NaOH HCOONa + C H 3 OH 22. Hell-Volhard- Zelinsky (HVZ) C H 3 COOH i) Cl 2 / Red Phosphorus ii) H 2 O C H 2 Cl COOH 23.Hoffmann bromamide degradation O H 3 C C NH 2 Br 2 NaOH H 3 C NH Carbylamine R-NH 2 + CHCl KOH R-NC + 3 KCl + 3 H 2 O N 2 + Cl - Cl 25. Sandmeyer. CuCl / HCl + N 2 N 2 + Cl - Cl 26. Gatterman Cu / HCl + N 2 27.Coupling Reaction N 2 + Cl - + H OH OH - N N OH ELECTRON DISPLACEMENT EFFECTS + I : O -, COO -, (CH 3 ) 3 C, (CH 3 ) 2 CH, CH 3 CH 2, CH 3 (electron donating) - I : NR + 3, SR + 2, NH + 3, NO 2, SO 2 R, CN, COOH, F, Cl, Br, I, OR, OH, NH 2 (e - withdrawing) + R ( + M ) : OH, NH 2, OR, NHR, X (electron donating) - R ( - M ) : NO 2, CN, CHO, COOH, COCH 3 (electron withdrawing)

138 DIRECTIVE INFLUENCE OF SUBSTITUENTS IN BENZENE RING (for electrophilic substitution reactions) EFFECT OF THE GROUP DIRECTING ACTIVATING / DEACTIVATING + I Ortho / Para Activating + I, + R Ortho / Para Activating - I < + R Ortho / Para Activating - I > + R Ortho / Para Deactivating - I Meta Deactivating - I, - R Meta Deactivating Example: - I > + R : - X, - CH=CH 2, -CH=CH-COOH, -CH 2 -Cl These groups are deactivating but exceptionally o / p directing due to +E effect by the attacking reagents electron density increases at -ortho and -para position. If two groups are present initially 1.When both the groups present in benzene ring are o/p directing than the order of influence : O - > NH 2 > NR 2 > OH > OCH 3 > NHCOCH 3 > CH 3 > X 2.When both the groups present in benzene ring are meta directing than the order of influence : (CH 3 ) 3 N+ > NO 2 > CN > SO 3 H > CHO > COCH 3 > COOH 3.When one group is o/p and another is m directing than o/p directing group takes priority

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140 Distinction By Single Chemical Test 1. All aldehydes ( R-CHO) give Tollens Test and produce silver mirror. RCHO + 2 [Ag(NH 3 ) 2 ] OH - RCOO Ag + 2H 2 O + 4 NH 3 Tollens Reagent silver ppt Note: HCOOH(methanoic acid ) also gives this test, ketones(rcor) do not give this test 2. All aldehydes (R-CHO) and ketones(rcor) give 2,4-DNP test RCOR + 2,4-DNP Orange ppt R-CHO + 2,4-DNP Orange ppt 3. Aldehydes and ketones having CH 3 CO- (keto methyl) group give Iodoform Test. Alcohols having CH 3 CH- group also give Iodoform Test. OH CH 3 CHO + 3I NaOH CHI 3 + HCOONa + 3 NaI + 3H 2 O Yellow ppt The following compounds give Iodoform Test: ethanol (C 2 H 5 OH), propan-2-ol (CH 3 CH(OH)CH 3 ),ethanal(ch 3 CHO), propanone(ch 3 COCH 3 ),butanone ( CH 3 COCH 2 CH 3 ), pentan-2-one (CH 3 COCH 2 CH 2 CH 3 ), acetophenone ( PhCOCH 3 ) 4. All carboxylic acids ( R-COOH) give Bicarbonate Test RCOOH + NaHCO 3 RCOONa + CO 2 + H 2 O Brisk effervescence 5. Phenol gives FeCl 3 Test C 6 H 5 OH + FeCl 3 (C 6 H 5 O) 3 Fe + 3 HCl (neutral) (violet color) 6. All primary amines (R/Ar -NH 2 ) give Carbyl Amine Test R-NH 2 + CHCl 3 + KOH(alc) R-NC + KCl + H 2 O offensive smell 7. Aniline gives Azo Dye Test ( Only for aromatic amines) C 6 H 5 NH 2 + NaNO 2 + HCl C 6 H 5 N 2 + Cl - ; then add β-naphthol orange dye 8. All alcohols (ROH) give Sodium (Na) metal test R-OH + Na R-ONa + H 2 (bubbles) 9. For esters (RCOOR) : Hydrolyses first. Then see the products ( acid & alcohol) and give a test to identify them 10. All alkenes (C=C) and alkynes (C C) decolorizes Br 2 water from red to colorless 11. Lucas Test to distinguish primary, secondary and tertiary alcohols Lucas reagent: ZnCl 2 /HCl 3 0 -alcohol + Lucas reagent immediate turbidity 2 0 -alcohol + Lucas reagent turbidity after sometime 1 0 -alcohol + Lucas reagent no turbidity

141 SAMPLE QUESTION PAPERS With BLUE PRINT & MARKING SCHEME BLUE PRINT S.NO. NAME OF UNIT VSA SA(I) SA(II) LONG TOTAL 1 SOLID STATE 2X2 4 2 SOLUTION ELECTROCHEMISTRY CHEMICAL KINETICS SURFACE CHEMISTRY GENERAL PRINCIPLE OF EXTRACTION P BLOCK THE d-& f-block ELEMENTS 9 CO-ORDINATION COMPOUNDS 10 HALOALKANES& HALO ARENES 11 ALCOHALS, PHENOLS & ETHERS. 12 ALDEHYDE, KETONES & CARBOXYLIC ACIDS X AMINES BIOMOLECULES POLYMERS CHEMISTRY IN EVERYDAY LIFE TOTAL

142 MODLE QUESTION PAPER CLASS XII SUBJECT CHEMISTRY TIME: 3Hrs M.M. 70 General Instructions: (i) All questions are compulsory. (ii) Questions 1 to 8 are short answer questions. Each carries one mark. (iii) Questions 9 to 18 are short answer questions. Each carries 2 marks. (iv) Questions 19 to 27 are also short answer questions. Each carries 3 marks (v) Questions 28 to 30 are long answer questions each carry 5 marks. (vi) There is no overall choice, however, an internal choice has been provided in one question of two marks, one question of three marks and all three questions of five marks each. You have to attempt only one of the given choices in such questions. (i) Use log table wherever necessary. Use of calculators is not permitted. 1. Adsorption of a gas on the surface of solid is generally accompanied by decrease in entropy still it is spontaneous process. Explain. 2. Square planer complexes with co-ordination number of four exhibit geometric al isomerism whereas tetrahedral complexes do not. Why? 3. How will you prepare methanol from formaldehyde without using a reducing agent? 4. Give the IUPAC name of CH 3 CH 3 CH 3 CH 2 C C CH 2 CH 2 OH Br Br 5. Write steps involved in the conversion of 1-propaneamine to acetone. 6. Why can not vitamin C be stored in our body. 7. Define Elastomers? 8. Define non-biodegradable detergent with one example. 9. Analysis shows that a metal oxide has empirical formula of M 0.96 O. calculate the percentage of M 2+ and M 3+ ion in the crystal. 10. Account for the following: a) some of glass object recovered from ancient monuments look like milky instead of being transparent b)zinc oxide is white but turn yellow on heating. Explain 11. What do you understand by board spectrum antibiotics? 12 Differentiate between Homopolymers and Copolymers with example. OR Describe the method for the preparation of neoprene?

143 13 The osmotic pressure of human blood is 7.65 atm at 37 C. For injecting Glucose solution it is necessary the glucose solution has same osmotic pressure as of human blood. Find the molarity of glucose solution having same osmotic pressure as of human blood. Describe the method for the preparation of neoprene? 14. Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration. 15 What is lanthanoid contraction? Why lanthanides are known as f-block elements? 16) Copper I compounds are white and diamagnetic but copper II compounds are coloured and paramagnetic. Why? 17) Complete the following reactions: (a) C 6 H 5 ONa + C 2 H 5 Cl > (b) CH 3 CH 2 CH 2 OH + SOCl > 18.) Why is aniline a weaker base than methylamine? 19 a) Define glycosidic linkage b) What are the main sources of vitamins? c ) Define the term Oligosaccharides? OR a) Explain the term mutarotation? b)what is difference between Reducing and non-reducing sugars or carbohydrates? 20 How are the following preparations carried out? (a) Salicylic acid from phenol.(b) n-propyl alcohol from ethene. 21 a) Explain the coupling reaction with example. b) Describe the Gabriel-phthalimide reaction. 22 a) Explain ideal and non-ideal solutions with respect to intermolecular interactions in a binary solution of A and B. b) Aquatic animals are more comfortable in cold water than in warm water. Explain? (a) What is the effect of temperature on adsorption? b) Why are zeolites called shape selective catalysts? c) Define tyndall effect 25 (a) What is leaching? (b)how can the ores ZnS and PbS be separated from a mixture using froth floatation process? 26 (a) Why does NH 3 has higher boiling point than PH 3? (b) Why is the ionization energy of group 15 elements higher than that of group 14elements. (C) Acidic strength increases in this order : HF<HCl<HBr<HI. Give reason.

144 27 (a) How does +2 oxidation state becomes more and more stable in the first half of the first row transition elements with increasing atomic number? (b)vanadium pentaoxide acts as a good catalyst. Why 28 a) The initial concentration of N 2 O 5 in the following first order reaction N 2 O 5 (g) 2 NO 2 (g) + 1/2O 2 (g) was mol L 1 at 318 K. The concentration of N2O5 after 60 minutes was mol L 1. Calculate the rate constant of the reaction at 318 K. b) A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is (i) Doubled (ii) Reduced to half OR a)the rate constants of a reaction at 500K and 700K are 0.02s 1 and 0.07s 1 respectively. Calculate the values of Ea and A. b)for a reaction, A + B Product; the rate law is given by, r = k [ A] 1/2 [B] 2. What is the order of the reaction? c) Identify the reaction order from the following rate constants. (i) k = L mol 1 s a) a) OR b) Describe the following(write reactions only) (i) Aldol condensation (ii) Cannizzaro reaction (iii)friedel-crafts reaction. 30 a) Why is ammonia a mild reducing agent while BiH 3 the strongest reducing agent among all the hydrides? b) Explain the brown ring test for nitrates. Give the equations involved for the reactions occurring during the brown ring test. OR a) Give reasons: (i) Size of oxygen atom exceptionally small (ii) SF 6 exceptionally stable (iii) PCl 3 fumes in atmosphere b) How does Cl 2 react with (i) cold and dilute NaOH (ii) hot and concentrated NaOH

145 Answer Scheme (Sample Question Paper) 1. Because adsorption is an exothermic process 1mark 2. Because relative position of ligands attached to central atom are same with respect to one another 1mark 3. Give cannizarro reaction. 1mark 4. 3,4-dibromo-3,4-dimethylhexan-1-ol. 1mark 5. CH 3 CH 2 CH 2 NH 2 mark.rule H+/H 2 o HONO CH 3 CH 2 CH 2 OH Cu powder 573k CH 3 CHOHCH 3 CH 3 COCH 3 Conc. sulphuric acid,443k CH 3 CH=CH 2 1mark 6. Because it is soluble in water so excreted in urine 1mark 7. Elastomers are rubber-like solids with elastic properties. In elastomers, the polymer chain is held together by weak intermolecular forces which allow the polymer to be stretched. e.g. buna-s, buna-n neoprene 1mark 8. Detergent containing branched hydrocarbon chain are not easily degraded by microorganism and hence are called non-biodegradable detergent e.g. sodium-4- (1,3,5,7-tetramethyloctyl) benzenesulphonate. 1mark 9. Ratio of M 2+ : M 3+ =96:100 Total charge on M 2+ and M 3+ (+2)X +3(96-X) =200 X=88% of M 2+ and M 3+ =91.7 and 8.3 respectively 1+1mark 10 a) Due to annealing over a number of years glass acquires some crystalline character b) because it loose oxygen ZnO Zn /2O 2 + 2e - Metal excess defect electron in interstitial voids the colour is yellow as the remaining colour of white light are absorbed by these electron 1+1mark 11 These are effective against several types of bacteria. For example tetracycline, chloramphenicol, Ofloxacin which are used as antibiotics. 1+1mark 12) Homopolymers: The addition polymers formed by the repeated addition of monomer molecules possessing double or triple bonds, are known as homopolymers. E.g., nch 2 = CH > ---(CH 2 CH 2 )---

146 Ethene polythene Copolymers: The polymers formed by addition polymerization of two different monomers are termed as copolymers. E.g., nch 2 = CH CH = CH 2 + nc 6 H 5 CH=CH > ---[CH 2 CH = CH CH 2 CH 2 CH(C 6 H 5 )] n marks Neoprene or poly chloroprene is formed by the free radical polymerization of chloroprene. OR 1+1mark 13. (Ans) = Or 7.65 = = = Molarity = mark 14. Conductivity : conductance of a one cm 3 of a solution is known as conductivity. Conductivity decrease with dilution i.e. decrease with decrease of concentration Molar conductance : Molar conductance is the conductance of a solution having one mole of electrolyte when whole of the solution is present between the two electrode. Molar conductance increase with dilution i.e. increase with decrease of concentration. 1+1mark 15. ) In the lanthanide series as the atomic number increases there is a progressive decrease in the size of the atoms and trivalent ions which is known as lanthanide contraction. The last electron enters in the f-orbital, so lanthanides are knowns as f- block elements 1+1mark 16. ) In copper I ion all orbitals are completely filled so its compounds are white and diamagnetic. The electronic configuration of copper II ion is 1s 2 2s 2 2p 6 3s 2 3p 6 3d 9. it has one unpaired electron so it is paramagnetic and forms blue coloured compounds. 1+1mark 17. ) (a) C 6 H 5 ONa + C 2 H 5 Cl C 6 H 5 -O-C 2 H 5 + NaCl Sodium Ethyl chloride Phenetolephenoxide (b) CH 3 CH 2 CH 2 OH + SOCl 2 CH 3 CH 2 CH 2 Cl + HCl + SO 2 Propanol-1 Thionyl choride 1-chloropropane 1+1marks 18. Aniline and methylamine both have nitrogen with lone pair of electron. In aniline the phenyl group is electron attracting. It tend to decrease the electron density on nitrogen atom and hence decreases electron releasing tendency of nitrogen. While in methylamine CH 3 - group is electron repelling in nature. It tends to increase the electron

147 density on the nitrogen atom and helps in electron releasing tendency of nitrogen. So, the aniline is a weaker base than methylamine. 1+1mark 19. a)the two monosaccharide units are joined together through an ethereal or oxide linkage formed by the loss of a molecule of H 2 O. Such a linkage between two monosaccharide units through oxygen atoms is called glycosidic linkage b)the main sources of vitamins are milk, butter, cheese, fruits, green vegetables, meat, fish, eggs, etc. c) Those carbohydrates which give 2-10 molecules of monosaccharides in hydrolysis mark OR a)mutarotation is the change in the specific rotation of an optically active compound with time, to an equilibrium mixture b)all those carbohydrates which contain aldehydic and ketonic group in the solution are called reducing carbohydrates while others which do not reduce these reagents are called non-reducing reagents 1+1+1marks 20) (a) Salicylic acid from phenol: - (b) n-propyl alcohol from ethane: - By oxo process CH 2 =CH 2 + (CO + H 2 ) CH 3 -CH 2 -CHO CH 3 -CH 2 -CH 2 -OH 3/2+3/2 21.a) Benzene diazonium chloride reacts with phenol in which the phenol molecule at its para position is coupled with the diazonium salt to form p-hydroxybenzene. This type of reaction is known as coupling reaction. b) In this reaction potassium phthalimide is reacted with an alkyl halide to get N-alkyl phthalimde. N-alkyl phthalimide on hydrolysis with dil. HCl under pressure. 3/2+3/2

148 22. a) ) For the given binary solution of A and B, it would be ideal if A-B interactions are equal to A-A and B-B interactions and it would be non-ideal if they are different to each other. The deviation from ideal behaviour will be positive if A-B interactions are weaker as compared to A-A and B-B. The deviation will be negative if A-B interactions are stronger as compared to A-A and B-B. b) This is because Kh (Henry constant) values for both N 2 and O 2 increase with increase in temperature indicating that the solubility of gases increases with decrease in temperature 2+1 marks 23. a) For correct two reasons (1+1 M) b) For Correct value 1M 24. a) Adsorption processes, being exothermic, decreases with increase in temperature. b) Zeolites are called shape selective catalysts because their catalytic action depends upon the size and shape of the reactant and the product molecules as well as on their own pores and cavities. c) ) Itis defined as the scattering of light by the colloidal particles present in a colloidal solution mark 25.a) Leaching is the process of extracting a substance from a solid by dissolving it in a liquid. In metallurgy leaching is used for the ores that are soluble in a suitable solvent. b) During the froth floatation process a depressant like NaCN is added to the tank. The depressant selectively prevents ZnS from coming to the froth but allows PbS to come to the froth and hence helps the separation of PbS with the froth. 3/2+3/2 marks 26.a) NH 3 has higher boiling point than PH 3 because of the presence of inter molecular hydrogen bonding in NH 3, as the electronegativity difference is quite high in case of Nand H. b) The ionization energy of group 15 elements is higher than that of group 14 elements because the elements of group 15 have extra stable half-filled p orbital configuration and their size is smaller due to the higher nuclear charge c) Acidic strength increases from HF to HI because, down the group, as the size of the halogen increases, the Bond dissociation enthalpy decreases and it becomes easier for that halogen atom to lose its H + ion marks 27. a) The sum of first and second ionization enthalpies increases with increasing atomic number so the standard reduction potentials become less and less negative. Hence the +2 oxidation state becomes more and more stable. b) Vanadium shows different-different oxidation states because it has vacant d- orbitals.so vanadium pentaoxide acts as a good catalyst. 2+1 marks

149 28 a) b) Reaction is of second order w.r.t. reactant i)therefore when concentration is double rate of reaction become four times rate =[2R] 2 =4[R] 2 ii) when concentration is half rate of reaction become ¼ of the initial rate. rate =[1/2R] 2 =1/4[R] marks OR a) b) 2+1/2 =2.5 c) (3+1+1 marks)

150 29. (a) mark OR