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IMPORTANT QUESTIONS – 2015-16 SUBJECT : CHEMISTRY CLASS XII 1 CHAPTER 1 SOLID STATE 1mk ‘Q’ Q1. ‘Crystalline Solids are anisotropic in nature’. What does this Statement mean? A 1. The statement means that some of the physical properties like electrical resistance or refractive index of Crystalline Solids show different values when measured along different directions in the same crystal. Q 2. Why does the presence of excess of lithium make LiCl crystals pink? A 2. The presence of excess of lithium makes LiCl crystals pink due to e-strapped in anionic vacancies (F centers). These electrons absorb some energy of the white light giving pink colour to LiCl crystal. Q3. What is meant by anti-ferromagnetism? What type of substances exhibit anti ferromagnetism? A3. Substance like MnO, MnO2 in which magnetic domains are oppositely oriented and cancel out each other’s magnetic moment exhibit anti ferromagnetism. Magnetic. Alignment of magnetic moments in antiferromagnetic substance:- OR Q3 What type of substance would make better magnets, ferromagnetic or ferromagnetic? A3. Ferromagnetic substance would make better magnets because when ferromagnetic substance is placed in magnetic field all the domains get oriented in the directionof the magnetic field and a strong magnetic effect is produced eg : Co, Ni Q4. In a compound nitrogen atoms (N) make ccp and metal atoms (M) occupy one third of the tetrahedral voids present. Determine the formula of the compound formed by M & N? A4. Let the no of nitrogen atoms (N) be x No of tetrahedral voids = 2x No of metal atoms = 2/3 x. Ratio of M:N = 2/3x : x Therefore the formula of the compound is M2 N3 2 Q 5. Classify each of the following as being either a p type or n type semi conductor? (i) (ii) A 5. Ge doped with In. B doped with Si. (i) P type semiconductor because when group 14 element is doped with group 13 element, an electron deficit hole is created. (ii) n type semiconductor because when group 13 element is doped wih group 14 element , free electrons will become available. Q 6. Write a distinguishing feature between a metallic solid and an ionic solid? A 6. Ionic Solids Metallic Solids In solid state ionic solids are Metallic Solids are good electrical electrical insulator :- ions are not conductors in solid state because of the free to move. Eg NaCl, CuSO4 etc presence of free electrons , Eg :- Copper, Iron etc 2 mk Q 1. Analysis shows that nickel oxide has the formula Ni 0.98 0 1.00 . What fractions of nickel exist as Ni2+ and Ni3+ ions? A1. The formula of the oxide is Ni 0.98 0 1.00 Let the number of O2- ions be 100. Then number of nickel ions = 98 Let the number of Ni2+ be x Then number of Ni3+ = 98 - x Since total charge on cations = total charge on anions. X x (2) + (98-x) x (3) = 100 x 2 2x + 294 – 3x = 200 x = 94 % of Ni2+ = 98 94 x 100 = 96% % of Ni3+ = 100 -96 = 4% Q 2. A compound forms hcp structure. What is the total number of voids in 0.5 mol of it? How many of these are tetrahedral voids? A2. No of atoms = 0.5 mol = 0.5 x 6.022 x 1023 = 3.011 x 1023 No of octahedral voids = no of atoms in hcp structure = 3.011 x 1023 3 No of tetrahedral voids = 2 x no of atoms in hcp structure = 2 x 3.011 x 1023 = 6.022 x 1023 = 6.022 x 10 23 + 3.011 x 10 23 = 9.033 x 10 23 Total no of voids Q3. Examine the given defective crystal:A+ BA+ B- BBA+ A+ BA+ B- BA+ A+ BA+ B- A+ Answer the following questions (i) What type of stoichiometric defect is shown by the crystal? (1/2) (ii) How is the density of crystal affected by this defect? (iii) What type of ionic substance show such defect? (1/2) (1) A 3. (i) Schotty defect is shown by the crystals, since equal number of cations and anions are missing from the crystal lattice. (ii) Due to this defect, the density of the crystal decreases. (iii) This defect is shown by those ionic substance in which cations and anions are of almost similar size eg :- NaCl, KCl etc. Q 4. If NaCl crystals are doped with 2 x 10 cation vacancies per mole? -3 mol percent of SrCl2 ,calculate the A 4. Doping of NaCl with 2 x 10 -3mol percent of SrCl2 means 100 moles of NaCl is doped with 2 x 10 -3 mole of SrCl2 or 1 mole of NaCl is doped with 2 x 10-5 mole of SrCl2 Each Sr2+ will occupy the place of Na+ and displace one Na+ from crystal lattice to create cation vacancies. Cation vacancies = Number of Sr 2+ ion added. = 2 x 10 -5 mol = 2 x 10 -5 x 6.022 x 10 23. = 12.046 x 10 18 mol -1 4 Q 5. Calculate the packing efficiency of a metal crystal for a simple cubic lattice? A 5. Packing efficiency in simple cubic lattice rr r a Volume of one atom x 100 Volume of cubic unit cell (a3) Since a= 2r for simple cubic Vol of one atom =4/3π r3 = 4/3 π r3 x 100 (2 r)3 = 4 x 3.14 x r3 x 100 3 x 8 x r3 = 52.36% ≈ 52.4% Q 6 (a) In reference to a crystal structure, explain the meaning of coordination number? (b) What is the number of atoms in a unit cell of (i) Face centered cubic structure? (ii) Body centered cubic structure? A 6 (a)The number of nearest neighbours of any constituted particle in the crystal lattice is called its coordination number. (b) Number of atoms in a unit cell of (i) Face centered cubic cell structure is 4. (ii) Body centered cubic structure is 2. 5 3mk Q1. In terms of band theory.What is the difference? (a) between a conductor and an insulator. (b) between a conductor and a semi conductor. Empty conduction band Filled (valence band) _____ Partially Over lappling band filled band Conductors A1 (a) In conductors, energy band is partially filled or it overlaps with a higher energy unoccupied conduction band. Due to this electrons can flow easily under an applied electric field and show conductivity In insulators, the gap between filled valence band and the next higher unoccupied band is large, hence electrons cannot jump to it and substance has very small conductivity and behaves as an insulator. Empty band (conduction band) Filled band (Valence band) (b) In semiconductors the gap between valence band and conduction band is small. Therefore some electrons may jump from valence band to conduction band and show some conductivity eg ‘Si’ and ‘Ge’ 6 Q2 ) A2) Aluminum crystallizes in a cubic close packed structure. Its metallic radius is 125 pm. (a) What is the length of the side of the unit cell? (b) How many unit cells are there in 1 cm3 of aluminum? (a) In a cubic close packed structure :4 r = √2 Given r = 125 x 10 -12 m a= √ or 2 √2 r a = 125 x 10 -12 x 2 x 1.414 m = 354 pm. (c) a3 = (354)3 x (10-12) 3 = 44.21 x 10-30 m 3 No of unit cells in 1 cm3 = Total Volume = Vol of one unit cell 10-6 44.21 x 10-30 = 2.261x 10 22 unit cells. Q3 How will you distinguish between the following pairs of terms :(a) Hexagonal close packing and Cubic close packing? (b) Crystal lattice and unit cell? (c) Tetrahedral void and Octahedral void? Hexagonal close packing AB AB—type packing is called hexagonal close packing. i.e spheres of the third layer are exactly aligned with those of the first layer . eg : ‘Mg’ and ‘Zn’ Cubic close packing ABC ABC—type packing is called cubic close packing. i.e spheres of the third layer are not aligned with those of 1st or 2nd. When 4th layer is placed its spheres are aligned with those of 1st layer. eg :- ‘Cu’ and ‘Ag’ Crystal lattice Unit cell The three dimensional The smallest portion of crystal lattice arrangement of constituent which when repeated in different particles in the space which directions, generates the entire lattice. represents how the constituent particles (atoms, ions or molecules) are arranged in a crystal 7 Tetrahedral void Octahedral void A tetrahedral void is surrounded An octahedral void is surrounded by by 4 spheres which lie at the six spheres and formed by a vertices of a regular tetrahedron. combination of 2 triangular voids of the 1st and 2nd layer There are 2 tetrahedral voids per There is one octahedral void per atom atom in a crystal in a crystal Q4 Account for the following :(i) Table salt, NaCl sometimes appears yellow in colour. (ii) FeO(s) is not formed in stoichiometric composition. (iii) Some of the very old glass objects appear slightly milky instead of being transparent. A4. (i) Yellow colour in NaCl is due to metal excess defect due to which unpaired electrons occupy anionic vacancies. These sites are called F centers. These electrons absorb energy from the visible region and transmits yellow colour. (ii) In the crystal of FeO, some of the Fe2+ cations are replaced by Fe3+ ions. Three Fe2+ ions are replaced by two Fe3+ ions to make up for the loss of positive charge. Thus there would be less amount of metal as compared to stoichiometric properties. (iii) Very old glass objects become slightly milky, because of heating during the day & cooling at nights i.e annealing. Over a number of years, glass acquires some crystalline character. Q5. The density of copper is 8.95 gcm-3. It has a face centered cubic structure. What is the radius of copper atom? (Atomic mass Cu = 63.5 g mol -1 . NA = 6.022 x 1023 mol -1) ? A5. Mass per unit cell = Atomic mass of Cu x 4 NA = ______63.5__x 4______ 6.022 x 1023 = 4.22 x 10 -22 g Volume of unit cell = Edge __Mass__ Density = 4.22 x 10 -22 8.95 = 4.7 x 10-23 cm3 = (Volume )1/3 = (4.7 x10 -23)1/3 = 3.61 x 10-8 cm or 361pm 8 For fcc, r = r = r = ___a___ 2√2 . 128 pm 9 CHAPTER 2 SOLUTIONS 1 mk Q 1. Define an ideal solution and write one of its characteristics? A1 The solutions which obey Raoult’s law over the entire range of concentration are known as ideal solutions. i.e PA = PoA For ideal solutions A and PB = PoB H mix = 0 and B. V mix = 0. eg ;- Solution of n – hexane and n- heptane Q 2. What is meant by reverse osmosis ? A2 The process in which the solvent flows from the solution into the pure solvent through the semi permeable membrane when a pressure higher than the osmotic pressure is applied on the solution is called reverse osmosis. Application :- The technique is used in the desalination of sea water. Q3. Define mole fraction. A3 The mole fraction of a component is the ratio of the number of moles of that component to the total number of moles of all the components present in the solution. For a binary solution consisting of 2 components A and B if nAis the number of moles of A and nB is the number of moles of B then. A= Q 4. __nA___ nA + nB B = _ nB___ nA + nB Define the term azeotrope? A4 The constant boiling mixture which distill out unchanged in their composition are called azeotropes. Eg :- A mixture of ethanol and water containing 95.4% of ethanol forms an azeotrope with boiling point 351.15 K. Q5. Explain boiling point elevation constant for a solvent / Define ebullioscopic constant? A5 Since Tb = Kb.m where m is molality When m = 1 Kb = Tb 10 Therefore Ebullioscopic constant is defined as the elevation in boiling point of a solution when 1 mole of a solute is dissolved in 1 Kg of solvent . Q6 What are isotonic solutions ? A6 The solutions of equimolar concentrations having same osmotic pressure at given temperature are called isotonic solutions. eg :- A 0.9% solution of pure NaCl is isotonic with human red blood cells. Q7 Explain why aquatic species are more comfortable in cold water rather than in warm water? A7 Aquatic species need dissolved oxygen for breathing. As solubility of gases decrease with increase of temperature, less oxygen is available in summers in lakes. Hence they feel more comfortable in winter when the solubility is higher. OR Q7 Why do gases nearly always tend to be less soluble in liquid as the temperature is raised? A7 Dissolution of gas in liquid is an exothermic process. Gas + Solvent Solution + heat. Therefore as per Lechatlier’s principle if the temperature is increased, equilibrium shifts backward i.e the solubility decreases. 2 mk Q1 How is the vapour pressure of a solvent affected when a non volatile solute is dissolved in it? A1 When a non volatile solute is added to a solvent, its vapour pressure decreases because some of the surface sites are occupied by solute molecules. Thus less space is available for the solvent molecules to vaporize. Q2. The depression in freezing point of water observed for the same molar concentrations of acetic acid, trichloro acetic acid and trifluoroacetic acid increases in the order as stated above. Explain? A2 As depression in f.pt ( Tf) is dependent on degree of dissociation (α) and fluorine exerts the highest – I effect. So trifluoro acetic acid is the strongest acid and ionizes to a greater extent while acetic acid ionises to the minimum extent. Thus greater the number of ions produced, greater is the depression in freezing point. 11 Q3. Differentiate between molarity and molality for a solution. How does a change in temperature influence their values? Or Q3. State the main advantage of molality over molarity as the unit of concentration? Molarity It is defined as the number of moles of solute dissolved in one litre of the solution Mathematically Molarity (M) =No of moles of solute x 1000 Vol of Solution (ml) Molality It is defined as the number of moles of solute dissolved in 1 Kg of the solvent Mathematically Molality (M) =No of moles of solute x 1000 Mass of Solvent (g) It decreases with increase in temperature It does not change with change in (as V α T) temperature Since molality does not change with a change in temperature therefore it is a better method to express the concentration of a solution. Q4. What is meant by colligative property. List any four factors on which colligative properties of a solution depend? A4. The properties of solutions which depend upon the number of solute particles and not upon the nature of the solute are known as colligative properties eg :- Osmotic pressure. Factors :(i) (ii) (iii) (iv) Number of particles of solute. Concentration of solution. Temperature. Association or dissociation of solute. An aqueous solution of sodium chloride freezes below 273K. Explain the lowering in freezing point of water with the help of a suitable diagram? Q5. A5. Freezing point of a substance is the temperature at which solid and liquid phases of a substance coexist i.e they have the same vapour pressure. As the vapour pressure of the solution is less than that of pure solvent For solution its vapour pressure will become equal to that of a solid solvent only at a lower temperature. liquid solvent Vapo ur ………………… pres sure solid solvent Tf Tfo Temperature solution 12 3 mk Q1. A sample of drinking water was found to be severely contaminated with chloroform CHCl3 , supposed to be carcinogen. The level of contamination was 15ppm (by mass)? A1. (i) Express this in percent by mass. (ii) Determine the molality of CHCl3 in water sample. 15 ppm means 15 parts in 106 parts by mass in the solution. i.e 106 parts by mass of solution= 15 parts of solute. or 100 parts (Mass %) = Mass of solvent 15 x 100 106 = 15 x 10 -4 = (106 -15) g 106 g = nB wA ≈ Therefore Molality where nB = no of moles of solute wA= Mass of the solvent Molality = 15/119.5X1000 106 = 1.25 x 10-4m Q2 State Raoult’s law for a solution containing volatile components. How does Raoult’s law becomes as special case of Henry’s law? A2 For a solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction? i.e PAα A or PA = P0A A and PBα B or PB = P0B B Where PA and PB are the partial pressure of A and B, P0A , P0B are the vapour pressure of pure component and Aand B are their mole fractions. If gas is the solute and liquid is the solvent then according to Henry’s law PA = KH A Thus Raoult’s laws and Henry’s law become identical except that their proportionality constants are different. 13 Q3 Some ethylene glycol is added to your car’s cooling system along with 5kg of water. If the freezing point of water-glycol solution is -150c, what is the boiling point of the solution? Kb = 0.52 K Kgmol-1 and Kf = 1.86 K Kg mol-1 A3 Tf = 15oc Kf = 1.86 k/m Molality = Tf Kf Tb = 15/1.86 = 8.06 m = = = Kb x m 0.52 x 8.06 4.190c Boiling point of pure water = 100oc Therefore Tb = Tb+ Tbo Tb = 100 +4.19 = 104.19oC Q4 Assuming complete dissociation. Calculate the expected freezing point of a solution prepared by dissolving 6g of Glauber’s salt, Na2SO4.10 H2O in 0.1 Kg mol-1 of water.Kf of water = 1.86 K Kg mol -1. A4 Kf WB MB = = = 1.86 K Kgmol-1 6g W A = 0.1 Kg mol-1 322 g mol-1 Since there is a complete dissociation Therefore i = 3 Tf = i K f WB MB x W A = 3 x 1.86 x 6 322 x 0.1 = 1.04o 14 Q5. Calculate the mass of a nonvolatile solute (molar mass 40g mol-1) which should be dissolved in 114 g octane to reduce its vapour its pressure to 80%? A5. Ps=80% of Po=0.80 Po No of moles of solute = W 40 No of moles of solvent (octane) = 114 114 Po– Ps Po = XB Po–0.8 Po Po = W /40 W/40 + 1 therefore or 0.2 W +1 40 0.8 W 40 or = = 1 W 40 =0.2 W = 10g Q6 19.5g of CH2FCOOH is dissolved in 500g of water. The depression in freezing point observed is 1oc. Calculate the Van’t Hoff factor and dissociation constant of fluoroacetic acid .Kffor water is1.86 K Kgmol-1. A6 W B = 19.5g W A = 500g Kf= 1.86K Kgmol-1 (ATf) obs = 1o. MB (obs) =1000 Kf W B W A∆Tf = 1000 X1.86 X 19.5 =72.54 g mol-1 500 X 1 MB (cal) = 14 + 19 + 45 = 78 g mol-1 L = M(cal) = _78__ MB (Obs) 72.54 = 1.0753 15 Calculation of dissociation const (K) Ka = _Cα2 1-α CH2FCOO -+ H+ CH2 FCOOH Initial C MOL L-1 C(1-α) i = C(1+α) C O Cα = = O Cα 1 + α or α = i - 1 1.0753 – 1 = 0.0753 Taking volume of solution as 500 ml C =19.5 x_1_x 1000 78 500 Ka = Cα2 1-α = 0.5 M = 0.5 x (0.0753)2 1-0.0753 Ka = 3.07 x 10-3 Q7. Non ideal solutions exhibit either positive or negative deviations from Raoult’s law. What are these deviations and why are they caused? Explain with one example for each type? A7 For non ideal solutions, vapour pressure is either higher or lower than that predicted by Raoult’s law. If it is higher the solution exhibits positive deviation and if it is lower it exhibits negative deviation from Raoult’s law. Positive Deviation When solute – solvent interactions are weaker than solute – solute or solvent solvent interactions, vapour pressure increases which result in positive deviation. Eg :- Ethanol +Acetone. for a solution showing +ve deviation PA> PoA XA and PB > PoB XB H mix = +ve Vmix = +ve Negative Deviation When solute solvent interactions are stronger than solute – solute or solvent – solvent interactions, vapour pressure decreases which result in negative deviation. eg :- Chloroform + Acetone. for a solution showing –ve deviation PA< PoA XA PB< P0B XB H mix = - ve Vmix = - ve For plots of non ideal solution showing +ve and –ve deviation refer NCERT pg 56 16 5 mk Q1. (i) What is Van’t Hoff factor? What types of values can it have if in forming the solution the solute molecules undergo (ii) (a) Dissociation? (b) Association? How many ml of 0.1M HCl solution are required to react completely with 1g of a mixture of Na2CO3 containing equimolar amounts of both ? (Molar Mass Na2CO3 = 106g and NaHC03 = 84g) A1. (i) Van’t Hoff factor (i) is defined as the ratio of the experimental value of colligative property to the calculated value of colligativeproperty . i Also, i= = Observed Colligative property Calculated Colligative property Total number of moles of particles after dissociation/ association Number of moles of particles before association/ dissociation Therefore , For (a) dissociation i> 1 And (b) Association i< 1 (ii) Let x g of Na2CO3 be present in the mixture So amt of Na HCO3 = 1 –x Moles of Na2CO3 in x g = x_ 106 Moles of NaHCO3 in (1-x) = _1-x_ 84 As mixture contains equimolar amounts of the two. x_ =1-x_ 106 84 106 – 106 x = 84x X = _106__ = 0.558g 190 Moles of Na2 CO3 = __0.558__ 106 = 0.00526 Moles of NaHCO3 = 1- 0.558 =0.442 84 84 = 0.00526 To calculate the moles of HCl required. Na2CO3+ 2 HCl 2 NaCl + H2O + CO2 17 NaHCO3+HCl NaCl + H2O + CO2 1 mole of Na2CO3 requires 2 moles of HCl Therefore 0.00526 mole of Na2CO3 requires 2 x 0.00526 moles or 0.0152 moles of HCl. 1 mole of NaHCO3requires 1 mole of HCl Therefore 0.00526 mole of NaHCO3 requires 0.00526 moles of HCl. Total moles of HCl required = 0.01052 + 0.00526 = 0.01578 moles To calculate volume of 0.1M HCl 0.1 mole of 0.1M HCl are present m 1000ml HCl Therefore 0.01578 mole of 0.1M HCl will be present in 1000 x 0.01578 0.1 = 157.8 ml Q2. (i) The molecular masses of polymers are determined by osmotic pressure method and not by measuring other colligative properties. Give two reasons? (ii) At 300k, 36g of glucose C6H12O6 present per litre in its solution has a pressure pf 4.98 bar. If the osmotic pressure of another glucose solution is 1.52 bar at the same temperature, calculate the concentration of the other solution? A2. (i) because :- The osmotic pressure method has the advantage over other methods (a) It uses molarities instead of molalities and it can be measured at room temperature. (b) Its magnitude is large as compared to other colligative properties. (ii) π1 = C1 RT, π2 = C2 RT Therefore _π1 = C1 π 2 C2 or 4.98 1.52 = 36/180 C2 or C2 = 0.061 moll-1 = 0.061 x 180 gl-1 = 10.98 gl-1 18 Q3 (i) Define the terms osmosis and osmotic pressure? (ii) An aqueous solution containing 12.48g of barium chloride in 1.0 kg of water boils at 373.0832 k. Calculate the degree of dissociation of barium chloride? (Given Kb for H2O = 0.52 Km-1 molar mass of Ba Cl2 = 208.34 gmol-1) A3. (i) The net spontaneous flow of the solvent molecules from the solvent to the solution or from a less concentrated solution to a more concentrated solution through a semipermeable membrane is called osmosis. Osmotic pressure :- The minimum excess pressure that has to be applied on the solution to prevent the entry of the solvent into the solution through the semipermeable membrane is called osmotic pressure. (iii) (iv) Given W 2 = 12.48g , W1 = 1 Kg = 1000g Tb (solution) = 373.0832 K Kb for H2O = 0.52 Km-1 and M2 = (Ba Cl2) = 208.34 Tb= Tb - Tbo = 373.0832 – 373 = 0.0832 K M2 (Observed) = Kb x W 2 x 1000 Tb x W 1 = 0.52 x 12.48 x 1000 0.0832 x 1000 = 78 g mol-1 For BaCl2 i = M2 (calculated) = M2 (observed) m = 3 as it gives 3 ions on dissociation α 208.34 = 2.67 78 = i-1 = 2.67 -1 = 1.67 = 0.835 m-1 3-1 2 = 83.5% Q4. (a) What type of deviation is shown by a mixture of ethanol and acetone? Give reason. (b) What do you expect to happen when RBC’s are placed in (i) 1% NaCl solution (ii) 0.5 % NaCl solution (c) Calculate the molarity of 68% (w/w) solution of nitric acid , if the density of the solution is 1.504 g ml-1 19 A4. (a) Ethanol and acetone shows positive deviation because on mixing the two the forces of attraction decreases and the vapour pressure increases. (b) (c) Since RBC’s are isotonic with 0.9% NaCl solution therefore (i) In 1% solution of NaCl they will shrink due to plasmolysis (ii) In 0.5% solution of NaCl they will swell or may even burst. Molarity = % x density x 10 Molar mass Mass % = 68 , d = 1.504 Molar mass of HNO3 = 63 gmol-1 Therefore Molarity = 68 x 1.504 x 10 63 = 16.23 M Q5. (a) State Henry’s law and mention two of its important applications (b) The partial pressure of ethane over a saturated solution containing 6.56 x 10-2g of ethane is 1bar. If the solution were to contain 5 x 10-2 g of ethane, then what will be the partial pressure of the gas. A5. (a) Henry’s law states that at a constant temperature , the solubility in a liquid is directly proportional to the pressure of the gas of a gas P = KH Applications :- To increase thje solubility of CO2 in soda water, the bottle is sealed under high pressure - There is a low concentration of oxygen in the blood and tissues of the people living at high altitudes due to which they feel weak and are unable to think clearly (anoxia). - (b) M = KH x P For the 1st case 6.56 x 10-2 = KH x 1 bar or KH = 6.56 x 10-2 g bar -1 In the 2nd case 5 x 10 -2 = (6.56 x 10-2) x P or P = 5 x 10 -2 6.56 x 10 -2 P = 0.762 bar 20 CHAPTER 3 ELECTRO CHEMISTRY 01 MARK Q1. Why does the conductivity of a solution decrease with dilution? A1. Conductivity of a solution is the conductance of ions present in a unit volume of the solution. On dilution the number of ions per unit volume decreases. Hence, the conductivity decreases. Q2) What does the negative sign the expression E0Zn2+ / Zn = -0.76 V mean? A2) Negative sign shows that zinc is more reactive than hydrogen. This means that when zinc electrode is connected to SHE, Zn will be oxidized to Zn2+ and H+ will be reduced to H2. Q3) State Kohlrausch laws of independent migration of ions? A3) It states that “ At infinite dilution, when the dissociation of the electrolyte is complete, each ion makes a definite contribution to the total molar conductivity of the electrolyte irrespective of the nature of the other ion with which it is associated. Q4) Write two advantages of H2O2 fuel cell over ordinary cell? or What advantage do the fuel cells have over primary and secondary batteries? or Name the type of cell which was used in Apollo space programme for providing electrical power? A4) (a) Fuel cells can be run continuously so long as the reactants are supplied, primary batteries become dead and secondary batteries take a long time for recharging . (b) Fuel cells do not cause any pollution H2–O2 fuel cell was used in Apollo Space programme for providing electrical power. Q5) Express the relation among the conductivity of a solution in the cell, the cell constant and the resistance of solution in the cell? A5) K= 1xl R A Where K = Conductivity l/A = Cell constant R = Resistance. 21 Q6) What is electrode potential? A6) It is defined as the tendency of an electrode to lose or gain electrons when it is in contact with solution of its own ions. Q7) Is it safe to stir AgNO3 solution with a copper spoon? Why or Why not ? Given :- Eo Ag+/Ag = 0.8 volt and Eo cu2+ /Cu = 0.34 volt A7) The given values of reduction potentials show that Cu is more reactive than Ag. i.e Cu reacts with AgNO3 solution. Hence it is not safe to stir AgNO3 solution with copper spoon. Q8) What is the role of ZnCl2 in a dry cell? A8) ZnCl2 combines with the NH3 produced to form the complex salt [Zn (NH3) 2 Cl2] as otherwise the pressure developed due to NH3 would crack the seal of the cell. Q9) State faraday’s first law of electrolysis ? A9) The mass of any substance (w) deposited or liberated at any electrode is directly proportional to the quantity of electricity (Q) passed through the electrolyte or W α Q W = ZQ Z = electrochemical equivalent 02 MARKS Q1. Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10. A1. H+ + e- ½ H2 Applying Nernst eqn EH+/1/2 H2 = EoH+ ½ H2 -0.0591 log _1_ n [H+] = 0 – 0.0591log _1_ 1 10-10 because pH = 10 means [H+] = 10 -10 M. = - 0 .0591 x -10 = + 0.591V Q2. Mention the reactions occurring at (i) Anode (ii) Cathode, during the working of a mercury cell. Why does the voltage of a mercury cell remain constant during its operation? 22 A2. At Anode :Zn (Hg) + 2OH- ZnO (s) + H2O + 2e- At Cathode HgO + H2O + 2e- Hg (l) + 2OH- ______________________________ Zn(Hg) + HgO (s) ZnO (s) + Hg(l) The voltage of a mercury cell remains constant during its life as the overall reaction does not involve any ion in solution whose concentration can change during its life . Q3. Explain why electrolysis of aqueous solution of NaCl gives H2 at cathode & Cl2 at anode? Write overall reaction? Given EoNa+ /Na = - 2.71 V Eo H2O/H2 = -0.83 V Eo Cl2/Cl- = + 1.36 V Eo H+/H2/H2O = + 1.23V A3. At Cathode the reaction with higher value of Eois preferred, therefore H2 is released at cathode, which is produced by dissociation of H2O. Therefore at Cathode H2O (l) + e- ½ H2 (g) + OH – (aq) At anode the reaction with lower value Eo is preferred, therefore water should oxidised to O2 but due to over potential of oxygen, chlorine is released at anode. At anode :- Cl- (aq) ½ Cl2 (g) + e- Net reactions :NaCl (aq) + H2 O (l) Na+ (aq) + OH- (aq) + ½ H2 (g) + ½ Cl2 (g) Q4. Calculate G for the reaction, Mg (s) + Cu2+ (aq) Given Eo cell = + 2.71V 1F = 96500 (Cmol -1) A4. For the reaction Mg(s) + Cu2+ (aq) n=2 Go= -nFEo cell = -2 x 96500 x 2.71 = - 523030 J mol-1 Mg 2+ (g) + Cu (s) Mg2+ (aq) + Cu (s) 23 Q5. Calculate the degree of dissociation of acetic acid at 298K given that λmo (CH3 COOH) = 11.7 Scm2 mol -1 λmo (CH3 COO-) = 40.9 Scm2 mol -1 λmo (H+) A5. = 349.1 Scm2 mol -1 Degree of dissociation α = λm λ 0m λ Mo = λMo (CH3 COO-) + λM (H+) = 40.9 + 349.1 = 390 Scm2 mol-1 α = λm λm = 11.7 390 o α = 3 x 10-2 Q6. A solution of Ni(NO3)2 is electrolyzed between platinum electrodes using a current of 5 A for 20 min. What mass of nickel will be deposited at the cathode? Given : Atomic mass of Ni = 58.7 gmol-1 1F = 96,500 (mol-1) A6. Given :- I = 5A Time (t) = 20 x 60 = 1200s Q = I x t = 5 x 1200 = 6000C. M=ZxIxt M = Eq. wtx I x t 96,500 M = __58.7__ x 6000 = 1.82g 2 x 96500 Q7. (a) Define electrochemical series . (b) Given that the standard electrode potentials (Eo) metals are :K+/K = -2.93V, Ag+/Ag = 0.8V, Cu 2+/Cu = 0.34V, Mg 2+/Mg = -2.37V, Cr 3+/Cr = - 0.74, Fe2+/Fe = -0.44V. Arrange these metals in an increasing order of their reducing power. 24 A7. (a) The arrangement of the various electrodes in order of their increasing values of standard reduction potentials is called electrochemical series. (b) Greater the negative value of the standard electrode potential (Eo) greater is the reducing power of the electrode . Thus the increasing order of reducing power is Ag+/Ag < Cu 2+/Cu < Fe2+/Fe < Cr 3+/Cr < Mg 2+/Mg < K+/K 03 MARKS Q1. Three electrolytic cells A, B,C containing solutions of ZnSO4 , AgNO3 and CuSO4 respectively are connected in series . A steady current of 1.5A was passed through them until 1.45 g of Ag were deposited at the cathode of cell B. How long did the current flow? What mass of copper and what mass of zinc were deposited in the concerned cells? (Atomic mass of Ag = 108 , Zn = 65.4 u and Cu = 63.5u) A1. Ag+ + e(1F) Ag 108g 108 g of Ag is deposited by 96500C of charge Therefore 1.45g of Ag will deposited by 96500 x 1.45 =1295.6 C 108 Q =It t = 1295.6 1.5 Cu2+ + 2e- = 863.73 s Cu 2 x 96500 c deposits Cu = 63.5 g Therefore 1295.6c charge will deposit Cu = 63.5 x 1295.6 2 x 96500 = 0.426g Zn2+ + 2e- Zn Amt of zinc deposited = = 65.4 x 1295.6 2 x 96,500 0.439g 25 Q2. What type of battery is lead storage battery? Write the anode and cathode reactions and the overall cell reaction occurring in the operation of a lead storage battery? A2. Lead storage battery is a secondary cell i.e it can be recharged by passing direct current through it and can be reused. Anode :- Spongy lead. Cathode :- Grid of lead packed with PbO2 Electrolyte :- 38% H2 SO4 At Anode :- Pb (s) + SO42- (aq) Pb SO4 + 2e- At Cathode :- Pb O2 (s) + SO42- + 4H+ (aq) + 2e- Pb SO4 (s) + 2H2O (l) Overall cell reaction Pb (s) + PbO2 (s) + 4H+ + 2SO42- 2 PbSO4 (s) + 2H2O(l) Q3. The resistance of 0.01M NaCl solution at 25oc is 200 Ω. The cell constant of the conductivity cell used is unity. Calculate the molar conductivity of the solution. A3. Given Resistance ® = 200 Ω Molarity of NaCl solution = 0.01M Cell constant (l/A) = 1cm-1 Conductivity (K) = _1_ x _l_ R A = __1__ 200 = 5 x 10-3 Ω -1 cm-1 Molar conductivity λM = 1000K Molarity = (5 x10-3) x 1000 0.01 = 500 Scm2 mol-1 26 Q4. A strip of nickel metal is placed in a 1 molar solution of Ni(NO3)2 and a strip of silver metal is placed in one molar solution of AgNO3. An electrochemical cell is created when the two solutions are connected by a salt bridge and the two strips are connected by wires to a voltameter? (i) Write the balanced equations for the overall reaction occurring in the cell and calculate the cell potential. (ii) Calculate the cell potential, E at 25oC for the cell, if the initial concentration of Ni(NO3)2 is 0.100 molar and initial concentration of AgNO3 is 1.00 molar. [Eo Ni2+/Ni = - 0.25V, Eo Ag+/Ag = 0.8V log 10-1 = -1] A4. Ni2+ (aq) + 2e- At anode :- Ni (s) At Cathode :- 2Ag+ (aq) + 2e- 2Ag (s) Overall rn = Ni (s) + 2 Ag+ (aq) Ni2+ (aq) + 2 Ag(s) Eo cell = Eo cathode – Eo Anode = + 0.8V – (-0.25V) = 1.05V Ecell = Eo cell – 0.0591 log [Ni2+] 2 [Ag+]2 Eo cell – 0.0591 2 = log 0.1 (1)2 = 1.05 – 0.0591 x -1 2 1.05 + 0.0295 = 1.0795V Q5. Determine the values of equilibrium constant (Kc) and reaction. Ni (s) + 2 Ag+ (aq) Ni2+ (aq) + 2Ag (s) Eo= 1.05V (1F = 96500 cmol-1) A5. Given Eo = 1.05V n =2 Therefore Go = -nf Eo Go = -2 x 96500 x 1.05 = -202.65 KJ mol-1 -nfEo = -2.303RT log Kc (because Go = -2.303 RT log Kc) G0 for the following 27 Log Kc = nEo 0.0591 = 2 x 1.05 = 35.5329 0.0591 Kc = antilog 35.5329 Kc = 3.411 x1035 Q6. Set up Nernst equation for the standard dry cell. Using this equation show that the voltage of a dry cell has to decrease with use? A6. Dry cell (Leclanche cell) consists of a zinc container which acts as anode and cathode is a graphite rod surrounded by powdered MnO2 and ‘C’ The space between the electrodes is filled by a moist paste of NH4Cl and ZnCl2. Zn 2+ (aq) + 2e- At anode :- Zn (s) At cathode :- 2MnO2(S) + 2NH4+(aq) + 2e- Mn2O3 (s) + 2NH3 (g) + H2O NH3 formed combines immediately with Zn2+ ions to form complex ion [Zn(NH3)2] 2+ Zn 2+(aq) + 2NH3(g) [ Zn (NH3)2] 2+ (aq) Overall reaction :Zn2+ (aq) + 2NH4+ (aq)+ 2MnO2(s) [ Zn(NH3)2] 2+ (aq) + Mn2O3 (s) + H2O Nernst equation for dry cell E cell = Eocell – 2.303 RT nF log [Zn (NH3)2]2+ [NH4+]2 Due to the presence of ions in the overall reaction, its voltage decreases with time. Q 7. What is the relationship between Gibbs free energy of the cell reaction in a galvanic cell and emf of the cell. When will the maximum work be obtained from a galvanic cell? A7. In a galvanic cell – Electrical work done = Electrical energy produced = = Quantity of electricity flowing x EMF for n moles of electrons transferred in any cell reaction. The quantity of electricity flowing = nf faradays. Therefore Electrical work done = nF Ecell. Also electrical work done = Decrease in free energy Therefore - ` r G = nF Ecell. Or - Go = nF E o cell 28 ΔrGo = –2.303 RT log Kc Hence, knowing Eocell, ΔG o can be calculated which in turn can be used for the calculation of the equilibrium constant Kc. –ΔG = Wmax. Hence, the decrease in free energy is equal to the maximum work that can be obtained from the cell. 5 MARKS Q1. (a) The chemistry of corrosion of iron is essentially an electrochemical phenomenon. Explain the reactions occurring during the corrosion of iron in the atmosphere? (b) Give reasons :(i) Alkaline medium inhibits the rusting of iron. (ii) Rusting of iron pipe can be prevented by joining it with a piece of magnesium. A1. (a) The chemistry of corrosion of iron is essentially an electrochemical phenomenon. At a particular spot of an object made of iron, oxidation takes place and that spot behaves as anode. At anode 2 Fe(s) 2 Fe2+ + 4e- ; EoFe2+/ Fe(s) = -0.44 V Electron released at anodic spot go to another spot on the metal and reduce oxygen in the presence of H+ ions (H+ ions obtained from H2CO3 which is formed due to dissolution of CO2 from air into water). This spot behaves as cathode for the reduction. At cathode O2 (g) + 4H+ (aq) + 4e-2H2O(l); EoH+/O 2/H20 = 1.23V Overall reaction 2Fe(s) + O2 (g) + 4H+ (aq) 2 Fe2+ (aq) + 2H2O (l) , E0cell = 1.67V 29 The ferrous ions are further oxidized to ferric ions by atmospheric oxygen . Reaction is as follows :2 Fe2+ (aq) + 2H2O (l) + ½ O2(g) Fe2O3 + x H2O Fe2O3 (s) + 4H+ (aq) Fe2O3 . xH2O Hydrated ferric oxide (b) (i) Rusting of iron takes place in presence of H+ and as alkaline medium neutralizes H+ , the rusting is inhibited. (ii) It is due to cathodic protection in which Mg metal is oxidized in preference to iron and acts as anode. Q2. (i) Predict the product of electrolysis in each of the following :(a) An aqueous solution of AgNO3 with platinum electrodes. (b) An aqueous solution of H2SO4 with platinum electrodes. (ii) Estimate the maximum potential difference needed to reduce Al2O3 at 5000C . The Gibbs energy change for the decomposition reaction 4/3 Al +O2 is 960 kJ. (F = 96500 C mol-1) 2/3 Al2O3 A2. (i) (a) Electrolysis of AgNO3 (aq) using Pt electrodes Ag+ (aq) + NO-3 (aq) AgNO3 (aq) At cathode Ag+ (aq) + e- Ag (s) O2 (g) + 4H+ (aq) + 4e- At Anode 2H2O (l) (1 ½ ) (b) Electrolysis of H2SO4 (aq) using Pt electrodes. 2H2O (l) 2H2(g) + O2 (g) At Cathode H2O + e- O2 + 4H+ + 4e- At Anode 2H2O (iii) ½ H2 + OH- Reaction involved 2/3 Al2O3 4/3 Al + O2 Here n = 4 Given , G = 960K J -960 kJ = - 4 x 96500 x E0 G = - nFE0 (1 ½ ) 30 E0 = ___960000 J____ 4 x 96500 Q3. = 2.48V ≈ 2.5V (2) (i) Define molar conductivity of a solution and explain how molar conductivity changes with change in concentration of solution for a weak and a strong electrolyte. (ii) The resistance of a conductivity cell containing 0.001M KCI solution at 298 K is 1500Ω. What is the cell constant if the conductivity of 0.001M KCI solution at 298K is 0.146 x 10-3 S cm-1? A3. Molar conductivity. It is defined as the conductivity of the solution which contains one mole of the electrolyte such that entire solution is in between the two electrodes kept one centimeter apart. Molar conductivity, λm = __K___ C where k = conductivity. c is concentration of solution Variation of conductivity and molar conductivity with concentration. CH3COOH (weak electrolyte) λm 2 (S cm mol-1) KCI (strong electrolyte) C1/2 (mol/L) ½ Conductivity and molar conductivity change with change in concentration of electrolyte. Conductivity always decreases with decrease in concentration for both weak as well as strong electrolytes. But molar conductivity increases with decreases in concentration. For strong electrolytes λm increases slowly with dilution but for weak electrolytes λm increases steeply on dilution, especially near lower concentrations. Molar conductivity increases with decrease in concentration because both the number of ions as well as the mobility of increase with dilution. (ii) Given, conductivity k = 0.146 x 10-3 S cm-1 Resistance , R = 1500Ω 31 Therefore Cell constant , G* = K x R = 0.146 x 10-3 x 1500 = 0.219 cm-1 Q4) (a) What is standard hydrogen electrode? Give reactions that occur at this electrode when it acts as a positive electrode in an electro chemical cell. (b) What is the function of salt bridge in an electrochemical cell. (c) Cu2+ + 2e- Cu, Eo= + 0.34 V Ag + + e- Ag, Eo = + 0.80 V (i) Construct a galvanic cell using the above data. (ii) For what concentration of Ag+ ions will the emf of the cell be zero at 25oc , if the concentration of Cu2+ is 0.01 M? A4) (a) It is a reference electrode used to measure the electrode potential of other electrode. The electrode potential of SHE is taken as zero. (b) (c) Salt bridge performs the following functions :(i) It completes the inner circuit by flow of ions. (ii) It maintains the electrical neutrality in the solution of half cells. (i) Cu (s) I Cu2+ (aq) I I Ag + (aq) I Ag (s) At Anode : Cu 2+ (aq) + 2e- Cu (s) At Cathode 2 Ag+(aq) + 2eCu (s) + 2Ag+ (aq) 2Ag (s) Cu2+ (aq) + 2Ag (s) Thus , n = 2 E cell = EoAg+/Ag - EoCu2+ /Cu = + 0.80 V – 0.34V = 0.46V Using Nernst equation to calculate concentration of Ag+ Ecell = Eo cell – 0.0591 log [ Cu2+] 2 [ Ag+] 2 or 0 = .46 V – 0.0591 log _0.01_ 2 [ Ag+] 2 or log 0.01_ [Ag+] 2 = +0.46 V x 2 = 0.0591 _0.92_ 0.0591 = 15.567 32 Q5A) = Antilog 15.567 = 3.690 x 1015 or log 0.01_ [Ag+] 2 or [Ag+] 2 = ___0.01___ = 3.688 x 1015 or [Ag+] = 1.65 x 10-9 mol L-1 (i) ___1__ x 10-17 = 0.271 x 10—17 = 2.71 x 10-18 3.688 State two advantages of H2 –O2 fuel cell over ordinary cell. (ii) Silver is electrodeposited on a metallic vessel of total surface area 900 cm3 by passing a current of 0.5A for two hours. Calculate the thickness of silver deposited. (Given : Density of silver = 10.5 g cm-3, Atomic mass of silver = 108 u, 1 F = 96500 C mol-1] A5A) (i) It has high efficiency and is eco –friendly. (ii) Use the following relation to calculate mass of silver deposited m=ZxIxt or m = _108 x 0.5 x 2 x 60 x 60 = 108 x 5 x 2 x 6 x 6 = 4.03g 96500 965x 10 Also, Mass = Volume X Density or 4.03 g = Volume x 10.5g cm-3 or Volume = 4.03 cm3 = 0.38 cm3 10.5 Volume = Area x Thickness Q5B) or 0.38 cm3 = 900 cm2 x Thickness or Thickness = 0.38 cm3 = 4.22 x 10-4 cm. 900 cm2 Give reasons for the following :(i) Aluminium metal cannot be produced by the electrolysis of aqueous solution of aluminium. (ii) A 5B) Equilibrium constant is related to Eo cell but not to E cell (i) It is because aluminium metal is more reactive that hydrogen and will react with H2O. (ii) When equilibrium is reached in the cell reaction. E equal to zero . However Eo cell is a constant quantity . Therefore, Eocell =RT ln Kc nF cell becomes 33 CHAPTER – 4 CHEMICAL KINETICS 01 MARK Q1) State a condition under which a bimolecular reaction is kinetically first order? A1) Bimolecular reaction becomes kinetically first order when one of the reactants is in excess. Q2) Write the rate equation for 2A + B A2) Rate = k [A]o [B]o or Rate = k. C if the order of the reaction is zero? Q3) The reaction between H2 (g) an O 2 (g) is highly feasible yet allowing the gases to stand at room temperature in the same vessel does not lead to the formation of water. Explain? A3) Activation energy of this reaction is very high. Therefore the reaction does not take place. or Q3) Oxygen is available in plenty in air yet fuels do not burn by themselves at room temperature. Explain? A3) Activation energy for combustion reactions of fuels is very high at room temperature therefore they do not burn by themselves. Q4) For a certain reaction large fraction of the molecules has energy more than the threshold energy, yet the rate of the reactions very slow. Why? A4) Apart form energy considerations, the colliding molecules should have a proper orientation for effective collision. It appears that this condition is not fulfilled in this case. That is why the reaction is very slow. Q5) Why is the probability of reaction with molecularity higher than three vey rare? A5) The probability of more than three molecules colliding simultaneously is very very small. Hence the probability of reaction with molecularity higher than three is very rare. Q6) Why does the rate of reaction generally decrease during the course of reaction? A6) Rate of the reaction depends upon the concentration of reactants. With the progress of the reaction, reactants convert into products. Thus the concentration of the reactant decreases and hence the rate of reaction decreases. Q7) Thermodynamic feasibility of a reaction alone cannot decide the rate of the reaction. Explain with the help of example? A7) For a reaction to take place, appropriate activation must be supplied to the reactant. If this not available, the reaction does not take place. For example thermodynamically, conversion of diamond to graphite is possible, but it requires an 34 activation energy which is not available at room temperature. Hence the reaction does not take place. Q8) Why in the redox titration of KMnO4 vs oxalic acid, we heat oxalic acid solution before titration ? A8) The reaction between KMnO4 and oxalic acid is slow. On raising the temperature, we provide the required activation energy and the rate of reaction is increases. Q9) Write the difference between instantaneous rate of a reaction and average rate of reaction? Instantaneous Rate of Reaction Average Rate of Reaction Instantaneous Rate of Reaction at It is the appearance of product or any instant of time is defined as the disappearance of reactants over a long rate of change in concentration of time interval any one of the reactants or products at that particular instant of time. For a reaction :A+2B C [ ] = =- [ ] = [ ] = [ ] =- [ ] = [ ] 02 MARKS Q1) (a) For a general reaction A B , plot of concentration of A vs time is given in the figure below. Answer the following questions on the basis of this graph? (i) What is the order of the reaction? (ii) What is the slope of the curve? (iii) What are the units of the rate constant? A Co nc .of A t B 35 (b) For the reaction A + B Products, the rate law Rate = k A] [B] 3/2. Can the reaction be elementary reaction? Explain. A 1) is : (a) (i) For a zero order reaction, a plot of [A] with t gives the above type of graph. Hence this is a zero order reaction. (ii) Slope of the curve = - k (iii) Unit of rate of constant for zero order reaction are mol L-1 s-1 (b) Had this been an elementary reaction, the order with respect to B would have been 1. But it is given by 3/2 in the problem. Hence it is not an elementary reaction. Q2) (a) Why molecularity is applicable only for elementary reactions and order is applicable for elementary as well as complex reactions? (b) Why can we not determine the order of a reaction by taking into consideration? A2) (a) A complex reaction proceeds through a number of elementary reactions. Number of molecules involved in each elementary reaction may be different i.e, the molecularity of each step may be different. Therefore, molecularity of overall complex reaction has no sense. On the other hand, order of a complex reaction is determined by the slowest step in its mechanism and is valid even in the case of complex reactions. (b) Balanced chemical equation often leads to incorrect order or rate law as the rate of reaction may not depend upon all the molecules of a reactant present in the balanced chemical equation. For eg :- Decomposition of NH3 on Pt surface is actually a zero order reactions which is only determined experimentally & cannot be predicted from the chemical equation. 2NH3 Q 3) Pt N2 + 3H2 (a) Illustrate graphically the effect of catalyst on activation energy. (b) Catalyst have no effect on the equilibrium constant. Why? 36 A 3) (a) ---------Pot enti al ene rgy -------Reactants Reaction path with catalyst Energy activation with catalyst Energy of activation without catalyst -----------Products Reaction coordinate The catalyst provides an alternate pathways or reaction mechanism by reducing the activation energy between the reactants and products and hence lowering the potential energy barrier as shown in the figure. A catalyst does not change the equilibrium constant of a reaction, rather it helps in attaining the equilibrium faster. That is, it catalyses the forward as well as backward reactions to the same extent so that the equilibrium state remains same but is reached earlier. (b) Q4) How can you determine the rate law of the following reaction ? 2NO(g) + O2(g) A4) 2NO2(g) The rate law can be determined by initial rate method. Keeping the concentration of one of the reactant constant and changing the concentration of the other, the effect on the rate of reaction is determined. for eg : for the given reaction (i) Keeping [O ] constant, if [NO] is doubled rate is found to become four times. This shows that. Rate α [NO]2 (ii) Keeping [NO] constant, if [O ] is doubled, rate is also found to become doubled . This shows that Rate α [O ] (iii) Hence overall rate law will be Rate = k[NO]2[O ] Q5) How does a change in temperature affect the rate of a reaction? How can this effect on the rate constant of a reaction be represented quantitatively? A 5) Rate of reaction increases with temperature. Temperature coefficient is the ratio of rate constants of reaction at two different temperatures differing by 10o. Temperature coefficient = It is observed that for a chemical reaction with rise in temperature by 10 o, the rate constant is nearly doubled. 37 O3 MARKS Q1. The half life of decay radioactive 14C is 5730 years. An archeological artifact containing wood had only 80% of 14C activity as found in a living tree. Calculate the age of the artifact? A1. Radioactive decay is a first order reaction. or k = t = . log [ ] [ ] . log [ ] [ ] For 80% activity, [ ] = 0.8[ ] Substituting in the above equation, we get t or . = t ⁄ . . = x . log [ ]o 0.8[ ]o x 5730 x 0.0969 = 1845 years . Thus, the artifact in 1845 years old. Q2. During nuclear explosion, one of the products is 90Sr with half-life of 28.1 years. If 1 µg of 90 Sr was absorbed on the bones of a newly born instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically? A2) Determine k, using the following relation : . k= = . . / . k= log Substituting the values in the equation of first order, we get After 10 years, or log = or N= . . = . . . . = log . . = 0.1071 or = Antilog 0.1071 = 1.279 = 0.7818 µg Thus, 0.7818 µg of 90Sr will remain after 10 years 38 . After 60 years, 60 = . orlog = or x 28.1 log . = . N= . = 0.6427 or . = Antilog 0.6427 = 4.392 = 0.227 µg. . Thus , 0.227 µg of 90Sr will remain after 60 years. Q3. For a decomposition reaction, the values of rate constant at two different temperatures are given below :k1= 2.15 x 10 L/(mol.s) at 650 K k2= 2.39 x 10 L/(mol.s) at 700 K Calculate the value of for the reaction. (log 11.11 = 1.046) (R = 8.314J K-1mol A3. Using the formula, log = = 2.15 x 10 L/mol at 650K. = 2.39 x 10 L/mol at 700K. Given, = log log − . . / . / log 11.12 = = . . = . . 1 . . x . . / 650 − 1 700 . . = 182.25 kJ Q4. The following data were obtained during the first order thermal decomposition of SO Cl at a constant volume :SO Cl (g) SO (g) + Cl (g) Experiment 1 2 Time 0 100 Total pressure/atm 0.4 0.7 39 Calculate the rate constant. (Given , log 4 = 0.6021, log 2 = 0.3010) A4) SO Cl (g) SO (g) + Cl (g) Initial pressure After time, t –p 0 0 p p Total pressure aftertime t, i.e. = – p+ p+ p = p= Thus, a = - and a –x = = +p – p= – pt+ =2 -( - ) - Substituting the values of a and (a-x) in equation, . k = log . k = log ( ) Calculation of rate constant (k), when t = 100 s Given, p = 0.4 atm and . Then, k = = . = = log log = = 0.7 atm . . ( log . . . . . ) = . log 4 x 0.6021 = 0.01387 1.387 x10 40 Q 5) Nitrogen pent oxide decomposes according to equation, 2N O (g) 4NO (g) + O (g). This first order reaction was allowed to proceed at 40 C and the data below were collected. [ Time (min) 0.00 20.0 40.0 60.0 80.0 ]M 0.400 0.289 0.209 0.151 0.109 A5) (i) Calculate the rate constant. Include units with your answer. (ii) Calculate initial rate of reaction. (iii) After how many minutes will [N O ] equal to 0.350 M? (i) k= . log [ ] [ ] [ ] = 0.400 M [ ] = 0.289 M, t = 20 min k= k= k= . log . [ . [ . ] ] x log 1.384 . x log 0.1414 0.01625 min = When t = 40 min, [ ] = 0.209M . k= . = . log . = . log 1.914 x 0.2819 = 0.01623 min Similarly, when t = 60 min, [ ]= 0.151 k= k= . . log . . x 0.4231 = 0.01623min k =0.01625 + 0.01623 + 0.01623 3 = 0.016236 min 41 (ii) Initial rate = k [N O ] = 0.016236 x 0.4 = 0.00649 mol L s (iii) . k= t= t= t= log [ ] [ ] . . . . . . . log . log 1.143 log 0.0581 t = 8.24 min. 05 MARKS Q1) (a) Explain the following terms : (i) (ii) (b) Rate of reaction Activation energy of reaction The decomposition of phosphine, PH , proceeds according to the following equation : 4PH (g) P (g) +6H (g) It is found that the reaction follows the following rate equation : Rate = k[PH ]. The half life ofPH is 37.9 sat120 C. A1) (i) (ii) How much time is required for 3/4th of PH to decompose? (a) (i) Rate of a reaction : Rate of a reaction may be defined as change in concentration of reactant or product in unit time. What fraction of the original sample of PH remains behind after 1 minute? (ii) Activation energy of a reaction : Difference in energy of the reactants and the activated complex in the reaction is called activation energy. Reactants must gain energy to reach the state of activated complex in order to give the products. (b) 4PH (g) P (g) +6H (g) Half – life of PH = 37.9 s Rate = k[PH ] This means that the reaction is of first order. For a first order reaction k= . or / . . = 0.01828496 42 (i) Here R = . Apply the reaction k = - log [ ] [ ] = Substituting the values in the equation above, we have 0.01828496 = or t = or t = . log [ ] [ . ] /4 log 4 . . x 0.60206 . = 75.83 s (ii) After 1 minute or 60 seconds k = or log [ ] [ ] = . log [ ] [ ] . Substituting the values, we havelog [ ] [ ] = . . = 0.4768 Taking antilogarithms , we have [ ] [ ] or [ ] = 2.992 = . = 0.3342 Q2) (a) All energetically effective collisions do not result in a chemical change. Explain with the help of an example? (b) Show that for a first order reaction, the time required for half the change (half –life period) is independent of initial concentration. A2)(a) All energetically effective collisions do not result in a chemical change because even if colliding molecules have energy greater than the threshold energy, they may not have proper orientation at the time of collision and no breaking of bonds in the reactant molecules and formation of new bonds to form product molecules may occur. For eg :The reaction between bromomethane and an alkali may or may not lead to the formation of methanol depending upon whether they have a proper orientation or not at the time of collision. 43 H Proper orientation OH + H H - H H H H C – Br HO C -- Br H H Intermediate C – Br + OH– CH3 OH + Br– Methanol Improper orientation H H H C – Br HO No product (Repel) A2)(b) For a first order reaction, rate constant, k= When, t = . log / , / = / = = / = . . . [ ] [ ] [ ]= log [ ] [ ] [ ] log 2 . . Thus for a first order reaction,half life period is constant & is independent of initial concentration of the reacting species. Q3)(a) What are pseudo first order reactions? Give one example of each reactions. (b) Hydrogen peroxide,H O (aq) decomposes to H O (l) and O (g) in a reaction that is first order in H O and has a rate constant. k = 1.06 x 10 min-1. (i) How long it will take for 15% of a sample of H O to decompose? (ii) How long it will take for 85% of the sample to decompose? A3)(a) Pseudo first order reaction The reaction which is bimolecular but has order one, is called pseudo first order reaction, e.g acidic hydrolysis of ester. H+ CH COOC H + H O CH COOH(l) + C H OH(l) 44 (b) (i) When 15% of a sample of H O is decomposed. For a first order reaction, k = . log [ ] [ ] Given, k = 1.06 x 10 min−1 [ ] = 100M, [ ] (after time, t) = 100 – 15 = 85M t = . . t= = (ii) = = (a) log 1.176 . . . = 152.9 min . When 85% of a sample of H O is decomposed, [R] = 100-85 = 15M t = Q4) log . . log . . log 6.667 . . . = 1790.25 min . Explain the following terms : (i) Order of a reaction. (ii) Molecularity of a reaction (b) The rate of a reaction increases four times when the temperature changes from 300 K to 320 K. Calculate the energy of activation of the reaction, assuming that it does not change with temperature. (R = 8.314 J K -1 mol -1). A4) (a) (i) Order of a reaction : The sum of the powers of the concentration of the reactants in the rate law expression is called the order of that chemical reaction. (ii) Molecularity of a reaction : The number of reacting species (atoms, ions or molecules) taking part in an elementary reaction, which must collide simultaneously in order to bring about a chemical reaction is called molecularity of a reaction. Thus there are unimolecular, bimolecular, trimolecular reactions when the numbers of molecules taking part are one, two or three. (b) Apply Arrhenius equation log = . 45 Substituting the values, we have = log or log 4 = or log 4 = or 0.6021 = or = (a) . . . x x 1.088 x 10 . . x 1.088 x 10 x 10 = 5.53 x 10 J = 55.3 kJ/mol. or Q5) . A reaction is first order in A and Second order in B. (i) Write the differential rate equation. (ii) How is the rate affected when concentrations of both A and B are doubled? (b) In a reaction between A and B , the initial rate of reaction( ) was measured for different initial concentrations of A and B as given below : A/mol L-1 B/mol L-1 /mol L-1 0.20 0.30 5.07 x 10 0.20 0.10 5.07 x 10 0.40 0.05 7.16 x 10 What is the order of reaction with respect to A & B ? A5 (a) = K [A] [B]2 (i) (Rate)1 (ii) If both [A] = [B] are doubled (Rate)2 = k [2A] [2B]2 = 8K [A] [B]2 (Rate)2 = 8K [A] [B]2 = (Rate)2 = 8 (Rate) (Rate)1 k [A] [B]2 i.e The rate increases 8 times (b) = k [A]α[B]β ( )1 = k x 5.07 x 10-5 = (0.20) α (0.30) β ( )2 = k x 5.07 x 10-5 = (0.20) α (0.10) β ( )3 = k x 1.43 x 10-4 = (0.40) α (0.05) β Dividing (i) by (ii) ( )1/( )2 = 1 = (0.30/0.10)β = (3)β …..(i) …..(ii) …..(iii) 46 Therefore β =0 Dividing (iii) by (ii) ( )3/( ) = . . = (0.40/0.20)α (0.05/0.10)β or 1.412 = 2α = 2α or log 1.412 = α log 2 or α = 0.1523/0.3010 = 0.5 Thus, order w.r.t A = 0.5 , order w.r.t B = 0 Because [β] = 0 47 CHAPTER 5 SURFACE CHEMISTRY 01 MARK Q1). What is meant by shape selective catalysis? A1) The catalytic reaction that depends upon the pore structure of the catalyst and the size of the reactant and product molecules is called shape selective catalysis. eg:- zeolites (eg ZSM 5) converts alcohol directly into gasoline. Q2) What are the dispersed phase and dispersion medium in milk? A2) Dispersed phase : Fat Dispersion medium : water Q3) What is Collodion? A3) Collodion is 4% solution of nitro cellulose in a mixture of alcohol and ether. Q4) Define Kraft temperature? A4) The formation of micelles from the ionic surfactant can take place only above a certain temperature which is called Kraft temperature. Q5) Why is adsorption always exothermic ? Or What is the sign of A5) H& S when a gas is adsorbed by adsorbent ? When a gas is adsorbed in the surface of solid, its entropy decreased i.e. S =-ve. From Gibbs Helmholtz equation : G= H–T S for the process to be spontaneous , G must be negative, which is possible only when H = -ve . Hence adsorption is always exothermic. Q6) Why is it essential to wash the precipitate with water before estimating it quantitatively? A6) Some amount of the electrolytes mixed to form the precipitate remains adsorbed on the surface of the particles of the precipitate. Hence it is essential to wash it with water before it quantitatively. Q7) What happens when persistent dialysis of colloidal solution is carried out? A7) The stability of a colloidal sol is due to the presence of a small amount of the electrolyte. On persistent dialysis, the electrolyte is completely removed, so the colloidal sol becomes unstable and gets coagulated. Q8) What causes Brownian movement in a colloidal solution? 48 A8) Brownian movement ie zigzag movement of the colloidal particles is due to hitting of these particles by the molecules of the dispersion medium with different forces from different directions. Q 9) Why it is important to have clean surface in surface studies? A 9) It facilitates the adsorption of species on the adsorbent. Q 10) What happens when gelatin is mixed with gold sol? A 10) Gold sol is a lyophobic sol. On addition of gelatin, the sol is stabilized.. Q 11) Gelatin which is a peptide is added in ice – creams. What can be it s role? A 11) Ice –creams are emulsion which get stabilized by emulsifying agents like gelatin. Q 12) What is the role of activated charcoal in gas mask used in coal mine? A 12) Activated charcoal adsorbs poisonous gases present in coal mine. Q 13) In what way is a sol different from a gel? A 13) Sols are colloidal solutions of solid dispersed in liquid while gels are colloidal solutions of liquid dispersed in solid. Q 14) Of NH3 and N2,which gas will be adsorbed more readily on the surface of charcoal and why? A 14) NH3 is adsorbed more readily as it is more easily liquefiable compared to N2, Moreover, NH3 molecule has greater molecular size. Q 15) How does it become possible to cause artificial rain by spraying silver iodide on the clouds? A 15) Clouds are colloidal in nature and carry a charge. On spraying silver iodide which is an electrolyte, the charge on the colloidal particles is neutralized. Clouds coagulate to form rain. Q 16) What is the role of diffusion in heterogeneous catalysis? A 16) The gas molecules diffuse onto the surface of the catalyst and get adsorbed. After the chemical change, the products formed diffuse away from the surface of the catalyst setting the surface free for other reactant molecules to adsorb on the surface and give the product. 02 MARKS Q1) Write the differences between physisorption and chemisorption with respect to the following :(i) Specificity (ii) Temperature dependence (iii) Reversibility (iv) Enthalpy change Criteria Physisorption Chemisorption 49 Specificity It is not specific in nature It is highly specific in nature Temperature dependence It decreases with increase in temperature. Thus, low temperature is favorable for physisorption Reversible in nature It increases with increase in temperature. Thus, highly temperature is favorable for chemisorption Irreversible in nature Low enthalpy of adsorption High enthalpy of adsorption Reversibility Enthalpy change Q2) Distinguish between homogeneous and heterogeneous catalysis. A2) When reactants and the catalysts are in the same phase (i.e. liquid or gas) the catalysis is known as homogeneous catalysis. 3 (g) O2 ( g ) NO O3 ( g ) 2 When reactants and the catalysts are in the different phase (i.e. liquid or gas) the catalysis is known as heterogeneous catalysis. (s ) 4NH3(g) + 5O2(g) Pt 4 NO ( g ) 6 H 2 O( g ) Q3) What is meant by coagulation of a colloidal solution? Describe briefly any three methods by which coagulation of lyophobic sols can be carried out? A3) The process of aggregating together the colloidal particles is called coagulation of the sol. It is also known as precipitation. Following are the three methods by which coagulation of lyophobic sols can be carried out. Q4) (i) Electrophoresis. In this process, the colloidal particles move towards oppositely charged electrodes and get discharged resulting in coagulation. (ii) Mixing of two oppositely charges sols. When equal proportions of positively charges sols are mixed, they neutralize each other resulting in coagulation. (iii) Dialysis. By this method, electrolytes present in sol are removed completely and colloid becomes unstable resulting in coagulation colloid is a heterogeneous system, e.g gold sol, sulphur sol, soap, etc. What are emulsions ? Discuss the role of an emulsifier in forming emulsion. A4) Emulsions are one of the types of colloidal system, in which both the dispersed phase and dispersion medium are liquids, e.g milk. 50 Role of emulsifier Emulsifying agents are added to emulsions to stabilize them. The emulsifying agent forms an interfacial film between suspended particles and the medium. For oil in water emulsions, the principal emulsifying agents are gums, proteins, natural and synthetic soaps. Q 5) A5) How are the following colloidal solutions prepared? (i) Sulphur in water (ii) Gold in water (i) Sulphur sol is prepared by the oxidation of H2S with SO2. SO2 + 2H2S oxidation 3S + 2H2O (Sol) (ii) Gold sol is prepared by Bredig’s arc process or by the reduction of AuCl3 with HCHO. 2AuCl3 + 3HCHO + 3H20 Reduction 2Au + 3HCOOH + 6HCI (Sol) Q6) What are enzymes? Write in brief the mechanism of enzyme catalysis. A6) Enzymes are biochemical catalysts whih are globular proteins and form macromolecular colloidal solution in water. The mechanism of enzyme catalysis may be explained on the basis of lock and key theory. Acc to the theory , Enzymes are highly specific due to the presence of active sites on their surface. The shape of active site of any given enzyme is such that only a specific substrate can fit in to it just as one key can fit into a particular lock enzyme catalyzed reactions takes place in two steps as follows :Step I :- E Formation of Enzyme – substrate complex + Enzyme Step II :- S Substrate (fast and reversible) Enzyme substrate complex Dissociation of enzyme substrate complex to form the products ES Enzyme Substrate Complex Q7) ES [EP] E + P (slow and rate determining) Enzyme Enzyme Product (Regenerated) association What do you mean by activity and selectivity of catalysts? Product 51 A7) Activity : Ability of a catalyst to accelerate chemical reactions is known as its activity . For example, Pt catalyses the combination of H2 and O2 to form water. It has been found that for hydrogenation reactions, the catalytic activity increases from Group 5 to Group 11 metals with maximum activity being shown by Group 7-9 elements of the periodic table. Selectivity : The ability of a catalyst to direct a reaction to yield a particular product is called its selectivity. Combination of CO and H2 yields different products with different catalysts as given below :CO (g) + 3H2(g) Ni CH4 (g) + H2O (g) CO (g) + H2(g) Cu CO (g) + H2(g) Cu I ZnO – Cr O 2 3 HCHO (g) CH3OH (g). Q8) Describe some features of catalysis by zeolites? A8) (i) Zeolites are hydrated alumino-silicates which have a three- dimensional network structure containing water molecules in their pores. (ii) On heating, water of hydration present in the pores is lost and the pores become vacant to carry out catalysis. (iii) The size of the pores varies from 260 to 740 pm. Thus, only those molecules can be adsorbed in these pores and catalyzed whose size fits these pores. Hence, they act as molecular sieves or shape selective catalysts. An important catalyst used in petroleum industry is ZSM -5 (Zeolite sieve of molecular porosity – 5). It converts alcohols into petrol by dehydrating them to form a mixture of hydrocarbons. ZSM -5 Alcohols Hydrocarbons Dehydration 03 MARKS Q1) Account for the following :(a) Medicines are more effective in the colloidal form/ colloidal gold is used for intramuscular injection. (b) Sky appears blue in colour. (c) Alum is added to purify water. OR Bleeding stops on rubbing moist alum on the cut surface. A1) (a) Medicines are more effective in the colloidal form because they have a large surface area and are easily assimilated in this form. (b) There are dust particles in the atmosphere. These dust particles are of colloidal size and scatter the light. Blue light coming from the sun is scattered and the sky appears blue. 52 (c) Alum coagulates colloidal impurities present in water. or Alum brings about and coagulation of blood and stops further bleeding. Q2) A 2) Explain clearly how the phenomenon of adsorption finds application in (i) Production of vacuum in a vessel (ii) Heterogeneous catalysis (iii) Froth floatation process in metallurgy. (i) Production of high vacuum. The traces of air can be adsorbed by charcoal from a vessel evacuated by a vacuum pump to give a very high vacuum. (ii) Heterogeneous catalysis. Adsorption of reactants on the solid surface of the catalyst increases the rate of reaction. (iii) Froth floatation process. A low grade sulphine ore is concentrated by this method using pine oil and frothing agent. The mineral particles become wet by oils while the gangue particles by water. Q3) What is an adsorption isotherm? Describe Freundlich adsorption isotherm. A 3) Adsorption isotherm. It is the variation in the amount of gas adsorbed by the adsorbent with pressure at constant temperature. Freundlich’s adsorption isotherm. It is an empirical relationship between the quantity of gas adsorbed by unit mass of solid adsorbent and pressure at a particular temperature. 1 x kp n ( n 1) m When, n 1, …. (i) x x kp or p m m Where x is the mass of gas adsorbed on mass m of the adsorbent at pressure p.k and n are constants which depend on the nature of the adsorbent and the gas at a particular temperature. Taking log in Eq. (i), gives log x 1 log k log p m n x on Y-axis and m log p on X-axis. If, it comes to be a straight line, the Freundlich isotherm is valid. The validity of Freundlich isotherm can be verified by plotting log 53 Q4) Explain the terms with suitable examples :(i) A4) Alcosol (ii) Aerosol (iii) Hydrosol (i) Alcosol :- It is a colloidal dispersion having alcohol as the dispersion medium. For eg :- Collodion, which is a colloidal sol, of cellulose nitrate in ethyl alcohol. (ii) Aerosol :- It is a colloidal dispersion of a liquid in a gas eg:- fog. (iii) Hydrosol :- IT is a colloidal dispersion of a solid in water as the dispersion medium.eg :- starch sol. Q5) Explain the following observations :(a) Cotrell’s smoke precipitator is fitted at the mouth of the Chimneys use is factories. (b) Physical adsorption is multilayered, while chemisorption is monolayered. (c) A white precipitate of silver halide becomes coloured in presence of the dye eosin. A5) (a) The charged collodial particles of carbon after coming in contact with oppositely charges electrode in Cottrell precipitator lose their charge and settle down at the bottom. (b) In physical adsorption, there are Weak Vander Waals forces. Therefore it forms multilayers. In Chemisorption, adsorbate is attached by chemical bond. There is a strong force of attraction. Therefore, only one layer is obtained. (c) Eosin is adsorbed on the surface of silver halide precipitate making it look coloured. 54 CHAPTER – 6 GENERAL PRINCIPLES OF EXTRACTION 01 MARK Q1. Name the chief ores of aluminium and zinc. A1. Chief ores of – Aluminium – Bauxite. A 0 (OH) 3- 2x[Where O<X<1] Zinc :- zinc blende (ZnS) Q2. What is the function of collectors in the froth floatation process for the concentration of ores? or What is the role of collectors in froth floatation process? A2. Collectors such as pine oils, fatty acids, xanthates etc, enhance non – wettability of the mineral particles. Q3. What is the role of flux in metallurgical processes? A3. Flux is a substance that chemically combines with gangue (earthy impurities) which may still be present in the roasted or the calcined ore to form an easily fusible material called the slag. Q4. What types of ores can be concentrated by magnetic separation method? A4. Those ores which are magnetic in nature and associated impurities are nonmagnetic in nature or vice-versa, are concentrated by magnetic separation method. Q5. Differentiate between a mineral and an ore? A5. Naturally occurring substances from which metal may or may not be extracted profitably is known as mineral. The minerals from which metal can be extracted chiefly, easily and profitably are known as ores. Q6. Why is it that only sulphide ores are concentrated by froth floatation process? or Why is the froth floatation method selected for the concentration of sulphide ores? A6. Only sulphide ores are concentrated by froth floatation method because sulphide ores are wetted preferentially by pine oil and impurities (gangue) are wetted by water. Q7. Why are sulphide ores converted to oxide before reduction? A7. It is more easy to reduce an oxide ore than a sulphide ore. This is why sulphide ores are roasted and first converted to its oxide form. 55 Q8. Although carbon and hydrogen are better reducing agents, but they are not used to reduce metallic oxides are at high temperature. Why? A8. This is because carbon and hydrogen react with metals to form carbides and hydrides respectively at high temperature. Q9. What is the role of CO in the extractive metallurgy of aluminium from its ore? or What role is played by CO in the getting pure alumina (Al O ) in the extraction of aluminium? A9. The aluminate in solution is neutralized by passing CO and hydrate Al O is precipitated 2 Na [Al (OH)4] (aq) + CO (g) Al O .xH O( ) + 2NaHCO (aq) Q10. Why is electrolytic reduction preferred over chemical reduction for the isolation of certain metals? A10. More reactive metals, i.e. alkali metals and alkaline earth metals are usually extracted by electrolysis of their fused salts because in electrolytic reduction, by applying external potential from outside source, reduction can be brought easily. Further, more reactive metals have more affinity with oxygen than that of carbon. That’s why electrolytic reduction is preferred over chemical reduction. 02 MARKS Q1. How is copper extracted from a low grade ore of it? A1. Copper is extracted by hydrometallurgy from low grade ores. It is leached out using acid or bacteria. The solution containing Cu is treated with scrap iron or H . Cu (aq) + H .(aq) Cu(s) + 2H (aq) Cu (aq) + Fe(s) Cu(s) + Fe (aq) Q2. How do we separate two sulphide ores by Froth floatation method? Explain with an example? Or State the role of depresent in froth floatation process? A2. Two sulphide ores can be separated by adjusting properties of oil to water or by using depressants which prevent one type of sulphide ore particles from forming the froth with air bubbles. Eg :-In case of an ore containing Zns and Pbs, the depresent NaCN is used. It forms complex with ZnS and thus prevents it from forming a froth while Pbs forms the froth and hence can be separated from ZnS. 4NaCN + ZnS Na [Zn CN)4] + Na S sodiumtetracyanidozincate(II)complex 56 Q3. State Write the reactions involved in the following process : (i) Leaching of bauxite ore to prepare pure alumina. (ii) Recovery of gold after gold ore has been leached with NaCN solution. A3. Leaching of alumina from bauxite ore Bauxite usually contains SiO , iron oxide and titanium oxide (TiO ) as impurities. Powered ore is digested with conc NaOH solution at 473-523Kand 35-36 bar pressure.Al O and SiO are dissolved in the solution while impurities do not. Al O (s) + 2NaOH (aq) + 3H O(l) 2Na [Al(OH)4](aq) Filtrate is neutralized by passing CO gas and hydrated Al O is precipitated.Some fresh alumina is also added to solution to induce precipitation. 2 Na [Al (OH)4] (aq) + 2CO (g) Al O .xH O (s) + 2NaHCO (aq). ppt Precipitate is filtered, washed, dried and heated to give back pure Al O . Al O .xH O (s) 1470KAl O ( ) +.xH O (g) Alumina Recovery of gold 4Au(s)+ 8CN (aq) +2H O (aq) + O (g) 4[Au(CN)2]2-(aq) + 4OH (aq) 22[Au(CN)2] (aq) + Zn (s) [Zn(CN)4 ] (aq) +2Au(s) Soluble gold complex In this reaction, zinc acts as a reducing agent. Q4. A4. Define the following terms : (i) Roasting (ii) Calcination (i) Roasting. The process of heating of metal ore below its melting point in the presence of air is called roasting. (ii) Calcination. The process of heating of metal ore in the absence of air is called calcination. Q5. Why is an external emf of more than 2.2V required for the extraction of CL from brine? A5. For the reaction 2Cl (aq) + 2H O (l) 2OH (aq) + H (g) + Cl (g) Value of is + 422kJ. Using the equation = -n , the value of comes out to be -2.2V. Therefore an external emf of more than 2.2V is required for the extraction of from brine. 57 Q6. A6. (a) What is the composition of copper matte? (b) What is the function of SiO in the metallurgy of copper? (a) Copper matte contains Cu s and FeS. (b) Silica(SiO ) is added in the reverberatory furnace during the extraction of Cu to remove impurities of iron oxide (FeO) present in the ore. Silica here acts as flux and reacts with iron oxide gangue to remove it as slag, iron silicate. FeO + SiO Iron oxide Q7. A7. Silica FeSiO Iron silica slag Give reason for the following :(i) Alumina is dissolved in cryolite instead of being electrolysed directly. (ii) Zinc oxide can be reduced to the metal by heating with carbon but . (i) Melting point of alumina is very high and it is a bad conductor of electricity. Cryolite is added to alumina to lower its melting point and to make it conductor of electricity. (ii) Reduction of zinc oxide is done by using coke ZnO +C 1673K Zn + CO Chromium oxide is reduced through alumina (thermic reduction process}. Cr O + 2Al Al O + 2Cr This is because, the free energy change for the formation of Cr O is more negative than that of ZnO. Q8. State the basis of refining a substance by chromatographic method. Under what circumstances is this method specially useful? A8. Chromatographic Method. This method is based on the principle that different components of a mixture are differently adsorbed on an adsorbent. Adsorbed components are removed by using suitable solvent (eluent). This method is very useful for the purification of the elements which are available in minute quantities and the impurities are not different in chemical properties from the element to be purified. Q9. What chemical principle is involved in choosing a reducing agent for getting the metal from its oxide ore? Consider the metal oxides, Al O and Fe O , and justify the choice of reducing agent in each case? A9. Thermodynamic factor helps us in choosing a suitable reducing agent for the reduction of a particular metal oxide to metal. From Ellingham diagram, it is evident that metals which have more negative f of their oxide can reduce those metal oxide for 58 which is less negative. Since, the free energy change for the formation of f less negative than CO, CO is a good reducing agent for it. + 3CO 823K 2Fe + 3 ,G = -ve [because f G( ) is less negative then f is G (CO)] The free energy change for the formation of Al O is highly negative (~ –1000 to –1100), thus, no reducing agent is suitable for its reduction. It is thus,reduced by electrolysis of its oxide. Q10. How is wrought iron different from steel? A10. Pig iron contains 4% C and trace elements like, S, P,Si and Mn as impurity. It is heated strongly (in excess of oxygen) in Bessemer converter. By this, volatile impurities like C,P,S are removed as CO, P O and SO repectively. Si amd Mn are also oxidised into their oxides which combine to form MnSiO (slag). The iron so obtained is called wrought iron. When, it is heated with 0.5% C, it gives steel. Thus, steel is less pure3 than wrought iron. 03 MARKS Q1. Explain the basic principles of following metallurgical operatins : (i) Zone refining (ii) Vapour phase refining (iii) Electrolytic refining A1. (i) Zone refining It is based on the fact that the impurities are more soluble in the melt than in the solid state of the metal. When one end of the impure metal rod is heated with the help of mobile heater, the molten zone moves forward along with impurities. In this way impurities are concentrated at the other end of the rod. This end is cut off. The process is repeated several times to obtain ultrapure metals for producing semiconductors. Ge,Si,Ga and In are purified by this method. Noble gas atmosphere Metal rod Induction –coil heaters moving as shown Molten zone Zone refining process 59 (ii) Vapour phase refining In this method, the metal is converted into its volatile compound and is collected elsewhere. It is then decomposed to give pure metal. So the two requirements are (a) the metal should form a volatile compound with an available reagent. (b) The volatile compound should be easily decomposable, for the recovery of metal. (iii) Electrolytic refining In this method, the impure metals is made anode and a strip of pure form of same metal is made cathode. Aqueous solution of salt of same metal is taken as electrolyte. On passing electric current, metal ions from the electrolyte are deposited at the cathode in the form of pure metal while an equivalent amount of metal dissolves from the anode and goes into the electrolyte solution as metal ions, i.e. pure metal is dissolved from anode and deposited at cathode through electrolyte in equivalent amount. Q2. Write down the reactions which occur in upper, middle and lower zones in the blast furnace during the extraction of iron from iron ore? A2. Reduction of iron oxide in blast furnace (i) Lower zone of the blast furnace C + O CO + Heat C + CO 2CO Coke is burnt to give temperature upto 2200K at lower part of the blast furnace. (ii) Middle zone of the blast furnace CO and heat move up in the furnace. The temperature range on the middle zone of the blast furnace is 900 – 1500 K. FeO + CO Fe + CO Limestone is also decomposed to CaO which removes silicate impurity of the one as slag. CaCO 1100K C + CO CaO + SiO CaO + CO 2CO CaSiO Slag 60 (iii) Upper zone of the blast furnace Temperature range in this zone is 500-800K. Following reactions take place in this zone. 3Fe O + CO 2Fe O + CO Fe O + 4CO 3Fe + 4CO Fe O + CO 2FeO + CO The iron obtained from blast furnace is known as pig iron. It contains about 4% carbon. Q3. (a) The reaction : Cr O + 2Al Al O + 2Cr ( Δ = - 421kJ) is thermodynamically feasible as is apparent from the Gibbs energy value. Why does it not take place at room temperature? (b) The value of Δf for formation of Cr O is -540kJ mol is - 827 kJ mol . Is the reduction Cr O possible with AI? and that of Al O A3. (a) In the given redox reaction, all the reactants and the products are solids at room temperature, therefore, there does not exist any equilibrium between the reactants and the products and hence the reaction does not occur at room temperature. The interpretation of is based on K ( = -RT in K) where K is the equilibrium constant. Where there is no equilibrium in the solid state and the value of K becomes insignificant. However, at high temperature, when chromium melts, the value of TSincreases. As a result, the value of proceeds rapidly. (b). r becomes more –ve and hence the reaction The two thermochemical equations can be written as : Al (s) + O (g) Al2 O3 (s) ; Cr (s) + O (g) Cr2 O3 (s): AI, Al O f f = -827kJmol = -540 kJ mol (i) (ii) Subtracting equation (ii) from equation (i), we get Al (s) + Cr2 O3 (s) Al2 O3 (s) + Cr (s) ; f = -287 kJ mol As f of the combined redox reaction is –ve, therefore , reduction of Cr O by AI is possible. 61 Q4. Write chemical reactions taking place in the extraction of zinc from zinc blende. A4. The various steps involved in the extraction of zinc from zinc blende are : (a) Concentration. floatation process. The ore is crushed and then concentrated by froth (b) Roasting. The concentrated ore is roasted in presence of excess of air at about 1200 K when zinc oxide (ZnO) is formed and SO is evolved. 2ZnS + 3O Zinc blende 1200K 2ZnO + 2SO Zinc oxide (c) Reduction Zinc oxide obtained in step (b) is mixed with powdered coke and heated to 1673 K in a fire clay retort when it is reduced to zinc metal. ZnO + C 1673K Zn + CO At 1673 K, zinc metal being volatile (b.p 1180K), distils over and is condensed. (d) Electrolytic refining The metal obtained as above is impure. It is purified by electrolytic method. Impure zinc is made the anode while cathode consists of a sheet of pure zinc. The electrolyte consists of ZnSO solution acidified with dil. H SO . On passing electric current, pure Zn is deposited on the cathode. Q5. (a) (b) Describe the principle involved in each of the following process :(i) Mond process for refining of nickel. (ii) Liquation method What is the role of graphite rod in the electrometallurgy of aluminium? A5. (a) (i) Mond process for refining of nickel. (vapour phase refining) is based upon the principle that nickel is heated in the presence of carbon monoxide to form nickel tetra carbonyl, which is a volatile complex. Ni+ 4CO 330-350 K Ni(CO)4 Nickel tetra carbonyl Then, the obtained nickel tetra carbonyl decomposed by subjecting it to a higher temperature (450 – 470K) to obtain pure nickel metal. Ni(CO)4 Ni + 4CO Nickel tetra Nickel carbonyl (ii) Liquation method In this method alow melting metal like tin can be made to flow on a sloping surface. In this way, it is separated from higher melting impurities. 62 (b) In the electrometallurgy of alumina, a fused mixture of alumina, cryolite and fluorspar (CaF2) is electrolysed using graphite as anode and graphite lined as cathode. During electrolysis, Al is liberated at the cathode while CO and CO2 are liberated at the anode. The relevant equations are as under : At cathode : At anode : Al 3 3e Al (l ) C (s) + O C (s) + 2O (melt) CO(g) + 2e (melt) CO (g) + 4e Oxygen evolved at the anode burns, the anode material (graphite). If instead of graphite, some other metal is used as the anode, then O liberated will oxidise the metal of the electrode. Since graphite is much cheaper than any metal, therefore, graphite is used as the anode. 63 CHAPTER – 7 P-BLOCK ELEMENTS 01 MARK 1) Ans: Why does PCl3 fume in moisture? PCl3 hydrolyses in the presence of moisture giving fumes of HCl. PCl3 + 3H2O - H3PO3 + 3HCl 2) PH3 forms bubbles when passed slowly in water but NH3 dissolves. Explain why? Ans:- Due to high electronegativity & small size of N, NH3 forms H- bonds with water and hence it is water soluble. On the other hand, due to its lower electronegativity and its bigger size. PH3 does not form H-bonds with H2O. As a result, it does not dissolve in H2O and hence escapes as bubbles. 3) Solid Phosphorus pentachoride behaves as an ionic compound. Explain why? Ans In the solid state PCl5 exists as an ionic solid [PCl4]+ [PCl6]-in which the cation [PCl4]+ tetrahedral and the anion [PCl6]-is octahedral. 4) Sulphur in vapour state exhibits paramagnetic behavior. Give reason. Ans In vapour state, sulphur partly exists as S2 molecule which has two unpaired electrons in the antibonding π-orbitals like O2 and hence, exhibits paramagnetism. 5) In solution of H2SO4 in water, the second dissociation constant, Ka2 is less than the first dissociation constant, Ka1. Explain. Ans Ka2 is less than Ka1 because the negatively charged HSO4- ion has much less tendency to donate a proton to H2O as compared to neutral H2SO4. 6) Draw the structure of SO2 molecule. Comment on the nature of two S-O bonds formed in it. Are the two S-O bonds in this molecule equal? Ans Structure of SO2molecule is given below: In SO2, sulphur is sp 2 hybridised. The molecule of SO2 is angular. It is resonance hybrid of the above two canonical forms. In SO2, both the S-O bonds are covalent and have equal strength due to resonance. 64 7) Ans Write the balanced chemical equation for the following reaction : Excess of SO2 reacts with sodium hydroxide solution. Chemical equation for sodium hydrogen sulphite. 2NaOH + SO2 - Na2SO3 + H2O Na2SO3 + H2O + SO2 - 2NaHSO3 Sodium hydrogen sulphite 8) Name two poisonous gases which can be prepared from chlorine gas. Ans Phosgene (COCl2), tear gas (CCl3NO2), mustard gas (ClCH2CH2SCH2 CH2Cl) all are obtained from chlorine gas.. 9) Ans Write two uses of ClO2 i) ClO2 is a powerful oxidizing agent & chlorinating agent ii) It is an excellent bleaching agent. 10) What inspired N Bartlett for carrying out reaction between Xe and PtF6? Ans Bartlett found that the first ionization enthalpy of molecular oxygen is almost similar with that of xenon. Thus, after preparing red coloured compound O2+[PtF6]-, he got inspired for carrying out reaction between Xe and PtF6 and made efforts to prepare Xe+[PtF6]- by mixing Xe and PtF6. 02 MARKS 1) Arrange the following NH3, PH3, A3H3, SbH3 BiH3 in order of – a) Increasing basic strength b) Increasing reducing character. Ans a) On moving down the group, size of the elements increases. As the size increases, tendency to attract a proton decreases and thus, the basic character decreases. Hence the increasing order of basic strength is BiH3< Sbh3< AsH3< Ph3< NH3 b) The stability of hydrides decreases from NH3 to BiH3 which can be observed from their bond dissociation enthalpy. Thus the reducing character of hydrides increases in the order – NH3< PH3< AsH3< SbH3< BiH3 2) Explain the following : i) NO2 readily forms a dimer. ii) BiCl3 is more stable than BiCl5. Ans i) NO2 is an odd molecule due to the presence of odd number of valence electrons. Hence, it readily dimerises to give more stable N2O4 molecule with even number of electrons. ii) Due to inert pair effect, +3 oxidation state of Bi is more stable than its +5 oxidation state. Thus, BiCl3 is more stable than BiCl5. 3) Draw the structures of white phosphorus and red phosphorus. Which one of these two types of phosphorus is more reactive and why? 65 Ans Structure of white phosphorus is given below : Structure of red phosphorus is given below: White phosphorus is more reactive because of its discrete tetrahedral structure and angular strain. 4) Ans Write the conditions to maximize the yield of H2SO4 by contact process. The key step in the production of H2SO4 is the oxidation of SO2 to SO3. SO2(g) + O2(g) 2SO3 (9) ∆ƒ Ho = - 196.6 KJmol-1 The reaction is exothermic & reversible and the forward reaction proceeds with decrease in volume. Therefore in accordance with Le chatliers; principle, to maximize the yield of SO3 & hence of H2SO4, a low temperature (720 K), a high pressure (2 bar) & V2O5 is used as a catalyst. 5) Assign reasons for the following. a) H2S is more acidic than H2O. b) Sulphur has a greater tendency for catenation than oxygen. Ans a) Due to decrease in (E-H) bond dissociation enthalpy down the group, acidic character increases. Thus, H2S is more acidic than H2O. b) Bond energy of S-S bond (213 kJ mol-1) is greater than O-O bond (138 kJ mol-1). Due to small size of oxygen atom, there is greater lp-bp repulsion in O-O, resulting in weakening of O-O bond more than in S-S bond. Therefore, the tendency of catenation in oxygen is lower than sulphur. 6) Draw the structure of each of the following. a) H2S2O8 b) H2SO4 66 Ans 7) Ans 8) Ans a) Structure of H2S2O8 is given below: b) The structure of H2SO4 Complete the following chemical equations:a) NaOH (hot & Conc.) + Cl2 → b) Cl2 + F2 (excess) → a) 3Cl2 + 6NaOH → 5NaCl + NaClO3 + 3H2O (hot & conc) b) Cl2 + 3F2 (excess) → 2ClF3 a) b) a) b) Which neutral molecule would be isoelectronic with ClO-? How are interhalogen compounds formulated and how are they prepared? ClO- has 26 electrons. A neutral molecule with 26 electrons is OF2. Interhalogen compounds are formulated as XX | XX 3| XX 5| , etc. The interhalogen compounds can be prepared by direct combination or by the action of halogen on lower interhalogen compounds. The product formed depends upon some specific conditions, eg: K Cl2 + F2 437 2ClF (Equal volume) K Cl2 + 3F2 573 2CIF3 (Excess) 9) Ans a) i) ii) i) ii) How are the following compounds prepared from XeF6? XeOF4 XeO3 XeF6 + H2O → XeOF4 + 2HF XeF6 + 3H2O → XeO3 + 6HF 67 9) b) XeF2 is linear molecule without a bent. Explain. Ans XeF2 is linear molecule. According to VSEPR theory, the three lone pairs will occupy the equalorial positions and two bond pairs will occupy axial positions to minimize lp-lp and lp-bp repulsions. 10) Explain Why – a) Noble gases have comparatively large atomic sizes. b) Noble gases form compounds with fluorine & oxygen only. Ans a) It is due to the reason that noble gases have only vander waals radii while others have covalent radii & by definition, Vander waals radii are larger than covalent radii. b) Fluorine and oxygen are the most electronegative elements and hence, are very reactive. Therefore, they form compounds with noble gases particularly with xenon. 03 MARKS: 1) What happens when – a) White phosphorus is heated with conc. NaOH solution in an inert gas atmosphere? b) Orthophosphorous acid is heated? c) PCl5 is heated. Ans a) When white phosphorus is heated with Conc.NaOH solution in an inert gas atmosphere, phosphine gas is produced. P4 + 3NaOH + 3H2O → PH3 + 3NaH2PO2 2) Ans b) On heating, H3PO3 disproportionate into H3PO4 and phosphine. 4H3PO3 → 3H3PO4 + PH3 c) When PCl5 is heated, it decomposes into PCl3 and Cl2. PCl5 → PCl3 + Cl2 Complete the following chemical reaction – i) I2 + conc.HNO3 → ii) HgCl2 + PH3 → iii) AgCl(s) + NH3 (aq) → i) I2 + 10HNO3 → 2HIO3 + 10NO2 +4H2O ii) 3HgCl2 + 2PH3 → Hg3P2 + 6HCl iii) AgCl(s) + 2NH3 (aq) → [Ag(NH3)2] cl (aq) 68 3) On heating compound (A) gives a gas (B) which is a constituent of air. This gas when treated with 3 mol of hydrogen (H2) in the presence of a catalyst gives another gas (C) which is basic in nature. Gas C on further oxidation in moist condition gives a compound (D) which is a part of acid rain. Identify compounds (A) to (D) and also give necessary equations of all the steps involved. Ans Compounds (A) to (D) are as follows : (A) = NH2NO2 (B) = N2 (C) = NH3 (D) = HNO3 The reactions are given as under : i) NH4NO2 N2 + 2H2O ii) N2 + 3H2 → 2NH3 iii) 4NH3 + 5O2 → 4NO + 6H2O iv) 2NO + O2 → 2NO2 v) 3NO2 + H2O → 2HNO3 + NO 4) On reaction with Cl2, phosphorus forms two types of halides A and B. Halide A is yellowish-white powder but halide B is colourless oily liquid. Identify A and B and write the formulas of their hydrolysis products. Ans A is PCl5 (It is yellowish-white powder) P4 + 10Cl2 → 4PCl5 B is PCl3 (It is a colourless oily liquid) P4+6Cl2 → 4PCl3 Hydrolysis products are formed as follows: PCl3 + 3H2O → H3PO3 + 3HCl PCl5 + 4H2O → H3PO4 + 5HCl 5) An amorphous solid “A” burns in air to form a gas “B” which turns lime water milky. The gas is also produced as a by-product during roasting of sulphide ore. This gas decolourises acidified aqueous KMnO4 solution and reduces Fe3+ to Fe2+. Identify the solid “A” and the gas “B” and write the reactions involved. Ans The solid A is sulphur, S8 and the gas B is SO2. The reactions are given as under: S8 + 8O2 8SO2 2MnO 4 + 5SO2 + 2H2O → 5SO42- + 4H+ + 2Mn2+ 2Fe3+ + SO2 + 2H2O → 2Fe2+ + SO42- + 2H+ SO2 is produced as a by product during roasting of sulphide ore. The sulphides are converted into oxides with the evolution of SO2. 6) Ans Complete the following chemical equation i) C + H2SO4 (conc.) → ii) SO3 + H2SO4 (conc.)→ iii) O3(9) + I-(aq) + H2O (e) → i) C + 2H2SO4 (conc.) → CO2 + 2SO2 + 2H2O ii) SO3 + H2SO4 (conc.) → H2S2O7 Oleum iii) O3(g)+ 2I (aq) + H2O(l) → 2OH-(aq) + I2(s) + O2(g) 69 7) Account for the following: a) Iron on reaction with HCl forms FeCl2& not FeCl3 . b) O2& F2 both stabilize higher oxidation states of metals but O2 exceeds F2 in doing so. c) ICl is more reactive than I2 ? Ans a) HCl reacts with Fe and produces H2. Fe + 2HCl → FeCl2 + H2 Liberation of hydrogen prevents the formation of ferric chloride Ans b) O2 and F2 both stabilize higher oxidation states of metal but O2 exceeds F2 in doing so due to ability of oxygen to form multiple bonds with metals. Ans c) Interhalogen compounds are more reactive than halogens (except fluorine) because | XX bond (I-Cl bond in question) in interhalogens is weaker than X-X bond (I-I bond) in halogens, except F-F bond. In other words, I-Cl bond is weaker than I-I bond. That’s why, ICl is more reactive than I2. 8) Ans (ii) a) i) ii) iii) i) Draw the structure of XeOF4 Molecule. HClO4 Molecule BrF3 70 (iii) 9) How are xenon fluorides XeF2, XeF4 and XeF6 obtained ? Ans These xenon fluorides are obtained by direct reaction between Xe and F2, under different conditions as shown below: , Xe (g) + F2 (g) ⎯⎯⎯⎯⎯⎯ XeF2 (s) (excess) , Xe (g) + 2F2 (g) ⎯⎯⎯⎯⎯⎯ XeF4 (s) (1:5 ratio) , Xe (g) + 3F2 (g) ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ XeF6 (s) (1: 20 ratio) 10) Account for the following – a) Helium is used in diving equipments. b) Structure of xenon fluorides cannot be explained by valence bond approach. c) Bleaching of flowers by chlorine is permanent while that by sulphur dioxide is temporary. Ans a) Helium is used as a diluent for oxygen in modern diving apparatus because of its very low solubility in blood. b) According to the valence bond approach, covalent bonds are formed by the overlapping of half-filled atomic orbitals. But xenon has fully-filled electronic configuration. Hence, the structure of xenon fluorides cannot be explained by VBT. c) Cl2 bleaches coloured material by oxidation. Cl2 + H2O → 2HCl + [O] hence bleaching is permanent. On the other hand, SO2 bleaches coloured material by reduction & hence bleaching is temporary SO2 + 2H2O → 2H2SO4 + 2[H] 71 05 MARKS 1) On heating lead (II) nitrate gives a brown gas “A”. The gas “A” on cooling changes to colourless solid “B”. Solid “B” on heating with NO changes to a blue solid ‘C’. Identify ‘A’, ‘B’ and ‘C’ and also write reactions involved and draw the structures of ‘B’ and ‘C’. Ans The gas A is NO2 The reactions are explained as under : 2Pb(NO3)2 2PbO + 4NO2 + O2 A (brown colour) 2NO2 ⎯⎯⎯⎯⎯⎯ N2O4 (Colourless solid) 2NO + N2O4 ⎯⎯⎯ 2N2O3 B C (Brown coloursolid) Structure of N2O4 Structure of N2O5 2) i) How would you account for the following ? a) NF3 is an exothermic compound whereas NCl3 is not. Explain. b) All the P-Cl bonds in PCl5 molecule are not equivalent. Explain why? c) Why nitrogen gas is very unreactive ? Ans a) In case of nitrogen, only NF3 is known to be stable. N-F bond strength is greater than F-F bond strength, therefore formation of NF3 is spontaneous. In case of NCl3, N-Cl bond strength is lesser than CL-Cl bond strength. Thus, energy has to be supplied during the formation of NCl3. 72 Ans b) PCl5 has a trigonal bipyramidal structure and the three equatorial P-Cl bonds are equivalent while the two axial bonds are different and longer than equatorial bonds. This is because the axial bond pairs suffer more repulsion as compared to equatorial bond pairs. Ans c) Nitrogen is chemically less reactive. This is due to the presence of a more stable triple bond in N2 molecule whereas phosphorus forms only P-P single bond. Therefore, phosphorus is more reactive than nitrogen. ii) Ans Draw the structures of the following a) (HPO3)3 b) H4P2O7 a) Structure of (HPO3)3 is given below: b) Structure of H4P2O7 is given below: 3) a) P4O6 reacts with water according to equation P4O6 + 6H2O → 4H3PO3, Calculate the volume of 0.1 M NaOH solution required to neutralize the acid formed by dissolving 1.1 g of P4O6 in H2O. 73 P4O6 + 6H2O → 4H3PO3 Ans H3PO3 + 2NaOH → Na2HPO3 + 2H2O] x 4 ---------------------------------------------------------------------------P4O6 + 8NaOH →4Na2HPO3 + 2H2O 1 mol 8 mol No. of moles in P4O6 = . 1 mol P4O6 requires = 8 mol NaOH . mol P4O6 requires = 8 ∗ = . . mol NaOH =0.04 mol NaOH Molarity of NaOH solution is 0.1 M 0.1 mol NaOH is present in = 1 litre 0.04 mol NaOH is present in = . x 0.04 litre = 400 ml. b) Give an example to show the effect of concentration of nitric acid on the formation of oxidation product. Ans Dilute and concentrated nitric acid gives different oxidation products on reaction with copper metal. 3Cu + 8HNO3 (dil.) → 3Cu(NO3)2 + 2NO + 4H2O Cu + 4HNO3 (conc.) → Cu(NO3)2 + 2NO2 + 2H2O NO and NO2 are the oxidation products obtained wilh dil. HNO3 and conc. HNO3 respectively. 4) Give reasons for the following. i) Decomposition of O3 molecule is a spontaneous process. ii) SF6 is inert towards hydrolysis. iii) (CH3)3 P=O exists but (CH3)3 N=O does not. iv) Oxygen has less electron gain enthalpy with negative sign than sulphur. v) SO2 is an air pollutant. Ans i) Decomposition of O3 is an exothermic process (ΔH = -ve) and occurs with increase in entropy (ΔS = +ve). These two effects reinforce each other which results in large negative Gibbs energy change. It favours its decomposition into oxygen. ii) In SF6, S is surrounded by 6 F- octahedrally. Therefore, attack of water molecule on S is sterically hindered. iii) Due to absence of d-orbitals,N cannot form pπ-dπ multiple bonds.As a result,N cannot expand it’s covalency more than four,but in R3N=O,N has it’s covalency 5,therefore,the compound R3N=O doesnot exist.On the other hand,due to the presence of d-orbitals in P,it forms pπ-dπ multiple bonds,therefore can expand it’s covalency more than four.As a result P can form (CH3)3 P=O.In this compound the covalency of P is 5. iv) The electron gain enthalpy of oxygen is less negative than sulphur due to its compact size. As a result of which, the electron repulsions in the relatively compact 2p subshell 74 are comparatively large hence the incoming electrons are not accepted with the same ease as in case of sulphur. v) SO2 is water soluble, therefore it dissolves in rainwater causing acid rain. Moreover, when released in air, it mixes with it and leads to several diseases like eye irritation, redness, asthma, bronchitis, etc. Thus, it is considered as an air pollutant. 5) a) Name the two most important allotropes of sulphur. Which one of the two is stable at room temperature? What happens when the stable form is heated above 370 K? Ans Two most important allotropes of sulphur are i) Rhombic sulphur (α-sulphur) ii) Monoclinic sulphur The stable form at room temperature is rhombic sulphur which transforms to monoclinic sulphur when heated above 369K 370K Rhombic sulphur Monoclinic Sulphur b) i) Ans i) ii) iii) Draw the structures of – H4S2O7 ii) O3 iii) S8 75 Crown Shape 6) a) i) ii) iv) How would you account for the following – The oxidizing Power of oxoacids of chlorine follows the order – HClO4< HClO3< HClO2< HClO The halogens are coloured. F2 is a stronger oxidizing agent than Cl2 b) Complete the following chemical equations. i) I2 + NaClO3→ ii) I2 + H2O + Cl2→ Ans a) As the oxidation number of halogen atom in oxoacid increases, its oxidizing power decreases. Therefore, HClO is least stable and gives [O] most easily, so its oxidizing power is greater than HClO4. b) All halogens are coloured due to the absorption of raditions in visible region which results in the excitation of outer electrons to higher energy level. By absorbing different quanta of radiation, they display different colours. c) Since Eo for F2/F- eleclrode is higher than that of Cl2/Cl- electrode F2 is more easily reduced than Cl2& is a stronger oxidizing agent than Cl2. Ans b) i) I2 + 2NaClO3→ 2NaI+2ClO2 + O2 ii) I2 + 6H2O + 5Cl2→ 2HIO3 + 10HCl Iodic acid. 2) a) How can you prepare Cl2 from HCl & HCl from Cl2. Write reactions only. b) Write the reactions of F2& Cl2 with water. Ans a) HCl can be oxidized to Cl2 by a number of oxidizing agents such as MnO2, KMnO4, K2Cr2O7, etc. Reaction with MnO2 is given below: MnO2 + 4HCl → MnCl2 + Cl2+ 2H2O Cl2 can be reduced to HCl by reaction of H2 in presence of diffused sunlight. H2 + Cl2 > 2HCl b) F2 being a stronger oxidizing agent oxidises H2O to O2 or O3. The reactions are given as under: 2F2 (g) + 2H2O (l) → 4H+ (aq) + 4F- (aq) + O2 (g) 2F2 (g) + 3H2O (l) → 6H+ (aq) + 6F- (aq) + O3 (g) 76 Cl2, on the other hand, reacts with H2O to form hydrochloric acid and hypochlorous acid as per the following equation : Cl2(g) + H2O (l) → HCl (aq) Hydrochloric acid 8) + HOCl (aq) Hypochlorous acid a) i) ii) iii) Give reason for the following. F2 is more reactive than ClF3 but ClF3 is more reactive than Cl2. H3PO2 is a stronger reducing agent than H3PO3. PCl5 is more covalent than PCl3. b) Complete the following chemical equation. i) XeF4 + O2F2→ ii) XeF4 + SbF5 → Ans a) i) Fluorine due to its small size, high electronegativity and low bond energy is more reactive than ClF3 but bond energy of C-Cl bond is higher than Cl-F bond, therefore ClF3 is more reactive than Cl2. ii) The structure of H3PO2 ,H3PO3&H3PO4 are as follows: The acids which contain P-H bond, have strong reducing properties. Hypophosphorous acid (H3PO2) contains two P-H bonds, whereas orthophosphorous acid (H3PO3) has one P-H bond. Hence, H3PO2 is a stronger reducing agent than H3PO3. iii) Since, pentavalent metal ion has higher polarizing power than trivalent metal ion. Thus, PCl5 is more covalent than PCl3. b) i) ii) XeF4 + O2F2 XeF6 + O2 XeF4 + SbF5→ [XeF3]+ [SbF6]- 77 CHAPTER – 8 d AND f BLOCK ELEMENTS 01 MARK Q1. Why EO values for Mn, Ni and Zn are more negative than expected? Ans. Negative values for Mn2+ and Zn2+ are related to the stabilities of half-filled and fully filed configurations respectively. For Ni2+, EO value is related to the highest negative enthalpy of hydration. Q2. Transition metals show high melting points. Why? Ans. The high melting points of transition metals are attributed to the involvement of greater number of electrons in the interatomic metallic bonding from (n – 1)d orbitals in addition to ns electrons. Q3. When Cu2+ is treated with KI, a white precipitate is formed. Explain the reaction with the help of chemical equation. Ans. Reduction of Cu 2+ to Cu+ takes place due to reaction with I- ions. This is given by the equation 2Cu2+ + 4I- Q4. Cu 2I2 + I2 Out of Cu2Cl2 and CuCl2, which is more stable and shy? Ans. CuCl2 is more stable than Cu 2Cl2. Greater stability of Cu2+ (aq) is due to more negative o 2+ + ∆hyd H of Cu (aq) than that of Cu (aq). Q5. Although fluorine is more electronegative than oxygen, but the ability of oxygen to stabilize higher oxidation states exceeds that of fluorine. Why? Ans. The electronic configuration of fluorine is 1s22s22p 5. Thus it can form only one bond as it has only one unpaired electron. Electronic configuration of oxygen is 1s22s22p63s23p4. It may be mentioned that oxygen also has vacant d-orbitals along with two 3p orbitals containing single electrons.Thus oxygen has greater ability to stabilize higher oxidation states. Q6. Although +3 oxidation state is the characteristic oxidation state of lanthanoids but cerium shows +4 oxidation state also. Why? Ans. The electronic configuration of Ce is -4f15d16s2. Usually 5d 1 and 6s2 electrons are lost by the lanthanoids in their reaction i.e., they exhibit +3 oxidation states. But Ce exhibit +4 oxidation state also because it gains extra stability by losing 4f1 electron because it will give rise to completely filled orbitals. Q 7. Explain why does the colour of KMnO4 disappear when oxalic acid is added to its solution in acidic medium. 78 Ans. KMnO4 acts as oxidizing agent. It oxidises oxalic acid to CO2 and is itself changed to Mn2+ ions which are colourless. The reaction is given as under:5 +2 + 16 (Coloured) →2 +8 + 10 (Colourless) Q8. The halides of transition element becomes more covalentwith increassing oxidation state of the metal. Why? Ans. As the oxidation state of the element increases, its charge increases. According to Fajan’s Rules, as the charge of the metal ion increases covalent character increases because the positively charged cation attracts the electron cloud on the anion towards itself. Q9. While filling up of the electrons in the atomic orbitals, the 4s orbital is filled before the 3d orbital but reverse happens during the ionization of the atom. Explain why? Ans. Filling of electrons is explained by (n+l) rule. Electrons will go to that level where the value of (n+l) is minimum. For 3d orbital n + 1 = 3 + 2 = 5 For 4s orbital n + 1 = 4 + 0 = 4 So the electrons will be filled first in 4s orbital and then in 3d orbital. When it comes to ionization of the atom, we need to compare the ionization enthalpy. Removal of electron from 4s requires less energy than removal of electron from 3d orbital. Q10. Reactivity of transition elements decreases almost regularly from Sc to Cu. Explain. Ans. Reactivity of an element is dependent on the value of ionization enthalpy. In moving from Sc, the first element to Cu, the ionization enthalpy increases regularly. Therefore, the reactivity decreases as we move from Sc to Cu. 02 MARKS Q 1. When a brown compound of manganese (A) is treated with HCl it gives a gas (B). The gas taken in excess reacts with NH3 to give an explosive compound (C). Identify compounds A, B and C. Ans. The compounds A, B and C are as follows:A = MnO2 B = Cl2 C = NCl3 The reaction are explained as under:+4 → + [A] 3 +2 [B] + → +3 Excess [C] Q2. Complete the following chemical equations:(i) ( )+ ( )+ ( )→ 79 (ii) Ans. (i) 8 ( )+ ( )+3 8 (ii) ( 2 ( )+ ( +6 )+6 +6 ( )+ )→ ()→ +2 ( ) + 14 ( )→ +7 Q3. When orange solution containing ion is treated with an alkali, a yellow solution is formed & when H+ ions are added to yellow solution, an orange solution is obtained. Explain why does this happen? Ans. This is due to the following inter conversion:OHCr2O72 CrO42 H+ Dichromate (orange) Chromate (Yellow) Q4. Assign a reason for each of the following observations:(i) The transition metals (with the exception of Zn, Cd and Hg) are hard and have high melting and boiling points. (ii) The ionization enthalpies (first and second) in the first series of the transition elements are found to vary irregularly. Ans. (i) The transition metals (except Zn, Cd and Hg) are hard and have high melting and boiling metallic points. They show strong metallic bonding. Greater the number of valence electrons stronger is the resultant bonding. (ii) Ionisation enthalpy increases with increase in nuclear charge along each series. However, its value for Cr is lower because of the absence of any change in d-configuration (d5) and the value for Zn is higher because it represents an ionization from the completely filled 4s level.. Q5. Assign reason for each of the following:(i) Transition elements exhibit paramagnetic behavior. (ii) Co 2+ is easily oxidized in the presence of a strong ligand. Ans. (i) Transition metal ions have unpaired electrons in d-orbitals (d1 to d9). Paramagnetism arises due to the presence of unpaired electrons. Therefore, they exhibit paramagnetic behaviour. (ii) In Co 2+, electronic configuration is 3d 7 there is one unpaired electron even after pairing occurs in the presence of a strong ligand. Hence, Co 2+ is oxidized to more stable Co 3+ 80 3 marks Q1. (a) Transition metals can act as catalysts because these can change their oxidation state. How does Fe(III) catalyse the reaction between iodide and persulphate ions? (b) Mention any three processes where transition metals act as catalysts. Ans. (a) Reaction between iodide and persulphate ions is: 2 I S 2 O82 Fe ( III ) I 2 2SO42 Role of Fe(III) ions:2Fe3 2I 2Fe 2 I 2 2 Fe 2 S 2 O82 2 Fe3 2SO42 Q2. (b) (i) Vanadium (V) oxide in contact process for oxidation of SO2 to SO3. (ii) Finely divided iron in Haber’s process in conversation of N2 and H2 to NH3. (iii) MnO2 in preparation of oxygen from KClO3. Assign suitable reasons for the following:(i) The Mn2+ compounds are more stable than Fe2+ towards oxidation to their +3 state. (ii) In the 3d series from Sc (Z=21) to Zn (Z=30), the enthalpy of atomization of Zn in the lowest. Ans. (iii) Sc3+ is colourless in aqueous solution whereas Ti3+ is coloured. (i) Electronic configuration of Mn2+ = [Ar] 3d5. Electronic configuration of Fe2+ = [Ar] 3d 6. It is known that half-filled and fully-filled orbitals are more stable. Therefore, Mn in (+2) state has a stable d5 configuration. Due to this Mn2+ is resistant towards oxidation to Mn3+. Also Fe2+ has 3d6 configuration and by losing one electron, its configuration changes to a more stable 3d5 configuration. Therefore, Fe2+ easily gets oxidized to Fe3+ oxidation state. (ii) Transition metals have high enthalpies of atmoisation due to strong metallic bonding and addition covalent bonding. Metallic bonding is due to their smaller size while covalent bonding is due to d-d overlapping. But zinc has completely filled d-orbitals so it cannot overlap with other atom to show d-d overlapping. (iii) Sc3+ is colourless because of d o – orbitals configuration here d-d transition is forbidden. But in Ti3+ due to presence of one electron in d-orbital (d1) d-d transition are possible (allowed). Q3. How would you account for the following? (i) Cr2+ is reducing in nature while with the same d-orbital configuration (d4) Mn3+ is an oxidizing agent. 81 (ii) In a transition series of metals, the metal which exhibits the greatest number of oxidation states occurs in the middle of the series. (iii) Ans. What is misch metal. Mention its two important uses. (i) Cr2+ is a reducing agent as its configuration changes from d4 to d 3, the latter having a half-filled t2g level. On the other hand, the change from Mn2+ to Mn3+ results in the halffilled (d5) configuration which has extra stability. Thus, it behaves as oxidizing agent. (ii) This is due to large number of unpaired electrons in d-orbitals in the middle of the series. (iii) Misch metal consists of lanthanoids (95%) iron, traces of S, C, Ca & Al. It is used to produce bullets. It is used in shells & lighter flints. Q4. What is meant by disproportionation? Give two examples of disproportionation reactions in aqueous solution. Ans. The disproportionation reactions are those in which the same substance gets oxidized as well as reduced. When a particular oxidation state becomes less stable relative to other oxidation states, one lower and one higher, it undergoes disproportionation. e.g. Mn (VI) becomes unstable relative to Mn (VII) and Mn (IV) in acidic solution. 6 7 4 3 Mn O42 4 H 2 Mn O4 Mn O2 2 H 2 O Many copper (I) compounds are unstable inaqueous solution and undergo disproportionation. 2Cu Cu 2 Cu Q5. Account for the following:(i) Oxidising power in the series VO 2 Cr2 O72 MnO4 . (ii) Actinoid contraction is greater from element to element than lanthanoid contraction. Ans. (iii) Oxoanions of a metal show higher oxidation state. (i) The oxidizing power of the species varies in the order:- VO 2 Cr2 O72 MnO4 This is explained on the basis of increasing stability of the lower species to which they are reduced. (ii) Actinoid contraction is greater from element to element than lanthanoid contraction due to poor shielding of 5f electrons compared to that of 4f electrons in lanthnoids. 82 (iii) Oxoanions of a metal shows higher oxidation state due to the ability of oxygen to form multiple bonds and its high electronegativity. 05 MARKS 1) When a chronite ore (A) is fused with sodium carbonate in free excess of air and the product is dissolved in water, a yellow solution of compound (B) is obtained. After treatment of this yellow solution with sulphuric acid, compound (C) can be crystallised from the solution. When compound (C) is trated with KCl, orange crystals of compound (D) crystallise out. Identify (A) to (D) and also explain the reactions. Ans The compounds A, B, C and D are given as under: A = FeCr2O4 B = Na2CrO4 C=Na2Cr2O72H2O D = K2Cr2O7 The reactions are explained as under : 4FeCr2O4 + 8Na2CO3 + 7O2→8Na2CrO4 + 2Fe2O3 + 8CO2 (A) (B) 2NaCrO4 + 2H+→ Na2Cr2O7 + 2Na+ + H2O Na2Cr2O7 + 2KCl → (C) K2Cr2O7 + 2NaCl (D) 2) When an oxide of manganese (A) is fused with KOH in the presence of an oxidizing agent and dissolved in water, it gives a dark green solution of compound (B). Compound (B) disproportionates in neutral or acidic solution to give purple compound (C). An alkaline solution of compound (C) oxidises potassium iodide solution to a compound (D) and compound (A) is also formed. Identify compounds (A) to (D) and also explain the reactions involved. Ans. The compounds (A), (B), (C) and (D) are given as under : (A) MnO2 (B) K2MnO4 (C) KMnO4 (D) KlO3 The reactions are explained as under: 2MnO2 + 4KOH + O2→ 2K2MnO4 + 2H2O (A) (B) 3MnO42- + 4H+→ 2MnO-4 + MnO2 + 2H2O (C) 2MnO-4 + H2O + KI → 2MnO2 + 2OH- + KIO3 (A) (D) 3) A violet compound of manganese (A) decomposes on heating to liberate oxygen and compounds (B) and (C) of manganese are formed. Compound (C) reacts with KOH in the presence of potassium nitrate to give compound (B). On heating compound (C) with conc. H2SO4 and NaCl, chlorine gas is liberated and a compound (D) of manganese along with other products is formed. Identify compounds (A) to (D) and also explain the reactions involved, Ans The compounds A, B, C and D are given as under: A = KMnO4 B = K2MnO4 C = MnO2 D = MnCl2 The reactions are explained as under KMnO4 K2MnO4 + MnO2 + O2 (A) (B) (C) 83 MnO2 + KOH + O2→2K2MnO4 + 2H2O MnO2 + 4NaCl + 4H2SO4→ MnCl2 + 2NaHSO4 + 2H2O + Cl2 4) On the basis of Lanthanoid contraction explain the following – i) Nature of bonding in La2O3 & Lu2O3 ii) Trends in the stability of oxosalts. iii) iv) v) Stability of the complexes of lanthanoids. Radii of 4dand 5dblock elements. Trends in acidic character of lanthanoid oxides. A 4) (i) As the size decreases covalent character increases. ThereforeLa2O3is more ionic andLu2O3is more covalent. 5) (ii) (iii) (iv) (v) As the size decreases from La to Lu, stability of oxosalts alsodecreases. Stability of complexes increases as the size of lanthanoids decreases. Radii of 4dand 5dblock elements will be almost same. Acidic character of oxides increases from La to Lu. i) Complete the following equations: a) Cr2O72 + 2OH-→ MnO4 + 4H+ + 3e-→ Account for the following : a) Zn is not considered as a transition element. b) Transition metals form a large number of complexes. c) The Eo value for the Mn3+ / Mn2+ couple is much more positive than that for Cr3+ / Cr2+ couple. b) ii) Ans i) a) Cr2O72- + 2OH-→ 2CrO42-+ H2O b) MnO4- + 4H+ + 3e-→ MnO2 + 2H2O ii) a) Zn has completely filled d-oritals in its atomic as well as in its common oxidation states (Zn2+ state). Therefore it is not regarded as transition element. b) Due to the comparatively smaller size of the metal ions, the high ionic charges and the availability of vacant d-orbitals for bond formation, transition metals form a large number of complex compounds. c) Mn3+ (3d4) is less stable than Mn2+ (3d 5), which has stable half-filled configuration. Cr3+ has stable t2g configuration, therefore, Cr3+ cannot be reduced to Cr2+. So, Eo vaule for the Mn3+ / Mn2+ couple is much more positive. 84 CHAPTER – 9 COORDINATION CHEMISTRY 01 MARK Q1. Define ambidentate ligand with an example ? A1. Ligand that can ligate through two different atoms eg NO - ion can ligate through N or through oxygen. Q2. Draw structures of geometrical isomers of [Fe (NH )2 (CN)4]NH A2. NH NH NC NC CN Fe NC Fe CN NC CN NH CN cis isomer trans isomer Q3. Calculate the magnetic moment of Mn A3. Mn Q4. Give the IUPAC name of [NiCl ]2- A4. Tetrachloridonickelate (II) Q5. Out of [Fe(C O )3]3- and [FeCL ]3-, which is more stable and why? A5. [Fe(C O )3]3-, as it is a chelate complex. 3d 45 μ= ion? ( + 2) = 5(7) = √35BM 02 MARKS Q1. What is spectro chemical series? What is the difference between a strong field ligand and a weak field ligand? A1. Arrangement of ligands in increasing orde of field strength is called spectro chemical series. A strong field ligand produces greater splitting in the d-orbitals, the complex is likely to be low – spin. A weak field ligand, on the other hand produces lesser splitting, and complex is likely to be high – spin. Q2. Using valence bond, approach, deduce the shape and magnetic character of [Co (NH ) 6]3+ ion. (Atomic no of Co = 27) 85 A2. 27Co CO 3d74s 3d64s o CO 3d 4s 4p Co3+ assumes d2sp3 hybridization and accepts six pairs of e-from 6NH moleculesShape – octahedral Magnetic nature – Diamagnetic Q3. How is coordination chemistry useful in medicine? Where is Wilkinson’s catalyst used? A3. Chelates are being used to beat metal poisoning. E.g EDTA is used to treat lead poisoning. Wilkinson’s catalyst [(Ph P)3 Rh Cl] is used for hydrogenation of alkenes. Q4. Write the structure of the complex Pentaamminenitrito – N – Cobalt (III). Which isomerism can this complex exhibit? A4. [Co (NH )5 (NO )] 2+, Linkage isomerism. Q5. A solution of [Ni(H2O)6]2+ is green , but a solution of [Ni(CN)4]2- is colorless. Why? A5. Ni 3d 8 4 s o H O is a weak ligand, so high-spin complex is formed (sp3d2), leaving two unpaired e-. CN- being a strong ligand, produces a low spin square planar (dsp2)complex leaving no unpaired e-. 03MARKS Q1. List the postulates of Werner’s Theory for coordination compounds on the basis of the theory, assign primary and secondary valencies to Pd in [Pd (NH )4]Cl2 A1. (a) In coordination compounds metals show two types of linkages (valences) – primary and secondary. (b) Primary valencies are ionisable and satisfied by –ve ions. (c) Secondary valencies are non-ionisable and satisfied by neutral or –ve ions. Secondary valency is equal to coordination number and is fixed for a metal. (d) The ions/groups bound by secondary linkages to the metal having characteristic spatial arrangements corresponding to different coordination numbers. [Pd (NH )4]Cl2 Primary valence - +2 Secondary valence – 4 86 Q2. Discuss the nature of bonding in metal carbonyls? A2. The M-C bond in metal carbonyls possess both s and p character. M-C sigma bond is formed by donation of lone pair of e on the carbonyl carbon into a vacant orbital of the metal. The M-C π bond is formed by donation of a pair of e from a filled d-orbital of metal into the vacant anti bonding π∗ orbital of CO. The metal – ligand bonding creates a synergic effect which strengthens the bond between CO and the metal Q3. Explain the following :- A3. (a) Low spin octahedral complxes of Ni aren’t known. (b) π- complexes are known only for transition elements. (c) CO is a stronger ligand than NH for many metals. (a) For octahedral complex, coordination number should be 6. Since Ni has 8 [Ar]3d 4s configuration, i.e only one d orbital is empty, it can form only high spin complexes. (b) Transition metals have empty d –orbitals to accommodate ligands. (c) Q4. donated by Due to small size and ability to form π-bonds alogwith normal sigma bonds. For the complex [Fe(en)2 Cl2]Cl, identify the following :(i) Oxidation number of iron. (ii) Hybrid orbitals and shape of the complex. (iii) Magnetic behaviour of the complex. (iv) Number of its geometrical isomers. (v) Whether there may be optical isomers (vi) Name of the complex. 87 A4. (i) x + 2(0) + 2 (-1) + (-1) = 0 oxidn no +3 (ii) d sp octahedral geometry (iii) paramagnetic (1 unpaired e-) (iv) Two geometrical isomers (CIs, trans) (v) Yes, due to polydentate ligends (vi) Dichloridobis – (ethane -1,2- diamine) iron (III) chloride. Q5. What will be correct order of wavelengths of absorption in the visible region for the following :[Ni(NO )6]4- , [Ni(NH )6]2+, [Ni(H O)6]2+ A5. The strength of ligands according to sepctrochemical series NO ->NH >H O Therefore splitting produced bythem in d-orbitals will also be in the same oreder (or O Will be maximum in NO -, followed by NH and H O). Hence , energy absorbed will be in the same order. Since E α , wavelength absorbed will be in the opposite order. i.e [Ni(NO )6]4-< [Ni(NH )6]2+< [Ni(H O)6]2+. 88 CHAPTER – 10 HALOALKANES AND HALOARENES 1 Mk Questions: 1) Write the IUPAC name of the following compound : CH3 H3C – C – CH2Cl CH3 Ans 1 – Chloro – 2, 2 – dimethylpropane 2) Ans Draw the structure of 2-bromopentane H3C – CH2 – CH2 – CH – CH3 Br 3) Write a chemical reaction in which iodide ion replaces the diazonium group in a diazonium salt. N2+ Cl- + KI(aq) . > Ans I + N2 + KCl 4) Out of CH3 CH Cl CH2 CH3 and CH3 CH2 CH2 CH2 Cl which one is hydrolysed more easily by aq. KOH? Ans CH3 CH – CH2 CH3 , as it’s a secondary halide | Cl 5) How can you convert methylbromide to methylisocyanide in a single step? Ans CH3Br . > CH3NC + Ag Br 2 Mks Questions 1) Ans What are ambident nucleophiles? Explain with an example. Ambident nucleophiles have two nuclophilic sites through which they can attack. Eg. [ΘC N ↔: C = NΘ] linking through C results in alkyl cyanides and through N results in isocyanides. 2) Write chemical equations when i) ethylchloride is treated with aq.KOH ii) chlorobenzene is treated with CH3COCl in presence of anhyd.AlCl3. 89 Ans i) C2H5Cl . > C2H5OH + KCl COCH3 ii) >Cl Cl 2-chloroacetophenone 3) Cl + HCl 4-chloroacetophenone Write the mechanism of the following reaction: N – BuBr + KCN Ans + H3COC , > n – BuCN CN- is an ambident nucleophile it can attack through C and N C-C bond is stronger than C-N bond, cyanide is formed. 4) Ans 5) Ans Give reasons: a) The order of reactivity of haloalkanes is RI>RBr>RCl b) Neopentyl chloride, (CH3)3C – CH3 Cl doesn’t follow SN2 mechanism. a) Larger the size of halogen atom, weaker the C-X bond, R group remaining same i.e; bond strength follows the order R-I <R-Br<R-Cl, hence reactivity is opposite. b) R – group being bulky, provides steric hindrance to the incoming nucleophile. Why does Wurtz reaction fail in case of tertiary halides? 3o halides prefer to undergo dehydrohalogenation in presence of a base like Na metal. CH3 CH3 | (CH3)3 C-Na+ + H – CH2 – C – Br > (CH3)3 CH + CH2 = C – CH3 | CH3 3 Mk Questions: 1) Ans Write a chemical test to distinguish between a) Chlorobenzene and benzyl chloride b) Chloroform and carbontetrochloride c) n-propyl bromide and isopropyl bromide a) Benzyl chloride gives a white ppt with AgNO3 Soln., chlorobenzene does not. b) Chloroform gives offensive smelling isocyanides, on heating with 1o amine and KOH, CCl4 doesn’t give this test. c) Isopropyl bromide, on treating with aq.KOH, gives Propan-2-ol which gives a yellow ppt. on heating wiith NaOH and I2 n- propyl bromide doesn’t answer this test 90 2) Complete the following reactions: i) ii) iii) Ans i) ii) iii) 3) Give reasons: a) p-dichlorobenzene has higher melting point than o- and m- isomers. b) Haloarenes are less reactive than haloalkanes towards nucleophilic substitution reactions. c) Alkyl chloride gives alcohol with aq.KOH, but alkene with alc. KOH. Ans a) p-isomer has more symmetric structure, so molecules can pack closely. b) The C-X bond has partial double – bond character due to resonance. Also, in haloalkanes, X is bonded to sp3 – C while in haloarenes, X is bonded to sp2 –C. sp3-C is less electronegative, releases electrons to halogen more easily. c) In aq.solution, KOH is almost completely ionized to give OH- ions, which being a strong nucleophile, gives a substitution product. An alcoholic soln. of KOH contains alkoxide ions (RO-), which being a strong base, preferentially eliminates an HCl molecule to form alkenes. 4) Convert: a) Ethene to ethanol b) Chlorobenzene to toluene c) Chlorobenzene to diphenyl Ans a) b) c) CH2 = CH2 + H2O . . > CH3CH2OH 91 5) What are enantiomers? Identify the asymmetric carbons in the following molecule:a b c d HOOC CH (OH) – CH (OH) COOH Ans Enantiomers are non-superimposable mirror images of an optically active compound. b and c are asymmetric carbon atoms as they are bonded to 4 different groups. 6) Write short notes on : a) Gatterman reaction b) Wurtz reaction c) Peroxide effect or Kharasch effect Ans a) Aromatic 1o amines produce benzene diazonium salts with HNO2 (produced in situ) at 273 K. Chloro arenes and bromoarenes can be prepared using Cu/HCl or Cu/HBr from these diazonium salts. b) Alkyl halides, when treated with sodium in presence of dry ether, produce an alkane with even number of C atoms. 2 R-X + Na E.g. 2 CH3 Br + Na > R _ R + 2NaX . > CH3 – CH3 + 2NaBr If a mix of two different alkyl halides is taken, then a mixture of alkanes is obtained which is difficult to separate. E.g. CH3 Br + C2H5Br > CH3-CH3 + C2H5 + CH3-C2H5 c) The addition of Hydrogen bromide to unsymmetrical alkenes in presence of a peroxide takes place in such a way that H goes to that C which has lesser H atoms and Br goes to the C with greater number of H atoms. CH3 – CH = CH2 + H Br ( ) > CH3 – CH2 – CH2 Br This rule doesn’t apply to addition of HCl or HI 7) Explain the following with an example each a) Swarts reaction b) Finkelstein reaction c) Hunsdiecker reaction Ans a) Flourination of hydrocarbons directly with F2 occurs explosively due to the large amount of energy released. Hence, they can be conveniently prepared indirectly by halogen exchange with chloro and bromoalkanes. CH3Br + AgF → CH3F + AgBr 2 C2H5Cl + .Hg2F2 → 2C2H5F + Hg2Cl2 92 b) Iodoalkanes can be easily prepared from chloroalkanes or bromoalkanes by heating with NaI in acetone. R-Cl + NaI R –Br + NaI > R – I + NaCl > R – I + NaBr c) Bromoalkanes can be prepared by refluxing the silver salt of a carboxylic acid with Bromine in CCl4. CH3CH2 COOAg + Br2 > CH3CH2Br + CO2 + AgBr The yield of alkyl bromide is 1o>2o>3o 8) Explain why. a) Vinyl chloride is unreactive in nuceophilic substation reaction b) 3-Bromocyclohexene is more reactive than 4-Bromocyclohexene in hydrolysis with aq.NaOH c) tert-butyl chloride reacts with aq.NaOH by SN1 mechanism, while n-butyl chloride reacts with aq.NaOH by SN2 mechanism. Ans a) Vinyl chloride is unreactive in nucleophikic substitution reactions due to reasanance which results in a partial double bond character of C-Cl bond which is difficult to break. b) 3-bromocyclohexene forms allyl carbocation which is more stable than carbocation formed by 4-Bromocycohexene c) Tertiary carbocation is stable, so tert-butyl chloride follows SN1 mechanism. Nbutyl chloride would form 1o carbocation which isn’t that stable. Hence it undergoes SN2 mechanism through formation of transition state. 9) Identify the products A and B formed in the following reaction: CH3CH2 CH = CH –CH3 + HCl → A + B Ans Since both the doubly-bonded C atoms have same number of H atoms, Markonikov’s rule becomes irrelevant. So, products will be → CH3 – CH2 – CH – CH2 – CH3 CH3CH2 CH2CH CH3 | (A) and | (B) Cl Cl 3-chloropentane 2-choropentane 10) Ans What happens when: a) n-butyl chloride is treated with alc.KOH b) Bromobenzene is treated with Mg in presence of dry ether. c) Chlorobenze is subjected to hydrolysis. a) CH3CH2 CH2CH2Cl - alc.KOH → CH3-CH2-CH=CH2+KCl + H2O But – 1 – ene 93 a) Br + Mg a) >Mg Br Cl2 + aq.NaOH → no reaction. CHAPTER – 11 ALCOHOLS, PHENOLS AND ETHERS 1 Mark Questions 1) A) Write the IUPAC name of CH3 CH= CH – CH- CH2 – CH3 | OH Hex – 4 –en – 3 –ol 2) A) Why is the boiling point of C2H5OH higher than that of CH3OH ? Due to more number of C atoms, van der waal’s forces increase. 3) A) Name the reagent used in bromination of phenol to 2,4, 6 – tribromophenol. Aqueous solution of bromine (Bomine water) 4) A) Name the alcohol used to prepare the ester CH3 – C OOCH(CH3)CH3 Propan-2-ol CH3-CH(OH)-CH3 5) A) Which is more volatile – o-nitrophenol or p-nitrophenol? o-nitrophenol,, as it has intramolecular H-bonding, whereas p-nitrophenol has intermolecular H-bonding. O-Nitrophenol with intra molecular hydrogen bonding 94 02 MARKS 1) A) How will you convert: a) Propene to propan – 2 – ol b) Ethyl chloride to ethanol a) CH3-CH = CH2 H2O/H+ CH3 – CH – CH3 | OH Propan – 2 – ol b) 2) CH3 -CH2- Cl aq – NaOH CH3 CH2 OH [o] CH3 - CHO ----------- ------- Ethanal CrO3/PCC Write the mechanism for the following reaction: CH3CH2OH A) ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ CH3CH2Br + H2O 1 o alc undergoes the reaction by SN2 mechanism Protonated Alcohol 3) A) How would you obtain i) Picric acid from phenol? ii) 2- Methylpropan-2-ol from 2- Methylpropene? i) 95 ii) 4) A) 5) A) Give reasons : a) Boiling points of alcohols decrease with increase in branching of the alkyl chain. b) Phenol does not give protonation reaction readily a) Because increase in branching of the alkyl chain reduces surface area, so intermolecular forces of attraction decrease. b) Because the lone pair on oxygen is delocalized over the benzene ring due to resonance, hence not available for protonation easily. Explain why cleavage of phenyl alkyl ether with HBr always gives phenol and alkyl bromide. Due to resonance, the O-C bond in phenyl alkyl ether has a partial double bond character, hence it is difficult to break. Also, phenoxide ion is stabilized by resonance, hence we get phenol and alkyl bromide. 03 MARKS 1) A) Write a short note on Williamson’s Synthesis. An alkl halide, on treating with a suitable sodium alkoxide gives an ether RONa + RX - R– O – R + NaX Symmetric and unsymmetric ethers can be prepared by this method Aryl alkyl ethers can be prepared as shown But there is no reaction if arylhalide and sodium alkoxide are taken Best yields of unsymmetrical ethers are obtained when alkyl halides are primary with 2o and 3 o alkyl halide, dehydrohalogenation occurs to give alkene. 96 2) Convert: a) Methyl magnesium chloride --2-methyl propan-2-ol b) Benzyl chloride --- Benzyl alcohol c) Phenol -- benzoquinone A) a) propanone b) c) 3) A) Give reasons: a) The boiling point of ethanol is higher than methoxymethane. b) Phenol is more acidic than ethanol c) O- and p-nitrophhenols are more acidic than phenol. a) Ethanol has intermolecular H-bonding while methoxymethane only has dipoledipole forces. b) In phenols, O atom acquires a partial positive charge due to resonance, this weakens O-H bond, release of H+ is easy. Also, phenoxide ion is stabilised by resonance. On the other hand, alkoxide ion is destabilized due to the +I effect of alkyl group. c) -NO2 group, being electron-withdrawing, stabilizes the o- and p- nitrophenoxide ions. 97 4) A) Describe the mechanism of hydration of ethene to yield ethanol. Protonation of Alkene Nucleophilic attack by water on the carbocation Deprotonation 5) Give equations for Reimer Tiemann reaction and Kolbe’s reaction. A) Reimer Tiemann reaction – Phenol reacts with chloroform in the presence of alkali to give o- and p-hydroxy aldehyde. Kolbe’s reaction – Phenoxide ion generated by treating phenol with NaOH undergoes electrophilic substitution with CO2 98 6) A) 7) A) 8) Arrange the following compounds in increasing order of their acid strength. Propan-1-ol, 2, 4, 6-Trinitrophenol, 3 – Nitrophenol, 3, 5- Dinitrophenol, phenol, 4-Methylphenol, 4-Methoxyphenol Propan-1-ol <4-Methoxyphenol<4-Methylphenol<phenol< 3-Nitrophenol <3, 5 – Dinitrophenol<2, 4, 6 – Trinitro phenol Propan – 1 ol has R group, which destabilizes the propoxide ion due to +I effect, hence weaker than all phenols. All e- donating groups reduce acidity of phenols by destabilizing phenoxide ion and ewithdrawing groups increase acidity. Methoxy group is stronger e- - donating groups compared to methyl group. Larger the number of e- withdrawing groups, greater the acidity. 2, 4, 6 – Trinitrophenol is strongest, followed by 3, 5 – Dinitrophenol and 3 – Nitrophenol. Give a simple chemical test to distinguish between: a) Phenol and cyclohexanol b) Propan-2-ol and Benzyl alcohol c) Propan-1-ol and 2-Methylpropan-2-ol a) Phenol produces a deep violet colour with neutral FeCl3 solution., cyclohexanol doesn’t. b) Propan-2-ol gives yellow ppt of iodoform with NaOH and I2, benzyl alcohol doesn’t. c) On treating with conc. HCl and anhydrous ZnCl2 at room temperature, 2methylpropan-2-ol produces turbidity immediately while propan-1-ol doesn’t produce turbidity at room temperature. ( ) Give the mechanism for the reaction: 2C2H5OH ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ C2H5O C2H5 A) Protonation of Alcohol Nucleophillic attack by Alcohol Deprotonation 99 9) An organic compound A (C6H6O) gives a characteristic colour with aq. FeCl3 Solution. when (A) is treated with CO2 and NaOH at 410 K under pressure, it gives compound (B) which on acidification gives compound (C). Compound (C) reacts with acetyl chloride to give (D), which is a popular pain killer. Deduce the strutures of A,B,C and D and explain all reactions A) (A) C6H6O must be phenol Phenol gives sodium salicylate with CO2 which on acidification gives salicylic acid. Salicylic acid forms aspirin with acetyl chloride. 10) A (a) Explain hydroboration-oxidation reaction with an example Diborane reacts with alkenes to give trialkyl boranes as addition product. This is oxidized by H2O2 to alcohol, in the presence of aq.NaOH (b) Identify A B C in the following reaction. CH3Br A B C ( ) ⎯⎯⎯ A ⎯⎯⎯⎯ B ⎯⎯⎯⎯⎯⎯⎯⎯ C CH3CN CH3CH2NH2 CH3CH2N2Cl (Unstable, so forms CH3CH2OH + N2 + HCl with Water 100 CHAPTER – 12 ALDEHYDES – KETONES AND CARBOXYLIC ACIDS 01 MARK 1) Write the structure of p-methylbenzaldehyde. A) 2) Complete the reaction : . + CH3COCl ⎯⎯⎯⎯⎯ A) Acetophenone 3) Arrange the following in increasing order of their acid strength (CH3)2 CHCOOH, CH3 CH2 CH(Br) COOH, CH3 CH(Br) CH2 COOH. A) (CH3)2 CHCOOH < CH3 CH(Br)CH2COOH<CH3CH2CH(Br)COOH 4) Write IUPAC name of A) 5 – Bromo -3- chlorobenzoic acid 5) A) Give a chemical test to distinguish between acetaldehyde and benzaldehyde. On heating with Na OH and I2, acetaldehyde gives a yellow ppt of iodoform, benzaldehyde doesn’t 101 02 MARKS 1) a) A) Convert: Ethanol to 1,2 Ethanediol a) (b) Phenol to acetophenone ( ) C2H5OH ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ CH2 = CH2 + H2O ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ Ethane – 1, 2 - diol b) 3) Give reasons: a) b) Monochloroethanoic acid is a weaker acid than dichloroethanoic acid. Benzoic acid is stronger than ethanoic acid A) a) Cl has – I effect, making release of protons easier. Also, it stabilises the carboxylate ion. Monochloroacetic acid has one Cl atom. While dichloroacetic acid has two Cl atoms. Hence it is a stronger acid. b) Benzene ring is e-withdrawing, while CH3 group is e- donating. Hence, release of protons is easier in benzoic acid than ethanoic acid. Also, the benzoate ion is much more stable than ethanoate ion due to e- - withdrawing benzene ring. 4) Describe Cannizaro’s reaction with example. A) Aldehydes without α-H undergo self-oxidation and reduction (disproportionation) on treatment with conc. Alkali. One molecule is reduced to alcohol while the other is oxidized to salt of corresponding carboxylic acid 102 5) Complete the equations A) a) a) ( CH3CHO ⎯⎯⎯ CH3CH2OH . b) CH3COCH3 ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ b) CH3 CH2CH3 03 MARKS 1) Explain Aldol condensation with an example. A) Aldehydes or ketones having at least one α-H undergo condensation to produce β-hydroxy aldehyde (aldol) in the presence of a dil. Base. . ⎯⎯⎯⎯⎯⎯ ∆( ) CH3CH(OH)CH2CHO ⎯⎯⎯⎯⎯⎯⎯ CH3CH=CH-CHO 3-Hydroxybutanal Eg. If two different molecules of aldehydes and/or ketones undergo aldol condensation, the reaction is called CROSS ALDOL REACTION. Eg.2CH3CHO 2) Give reasons a) There are two –NH2 groups in semicarbazide, but only one is involved in the formation of semicarbazones: b) Cyclohexanone forms cyanohydrins in good yield, but 2,2,6Trimethylcyclohexanone doesn.t. c) Aldehydes are more reactive than ketones towards nucleophilic addition. A) a) One of the NH2 groups is involved in resonance, hence it cannot act as a nucleophile. Other one is free, it can act as a nucleophile. b) CN-can attack easily without any steric hindrance in cyclohexanone. Lot of CH3 groups offer too much steric hindrance. 103 c) Aldehydes have one R group and ketones have two.Two R groups cause more steric hindrance and greater +I effect reduces the electrophilicity of the carbonyl carbon. 3) How will you convert: a) Propanone to propane b) Benzoyl chloride to benzaldehyde c) Ethanal to but-2-enal A) a) b) c) 4) How will you bring about the following conversions? a) Benzene to Benzophenone b) Propyne to acetone c) p-nitrotoluene to p-nitrobenzaldehyde Ans 4 a) b) 104 c) 5) An organic compound A contains 69.77% carbon, 11.63% hydrogen and rest oxygen. The molecules mass of the compound is 86. It doesn’t reduce Tollen’s reagent but forms an addition compound with sodium hydrogen sulphite and gives positive iodoform test. On vigorous oxidation, it gives ethanoic abd propanoic acids. What is the structure of compound A? Ans Element C H O % Molar Mass 69.77 12 11.63 1 18.6 16 Moles 5.81 11.63 1.16 Simple ratio 5 10 1 Empirical formula C5H10O Empirical formula mass = 5(12) + 10(1) + 1(16) = 86 n= =1 Moecules formula C5H10O Since it doesn’t give Tollen’s test but gives positive iodoform test. It is a methyl ketone i.e it has – C – CH3 group. Since on oxidation it gives ethanoic and propanoic acid, it must be pentan-2-one. CH3 CH2 CH2 CO CH3→CH3COOH + CH3 CH2 COOH Pentan-2-one 05 MARKS 1) Ans a) i) ii) Draw the structures of: 5-Chloro-3methyl-pentan-2-one p-nitropropiophenone b) i) ii) iii) Give simple chemical tests to distinguish between: Ethanal and propanal Phenol and benzoic acid Benzaldehyde and acepopheone a) i) ii) 105 b) 2) i) CH3 CHO gives a yellow ppt of iodoform with NaOH/I2 (positive iodoform test) CH3CH2 CHO doesn’t answer this test. ii) Phenol produces deep violet colour with aq-neutral FeCl3 Soln., benzoic acid doesn’t answer the test. iii) Acetophenone produces yellow ppt on heating with NaOH+I2, benzaldehyde doesn’t. Identify A,B,C,D and E in the following sequence: A ⎯⎯ ⎯⎯⎯⎯ B ⎯⎯⎯⎯⎯⎯⎯⎯ / C ⎯⎯⎯⎯⎯⎯⎯⎯⎯ D + E A) A → CH4, B → HCOONa, C → C6H5COOOCH, D → CHO E → C6H5COOH 3) An organic compound A (C3H2O) on treatment with Cu at 573 K gives B. B doesn’t reduce Fehling’s solution. But gives a yellow ppt. of compound C with I2/NaOH. Deduce the structures of A, B and C. A) Since compound B gives iodoform test, it must be having –CO-CH3 group i.e it is a ketone. Since B is obtained by dehydrogenation of A, A is a 2o alcohol A is R-CH – CH3 | OH Comparing with the given molecules formula, R is –CH3 A is CH3-CH-CH3 | OH Propan-2-ol The reactions are: O CH3-CH-CH3 > CH3-C-CH3 . > no reactions OH Acetone (A) (B) CH3 COCH3 + 3I2 + 4NaOH → CH3COONa + 3NaI+CHI3 + 3H2O (C) So, A = CH3CHCH3 Propan-2-ol | OH B = CH3COCH3 Propan-2-one C = CHI3 Iodoform 106 4) a) An organic compound A (C3H6O) is resistant to oxidation but forms a compound B (C3H8O) on reduction. B reacts with H Br to form a bromide ‘C’ which on treatment with alcoholic KOH forms an alkene D (C3H6). Deduce the structures A,B,C and D. b) Carboxylic acids contain a carbonyl group, but do not show the nucleophilic addition reaction like ketones or aldehydes. Why? Ans a) CH3-CO-CH3 . > CH3-CH-CH3 OH Propan-2-ol ‘B’ Acetone ‘A’ H Br CH3-CH=CH2< . CH3-CH-CH3 . ‘D’ Propene b) Br 2-Bromo propane ‘C’ This is due to resonance (for structures refer NCERT) The carboxyl carbon of the resonance hybrid is less positive and hence less electrophilic than carbonyl ‘C’ of aldehydes and ketones 5) a) Explain the following : i) Cyanohydrin ii) Acetal iii) 2,4 – DNP derivative b) Complete the following: i) CH2CH3 . > COOH ii) iii) Ans COOH > C6H5CHO > a) i) Formed addition of HCN to aldehydes or ketones C O H CN C OH CN Cyanohydrin ii) Formed on addition of two moles of alcohol to an aldehyde or ketone. C = O + 2 R OH DryHC lg as Acetal C OR OR + H2O 107 iii) Addition product of aldehydes or ketones with 2,4 – Dinitro phenyl hydrazine in weakly acidic medium. NO2 NO2 C = O + H2N NH b) NO2 i) ii) iii) C6H5CH = NNHCONH2 Benzaldehyde semicarbazone C = NNH NO2 108 CHAPTER – 13 AMINES 01 MARK 1) Ans Draw the structure of prop-2-en-1-amine H2C = CH – CH2 – NH2 2) Give a chemical test to distinguish between ethylamine and aniline. Ans They can be distinguished by Azo dye test. On treating with HNO2 (NaNO2+HCl) followed by alkaline soln. of 2-naphthol (Temp – O-5 oC) Aniline forms an orange dye, while ethylamine only gives ethanol and N2. 3) Arrange the following in increasing order of their solubility in water : C6 H5NH2, (C2H5)2NH, C2H5NH2 Ans C6H5NH2<(C2H5)2NH<C2H5NH2 4) Out of CH3NH2 and (CH3)3 N, which one has higher boiling point? Ans CH3NH2 due to its ability to form Intermolecular H-bonds (CH3)3N has no H atoms bonded to N so cannot participate in H bonding. 5) Why is the pkb of aniline greater than methylamine? Ana In aniline, lone pair of N is delocalized over the benzene ring. reducing basicity. In CH3NH2, +I effect of –CH3 group increases basicity of CH3NH2. 02 MARKS 1) Explain Gabriel Pthalimide synthesis. Ans It is used to prepare pure 1o amine. Aromatic 1 o amines can’t be prepared by this method because aryl halides do not undergo nucleophilic substitution with the anion formed by pthalimide 2) Ans Give reasons: a) Diazonium salts of aromatic amines are more stable than those of aliphatic amines. b) Amines are more basic than alcohols of comparable molecular masses. a) Resonance stabilization of diazonium salts of aromatic amines 109 b) N is less electronegative than O, lone pair of electrons are easily available Such resonance stabilization is not possible in diazonium salts of aliphatic amines. 3) Identify A and B in each of the following: a) C2H5Cl >A b) C6H5NH2 Ans >A a) A → C2H5CN, Prppanenitrile b) A → C6H5N2+ ClBenzene diazonium Chloride 4) Ans / >B > B B → C2H5CH2NH2 propanamine B→ Complete the reactions a) C6H5N2Cl + H3PO2 + H2O → b) C6H5NH2 + Br2 (aq) → a) C6H6. + N2 + H3PO3 + HCl b) 2, 4, 6 – Tribromoaniline 5) Ans b) Convert a) Nitrobenzene to phenol a) b) aniline to chlorobenzene 110 03 MARKS 1) Ans Give the structures of A, B, C in the following: a) CH3Br >A >B >C b) CH3COOH >A a) A → CH3CN, B → CH3CH2NH2, C → CH3CH2OH b) A → CH3CONH2, >B B → CH3NH2 >C C → CH3NC 2) Write short notes on : a) Coupling reaction b) Ammonolysis Ans a) Benzene diazonium salts react with e- rich aromatic compounds such as phenols and amines to form azo compounds, which are often coloured and are used as dyes. E.g b) An alkyl or benzyl halide, on reaction with ethanolic soln. of NH3 undergoes nucleophilic substitution reaction in which the halogen atom is replaced with –NH2 group. This cleavage of C-X bond by ammonia molecule is called AMMONOLYSIS. NH3 + R – X → RN+H3 XSubstituted ammonium salt RNH2 . > R2NH > R4N+X- > R3N Quarternary Ammonium salt Disadvantage : Mixture of 1 , 2 , 3 amines and quarternary ammonium salt is obtained. Order of reactivity of halides with amines is RI > R Br > RCl o 3) o Write the main products of the following reactions: a) CH3CH2NH2 b) a) CH3CH2OH b) SO2NH C2H5 d) CH3NH2 > SO2Cl + C2H5NH2 c) CH3CONH2 Ans o > 111 4) Ans Give reasons: a) Amines are basic while amides are neutral. b) CH3NH2 in water reacts with ferric choride to precipitate Fe (OH)3 c) Reactivity of –NH2 group gets reduced in acetanilide. a) In amines, R group increase e- density on N due to +I effects, whereas in amides, O R – C – group is electron withdrawing. b) CH3NH2 + H2O → CH3NH3 + + OHFeCl3 + 3OH- → Fe(OH)3(↓) + 3ClReddish - brown O c) The lone pair on N is involved in conjugation with R-C-group O OC6H5NH – C – CH3 ↔ C6H5 N+H = C – CH3 5) An aromatic compound ‘A’ on treatment with ammonia followed by heating forms compound ‘B’, which on heating with Br2 and KOH forms a compound ‘C’ (C6H7N). Give the structures of A, B and C and write the reactions involved. Ans ‘A’ must be benzoic acid A 6) Ans B C Convert i) Benzene diazonium chloride to nitrobenzene ii) Nitrobenzene to aniline iii) Aniline to benzonitrile N 2Cl HBF4 i) Sn ii) iii) NO2 NH2 HCl NaNO2 N 2 BF4 Cu NO2 112 7) Give reasons a) Although amine group is O, P- directly in aromatic substitution reaction, aniline on nitroation gives some amount of m-nitro aniline. b) Aniline doesn’t undergo Friedel-Craft reaction. Ans a) Nitration is usually carried out with conc. HNO3+ con.H2SO4 In presence of these acids, some amount of aniline undergoes protonation to form anilinium ion, so the reaction mixture consists of aniline and anilinium ion, -NH2 group in aniline is o- and p- directing and activating while the –NH3 group in anilinium ion is m-directing and deactivating. Nitration of aniline mainly gives p-nitroaniline (due to steric hindrance at o- position) nitration of anilinium ion gives mnitroaniline. b) Aniline being a lewis base reacts with Lewis acid AlCl3 to form a salt. Cl3. As a result N acquires a positive charge 8) C6H5NH2+Al- Describe Hinsberg’s test to distinguish between 1 o,2 o and 3o amines. Ans The amine is shaken with benzenesulphonyl chloride (Hinsberg reagent) in presence of aq.KOH soln. Ans 113 9) Ans How will you convert 4-Nitrotoluene to 2-Bromobenzoic acid? 10) a) Predict, with reasons, the order of basicity of the following compounds is gases phase. (CH3)3 N, (CH3)2 N H, CH3NH2 , NH3 b) Describe Carbylamine reaction. Ans a) In gaseous phase, solvation effects are missing, Hence, greater the number of alkyl groups, greater +I effect and stronger the base. (CH3)3 N > (CH3)2 NH > CH3NH2> NH3 b) Aliphatic and aromatic primary amines on heating with CHCl3 and ethanolic KOH form isocyanides or carbylamines. which are foul-smelling substances. Secondary and tertiary amines do not show this reaction. Hence this reaction is used as a test for primary amines. R-NH2 + CHCl3 + 3KOH ⎯ RNC + 3KCl + 3H2O Carbylamine or Isocyanide 114 CHAPTER – 14 BIOMOLECULES 01 MARK 1) What are polypepticles? A) Polymers of amino acids having peptide linkage (– CONH –) 2) Which vitamin’s deficiency causes pernicious anemia? A) Vitamin B12 3) What is the biological effect of denaturation of proteins? A) The protein molecule uncoils from an ordered and specific conformation into a more random conformation. Primary structure remains undisturbed. 4) In what sense are the two strands of DNA not identical but complementary to each other? A) If one strand has the bases A T C G, the other strand has T A G C, i.e. A can only bond with T and C can pair with G. Hence the strands are not identical, but complementary. 5) Show the reaction of Glucose with Tollen’s reagent. 02 MARKS Q1. What is meant by (a) Pyranose structure of glucose (b) Glycosidic linkage 115 Ans 1 a) Six-membered cyclic structure of glucose, in analogy with pyran b) Two monosaccharide units are linked through oxygen atom accompanied by loss of a water molecule. This linkage is called glycosidic linkage. Q2) What is the difference between α – form of glucose and β-form of glucose? Ans These two forms differ from each other in orientation of –OH group at C-1. α – form is obtained by crystallization from concentrated solution of glucose at 303K while β – form (melting point 423 K) is obtained by crystallization from hot saturated soln. at 371 K. Q3) Mention the type of linkages responsible for the formation of the following: i) Primary structure of proteins ii) Cross-linking of polypeptide chains iii) α – helix formation iv) β – sheet structure. Ans i) ii) iii) iv) Q4) Ans Why are carbohydrates generally optically active? Due to presence of chiral or asymmetric carbon atom and absence of plane of symmetry. Peptide linkage H-bond, sulphide linkage, van der waal’s forces H-bond Intermolecular H-bonds. Q5) Give reactions characteristic of –CHO group, but not by glucose, as in it, free – CHO group is absent. 116 Ans i) No reaction with Schiff’s reagent CHO | (CHOH)4 + Schiff’s reagent → no reaction | CH2OH ii) No reaction with NaHSO3 and NH3 CHO | (CHOH)4 + NaHSO3/NH3 → no reaction | CH2OH 03MARKS Q1) Ana Define : a) Invert sugar b) Vitamins c) Nucleosides a) Hydrolysis of sucrose brings about a change in sign of rotation from dextro (+) to laevo (-), hence it is called invert sugar. b) Organic compounds which can’t be produced by the body and must be supplied in small amounts in diet to perform specific biological functions for normal health, growth and maintenance of body. c) A unit formed by attachment of a base to I’ position of sugar Q2) Ans Differentiate between fibrous and globular proteins. FIBROUS PROTEINS 1. Fibre – like structure 2. Water – insoluble e.g.keratin, myosin, fibrin etc. 3. Stable to moderate changes in temperature and pH. GLOBULAR PROTEINS 1. Polypeptide chains coil around to give a spherical shape. 2. Water soluble e.g.insulin, haemoglobin, enzymes, hormones. 3. Very sensitive even to small changes in temperature and pH. 117 Q3) What happens when D-Glucose is treated with a) HI b) Br2 water c) HNO3 Ans 3 (a) C H (b) (c) 6 +HI CH – (CH )6- CH n-Hexane CHO COOH (CHOH)4Br2 H2O (CHOH)4 CH OH Glucose CH OH Gluconic acid CHO HNO3 CH OH (a) (Mild Oxidation) COOH (CHOH)4 Q7. (Reduction) (CHOH)4 (Strong Oxidation) COOH (Saccharic acid) Define Zwitter ions with examples? (b) What is the difference between essential amino acids and non-essential amino acids? A7. (a) Amino acids contain -NH and –COOH groups. These two groups interact by transferring a proton from carboxyl group to amino group within the molecule. Hence, a dipolar ion called ZWITTER ION is formed. NH R (b) NH CH S No 1. 2. COOH R Essential Amino Acids Cannot be synthesized by the body. Therefore should be supplemented through diet. Eg Valine, leucine + CH COO- Non-Essential Amino Acids Can be synthesized by the body. E.g Glycine, alanine Q8. Which is the sugar present in milk? How many monosaccharaides are present it? What are such oligosaccharides called? A8. Lactose, two monosaccharaide units oligosaccharides are called DISACCHARIDES. (glucose and galactose) such 05 MARKS Q1. Discuss the structure of proteins in detail? What is the difference between helix and β-pleated sheet structures of proteins? A1. α- (a) Specific sequence in which various amino acids in a protein are linked to one-another – primary structures. 118 (b) The conformation adopted by these polypeptide chains as a result of Hbonding – SECONDAY STRUCTURES. (i) α- Helix. Formed by intra molecular H – bonds, causing the polypeptide chain to coil – up into a spiral structure or right handed helix Eg . Fibrous protein like keratin and myosin. (ii) β-pleated sheet structures Polypeptide chains lie side by side held by intermolecular H-bonds , forming sheets. These sheets can then be stacked one over the other forming a 3-D structure. This structure resembles pleated folds of drapery, hence also called β-pleated sheet structures. (c) Tertiary Structure. The secondary structure is further arranged, leading to flowing two possibilities. (i) FIBROUS PROTEIN.The long linear protein chains form thread like structure. These are insoluble in water and have β-pleated structure. (ii) GLOBULAR PROTEINS. Different segments of the protein fold up to give the entire molecule a spherical shape. The folding involves various interactions between the side-chains-such as, vander waal’s interactions, disulphide bridges ,hydrogen bonding etc. (d) Quaternary Structure. Some proteins exist as assembly of two or more polypeptide chains called SUBUNITS or PROMOTERS. These subunits may be identical or different and are held together by H-bonds, electrostatic and van der waal’s interactions. The quaternary structure refer to the determination of number of subunits and their spatial arrangement w.r.t each other in an aggregate protein molecule. Primary Structure Secondary Structure Tertiary Structure Quarternary Structure 119 Q2. (a) RNA? Write important structural and functional differences between DNA and S Structural Differences No 1. Sugar present is 2 – deoxy –D-(-) Sugar present is D-(-) ribose ribose 2. Cytosine and thymine are the Cytosine and Uracil are the pyrimidine bases pyrimidine bases 3. Double- stranded α- helix Single- stranded α- helix 4. Very large molecules ; molecular Smaller molecules with mass may vary from six – sixteen molecular mass ranging million from 20,000- 40,000µ S No 1. 2. Q2. Functional Differences Unique property of replication Usually doesn’t replicate Responsible for transmission of Responsible hereditary characteristics synthesis protein (b) What deficiency diseases are caused due to lack of vitamins A,B ,B and K in human diet? S No Q3. for Vitamins Deficiency disease 1. A Xerophthalmia , night blindness 2. B Beriberi, loss of appetite 3. B Anaemia, general weakness 4. K Hemorrhage , slow blood clotting (a) An optically active amino acid (A) can exist in three forms depending on the of the medium. If the molecular formula of (A) is C H NO , write (i) Structure of (A) in aqueous medium. What are such ions called? (ii) In which medium will the cationic form of (A) exist? (iii) In alkaline medium, towards which electrode will compound (A) migrate? A3. (a) (i) CH CH NH2 (A) COOH In aqueous medium, CH NH CH COO- + ZWITTER ION 120 (ii) In acidic medium, CH NH (ii) COO-+ H+ CH CH In alkaline medium CH + NH + + COOH (cationic form) COO- + OH-CH CH NH CH + COO- CH NH2 Anionic form Therefore the compound in alkaline medium will migrate towards anode. (b) What are reducing sugars ? How are proteins related to amino acids? A3 (b) Carbohydrates with free aldehyde group that can reduce Fehling’s solution and Tollen’s reagent are called reducing sugars. All monosaccharaides are reducing sugars. Also, those disaccharides where the linkage between two monosaccharaides is not through aldehydic group are also called reducing sugars. Q4. (a) Write four characteristics features of enzymes. Name a disease caused by the deficiency of a particular enzyme? A4. (b) What is a peptide bond? Explain its formation with an e.g? (a) (i) Enzymes are specific in their action. (ii) Work at specific pH & moderate temperature. (iii) Activity is high. (iv) Catalyse biochemical reactions and their deficiency can cause diseases. Eg Phenylketane urea is caused by deficiency of phenylalanine hydroxylase and Albinism is caused by deficiency of tyrosinase. (b) Proteins are polymers of α-amino acids connected to each other by peptide linkage. Chemically peptide linkage is an amide formed between – COOH group and –NH2 group. Q5. (a) Two samples of DNA, A and B have melting points 340K and 350K respectively. What conclusion can you draw from these data regarding their base content? (b) Discuss the composition of starch. 121 A5. (a) CG base pair has 3 H- bonds and AT base pair has two H – bonds. Therefore CG base pair is more stable than AT base pair. Since sample B has higher melting point than sample A, therefore sample B has higher CG content than sample A. (b) Starch is made of two components (i) Amylose (15 – 20%) – Water – soluble, linear polymer of α-D Glucose in which C of one glucose unit is attached to C of the other through α – glycosidic linkage. Aqueous solution of amylose gives blue colour with iodine solution. (ii) Amylopectin (30-85%) – Highly branched polymer, large number of short chains containing 20-25 glucose units joined through α – glycosidic linkage involving C of one glucose unit and C of another. The C of terminal glucose unit in each chain is further linked to C of some other glucose unit in the next chain through C - C α – glycosidic linkage. 122 CHAPTER – 15 POLYMERISATION 01 MARK Q1. Define polymerization? A1. The process of formation of a macromolecule from its repeating units (monomers). Q2. Is (CH – CH)na homo polymer or copolymer? A2. Homopolymer with Q3. Write the distinguishing feature between homo polymers and copolymers? CH = CH as monomer. A3. Homopolymers. These polymers are formed by the polymerisation of single monomeric species, e.g. formation of polythene from ethene. Copolymers. These polymers are formed by the polymerisation of different monomeric species, e.g. formation of buna-S from styrene and buta-1, 3-diene. Q4. Write the name and structure of a commonly used initiator in free – radical addition polymerization. o o A4. Benzoyl peroxide C H -C – O-O-C-C H Q5. Based on molecular forces, what type of polymer is neoprene? A5. Elastomer (weakest intermolecular forces). 02 MARKS Q1. What is a biodegradable polymer? Give an example of biodegradable polyester? A1. The natural polymer disintegrates by itself or by microorganisms within a certain period of time. Eg PHBV (Poly-β-Hydroxy butyrate-co-β- hydroxy valerate). Q2. What are the differences between thermoplastic and thermosetting polymers? A2. Thermoplastic polymers are linear, can be softened by repeated heating and hardened on cooling, so can be used again and again without change in chemical composition. E.g polythene. Thermosetting polymers: When heated in a mould, they under permanent change in their chemical composition to give a hard mass. Thus, they can be heated only once and then set into a solid which can’t be remoulded e.g. Bakelite. Q3. (a) What does 6,6 in nylon – 6,6 mean? (b) On free radical polymerization of chloroprene, which polymer is obtained? 123 A3) (a) The monomers – adipic acid and hexamethaylene diamine – both have 6 C atoms each. (b) Neoprene Q4. A natural linear polymer of 2- methyl – 1,3 butadiene becomes hard on treatment with sulphur between 373 to 415K and –S-S bonds are formed between chains. Write the structure of product of this treatment? A4. CH3 CH C CH S S H C C CH CH CH CH Q5. What is the repeating unit in the condensation polymer obtained by combining HO C CH2CH CO H (succinic acid) and H NCH CH NH ? A5. -(NH CH CH NH CO CH CH CO---) n 03 MARKS Q1. How are polymers classified into different categories on the basis of intermolecular forces? Give one example of a polymer of each of these categories? A1. There are 4 types of polymers based in intermolecular forces. (i) Fibres – Strongest intermolecular forces, thread-like, crystalline ; Nylon (ii) Elastomers – Weakest intermolecular forces, stretchable e.g natural rubber. (iii) Thermoplastics – No cross – linkages become soft on heating, eg polyethene. (iv) Thermosetting polymers – undergo extensive cross –linking on melting eg Bakellite. Q2. Distinguish between addition (chain- growth) and condensation (step-growth) polymers? 124 A2. S No 1. 2. 3. 4. Q3. Addition Polymerisation Condensation Polymerisation Formed by repeated addition of Formed by condensation reactions monomer units between bi-functional componds No removal of any molecules Removal of small molecules like H O, NH Also known as chain-growth Also known as step- growth polymers and the chain grows by polymers as each step is free – radical mechanism independent of the other. Eg :- Polythene, PVC Nylon 6,6 , Dacron Write chemical equations for the synthesis of (i) Nylon – 6 (ii) Nylon – 6,6 (iii) O A3. (i) [O] Polythene o NH OH H2N(CH )5COOH cyclohexane cylohexanone cylohexanone caprolactum polymer risation o (NH - (CH )5 C)n Nylon – 6 (ii) (iii) n(CH2=CH2) - (CH2 – CH2–)n Q4. What is the purpose of vulcanization of rubber? A4. The main purpose of vulcanization of rubber is to improve upon its properties. Natural rubber becomes soft at high temperature (.335K) and brittle at low temperature (<283) and shows high water absorption capacity. It is soluble in non-polar solvent and resistant to oxidizing agents. To improve upon these properties, raw rubber is mixed with an additive such as zinc oxide and heated between 373K and 415K. Sulphur forms cross-links at the reactive sites of double bonds and rubber gets stiffened. The probable structures of vulcansied rubber. 125 CH CH C H C CH CH CH S S C CH CH C = CH - CH S CH CH CH = CH - CH2 CH Q5. A5. Write the structure and one use of each of the following :(i) PVC (ii) Urea – formaldehyde resin. (ii) Bakelite (i) (CH CH)n use : pipes and raincoats Cl (ii) (NH C NH CH )n use : making unbreakable crockery O OH (iii) CH OH CH O OH CH CH O O CH CH CH H2O CH CH O O O OH OH OH Use : electrical switches, handles of utensils 126 CHAPTER – 16 CHEMISTRY IN EVERY DAY LIFE 01 MARK Q1. Define narrow spectrum antibiotics with an example? A1. Antibiotics that kill or inhibit a short range of gram positive or gram negative bacteria are called narrow spectrum antibiotics. For example – Penicillin G. Q2. Name a substance that can be used as an antiseptic as well as disinfectant? A2. Phenol can be used as both .A 0.2% soln of phenol is an antiseptic, while its 1% is disinfectant. Q3. Why is bithional added to soap? A3. Bithional acts as an antiseptic agent and reduces the odour produced by bacterial decomposition of organic matter. Q4. What makes detergents non-biodegradable? A4. Branching in the hydrocarbon chain. 02 MARKS Q1. A1. Explain the following with an example? (a) Antifertility drugs (b) Antibiotics (a) Drugs which help in birth-control eg norethindrone (b) Antibiotics are substances produced wholly or partially by chemical synthesis, which in low concentrations inhibit the growth or destroy microorganisms by intervening in their metabolic processes. Q2. Give reasons :(a) Soaps don’t work well in hard water. (b) Synthetic detergents are better than soaps. A2. (a) Mg and Ca ions of hard water react with soap to form magnesium and calcium salts which are insoluble in water and form scum. (b) Q3. A3. Synthetic detergents can work even with hard water. Mention one important use of: (i) Equanil (ii) Sucralose. (i) Equanil is a tranquilizer. (ii) Sucralose is an artificial sweetener. 127 Q4. A4. Q5. (i) Why is the use of aspartame limited to cold foods and drinks? (ii) Give two examples of macromolecules that are chosen as drug targets. (i) It is unstable at high temperature. (ii) Nucleic acids/proteins/carbohydrates/lipids. Classify the following into antihistamine, antacid, tranquiliser, antibiotic. Drug: penicillin, meprobamate, terfenadine, and ranitidine. A5. Pencillin – antibiotic, Meprobamate – tranquiliser, Terfenadine – antihistamine, Ranitidine – Antacid. 03 MARKS Q1. What are analgesics? How are they classified? When are they recommended for use ? A1. Chemical substances used to relieve pain are called analgesics. Types of analgesics : (i) Non-narcotic drugs – Non-habit forming, effective in relieving or preventing heat attack, skeletal pain, viral inflammation. (ii) Narcotic drugs – Habit forming produce sleep, unconsciousness, convulsions. Useful in post – operative pain, cardiac pain, terminal cancer. Q2. What are antipyretics? Given an example. Can it play any other role? A2. Antipyretics reduce body temperature under conditions of fever. E.g Paracetamol. They can also act as analgesics. Q3. Define the following and give an example of each? (a) Antacids (b) Sulpha drugs (c) Antioxidants A3. (a) Antacids are drugs that are administered into the body to neutralize excess acid produced in the stomach. E.g ranitidine. (b) Derivatives of sulphanilamide have great antibacterial powers, e.g sulpha diazine, and sulphapyridine. (c) Antioxidants retard the action of oxygen on food and thus prevent the decomposition of foods. Eg (BHA) Q4. Explain the mechanism of drug – enzyme interaction. 128 A4. Drugs can block activities of enzymes in the following two manners. (a) Drugs can block activities of enzymes in the following two manners (i) Drug might compete with the substrate for attachment on the active sites of enzymes. Such drugs are called competitive inhibitions. (ii) Some drugs don’t bind to the active site instead they bind to a different site on the enzyme called allosteric site. This action changes the shape of the active site to an extent that the substrate doesn’t recognize it. Q5. How do receptors transfer message to the cell? A5. Message between two neurons or between neurons and muscles is communicated through chemical substances called chemical messengers. They are received at the binding site of the receptor protein. To accommodate these chemical messengers shape of receptor protein changes a little and messenger gives the message. After the message is given, chemical messenger departs and active site of receptor protein returns to its original shape. 129 VALUE – BASED QUESTIONS 1) Mr Arora and family were having a tough time. It had been raining all night and the sewer water was moving back in their toilets. Mr Arora thinks it is due to polythene bags choking the sewer lines. His wife wonders why these bags don’t dissolve in water. Mr Arora explains that these bags are non-biodegradable. He says that cloth bags must be used for shopping to minimise use of polythene bags.Answer the following questions: (a) Is there any biodegradable polymer? What are its uses? (b) What values were displayed by Mr. Arora? A) a) PHBV [Poly-β-Hydroxybutyrate-Co-β-hydroxyvalerate] is biodegradable. It is a co-polymer of 3- Hydroxybutanoic acid and 3-Hydroxypentanoic acid. It is used in speciality packing, orthopaedic devices and in controlled release of drugs. b) Mr. Arora displays concern for environment. He finds the polythene bags a serious threat to the environment. He must follow his ideals. 2) Shyam wants to wash his woolen clothes. His grandmother suggested that he washes the clothes in warm water. Shyam follows her advice and finds that the clothes become cleaner that how they were, when washed in cold water! Answer the following questions: a) b) Why is it easier to wash the clothes in lukewarm water? What values are associated with this information? A) a) Soaps work through micelle-formation. But the optimum temperature (KRAFT TEMPERATURE) at which this happens is attained in lukewarm water. b) Be humble to accept a scientific fact form an elderly person. 3) During Chemistry practical’s, the students were told to add reagents very carefully and not to unnecessarily touch or play with the chemicals. Even then, a student quietly removed the labels from bottles of 2-propanol and 1-propanol. Nikhil, a responsible student, reported the matter to the teacher. Answer the following questions: a) What values are expressed by Nikhil? b) The teacher easily found out the correct alcohols by performing a simple test. Write the chemistry of the test performed. A) a) Nikhil shows very responsible behavior. He not only obeys the teacher, but also reports the matter, showing concern for others. b) . ( , ) CH3-CH(OH)-CH3 ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ CH3-CH(Cl)-CH3 Propan-2-ol turbidity after 5-mins (Secondary Alcohol) . , ( ) CH3CH2 CH2OH ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ CH3CH2CH2Cl Primary Alcohol turbidity appears only on heating 130 4) Fermentation of sugars gives a mixture containing an alcohol A and water. The alcoholwater mixture on fractional distillation gives an azeotropic mixture. The azeotropic mixture is composed of 95.6% alcohol and 4.4% water by mass. The mixture is commonly called rectified spirit. To refrain people from drinking rectified spirit, industrial alcohol is denatured. Some countries now use alcohol A as an additive in gasoline since it is a cleaner fuel.Answer the following questions: a) Identify alcohol A. Write reactions involved in fermentation of a sugar. b) What is denaturation of alcohol? Why is industrial alcohol denatured? c) Would you support the use of alcohol ‘A’ as an additive in gasoline in India, although it may result in decrease in production of sugar? d) What are the values associated with your decision? A) a) A Ethanol C2 H5OH C12H22O11+ H2O ⎯⎯⎯⎯⎯⎯⎯⎯ C6H12O6 + C6H12O6 Glucose Fructose C6H12O6 ⎯⎯⎯⎯⎯⎯ 2C2H5OH + 2CO2 Glucose or Fructose b) Making industrial alcohol unfit for drinking by adding CuSO4 (to give it colour) and pyridine ( a foul smelling liquid) is called Denaturation of Alcohol. Ethanolis required for various chemical industries. To stop people from drinking it, it is denatured. c) Gasoline need of our country is met by importing it which increase our fiscal deficit. To meet energy requirements and reduce oil-consumption, ethanol is used as an additive to gasoline. But it must be regulated by the Government. d) Concern for energy crisis and economy of the country. 5) There is growing interest in the use of chelate therapy in medicinal chemistry. An example is the treatment of problems caused by the presence of metals in toxic proportions in plant/animal systems. Detection of cations through coloured complex formation is done in qualitative analysis.Answer the following questions: A) a) Name the chelating agents that can remove copper, iron and lead from water. b) Name the compound that inhibits the growth of tumours. c) Recent studies show that cis-platin can cause serious side-effects, including severe kidney damage. What is the alternative medicine? a) Copper, iron - D – penicillamine desferrioxime D lead EDTA b) 131 c) cis-platin is being replaced by trans-isomer. 6) Vitamin C is water-soluble and can’t be synthesized by humans, monkeys and guineapigs. It is also required for the synthesis of collagen, which is the structural protein of skin, tendons, connective tissue and bone.Answer the following questions: a) b) c) A) Write structure of vitamin C and the functional groups therein. Is it acidic in nature? What is its common name? What is the oxidation product of vitamin C? a) b) It is acidic due to enolic -OH group. Its common name is ascorbic acid. c) 7) Deepti and her father went to the market to buy some mosquito-killer chemical. Deepti stopped her father from buying DDT.Answer the following questions: a) b) A) Why is Deepti’s action justified? What are the values shown by Deepti? a) i) Many species of insects have developed a resistance to DDT. ii) It is highly toxic for aquatic life. iii) Stability of DDT makes it difficult to disintegrate or degrade. b) Concern for environment, application of knowledge to everyday situations, awareness and willingness to educate others. 132 8) Dentist Dr Verma opened a clinic in a village and found many people suffering from bleeding gums. He spread awareness amongst villagers about some inexpensive steps to prevent the problem.Answer the following questions: a) b) c) d) A) What disease were the villagers suffering form? What causes this disease? What is the cure for this disease? What values are displayed by Dr Verma? a) Scurvy b) Deficiency of vitamin C c) Vitamin C in the form of citrus fruits like orange, lemon, amla should be taken on a regular basis. d) Dr Verma displays sensitivity to the villagers, who may not be able to afford expensive medicines, but simple fruits should be easily available. He also shows social responsibility, a duty of a doctor. 9) At a chemist shop, you see a person pleading with the shopkeeper to give him sleeping pills, without a doctor’s prescription. The chemist refuses, stating that the pills can be misused. Answer the following questions: a) Should the shopkeeper give the customer the pills without a doctor’s prescription? Why? b) What alternative approach can the shopkeeper take to help the customer without giving him the pills? Which values can be reflected in this alternative approach? A) a) No, sleeping pills should not be given without prescription. An overdose of these pills can even cause death. b) He can suggest some good doctor nearby, convincing the customer to seek a doctor’s advice.Value : Professional ethics, responsibility, human approach. 10) Rohini wanted to give her baby some medicine. She added boiled and cooled water as per instructions and shook the bottle well before giving the medicine to the baby.Answer the following questions: A) a) b) What values are associated with selling medicine in this form? Why did she shake the contents? Name the process. a) The medicines sold in anhydrous form have higher shelf life and can be stored for a longer time. This is a way of being thrifty by not wasting the available resources. Rohini shook the contents of the bottle to change the content into the form of Sol because adsorption of medicines are easier in the colloidal form. The process involved is Peptisation b) 133 11) I work in a lab where aldehydes are prepared by the following method : / RCOCl + H2 ⎯⎯⎯⎯⎯⎯⎯⎯ + My mother, an organic chemistry professor, advised me against this method and suggested another one.Answer the following questions: c) Why did my mother advise me against this method? d) What could be the other method? Can you judge it for environment and economically? A) 12) a) BaSO4 harms the environment and can be toxic for us. b) CH2 = CH2 + O2 ⎯⎯⎯⎯⎯⎯⎯⎯ It is economical, but heavy metals like Pd and Cu drain underground and contaminate underground water. ( )/ ( ) Mrs Sharma lives near a lake along with 5 neighbours. Initially the lake was clear and beautiful, but now it has a lot of phytoplanktons. One day Mrs Sharma’s daughter, after returning from school, advised the ladies to use natural soaps for washing clothes, not detergents. Mrs Sharma and the others follow her advice and get the lake cleared gradually! Answer the following questions: a) What values are displayed by Mrs Sharma and her daughter? b) Why did the phytoplanktons grow excessively whie they were using detergents? A) a) b) Mrs Sharma – Environmental awareness, giving due importance to educated person’s advice. Daughter – Knowledge of chemistry, environmental awareness, application of knowledge to real-life situation. Synthetic detergents contains phosphates which lead to overgrowth of vegetation or phytoplanktons. 13) Rajiv saw policemen stopping vehicles and asking drivers to blow air into an instrument. His father told him that they were checking whether the driver had consumed alcohol or not. Answer the following questions: A) a) b) c) What is the instrument called? What values are the police trying to inculcate? How does the instrument test whether or not a driver is drunk? a) b) c) Alcometer Social responsibility, awareness of risks involved due to drunken driving Alcometer uses K2Cr2O7 which changes colour on reacting with alcohol. Depending on the amount of alcohol, the colour range varies. 2Cr2O7-2 + 3C2H5OH + 16 H+ 4Cr3+ + 3CH3COOH + 11 H2O (orange) blue green 134 14) These days, people watch “Solar Eclipses” through UV-protected sun-glasses. Earlier, people avoided eclipses, as they were believed to be evil.Answer the following questions: a) b) A) c) Which transition metal oxide is used in making UV protected lens? By watching the eclipse through UV protected lens, which value are people displaying? Which rays can damage the eyes while watching solar eclipse with naked eyes? a) b) c) Neodymium and praseodymium oxides (f-block) Scientific attitude UV rays 15) Sameer and his friends decide to play Holi with natural colours. When some people try to play with synthetic colours, Sameer explains how they could be harmful, cause skin allergies etc. Answer the following questions: A) a) b) c) Mention the values shown by Sameer. Write the reaction for preparation of an azo (synthetic) dye. Name two pigments present in natural colours. How are natural colours prepared? a) Sensibility, awareness and courage b) c) Chlorophyll and carotene. Natural colour can be prepared from leaves and flowers. 16) Zahira’s brother took cough syrup one day, even though he was healthy. After sometime his head started aching and skin started itching. Zahira insisted on her parents to take him to a doctor.Answer the following questions: a) b) c) A) Mention the values shown by Zahira? Why did her brother’s body start itching? Why was it necessary to go to the doctor? a) Critical thinking, keeping calm in crisis b) Due to an allergic reaction c) Only a doctor can prescribe the correct medicine. Incorrect drug or dosage can be fatal. 135 17) Nita’s mother got easily tired and was lethargic. Nita took her to a doctor, who diagnosed her with pernicious anemia. Nita took good care of her mother till she recovered. Answer the following questions: A) a) b) c) Which vitamin-deficiency causes pernicious anemia? Name the sources that can provide this vitamin Mention the values shown by Nita. a) b) c) Vitamin B12 Cheese, eggs, meat, milk and yogurt. Caring, helpful, keen observation