UNIT-1 SOLIDS QUESTIONS CARRYING ONE MARK: 1. Which type of solid is anisotropic in nature? Ans: Crystalline solids are anisotropic in nature 2. Which type of solids is called as super cooled liquids or pseudo solids? Ans:Amorphous solids are called super cooled liquids 3. A solid has a sharp melting point, and then to which type of solids does it belong? Ans: Crystalline solids 4. Which type of solids has long range orderly arrangement of constituent particles? Ans: Crystalline solids 5. Sodium chloride and quartz belong to which type of solid? Ans: Crystalline solids 6. A solid shows different values for refractive index when measured in different directions. - Identify the type of solid Ans: Crystalline solids 7. When a solid is cut with a sharp edged tool, they cut into two pieces and the newly generated surfaces are plain and smooth. – Identify the type of solid. Ans: Crystalline solids 8. Which type of force of attractions is present between the molecules in polar molecular solids? Ans: Dipole –dipole interactions 9. Which type of force of attractions is present between the molecules in non-polarmolecular solids? Ans: London forces or Dispersion forces

 

 

10. Which type of force of attractions is present between the particlesin ionic solids? Ans: Electrostatic force of attraction or coulombic force of attraction 11.Solid SO2 and solid NH3belong to which type of molecular solids? Ans: Polar molecular solids 12. What is crystal lattice? Ans: The regular three dimensional array of lattice points in space is called crystal lattice 13. What is a unit cell? Ans: It is the smallest repeating unit which when arranged in three dimension gives the crystal lattice. 14. How many types of primitive unit cells are present? Ans: Three types 15. What is a primitive cubic unit cell? Ans: The cubic unit cell in which the particles/atoms are present only at the eightcorner of the cube is called primitive cubic unit cell. 16. Define the co-ordination number of a particle in solids. Ans:It is the total number of nearest neighboring particles to a given particle. 17. What is the number of octahedral voids generated, if the number of close packed spheres is N? Ans: N 18. What is the number of tetrahedral voids generated, if the number of close packed spheres is N? Ans: 2N 19. What is the co-ordination number of a particle in a tetrahedral void? Ans:Four 20. Among Schottky and Frenkel defect, which type of defect decreases the density of the crystal? Ans: Schottky defect

 

 

21. What are point defects? Ans: Deviations from the ideal arrangement around a particular point or an atomin a crystalline solid 22. What are F-centers? Ans: The anionic sites occupied by the unpaired electrons are called F- centre. 23. To which colour potassium chloride crystal turns, when excess potassium ionis present? Ans: Violet 24. Name the type of non-Stoichiometric defect observed when white ZnO turns yellow on heating. Ans: Metal excess defect 25. Name the non-Stoichiometric defect responsible for the composition of ferrous oxide to be Fe0.95O1. Ans:Metal deficiency defect 26. Which type of point defect is observed when NaCl containing little SrCl2 is crystallized? Ans:Impurity defect 27. Which defect is also called as dislocation defect? Ans:Frenkel defect 28. What is doping? Ans: The process of increasing the conductivity of an intrinsic semiconductors by adding asuitable impurity is called doping 29. What type of semiconductors are obtained when silicon doped with boron impurity? Ans: p-type semiconductor 30. Name the unit used to measure magnetic moment. Ans: Am2( 1Bohr magneton= 9.27x10-24 Am2) 31. What are diamagnetic substances? Ans: These are the substances which are repelled by the magnetic field

 

 

32. What are ferromagnetic substances? Ans: These are the substances which are strongly attracted by the magnet 33. How body diagonal and radius of a sphere(r) are related in bcc unit cell? Ans:4r =√2a 34. Give an example for Ferromagnetic substance. Ans: Fe Co Ni Gd CrO2

35. Give an example for Diamagnetic substance. Ans: H2O, NaCl, and C6H6

QUESTIONS CARRYING TWO MARKS: 1. How crystalline solids differ from amorphous solids in their melting point? Ans:Crystalline solids have sharp melting point whereas amorphous solid do not have a sharp melting point 2. Write any two differences between crystalline solids and amorphous solids?

Crystalline solid Amorphous solid

3-D long range orderly arrangement of particles

No orderly arrangement of constituent particles

Sharp Melting point Do not have sharp M P ( Softening temperature)

True solids having definite shape Pseudo solids having irregular shapes They have a well-defined cleavage planes

Do not have cleavage planes

Anisotropic in nature Isotropic in nature 3. What is meant by anisotropy? What type of solids show this nature? Ans: The physical properties like refractive index, coefficient of thermal expansion, when measured in different directions gives different value for a crystalline solid hence it is anisotropic in nature. Crystalline solids 4. What are the nature of particles and the force of attractions between the particles in non-polar molecular solids? Ans: In a non-polar molecular solids, the constituent particles are non-polar moleculeslike H2, Cl2, I2 and even atoms like Ar, Ne, Xe etc. The nature of force of attraction is weak dispersion force or London force.

 

 

5. What are the nature of particles and the force of attractions between the particles in polar molecular solids? Ans: In a polar molecular solids, the constituent particles are formed by polar covalent bond like HCl, SO2. The nature of force of attraction is dipole-dipole attractions 6. What are the nature of particles and the force of attractions between the particles in hydrogen bonded molecular solids? Ans: In a hydrogen bonded molecular solids, the constituent particles are polar molecules capable of forming hydrogen bond like water. The nature of force of attraction is hydrogen bonding 7. What are point defects? Mention the types Ans: Point defects are the irregularities in the arrangement of constituent particlesaround a point or a lattice site in a crystalline substance. These are of three types.

1. Stoichiometric defects. 2. Non-stoichiometric defect

3. Impurity defect.

8. What are the differences between Schottky and Frenkel defect? Schottky defect

a. Shown by ionic solidscontaining similar-sized cationsand anions (having high coordination number)

b. An equal number of cations and anions are missing to maintain electrical neutrality

c. Decreases the density of the substance d. Example, NaCl, KCl , CsCl, and AgBr

Frenkel defect

a. Shown by ionic solids containing largedifferences in the sizes of ions, (having less coordination number)

b. Created when the smaller ion

(usually cation) is dislocated from its normal site to an interstitial site

c. No change in density of the crystal. creates a vacancy defect as well as an interstitial defect .Also known as dislocation defect

d. Example: AgCl, AgBr, AgI and ZnS

9. What are the nature of particles and the force of attractions between the particles in ionic solids? Ans: The nature of the particles is ions (both cation and anion). The nature of the force of attraction is electrostatic force or coulombic force 10. What are the nature of particles and type of bonding in network solids? Ans: The nature of the particles is atoms. The bonding is covalent bond.

 

 

11. Classify the following into polar and non-polar molecular solids: Ar, HCl, I2 and SO2 Ans: Non-polar molecular solids: Ar, I2 Polar molecular solids:HCl, SO2 12. Calculate the number of particles present per unit cell in an FCC unit cell. Ans:Contribution of corner particle = 8 x 1/8 = 01 Contribution of a particle at the centreof face = 6 x ½ = 03 Total number particle /unit cell = 04 13. Calculate the number of particles present per unit cell in a BCC unit cell. Ans:Contribution of corner particle = 8 x 1/8 = 01 Contribution of a particleat the centre = 1 x 1 = 01 Total number particle /unit cell = 02 14. Calculate the number of particles present per unit cell in a simple cubic unit cell. Ans:Contribution of corner particle = 8 x 1/8 =01 Total number particle /unit cell = 01 15. Mention the two characteristics of a unit cell. Ans: Two characteristics of unit cells are a. Edge length b. Axial angles 16. What is the relation between edge length (a) and radius of the sphere (r) infcc unit cell? What is itspacking efficiency? Ans: The relationship between edge length and radius of the sphere are a=2   2  r Packing efficiency is 74% 17. What is the relation between edge length (a) and radius of the sphere (r) in bcc unit cell? What is its packing efficiency?

Ans: The relationship between edge length and radius of the sphere are a=!!!

Packing efficiency is 68 % 18. How many tetrahedral and octahedral voids is present, if the number of sphere is N? Ans: The number of tetrahedral void is 2N The number of octahedral void is N 19. Explain Schottky defect. Give an example. Ans: The defect which arises due to missing of equal number of cations and anions from the crystal lattice is called Schottky defect. Ex. NaCl, KCl ,CsCl, AgBr

 

 

20. Explain Frenkel defect. Give an example. Ans: The defect in which an ion (generally cation) leaves the original site and occupies the interstitialsite is called Frenkel defect. E. AgCl, AgBr, AgI 21. How Schottky defect and Frenkel defect affect the density of the crystal? Ans: In Schottky defect density of the crystal decreases. In Frenkel defect the density of the crystal remains same. 22. Mention the two types of Non-stoichiometric defects in solids? Ans: Metal excess defect and metal deficiency defect. 23. What is F- center? What colour is imparted to the NaCl crystal, due to the presence of excess sodium? Ans: The anionic sites occupied by the unpaired electrons are called F- Centre The colour of NaCl crystal is Yellow 24. Write the formula to calculate the density of the unit cell and explain the terms. Ans: z = number of particles present per unit cell

d = 𝒛𝑴𝒂𝟑𝑵𝑨

               M = Molecular mass , d = density NA = Avogadro’s number

a = Edge length. 25. What are n-type and p-type semiconductors? Ans: n-type semiconductor is obtained by doping of the crystal of a group 14 element such as Si or Ge, with a group 15 element such as P or As(pentavalent). Conductivity increases due to negatively charged electrons. p-type semiconductor is obtained by doping of the crystal of a group 14 element such as Si or Ge, with a group 13 element such as B, Al or Ga( trivalent). Conductivity increases as a result of electron hole 26. An ionic compound is formed by two elements A and B. The cat ions A are in ccp arrangement and those of anions B occupy all the tetrahedral voids. What is the simplest formula of the compound? Ans: Since cations are in ccp arrangement, the total number cat ions A = 4 The number of tetrahedral voids is double the number of particles = 8 All the tetrahedral voids are occupied by anions B. The number of elements of B = 8 Hence the formula of the ionic compound is A4B8 or AB2

 

 

27. A compound is formed by two elements X and Y. The element X forms ccp and atoms of Y occupy 1/3 rd of tetrahedral voids. What is the formula of the compound? Ans: Since element X are in ccp arrangement, the number of X per unit cell = 4 The number of tetrahedral void = 8 But only 1/3 rd is occupied by Y, therefore 8 x1/3 = 8/3 Hence the formula of the compound is X4Y8/3 = X12Y8 or X3Y2

28. Gold(atomic radius=0.144nm)crystallizesin a face centered unit cell. What is the length of the side of the cell? Ans: For FCC the edge length and radius of sphere arerelated by the equation, r = 0.144nm a=2   2  r a = ? = 2 2    x  0.144  nm = 2x1.414 x 0.144 = 0.40723nm. 29. Silver forms ccp lattice and X- ray studies of its crystals show that the edge lengthof its unit cell is 408.6pm. Calculate the density of silver (atomic mass = 107.9 u) Ans:

d = !"!!!!

d= 4 x 107.9/(4.08)3 x10-24 x 6.022 x1023 d = 431.6/40.899 d = 10.5528g/cm3

30. X- ray diffraction studies show that copper crystallizes in an fcc unit cell with cell edge of 3.6 x10-8cm. In a separate experiment, copperis determined to have a density of 8.92g/cm3,calculate the atomic mass of copper. Ans:

d = !"!!!!

M = d a3 NA/Z = 8.92 x(3.6)3x10-24 x 6,022 x1023/4 = 250.61/4 M = 62.6525 u 31. The edge of fcc unit cell of platinum is 392 pm and density is 21.5 g/cm3, calculate the Avogadro number. Ans:

d = !"!!!!

NA = Z x M/ d a3 = 4 x 195.08/21.5 x (3.92)3x 10—24

= 780.32/1295.08 x10—24

NA= 6.025 x1023

 

 

32. A unit cell of sodium chloride has four formula units. The edge length of the unit cell is 0.564 nm. What is the density of sodium chloride? Ans:

d = !"!!!!

d = 4 x 58.5/(5.64)3 x10-24 x 6.022 x1023 d = 234/108.038 d = 2.165 g/cm3

33. A body centered cubic element having density 10.3 g/cm3, has a edge length of 314pm. Calculate the atomic mass of the element (Avogadro’s number= 6.023x1023/mol) Ans:

d = !"!!!!

M = d x a3 xNA/Z = 10.3 x (3.14)3x 10-24 x 6.022 x1023/2 M = 96.01u 34. Calcium metal crystallizes in a face centered cubic lattice with edge length of 0.556nm. Calculate the density of the metal. (Atomic mass of calcium = 40g/mol and Avogadro number= 6.022 x1023mol-1) Ans:

d = !"!!!!

d = 4 x 40/(5.56)3 x10-24 x 6.022 x1023 d = 160/103.50 d = 1.54 g/cm3

35. Copper crystallizes into afcc lattice with edge length 3.61 x10-8cm. Calculate the density of the of the crystal (Atomic mass of copper =63.5g/mol and Avogadro number= 6.022 x1023mol-1 ) Ans:

d = !"!!!!

d = 4 x 63.5/(3.61)3 x10-24 x 6.022 x1023 d = 254/28.33 d = 8.9 g/cm3

 

 

36. Silver crystallizes in a face centered cubic structure. If the edge length is 4.077 x10-8cm and density is 10.5 g/cm3, calculate the atomic mass of silver. Ans:

d = !"!!!!

M = d a3 NA/Z = 10.5 x (4.077)3x10-24 x 6,022 x1023/4 = 103.57/4 The atomic mass of silver M = 107.09 u 37. The density of Li atoms is 0.53g/cm3.The edge length of Li is 3.5 A0. Find out the number of Li atoms in a unit cell (N0= 6.022 x1023/mol& M= 6.94)

Ans:d = !"!!!!

Z = d x 𝑎!𝑁!/𝑀 = 0.53 x (3.5)3 x10–24x 6.022 x1023/6.94 = 2 The number of lithium atoms in unit cell is 2

Questions carrying THREE marks 1. Calculate the packing efficiency in simple cubic unit cell

Edge length of the cube = a = 2r

Volume of the cubic unit cell= a3 = (2r)3= 8r3

volume of one particle(sphere) = !!𝝅r3

The number of particles per unit cell =1

Total volume occupied by one sphere =!!𝜋𝑟!

Packing efficiency= !"#$%  !"#$%&  !""#$%&'  !"  !"#  !"!!"!!"#$%&  !"  !"#$!  !"#$  !"##

× 100

= !!!!

!×!

!  !!×100

= 52.4%

 

 

2. Calculate the packing efficiency in face centered cubic unit cell

edge length of the cube be ‘a’ In �ABC AC2 = BC2 + AB2

b2 = a2 + a2 b2 = 2a2 b = 2   a

Let the radius of the atom = r

Length of the diagonal of �ABC, b= 4r

2  a = 4r

a = 2 2  r

Edge length of the cube =a=2   2  r Volume of the cubic unit cell= a3 = 2   2  r

!

volume of one particle(sphere) = !!𝝅r3

The number of particles per unit cell =4

Total volume occupied by four spheres = 4 × !!𝜋𝑟!

packing efficiency= !"#$%  !"#$%&  !""#$%&'  !"  !"#$  !"!!"!#!"#$%&  !"  !"#$!  !"#$  !"##

× 100

= !!!!

!×!

! !!!  ×    100

= !"! !!

!

!"   !!!  ×    100 = 74 %

3. Calculate the packing efficiency in body centered cubic unit cell

In �ABG, b2 = a2 + a2 ⇒ b2 = 2 a2

In, �AGD, C2 = a2 + b2 C2 = a2 + 2a2

C2 = 3a2⇒ C = 3𝑎

Radius of the atom = r. Length of the body diagonal, C=4r 3𝑎 = 4r

a = !!!

Edge length of the cube =a=!!!

Volume of the cubic unit cell= a3 = !!!

!

volume of one particle(sphere) = !!𝝅r3

The number of particles per unit cell =2

Total volume occupied by two spheres = 2 × !!𝜋𝑟!

packing efficiency= !"#$%  !"#$%&  !""#$%&'  !"  !"#  !"!!"!#!"!"#$  !"  !"#$!  !"#$  !"##

× 100

= !!!!

!×!!!!!  ×  100

= !!!!

!

!"! !

!!  ×  100 = 68%

 

 

4.Based on band theory explain conduction in metals, insulators and semiconductors Conduction of electricity in metals: In metals, the valence shell is partially filled, so this valence band overlaps with a higher energy unoccupied conduction band so that electrons can flow easily under an applied electric field. Conduction of electricity in insulators: In insulators, the valence shell is empty, so the gap between the valence band and conduction band is very large. so that electrons cannot flow under an applied electric field. Conduction of Electricity in Semiconductors In semiconductors, the gap between the valence band and conduction band is so small that some electrons may jump to the conduction band. Electrical conductivity of semiconductors increases with increase in temperature. Substances like Si, Ge show this type of behaviour, and are called intrinsic semiconductors. 5. How are solids classified on the basis of the force of attraction? Ans: a. Molecular solids: Particles are held by a. London forces (in non-polar solids) ex : Benzene, Argon, P4O10, I2, P4

b. Dipole - dipole interaction ( in polar solids) ex: Urea, Ammonia c. Hydrogen bonding (in hydrogen bonded solids) ex: ice

b. Ionic solids a. Particles are held by ionic bond b. Conduct electric current in aqueous solution or molten state c. Examples: NaCl, MgO, ZnS d. In solid state, ions are held together by strong electrostatic forces and are

not free to move about within the solid. Hence, ionic solids do not conduct electricity in solid state. However, in molten state or in solution form, the ions are free to move and can conduct electricity. c. Covalent or network solids: a. Particles are held by covalent bonding. Examples SiO2 (quartz), diamond, d. Metallic solids:

a. Particles are held by metallic bond. b. These are electrical conductors, malleable, and ductile. Examples: Fe, Cu,

6. What are point defects? Explain the types. Ans: Point defects are the irregularities in the arrangement of constituent particles around a point or an atom in a crystalline substance. These are of three types.

1. Stoichiometric defects: Do not disturb stoichiometry of the solid. These are also called intrinsic or thermodynamic defects

Ex : Frenkel defect, Schottky defect 2. Non-stoichiometric defects: This defect alters the stoichiometric ratio of the

constituent elements i) Metal excess defect

a. Metal excess defect due to anionic vacancies: b. Metal excess defect due to the presence of extra cations at interstitial sites:

ii) Metal deficiency defect a. By cation vacancy

3. Impurity defect.

 

 

7. What are diamagnetic, paramagnetic and ferromagnetic substances? 1. Paramagnetic substance: The substance which are attracted by the magnet. The magnetic character is temporary and is present as long as the external magnetic field is present. Ex; O2, Cu2+, Fe3+, Cr3+ NO. 2. Diamagnetic substance: The substance which are weakly repelled by the magnetic field TiO2, H2O,NaCl.This property is shown by those substance which contain fully –filled orbitals (no unpaired electrons) 3. Ferro magnetic substance: The substance which are strongly attracted by the magnet. They show permanent magnetism even in the absence of magnetic field. Ex : Fe Co Ni Gd& CrO2 8. An element with molar mass 2.7 x 10-2 kg/mol forms a cubic unit cell with edge length 405pm. If its density is 2.7 x 103 kg/m3, what is the nature of the cubic unit cell Ans:

d = !"!!!!

Z = d x 𝑎!𝑁!/𝑀 = 2.7 x103 x(405)3 x 10—27 x 6.022 x1023/2.7 x 10-2 = 4 Since there are 4 atoms of the element present per unit cell. Hence, the cubic unit cell must be face centered or cubic close packed structure (ccp) 9. Niobium crystallises in body-centered cubic structure. If density is 8.55g/cm3, calculate atomic radius of niobium, given that its atomic mass is 93 u.

Ans: d = !"!!!!

a3 = !"!  !!

= 2 x 93/8.55 x6.022 x1023 = 36.1 x106

a = (36.1)1/3 x102

=330 pm

For BCC r = !!

a

r = !!

x 330

r = 143pm 10. An element has a body-centered cubic (bcc) structure with cell edge of 288pm. The density of the element is 7.2 g/cm3. How many atoms are present in 208g of the element?

 

 

Ans:

d = !"!!!!

M = d a3 NA/Z = 7.2 x (2.88)3x10-24 x 6,022 x1023/2 = 103.57/2 M = 51.78 u 51.78 g (1mole) contains 6.022 x1023 atoms Therefore 208g contains 4.01 x 6.022 x1023 = 24.187 x1023 atoms.

nBMV

=

=A

nBMW

Unit -2

THEORY OF DILUTE SOLUTIONS

1) What  is  solution?                     [1]  

A:  It  is  a  homogenous  mixture  of  two  or  more  compounds.  

2) What  is  dilute  solution?                     [1]  

A:  It  is  a  solution  in  which  solute  concentration  is  very  less.  

3) Give  an  example  for    solid-­‐solid  solution               [1]  A:  Copper  dissolved  in  gold.  4) Give  an  example  for  gas-­‐gas  solution                 [1]  A:  Mixture  of  oxygen  and  nitrogen  gases.  5) Give  an  example  for    gas-­‐solid  solution               [1]  A:  Solution  of  hydrogen  in  palladium.  6) Give  an  example  for  liquid-­‐solid  solution               [1]  A:  Amalgam  of  mercury  with  sodium.  7) Give  an  example  for    liquid-­‐liquid  solution               [1]  A:  Ethanol  dissolved  in  water.  8) Give  an  example  for    solid-­‐gas  solution               [1]  

A:  Camphor  in  nitrogen  gas.  9) Define  mole  fraction  and  give  the  equation  to  calculate  it.         [2]  

A:Mole  fraction  is  the  ratio  of  number  of  moles  of  one  component  to  the  total  number  of  moles  

of  all  the  components  in  the  solution.  

AB

nAXn nA

=+

 

BB

nBXn nA

=+

 

10) Define  molarity  and  give  the  equation  to  calculate  it.           [2]  A:  Number  of  moles  of  the  solute  present  per  liter  solution  is  known  as  molarity.  

 

 

11) Define  molality  and  give  the  equation  to  calculate  it.           [2]  A:  Number  of  moles  of  the  solute  present  perkgsolvent  is  known  as  molality.  

12) Define  the  term  solubility  of  a  substance.               [1]  A:  Solubility  of  a  substance  is  its  maximum  amount  that  can  be  dissolved  in  a  specified  amount  of  solvent  at  a  specified  temperature  

13) State  Henry’s  law.                     [2]  A:  Henry’s  Law:  At  constant  temperature  solubility  of  a  gas  in  a  liquid  is  directly  proportional  to  the  partial  pressure  of  gas  present  above  the  solution.  OR     At  constant  temperature  the  partial  pressure  of  the  gas  in  vapor  phase  (p)  is  proportional  to  the  mole  fraction  of  the  gas  (x)  in  the  solution.  

Mathematically p ∝ x ; p = KH x.

Where KH is Henry’s law constant. KH depends on the nature of the gas.  

14) Write  the  plot  which  shows  relation  between  partial  pressure  of  a  gas  v/s  its  mole  fraction.                         [2]  A:    

 

 

 

 

 

 

 

15) Mention  the  factors  affecting  solubility  of  a  gas  in  liquid.           [2]  A:  1.  Temperature  2.  Pressure  

16) Explain  how  temperatures  effect  the  solubility  of  a  gas  in  liquid.       [2]  A:  Solubility  of  gases  in  liquid  decreases  with  rise  in  temperature.  According  to  Le  Chatelier’s  Principle,as  dissolution  is  an  exothermic  process,  the  solubility  should  decrease  with  increase  of  temperature.    

17) Explain  how  pressure  effects  the  solubility  of  a  gas  in  liquid.         [1]  A:    The  solubility  of  gases  increases  with  increases  of  pressure.  

18) Mention  the  applications  of  Henry’s  law.               [3]  A:  (a)  To  increase  the  solubility  of  CO2  insoft  drink  and  soda  water,  the  bottle  is  sealed  under  high  pressure.  

Mole  fraction.  

Partial  pressure  of  a  gas

 

(b)  To  avoid  bends,  as  well,  the  toxic  effects  of  high  concentration  of  nitrogen  in  the  blood,  the  tanks  used  by  scuba  divers  are  filled  with  air  dilute  with  helium.  (c)  At  high  altitudes  the  partial  pressure  of  oxygen  is  less  than  that  at  the  ground  level.  This  leads  to  low  concentrations  of  oxygen  in  the  blood  and  tissues  of  people  living  at  high  altitudes  or  climbers.      

19) State  Raoult’slaw  of  liquid-­‐liquid  dilute  solutions.             [2]  A:  The  partial  vapour  pressure  of  each  component  of  the  solution  is  directly  proportional  to  its  mole  fraction  present  in  solution.  Thus,  for  component  1  

P1  ⍺  x1  And     p1  =  p10x1  

20) What  are  ideal  solutions?                   [1]  A:  The  solution  which  obey  Raoul’s  law  over  the  entire  range  of  concentration  are  known  as  ideal  solution  

21) Mention  the  characters  of  ideal  solutions.               [3]  A:    

                       22) What  are  non-­‐ideal  solutions?                   [1]  

A:  When  a  solution  does  not  obey  Raoult’s  law  over  the  entire  range  of  concentration,  then  it  is  called  non-­‐ideal  solution.  

23) Mention  the  types  of  non-­‐ideal  solutions.               [1]  A:  There  are  two  types  (a)  Non-­‐ideal  solution  with  positive  deviation  from  Raoult’s  law  (b)  Non-­‐ideal  solution  with  negative  deviation  from  Raoult’s  law    

24) Give  an  example  for  non-­‐ideal  solution  with  positive  deviation  from  Raoult’s    law.    [1]  A:  Mixtures  of  ethanol  and  acetone  

Ideal  I.  It  obeys  Raoults  law  is  obeyed  at  all  temperature  and  concentration  P  =  PA  +  PB  

II. ∆  V  mix  =  O  i.e.,  there  is  no  change  in  volume  on  mixing  

III. ∆Hmix    =  O  i.e.,  there  is  no  enthalpy  change  when    ideal  solution  formed  

IV. It  doesn’t  form  azeotropic  mixture  V. Force  of  attraction  between  A―A,  B―B is similar

as A―B    

25) Give  an  example  for  non-­‐ideal  solution  with  negative  deviation  from  Raoult’s    law.   [1]    

A:  An  example  of  this  type  is  a  mixture  of  phenol  and  aniline.    26) What  are  azeotropes?  Give  example.                 [2]  

A:  Azeotropes  are  binary  mixtures  having  the  same  composition  in  liquid  and  vapour  phase  and  boil  at  a  constant  temperature.  For  example:  ethanol-­‐water  mixture    

27) State  Raoult’s  law  of  relative  lowering  of  vapour  pressure.         [1]  A:  Relative  lowering  of  vapour  pressure  is  equal  to  the  mole  fraction  of  the  solute.    

28) Define  colligative  property.                   [1]  A:  The  properties  depend  on  the  number  of  solute  particles  irrespective  of  their  nature  relative  to  the  total  number  of  particles  present  in  the  solution.  Such  properties  are  called  colligative  properties    

29) Mention  four  colligative  properties  of  dilute  solutions.           [2]  A:  Relative  lowering  of  vapour  pressure  

I. Elevation  in  Boiling  point  

II. Depression  in  Freezing  point  

III. Osmotic  pressure  

30) Define  the  term  relative  lowering  of  vapour  pressure.           [2]  A:  It  is  the  ratio  of  lowering  of  vapour  pressure  to  the  vapour  pressure  of  the  pure  solvent  o

o

P P Relativeloweringof V.PP−

=  

31) What  is  elevation  in  boiling  point?                 [1]  A:  Elevation  in  boiling  point  is  the  difference  between  the  boiling  point  of  the  solution  

containing  non-­‐volatile  solute  and  the  boiling  point  of  the  pure  solvent  

∆Tb  =  T  –  To  

32) Give  the  relation  between  elevation  in  boiling  point  and  molecular  mass  of  solute.   [2]  A:    

ΔTb = Kb 2

1 2

w 1000w M××

 

Where w2 is mass of solute, w1 is the mass of the solvent; M2 is molar mass of the solute

 33) Give  the  S.I.unit  of  ebullioscopic  constant  or  boiling  point  elevation  constant  or  molal  

elevation  constant.                     [1]  A:  The  unit  of  Kb  is  K  kg  mol-­‐1    

34) What  is  depression  infreezing  point?                 [1]    A:  It  is  the  decrease  in  the  freezing  point  of  solution  when  non-­‐volatile  solute  is  added  into  

solvent.  

35) Give  the  relation  between  depression  infreezing  point  and  molecular  mass  of  solute.[2]  A:    

ΔTf = Kf

2

2

1

wMw1000

∴ M2 = f 2

f 1

K 1000 wT w× ×Δ ×

where M2 is molar mass of the solute.

Note: Values of Kf and Kb of the solvent depends on their molecular mass and ΔHfusion and ΔHvap of the solvent respectively.  

36) Give  the  S.I.unit  of  cryoscopic  constant.               [1]    A:  The  unit  of  Kf  is  K  kg  mol-­‐1    

37) Draw  the  plot  showing  elevation  in  boiling  point  in  a  solution.         [2]  A:    

         

 

 

        ―∆Tb  ―  

 

    Temperature/K  

     

Vapour  pressure  

Tb  Tb0  

Solution  Solvent  

38) Draw  the  plot  showing  depression  in  freezing  point  in  a  solution.       [2]  

 

 

 

 

 

 

      ―∆Tf―  

 

    Temperature/K  

39) Define  osmosis.                       [1]  A:  The  process  of  movement  of  solvent  particles  from  lower  concentration  to  higher  

concentration  through  semi-­‐permeable  membrane  to  attain  equilibrium  is  called  osmosis.  

 

 

 

 

 

 

40) What  is  osmotic  pressure  and  give  its  relation  with  concentration  of  solution.   [2]  A:  The  amount  of  external  pressure  required  to  stop  the  osmosis.  � =  CRT    

Where:  �=  osmotic  pressure,  R  =  gas  constant,  T  =  temperature,  C  =  concentration  of  solution.    41)  What  are  isotonic  solutions?                   [1]  

A:  Two  different  solutions  having  sameosmotic  pressure  are  called  isotonic  solutions  

42) What  are  hypertonic  solutions?                 [1]  A:  The  solution  having  more  osmotic  pressure  than  other  

43) What  are  hypotonic  solutions?                 [1]  A:  The  solution  having  less  osmotic  pressure  than  other  

Tf   Tfo  

Vapour  pressure  

Solution  

Liquid  solvent  

Frozen  solvent  

44) Explain  the  application  of  reverse  osmosis  in  desalination  of  water.       [2]  A:  When  pressure  more  than  osmotic  pressure  is  applied,  pure  water  is  squeezed  out  of  the  sea  water  through  the  membrane.  A  variety  of  polymer  membranes  are  available  for  this  purpose.  The  pressure  required  for  the  reverse  osmosis  is  quite  high.  A  workable  porous  membrane  is  a  film  of  cellulose  acetate  placed  over  a  suitable  support.  Cellulose  acetate  is  permeable  to  water  but  impermeable  to  impurities  and  ions  present  in  sea  water.  

45) What  is  reverse  osmosis?                   [1]  A:  Movement  of  solvent  particles  from  higher  

concentration  to  lower  concentration  through  a  semi  

permeable  membrane,  when  pressure  is  applied  

greater  than  osmotic  pressure  

 

 

 

46) What  is  abnormal  molar  mass?                 [1]  A:  A  molar  mass  that  is  either  lower  or  higher  than  the  expected  or  normal  value  is  called  as  abnormal  molar  mass.  

47)  Define  Vant  hoff  factor  

Van’t Hoff factor ‘i’ to account for the extent of association or dissociation of a solute in a solvent is

i = Normal molar massAbnormal molar mass

or

i = observed colligative propertycalculated colligative property

or

i = total number of moles of particles after association or dissociationNumber of moles of particles before association or dissociation

48) What  is  the  value  of  i  for  NaCl.                   [1]  A:  2    49) What  is  the  value  of  i  for  K2SO4.                 [1]  A:  3  

50) What  is  the  value  of  i  for  sugar.                 [1]  A:  1  51) What  is  the  value  of  i  for  glucose.                 [1]  A:  1  52) On  what  factor  the  colligative  property  depends  on.           [1]  A:  It  depends  on  number  of  moles  of  solute  particles  but  not  on  the  nature  of  the  solute.  53) Write  the  mathematical  equation  of  Raoults  law  in  case  of  non-­‐volatile  solute.   [1]  

A:  If  one  of  the  components  (solute)  is  non-­‐volatile  then  the  equation  of  Raoults  law  is.  

PB=  O  

P  =  PA  +  PB  

P  =  PA  +  O  

P  =  PA  

 

54) Write  the  differentiate  between  non-­‐ideal  solutions  with  positive  deviation  and  negative  deviation  from  Raoult’s  law                 [2]  

 

 

 

 

 

 

 

 

55) Define  lowering  of  vapour  pressure?                 [1]  

A:  It  is  defined  as  the  difference  between  the  vapor  pressure  of  the  solvent  in  pure  state  and  the  

vapour  pressure  of  the  solution    

∆P  =  Po  –  P    

56) State  Roult’s  law  of  relative  lowering  of  vapour  pressure           [1]  

A:  It  states  that  the  relative  lowering  of  vapour  pressure  is  equal  to  the  mole  fraction  of  the  

solute    

 

Positive  deviation    

(a)In  this  solution  solvent  –  solute  interaction  is  weaker  than  solvent  –  solvent,  solute-­‐solute  interactions    (b)  P  >  PA  +  PB  (c)  ∆  V  >  O  (d)  ∆H  =  positive  (e)      It  forms  azeotrope  with  minimum  boiling  point    

Negative  deviation    

(a)      In  this  solution  solvent  –  solute  interaction  is  stronger  than  solvent  –  solvent,  solute-­‐solute  interactions  (b)  P  <  PA  +  PB  (c)    ∆  V  <  O  (d)  ∆H  =  negative  (e)It  forms  azeotrope  with  maximum  boiling  point      

P  =  PAO.  XA  

57) Why  sea  water  freezes  below  00C?                 [1]  

 

A:  Sea  water  freezes  below  00C  due  to  the  presence  of  the  non-­‐volatile  solute  dissolved  in  the  

water.  

 

58) Derive  the  equation  to  calculate  molecular  mass  of  unknown  solute  using  Raoult’s  law  

of  relative  lowering  of  V.P                   [3]  

A:  According  to  Raoult’s  law  relative  lowering  of  vapour  pressure  is  equal  to  the  mole  fraction  

of  the  solute.  o

Bo

P P XP−

=  

o nP P Bo n nP BA

−=

+  

nB<<<nA  for  dilute  solution  

So  we  can  neglect    nB  in  denominator    o nP P Bo nP A

−=  

B

B

A

A

Wo MP Po WP

M

−=  

o MWP P B Ao W MP BA

−=  

B AB

A

oW .M PM oW P P⎛ ⎞

= ⎜ ⎟−⎝ ⎠

 

 

 

 

Numerical  problems  

1. A  solution   containing  2.56  g   sulphur   in  100  g  CS2  gave  a   freezing  point   lowering  of  0.383  K.  Calculate  the  molar  mass  of  sulphur  molecules.  Given  Kf  of  CS2  =  3.83  K  kg  mol−1.  

Ans.  ΔTf  =  0.383  K,     Kf  =  3.83  K  kg  mol−1  

ΔTf  =  Kf  ×  m     ;   ΔTf  =  Kf  ×  

2

2

1

WMW1000

 

M2  (molar  mass  of  sulphur  molecules)  =   2.56 1000 3.83100 0.383× ×

×  =  256  g  mol−1  

2. 100  g  of  water  has  3g  of  urea  dissolved  in  it.  Calculate  the  freezing  point  of  the  solution.  Kf  for  water  =  1.86  K  kg  mol−1,  molar  mass  of  urea  =  60  g  mol−1,    freezing  point  of  water  =  273.15  K  (0°C)  

Ans.  ΔTf  =  Kf  ×  m       ;   ΔTf  =  1.86  ×  

2

2

1

WMW1000

    ΔTf  =  1.86  ×3 100060 100×

×  =  0.93  

  ΔTf  =   0f fT T−         ∴  Tf  =  273.15  −  0.93  =    272.22     or     −0.93°C  

3. Human   blood   has   osmotic   pressure   of   7.2   atm   at   body   temperature   of   37°C.   Calculate   the  molar  concentration  of  solute  particles  in  blood.    Given  R  =  0.0821  L  atm  K−

1.  

Ans.    π  =  CRT     ;     C  =  RTπ       T  =  273  +  37  =  310  K  

C  (molar  concentration)  =   7.20.0821 310×

 =  0.2828  M  

4. Vapour  pressure  of  benzene  is  200  mm  of  Hg.  2g  of  a  non-­‐volatile  solute  in  78  g  benzene  has  vapour   pressure   of   195   mm   of   Hg.   Calculate   the   molar   mass   of   the   solute.   Molar   mass   of  benzene  =  78  g  mol−1.  

Ans.   2

1

nP PP n°−

=°

 ;  

2

2

1

1

WMP PWPM

°−=

°     200 195

200−  =   2

2M7878

;      

  Molar  mass  of  solute  (M2)  =  200 25×  =  80  g  mol−1  

5. 500  g   of  water   containing   27   g   of   a   non-­‐volatile   solute  will   boil   at   100.156°C.   Calculate   the  molar  mass  of  the  solute.  Given  boiling  point  of  water  =  100°C,  Kb  =  0.52  K  kg  mol−1.  

Ans.  ΔTb  =  Kb  ×  m   ;   ΔTb  =  Kb  ×  

2

2

1

WMW1000

 

  Molar  mass  of  solute  (M2)  =  0.52 27 1000500 0.156× ×

×  =  180  g  mol−1.  

   

Unit  3  

Electrochemistry  

One  mark  questions  

1. What  is  an  electrolyte?  An  electrolyte  is  a  compound  which  conducts  electricity  either  in  its  aqueous  solution  or  in  its  molten  state.  e.g    Acids    HCl,  CH3COOH,  HNO3  

                   Bases    NaOH,  NH4OH                      Salts        CuSO4,  NaCl    etc      

2. Define  conductivity  of  an  electrolytic  solution.  Conductivity  of  a  solution  of  an  electrolyte  is  the  conductance  of  a  solution  placed  between  two  electrodes  each  of  one  square  meter  area  kept  at  a  distance  of  1  meter  apart.    

3. Write  the  S.I  unit  for  conductivity.  SI  unit  for  conductivity  is  Sm-­‐1.    

4. Give  the  S.I  unit  for  molar  conductivity.  Sm2  mol-­‐1  

 5. State  Kohlrausch    Law.  

The  limiting  molar  conductivity  of  an  electrolyte  can  be  represented  as  the  sum  of  the  individual  contributions  of  the  anion  and  cation  of  the  electrolyte.    

6. Define  electrode  potential.  The  potential  difference  developed  between  the  electrode  (metal)  and  the  electrolyte  (solution  containing  its  own  ions)  when  both  the  metal  and  the  solution  are  in  equilibrium  is  called  electrode  potential.    

7. Define  standard  electrode  potential.  Standard  electrode  potential  is  the  electrode  potential  when  the  concentrations  of  all  the  species  involved  is  unity  (1M)  and  if  a  gas  is  involved  its  pressure  should  be  1  bar.        

8. Write  Nernst  Equation.  

⎡ ⎤⎣ ⎦n+ n+

o10 n+( M / M) ( M / M)

0.059 1E = E - logn M

 

 

9. State  Faradays  second  law  of  electrolysis.  The  amounts  of  different  substances  liberated  by  the  same  quantity  of  electricity  passing  through  the  electrolytic  solution  are  proportional  to  their  chemical  equivalent  weights.  

10. Define  cell  potential.  Cell  potential  is  the  potential  difference  between  the  two  electrodes  of  the  galvanic  cell.    

11. Define  EMF  of  the  cell.  It  is  the  difference  between  the  electrode  potential  of  the  cathode  and  anode  when  no  current  is  drawn  through  the  cell.    

12. What  is  Fuel  cell?  Galvanic  cells  that  are  designed  to  convert  the  energy  of  combustion  of  fuels  like  hydrogen,  methane  etc  directly  into  electrical  energy  are  called  fuel  cells.    

13. Give  a  method  to  prevent  rusting.  Rusting  may  be  prevented  by  barrier  protection  like  painting,  metal  plating  etc.    

14. Write  the  relationship  between  cell  potential  and  Gibb’s  energy  

                                                                   o o

r cellG nFEΔ = −  

15. Write  the  relationship  between  equilibrium  constant  and  Eocell  

0.059 logocell c

VE Kn

=    

2  mark  questions  

1. What  are  redox  reactions?  Give  an  example.  Reactions  in  which  both  oxidation  and  reduction  taken  place  simultaneously  are  called  redox  reactions.  

e.g         ⎯⎯→2+ 2+Zn+Cu Zn +Cu    In  this  Zn  is  oxidised  to  Zn2+                            Cu2+  is  reduced  to  Cu    2. Mention  any  two  factors  on  which  the  conductivity  of  an  electronic  conductor  depends.  

The  electronic  conductance  depends  on    (i) The  nature  and  structure  of  the  metal  (ii) The  number  of  valence  electrons  per  atom.  (iii)  Temperature  (it  decreases  with  increase  in  the  temperature)  (any  two)  

 3. Mention  any  two  factors  on  which  the  conductivity  of  an  electrolytic  conductor  depends.  

The  conductivity  of  electrolytic  solution  depends  upon    (i)    The  nature  of  the  electrolyte  

(ii)    Size  of  the  ions  produced  and  their  solvation.  (iv) The  nature  of  the  solvent  and  its  viscosity.  

(iv)    Concentration  of  the  electrolyte  and    (v)  Temperature  (increases  with  increase  in  temperature  (any  two)    

4. Give  two  difference  between  the  conductivity  of  an  electronic  conductor  and  electrolytic  conductor.    1.  On  passing  direct  current  composition  of  electronic  conductor  does  not  change  but  that  of              electrolytic  conductor  changes.  2.  On  increasing  the  temperature  in  case  of  electronic  conductor  conductivity  decreases  in  case  of              electrolytic  conductor  conductivity  increases.    

5. What  is  a  strong  electrolyte?  Give  an  example.  A  strong  electrolyte  is  an  electrolyte  that  dissociates  completely  into  ions  at  moderate  concentrations  of  its  aqueous  solution  Ex:  acids  HCl,  H2SO4,  HNO3  Base  NaOH,  KOH  Salts  NaCl,  CuSO4  (any  salt)    

6. What  is  a  weak  electrolyte?  Give  an  example.  A  weak  electrolyte  is  an  electrolyte  that  dissociates  partially  into  ions  in  its  aqueous  solution.  Ex:  CH3COOH,  NH4OH    

7. Define  molar  conductivity.  How  is  it  related  to  conductivity?  Molar  conductivity  of  a  solution  at  a  given  concentration  is  the  conductance  of  the  volume  V  of  a  solution  containing  one  mole  of  electrolyte  kept  between  two  electrodes  with  area  of  cross  section  A  and  distance  of  unit  length.      It  is  represented  by  λm  

λm  =  kv    where  k  is  conductivity  and  v  is  volume  of  the  solution  containing  1  mole  of  the  electrolyte                                                                                                    or  If  λm    is  in  Sm2mol-­‐1  and  k  in  Sm-­‐1    

                                                              mkλ =

1000C          where  C  is  conc.  in  mol  L-­‐1  

                                                                                         or                                  When  λm        is  in  S  cm2mol-­‐1  and  k  is  in  Scm-­‐1  

                                                                                      m1000kλ =C

 

 8. How  does  conductivity  of  a  solution  change  with  change  in  concentration  of  the  solution?  Give  

reason.    Conductivity  of  a  solution  decreases  with  decrease  in  concentration  of  the  solution  due  to  decrease  in  the  number  of  ions  per  unit  volume  of  the  solution.        

9. Define  limiting  molar  conductivity.  Write  the  relationship  between  molar  conductivity  and  limiting  molar  conductivity.  Limiting  molar  conductivity  is  the  molar  conductivity  of  a  solution  when  concentration  approaches  zero  or  molar  conductivity  at  infinite  dilution.  

1o 2

m mλ = λ - AC  where  λm  is  molar  conductivity  and  λom  is  limiting  molar  conductivity,  C  is  

concentration  in  mole/L  and  A  is  constant  which  depends  on  nature  of  the  electrolyte,  solvent  and  temperature.    

10. Draw  a  graph  of  molar  conductivity  verses  square  root    of  the  molar  concentration  for  KCl  and  CH3COOH  mentioning  clearly  each.    

                                                                                       

11. How  is  limiting  molar  conductivity  for  a  strong  electrolyte  found  out  by  extrapolation  method?  Prepare  four  solutions  of  given  strong  electrolyte  of  different  concentrations.  Measure  the  conductivities  of  each  solutions  using  conductivity  cell  and  calculate  the  molar  conductivities  of  each  solution.  Plot  a  graph  of  molar  conductivity  verses  square  root  of  the  molar  concentration  for  these  solutions.  A  straight  line  is  obtained  which  is  to  be  extrapolated  back  so    as  to  touch  the  vertical  axes  .This  point  of  intersection  on  the  vertical  axes  gives  the  limiting  molar  conductivity.    

12. State  and  illustrate  Faradays  first  law  of  electrolysis.  The  amount  of  chemical  reaction  which  occurs  at  any  electrode  during  electrolysis  by  a  current  is  proportional  to  the  quantity  of  electricity  passed  through  the  electrolyte  either  through  its  aqueous  solution  or  molten  state.    If  w  is  the  mass  of  the  substance  deposited  and  Q  is  the  current  passed  in  coulombs  

                                   w  ∝    Q  But  Q  =  I  t    where  I  is  the  current  strength  in  ampere  and  t  is  time  in  seconds.    13. Conductivity  of  0.01  M  NaCl  solution  is  0.12  Sm-­‐1.  Calculate  its  molar  conductivity.  

-2 2m

k 0.12λ = = =1.2 ×10 Sm / mol1000C 1000 × 0.01

 

   

14.  The  molar  conductivity  of  0.1M  nitric  acid  is  630  S  cm2  /mol.  Calculate  its  conductivity.    

                     m

-1

1000kλ =C1000k630 =0.1

630×0.1∴ k = = 0.063 Scm1000

 

 15. A  solution  of  Ni(NO3)2  is  electrolysed    between  platinum  electrodes  using  a    current  of  5  amperes  for  

20  minutes.    What  mass  of  nickel  is  deposited  at  the  cathode?  (Mol  mass  of  Ni  =  58.7)                                                                                                          Q  =  I    t      

                                                                                                             =  5×20×60  =  6000C  

                                               

⎯⎯⎯⎯⎯→ 2+ -Ni + 2e Ni 2 × 96500C 193000C 58.7g

 

For  193000C  of  electricity  mass  of  nickel  obtained                                                      =  58.7g  

For  6000C  of  electricity          6000 × 58.7 =1.812g193000

 

 16. How  long  it  will  take  for  the  deposition  of    0.2g  of  silver  when  silver  nitrate  solution  is  electrolysed  

using  0.5  ampere  of  current  (Mol  mass  of  Ag  =  108)      

⎯⎯→+ -Ag + e Ag 96500C 108g

 

For  108g  of  silver  to  be  deposited  current  required  is  96500C.  For  0.2g  of  Ag      

But  Q  =  I  t        0.2 × 96500 =178.7C = Q

108  

Q 178.7t = = = 357.4 seI 0.5

 

17.    The  cell  in  which  the  following  reaction  occurs   3 2( ) ( ) ( ) 2( )2 2 2aq aq aq sFe I Fe I+ − ++ ⎯⎯→ +  

Has  Eocell  =  0.236V  at  298K.  Calculate  the  standard  Gibb’s  energy  and  the  equilibrium  constant  for  the  cell  reaction.  

               n  =  2                   Δ.Go  =  -­‐nFEo                                                    =  -­‐  2×96500×0.236                                                                                                          =  -­‐  45548  J  

                                                                                 0.059 logcellE Kn

=  

                                                                             0.0590.236 log2

K=  

                                                                               2 0.236log 80.059

K ×= =  

Taking  the  antilog    K  =  108    

 18. Write  the  reaction  taking  place  at  cathode  and  anode  when  aqueous  solution  of  copper  sulphate  is  

electrolysed  using  copper  electrodes.  

                                                   2

( ) ( )

2( ) ( )

t 2

t 2

anode

cathode

oxdns aq

rednaq s

A Cu Cu e

A Cu e Cu

+ −

+ −

⎯⎯⎯→ +

+ ⎯⎯⎯→  

Thus  copper  from  anode  dissolves  and  an  equivalent  amount  of  pure  copper  is  deposited  on  cathode.  This  technique  is  used  in  electrolytic  refining  of  crude  copper.      

19. Write  the  reaction  taking  place  at  anode  and  cathode  when  molten  NaCl  is  electrolysed.  When  molten  sodium  chloride  is    electrolysed  using  inert  electrodes  

                                                    22 2t anode

oxdn

redn

A Cl Cl eAt cathode Na e Na

− −

+ −

⎯⎯⎯→ +

+ ⎯⎯⎯→  

Thus  chlorine  gas  is    liberated  at  anode  and  Sodium  metal  is  formed  at  cathode.    

20. Write  the  reaction  taking  place  when  aqueous  solution  of  NaCl  is  electrolysed.  When  aqueous  solution  of  NaCl  is  electrolysed,  

                                                           2

NaCl Na Cl

H O H OH

+ −

+ −

⎯⎯→ +

+à àÜá àà                                                                    

The  reaction  taking  place  at  cathode  is    

                                                              2( )1 2aq gH e H+ −+ ⎯⎯→  

The  reaction  taking  place  at  anode  is    

                                                  ⎯⎯→- -aq 2 aq

1Cl Cl + e 2

 

 21. What  is  a  primary  battery/cell  ?Give  an  example.  

Primary  battery  is  one  in  which  reaction  occurs  only  once  and  cannot  be  recharged.  Eg  Dry  cell  or  Leclanche  cell  and  Mercury  cell    

22. What  is  a  secondary  battery/cell  ?  Give  an  example.  Secondary  battery  is  one    which  can  be  recharged  by  passing  current  through  it  in  opposite  direction,so  that  it  can  be  Reused.  

Eg:    Lead  storage  battery  and  Nickel  cadmium  cell.    

23. EoCu  =  +0.34V  and  EoZn  =-­‐0.76V.  Daniel  cell  is  obtained  by  coupling  these  two  electrodes.  (i)  represent  the  cell  symbolically  (ii)  calculate  the  EMF  of  the  cell    (i)  Daniel  cell  can  be  represented  as  

                                                                                                     Zn/  Zn2+  (aq)  ||  Cu2+  

(aq)  /Cu    (ii)  EMF  of  Daniel  cell  Eo  cell  =  EoR  -­‐  EoL                                                                                                          =  EoCu  -­‐  EoZn  =  0.34-­‐(-­‐0.76)                                                                                                        =  1.10V    24. Calculate  the  molar  conductivity  of  a  solution  of  MgCl2  at  infinite  dilution  given  that  the  molar  ionic  

conductivities  of     2+ -o 2 -1 o 2 -1( Mg ) ( Cl )λ =106.1 Scm mol and λ = 76.3 Scm mol  

                                                                                                         

2+ -2

o o oMgCl Mg Cl

2 -1

λ = λ + 2λ= 106.1+ 2( 76.3)= 258.7 Scm mol

 

 25. The  resistance  of  a  conductivity  cell  containing  0.001  M  KCl  solution  at  298K  is  1500Ω .  What  is  the  

cell  constant  if  the  conductivity  of  0.001M  KCl  solution  at  298K  is  0.146×10-­‐3  Scm-­‐1?    Cell  constant  G*=  Rk                                                            =resistance  ×  conductivity                                                          =0.146×10-­‐3  Scm-­‐1×1500S-­‐1                                                          =  0.219  cm-­‐1  

 Question  carrying  3  or  4  marks  

 1. Explain  the  construction  of  Daniel  cell.  Write  the  reaction  taking  place  at  anode  and  cathode  and  the  net  

cell  reaction.  (3  mark)  To  prepare  Daniel  cell  get  a  zinc  electrode  by  dipping  zinc  rod  in  1M  ZnSO4  solution.  Get  a  copper  electrode  by  dipping  a  copper  plate  in  1  M  CuSO4  solution.  Couple  these  two  electrodes  using  a  salt  bridge  to  get  Daniel  cell.  Reactions  taking  place    

                               

2

2

2 2( ) ( ) ( ) ( )

t anode 2 cathode 2

oxdn

redn

s aq aq s

A Zn Zn eAt Cu e CuNet cell reaction Zn Cu Zn Cu

+ −

+ −

+ +

⎯⎯⎯→ +

+ ⎯⎯⎯→

+ ⎯⎯→ +  

 

2. With  a  labeled  digram  explain  standar  hydrogen  electrode.  Represent  it  symbolically.  Write  the  reduction  reaction  at  the  anode.  What  is  its  electrode  potential?    (4  marks)                                                                                        

 It  consists  of  a    platinum  electrode  coated  with  platinum  black.  The  electrode  is  dipped  in  1M  HCl.  Pure  hydrogen  gas  is  bubbled  through  it  under  a  pressure  of  1  bar.  S.H.E  is  represented  as                                                                                                        Pt(s)  |H2  (g)(1bar)  |H+

(aq)(1M)          The  reduction  reaction  taking  place  is    

                                                                                        21( ) ( )2

H aq e H g+ −+ ⎯⎯→  

S.H.E  is  assigned  an  electrode  potential  of  0.0  V  at  all  temperatures.      

3. Explain  the  use  of  standard  hydrogen  electrode  in  measuring  the  standard    electrode  potentials  of  copper  and  zinc  electrode  (4  mark)  Construct  a  standard  electrode  of  the  given  metal  by  dipping  the  pure  metal  in  1M  solution  of  its  own  ion  at  25o  C  Couple  this  standard  electrode  with  SHE  using  a  salt  bridge  to  get  galvanic  cell.  Measure  the  emf  of  the  cell  using  suitable  instrument  like  potentiometer.                                Eo  =  EoR  –  EoL    One  of  the  electrodes  of  the  cell  is  SHE  and  its  electrode  potential  is  0.0V.  So  the  electrode  potential  of  the  given  electrode  will  be  the  emf  of  the  cell  in  magnitude.  If  reduction  takes  place  at  the  given  electrode  its  Eo  will  be  +ve  but  if  oxidation  takes  place  at  the  given  electrode  is  Eo  will  be  –ve.    e.g  if  SHE  is  coupled  with  standard  copper  electrode  reduction  takes  place  at  copper  electrode  cell  can  be  represented  as      Pt  (s)  |H2(g.  1bar)|H+

(aq1M)||Cu2+(aq.1M)|Cu  

2+ +2

o o ocell Cu / Cu H / HE = E - E  

2+ 2+o o

Cu / Cu Cu / Cu0.34 = E - 0 ∴ E = 0.34V          

 If  SHE  coupled  with  standard  zinc  electrode  oxidation  takes  place  at  zinc  electrode.  Cell  can  be  represented  as    

2( .1 ) ( ) ( ) 2( .1 ) ( .1 )/ aq M s s g bar aq MZn Zn Pt H H+ +⏐⏐ ⏐ ⏐  

22/ /

o o oH H Zn Zn

E cell E E+ += −  

2 2 //0.76 0 0.76o

ZnZn Zn ZnE E V+ += − ∴ = −  

 4. How  is  Kohlrausch  law  helpful  in  finding  out  the  limiting  molar  conductivity  of  a  weak  electrolyte?  (3  m)  

Let  us  try  to  calculate  λom  for  a  weak  electrolyte  CH3COOH.  Select    three  strong  electrolytes  whose  λo

m  can  be  found  by  extrapolation  method  in  such  a  way  that  if  we  subtract  λo

m  for  one  electrolyte  from  the  sum  of  λo

ms  of  the  remaining  two  electrolyte  λom  for  CH3COOH  can  be    obtained.  The  three  electrolytes  to  be  

selected  are  CH3COONa,  HCl  &  NaCl  

                                   3 3

o o o oCH COOH CH COONa HCl NaClλ = λ + λ - λ  

 5. The  values  of  limiting  molar  conductivities  (λo

m)  for  NH4Cl,  NaOH  and  NaCl  are  respectively  149.74;  248.1  and  126.4  Scm2mol-­‐1.  Calculate  the  limiting  molar  conductivity  of  NH4OH      (3M)  

                                                                       4 4

o o o oNH OH NH Cl NaOH NaClλ = λ + λ - λ  

                                                                                                         =  149.74+248.1-­‐126.4                                                                                                            =  271.44  Scm2  mol-­‐1    

6. Calculate  the  equilibrium  constant  for  the  reaction  at  298K   2( ) ( ) ( )2 ( ) 2s aq sCu Ag aq Cu Ag+ ++ ⎯⎯→ +  

Given  that  Eo  Ag+/Ag  =  0.80V  and    Eo(Cu2+/Cu)  =  0.34V  

                                                     0.059 logo

cell cE Kn=  

                                                log0.059

o

cnE cellK∴ =  

                                                            2( / ) ( / )o o ocell Ag Ag Cu Cu

E E E+ += −  

                                                                                 =0.80-­‐0.34=0.46V  

                                                     2 0.46log 15.590.059cK×

= =  

                         Taking  the  antilog  Kc  =3.92×1015          

7. In  Leclanche  cell  (dry  cell)  what  are  anode  and  cathode?  What  is  the  electrolyte  used?  Write  the  reactions  at  each  electrode.  What  is  the  role  of  zinc  chloride?    It  consists  of  a  zinc  container  as  an  anode.  A  graphite  rod  surrounded  by  a  mixture  of  manganese  dioxide  and  carbon  powder  is  cathode.      The  space  between  the  electrodes  is  filled  with  electrolyte  a  moist  paste  of  ammonium  chloride  and  zinc  chloride        

Reaction  taking  place  

⎯⎯→ 2+ -( s)At anode Zn Zn + 2e  

⎯⎯→+ -2 4 3At cathode MnO + NH + e MnO( OH) + NH  

NH3  produced  in  the  reaction  forms  a  complex  with  Zn2+  to  form  [Zn(NH3)4]2+.    

8. What  are  the  anode  and  cathode  of  lead  acid  battery?  What  is  the  electrolyte?  Write  the  reactions  taking  place  at  anode  and  cathode  and  the  overall  reaction  during  discharging  of  the  battery.  (3  M)  It  consists  of  lead  anode  and  a  grid  of  lead  packed  with  lead  dioxide  (PbO2)  as  cathode.    Electrolyte  is  38%  solution  of  sulphuric  acid.  The  reactions  taking  place  when  the  battery  is  in  use  are    

                                                   2

( ) 4 ( ) 4( )

22 4 ( ) ( ) 4 ( ) 2 ( )

2

( ) 4 2 2 s aq s

aq aq s l

Anode Pb SO PbSO e

Cathode PbO s SO H e PbSO H O

− −

− + −

+ ⎯⎯→ +

+ + + ⎯⎯→ +  

The  overall  reaction  is   ( ) 2 ( ) 2 4 ( ) 4 ( ) 2 ( ) 2 2 2 s s aq s lPb PbO H SO PbSO H O+ + ⎯⎯→ +  

 9. In  Hydrogen  oxygen  fuel  cell  (i)  Draw  the  schematic  diagram  mentioning  the  anode  and  cathode.  What  is  

the  electrolyte?  Write  the  reaction  taking  place  at  each  electrodes  and  the  net  cell  reaction.  (4M)  

                                                                                                                   In  this  hydrogen  and  oxygen  gases  are  bubbled  through  porous  carbon  electrodes  into  concentrated  aqueous  sodium  hydroxide  solution.  Catalyst  like  finely  divided  platinum  or  palladium  is  incorporated  into  the  electrodes  for  increasing  the  rate  of  electrode  reaction    Reaction  taking  place  are  

⎯⎯→

⎯⎯→

- -2 ( g) 2 ( l) ( aq)

- -2( g) ( aq) 2 ( l)

Cathode O + 2H O + 4e 4OHAnode 2H + 4OH 4H O + 4e

 

Overall  reaction  is    

⎯⎯→2 ( g) 2 ( g) 2 ( l)2H + O 2H O  

 

10. What  is  corrosion?  During  rusting  of  iron  write  the  anodic  and  cathodic  reactions.  Give  the  composition  of  rust.  (3M)  When  a  metal  is  exposed  to  the  atmosphere  it  is  slowly  attacked  by  the  constituents  of  the  environment  as  a  result  of  which  the  metal  is  slowly  lost  in  the  form  of  its  compound  .  This  is  called  corrosion.      Reaction  taking  place  are  

⎯⎯→

⎯⎯→

2+ -( s)

-2 ( g) 2 ( l)

At Anode 2Fe 2Fe + 4eAt Cathode O + 4H +( aq)+ 4e 2H O

 

H+  are  produced  from  H2CO3  formed  due  to  dissolution  of  carbon  dioxide  from  air  into  water  The  Fe2+  ions  are  further  oxidised  by  atmospheric  oxygen  to  ferric  ion  which  are  ultimately  converted  to  hydrated  ferric  oxide  called  rust.  Composition  of  rust  is  (Fe2O3.xH2O).      

11. A  conductivity  cell  when  filled  with    0.01M  KCl  has  a  resistance  of  747.5  ohm  at  25oC.  When  the  same    cell  was    filled  with  an  aqueous  solution  of    0.05M  CaCl2  solution  the  resistance    was  876  ohm.  Calculate    (i)    Conductivity  of  the  solution  (ii  )Molar  conductivity  of  the  solution            (given  conductivity  of  0.01M  KCl  =  0.14114  sm-­‐1)                                (3M)    Cell  constant  G*  =  Rk                                                              =  747.5×0.14114                                                            =0.105.5m-­‐1    

-1-1cell constant 105.5mConductivity k = = = 0.1204Sm

R 876 ohm  

2 -1m

k 0.1204Molar conductivity λ = = = 0.00241sm mol1000C 1000 × 0.05

 

12. The  electrical  resistance  of  a  column    of  0.05M  NaOH  solution  of  diameter  1cm  and  length  50cm  is  5.55×103  ohm.  Calculate  its    (i)  resistivity    (ii)  conductivity    (iii)  molar  conductivity                                    (3M)  

Cell  constant   o lG =a  

l  =  50  cm  Diameter  =  1  cm  ∴  radius  =  0.5  cm  Area  of  cross  section  A  =  πr2  =  3.14×(0.5)2      =  0.785  cm3  

 

                                                                                                                                                                    * -150G = = 63.694 cm0.785

 

-21 1Resistivity ρ = = = 87.135 Ωk 1.148 ×10

 

m

-2

2 -1

1000kMolarconductivityλ =C1000 ×1.148 ×10 =

0.05 = 229.6 S cm mol

 

13. Calculate  the  emf  of  the  cell  in  which  the  following  reaction  takes  place.  2

( ) ( )2 (0.002 ) (0.160 ) 2s sNi Ag M Ni M Ag+ ++ ⎯⎯→ +  

 Given  that  Eocell  =  1.05V    

2 2( )

10 2( )

[ ][ ]0.059 log2 [ ][ ]

socell cell

s

Ni AgE E

Ni Ag

+

+= −  

But  [M]  for  any  element  is  taken  as  unity    

⎡ ⎤⎣ ⎦

⎡ ⎤⎣ ⎦

2+o

cell cell 10 2+

Ni0.059E = E - log2 Ag

 

( )20.059 0.160=1.05 - log2 0.002

 

=  0.914V                  

Unit-4 CHEMICAL KINETICS

Number of Hours of Teaching-9 Marks allotted-8

In part-D, 5mark question is split preferable in the form of 3+2 Definition :- The branch of chemistry which deals with study of reaction rate and their mechanism is called chemical kinetics Rate of a chemical reaction Q. 1 What is rate of reaction? (1m) Ans: Change in molar concentration of reactant or product in per unit time is called rate of reaction. Types of rate of reactions For reaction R à P Average rate = decrease in conc. R Of reaction time taken rav = - ∆[R] ∆t Average rate = increase in conc. Of P Reaction time taken rav = + ∆[P] ∆t

8

Marks

Part –A 1x1=1

Part –B 1x2=2

Part-C 1x5=5

Q2:- For the reaction RàP, the conc. of reactant changes from 0.03M to 0.02M in 25 min. calculate average rate of the reaction using the unit of time in seconds. rav= - ∆[R]= - (0.02-0.03 ) ∆t 25x60 =-[-0.01] 1500 = 6.66x10-6 M/s Q3: What is the SI Unit of rate of reaction ? (1m) Ans: Mol /L /s

Factors influencing Rate of reaction

Q4 :- Mention any two factors which influence the rate of reaction . 2M Ans 1) Pressure or conc. of reactants 2) temperature 3) catalyst.

Dependence of rate on concentration . Q5.) What is rate law ? (1 m) Ans: Representation of rate of reaction in terms of concentration of reactants is called rate law. Rate expression and rate equation Q.6) Define rate equation or rate expression (2m) Ans: Expression in which reaction rate is given in terms of molar conc. of reactants with each term raised to some power which may or may not be same as the stoichiometric coefficient of the reacting species in a balanced chemical equation. Q.7) Define rate constant of a reaction. (1m) Ans: Rate constant is equal to rate of reaction when the product of the molar conc. of reactants is unity.

Order of a Reaction Q.8) Define order of a reaction. 1M Ans: Sum of the powers of the concentration of the reactants in the rate equation is called order of reaction. Q.9) Calculate the overall order of a reaction which has the rate expression. 1M Rate= K [A]1/2 [B]3/2 Ans: Order of reaction = 1/2 + 3/2

= 2

Q.10) What is elementary reaction ? (1m) Ans: Reaction taking place in one step is called elementary reaction.

Q.11)What are complex reactions?(1m) Ans: Reactions taking place in more than one step are called complex reaction. Q.12) What is SI Unit of rate constant of nth order reaction ? (1m) Ans: (mol)1-n . Ln-1.s-1 Q.13) What is SI unit of rate constant of zero order reaction? (1m) Ans: Mol/L/s Q.14) What is the order of reaction whose unit of rate constant and rate of reaction are same ?

(1m) Ans: Zero order. Q.15) Identify the reaction order from the rate constant K=2.3x10-5 mol-1.L.S-1 (1m) Ans: Comparing the unit of rate constant with general unit Mol-1.L.S-1 with ( Mol)1-n.Ln-1.S-1 1-n= -1 n=2

Molecularity of a reaction Q.16) Define molecularity of a reaction . (1 m) Ans: The number of reacting species taking part in an elementary reaction which must colloid simultaneously in order to bring about a chemical reaction is called molecularity of reaction. Q.17) In a complex reaction which step controls the overall rate of reaction and

what is it called? (2m) Ans: Slowest step, which is called rate determining step. Q.18) The conversion of molecules X to Y follows second order kinetics .If conc. of X Increased to three times,how will it affect the rate of formation of Y ? (1m) Ans: Increased rate =( Increased conc.)n

=32 =9 Rate of formation of Y increases by 9 times Integrated rate equations Q.19) Derive rate constant of zero order reaction (3m) Ans: Consider a zero order reaction R--> P Rate =-d[R] = K[R]o dt = - d[R] = K dt = d[R] = -kdt -------(1)

Integrating equation (1) both sides [R]= -kt+I ----------(2) Where “I” is integration constant At t=0 [R]=[R]o where [R]o is initial concentration of reactant. ∴Eqn (2) becomes I=[R]o Substituting I in eqn-------- (2) [R] = -Kt + [R]o -Kt = [R]-[R]o Kt = [R]o-[R] K = [R]o-[R] t Q:20) Derive integrated rate equation for first order reaction? (4m) Ans-Consider a first order reaction. RàP Rate = - d[R] = K[R] dt d[R]= - K[R] dt d[R] = - K. dt. -------(1) [R] Integrating eqn.(1)on both side ln [R] = - Kt + I --------(2) Where “I” is integration constant At t=o [R]=[R]o which called initial Concentration reactant Substituting the values in ln[R]o= I Equation (2) can be written as ln[R] = -Kt + ln[R]o Kt = ln[R]o – ln [R] Kt = ln [R]o [R] Kt =2.303 log[R]o [R] K= 2.303 x log[R]o t [R] Log [R]o Slope = K [R] 2.303. Or o time K= 2.303xSlope

Half life of a reaction Q21)Define halfe life of a reaction . (1m) Ans: The time in which the conc.of a reactant is reduced to one half of its initial conc. is called half life of a reaction (t1/2) Q:22) Show that half life of a zero order reaction is directly proportional to initial concentration of reactant from integrated rate equation. OR Derive the relation between half life and rate constant of zero order reaction .(2m). Ans:-Rate constant of zero order reaction is K= [R]o – [R] t At half life t =t ½ & [R] = ½ [R]o .: K= [R]o – ½ [R]o t1/2 K=[R]o 2t½ t ½ =[R]o 2 K

OR t ½ ∝ [R]o

Q.23) Show that half life of a first order reaction is independent of initial Conc. of reactant from integrated rate equation (2m) Or Derive the relation between half life of a first order reaction and its rate constant . (2m) Ans: Rate constant of first order reaction is K= 2.303 x log [R]o t [R] At half life t=t ½ ,[R]=[R]o 2 .: K = 2.303 x log [R]o t½ [R]o/2 K =2.303 x log 2 t ½ K= 2.303 x0.3010 t½ t½ = 0.693 K

K= 2.303 x0.3010 t½ t½ = 0.693 K

Q.24)A first order reaction is found to have a rate constant 5.5x10-14 /s .Calculate the half life of the reaction (2m) Soln. : K= 5.5x10-14 /s t½ =? t½ = 0.693 K = 0.693 5.5x10-14 t½ = 1.26x1013 sec Q:25) Show that the time required for 99/. Completion of a first order reaction is twice the time required for the completion of 90% of reaction( 4m) I set : [R]o= 100, [R]=[100-90]=10 t=t90% IIset : [R]o= 100 [R]= [100-99]1 t=t99% To be proved t99%= 2t90% K= 2.303 x log [R]o t [R] Sub. I set values . K= 2.303x log 100 t 90% 10 K = 2.303x log 10 t90% K= 2.303 X 1 - (1) t90% Substituting II set values K=2.303 x log 100 t99% 1 K= 2.303 x 2 --------------(2) t99% Comparing equations (1) & (2) 2.303x 1 = 2.303x2 t90% t99% t99% = 2t90%

Pseudo first order reaction

Q:26 Define pseudo first order reaction . Give an example. (2m) Ans: Chemical reactions which are not first order but behave as fist order reaction under suitable conditions are called pseudo first order Reactions. Ex: Inversion of cane sugar. C12 H22O11+H2O àC6H12O6 + C6H12 O6

Temperature dependence of the rate of a reaction

Q:27)How does rate of reaction vary with temperature? (1m). Ans: Rate of reaction increases with increase of temperature.

Q:28) What happens to the rate constant of a reaction when temperature is increased by 10o.? Ans: Rate constant increases nearly by two times. Q.29) Write Arrhenius equation which relates the rate constant , activation energy and temperature . (1m) Ans K= A e-Ea/RT

Energy of activation

Q.30) Define energy of activation (1m). Ans: The minimum energy required for the reactants to form activated complex is called Activation energy. Q.31) How is activation energy related to rate of reaction? (1m) Ans: Rate of reaction is inversely proportional to activation energy. ie r ∝ 1 Ea Q.32) How is activation energy affected by presence of positive catalyst? (1m) Ans: Activation energy of a reaction decreases in presence of catalyst. Q.33) On increasing 100K temperature rate of reaction becomes double, explain from the max well Boltzmann distribution curve. (2m)

On increasing 100K temperature, substance Increases the fraction of molecules double,hence rate of reaction doubles. Q:34) How does positive catalyst increases the rate of reaction? (2m) Ans positive catalyst decreases the activation energy by changing the Path of the reaction,which increases the rate of reaction

Collision theory of chemical reactions.

Q:35) What is effective collision? How is it related to rate of reaction? (2m). Ans. Collision in which molecules colloid with sufficient kinetic energy and proper orientation so as to form products is called effective collision. It is directly proportional to the rate of reaction. Q:36)How is activation energy calculated by plotting graph ln K against 1/T ? (2m) Q:37)Write Arrhenius equation at different Temperature and rate constants. Ans: log K2/K1 = Ea X T2-T1 2.303RT T1 T2 Q:38)The rate constants of a reaction at 500K. and 700K are 0.02s-1 and 0.07s-1respeetively calculate the activation energy. (3m) Ans: log K2/K1 = Ea X T2-T1 2.303RT T1 T2 log 0.07 = Ea x 700- 500 0.02 2.303x8.314 500x 700 0.544 = Ea x 5.714 x10-4 19.15 Ea= 0.544x19.15 5.714 x10-4 Ea= 18230.8 J = Ea= 18. 2308 KJ.

UNIT -5

SURFACE CHEMISTRY A. Short answer questions carrying 1 mark

1. What is adsorption

A surface phenomenon wherein there is accumulation of molecules on the surface (than in the bulk) of a solid or a liquid.

2. Why solids in finely divided state are good adsorbent?

Solids in finely divided state have large surface area, as surface area increases adsorbing power increases.

3. What is desorption?

The process of removing an adsorbed substance from a surface on which it is adsorbed is called desorption.

4. Name the substance used to decolour the solution of raw sugar.

Animal charcoal.

5. Name of the phenomenon in which both the adsorption and desorption takes place simultaneously.

Sorption

6. Why is adsorption always exothermic?

During adsorption there is always decrease in residual forces on the surface, hence adsorption is always exothermic.

Or There is decrease in surface energy which appears as heat, hence adsorption is always

exothermic. 7. Name catalyst used in the conversion of alcohols into gasoline (petrol)

Zeolite ZSM-5 (Zeolite Sieve of molecular porosity-5)

8. Name the colloidal system in which dispersed phase is solid and dispersion medium is liquid

Sol

9. Name the dispersed phase in gel

Liquid

10. Give an example for oil in water emulsion

Milk, Vanishing cream

11. What type of colloidal emulsion is present in butter

Water in oil (W/O)

12. What is the dispersion medium in gel?

Solid

13. Between Na2SO4 and Na3PO4 which has greater power to coagulate a positively charged colloid?

Na3PO4

14. Alum is added to muddy drinking water. Why?

Alum is added to muddy drinking water to coagulate

15. What is the dispersed phase in milk?

Oil or liquid

16. A liquid is dispersed in a gas. Name the type of colloid obtained.

Liquid aerosal

17. Name the instrument designed by Zigmondy.

Ultramicroscope

18. Movement of the dispersion medium in an electric field by preventing the movement of colloidal particles by suitable method.Name the phenomenon

Electroosmosis

19. The process by which colloidal particles aggregate, become bigger and settle down. Name the phenomenon

Coagulation

20.What happens when an electrolyte is added to lyophobic sol? Coagulation or precipitation 21. Name the phenomenon, when an electrolyte having a common ion is added to

freshly prepared precipitate? Peptization

B. Answer questions carrying 2 marks

ADSORPTION

1. What are adsorbate and adsorbent? Give an example.

Molecules (substances) that accumulates on the surface is called adsorbate. The material on the surface of which adsorption takes place is called adsorbent.

Example: Ni adsorbs H2. Ni is the adsorbate, H2 is the adsorbent

2. Give two examples for adsorption.

i) When animal charcoal is added to methylene blue, charcoal adsorbs the dye. ii) Air becomes dry in the presence of silica gel because silica gel adsorbs water

molecules on the surface iii) A small pillow of silica gel in a box adsorbs moisture in the box keeps the air

dry. (Any two)

3. Give differences between adsorption and absorption.

Adsorption Absorption 1. A substance gets

concentrated on the surface of a solid or liquid.

2. It increases with increase in surface area.

Example: adsorption of water by silica gel.

A substance gets uniformly distributed through the bulk of solid or liquid.

It remains unaffected by increase in surface

area. Example: Absorption of water by anhydrous

CaCl2.

4. Of SO2 (critical temperature 630K) and CH4 (critical temperature 190K) which gas will be adsorbed readily on the surface of 1 gram of activated charcoal. Justify the answer.

SO2 gas

Easily liquefiable gases with higher critical temperature are readily adsorbed as the theVander Waal’s forces are stronger near critical temperature.

5. What is the effect of temperature on physical and chemical adsorption?

Physical adsorption decreases with increase in temperature. Chemical adsorption increases with increase intemperature.

6. Mention any two applications of adsorption.

i) In the production of high vaccum

ii) In gas mask, to adsorb poisonous gases

iii) In the separation of noble gases using activated charcoal

iv) Removal of colouring matter from solutions

v) In adsorption chromatography to analyse a given

CATALYSIS

1. What is catalysis? Give an example.

A substance that accelerates the rate of a reaction without itself remaining unchanged chemically and quantitatively is a catalyst. The phenomenon is catalysis.

E.g.: 2KClO3 2MnO⎯⎯⎯→2KCl + 3O2 MnO2 is a catalyst.

2. What are promoters and poisons with respect to a catalytic process?

Promoters are substance that increases the activity of a catalyst. E.g.: In Haber’s process molybdenum acts as a promoter for iron used as a catalyst. A catalytic poison is one that decreases the efficiency or activity of a catalyst. E.g.: In Haber’s process CO if present in the mixture of H2 and N2, poisons the iron catalyst.

3. What is homogeneous catalysis? Give an example.

When reactants and catalyst are in the same phase the process is homogeneous catalysis. E.g.:

a) 2SO2(g) + O2(g) ( )gNO⎯⎯⎯→ 2SO3(g)

Here the reactants (SO2 and O2) and catalyst (NO) are all gases.

b) Acid hydrolysis of cane sugar is also an example for homogeneous catalysis. Here the reactants sugar solution, water and the catalyst dil. HCl are in the same phase (aqueous solution)

C12H22O11(aq) + H2O(l) H+

⎯⎯→C6H12O6 + C6H12O6 (both are in aq solution) Sucrose glucose fructose

4. What is heterogenous catalysis? Give an example.

A catalytic process in which reactants and catalyst are in different phases are known as heterogenous catalysis.

E.g.: 1. N2(g) + 3H2(g) (s)Fe⎯⎯⎯→2NH3 Here the reactants are gases, catalyst iron is a solid 2. Vegetable oil (l) + H2(g) ( )sNi

⎯⎯⎯→Vanaspathi ghee Here reactants and catalyst are in different phases.

5. Write a note on

a) activity b) selectivity of solid catalysts.

a) Activity: The activity (efficiency) of a solid catalyst depends on how strongly the reactants are chemisorbed on it. It is found that elements (metals) in group 7-9 of the periodic table show greater catalytic activity for hydrogenation reactions.

E.g.: 2H2(g) + O2(g) Pt⎯⎯→2H2O (l)

b) Selectivity: For a given set of reactants, different catalyst may yield different products. This is selectivity of a catalyst.

E.g.: CO(g) + 3H2(g) Ni⎯⎯→CH4(g) + H2O(g) CO(g) + H2(g) Cu⎯⎯→H−CHO

Ni is selective to convert water gas to CH4 whereas Cu converts water gas into formaldehyde. In otherwords Ni catalyses the conversion of water gas to CH4 but cannot catalyse to convert water gas to formaldehyde.

Catalyst is highly selective in nature i.e a given substance can act as a catalyst only in a particular reaction and not for all the reactions.

6. What is shape selective catalysis? Give an example.

A catalytic reaction that depends on pore structure of the catalyst and size of the reactant and product molecules is called shape selective catalysis. E.g.: zeolites.

7. Write a note on zeolites as shape selective catalysts.

Zeolites are aluminosilicates with 3D nework of Al-O-Si frame with honey comb like structure. This structure makes them to act as shape selective catalyst depending on pore size in them and on the size of reactant and products. Many zeolites are synthesized for selective catalytic activity.

E.g.: 1) Zeolite ZSM-5 (Zeolite Sieve of molecular porosity- 5) converts alcohols into gasoline (petrol) by dehydrating alcohols.

2) Many zeolites are used in petroleum industry in cracking of hydrocarbons and in isomerisation.

8. What are enzyme catalysis or biochemical catalysis? Give an example for enzyme catalysis.

Enzymes are proteins, which catalyse large number of reactions that maintain life processes in both plants and animals. Hence they are biochemical catalysts and the phenomenon is called as biochemical catalysis.

Inversion of cane sugar in the presence of enzyme invertase into glucose and fructose

9. Give two examples for enzyme-catalysed reaction.

a) Conversion of starch into maltose 2(C6H10O5)n (aq) + nH2O (l) diastase enzyme⎯⎯⎯⎯⎯→nC12H22O11 (aq)

Starch maltose b) Urea into ammonia and carbon dioxide NH2CONH2(aq) + H2O (l) urease⎯⎯⎯→2NH3(g) + CO2(g)

c) In human beings enzyme pepsin converts proteins into peptides and pancreatic trypsin enzyme converts proteins into amino acids.

d) Milk is converted into curds by lactobacilli enzyme present in the curd which is added to milk.

10. Mention characteristics of enzyme catalysis.

a) Their efficiency as catalyst is very high. b) They are highly specific in their action. c) They are highly active at optimum temperature and optimum pH. d) Their activity increases in presence of activatiors and coenzyme. e) Their activity decreases in presence of inhibitors and poisons.

11. Write the two steps involved in mechanism of enzyme catalysis.

a) An enzyme binds to the substrate to form an activated complex: E + S ⎯⎯→ES* b) Decomposition of activated complex to form the product: ES* ⎯⎯→E + P.

COLLOIDS

1. What is a colloid?

Colloid (Colloidal system or solution) is a heterogeneous system in which one substance is dispersed as very fine particle in another substance called dispersion medium. The size of the particle is larger than the true solution but smaller than the suspended particle i.e their diameter ranges between 1nm to 1000nm.

2. Name the 2 phases of colloidal system

Dispersion medium and dispersed phase (colloidal particles)

3. What is Dispersion Medium and dispersed phase for a colloid? Give an example.

The continuous medium in which the colloidal particles are dispersed is called Dispersion Medium. The discontinuous phase which the substance is dispersed as colloidal particles is called dispersed phase. Eg: Milk is a colloid in which fat globules form the dispersed phase, water is the dispersion medium.

4.Classify the colloids based on the physical state Based on the physical state of dispersed phase and dispersion medium colloids are

classified into 8 types

Dispersed Phase

Dispersion Medium

Name of the Type

Example

solid solid Solid sol Ruby glass, gems solid liquid sol Ink, Paint, Gold Sol solid gas Solid aerosol Dust, Smoke, Soot in Air, liquid solid gel curds, jam, silica gel, butter

liquid liquid emulsion Milk, Cream, Cod Liver Oil liquid gas Liquid aerosol Fog, Mist, Cloud gas solid Solid foam Foam rubber, Pumice stone gas liquid foam Shaving cream, soap lather

5. What is a Sol? Give an example.

It is a colloid wherein the dispersed phase is a solid and dispersion medium is a liquid. Eg: Sulphur dispersed in Water. Sulphur (solid) is the dispersed phase, water is the dispersion medium.

If the dispersion medium is water, alcohol and benzene, sol is called aqua sol (hydrosol), alcosol and benzosol.

6. How are colloids classified based on the affinity of the dispersed phase towards dispersion medium

Based on the affinity of dispersed phase towards dispersion medium, sols are classified as lyophilic and lyophobic sols (colloids).

7. What is lyophylic sol? Give an example

Lyophilic sol (colloid): These are sols in which the (colloidal particle) dispersed phase has affinity towards dispersion medium. (Intrinsic colloids). If the dispersion medium is separated from the dispersed phase, these sols can be formed by remixing them. Hence these are called reversible sols. Eg: Starch dispersed in water Albumin dispersed in water Gum or gelatin in suitable solvent.

8. What is lyophobic sol? Give an example Lyophobic sol (colloid): These are sols in which the dispersed phase has no affinity towards the dispersion medium (extrinsic colloids). Once precipitated or if the dispersion medium is separated from the dispersed phase, these sols cannot be formed by remixing them. Hence these are called irreversible sols. Eg: Sulphur dispersed in water, gold sol.

9.Distinguish between lyophillic to lyophobic sols (Any two)

Property Lyophilic Lyophobic 1 Affinity towards

dispersion medium High affinity Low affinity

2 Method of preparation Easily formed on mixing or heating the dispersed phase with dispersion medium

Special methods are used

3 Stability stable Highly unstable 4 Reversibility Reversible Irreversible 5 Solvation of colloidal

particles. Highly Solvated Not solvated

6 Addition of electrolyte Does not easily coagulate

Gets easily coagulated

10.Classify the colloids based on type of particles of the dispersed phase

Multimolecular colloid, Macromolecular colloid, associated colloid (micelles)

11.Write a note on Multimolecular colloids with an example

If large number of atoms or smaller molecules of a substance aggregate together to form particles having size in the colloidal range then the colloidal system is known as multimolecular colloid. E.g.: A gold sol contains colloidal gold particles each made up of large number of gold atoms. Sulphur sol contains an aggregate of thousands of S8 sulphur molecules.

12.Write a note on Macromolecular colloids with an example

Macromolecule (polymers) in a suitable solvent form solutions in which the size of the macromolecules may be in the colloidal range and the system is known as macromolecular colloids. E.g.: starch, cellulose, enzymes, proteins, nylon, polystyrene in a suitable solvent.

13. Write a note on Associated colloids /micelles with an example

Some substances at low concentration behave as strong electrolytes (true solution), but at higher concentrations aggregate to form colloidal particles. Such substances form associated colloid. Aggregate of molecules thus formed is called a micelle. Formation of micelle takes place if i) the temperature is above Kraft temperature (TK) ii) concentration is greater than critical micelle concentration (CMC).

If an associated colloid (micelle) is diluted, it behaves as a strong electrolyte. E.g.: Surface active agents like soaps and detergents form associated colloids. These have both lyophilic and lyophobic groups. For soaps, critical micelle concentration is 10−

4 to 10−

3 mol

14. Write equations for the preparation of Sulphur sol and Ferric hydroxide sol

I. SO2 + 2H2S oxidation⎯⎯⎯⎯→ 3S (sol) + 2H2O 2. FeCl3 + 3H2O hydrolysis⎯⎯⎯⎯→ Fe(OH)3(sol) + 3HCl 15. How is a metal sol prepared by Bredig’s arc

process?

This process involves both dispersion and condensation. Sol of metals like gold, platinum and silver can be prepared by this method. Two

metal

electrodes of a metal are dipped in water and an electric arc is struck between them. Intense heat of the arc causes the metal to vapourise.The vapours condenses to form metal particles of colloidal size. Thus metal sol is obtained.

16. What is peptization? Give an example

The process of converting a freshly prepared precipitate (suspension) into a colloid by adding a electrolyte having a common ion is called peptization. The electrolyte added is called a peptizing agent. During peptisation, the precipitate adsorbs one of the ions (positive or negative) of the electrolyte. This causes the precipitates to break into smaller particles of colloidal size.

17. What is electro dialysis?

The process of increasing the rate of dialysis, under the influence of an electric field is called electro dialysis. The process can be used iff the impurity is an electrolyte. In presence of the electric field, the ions diffuse faster (through parchment paper) towards the oppositely charged electrodes.

18. Write a note on ultrafiltration

It is a process of separating collidal particles from the solvent (dispersion medium) and all other soluble solutes present in collidal solution using specially prepared ultrafilters. An ultra filter paper (made by soaking filter paper in 4% nitro cellulose in alcohol and ether and later hardening it by using formaldehyde) allows all other particles except the colloidal particles to filter (pass) through it. To speed it up pressure or suction can be applied. The colloidal particles left on the ultrafilter paper are then stirred into fresh dispersion medium to get the pure colloidal solution.

19. Describe Tyndall effect Scattering of light by colloidal particles in the medium is called Tyndall effect. The path of light in the colloidal medium becomes visible when observed at right angles. The illuminated path within the medium is called Tyndall cone.

20. Write two conditions in which tyndal effect can be clearly observed.

Tyndall effect is clearly observed when a) size of colloidal particles matches with the wave length of light used b) there is large difference in refractive index between dispersed phase and medium

21. What is Brownian movement? How is it caused?

Zig– Zag movement of colloidal particles in a medium is called Brownian movement.

Reason: Particles of the medium are very small and are moving randomly in all directions. They collide with the colloidal particles and transfer their kinetic energy. Colloidal particles move slowly and randomly due to unequal bombardments by the particles of the medium. This is seen as “Brownian Movement”. “This property is a direct proof for the concept that liquid state of matter is made up of small molecules, which are in random motion, does not allow the particles to settle and is responsible for the stability of the sols”

22. Classify the following colloids into positively and negatively charged sols

Al (OH)3, CdS, AS2S3, Fe (OH)3, gum, clay, basic dyes, Sols of acidic dyes, sols of starch, and metallic sulphides, sols of metals (Ag, Au), haemoglobin.

Positively charged sol negatively charged sol

Al (OH)3, Fe (OH)3, basic dyes, haemoglobin,

CdS, AS2S3, Sols of acidic dyes, sols of starch, gum, clay, and metallic sulphides and sols of metals (Ag, Au)

23. How do colloidal particle become charged or acquire charge? The charge on the colloidal particles may be due to (i) preferential adsorption of ions from the medium or (ii) due to electron capture by sol particles during electrodispersion of metals.(iii)formation of electrical double layer

24.What is Electrophoresis.

Movement of electrically charged colloidal particles towards their oppositely charged electrodes when the colloid is placed in an electric field is electrophoresis. Positively charged particles move towards cathode and negatively charged particles move towards anode.

25. Mention any two methods of Coagulation of lyophobic sol

i) Electrophoresis ii) Mixing of two oppositely charged sols. E.g.: positively charged Fe(OH)3 sol with negatively charged As2S3 sol iii) Continuous dialysis iv) Addition of electrolyte v) By boiling

26. State and illustrate Hardy- Schulze rule.

Higher the valency of the flocculating ion added, greater is the coagulating power of the ion.

Ex (1): In the coagulation of negatively charged sol (As2S3) the coagulation power of the positively charged active ion is Na+ < Ba+2 < Al+3.

Ex (2): In the coagulation of positively charged sol [Fe (OH)3] the coagulating power of the negatively charged active ion is Cl− < SO4−

2 < PO4−3 < [Fe(CN)6]4

−.

Note: Higher the charge on the flocculating ion, lesser is the amount of the electrolyte required to coagulate a sol.

27. Difine coagulating value or flocculating value

The minimum concentration of electrolyte in millimoles per litre required to cause precipitation of a sol in 2 hours is called coagulating value. Smaller the coagulating value, higher is the coagulating power of the ion.

28. What is protective action of a sol? Give an example.

The property of a lyophilic sol by which it protects the lyophobic sol from precipitation, even upon adding an electrolyte to it, is called protective action of lyophilic sol. Lyophilic sol particles form a coat or layer around the lyophobic sol and hence protect them from the action of the electrolytes.

29. What are Emulsions? Give an example

A liquid in a liquid colloid is called an emulsion. If two immiscible liquids are shaken well, a dispersion of one liquid in the other, an emulsion is obtained.

Eg: Milk, butter, vanishing cream

30.Write a note on formation of delta region.

River water flowing towards the sea picks up many colloidal particles (clay, mud, humus, slit) with it. These particles are negatively charged. When the river water meets the sea, the electrolytes (salts like NaCl, MgSO4 etc) in the sea causes the coagulation of these colloidal particles. Thus clay, mud, humus gets precipitated and scattered at these places to form delta region.

31.Write the application of colloids in purification of smoke using Cottrell precipitator

Smoke (Colloidal dispersion of solid in gas) from industries contains carbon, dust, soot and many others as colloidal particles. To remove these, electrostatic precipitator called cottrell precipitator is used. The precipitator consists of metal plates attached to a high potential. As the smoke enters the precipitator, the charged colloidal particles

gets neutralized and precipitated on the metal plates. Gases free from colloidal impurities are led into chimney.

32. Write the application of colloids in the Purification of drinking water

Drinking water if muddy contains negatively charged clay, sand, mud as colloidal particles dispersed in it. When alum is added to this, Al+3 ions of the alum causes the coagulation of the negatively charged muddy colloidal particles which settle down as a precipitate. The upper layers of clear clean water are decanted. Thus water gets purified.

C. Questions carrying 3 marks

1. What happens to ΔH, ΔS and ΔG during the process of adsorption?

i) Adsorption is always an exothermic process, because there is decrease in surface energy. ∴ ΔH is negative (enthalpy decreases).

ii) When a gas is adsorbed on a liquid or solid, freedom of movement of gas molecules decreases. ∴ entropy decreases. ΔS is negative.

iii) Adsorption is a spontaneous process hence ΔG must be negative. ΔG = ΔH − TΔS. For adsorption ΔH = negative, ΔS = negative. Therefore ΔH must be more negative than TΔS being positive so that ΔG becomes negative.

2.Write any three differences between two types of adsorption of gases on solids. Physiosorption (physical adsorption) Chemisorption (chemical adsorption) 1. Accumulation of gas on a solid due

to weak van der Waal’s forces. 2. This is not specific, as force

between adsorbate and adsorbent is van der Waals forces which is universal.

3. The process is reversible. 4. Gases that can be easily liquefied

(high critical temperature) are readily absorbed.

5. Enthalpy of adsorption is low, as the forces involved are weak (ΔH is negative but low)

6. Adsorption decreases with increase in temperature. Low temperature favours better adsorption.

7. Under high pressure, it leads to multimolecular layers of adsorption.

1. Accumulation of gas on a solid due to chemical bond (covalent or ionic)

2. It is highly specific as there is chemical bonding between adsorbate and adsorbent.

3. Process is irreversible. 4. Gases that can form chemical

compounds with adsorbent are specifically adsorbed.

5. Enthalpy of adsorption is high, as the forces involved are strong (ΔH is negative, very high)

6. Adsorption process involves high energy of activation, therefore increases with increase in temperature.

7. It leads to unimolecular layer of adsorption even at high pressure.

(Any 3 of the above)

3. Classify the following colloids to their respective type of colloids a. Smoke b. Cod liver oil c. gems. a. Smoke-Solid aerosal b. Cod liver oil-Emulsion c. gems-solid sol 4.Write the mechanism of micelle formation considering soap as an example

Soap is sodium or potassium salt of higher fatty acid RCOO−Na+.

In water RCOO−Na+ dissociates into RCOO− and Na+. RCOO− has two parts. R is long hydrocarbon chain and is a non-polar tail (hydrophobic). COO− is polar-ionic head (hydrophilic).

At low concentration COO− group will be dissolved in water and R chains away from water and remain at the surface.

At critical micelle concentration, the anions are pulled into the water. They aggregate to form spherical shape in which hydrocarbon chains point to the interior and COO− projects outwards of the sphere. Such an aggregate is called a

micelle.

5. Write a note on Cleansing action of soap

a. It is due to formation of micelle by soap.

b. Soap molecules form a micelle around oil droplet (dirt) in such a way that hydrophobic R is in the oil and hydrophilic −COO− projects out into water.

c. The oil droplet thus gets pulled into water and gets detached from dirty cloth (material to be washed).

d. Soap thus helps in emulsification of oil and fat in the dirt, which is then washed away with water.

a) Grease on cloth

b) Stearate ions arranging around the grease droplet and

c) grease droplet surrounded by stearate ions (micelle formed)

6. Write a note on Dialysis.

A process of purifying a lyophobic sol by removing particles of true solution (ions or molecules) by their preferential diffusion through parchment paper or animal membrane is called Dialysis. The membrane is called a dialyser. Particles of true solution pass through the membrane but not the colloidal particles.

Process: The sol to be purified is taken in a parchment bag. The bag is suspended in a tank, in which water is circulated. Particles of true solution diffuse out from the bag. Water flowing in the tank carries away these particles. The sol gets purified and stabilized.

7. Describe how colloidal particles acquire charge by preferential adsorption of ions The colloidal particle in a lyophobic sol tends to adsorb cations or anions from the medium and hence become positively or negatively charged sols. They show a preference to adsorb a common ion from the medium.

Ex: (a) when potassium iodide solution is slowly added to silver nitrate solution, the silver iodide sol formed adsorbs Ag+1(present in plenty) and becomes positively charged. (AgI/ Ag+1)

Ex: (b) When silver nitrate solution is slowly added to potassium iodide solution, silver iodide sol formed adsorbs I-1 (present in plenty)ions from the medium and becomes negatively charged sol (AgI / I-1).

8. Mention two types of emulsion. Give example for each

i) Oil in water or ii) water in oil emulsion.

For oil in water emulsion, water is the dispersion medium, oil the dispersed phase. E.g.: milk, vanishing cream. In milk, liquid fat is dispersed in water.

For water in oil emulsion, water is the dispersed phase, oil is the dispersion medium. E.g.: butter, cream.

UNIT- 6 PRINCIPLES AND PROCESSES OF EXTRACTION OF METALS.

I. ONE MARK QUESTIONS:

1. Name an important ore of Aluminium .

Ans: Bauxite

2. Give the composition of copper pyrites. A: CuFeS2

3. What is meant by concentration of ores?

A: The process of removal of earthy impurities from the ore.

4. Name the electrolyte used in the extraction of aluminium. A: Molten Al2O3 + Cryolite + CaF2

5. Sulphide ores are roasted before reduction. Why?

A: To convert sulphides to oxides so that reduction is easy.

6. What are the products formed when calcium carbonate is calcined? A: CaO + CO2

7. Give the composition of copper matte.

A: Cu2S + FeS

8. How is FeO removed during the extraction of copper? A: It is removed as ironsilicate FeSiO3 using SiO2.

9. What do you mean by blister copper? A: The solidified copper obtained has blistered appearance due to the evolution of SO2 and so it is called blister copper.

10. During froth floatation process, name the component that comes along with the froth. A: Ore

11. Why do we add collectors during froth floatation? A: To enhance non-wettability of ore particles by water.

12. Haematite ore particles are heavier than gangue. Suggest a suitable method for its concentration. A: Gravity separation

13. What is the importance of roasting and calcination. A: This is done to get the metal in its oxide form so that reduction can be done easily.

14. Give an example of a metal that can be extracted by electrolytic method. A: Aluminium ( Or Sodium, magnesium)

15. In the extraction of aluminium carbon anodes are replaced regularly. Why? A: Because the carbon gets worn out as the oxygen liberated reacts with it to form CO2.

II. TWO MARK QUESTIONS

1. What is the role of (i) lime stone in iron extraction and (ii) cryolite in aluminium extraction.? A: (i) Removes silica impurity as slag calcium silicate (ii) Cryolite increases conductivity and reduces melting point of Al2O3.

2. Give the chemical reactions involved in (a) Iron extraction 2 marks A: C + O2 →CO2 ; CO2 + C →2CO, Fe2O3 + CO →2FeO + CO2 ; FeO + CO →Fe + CO2

CaCO3 →CaO + CO2, ; CaO + SiO2 →CaSiO3. (b) Aluminium extraction 2 marks A: Cathode: Al3+ (melt) + 3e� � Al(l) Anode: C(s) + O-2(melt) � CO(g) + 2e� C(s) + 2O2-(melt)� CO2 (g) + 4e� Overall reaction: 2Al2O3 + 3C � 4Al + 3CO2 (c) Copper extraction 2 marks A: 2FeS + 3O2 � 2FeO + 2SO2

FeO + SiO2 � FeSiO3 2Cu2S + 3O2 � 2Cu2O + 2SO2 2Cu2O + Cu2S � 6Cu + SO2

3. How is zinc obtained from ZnO?

A: ZnO is heated with coke at 1673K. Zinc and carbon monoxide are formed. ZnO + C � Zn + CO

4. Give equations for the extraction of gold using NaCN. A: 8NaCN (aq)+ 4Au(s) + 2H2O(aq) + O2(g) � 4Na[Au(CN)2] (aq) + 4NaOH(aq) 2Na[Au(CN)2] (aq) + Zn(s) � 2Au(s) +Na2[Zn(CN)4](aq).

5. Give the principles involved in (i) zone refining (ii) liquation A: (i) Zone refining: The impurities are more soluble in the melt than in the solid state of the metal.

(ii) Liquation: melting point of metals is lower than the impurities.

6. What are the requirements for the compound to be purified by vapour phase refining? A: (i) The metal should form a volatile compound with an available reagent, (ii) The volatile compound should be easily decomposable.

7. How do you refine nickel by Mond’s process? A: In this process, nickel is heated in a stream of carbon monoxide forming a volatile complex, nickel tetracarbonyl:

The carbonyl is subjected to higher temperature so that it is decomposed giving the pure metal:

8. How do you remove oxygen and nitrogen impurities from Zirconium.

A: By van Arkel method: The crude metal is heated in an evacuated vessel with iodine. The metal iodide volatilises Zr + 2I2 � ZrI4 The metal iodide is decomposed on a tungsten filament. The pure metal is thus deposited on the filament. ZrI4 � Zr + 2I2

III. THREE MARK QUESTIONS:

1. Explain the concentration of bauxite ore.

A: Bauxite ore is concentrated by leaching. The steps involved are i) Bauxite is concentrated by digesting the powdered ore in a concentrated solution of

sodium hydroxide at 473-573 K and 35 bar pressure. Al2O3 is leached as sodium aluminate.

ii) Aluminate solution is neutralised by passing CO2. Hydrated Al2O3 is precipitated by seeding. iii) Hydrated Al2O3 is filtered, dried and heated to get pure Al2O3.

2. Write the equations involved in leaching of alumina.

A: Al2O3(s) + 2NaOH(aq) + 3H2O(l) → 2Na[Al(OH)4](aq) 2Na[Al(OH)4](aq) + CO2(g) → Al2O3.xH2O(s) + 2NaHCO3 (aq) Al2O3.xH2O(s) ∆ Al2O3 + xH2O

3. Two reactions are given below, which of these two happen at a temperature above 1073K. Justify. FeO + C � Fe + CO --------(A) FeO + CO � Fe + CO2--------(B) A: FeO + C � Fe + CO happens at a temperature above 1073K. In the Ellingham diagram we can see that the C,CO line goes below while CO,CO2 goes above FeO line at temperature above 1073K. So, C is the reducing agent.

4. How do you extract aluminium from bauxite ore? A: Diagram.

Aluminium is extracted from bauxite ore by Hall-Heroult process. the electrolyte is purified Al2O3 +Na3AlF6 + CaF2 . Na3AlF6 + CaF2 lowers the melting point of the mix and brings conductivity. The fused matrix is electrolysed. Steel cathode and graphite anode are used. The overall reaction may be taken as: 2Al2O3 + 3C � 4Al + 3CO2

5. How do you extract iron from roasted haematite ore. Explain with diagram. A: Diagram

Iron extraction is carried out in blast furnace and different reactions takes place at different temperatures. Ore, limestone and coke are fed into the furnace. Hot air is blown from the bottom. Coke is burnt to give high temperature. At the top of the furnace at lower temperature iron oxide is reduced to iron by carbon monoxide and at high temperatue at the botton iron oxide is reduced by carbon. Calcium carbonate forms CaO which removes silica as calcium silicate.

6. Name the methods used in the refining of

(a) tin (b) copper (c) germanium (d) Titanium A: (a) tin- liquation (b) copper- electrolytic refining (c) germanium-zone refining (d) Titanium-van Arkel refining Or vapour phase refining

7. Explain van Arkel method of refining of zirconium. A: The crude metal is heated in an evacuated vessel with iodine. The metal iodide volatilises Zr + 2I2 � ZrI4 The metal iodide is decomposed on a tungsten filament. The pure metal is thus deposited on the filament. ZrI4 � Zr + 2I2

8. Explain magnetic separation method of concentration of ore.

A: Priniciple: This is based on differences in magnetic properties of the ore and the gangue. The powdered ore is carried on a conveyer belt which passes over a magnetic roller. Magnetic substances stick to the roller while non magnetic

substances fall and form a heap. Once the roller moves the magnetic substances come out of the influence of the magnetic roller and fall off forming a separate heap.

9. Give the principle of froth floatation process. How can we separate ZnS and PbS present in an ore using froth floatation process? A: Principle: This is based on the differences in the wetting properties of ore and gangue.

ZnS and PbS present in the ore can be separated by using depressants like NaCN. It selectively prevents ZnS from coming to the froth but allow PbS to come with the froth.

Unit 7

p- BLOCK ELEMENTS:

1. Classify the following 1 5 th group p-block elements in to nonmetals/metalloids

/metal.

1)Nitrogen 2) Phosphorus 3) Arsenic 4) Antimony 5) Bismuth 1M each

Answer:

Nonmetals: Nitrogen and Phosphorus

Metalloids: Arsenic and Antimony

Metal: Bismuth

2. Write the formula of 1M each

1) chile salt petre 2) Indian salt petre 3)apatite mineral 4)chlorapetite 5) Fluorapetite

Answer: 1) NaNO3 2) K NO3 3) Ca9(PO4)6 CaX2 4) Ca9(PO4)6 CaCl2 5)

Ca9(PO4)6 CaF2

3. Write the valence shell electronic configuration of 15th group elements. 1M

Answer; ns2 np3

4. There is a considerable increase in covalent radius from N to P. However, from As to Bi

only small increase in covalent radius is observed. Give reason. 1M

Answer: This is due to the presence of completely filled d and/or f orbital in heavier

members.

5. Ionization enthalpy decreases down the group 15. Give reason. 1M

Answer: Due to gradual increase in atomic size.

6. The ionization enthalpy of the group 15 elements is much greater than that of group

14 and group 16 elements in the corresponding periods. Give reason. 1M

Answer: Because of the extra stable half-filled p orbital electronic configuration and smaller

size.

7. How does electronegativity of 15th group elements varies down the group? 1M

Answer: decreases

8. Mention the common Oxidation states of p block elements. 1M

Answer: common ox.state of these elements are-3,+3 and +5

9. How is stability of oxidation states of 15th group elements varies? 1M

Answer: -3 in their covalent compounds. In addition to the −3 state, N and P also show −1

and −2

All the elements present in this group show +3 and +5 oxidation states. Stability of +5 ox.

State decreases and that of +3 ox. State increases due to inert pair effect.

10. Nitrogen atom has five valence electrons but it does not form NCl5. 1M

Answer: Because of absence of d-orbitals it can’t expand its covalency from 3 to 5.

11. Nitrogen does not form pentahalides.Why? 1M

Answer: Nitrogen with n = 2, has s and p orbitals only. It does not have d orbitals to expand

its covalence beyond four. That is why it does not form pentahalide.

12. Why is Nitrogen an inert gas? 1M

Answer: Nitrogen exists as triply bonded diatomic non polar molecule. Due to short

internuclear distance between two nitrogen atoms the N ≡ N bond strength is very high. It

is, therefore, very difficult to break the bond.

13. Why nitrogen exhibits anomalous behavior? 2M

Answer: Due to smaller size, high electronegativity, high ionisation enthalpy and non-

availability of d-orbitals, nitrogen shows anomalous behavior.

14. Mention any three anomalous properties of nitrogen. 3M

Answer; (any three of the following)

1)Nitrogen forms pπ – pπ bonds where as other members not.

2) nitrogen exists as diatomic molecule with a triple bond

3)The single N−N bond is weaker than P−P bond due to small bond length.

4) lower catenation tendency.

5)cannot form dπ – pπ bonds like phosphorus.

15. Why R3P=O exist but R3N=O does not? 1M

Answer: Due to the absence of d orbitals in valence shell of nitrogen, nitrogen cannot form

d π–p π bond. Hence R3N=O does not exist.

16. Catenation property of nitrogen is less than phosphorus. Why? 1M

Answer: Due to strong pπ–pπ overlap in Nitrogen and weaker N-N bond than the single P-

P bond.

17. Write the formula of hydrides formed by 15th group elements? 1M

Answer: EH3

18. How does the stability of 15th group metal hydride varies down the group? 1M

Answer:The stability of hydrides decreases on moving down from NH3 to BiH3.

19. Why is NH3 basic while BiH3 is only feebly basic. 1M

Answer: NH3 is basic due to smaller size & high electro negativity of Nitrogen.

20. Ammonia has higher boiling point than Phosphine. Explain. 1M

Answer: Ammonia (NH3) form hydrogen bond but Phosphine (PH3) does not. Hence boiling

point of ammonia is higher than that of phosphene.

21. Write the formula of two types of oxides formed by 15th group elements? 1M

Answer: E2O3 and E2O5

22. Out of E2O3 and E2O5 which is acidic? 1M

Answer: E2O5 ( oxide with higher oxidation state is more acidic)

23. How does the acidic characters of 15th group metal oxides varies down the group? 1M

Answer; The acidic character decreases on moving down a group.

24. Write the increasing order of acidic character of N2O5, P2O5, As2O5 and Sb2O5 1M

Answer: N2O5 >P2O5 >As2O5 > Sb2O5

25. How is dinitrogen prepared in the laboratory? 2M

Answer: In the laboratory, dinitrogen is prepared by treating an aqueous solution of

ammonium chloride with sodium nitrite.

NH4Cl (aq) + NaNO2 (aq) → N2 (g) + 2H2O (l) + NaCl (aq)

26. How is dinitrogen prepared from ammonium dichromate? 2M

Answer: thermal decomposition of ammonium dichromate gives dinitrogen.

(NH4)2Cr2O7 → N2 + 4H2O + Cr2O3

27. How does dinitrogen reacts with Mg? 2M

Answer: Dinitrogen reacts with Mg to form magnesium nitride.

N2 +3 Mg → Mg3N2

28. For the manufacture of ammonia by Haber’s process, write flow chart and balanced

equation along with conditions? 3M

Answer: On large scale, obtained by Haber’s process\

Optimum condition:

Pressure = 200 × 105 Pa (about 200 atm)

Temperature ∼ 700 K

Catalysts used − Iron oxide with small amounts of K2O and Al2O3 to increase the rate

of attainment of

equilibrium.

29. How does ammonia react with zinc sulphate? 2M

Answer: Ammonia reacts with zinc sulphate to form white precipitate of zinc hydroxide.

ZnSO4(aq) + 2NH4OH(aq) → Zn(OH)2(s) + (NH4)2SO4(aq)

30. How does ammonia react with cupric ion? 2M

Answer: With Cu2+ ion Ammonia acts as lewis base and forms deep blue colored

cuprammonium complex.

Cu2+(aq) + 4NH3(aq) → [Cu(NH3)4]2+ (aq)

(blue) (deep blue)

31. How is Nitric acid manufactured by Ostwald process? 3M

Answer: Nitric acid in Ostwald process manufactured by the oxidation of ammonia.

Nitric oxide thus formed combines with oxygen giving NO2.

2NO ( g ) + O2 ( g )→2NO2 ( g )

Nitrogen dioxide so formed, dissolves in water to give HNO3.

3NO2 ( g ) + H2O ( l ) → 2HNO3 ( aq ) + NO ( g )

Dilute nitric acid on distillation followed by dehydration using conc. sulphuric acid gives

98% nitric acid.

32. How is nitric acid prepared in laboratory? 2M

Answer: Nitric acid is prepared in the laboratory by heating KNO3 or NaNO3 with

concentrated H2SO4 in glass retort.

NaNO3 + H2SO4 → NaHSO4 + HNO3

33. How does dilute nitric acid with copper? 2M

Answer: Dil. Nitric acid reacts with copper to form cupric nitrate with the liberation of

nitric oxide.

3Cu + 8 HNO3(dilute) → 3Cu(NO3)2 + 2NO + 4H2O

34. How does concentrated nitric acid with copper? 2M

Answer: Conc.Nitric acid reacts with copper to form cupric nitrate with the liberation of

nitrogen dioxide.

Cu + 4HNO3(conc.) → Cu(NO3)2 + 2NO2 + 2H2O

35. How does dilute nitric acid with zinc? 2M

Answer: Dil. Nitric acid reacts with zinc to form zinc nitrate with the liberation of nitrous

oxide.

4Zn + 10HNO3(dilute) → 4 Zn (NO3)2 + 5H2O + N2O

36. How does concentrated nitric acid with zinc? 2M

Answer: Conc.Nitric acid reacts with zinc to form zinc nitrate with the liberation of

nitrogen dioxide.

Zn + 4HNO3(conc.) → Zn (NO3)2 + 2H2O + 2NO2

37. How does concentrated nitric acid with iodine? 2M

Answer: Conc.Nitric acid oxidizes iodine to form iodic acid.

I2 + 10HNO3 → 2HIO3 + 10NO2 + 4H2O

38. How does concentrated nitric acid with carbon? 2M

Answer: Conc.Nitric acid oxidizes carbon to carbon dioxide

C + 4HNO3 → CO2 + 2H2O + 4NO2

39. What is passivity? 1M

Answer: Some metals like aluminium and chromium do not dissolve in concentrated

nitric acid due to the formation of a protective layer of oxide on the surface of the

metal. This phenomena is called passivity of metals.

40. explain Brown ring test with equations . 3M

Answer: Dilute FeSO4 solution is added to an aqueous solution of nitrate ion.

concentrated H2SO4 is then added along the sides of the test tube. A brown ring is

observed at the interface between the solution and H2SO4 layers indicates the presence of

nitrate ion in the solution.

NO3- + 3Fe2+ + 4H+ → NO + 3Fe3+ + 2H2O

[Fe (H2O)6 ]2+ + NO → [Fe (H2O)5 (NO)]2++ H2O

(brown)

41. Write the resonance structurs of a) NO b) NO2 c) N2O5 1M each

Answer:

a) Structure of NO:

b) Structure of NO2 :

c) Structure of N2O5 :

42. Distinguish between white and red phosphorus. 2M

Answer: (any two)

White phosphorus Red Phosphorus

It is a soft and waxy solid. It is a hard and crystalline solid.

It is poisonous. It is non-poisonous.

It is insoluble in water but soluble in

carbon disulphide.

It is insoluble in both water and carbon

disulphide.

Highly reactive It is relatively less reactive.

In both solid and vapour states, it exists as

a P4 molecule.

It exists as a chain of tetrahedral P4 units.

Less stable More stable

43. How is Phosphine prepared in the laboratory form white phosphorous? 2M

Answer: In the laboratory phosphene is prepared by heating white phosphorus with

concentrated NaOH solution in an inert atmosphere of CO2.

P4 + 3NaOH + 3H2O → PH3 + 3NaH2 PO2

44. Give a reaction to support Basic nature of phosphine. 2M

Answer: Phosphine react with hydrogen bromide and forms phosphonium bromide.

PH3 + HBr → PH4 Br

45. How is Phosphorous trichloride is obtained from phosphorous and chlorine? 2M

Answer: Phosphorus tri chloride is obtained by passing dry chlorine over heated white

phosphorus.

P4 + 6Cl2 → 4PCl3

46. How is Phosphorous pentachloride is obtained from phosphorous and chlorine? 2M

Answer: Phosphorus pentachloride is prepared by the reaction of white phosphorus

with excess of dry chlorine.

P4 + 10Cl2 → 4PCl5

47. How does Phosphorous trichloride react with water? 2M

Answer: PCl3 hydrolyses in the presence of moisture to give phosphorus acid.

PCl3 + 3H2O →H3PO3 + 3HCl

48. How does Phosphorous pentachloride react with water? 2M

Answer: PCl5 in presence of water hydrolyses to POCl3 and finally gets converted to

phosphoric acid.

PCl5 + H2O → POCl3 + 2HCl

POCl3 + 3H2O → H3PO4 + 3HCl

49. Write the formula, structure ,reducing property and basicity of Hypophosphorus acid.

3M

Answer:

Formula- H3PO2

Reducing property: Reducing agent as it contains two P – H linkage

structure:

Basicity: one( as it contains only one P-OH linkage)

50. Write the formula, structure ,reducing property and basicity of Orthophosphorus acid.

3M

Answer:

Formula- H3PO3

Reducing property: Reducing agent as it contains one P – H linkage

structure:

Basicity: Two( as it contains two P-OH linkage)

51. Write the formula, structure ,reducing property and basicity of Orthophosphoric acid.

3M

Answer:

Formula- H3PO4

Reducing property: Not a Reducing agent as it does not have P – H linkage

structure:

Basicity: Three( as it contains three P-OH linkage)

52. How do you account for the reducing behavior of H3PO2 on the basis of its structure?

1M

Answer: In H3PO2, two H atoms are bonded directly to P atom which imparts reducing

character to the acid.

53. Classify the following 1 6 th group p-block elements in to nonmetals/metalloids /

metal.

1)Oxygen 2) Sulphur 3) selenium 4) Tellurium 5) Polonium 1M each

Answer:

Nonmetals: Oxygen and Sulphur

Metalloids: selenium and Tellurium

Metal: Polonium

54. Name the 16th group p-block element which is radioactive in nature. 1M

Answer: Polonium

55. Write the valence shell electronic configuration of 16th group elements. 1M

Answer; ns2 np4

56. Mention the Oxidation state of oxygen. 1M

Answer; Oxygen exhibits the oxidation state of−2 in metal oxides, −1 (H2O2), zero (O2 and

O3) and +2 (OF2).

57. Write a note on Anomalous Behavior of Oxygen. 3M

It is due to its Small size, High electronegativity and absence of d-orbitals oxygen shows

anomalous properties.

1. Strong hydrogen bonding is present in H2O, which is not found in H2S.

2. Absence of d-orbitals in oxygen limits its covalence to four and in practice rarely

exceeds two. On the other hand, other elements of the group can expand their

covalence beyond four.

58. How is Oxygen obtained from potassium chlorate? 2M

Oxygen can be obtained by heating potassium chlorate in presence of MnO2.

59. Write the chemical equation for the reaction of Oxygen with aluminum /carbon

/methane. 1M each

Answer:

60. What are acidic oxides? What type of oxides are acidic in nature? Give example. 3M

Answer: Acidic oxides are those oxides which combine with water to give an acid.

Non-metal oxides and Oxides of some metals in higher oxidation state are acidic in nature.

Example for non metal acidic oxides− SO2, Cl2O7, CO2, N2O5

Examples for metal oxides which are acidic − Mn2O7, CrO3, V2O5

61. What are basic oxides? What type of oxides are basic in nature? Give example. 3M

Answer: Basic oxides are those oxides which combine with water to give bases.

Metal oxides are basic in nature.

Examples for metal oxides which are basic- Na2O, CaO, BaO

62. What are amphoteric oxides? Give example. 2M

Answer: Amphoteric oxides are those oxides which show the characteristics of both acidic

as well as basic oxides .

Example − Al2O3

63. Illustrate amphoteric nature of Al2O3 with suitable reactions. 2M

Answer:

64. What are neutral oxides? Give example. 2M

Answer: Neutral oxides arethose oxides which are neither acidic nor basic.

Examples − CO, NO, N2O

65. How is Ozone prepared from oxygen? 2M

Answer: A slow dry stream of oxygen is passed through a silent electrical discharge. Oxygen

partially gets converted into ozone.

66. Why is high concentrations of ozone can be explosive? 2M

Answer: High concentrations of ozone can be explosive because the decomposition of O3 to

O2 results in the liberation of heat (ΔH = − ve) and an increase in entropy (ΔS = + ve), leading

to large negative value of ΔG.

67. How does ozone react with PbS? write equation. 2M

Answer: Ozone oxidizes lead sulphide to lead sulphate.

PbS + 4O3 → PbSO4 + 4 O2

68. How does ozone react with NO? write equation. 2M

Answer: Nitric oxides reacts with ozone to give nitrogen dioxide and oxygen

O3 + NO → NO2 + O2

69. In the preparation of H2SO4 by Contact Process, why is SO3 not absorbed

directly in water to form H2SO4? 1M

Answer: SO3 is not dissolved in water directly as the process is highly exothermic &

the H2SO4 obtained is in the form of a mist which cannot be condensed easily.

70. Which form of the sulphur is stable at room temperature? 1M

Answer:Rhombic sulphur. ( α sulphur)

71. Which form of the sulphur is stable above 369K? 1M

Answer:Monoclinic sulphur( β sulphur)

72. Explain the laboratory method of preparation of SO2 from 2

3SO . 2M

Answer: sulphites are treated with dil H2SO4 to get SO2

i) SO3-(aq) +2H+(aq) H2O +SO2

73. , What happens when Sulphrur dioxide is treated with (2 Marks)

i)NaOH

ii) Cl2

Answer:i) 2NaOH +SO2 Na2SO3 +H2O

Na2SO3 + H2O 2NaH SO3

ii) SO2+ Cl2 SO2Cl2( Salphuryl chloride)

74. Give any two reactions to show that SO2 is a reducing agent. 2M

Answer: 2Fe3++ SO2+ 2 H2O2Fe2+ + SO4 2- +4H+

5SO2+2MnO4- + 2H2O 5SO4 2- +4H++2Mn2+

75. How is the presence of SO2 detected?

Answer: SO2 discharges pink colour of KMnO4 due to the reaction

5SO2+2MnO4- +2H2O 5SO4 2-+4H++2Mn2+

76. Draw the structure of i) Sulphurus acid ii) Sulphuric acid (iii)peroxo sulphuric acid iv)

pyrosulphuric acid( oleum). 1M each

Answer:(i) (ii) (iii) (iv)

77. Name the catalyst used in the manufacture of sulphuric acid by contact process

1M

Answer: V2O5

78. Write chemical equations in the manufacture of sulphuric acid by contact process with

the conditions required. (3 Marks)

Answer: 2SO2 +O2 2SO3 At 720K, temperature & 2 bar pressure.

SO3 + H2SO4 H2S2O7

79. Explain the manufacture of H2SO4 by contact process from purified SO2. 3M

Answer: Purified SO2 is passed through catalytic converter containing V2O5 at

720K,and 2 bar pressure. SO3 is obtained.

2SO2 +O2 2SO3

SO3 obtained is dissolved in to get oleum in absorption tower.

SO3 + H2SO4 H2S2O7

Oleum is carefully diluted with water to get sulphuric acid.

80. Draw the flow chart for manufacture of H2SO4 by contact process 2M

81. What happens when Concentrated H2SO4 is added to 2M Each

i)CaF2.

ii) Sugar.

Answer: i) CaF2+ H2SO4 CaSO4 + 2HF

ii) C12H22O11 12C + 11 H2O( Charring of sugar- Dehydrating property)

82. Give an example to show that Conc H2SO4 is a strong oxidizing agent. 1M Each

Answer: Cu + 2 H2SO4( Hot ,Conc) CuSO4 + SO2 + 2H2O

3S + 2 H2SO4( Hot ,Conc) 3 SO2 + 2H2O

C + 2 H2SO4( Hot ,Conc) CO2+ 2SO2 + 2H2O

83. Name the halogens 1M

Answer:Flourine, Chlorine,bromine,iodine,asyatine.

84. Which is the radioactive halogen? 1M

Answer: Astatine

85. Name the halogen present in sea weeds. 1M

Answ: Iodine.

86. Write the outermost electronic configuration of halogens. 1M

Answer: ns2np5.

87. Give reason (1 M each)

i) Halogens have very high ionization enthalpy in the corresponding period.

ii) Halogens have Maximum negative electron gain enthalpy in the corresponding

period

iii) Negative electron gain enthalpy of fluorine is less than that of chlorine.

iv) Enthalpy of dissociation of F2 is less than Cl2.

v) Fluorine is stronger oxidizing agent than chlorine.

vi) Fluorine exhibits only -1 oxidation state.

Answer:i) Due to the ns2np5 configuration, they have little tendency to loose

electrons.

ii) They have only one electron less than the stable noble gas configuration.

iii) Due the very small size of fluorine atom.

iv) Due the very small size of fluorine.

v) Due to the high electro negativity of fluorine atom it readily accepts an electron.

vi) Due to non availability of d- orbital.

88. Write the chemical equation 1M Each

i) When F2 is treated with Cl-,Br- & I-

ii) When Cl2 is treated with Br- & I-

iii) When Br2 is treated with I-

iv) When F2 is treated with H2O

v) When Cl2 is treated with H2O

Answer: i) F2 +2X- 2F- +X2 ( X= Cl, Br, or I)

ii) Cl2 +2X- 2Cl- +X2 ( X= Br, or I)

iii) Br2 +2I- 2Br- +I2 ( X= Cl, Br, or I)

iv) 2F2 + 2H2O 4H+(aq) +4F-(aq) +O2.

v) 2Cl2 + 2H2O 4HCl(aq) +HOCl(aq)

89. Mention the three reasons for the anomalous behavior of fluorine. 3M

Answer: Due to its small size, highest electro negativity, low F—F bond dissociation

enthalpy & non- availability of d- orbitals in the valence shell of fluorine.

90. Give any three examples to show anomalous behavior of fluorine. 3M.

Answer: i) ionisation enthalpy, electronegativity, electrode potential are higher for F

ii) Ionic & covalent radii, m.pt, b.pt, bond dissociation enthalpy,electron gain

enthalpy lower than expected.

iii) F forms only one halo acid

iv) HF is liquid, other hydrogen halides are gases.

91. How is chlorine prepared from KMnO4. Write the chemical equations involved. 2M

Answer: By the action of HCl on KMnO4,

2KMnO4 + 16 HCl 2KCl +2MnCl2 + 8H2O +5Cl2

92. What happens when Concentrated chlorine is treated with

i)Alluminium ii) sulphur S8

iii) H2S iv) excess of NH3 v) cold & dilute NaOH

vi) hot & concNaOH vii) Dry slaked lime. 1M each

Answer: (i)with Al: 2Al + 3 Cl2 2AlCl3

(ii) with S: S8 +4 Cl2 2S2Cl2

(iii) With H2S: H2S +Cl2 2HCl+S

(iV) With NH3: 8NH3 +3Cl2 6NH3Cl + N2

( excess)

NH3 + 3Cl2 6NCl3 + 3HCl

( excess)

(v) With NaOH:

2NaOH+Cl2 NaCl + NaOCl+H2O

( cold & dil) (hypochlorite)

(vi) 6NaOH+3Cl2 5NaCl + NaOCl3+3H2O

( hot & conc) ( chlorate)

(vii) With Ca(OH)2 :

2Ca(OH)2+2Cl2Ca(OCl)2+CaCl2 +2H2O

(dry slaked lime) ( bleaching powder)

93. Give any one example for oxidizing property of chlorine with FeSO4, Na2SO3.

(2M each)

Answer:2 FeSO4+H2SO4 + Cl2 Fe2(SO4)3+ 2HCl

( Ferrous) ( Ferric)

Na2SO3 + Cl2+H2O Na2SO4 + 2HCl

( Sulphite) ( Slphate)

94. Give the reason for the bleaching action of chlorine. 1M

Answer: Due to the oxidation

Cl2+H2O 2HCl + O

Coloured sub + O Colourles substance.

95. Give the composition of bleaching powder. 1M

Answer: Ca(OCl)2.CaCl2. Ca(OH)2 .2H2O.

96. How is HCl is prepared in the laboratory? 2M

Answer: NaCl + H2SO4 NaHSO4+ HCl at 420K

NaHSO4+ NaCl Na2SO4 + HCl at 823K

HCl is dried using Conc H2SO4

97. Give the composition of aqua regia? Write the ionic equation when it is treated with

gold/ platinum. 3M

Answer: Aqua regia: 3:1 part conc HCl & conc HNO3

Dissolves noble metals

Au +4H + +NO3- +4Cl- AuCl4

- +NO + 2 H2O

3Pt+16H ++4NO3-+18Cl-3PtCl6

-+4NO + 8 H2O

98. What happens when hydrochloric acid is treated with NH3 1M

Answer: NH3 +HCl NH4Cl ( White fumes)

99. Write the structure of

i) Hypochlorus acid ii) Chlorus acid iii) Chloric acid

iv) Perchloric acid. 1Meach

Answer:

Answer:

100. What are interhalogen compounds? Give an example. Why they are more reactive

than individual halogen. 3M

Answer:When two different halogen atoms react inter halogen compounds are

formed. Eg: ClF3, ICl, BrF5 ,lF7 Reactivity is more compared with halogens because X-

X’ bond is weaker than X-X bond in pure halogens.

101. How is following interhalogen compound prepared?

i) ClF3 ii) ICl3 iii) BrF5 1Meach

Answer: 473K

i) Cl2+F2 2ClF3

ii) I2+Cl2 2ICl

iii) Br2+5F2 2BrF5

( excess)

102. Name i) the radioactive noble gas ii) most abundant noble gas. 1M

Answer: i) Radon ii) Argon

103. Why noble gases are chemically inert? 1M

Answer: Stable completely filled orbitals are there.

104. Why noble gases have maximum ionization enthalpy in the corresponding period.

1M

Answer: Stable completely filled orbitals are there

105. Why noble gases have positive electron gain enthalpy . 1M

Answer:Stable completely filled orbitals are there

106. Which is the first noble gas compound synthesized? 1M

Answer: Xe+PtF6-

107. Who prepared first noble gas compound? 1M

Answer: Neil Bartlett

108. Write the chemical equations to prepare following compounds with the conditions

required.

i) XeF6 ii) XeO3 iii) XeO2F2. 1M each

573K, 60-70 bar

Answer: i) Xe(g) + 3F2(g) 3XeF6(s)

ii) XeF6 + 3H2O XeO3 + 6HF

iii) XeF6 + 2H2O XeO2F2 + 4HF

109. Write/ Name the structure of

i) Xe F2 ii) XeF4 iii) XeF6 iv) XeOF4 v) XeO3. 1M each

Answer: i) linear ii) sqare panar iii) Distorted octa hedral

iV) Square pyramidal) trigonal pyramidal

110. Noble gases have very low boiling point .Why? 1M

Answer:They are mono atomic due to weak dispersion forces, hence have low boiling

points.

Unit 8

The d- and f- Block Elements

I. Answer the following questions. Each question carries one mark

1. Define transition elements. Ans. Transition element is defined as the one which has incompletely filled d orbitals in its ground state or in any one of its oxidation states. 2. What is the position of the d block elements in the periodic table?

Ans. The d block elements are in the middle of s and p blocks, comprising the groups 3 to 12.

They are the four rows of elements in the periods 4th (3d series), 5th (4d series), 6th ( 5d series)

and 7th ( 6d series).

3. Zinc, cadmium and mercury of group 12 are not regarded as transition metals, Why ? Ans. Zinc, cadmium and mercury of group 12 have full d10 configuration ( d orbitals are completely filled ) in their ground state as well as in their common oxidation states and hence, are not regarded as transition metals 4. Why d- block elements are named as ‘transition elements ‘ ? Ans. The d–block elements occupies the middle of the periodic table and their properties are transitional between s– and p– block elements. 5.Write the general electronic configuration of d block elements.

Ans. [ Noble gas] (n-1)d1-10ns1-2

6. Write the general outer electronic configuration of d- block elements. . Ans. The general outer electronic configuration of d- block elements is (n-1)d1–10 ns1–2

7. Write the general electronic configuration of f- block elements. Ans. The general electronic configuration of f- block elements (Lanthanoids) is [Xe] 4f1 – 14 5d 0- 1 6s2 8.Name a member of the lanthanoid series which is well known to exhibit +4 oxidation state. Ans. Cerium 9. The outer electronic configuration of Cr is 3d5 4s1 instead of 3d44s2, why? Ans. Half filled (3d5) orbitals are relatively more stable, hence one electron of 4s orbital jumps to 3d orbital. 10. The outer electronic configuration of Cu is 3d10 4s1 instead of 3d94s2 , why? Ans. Completely filled (3d10 ) orbitals are relatively more stable, hence one electron of 4s orbital jumps to 3d orbital.

11. Account for high melting point and boiling points of transition metals. Ans. The melting and boiling points of transition metals are high because of the involvement of greater number of electrons from (n-1)d orbitals in addition to the ns electrons in the inter atomic metallic bonding. 12.What is the trend in melting points of transition metals in a series? Ans.The melting points of the transition metals in a series rise to a maximum at the middle of the series (i.e. Cr or Mo or W - element with d 5 configuration ) and fall regularly as the atomic number increases. 13.Why do transition metals have higher enthalpies of atomization?

Ans. Involvement of a large number of unpaired electrons of d orbitals favour stronger inter

atomic interactions resulting in stronger bonds between the atoms of a metal and higher

enthalpies of atomization.

14.Name one 3d series elements, that do not show variable oxidation states.

Ans. Sc (+3)

15.Transition metals exhibit variable oxidation states in its compounds, why? Ans. Transition metals exhibit variable oxidation states in its compounds due to the

availability of both ns & (n – 1 ) d electrons for bond formation.

16. Name 3d series metal which shows highest oxidation state.

Ans. The highest oxidation state shown by 3d series transiNa

tion metals is +7 by Mn

17. Name a metal in the 3d series of transition metals which exhibit +1 oxidation state

most frequently.

Ans. copper

18.What is the trend in oxidation state of transition metals ? Ans. The oxidation state increases with increase in atomic number & reaches a maximum in the middle and then decreases. 19. 3d series transition metals exhibit +2 as the most common oxidation state (except Sc) why? Ans. The +2 oxidation state, which commonly occurs for nearly all the transition metals is due to the loss of their outer 4s electrons

20. Why transition metals and their compounds shows paramagnetic behavior ?

Ans. The transition metal ions are generally containing one or more unpaired electrons in them & hence their compounds are generally paramagnetic. 21. Name an of alloys of transition metals with non transition metals.

Ans. Brass ( Cu & Zn) or Bronze ( Cu & Sn)

22.What is the action of neutral or faintly alkaline permanganate solution on iodide ?

Ans. Alkaline permanganate solution oxidize iodide to iodate.

23. What happens when potassium permanganate is heated to 513 K ?

Ans. Potassium permanganate decomposes at 513K to potassium manganate, manganese

dioxide and oxygen.

24. What is the principal oxidation state exhibited by the lanthanoids? Ans. The principal oxidation state of lanthanoids is +3.

25.Write the spin-only formula used to calculate the magnetic moment of metal ions. Ans. The magnetic moment is determined by using the spin only formula,

where n is the number of unpaired electrons and μ is the magnetic moment in units of Bohr magneton (BM). 26. Why is Sc3+ (or Zn2+ ) diamagnetic?

Ans. Sc3+(Z=21) 3d0 no unpaired electron, n=0, μ=0.

(or Zn2+(Z=30) 3d10 no unpaired electron, n=0, μ=0)

27. What is the most common oxidation state of lanthanoids and actionoids?

Ans. The most common oxidation state of lanthanoids and actionoids is +3. 28.What is Actinoid contraction? Ans. There is a gradual decrease in the size of atoms or M3+ ions across the series. This is known as the actinoid contraction. 29.Actinoid contraction is more than lanthanoid contraction. Give reason.

Ans. The actinoid contraction is, more than lanthanoid contraction due to poor shielding by 5f electrons from nuclear charge. 30. Actionoids show larger number of oxidation states than lanthanoids. Why? Ans. In actinoids 5f, 6d and 7s levels are of comparable energies ,hence electrons from these orbitals are available to lose or share.

31. Give one use of Mischmetall . Ans. Mischmetall is used in Mg-based alloy to produce bullets, shell and lighter flint. 32. Why transition metals forms alloys readily?

Ans. Transition metals readily form alloys with other transition metals because of their similar radii 33. Give one use of transition metal alloy.

Ans.Ferrous alloys containing chromium, vanadium, tungsten, molybdenum and manganese

are used for the production of a variety of steels.

II. Answer the following questions. Each question carries TWO marks. 34. Name two characteristic properties exhibited by d – block elements due to their partly filled d orbitals. Ans. The characteristic properties exhibited by d – block elements due to their partly filled d orbitals are variable; (i) Oxidation states (ii) Formation of coloured ions. 35. Name two typical metallic properties displayed by transition elements. Ans.High tensile strength, ductility malleability, high thermal and electrical conductivity and metallic luster etc. 36. What are interstitial compounds? Give example. Ans. Interstitial compounds are those which are formed when small atoms like H, C or N are trapped inside the crystal lattices of transition metals. Example; TiC, Mn4N, Fe3H, VH0.56 and TiH1.7, etc. 37. Give any two physical characteristics of interstitial compounds. Ans.Two physical characteristics of interstitial compounds are:

(i) They have high melting points, higher than those of pure metals.

(ii) They are very hard and they retain metallic conductivity.

38.Calculate the ‘spin only’ magnetic moment of M2+ (aq) ion (Z = 27).

Ans. M (z= 27 , 3d7 4s2) M+2 (3d7 4s0) hence it has 3 unpaired electrons n= 3

= 3(3+2) = 3.87 BM 39. The second ionisation enthalpy is high for Cr and Cu , why?

Ans: The second ionisation enthalpy is unusually high values for Cr and Cu because when

M+ ion ionize to M+2 ion , the d5 and d10 configurations of the M+ ions (i.e Cr+ or Cu+) are

disrupted, with considerable loss of exchange energy .

40. Why first ionisation enthalpy of Cr is lower than that of Zn ?

IE1 of Cr is lower, because removal of an electron from Cr does not change the d (3d5 4s1

to 3d5 4s0 ) configuration . Cr (z= 24 , 3d5 4s1) Cr+ (3d5 4s0) ------ IE1

IE1 value for Zn is higher, because removal of electron from 4s level needs more energy.

Zn (z= 30 , 3d10 4s2) Zn+ (3d10 4s1) ------ IE1

IE1 (Zn) > IE1 (Cr)

41.Give two characteristics of transition metal alloys.

. The alloys are hard and have high melting points.

42.What is the action of heat on potassium permanganate ? Give equation.

Ans. It decomposes at 513K to potassium manganate, manganese dioxide and oxygen.

2KMnO4 K2MnO4 + MnO2 + O2

43. What is the action of neutral or faintly alkaline permanganate solution on iodide

?Give equation.

Ans. Alkaline permanganate solution oxidize iodide to iodate.

I- + 2MnO4- + H2O IO3

- + 2MnO2 + 2OH-

44. Explain the oxidising action of acidified potassium dichromate on (iron(II) salts) Fe +2 ions and write the ionic equations for the reaction. Ans. Acidified potassium dichromate oxidises iron(II) salts to iron(III). Cr2O7

2− + 14H+ + 6Fe2+ → 2Cr3+ + 6Fe3+ + 7H2O 45. The transition metals generally form coloured compounds, why? Ans. The compounds of transition elements shows colour due to presence of unpaired electron & ability to undergo d-d transition. When an electron from a lower energy d orbital is excited to a higher energy d orbital, the energy of excitation corresponds to the frequency of light absorbed. This frequency generally lies in the visible region. 46. Give reason “ transition metals and their many compounds acts as good catalysts”. Ans. Transition metals and their many compounds acts as good catalysts,it is due to (i) partially filled (n-1) d orbital (ii)variable oxidation state and provide a suitable surface for the reaction to take place. 47.Explain giving reason “transition metals form a large number of complex compounds”.

Ans.Transition metals form a large number of complex compounds due to (i) Small size & high charge density of the ions of transition metals. (ii) presence of vacant d orbitals of suitable for bond formation. 48. What is the effect of increasing pH on a solution of potassium dichromate? Ans.On increasing the pH of the solution Potassium dichromate (orange) becomes potassium chromate (yellow) Cr2O7

2- + 2OH- → 2CrO42- + H2O

49.What happens when H2S is passed into potassium dichromate in acidic medium? Give the equation. Ans. H2S gets oxidized to sulphur Cr2O7 −2 + 14H+ + 6e− → 2Cr+3 + 7H2O 3H2S → 6H+ + 3S + 6e− Cr2O7

−2 + 3H2S + 8H+ → 2Cr+3 + 7H2O + 3S

50. What is ‘disproportionation’ of an oxidation state ? Give one example of disproportionation reaction in aqueous solution. Ans. A particular oxidation state , which is relatively less stable compared to other oxidation states , under goes disproportion. Manganese (VI) which is relatively less stable changes over to manganese (VII) and manganese (IV) in acid solution.

3 MnO4-2 + 4H+ MnO2 +2MnO4

-+ 2 H2O 51. What is lanthanoid contraction? Write any one consequence of lanthanoid contraction. Ans. Steady decrease in the size of lanthanides with increase in atomic number is known as lanthanoid contraction. Due to lanthanoid contraction radii of members of 3rd transition series are very much similar to corresponding members of 2nd series. 52.Write any two consequences of lanthanoid contraction. Ans. Two consequences of lanthanoid contrations are (i) The radii of the members of the third transition series to be very similar to those of the corresponding members of the second series. Ex. The almost identical radii of Zr (160 pm) and Hf (159 pm) & Nb (146pm) & Ta (146pm) (ii) Difficulty in separation of lanthanoids due to similarity in chemical properties. 53. Name the two series of f-block.

Ans. The f-block consists of the two series, lanthanoids (the fourteen elements following lanthanum) and actinoids (the fourteen elements following actinium.

54. The chemistry of actionoids is more complicated than lanthanoids. Why?

Ans. The actinoids are radioactive elements having half lifes varying. Some members can be prepared only in nanogram quantities. These facts render their study more difficult. 55.Write two comparisons of variability in oxidation states of transition metals and non transition elements (p- block elements) ? Ans. 1. In transition elements , variable oxidation state differ from each other by unity, whereas in case of non transition elements , oxidation state differ by units of two.(For example Fe exhibits o.s of +2 and +3 . similarly copper exhibits two o.s of +1 and +2 . on the other hand, Sn, Pb exhibit o.s of +2 and +4.) 2. In transition elements, higher o.s are more favoured in elements of higher atomic mass, whereas in p-block elements lower o.s are favoured by heavier members ( due to inert pair effect, For example Mo(VI) and W(VI) are more stable than Cr(VI). On the other hand Pb(II) is more stable than Sn(II)) 56. What happens when

(a) A lanthonoid reacts with dilute acids ? (b) A lanthonoid reacts with water?

Ans.(a) When lanthonoid reacts with dilute acids , it liberates hydrogen gas.

(b)When lanthonoid reacts with water , it forms lanthanoid hydroxide and liberate hydrogen gas.

57. What is the gas liberated when i) crystals of potassium permanganate is heated to 513K ? ii) acidified potassium permanganate is treated with oxalate ion at

333K? Ans. i) When crystals of potassium permanganate is heated to 513K Oxygen (O2) gas is liberated . ii) Acidified potassium permanganate when treated with oxalate ion at 333K liberates Carbon dioxide (CO2) gas.

58. What is the composition of mischmetall? Give its one use.

Ans. The composition of mischmetall is lanthanoid metal (~ 95%) and iron (~ 5%) and traces of S, C, Ca and Al. Mischmetall is used in Mg-based alloy to produce bullets, shell and lighter flint 59. Show the interconversion of chromate and dichromate The chromates and dichromates are interconvertible in aqueous solution depending upon pH of the solution. At pH less than 7: Chromate (yellow) on adding acid becomes dichromate (orange) 2CrO4

2- + 2H+ → Cr2O72- + H2O

At pH more than 7 : Dichromate (orange) on adding base becomes Chromate (yellow) Cr2O7

2- + 2OH- → 2CrO42- + H2O

60.How does the neutral or faintly alkalline potassium permanganate solution react with (a) Iodide (b) thiosulphite? Write the ionic equations for the reactions In neutral or faintly alkaline solutions: (a) The oxidation of iodide to iodate: 2MnO4

− + H2O + I− ——> 2MnO2 + 2OH− + IO3−

(b) Thiosulphate is oxidised almost quantitatively to sulphate: 8MnO4

– + 3S2O32– + H2O ——> 8MnO2 + 6SO4

2– + 2OH–

III. Answer the following questions. Each question carries THREE marks 61.Name the metal of the 1st row transition series that i) has highest value for magnetic moment ii) has zero spin only magnetic moment in its +2 oxidation state. iii) exhibit maximum number of oxidation states.

Ans. i) Chromium ii) Zinc iii) Manganese

62.Transition metals form a large number of complex compounds.Give reason.

Ans. Transition metals for complex compounds due to,

i) small sizes of metal cations

ii) their ionic charges and

iii) availability of d orbitals for bond formation.

63.Explain the trend in atomic size of 3d series of transition elements with reason. Ans. With increase in atomic number in 3d series - atomic size decreases (Sc to Cr) , then remain almost constant (Cr to Cu) and increase slightly at the end (Cu to Zn). Reason: In the beginning of the series the screening (or shielding effect) effect of a d electron is not that effective, hence the net electrostatic attraction between the nuclear charge and the outermost electron increases , hence atomic size radius decreases. In the middle of the series, increase in nuclear charge and increase in screening effect balance each other. So atomic radii become almost constant. Increase in atomic radii towards the end is due to the electro – electron repulsions causes the expansion of electron cloud. 64. Explain trend in Ionisation Enthalpies of 3d series of transition elements . Ans.Ionisation enthalpy increase along each series of the transition elements from left to right. However many small variations, IE of Chromium is lower because removal of an electron from Chromium does not change in the d (3d5 4s1 to 3d5 4s0 ) configuration. I.E value for Zn (3d10 4s2) is higher because an electron is removed from 4s level which needs more energy. 65. How is potassium dichromate prepared from iron chromite ore?

Ans.Potassium dichromate is manufactured from chromite ore (FeCr2O4).

(i) Chromite ore is fused (FeCr2O4) with sodium or potassium carbonate in free access of air to get sodium chromate . 4FeCr2O4 + 8Na2CO3 + 7O2 → 8Na2CrO4 + 2Fe2O3 + 8CO2

(ii) The yellow solution of sodium chromate is filtered and acidified with sulphuric acid to give a solution from which orange sodium dichromate, Na2Cr2O7. 2H2O can be crystallised. 2Na2CrO4 + 2H+ → Na2Cr2O7 + 2Na+ + H2O

Potassium dichromate prepared by treating the solution of sodium dichromate with potassium chloride. Na2Cr2O7 + 2KCl → K2Cr2O7 + 2NaCl Orange crystals of potassium dichromate crystallise out.

66. Describe the preparation of potassium permanganate from manganous dioxide. Potassium permanganate is prepared by fusion of MnO2 with an alkali metal hydroxide and an oxidising agent like KNO3. This produces the dark green K2MnO4 which disproportionates in a neutral or acidic solution to give permanganate. 2MnO2 + 4KOH + O2 → 2K2MnO4 + 2H2O 3MnO4

2– + 4H+ → 2MnO4– + MnO2 + 2H2O

67.How does the acidified permanganate solution react with (a) iron(II) ions (b) oxalic acid and (c) hydrogen sulphide ? Write the ionic equations for the reactions In acid solutions: (a) Fe2+ ion (green) is converted to Fe3+ (yellow): 5Fe2+ + MnO4

– + 8H+ ——> Mn2+ + 4H2O + 5Fe3+ (b) Oxalate ion or oxalic acid is oxidised at 333 K: 5C2O4

2− + 2MnO4- + 16H+ ——> 2Mn2+ + 8H2O + 10CO2

(c) Hydrogen sulphide is oxidised, sulphur being precipitated: H2S —> 2H+ + S2− 5S2− + 2MnO4

− + 16H+ ——> 2Mn2+ + 8H2O + 5S

UNIT 9 Topic: Coordination Compounds

1. State the postulates of Werner’s theory of coordination compounds. 3 Marks

Postulates:

1. Central metal ion in a complex shows two types of valences - primary valence and

secondary valence.

2. The primary valence is ionisable and satisfied by negative ions.

3. The secondary valence is non ionisable. It is equal to the coordination number of the

central metal ion or atom. It is fixed for a metal. Secondary valences are satisfied by

negative ions or neural molecules (ligands).

4. The primary valence is non directional. The secondary valence is directional. Ions or

molecules attached to satisfy secondary valences have characteristic spatial

arrangements. Secondary valence decides geometry of the complex compound.

2. What are the limitations of Werner’s theory of coordination compounds?

3 Marks

This theory fails to explain why,

a) a few elements have the property to form coordination compounds

b) the bonds in coordination compounds have directional properties

c) coordination compounds have characteristic magnetic and optical properties.

3. Write one difference between double salts and complex salts with respect to

their ionisation. Give one example for each type of salt. 2 Marks

Double salt Complex salt

Double salt is stable only in solid state, but

dissociate into simple ions completely in solution state.

Complex salt is stable both in solid and

solution state and does not dissociate completely in solution state.

E.g: KCl.MgCl2.6H2O E.g: K4[Fe(CN)6]

4. Define Coordination entity of coordination compounds. 2 Marks

A coordinate entity constitutes a central metal atom or ion bonded to a fixed number of

ions or molecules (ligands).

e.g: [Fe(CN)6]4-.

5. What is central metal ion in a coordination compound? Give an example.

2 Marks

The metal atom or ion in a coordination entity to which, a fixed number of ions or

molecules (ligands) are bound in a definite geometrical arrangement around it is called

central metal ion. It is a Lewis acid.

e.g: In K4[Fe(CN)6] , central metal ion is Fe2+.

6. What are ligands? Give an example. 2 Marks

The ions or molecules bonded to the central metal atom or ion in a coordination entity

are called ligands. Ligands are Lewis bases.

e.g: In [Ni(CN)4]2-, ligand is CN- ion.

7. Define the term coordination number of a central metal atom or ion in a

complex compound. 2 Marks

The coordination number of central metal atom or ion in a complex is the number of

ligand donor atoms to which the metal is directly bonded.

In K4[Fe(CN)6] , coordination number of Fe2+ is 6.

8. Define coordination sphere of coordination compounds. 2 Marks

The central metal atom or ion and the ligands of the complex compound are written

within square bracket. This is called coordination sphere of coordination compounds.

9. What are homoleptic complexes? Give an example. 2 Marks

Homoleptic complexes are the complexes in which central metal ion or atom is bound to

only one type of donor groups.

e.g: K4[Fe(CN)6]

10. What are heteroleptic complexes? Give an example. 2 Marks

Homoleptic complexes are the complexes in which central metal ion or atom is bound to

more than one type of donor groups.

e.g: [Co(NH3)5Cl]SO4

11. Classify the following ligands into unidentate, didentate and polydentate

ligands. NH3, EDTA, oxalate. 3 Marks

NH3 Unidentate

EDTA Polydentate

Oxalate Didentate

12. Give the IUPAC name for the following compounds.

a) K4[Fe(CN)6] potassium hexacyanidoferrate(II)

b) [Cu(NH3)4] SO4 tetramminecopper(II) sulphate

c) [Co(NH3)5Cl]SO4 pentamminechloridocobalt(III) sulphate

d) K3[Fe(C2O4)3] potassium trioxalatoferrate(III)

e) [CoCl2(en)2]+ dichloridobis(ethane-1,2-diamine)cobalt(III)

f) [Co(NH3)5(NO2)]Cl2 pentamminenitrito-N-cobalt(III) chloride

g) [Co(NH3)5(ONO)]Cl2 pentamminenitrito-O-cobalt(III) chloride

h) [Ni(CO)4] tetracarbonylnickel(0)

13. What are ambidentate ligands? Give one example. 2 Marks

Ligands which have two donor atoms, but can bond to central metal atom or ion through

only one donor atom are called ambidentate ligands.

e.g: NO2-, SCN

- etc.,

14. What is geometrical isomerism in complexes? Give an example. 2 Marks

It is a phenomenon in which two complex compounds have the same molecular formula

and same chemical bonds, but different geometrical arrangement of the ligands. The two

forms are called cis and trans forms.

e.g: [Pt(NH3)2Cl2]

Cl NH3

Pt Cl NH3

Cis isomer

Cl NH3

Pt NH3 Cl

Trans isomer

15. Explain optical isomerism in complexes with an example. 2 Marks

It is a phenomenon in which two complex compounds have the same molecular formula

and same chemical bonds but cannot be superposed on one another. They differ in the

rotation of plane of plane polarised light. These isomers are called enantiomers. The two

forms are called dextro (d) and laevo (l).

e.g:

16. What is linkage isomerism? Give an example. 2 Marks

Linkage isomerism is type of isomerism in which two complex compounds differ in the

donor atoms for ligands (different ligating atoms). It is shown by complex compounds

containing ambidentate ligands.

e.g: [Co(NH3)5NO2]Cl2 and [Co(NH3)5(ONO)]Cl2

17. Indicate the type of isomerism in the following set of complex compounds.

a) [Co(NH3)5Cl]SO4 and [Co(NH3)5 SO4]Cl

b) [Co(NH3)5(SCN)]Cl2 and [Co(NH3)5(NCS)]Cl2 2 Marks

a) [Co(NH3)5Cl]SO4 and [Co(NH3)5 SO4]Cl Ionisation isomerism

b) [Co(NH3)5(SCN)]Cl2 and [Co(NH3)5(NCS)]Cl2 Linkage isomerism

18. Explain coordination isomerism in complexes. Give one example.

Coordination isomerism is type of isomerism due to interchange of ligands between

cationic and anionic entities of different metal ions present in a complex.

e.g: [Co(NH3)6] [Cr(CN)6] and[Cr(NH3)6] [Co(CN)6]

19. Explain ionization isomerism in complexes. Give an example.

Ionisation isomerism is type of isomerism in which two complex compounds produce

different ions in solution form.

e.g: [Co(NH3)5Cl]SO4 and [Co(NH3)5 SO4]Cl

20. What is solvate isomerism? Explain with an example.

Solvate isomerism is a type of isomerism in which complex compounds differ in the

number of water molecules acting as ligands and water of hydration.

e.g: [Cr(H2O)6]Cl3 and [Cr(H2O)5Cl]Cl2. H2O

21. Explain the formation of [CoF6]3- . Is this complex paramagnetic? 3 Marks

Co, Z=27 [Ar]3d7 4s2 4p0 4d0

Co3+ [Ar] 3d6 4s0 4p0 4d0

It undergoes sp3d2 hybridisation.

When F - ligand attacks the central metal ion, Co3+

This complex uses outer d orbital (4d) for hybridisation. It is an outer orbital

complex. It has unpaired electrons. [CoF6]3- is paramagnetic. This complex is called high

spin or spin free complex.

22. Give the geometry, hybridization and magnetic property of [Co(NH3)6]3+ based

on VBT. 3 Marks

Co, Z=27 [Ar]3d7 4s2 4p0

Co3+ [Ar] 3d6 4s0 4p0

When NH3 ligand attacks the central metal ion Co3+ , pairing of electrons in 3d orbital

occurs against Hund’s rule. Co3+ can undergo d2sp3 hybridization.

This complex uses inner d orbital (3d) for hybridisation. It is an inner orbital complex. It

has no unpaired electrons. [Co(NH3)6]3+ is diamagnetic. This complex is called low spin or

spin paired complex.

It has octahedral geometry.

23. Using VBT, explain the type of hybridization, geometry and magnetic property

of [NiCl4]2-. 3 Marks

Ni, Z =28 [Ar] 3d8 4s2 4p0

Ni2+, [Ar] 3d8 4s0 4p0

It undergoes sp3 hybridisation.

Four pair of eelctrons from 4 Cl- ions

This complex has unpaired electrons. It is paramagnetic.

[NiCl4]2- has tetrahedral structure.

24. Explain the hybridization, geometry and magnetic property in the complex

compound [Ni(CN)4]2-. 3 Marks

Ni, Z =28 [Ar] 3d8 4s2 4p0

Ni2+, [Ar] 3d8 4s0 4p0

When CN- ligand attacks the central metal ion Ni2+, unpaired electrons in 3d orbital are

paired up against Hund’s rule. It undergoes dsp2 hybridisation.

Four pair of electrons from 4 CN- ligands.

This complex has no unpaired electrons. It is diamagnetic.

[Ni(CN)4]2- has square planar structure.

25. What are the limitations of Valence Bond Theory? 2 Marks

This theory,

a) has a number of assumptions

b) does not give quantitative interpretation of magnetic data

c) does not explain colour shown by complexes.

d) could not predict tetrahedral or square planar shape for the coordination number 4

e) could not distinguish between strong and weak ligands.

26. Why [CoF6]3- is called an outer orbital complex? 1 Mark

In this complex, Co3+ uses outer d orbital (4d) for hybridization. Therefore it is called an

outer orbital complex.

27. What are inner orbital complexes? Give an example. 2 Marks

Inner orbital complex is one where the central metal ion uses inner d orbital {(n-1)d

orbital} for hybridsation.

e.g: [Co(NH3)6]3+

28. Explain the salient features of crystal field theory. 2 Marks

This theory considers ligands as point charges in case of anionic ligands and dipoles in

case of neutral molecules. The bond formed between central metal ion and the ligands is

purely ionic.

29. What is crystal field splitting? Explain crystal field splitting in octahedral

entities using energy level diagram. 3 Marks

In an isolated gaseous central metal atom or ion, all the five d orbitals are having same

energy. i.e they are degenerated. In the presence of attacking ligands, it becomes

asymmetric and the d orbitals lose degeneracy, resulting in splitting of d orbitals. This is

called crystal field splitting.

In an octahedral complex, six ligands surround the central metal ion. dx2 – y2 and dz2

orbitals (called eg set) are directed along the direction of ligands and experience more

repulsion. They have more energy. dxy, dyz and dzx orbitals (called t2g set) are directed

between the axes of attacking ligands and experience lesser repulsion by the ligands.

They have lesser energy. The energy separation between two split sets is denoted as ∆o .

The energy of eg orbitals increase by 3/5 ∆o and that of t2g set decrease by 2/5 ∆o .

30. What is spectrochemical series? Arrange the following ligands in the increasing

order of their field strength.

Br-, I-, H2O, CO, F- 2 Marks

Spectrochemiccal series is the arrangement of ligands in the order of increasing field

strength.

Correct order for the given set is,

I- < Br- < F- < H2O < CO

31. Draw a figure to show the splitting of d orbitals in a tetrahedral crystal

field. 2 Marks

UNIT.10

HALOALKANES AND HALOARENES

ONE MARKS QUESTIONS

1. What are haloalkanes? [1]

A: Haloalkane is a derivative obtained by replacing hydrogen atom of alkane by halogen atom.

2. What is the hybridization of the carbon attached with vinylic halides (or) aryl halides?

[1]

A: sp2hybridisation.

3. Among phosphorus trihalides which halides are generated insitu ? [1]

A: PBr3, PI3

4. Free radical halogenation of hydrocarbons is not a best method to prepare haloalkanes.

Why? [1]

A: Because this method gives mixture of isomeric mono and poly haloalkanes, which is difficult

to separate as pure compounds.

5. Why fluoro compounds cannot be prepared from electrophilic substitution reaction?

[1]

A: Due to high reactivity of fluorine.

6. What happens to the boiling point of isomeric haloalkanes with increase in

branching?[1]

A: decreases.

7. How density and atomic mass of halogen atoms in haloalkanes are related? [1]

A: Directly

8. Why tertiary alkyl halide undergoes SN1 reaction very fast? [1]

A: Because of the high stability of tertiary carbocation

9. What is the order of reactivity of alkyl halides towards SN1 and SN2 reaction [1]

A: Towards SN1 reaction order of reactivity is 30> 20 > 10

Towards SN2 reaction order of reactivity is 10> 20> 30.

10. Allylic and benzylic halides are highly reactive towards SN1 reaction. Why? [1]

A: Because the carbocation formed from allylic and benzylic halides is more stable due to

resonance.

11. Between SN 1and SN2 reaction which one proceeds with complete stereo chemical

inversion? [1]

A: SN2 reaction.

12. Between SN1 and SN2 reactions which one proceeds with racemisation?

[1]

A: SN1 reaction.

13. What are optically active compounds?

[1]

A: Certain compounds rotate the plane polarized light when it is passed through their solutions

are called as optically active compounds.

14. What are dextro rotatory compounds?

[1]

A: The compound which rotates plane polarized light in clockwise direction is called as dextro

rotatory compound.

15. What is a laevo rotatory compound? [1]

A: The compound which rotates plane polarized light in anticlockwise direction is called as laevo

rotatory compound.

16. What are optical isomers?

[1]

A: The dextro and laevo rotatory isomers of a compound are called as optical isomers.

17. What is asymmetric carbon (or) stereocentre [1]

A: A carbon atom attached with four different substituent groups is called as asymmetric carbon

(or) stereocentre.

18. What are chirals? [1]

A: The objects which are non-superimposable on their mirror image are said to be chirals.

19. What are achiral molecules? [1]

A: The molecules which are, superimposable on their mirror images are called achiral

molecules.

20. Between propan – 2- ol and butan – 2 – ol, identify the chiral molecule?

[1]

A: Butan – 2- ol.

21. What are enantiomers? [1]

A: The stereo isomers related to each other as non-superimposable mirror images are called

enantiomers.

22. What are racemic mixtures? [1]

A: A mixture containing two enantiomers in equal proportions will have zero optical rotation, are

called as racemic mixture (or) racemic modification.

23. What is racemization? [1]

A: The process of conversion of enantiomer into a racemic mixture is known as racemization.

24. Between α and β- hydrogen which one is removed during dehydrohalogenation of alkyl

halides? [1]

A: β - Hydrogen.

25. What are organo-metallic compounds? [1]

A: Most organic chlorides, bromides and iodides react with certain metals to give compounds

containing carbon-metal bonds are known as organo – metallic compounds.

26. Write the general formula of Grignard reagent? [1]

A: RMgX.

27. Name the product formed when Grignard reagent treated with water?

[1]

A: Alkanes

28. What is the hybridization of carbon atom in C-X bond of alkyl halides? [1]

A: sp3hybridisation

29. Mention the hybridization of carbon atom in C-X bond of aryl halides? [1]

A: sp2hybridisation

30. Why SN1mechanism is ruled out in haloarenes? [1]

A: In case of haloarenes, the phenyl cation formed as a result of self-ionization will not be

stabilized by resonance and therefore, SN1mechanism is ruled out.

31. What happens to the reactivity of haloarenes towards nucleophilic substitution when

electron withdrawing group present at ortho or para position? [1]

A: Increases.

32. What is wurtz-fitting reaction? [1]

A: A mixture of an alkyl halide and aryl halide gives an alkylarene when treated with sodium in

dry ether and is called as wurtz-fitting reaction.

33. Expand DDT [1]

A: Dichlorodiphenyltrichloro ethane

34. Among chloral, chloroform, DDT, carbontetrachlorides , which has more number of

chlorides ?

A;DDT

[1]

Cl

Cl

H

Cl

Cl

Cl

O2N

OH

NO2

NO2

BrCH3

H3CH2CH

CH3

H CH2CH3+ Br

Θ

2-bromobutane

secondary haloalkane C

R

H

R X

primary haloalkane C

H

H

R X

35. Identify the product.

A:

36.Write the structure of DDT [1]

A:

TWO MARKS QUESTIONS

37. Explain the classification of monohaloalkanes containing Sp3C – X bond

[2]

A: These are classified into three types

X

X X

H3C

X

Tertiary haloalkane C

R

R

R X

X CH2

X

CH2 X

38. What is the difference between allylic acid and benzylic halides

[2]

A:

39. What are vinylic halides? Give example?

[2]

A: These are the compounds in which the halogen atom is bonded to a Sp2-hybridised carbon

atom of a carbon-carbon double bond.

40. What are aryl halides? Give example?

[2]

A: These are the compounds in which the halogen atom is bonded to the Sp2hybridised carbon

atom of an aromatic ring.

Allylichalide

These are the compounds where the halogen atom is bonded to a Sp3-hybridised carbon atom next to carbon-carbon double bond.

Benzylic halides

These are the compounds in which the

halogen atom is bonded to an Sp3-hybridised

carbon atomnext to an aromatic ring.

Br

Br

Br

C

H3C

CH3

H3C CH2

Cl

CH2Cl

C+

X -

CH3 CH

Cl

Cl

CH2 CH2

ClCl

41. Write the IUPAC name of the following compounds [2]

a) b) c)

d) CH2=CH-Cl

A: a) 1, 3, 5 – tribro,obenzece

b) 1, chloro 2, 2 – dimethyl propane

c) Chloro phenyl methane

d) Chloroethene

42. Explain nature of C-X bond in haloalkanes? [2]

A: Haloalkanes are polar in nature because halogen atoms are more electronegative than

carbon as a result the carbon atom bears a partial positive charge whereas the halogen atom

bears a partial negative charge

43. What are geminal halides? Give example? [2]

A: These are the dihalo compounds where two same halogen atoms are present on the same

carbon atom

Ex:

44. What are vicinal halides? Given example? [2]

A: These are the dihalo compounds where two same halogen atoms are present on the adjacent

carbon atoms

Ex:

CH3

+ Cl2Fe

dark

CH3

Cl

TolueneO-chloro toluene

+

Cl

CH3

P-chloro toluene

N2

+X

-X

Cu2X2 + N2

Bebzene diazonium halide haloarene X = Cl, Br

45. R-OH + A R – Cl + H2O

Identify A and B in the given reaction [2]

A: R – OH + HCl R-Cl + H2O

A = HCl,

B = ZnCl2

46. How do you prepare chloroalkane using reaction between alcohol and phosphorus

pentachloride?

[2]

A: R – O – H + PCl5 R – Cl + HCl + POCl3

47. How do you prepare chloroalkane using alcohol and thionylchloride? Mention

advantage of this reaction?

[2]

A: R – OH +SOCl2 R – Cl + SO2↑ + HCl↑

48. Explain the reaction between toluene and chlorine?

[2]

A:

49. Why iodination of arenes by electrophilic substitution requires an oxidizing agent?

[2]

A: Because iodination of arenes is a reversible reaction due to formation of biproduct HI and

presence of an oxidizing agent oxidises the HI formed, there by prevents reversible reaction.

50. Explain sandmeyer’s reaction to prepare haloarenes (or) how do you convert benzene

diazonium salt into haloarenes?

[2]

A:

ZnCl2

B

N2

+Cl

-I

KI+ N2

+ CH3CH3 CH3

I

In this reaction iodination of benzene diazonium halide does not requires cuprous halide

51. Identify major product in the given reaction and give reason?

[2]

CH3 – CH = CH2 + H – I CH3 – CH2– CH2– I

Iodopropane 2- Iodo propane

A: 2- Iodo propane

Because 2- Iodo propane involves stable 2°- carbocation.

52. What happens when ethane reacts with bromine in presence of CCl4. Write the reaction

[2]

A: This reaction gives 1.2- dibromoethane product.

CH2 = CH2 + Br2 BrCH2 – CH2Br

53. What is finkelstein reaction? Give its general reaction?

[2]

A: The reaction in which an alkyl chloride (or) bromides reacts with sodium iodide in dry

acetone gives alkyl iodides is called finkelstein reaction.

R – X + NaI R – I + NaX

X= Cl, Br

54. How do you get fluoro methane from chloro (or) bromo methane and name the

reaction? [2]

A: CH3 – Br + AgF CH3 – F + AgBr

Bromomethane. Silver fluoride fluoro methane

This reaction is called as swart’s reaction.

55. Why the boiling point of halides are higher than hudrocarbons of comparable molecular

mass [2]

A: Because of greater polarity as well as higher molecular mass as compared to the parent

hydrocarbon, the inter molecular forces of attraction are stronger in the halogen derivatives.

CCl4

Nu C

X

C Nu + X

OH +

Cl OH Cl

H

HO

H H HH H H

H

+ Cl

H

(Slow step)r [OH ] [CH 3 - Cl]

order = 1 + 1 = 2

56. Arrange the halides R-Br, R-Cl, R-F, R-I containing same alkyl group in the increasing

order of their boiling point [1]

A: R-F < R – Cl< R-Br < R – I .

57. In isomeric dihalobenzenes, the para - isomers has high melting point than ortho and

meta – isomers. Why? [2]

A: Because para – isomers are symmetric in nature and fits in crystal lattice better as compared

to ortho and meta – isomers.

58. Haloalkanes are less soluble in water even though they are polar in nature. Why?

[2]

A: For a haloalkane to dissolve in water, energy is required to overcome and break the hydrogen

bond between water molecules. Less energy is released when new attractions are set up

between the haloalkane and water molecules as these are not as strong as the original

hydrogen bonds in water.

59. Whyhaloalkanes undergoes nucleophilic substitution reaction?

[2]

A: In haloalkanes due to difference in electronegativity of carbon and halogen, carbon bears

partial positive charge and attracts nucleophile and halogen atom departs as halide ion

60. Write SN2 mechanism [2]

61. In SN2 reaction transition state cannot be isolated. Why? [2]

CH3 - CH2 - CH = CH - CH3 CH3 - CH2 - CH2 - CH - CH2 HC3 - CH2 - CH2 - CH = CH2OH OHΘ Θ

Br

HPent - 2 - ene (81%) 2 - Bromo pentane Pent - 1 - ene (19%)

CH3H CH2

CH2

OH

CH3

+ HClheat

CH3H CH2

CH2

Cl

CH3

+ H - OH

(-) - 2 methylbutan -1-of (+) -1-chloro-2-methylbutane

A: In the transition state, the carbon atom is simultaneously bonded to incoming nucleophile and

the outgoing leaving group; as a result carbon atom in transition state is simultaneously bonded

to five atoms and therefore is unstable.

62. Why tertiary haloalkanes are less reactive towards SN2 reaction?

[2]

A: Because SN2 reaction requires the approach of the nucleophile to the carbon bearing the

leaving group, the presence of bulky substituent on (or) near the carbon atom like in 30 –

haloalkane have a dramatic inhibiting effect.

63. What is retention of configuration? Give example [2]

A: Retention of configuration is the preservation of integrity of the spatial arrangement of bonds

to an asymmetric centre during a chemical reaction.

64. Write the reaction between 2-bromopentane with alcoholic solution of potassium

hydroxide and mention the major product in thereaction?

[2]

A:

Pent-2-ene is the major product.

65. During dehydrohalogenation of 2-bromopentane, Pent-2-ene is the major product.

Why? [2]

A: According to saytzeff rule alkene which has the greater number of alkyl groups attached to

the doubly bonded carbon atoms are stable.

66. How do you prepare ethyl magnesium bromide from ethyl bromide? [2]

A: CH3CH2Br + Mg CH3CH2Mg Br

Dry Ether

:X: x+

X+

X+..

Θ

Θ

Θ

Cl ClNO2

Cl

NO2

+Conc. HNO3

CONC. H2SO4

Chlorobenzene 1 - chloro - 2 - nitro

- benzene (minor) 1 - chloro - 4 - nitrobenzene

(major)

Cl ClCH3

Cl

CH3

+

anhydrous

Chlorobenzene 1 - chloro - 2 - methyl

- benzene (minor) 1 - chloro - 4 - methylbenzene

(major)

AlCl3+ CH3 - Cl

Chloromethane

Cl

(i) NaoH, 623 K, 300 atm

(ii) H +

OH

:Cl: Cl+

Cl+

Cl

Θ

Θ

Θ

..

67. Aryl halides are extreamely less reactive towards nucleophilic substitution reaction

than alkyl halides. Why? [2]

A: In haloarenes, the electron pairs on halogen atom are in conjugation with π- electrons of the

ring and C-Cl bond aquires partial double bond character due to resonance. As a result, the

bond cleavage in haloarene is difficult than haloalkane and therefore, they are less reactive

towardsnucleophilic substitution reaction.

68. How do you convert chlorobenzene to phenol? [2]

A:

69. Write the resonance structure to show the halogenatom present on the haloarenes

areortho and para directing group? [2]

A:

70. Write the reaction involved in the nitration of chlorobenzene? [2]

A:

71. Explain methylation of chlobenzene and name the reaction? [2]

X

+ 2Na + 2NaXdry ether2

Diphenyl

X

+ 2Na + R - X

R

+ 2NaXdry ether

haloarene haloalkane alkylarene

(CH3)3 CBr CH3

CH3 CH3

+ + Br (slow)Θ

CH3

CH3 CH3

+ + OHΘ

(CH3)3 COH (fast)

72. Write general equation of wurtz-fitting reaction? [2]

A:

73. Explain fitting reaction with general equation? [2]

A: Two arylhalides reacts with sodium in presence of dry ether gives diphenyl. This reaction is

called as fitting reaction.

74. What are freons? Give an example [2]

A: The fluro, chloro compounds of methane or ethane collectively called as freons

Ex: CCl2 F2

THREE MARKS QUESTIONS

75. Explain SN1machanism [3]

A: (CH3)3CBr + ŌH (CH3)3 C OH + BrΘ

2-bromo-2-methylpropane 2-methylpropan – 2 -ol

Step (1): formation of carbocation

Step (2): attack of nucleophile on carbocation

H Br

CH3

C6H13

+ OH OH

CH3

H

C6H13

+ BrΘ Θ

BrCH3

H3CH2CH

CH3

H CH2CH3+ Br

Θ

2-bromobutane

CH3

CH2CH3CH3

Θ

OHHO HCH2CH3

CH3 OHΘ

OH

CH3

HH3CH2C

(+) - Butan - 2 - ol (-) Butan - 2 - ol

Rate of the reaction depends on slow step of the reaction

r α [(CH3)3C Br]1

Order = 1

76. With example explain SN2reactions of optically active halides areaccompanied by

inversion of configuration.

[3]

A: SN2 mechanism of optically active halides has the inverted configuration because nucleophile

attaches itself on the side opposite to the one where the halogen atom is present. For example

when (-) – 2 – bromooctane is allowed to react with sodium hydroxide, (+) – octan – 2 – ol is

formed with the – OH group occupying the position opposite to what bromide had occupied.

77. SN1 reaction of optically active halides accompanied by racemisation. Explain with

example?[3]

A: SN1 reaction of optically active halides accompanied by racemization (but not 100%

racemization) because carbocation formed in the slow step being sp2hybridised is planar. As a

result the attack of the nucleophile may be accomplished from either side resulting in a mixture

of products, one having the same configuration and the other having opposite configuration.

Ex:

(Please make correction of CH3 as H in the transition state)

1

Unit 11. Alcohols, Phenols and Ethers

One mark questions

1. Name the alcohol which is used for polishing wooden furniture.

Ans: Ethanol

2. What are alcohols?

Ans: Hydroxyl derivatives of aliphatic compounds are called alcohols.

3. What is the IUPAC name of ?

Ans: Ethane-1, 2-diol

4. Write the structure of 2-methyl cyclopentanol.

Ans:

5. Name the simplest hydroxyl derivative of benzene.

Ans: Phenol

6. What is the IUPAC name of Resorcinol?

Ans: Benzene-1, 3-diol

7. What is the common name of CH3OC2H5?

Ans: Ethylmethyl ether

8. Write the formula of anisole.

Ans: C6H5OCH3 or

9. What is the IUPAC name of anisole?

Ans: Methoxybenzene.

10. Write the IUPAC name of CH2 = CH CH2OH

Ans: prop-2-en-1-ol

11. Why is the bond angle in alcohols is slightly less than the tetrahedral

angle?

Ans: It is due to the repulsion between the unshared electron pairs of oxygen atom.

12. Why is the bond angle slightly greater than the tetrahedral angle in

ethers?

Ans: It is due to the repulsive interaction between the two bulky R groups or alkyl

groups.

13. Name the product obtained when propene is subjected to acid catalysed hydration.

Ans: Propan-2-ol or 2-propanol

14. In the reaction, H

2 2 2H C CH H O X

Identify X.

Ans: Ethanol

15. In a reaction, 2 2

diporane

3 2 H O /NaOHCH CH CH X Name the product X formed in the

reaction.

Ans: Propan-1-ol.

16. Write the chemical name of cumene.

Ans: Isopropyl benzene.

17. The boiling point of alcohols is much higher than ethers and other classes of

compounds with similar molecular masses. Give reason.

Ans: Due to intermolecular hydrogen bonding in alcohols.

18. Give reason: Lower alcohols are soluble in water.

Ans: Due to the formation of hydrogen bonds with water molecules.

19. Name the compound which is also known as carbolic acid.

Ans: Phenol

20. Name the method by which O-nitrophenol and p-nitrophenol are separated.

Ans: By steam distillation the two isomers are separated.

21. Ether is soluble in water. Give reason.

Ans: Ether is soluble in water because oxygen of ether form hydrogen bonds with

water molecule.

Two Mark Questions

1. What happens when an aldehyde is reduced? Write the general reaction OR explain

the reduction of aldehydes.

Ans: Aldehydes on reduction by hydrogen in presence of catalyst like finely divided

Nickel or platinum give the respective primary alcohols.

Ni

2 2RCHO H RCH OH

Or

Aldehydes on reduction in presence of sodium borohydride (NaBH4) or lithium

aluminium hydride (LiAlH4) yield the respective primary alcohols.

4

4

NaBH or

2LiAlHRCHO 2(H) RCH OH

2. An aldehyde reacts with Grignards reagent forming an inter immediate product

which on hydrolysis gives primary alcohol. Name the aldehyde and write the

chemical equation.

Ans: The aldehyde is methanal or formaldehyde.

Reaction: HCHO + RMgX RCH2OMgX 2H ORCH2OH+MgX(OH)

3. How is phenol prepared from aniline? Write the equation.

Ans: Aniline is treated with nitrous acid in presence of HCl at 273-278 K, when

benzene diazonium chloride is obtained. Which on warming with water or treating

with dilute acids gives phenol.

2 2NaNO H O

6 5 2 6 5 6 5 2HCl warmC H NH C H N NCl C H OH N HCl

4. What is meant by hydroboration – oxidation reaction. Illustrate with an example.

Ans: Diborane reacts with alkenes to give trialkyl boranes which is oxidized to

alcohol by hydrogen peroxide in presence of sodium hydroxide.

Reaction

2 2

3 2 3 2 3 2 2 3

OH

3 2 2 33H O

3CH CH CH (BH ) (CH CH CH ) B

3CH CH CH OH B(OH)

5. Give two reactions that show acidic nature of phenol.

Ans: Reaction (1) C6H5OH + Na C6H5ONa + H2

(2) C6H5OH + NaOH C6H5ONa + H2O

These two reactions prove that phenol is acidic.

6. Name the following reaction and predict the product X obtained.

H2SO4

R’COOH + RO-H X+ H2O

Ans: The name of the reaction is esterification and product X is an ester with the

formula RCOOR.

7. When phenol is treated with acid chloride in presence of pyridine base, what is the

product obtained. Write the equation.

Ans: The reaction is pyridine

6 5 6 5C H OH RCOCl C H OCOR HCl

The product is an ester.

8. Explain the dehydration of ethanol with equation.

Ans: Ethanol undergoes dehydration by heating it with conc. H2SO4 at 443 K.

forming ethene.

2 4conc H SO

3 2 2 2 2443 KCH CH OH CH CH H O

9. Explain the dehydration of a secondary alcohol with equation. OR

How is isopropyl alcohol converted to propene by dehydration reaction?

Ans: Secondary alcohols like isopropyl alcohol undergo dehydration on heating with

85% phosphoric acid at 440 K. forming an alkene (propene)

Reaction 3 485% H PO

3 3 3 2 2440 KCH CH CH CH CH CH H O

|OH

10. Explain the dehydration of tertiary alcohols.

Ans: Tertiary alcohols undergo dehydration when heated with 20% H3PO4 at 358 K

forming the respective alkene.

Reaction:

11. Complete the following reactions:

(a) Cu

2 573RCH OH X (b)

Cu

573X R C R

||O

Name X in both the reactions.

Ans:

(a) Cu

2 573RCH OH RCHO X-Aldehyde

(b) Cu

573R CH OH R C R

|||OR

X = Secondary alcohol

12. Explain the reaction of phenol with dil. nitric acid at 298 K. Write equation.

Ans: Phenol reacts with dil. HNO3 at 298 K forming O-nitrophenol and p-

nitrophenol respectively.

13. How do you convert phenol to picric acid? Explain with equation.

Ans: Phenol reacts with concentrated nitric acid forming picric acid or 2, 4, 6-

trinitro phenol.

14. Explain the bromination of phenol forming ortho and para bromophenols with

equation.

Ans: Phenol reacts with bromine in CS2 at 273 K forming ortho – and para

bromophenols respectively.

15. How is phenol converted to 2, 4, 6-tribromophenol? Explain with equation.

Ans: Phenol reacts with bromine water forming a white ppt of 2, 4, 6-trinitrophenol

16. Explain Kolbe’s reaction with equation. OR What happens when sodium phenate is

treated with carbon dioxide? Write equation and name the reaction.

Ans: Sodium phenate is treated with carbon dioxide and the product on acidification

forms salicylic acid. This reaction is called Koble’s reaction.

17. How is phenol converted to benzene? Write the equation.

Ans: Phenol is converted to benzene on heating with zinc dust.

18. Explain Reimer – Tiemann reaction with equation.

Ans: Phenol is treated with chloroform and sodium hydroxide solution. The product

on acidification forms salicyl aldehyde.

19. Explain the oxidation of phenol with equation.

Ans: Phenol undergoes oxidation with acidified sodium dichromate forming

benzoquinone.

20. How is diethyl ether or ethoxy ethane prepared from ethanol? Write equation.

Ans: Ethanol is heated with conc. H2SO4 to 413 K when ethoxy ethane is obtained.

2C2H5OH C2H5OC2H5+H2O

21. Explain Wilhamson synthesis with equation.

Ans: An alkyl halide reacts with sodium alkoxide forming the respective ethers.

By this method both symmetrical and unsymmetrical ethers can be prepared.

R X R ONa R O R NaX

22. Identify A and B in the following reactions and name the product obtained.

(A)

(B)

Ans: (A)

(B)

23. Explain the reaction of anisole with HI. Write the equation.

Ans: Anisole reacts with HI forming phenol and methyl iodide.

6 5 3 6 5 3C H O CH HI C H OH CH I

24. Explain the bromination of anisole with equation.

Ans: Anisole (methoxy benzene) undergoes bromination with bromine in ethanoic

acid in absence of FeBr3 catalyst forming O-bromoanisole and p-bromoanisole

respectively.

25. Explain the Friedel crafts reaction of anisole with equation.

Ans: Anisole reacts with chloromethane in presence of anhydrous aluminium

chloride as catalyst forming 2-methoxy toluene and 4-methoxy toluene.

OR

Anisole reacts with acetyl chloride in presence of anhydrous aluminium chloride

forming 2-methoxy acetophenone and 4–methoxy acetophenone.

26. Explain the reaction of anisole with a mixture of conc. H2SO4 and conc. HNO3 or

Explain the nitration of anisole with equation.

Ans: Anisole reacts with a mixture of conc. Sulphuric acid and conc. Nitric acid

forming ortho nitro anisole and paranitroanisole.

III. Three Mark Questions

1. Give three reasons that phenols are more acidic than alcohols.

Ans: (1) In phenol, the OH group is attached to sp2 hybridised carbon which is

more electronegative, hence the OH bond becomes more polar.

(2) Due to resonance is phenol, oxygen gets a positive charge and this increases the

polarity of the O H bond.

(3) Delocalisation of negative charge in phenoxide ion makes phenoxide ion more

stable than phenol favouring the ionization of phenol.

2. Explain the mechanism of dehydration of ethanol to ethane.

Ans: The dehydration of ethanol to ethane occurs in the following three steps, when

heated with conc. H2SO4 at 443 K.

2 4conc H SO

3 2 2 2 2443 KCH CH OH CH CH H O

Page 1

UNIT 12 ALDEHYDES KETONES AND CARBOXYLIC ACIDS

1) What are aldehydes ? 1 Aldehydes are the organic compounds containing carbonyl group,linked with one hydrogen and one alkyl /aryl group.

2) What are carboxylic acids? 1 Carboxylic acids are the organic compounds containing carboxyl(-COOH) group/s

3) Between aldehyde and ketones which one is confirmed using Tollen’s reagent. 1 Aldehyde.

4) Between aldehyde and ketones which one is confirmed using Fehling’s solution.. 1 Aldehyde.

5) Write the IUPAC name of the compound.CHO-CH2-CH(CHO)-CH2-CHO. 1 Propane-1,2,3-tricarbaldehyde.

6) The boiling point of aldehydes and ketones are higher than hydrocarbons and ethers of comparable molecular mass. Why. 1 Because in aldehydes and ketones there is a weak molecular association arising out of dipole-dipole interaction.

7) Arrange the following compounds in the increasing order of their acidic strength. HCOOH, CH3COOH, CH3CH2COOH. 1 CH3CH2COOH <CH3COOH< HCOOH.

8) Arrange the following compounds in the decreasing order of their acidic strength. HCOOH, CH3COOH, C6H 5COOH. 1 HCOOH> C6H 5COOH> CH3COOH.

9) Arrange the following compounds in the increasing order of their acidic strength. Cl-CH2COOH, Br-CH2COOH ,F-CH2COOH 1 Br-CH2COOH <Cl-CH2COOH<F-CH2COOH .

10) Name the reagent used in the Stephen reaction. 1 Stannous chloride in presence of HCl.

11) Explain the nature of carbonyl group in aldehydes and ketones. 2

The carbon-oxygen double bond is polarised due to higher electronegativity of oxygen relative to carbon. Hence, the carbonyl carbon is an electrrophilic and carbonyl oxygen is a nucleophilic centre.

12) Identify the product and name of the reaction. 2

Benzaldehyde Rosenmunds reduction

Page 2

13) How do you prepare aldehydes from alkane nitrile? Write the general reaction and name of the reaction. 2

Stephen reaction

14) What happens when toluene treated with chromyl chloride in CS2 solvent upon hydrolysis? Write chemical equation. 2

Benzaldehyde is obtained

15) Write the general reaction to prepare ketones from acyl chloride. 2

16) Name the functional group obtained when benzene reacts with acetyl chloride in presence of anhydrous AlCl3 2

Functional group is Ketone

17) Explain the mechanism of addition of HCN to aldehyde . 3

Aldehydes and ketones reacts very slowly with pure HCN. Hences,it is catalysed by base and generated cyanide adds to carbonyl compound to yield corresponding cyanohydrins.

18) How do aldehydes and ketones react with ammonia/ hydroxylamine/ hydrazine/ phenyl hydrazine

/semicarbazide? (Each one carries 2 marks.)

19) Identify the following reaction. 1

Page 3

Clemmensons reduction

20) Name the products obtained when aldehydes are oxidized. 1

Carboxylic acid

21) Write Wolff-Kishner reduction equation. 2

22) What is Tollen’s reagent. 1

Tollen’s reagent is ammonical silver nitrate solution.

23) What is Fehling’s solution. 1

Fehling’s solution is a mixture of alkaline copper sulphate solution and soium-potasium tartarate solution.

24) How do aldehydes and ketones react with sodium bisulphite? 2

25) Why are α- hydrogen of aldehyde are acidic in nature? 2

26) What is Aldol condensation reaction and explain this reaction by taking ethanol as example. 3

Page 4

27) How is propanone converted into 4-methylpent-3-en-2-one? 2

28) Explain the reaction between benzaldehyde and acetophenone in presence of dilute base and identify the name of the reaction. 3

C6H5CHO + C6H5COCH3 C6H5CH=CH-CO-C6H5 Benzaldehyde acetophenone 1,3-diphenylprop-2-en-1-one. This reaction is called as cross-aldol condensation reaction.

29) Write the reaction involved when two molecules of methanal reacted each other in presence of concentrated base. Name the reaction.

3

Cannizzaro’s reaction

30) How is toluene/ propyl benzene converted into benzoic acid? 2

Toluene

benzoic acid.

Page 5

31) How are carboxylic acids obtained from alkyl nitrile? Give example. 2

32) How do you prepare carboxylic acid obtained from Grignard reagent? 2

33) How is benzoic acid obtained from ethyl benzoate? 2

34) Identify the product in the following reaction. 1

35) Mention any two uses of acetic acid

2

36) Identify the product in the following reaction. 1

m-Nitrobenzaldehyde

37) What is formalin? Mention its uses. 2

40 % aqueous solution of formaldehyde is called as formalin.It is used for preservation of biological specimen

Page 6

39) What happenes when carboxylic acid reacts with PCL5/PCl3/SOCl2. 2

40) How do acetic acid react with ammonia. 2

41) Write the equation involving the reaction between benzoic acid and ammonia. 2

42) How do you convert benzene-1,2-dicarboxylic acid into phthalimide. 3

43) Name the product obtained when sodium acetate treated with sodalime. 1

Methane

44) Write the general reaction of Hell-Volhard-Zelinsky reaction. 2

45) Explain nitration reaction of benzoic acid. 2

46) Write the name of the following reaction. 1

38) How are carboxylic acids obtained from alcohols? 2

Page 7

Gutterman-koch reaction

1

Unit 13-NITROGEN CONTAINING ORGANIC COMPOUNDS

Two marks:

1. Name the product obtained when a nitrile is reduced by H2/Ni, . Give the equation.

Ans. Primary amine: RCN 2H /Ni,RCH2NH2.

2. How is nitrobenzene converted into aniline. Give the equation.

Ans. By reduction using Sn/ HCl

3. What are A and B?

Ans. A is RCH2NH2, B is RNH2

4. Give the equation which will be an example for Hofmann bromamide reaction.

Ans. CH3CONH2 + Br2 + 4NaOH CH3NH2 + 2NaBr + Na2CO3 + 2H2O

Acetamide methanamine

5. Gabriel phthalimide synthesis is used to prepare which class of organic compound?

Aniline cannot be prepared by this method. Give reason.

Ans. 1° aliphatic amine

Aryl halides are not reactive towards nucleophilic substitution reaction.

6. Name the reaction by which a 1° amine is prepared from an amide having one carbon

atom more than 1° amine. Give the general equation.

Ans. Hofmann bromamide degradation reaction

RCONH2 + Br2 + 4KOH 2NH2 + 2KBr + K2CO3 + 2H2O

7. Between CH3CH2CH2NH2 and (CH3)3N, which has higher boiling point and why?

Ans. CH3CH2CH2NH2 has higher boiling point. CH3CH2CH2NH2 has more H atoms on N

to form intermolecular hydrogen bonding.

8. Give reason: i) Amines have lower boiling point than alcohol of same molar mass.

ii) Amines are insoluble in water.

Ans. i) Nitrogen in amines is less electronegative than oxygen in alcohol. Hence amines do

not form H-bonds among them.

ii) Amines do not form H bonds with water.

9. Amines are both Bronsted base and Lewis base. How?

Ans. Amines can accept H+, hence Bronsted bases. Amines can donate a pair of electrons

hence Lewis bases.

2

10. Give reason: aniline is a weaker base than ammonia but methanamine is a stronger

base than ammonia.

Ans. Aniline is weaker base because the pair of electron on nitrogen gets delocalized

towards benzene ring. Methanamine is stronger base, because CH3 group is electron

releasing group and makes pair of electrons on nitrogen more available for protonation.

11. Arrange 1°, 2°, 3°methylamines in decreasing order of their base strength i) in gaseous

phase ii) in aqueous medium.

Ans. i) (CH3)3N > (CH3)2NH > CH3NH2

ii) (CH3)2NH > CH3NH2 > (CH3)3N

12. Name two factors that effect the basic strengths of 1°, 2°, 3° methyl amines in water.

Ans. i) Solvation (hydration) ii) steric hindrance

13. What is the final product obtained when 1° amine is alkylated? Give its general formula.

Ans. Quaternary ammonium salt: 4R N X

14. Give equation for the reaction between ethanamine and acetylchloride. Name the

product obtained.

Ans. C2H5NH2 + CH3COCl CH3CONHC2H5 + HCl

N-ethylacetamide

15. What is benzoylation of 1° amine? Give the equation with methanamine.

Ans. Reaction of amine with benzoyl chloride is benzoylation.

CH3NH2 + C6H5COCl C6H5CONHCH3 + HCl

16. Name the family of compounds that answers carbylamine test. Give the equation.

Ans. 1° amine

RNH2 + CHCl3 + alc. 3KOH RNC + 3KCl + 3H2O

17. How does a 1° aliphatic amine react with nitrous acid? Give the equation.

Ans. 1° aliphatic amine reacts with nitrous acid to form respective alcohol.

RNH2 + HNO2 2NaNO /HCl

2R N Cl

2H OROH + HCl + N2.

18. Name the reaction by which aniline is converted into phenyl isocyanide. Give the

equation.

Ans. Carbylamine reaction

C6H5NH2 + CHCl3 + Alc.3KOH C6H5NC + 3KCl + 3H2O

19. Complete the following equations:

i) CH3NH2 + CH3COCl HCl + ____________

ii) RNH2 1 mole R-X

HX __________

Ans. i) CH3CONHCH3 ii) RNHR

3

20. Identify the main organic product in the following reactions:

i) C6H5NH2 + HNO2 2NaNO /HCl

0 C ________

ii) C6H5CONH2 2Br /NaOH __________

Ans. i) C6H5N2Cl ii) C6H5NH2

21. What is benzene sulphonyl chloride also known as? An amine with benzene sulphonyl

chloride forms a compound insoluble in an alkali. Identify the class of the amine.

Ans. Hinsberg’s reagent. 2° amine.

22. How does Hinsberg’s reagent help to distinguish 1° amine and a 2° amine? Explain.

Ans. The given amine is treated with Hinsberg’s reagent. If the product formed is soluble in

an alkali, the amine is 1°. If the product formed is insoluble in an alkali, the amine is 2°.

23. Complete the following equations:

24. What is the significance of acetylation of aniline before nitrating it?

Ans. When aniline is treated with concentrated HNO3, much of the aniline gets oxidized,

aniline gets protonated and the major product is meta-nitroaniline. Hence to avoid all

this aniline is acetylated. Acetylated aniline, avoids oxidation of aniline and controlled

nitration yields p-nitro aniline as the major product.

25. Give reasons: i) aniline does not undergo Friedel-Crafts reaction. ii) aniline with

concentrated HNO3 forms meta nitro compound in significant amounts.

Ans. i) Aniline reacts with AlCl3 to form a salt, which makes nitrogen of aniline to get a

positive charge, which becomes a strongly deactivating group.

ii) Aniline with conc. HNO3 forms anilinium ion which is meta directing.

26. What is diazotization? Give the general formula of a diazonium salt.

Ans. Conversion of 1° aromatic amine into diazonium salt is diazotization.

General formula: 2Ar N X

or 2R N X

, where R = Ar

Ans.

4

27. How is benzene diazonium chloride prepared from aniline? Give the equation.

Ans. It is prepared by the reaction of aniline with nitrous acid (NaNO2/ HCl) at 0°C

C6H5NH2 + NaNO2 + 2HCl 0 C 6 5 2C H N Cl

+ NaCl + 2H2O

28. What is Sandmeyer’s reaction? Give an example.

Ans. Replacement of diazonium group by Cl/ Br

in presence of Cu(I) ion.

E.g.: 2Ar N X

2 2Cu Cl /HCl

ArCl + N2

29. Name the organic products obtained in the following reactions:

i) 2Ar N X

CuCN/KCN __________ + N2

ii) 2Ar N Cl

+ H3PO2 + H2O _____ + N2 + CH3CHO + HCl

Ans. i) Aryl cyanide ii) Benzene

30. How is a diazonium salt converted into iodobenzene? Give the equation.

Ans. By treating diazonium salt with potassium iodide.

2Ar N X

+ KI ArI + KCl + N2

iodobenzene

31. Give an example for a coupling reaction with an equation.

Ans. Benzene diazonium chloride reacts with phenol to form p-hydroxyazobenze. This is an

example for coupling reaction.

32. How is benzene diazonium chloride converted into an azo dye? Give an example for an

azo dye.

Ans. Azo dyes are the products obtained when reaction of benzene diazonium chloride with

phenol or aniline takes place with retention of diazo group. E.g.: benzene diazonium

chloride couples with aniline to form an azo dye p-amino azo benzene (yellow dye)

33. Mention the importance of diazonium salt in synthetic organic chemistry.

Ans. i) Aryl fluoride and iodides that cannot be prepared by direct halogenation can be

synthesized.

ii) It helps to introduce many functional groups into aromatic ring, which cannot be

done by direct methods.

THREE MARKS:

34. Identify the X, Y, Z in the following:

Ans. X is aniline, Y is benzene diazonium chloride, Z is iodobenzene

5

35. Give equations for the preparation of methylamine (methanamine) by Gabriel-

phthalimide synthesis.

36. RCN 2H /Ni,

X 3CHCl /Alc.KOHY. Y is a three carbon compound. What is R in RCN, X

and Y?

Ans. R = CH3, X = CH3CH2NH2, Y = CH3CH2NC

37. Give equation for the conversion of aniline into 4-bromoaniline.

38. An organic compound with formula C2H7N does not answer carbylamine test, but give a

product that is insoluble in an alkali, with Hinsberg reagent. Give the IUPAC name of X

and to what class of organic compound does it belong to?

Ans. X is CH3NHCH3. IUPAC name : N-methylmethanamine. It is a 2° amine.

39. X 2NaNO /HCl

0 CY warmZ. Y + Z orange dye (p-hydroxyazobenzene). What are X,

Y and Z?

Ans.

Ans.

Ans.

Unit 14 BIOMOLECULES

1 What are carbohydrates? Give examples 2

Carbohydrates are polyhydroxy aldehydes or ketones or the substances which gives these upon hydrolysis. Example: glucose fructose maltose lactose sucrose starch cellulose glycogen etc.

2 How are carbohydrates classified? 3

Carbohydrates

Reducing sugars

Sugars

Non sugars Non reducing

sugars

Aldoses Mono saccharaides

Oligo saccharides

Poly sacchrides

Ketoses

Trio

ses

Tetr

ose

s

Pe

nto

ses

Hex

ose

s

di

tri

tetr

a

pen

ta

hex

a

hep

ta

oct

a

nan

o

dec

a

3 What are sugars and non-sugars? 2

Sugars are the carbohydrates; soluble in water crystalline in nature and sweet in taste Example glucose fructose maltose lactose etc. non-sugars are carbohydrates; insoluble in water, amorphous in nature and tasteless. Example :starch cellulose glycogen etc.

4 What are reducing sugars? Give example 2

The sugars which can reduce Tollen’s reagent, Benedict’s reagent and Fehling’s reagent are reducing sugars. These contain a free hydroxyl group on anomeric carbon. Example glucose fructose maltose lactose

5 What are non-reducing sugars? Give example (Is sucrose a reducing sugar or not? Give reason.) 2

The sugars which cannot reduce Tollen’s reagent, Benedict’s reagent and Fehling’s reagent are non-reducing sugars. These do not contain a free aldehydic group(aldehydic groups are bonded). Example : sucrose

6 What are monosaccharaides? Give examples 2

Monosaccharaides are the simple sugars which do not undergo hydrolysis. Example : glucose fructose, Galactose

7 What are oligosaccharides? Give examples 2

Oligosaccharides are the sugars which undergo hydrolysis to give 2 to 10 monosaccharaide units. Example: maltose lactose sucrose etc.

8 What are disaccharides? Give examples 2

Disaccharides are the sugars which undergo hydrolysis to give 2 monosaccharaide units. Example: maltose lactose sucrose etc.

9 What are polysaccharides? Give examples

Polysaccharides are the carbohydrates which undergo hydrolysis to give more than 10 (many) monosaccharaide units. Example: starch, cellulose, glycogen etc.

10 Give an example of aldohexose 1

Glucose or Galactose

11 Give example of ketohexose 1

Fructose

12 How is glucose prepared? 2

13 Elucidate the structure of glucose 5

(i) Molecular formula − C6H12O6

(ii) Suggestion of straight chain

(iii) Confirmation of carbonyl (> C = O) group

(iv) Confirmation of the presence of carbonyl

group as aldehydic group

(v) Confirmation of the presence of five −OH

groups

(vi) Indication of the presence of a primary alcohol

The correct configuration of glucose is given by

Kiliyanissyntesis

14 Gluconic acid on oxidation with HNO3 gives saccharic acid. What does it indicate about the structure 1

of glucose?

Confirmation of the presence of primary alcoholic group

15 Mention the structural features of open chain structure of glucose 2

It has 1 aldehyde group, 1 primary alcohol group and 4 secondary alcoholic groups

16 Mention the structural features of open chain structure of fructose 2

It has 1 ketone group, 2 primary alcohol group and 3 secondary alcoholic groups

17 Mention demerits of open chain structure of glucose 3

The following reactions of glucose cannot be explained by its open-chain structure.

1. Aldehydes give 2, 4-DNP test, Schiff’s test, and react with NaHSO4 to form the hydrogen

sulphite addition product. However, glucose does not undergo these reactions.

2. The penta-acetate of glucose does not react with hydroxylamine. This indicates that a free

−CHO group is absent from glucose.

3. Glucose exists in two crystalline forms, α and β.

18 How do you explain the absence of aldehyde group of the pentaacetate of D – glucose? 2

The aldehyde group is involved in formation of cyclic hemiacetal with secondary alcoholic group of 5th carbon. In pentaacetate of D – glucose, all 5 -OH groups are acetylated, therefore, it does not form an open chain structure, and does not react with NH2OH. This fact indicates absence of aldehyde group in glucose. But, D-glucose reacts with hydroxylamine (NH2OH) to form an oxime because of the presence of aldehydic (−CHO) group or carbonyl carbon. This happens as the cyclic structure of glucose forms an open chain structure in an aqueous medium, which then reacts with NH2OH to give an oxime.

19 What is glycosidic bond / linkage? 1

Glycosidic linkage − Linkage between two monosaccharide units through oxygen atom

20 Name the sugar present in cane sugar 1

Sucrose

21 What are the expected products of hydrolysis of sucrose 1

α –glucose and β - fructose

22 What are the expected products of hydrolysis of lactose 1

Β – Galactose and β- glucose

23 Name the sugar present in milk sugar 1

Lactose

24 Name the components of starch 1

Amylose and amylopectin

25 Name water soluble component of starch 1

Amylose

26 Name water insoluble component of starch 1

Amylopectin

27 Name the storage polysaccharide in plants 1

Starch

28 Name the storage polysaccharide in animals 1

Glycogen( animal starch)

29 Name the structural polysaccharide in plants 1

cellulose

30 Write Haworth structure for α glucose / monomer in cellulose. (β glucose) / α fructose /β fructose 2

31 Write Haworth structure of sucrose/ maltose / lactose 2

Structure of sucrose:

Structure of Maltose:

Structure of Lactose

32 Why cellulose cannot be used as food by human beings? 1

Human saliva do not contain the enzyme that can hydrolyses β 1-4 linkages present in cellulose

33 What is glycogen? How does it differ from starch 3

Glycogen is a polymer of α – glucose linked by α 1-4 glycosidic bond and α 1-6 glycosidic bond at

the point of branching

starch glycogen

Storage polysaccharide in plants Storage polysaccharides in animals

Made of two compenents 1) amylose 2) Made of one component

amylopectin

Amylopectin has branched structure. The

frequency of branching is at every 30 glucose

units

Glycogen has branched structure. The

frequency of branching is at every 10 glucose

units

34 Mention two differences between starch and cellulose 2

starch cellulose

Storage polysaccharide in plants Structural polysaccharides in plants

Made of two compenents 1) amylose

2) amylopectin

Made of one component

Amylose is linear chain of α – glucose linked by

α 1-4 glycosidic bond

Amylopectin has branched structure. The

frequency of branching is at every 30 glucose

units

cellulose is linear chain of β – glucose linked by

β 1-4 glycosidic bond

35 Name the products obtained when proteins are hydrolysed? What do you understand by this

reaction?

2

Proteins upon hydrolysis form amino acids. This indicates that proteins are made of amino acids

36 What are amino acids? How many naturally occurring amino acids are present in proteins 2

These are the organic compounds containing both amino and carboxyl group on α carbon atom.

These are the building blocks(monomers) of proteins. There are 20 naturally occurring amino

acids

37 Write the general structure of amino acids 1

38 Write the structure of an optically inactive aminoacid 1

39 Name an amino acid containing sulphur 1

Cysteine ,methionine

40 Name an amino acid which is acidic 1

Aspateric acid, Glutamic acid

41 Name an amino acid which is basic 1

Glutamine,Lysine

42 Name an amino acid which contains heterocyclic nucleus 1

Proline,histidine

43 How amino acids are classified based on dietary requirement? 2

Based of dietary requirement they are classified into essential and Non-essential amino acids:

Essential amino acids: Amino acids that cannot be synthesised in the body, and must be obtained

through diet Example − Valine, leucine, isoleucine

Non-essential amino acids: Amino acids that can be synthesised in the body

Example − Glycine, alanine, glutamic acid Non-essential amino acids:

44 What is zwitter ion? Write its general structure 2

These are the amino acid dipolar ions, carrying both positive and negative charges. These moves

neither towards cathode nor towards anode in electric field

45 What is isoelectric point 1

The pH at which amino acids acts as zwitter ions in aqueous solution is called isoelectric pH / point

46 What is peptide bond? How is it formed? 2

It is the amide bond present between two amino acids units in peptides and protein. It is formedby

eliminating on molecule of water from α−COOH group and α −NH2 group of two amino acid

47 What is poly peptide? 1

Poly peptides are the polymers of (n)amino acids containing 10 to 50 amino acids in chain linked by

(n-1) peptide bonds

48 How many peptide bonds are present in a pentapeptide?

4

49 What are proteins? 1

proteins’ are the polymers of (n)amino acids containing more than 50 amino acids in chain linked by

(n-1) peptide bonds

50 Name a hormone which controls the carbohydrate metabolism. 1

insulin

51 How are proteins classified based on their molecular shape and solubility? 3

Based on the molecular shape, proteins are classified into two types

Fibrous proteins, polypeptide chains run parallel and are held together by hydrogen and disulphide

bonds. These are insoluble in water. These are also called structural proteins

Example: keratin (hair and nail), actin and myosin ( muscles) and collagen( cartilage)

Globular proteins In Polypeptide chains coil around, giving a spherical shape. These are soluble in

water. These are also called functional proteins. Example: albumin, globulin etc

52 Write a note on structure of proteins 4

Structures and shapes of proteins are studied at four different levels: primary, secondary, tertiary

and quaternary.

Primary structure of proteins: Contains one or more polypeptide chains, and each chain has amino

acids linked with each other in a specific sequence. This sequence of amino acids represents the

primary structure of proteins.

Secondary structure of proteins: Shape in which a long polypeptide chain can exist; two types of

secondary structures: α-helix, β-pleated sheet, stabilised by hydrogen bonds

α- Tertiary structure of proteins: Overall folding of the polypeptide chains; results in fibrous and

globular proteins; secondary and tertiary structures of proteins are stabilised by hydrogen bonds,

disulphide linkages, van der Waals forces and electrostatic forces.

Quaternary structure of proteins: Spatial arrangement of subunits, each containing two or more

polypeptide chains

53 What is denaturation of proteins? 2

Denaturation means loss of biological activity of proteins due to the unfolding of globules and

uncoiling of helix. Denaturation takes place due to action of heat, addition of electrolytes etc

Example − Coagulation of egg white on boiling, curdling of milk

54 What are enzymes? Give example 2

Enzymes are biocatalysts. Specific for a particular reaction and for a particular substrate

For example, maltase catalyses hydrolysis of maltose

55 What are vitamins? How are these classified? 3

Vitamins are micronutrients that take part in metabolic process, to produce energy and growth.

These are classified as water soluble vitamins (vitamin B complex and vitamin C). fat soluble

vitamins ( vitamin A,D,K,E,)

56 Mention the source and deficiency syndrome of vitaminA 1

Sources Deficiency diseases

Fish liver oil, carrots, butter and milk Xerophthalmia, night blindness

57 Mention the source and deficiency syndrome of Vitamin B12 1

Sources Deficiency diseases

Meat, fish, egg and curd Pernicious anaemia

58 Mention the source and deficiency syndrome of Vitamin C 1

Sources Deficiency diseases

Citrus fruits, amla and green leafy vegetables Scurvy

59 Mention the source and deficiency syndrome of Vitamin D 1

Sources Deficiency diseases

Exposure to sunlight, fish and egg yolk Rickets and osteomalacia

60 Name the products when nucleic acids are hydrolysed step wise 2

Nucleic acids → nucleotides

Nucleotides → nucleosides + phosphoric acid

Nucleosides → pentose sugar + heterocyclic bases (purine and pyrimidine)

61 How nucleoside and nucleotide are formed? 2

1) Nucleoside is formed when N-base gets attached to 1 position of pentose sugar.

N-base + Pentose sugar nucleoside

2) Nucleotide is formed when nucleoside is linked to phosphoric acid at 5th

position of

pentose sugar.

Nucleoside + H3PO4 nucleotide

62 What are nucleic acids? 1

Nucleic acids are the polymers of nucleotides linked by 3-5 phosphodiester bond

63 What are the differences between DNA and RNA 3

DNA RNA

Contains de- oxy ribose sugar Contains ribose sugar

Bases are A,G,C,T Bases are A.G.C.U

Has double helical structure Has single stranded structure

Present in nucleus of the cell Present in cytoplasm

Hereditary material Involved in protein synthesis

Messenger RNA (m-RNA)

Ribosomal RNA (r-RNA)

Transfer RNA (t-RNA)

64 Name a. The sugar moiety present in DNA b. Nitrogenous base present only in DNA, but not in RNA.

a) de- oxy ribose sugar

b)Thymine

65 Write the structure of ribose sugar / deoxy-ribose sugar 2

66 Name any 3 Biological functions of nucleic acids 3

1. DNA is chief chemical as reserve genetic information.

2. DNA is chiely responsible for identity of a species.

3. DNA is capable of self replication during cell division.

4. Important function of RNA is in protein synthesis in the cells. Message for the protein

synthesis is in DNA but various RNAs take part in protein synthesis.

67 What are hormones? Give an example for each type of hormone

a) Polypeptide hormones

b) Amino acid derivatives

c) Steroid hormones

Hormones are biochemical messengers produced by endocrine glands.

a) Polypeptide hormones ----- insulin/ glucagons

b) Amino acid derivatives----- Thyroxine/Epinephrine

c) Steroid hormones--- Testosterone/Estradiol/progesterone

68 Write the function of the following hormones :

a) Insulin

b) Thyroxine

c) Estrogen and androgen

a) Insulin: Maintains blood sugar level

b) Thyroxine: Growth and development

c) Estrogen and androgen: Development of secondary sex characters

UNIT 15

POLYMERS

Polymers in Greek means, poly means many and mer means unit or part. Polymers means many

units or parts.

1. What are polymers? 1M

A large number of simple repeating units linked together through covalent bond are called

polymers. They are also called as macromolecules

2. What is a monomer? 1M

The simple molecule which combine to form polymer are called monomers.

3. What is polymerisation ? 1M

The process by which monomers are converted into polymer is called polymerisation.

Classification of polymers:

Classification based on Source:

4. What are natural polymers? Give example. 2 M

The polymers which are found in nature i.e in plants and animals are called natural

polymers.

Ex: proteins, Nucleic acid , starch, cellulose, rubber

5. What are semi synthetic polymers? Give examples. 2M

Chemically modified natural polymers are called semi synthetic polymers.

Ex: Cellulose acetate (rayon), cellulose nitrate, valcanised rubber.

6. What are synthetic polymers? Give examples. 2M

Synthetic polymers are man –made polymers synthesized in the Laboratories or industries

used in daily life.

Ex: Polythene, poly vinyl chloride, nylon, terylene, Teflon bakelite

Classification based on structure of polymer:

7. What is Linear polymer? Give example.

In Linear polymer, the monomer units are linked together to form Long straight chains of polymer molecule

Ex: polythene, p v c, nylon, polyester, poly styrene

8. What is branched chain polymer? Give example

In branched chain polymer, the monomer unit combines to produce the Linear chains

having some branches.

Ex: Low density poly then, starch, glycogen etc.

9. What are cross linked or network polymer ? Give examples.

Cross- linked polymers are formed from monomer units containing two or more functional

Group. They contain strong covalent bond between various linear polymer chains.

Ex: Bakelite, melamine, urea –formaldehyde etc.

Classification based on mode of polymerization

10. What is addition polymerization? Give examples

A polymer formed by the addition of repeating monomer units possessing double or triple

bond without elimination of by product molecule during polymerization is called addition polymer.

Ex: polythene , poly propene

Low density

polyethene (LDPE) n(CH2=CH2) (CH2CH2)n

Electrical

insulator, toys,

squeeze bottles

HDPE (high

density polyethene)

n(CH2=CH2)

(CH2CH2)n

Buckets,

dustbin, pipes

Teflon (polytetra

fluroethene) nCF2 = CF2 (CF2CF2)n

Non-stick

cookware, oil

seals, gaskets

Polyacrylonitrile

(orlon)

Substitute for

wool

(Any one example)

11. What are homo polymer? Give example

Addition polymers formed by the polymerization of one type of monomers are called homo

polymer

2

2000 atm, 200°C

peroxide or O

4 2 5 3

Ziegler-Natta catalyst

TiCl -Al(C H ) ,6 atm, 60°C

free radical

initiator

Ex: Polythene (monomer unit in ethene)

12. What are co polymers? Give on example

Addition polymers formed by the polymerization of two different monomer units are called co-polymer.

Ex: Buna-S, Buna-N, Nylon 6,6 etc.

13. What is Co- polymerization

It is polymerization reaction in which a mixture of more than one monomeric species is allowed to polymerize and form a co polymer.

14. What is condensation polymerization? Give examples

A polymer formed by the condensation of two different bifunctional or trifunctional

monomers with the elimination of simple molecules like water, methanol ammonia is called

condensation polymerization.

Ex: Nylon 6,6

Polymer polymerization uses

Terylene/

Dacron (a

polyester

Terywoo

l, terycot

fabrics,

safety

glass

Nylon 6, 6 (a

polyamide)

Textiles,

bristles

for

brushes

Nylon-6

Tyre

cords,

fabrics,

ropes

Bakelite

(phenol

formaldehyd

e)

Or

Combs,

handles

of

utensils,

electrical

switches

Melamine

formaldehyd

e

Crockery

Classification based on molecular forces:

15. What are elastomers? Give examples

Elastomers are rubber like solid with elastic properties. In these the polymer chains are held by

weakest intermolecular forces. The weak binding forces permit the polymer to be stretched.

Ex: vulcanized rubber, Buna-S, Buna-N, neoprene etc.

16. What are fibers? Give examples

Fibres are thread- like polymer possessing high tensile strength and high modulus. These

characterization are due to strong intermolecular forces like hydrogen bonding which result in close

packing of chain impart crystalline structure to the polymer.

Ex: Nylon 6, 6, terylene, Nylon 6, silk etc

17. What is thermoplastic polymer? Give example

Thermoplastic are linear or slightly branched polymers which can be repeatedly

softened on heating and hardened on cooling.

Ex: polythene, polypropene , pvc, polystyrene, Teflon etc.

18. What are thermosetting polymers? Give Examples

Thermosetting polymers are cross linked and heavily branched molecules. On heating they

undergo extensive cross linking and become hard and infusible. These cannot be reused.

Ex: Bakelite, urea formaldehyde resin, etc.

19. Name the monomer and write the partial structure of polythene?

Monomer of polythene - Ethene or Ethylene

Partial structure - (CH2CH2)n

20. Name the monomer and write the partial structure of Nylon-6 ?

Monomer of Nylon-6 - caprolactum

Partial structure - [-CO-(CH2)5-NH-]n

21. Name the monomers and write the partial structure of Nylon- 6,6

Monomer of Nylon-6,6 - Hexamethylene diamine and Adipic-acid

Partial structure - [-OC-(CH2)4-CONH-(CH2)6-NH-]n

22. Name the monomers and write the partial structure of terylene (Dacron)

Monomer of terylene - Ethylene glycol and terephthatic-acid

Partial structure -

23. Name the monomer and write the partial structure of Bakalite ?

Monomer of Bakalite - Phenol and formaldehyde

Partial structure -

RUBBER:

24. Name the monomer present in natural rubber.Write the partial structure

Natural rubber is a polymer of cis-2-methyl-1,3-butadiene (isoprene). ( cis-poly-isoprene). Its

partial structure is

25. Define Synthetic rubber? Give one example

Synthetic rubber is defined as any valcanisable rubber like polymer capable of getting

stretched to twice its length and returns to its original length, size and shape when the stretching

force is withdrawn

Ex: Neoprene, Buna-S, Buna-N

26. What is Valcanisation ?

The process of heating natural rubber with sulphur or sulphur containing compounds at

about 415k for a few hours in order to give strength and elasticity to natural rubber is called

valcanisation.

27. Explain the preparation of Neoprene? Write the equation.

When chloroprene (2-chloro-1,3-butadiene) is heated in the presence of peroxide catalyst,

polychloroprene or neoprene is formed

28. Explain the preparation of Buna-N?

When 1,3-butadiene and acrylonitrile are heated in the presence of peroxide catalyst, Buna-N

is formed

29. What is bio-degradable polymer? Give example

Bio-degradable polymer are those which contain functional groups similar to the functional

groups present in bio-polymers

Ex: 1. Polyhydroxybutyrate-co-hydroxyvalerate (PHBV )

2. Nylon-2-Nylon-6

H2N CH2 COOH + NH2 (CH2)5 COOH (HNCH2 CO NH (CH2)5 CO)n

glycine aminocaproic acid Polyamide

30. What is non bio-degradable polymer? Give example

A large number of synthetic polymers are resistant to the environmental degradation

processes and responsible for the accumulation of polymers solid waste materials and cause

environmental problems are called Non-biodegradable polymers.

Ex: polythene, Nylon, terylene etc

Chapter-16

Chemistry in Everyday Life

1. Sleeping pills are recommended by doctors to the patients suffering from sleepness

but it is not advisable to take their doses with out consultation with the doctor.

Why?

Ans. Sleeping pills contain drugs which may be tranquilizers or anti-depressants . They affect

the nervous system and induce sleep. However, if these doses are not properly controlled, they

may create havoc. They even adversely affect the vital organs of the body. It is advisable to

take these sleeping pills under the supervision of a doctor.

2. “Ranitidine is an antacid” With reference to which classification, has this statement

been given?

Ans. Ranitidine is labelled as antacid since it is quite effective in neutralizing the excess of

acidity in the stomach. It is sold in the market under trade name Zintac.

3. Why do we require artificial sweetening agents?

Ans. The commonly used sweetening agent i.e., sucrose is a carbohydrate with molecular

formula C12H22O11. Since it has high calorific value, it is not recommended to the patients,

diabetics in particular which require low calorie diet. Most of the artificial sweeteners are better

than sucrose but hardly provide any calories to the body. These are being used as substitutes of

sugar.

4. Write Chemical equation for preparing sodium soap from glyceryl oleate and

glyceryl palmitate. Structural formulas of these compounds are given:

(i) (C15H31COO)3 C3H5(Glyceryl palmitate) (ii) (C17H33COO)3C3H5 (Glyceryl oleate)

Ans. CH2 OCOC15H31 CH2OH

CHOCOC15H31 + 3NaOH CHOH + 3C15H31COONa

CH2OCOC15H31 CH2OH Sod. Palmitate (soap))

(C15H31COO) 3C3H5 Glycerol

Glyceryl palmitate

CH2OCOC17H33 CH2OH

CHOCOC17H33 + 3 NaOH CHOH + 3C173 COONa

CH2OCOC17H33 CH2OH Sod. oleate (Soap) (C17H33COO)3C3H5 Glycerol

Glyceryl oleate

5. Label the hydrophilic and hydrophobic parts in the following molecule which is a

detergent. Also identify the functional groups present.

C9H19 O(CH2CH2O)xCH2CH2 OH

(x = 5to 10)

Ans: C9H19 O(CH2CH2O)xCH2CH2OH

Hydrophobic part Hydrophilic part

Detergents are esters formed by the combination between carboxylic acid and polyethylene

glycol.

6. Why do we need to classify the drugs in different ways?

Ans Drugs are to attack different targets which are the biomolecules from which our body is

made.Moreover, the drugs also differ in action. Therefore, there is a genuine necessity to classify

the drugs in different ways.

7. Explain the following as used in medicinal chemistry (a) Lead compounds (b)

Target molecules or drug targets.

Ans. (a) Lead compounds are the compounds which are effective in different drugs. They have

specific chemical formulas and may be extracted either from natural sources (plants and animals)

or may be synthesized in the laboratory.

(b) Target molecules or drug targets.

Ans. The different macromolecules or biomolecules , which are drug targets are carbonates,

proteins, enzymes, nucleic acids. Out of these , enzymes are the most significant because their

deficiency leads to many disorders in the body.

8 Why the medicines should not be taken without consulting doctors?

Ans. No doubt medicines are panacea for most of the body ailments. But their wrong choice and

overdose can cause havoc and may even prove to be fatal. Therefore, it is of utmost importance

that the medicines should not be given without consulting doctors.

9. Define the term chemotherapy.

Ans: Chemotherapy means the treatment of the disease by means of chemicals that have specific

effect upon the disease causing micro-organisms without harming the friendly micro-organisms or

bacterias which the body needs.

10. Which forces are involved in holding the drugs to the active sites of enzymes?

Ans. These are different inter-molecular forces like dipolar forces, Hydrogen bonding , van der

Waals’ forces etc..

11. Antacids and antiallegic drugs interfere with the function of histamines but donot

interfere with the function of each other . Explain.

Ans They donot interfere with the functioning of each other because they work on different

receptors in the body .

12. Low level of noradrenaline is the cause of depression. What type of drugs are needed to cure this problem? Na,me two drugs.

Ans: Low level of noradrenaline which acts as a neurotransmitter reduces the signal sending ability to the nerves and the patient suffers from depression. Antidepressants are needed to give relief from depression. These are also called tranquilizers or neurologically active drugs. The two specific drugs are iproniazid and phenelzine.

13. What is meant by the term ‘broad spectrum antibiotic’? Explain

Ans. Broad spectrum antibiotics are drugs which are effective against a large number of harmful micro-organisms causing diseases.

14. Why are cimetidine and ranitidine better antacids than sodium bicarbonate or magnesium or aluminium hydroxides ?

Ans. Both sodium bicarbonate and hydroxides of magnesium or aluminium are very good antacids since they neutralise the acidity in the stomach. But their prolong use can cause the secretion of excessive acid in the stomach . This may be quite harmful and may lead to the forrmation of ulcers Both cimetidine and ranitidine are better salts without any side effect.

15. Name a substance which can be used as an antiseptic as well as disinfectant.

Ans. About 0.2 percent solution of phenol can act as antiseptic whereas about 1.0 percent solution of the same can act as disinfectant.

16. What are the main constituents of dettol?

Ans. The main constituents of antiseptic dettol are chloroxylenol and terpenol.

17. What is tincture of iodine? What is its use?

Ans: Tincture of iodine is a dilute solution of iodine ( 2 to 3 percent ) prepared in ethanol. It is a powerful antiseptic particularly in case of fresh wounds.

18. Why is use of aspartame restricted to cold foods and drinks?

Ans: Aspartame is a very good sweetener for foods and drinks. But its use is restricted to cold stuff only. In case these are hot, the sweetener may decompose and it may not be effective any more.

19. Name the sweetening agent used in the preparation of sweets for a diabetic patient.

Ans: Saccharine is the well known sweetening agent which is more than 550 times sweet as compared to sucrose ( or sugar). It is commonly used in the preparation of sweets for diabetic patients. Actually, it is not a carbohydrate. Now better sweetening agents are also available.

20. What problem arises by using alitame as artificial sweetener?

Ans: Alitame is no doubt, a very potent sweetener. Its sweetening capacity is more than 2000 times as compared to ordinary cane sugar or sucrose. But sometimes, it becomes quite difficult to control the sweetness level in the food which is actually desired. 21. Why are detergents called soapless soaps?

Ans: Detergents are called soapless soaps becauses they resemble soaps in their cleansing action but they donot contain the usual chemical contents of soaps i.e., sodium or potassium salts of long chain fatty acids. In other words, we can say that they behave as soaps without being actually soaps.

22. What are biodegradable and non-biodegradable detergents? Give an example of each.

Ans: Detergents are non-biodegradable in the sense that they cannot be degraded or decomposed by the micro-organisms. They mix with water present in rivers, ponds, lakes etc. as such without getting decomposed and thus cause pollution problems. The biodegradable detergents are the ones which can be degraded. These are being synthesised by reducing the branching of the chain. Sodium n-dodecylbenzene sulphonate is a biodegradable detergent. Even soaps act as biodegradable detergents.

23. Why do soaps not work in hard water?

Ans: Soaps are water soluble sodium or potassium salts of higher fatty acids like palmitic acid

( C 15 H31 COOH), oleic acide ( C17 H33COOH ) and stearic acid ( C17 H35 COOH). Hard water contains certain calcium and magnesium salts which combine with soaps to form corresponding magnesium compounds These being insoluble, get separated as curdy white precipitates resulting in wastage of soap.

24. Can you use soaps and synthetic detergents to check the hardness of water?

Ans: Soaps can be used to check hardness of water as they will form insoluble precipitates of calcium and magnesium salts on reacting with hard water. Since detergents donot form any precipitate they cannot check hardness of water.

25. If water contains dissolved calcium bicarbonate, out of soaps and synthetic detergents, which one will you use for cleansing clothes?

Ans: Calcium bicarbonate makes water hard. Soap (RCOONa) will react with the salt to form corresponding calcium salt which will be precipitated and wasted. The synthetic detergents are chemically different from soaps. They will not react with the calcium bicarbonate and can be used for cleaning dirty clothes without being precipitated. In other words, there will be no wastage when the detergents are used.

3RCOONa + Ca(HCO3 )2 (RCOO)2Ca + 2NaHCO3 (Soap) 26. Label the hydrophilic and hydrophobic parts in the following compounds.

a) CH3(CH2)10 CH2OSO3 Na # (b) CH3(CH2) 15 N + (CH 3)3 Br

C) CH3 (CH2)16 – COO(CH2CH2O)n CH2CH20H. Ans: (a) CH3(CH2)10CH2OSO3

-Na + B) CH3(CH2)15 – N + ( CH3 )3 Br –

(Hydrophobic ) ( Hydrophilic ) (Hydrophobic ) (Hydrophilic) O (c) CH3 (CH2) 16 – C – O(CH2 CH2O)n CH2CH2OH (Hydrophobic ) (Hydrophilic)

13. Low level of noradrenaline is the cause of depression. What type of drugs are needed to

cure this problem? Na,me two drugs.

Low level of noradrenaline which acts as a neurotransmitter reduces the singal sending ability to

the nerves and the patient suffers from depression. Antidepressants are needed to give relief from

depression. These are also called tranquilizers or neurologically active drugs. The two specific

drugs are ipronaizine and phenelzine.

14. What is meant by the term ‘broad spectrum antibiotic’? Explain

Ans. Broad spectrum antibiotics are druga which are effective against a large number of harmful

micro-organisms causing disease.

15.

Why are cimetidine and ranitidine better antacids then sodium bicarbonate or magnesium or

aluminium hydroxides?

Ans. Both sodium bicarbonate and hydroxides of aluminium are very good antacids since they

neutralize the acidity in the stomach. But their prolong use can cause the secretion of excessive

acid in the stomach . This may be quite harmful and may lead to the foirmation of ulcers Both

cimetidine and rantidine are better salts without any side effect.

16. Name a substance which can be used as an antiseptic as well as disinfectant.

Ans About 0.2 percent solution of phenol can act as antiseptic whereas about 1.0 percent solution

of the same can act as disinfectant.

17. What are the main constituents of dettol?

Ans. The main constituencnts of antiseptic dettol are chloroxylenol and terpenol.

18. What is tincture of iodine? What is its use?

Ans: Tincture of iodine is a dilute solution of iodine ( 2 to 3 percent ) prepared in ethanol. It is a

powerful antiseptic particularly in case of fresh wounds.

19. Why is use of aspartame restricted to cold foods and drinks?

Ans: Asparatame is a very good sweetener for foods and drinks. But its use is restricted to cold

stuff only. In case these are hot, the sweetener may decompose and it may not be effective any

more. For more details, consult section 16.16.

20. Name the sweetening agent used in the preparation of sweets for a diabetic patient.

Ans: Saccharine is the well known sweetening agent which is more than 550 times sweet as

compared to sucro9se ( or sugar). It is commonly used in the preparation of sweets for diabetic

patients. Actually, it is not a carbohydrate. Now better sweetening agents are also available.

21. What problem arises by using alitame as artificial sweetener?

Ans: Alitame is no doubt, a ver4y potent sweetener, Its sweetening capacity is more than 2000

times as compared to ordinary cane sugar or sucrose. But sometimes, it becomes quite difficult to

control the sweetness level in the food which is actually desired.

22. Why are detergents called soapless soaps?

Ans: Detergents are called soapless soaps becauses they resemble soaps in their cleanising action

but they donot contain the usual chemical contents of soaps i.e., sodium or potassium salts of long

chain fatty acids. In other words, we can say that they behave as soaps without being actually

soaps.

23. What are biodegradable and non-biodegradable detergents? Give an example of each

Ans: Detergents are non-biodegradable in the sense that they cannot be degraded or decomposed

by the micro-organisms. They mix with water present in rivers, ponds, lakes etc. as such without

getting decomposed and thus cause pollution problems. The biodegradable detergents are the

ones which can be degraded. These are being synthesized by reducing the branching of the chain.

Sodium n-didecylbenzene sulphonate is a biodegradable detergent. Even soaps act as

biodegradable detergents.

24. Why do soaps not work in hard water?

Ans: Soaps are water soluble sodium or potassium salts of higher fatty acids like palmitic acid ( C

H COOH) oleic acide ( C17 H33COOH ) abnd stearic acid ( C17 H35 COOH) Hard water

contains certain calcium and magnesium salts which combine with soaps to form corresponding

magnesium compounds These being insoluble, get separated as curdy white precipitates resulting

in wastage of soap.

25. Can you use soaps and synthetic detergents to check the hardness of water?

Ans: Soaps can be used to check hardness of water as they will form insoluble precipitates of

calcium and magnesium salts on reacting with hard water. Since detergents donot form any

precipitate they cannot check hardness of water.

26. If water contains dissolved calcium bicarbonate, out of soaps and synthetic detergents,

which one will you use for cleansing clothes?

Ans: Calcium bicarbonate makes water hard. Soap (RCOONa) will react with the salt to form

corresponding calcium salt which will be precipitated and wasted. The synthetic detergents are

chemically different from soaps. They will not react with the calcium bicarbonate and can be used

for cleaning dirty clothes without being precipitated. In other words, there will be no wastage

when the detergents are used.

3RCOONa + Ca(HCO3 )2 (RCOO)2Ca + 2NaHCO3

27. Label the hydrophilic and hydrophobic parts in the following compounds.

a) CH2(CH2)10 CH2OSO3 Na # (B) CH3(CH2) 15 N + (CH 3)3 Br –

C) CH3 (CH2)16 – COO(CH2CH2O)NCH2CH20H.

Ans: (a) CJH3(CH2)10CH20SO3-Na + B) CH3(CH2)15 – N + ( CH3 )3 Br –

(Hydrophobic ) ( Hydrophilic ) (Hydrophobic ) (Hydrophilic)

(c) CH3 (CH2) 16 – C – O(CH2 CH2O)N CH2CH2OH

(Hydrophobic ) (Hydrophilic)