108inter 2nd year chemistry e

Chemistry – II Paper – II

Section – A Max M: 60 I Answer all the questions 10 × 2 = 20 M

1. Define mole fraction. Give its units

2. Give the products formed at the platinum electrode when aqueous solution of K2SO4 is electrolyzed. 3. a) How is aluminium useful in the extraction of Chromium and manganese from their oxides? b) What is the role of graphite rod in the metallurgy of aluminium? 4. Name of the following. a) most abundant noble gas in atmosphere b) radio active noble gas c) noble gas with least boiling point. d) noble gas forming large number of compounds e) noble gas not present in atmosphere. 5. What is the molecular shape of I ion and what is the type of hybridization in it? 6. a) What is a chealate ligand? Give example b) How many moles of Agcl is precipitated when 1 mole of Cocl3 is treated with AgNo3 solution. 7. a) What are Ambident nucleophiles? b) What is Racemic mixture? 8. Write the structure and name of the amine produced by Hoffman degradation of Benzamide. 9. Write the Coupling reactions of Benzene diazonium chloride with phenol.

10. Which compound in each of the following with react faster in SN2 reaction with OH-

i) CH3Br (or) CH3I ii) (CH3)3ccl (or) CH3Cl

II Answer any six questions. Each carries four marks 6 × 4 = 24 M

11. a) Write the difference between a conductor and an insulator in terms of band theory. b) Write the difference between a conductor and a semi conductor in terms of band theory. 12. Define molarity? Calculate the molarity of a solution containing 5 gr of NaoH in 450 ml of solution. 13. Describe the purification of colloidal solutions by the phenomenon of dialysis with a neat diagram. 14. Explain briefly the extraction of aluminium from bauxite. 15. Write the postulates of Wermer’s theory of coordination compounds and explain with two examples? 16. How are the following prepared and write their uses? a) Neoprene b) Buna - S 17. Explain the structure of primary, secondary, tertiary and quaternary proteins. 18. What are Cationic and anionic detergents? Give examples for each type.

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Number of Moles of the Component

Total number of Moles of all the Components in the Solution

III Answer any two of the following. Each question carries 8 marks 2 × 8 = 16 M

19. a) What are electro chemical cells? How are they constructed? Explain the working of the different types of galvanic cells. b) Determine the values of Kc for the following reaction

Ni(s)+2Ag(eq)+ Ni(eq)

2+ + 2Ag(s) ∈( ) = 1.05V

20. How is ammonia manufactured by Haber’s process? Explain the reactions of ammonia with a) ZnSo4(eq) b) CuSo4(eq) c) Agcl(s)

21. How do you distinguish the following pairs of compounds? (i) propanol and propanone (ii) Acetophenone and Benzophenone (iii) phenol and Benzic Acid (iv) pentan – 2 – one and pentan – 3 – one .

Section – A 1. Define mole fraction. Give its units. A. Mole fraction is the ration of number of Moles of one Component to the total number of moles of all the Components in the Solution. Mole fraction of a Component = 2. Give the products formed at the platinum electrode when aqueous solution of K2SO4 is electrolyzed.

A. In aqueous Solution of K2So4 the following reactions occur.

K2So4 2K++ So42-

3. a) How is aluminium useful in the extraction of Chromium and manganese from their oxides? b) What is the role of graphite rod in the metallurgy of aluminium? A. Aluminium is a reducing agent because of its electro positive character, in the extraction of chromium and manganese from their Oxides

Cr2O3 + 2Al Al2O3 + 2Cr + heat

3Mn3O4 + 8A/ 4Al2O3 + 9Mn + heat

b) Graphite acts as a reducing agent and reduces Al2O3 to A/

2Al2O3 + 3C 4A/ + 3Co2 Graphite acts as a node.





I

I

I

• •

• •

• •

4. Name of the following. a) most abundant noble gas in atmosphere b) radio active noble gas c) noble gas with least boiling point. d) noble gas forming large number of compounds e) noble gas not present in atmosphere. A. a) Argon b) Radon c) Helium d) Xenon e) Radon 5. What is the molecular shape of I ion and what is the type of hybridization in it? A. I is linear

Central Iodine atom in Sp3d hybridization

6. a) What is a chealate ligand? Give example b) How many moles of Agcl is precipitated when 1 mole of Cocl3 is treated with AgNo3 solution. A. a) Chelate lig: - A ligand which can form ring type complex is called chelate ligand.

Ethylendiamine H2N – CH2 – CH2 – NH2 is a chelate ligand.

It can from ring type of complex.

b) Three Moles of Agcl is precipitated when 1 Mole of Cocl3 is treated with AgNo3 Solution.

7. a) What are Ambident nucleophiles? b) What is Racemic mixture? A. a) Ambident nucleophiles: Groups like Cyanides and nitrites which possess two nucleophilic centres are called Ambident nucleophiles. b) Racemic mixture: A mixture containing two enantiomers in equal proportion will have zero Optical rotation, as the Rotation due to one isomer will be exactly Cancelled by the Rotation due to the other isomer, Such Mixture is known as Racemic mixture. 8. Write the structure and name of the amine produced by Hoffmann degradation of benzamide

A. Benzamide is C6H5CoNH2 which contains seven carbon atoms. On Hoffmann degradation it gives an

aromatic primary amine containing six carbon atoms only.





9. Write the Coupling reactions of Benzene diazonium chloride with phenol A. Diazonium Salts react with aromatic amines and phenol to give azo compounds having the general formula. Ar – N = N – Ar! The reaction is known as Coupling reaction. The Coupling of benzene diazonium chloride with phenols is carried out in Mild alkaline solution and with amines in weakly acidic Medium. Coupling with Phenol: -

10. Which compound in each of the following with react faster in SN2 reaction with -OH

i) CH3Br (or) CH3I ii) (CH3)3ccl (or) CH3Cl

A. (i) CH3I will react faster than CH3Br in SN2 reaction with –

OH because iodine is a better leaving group

because of its larger Size. (ii) Of the two compounds CH3Cl and (CH3)3ccl, the former i.e,

CH3Cl will react faster in SN2 reaction with –

OH because there are only three small hydrogen atoms on carbon,

(CH3)3ccl is a tertiary alkyl halide. It is least reactive under SN2 conditions because the bulky groups hinder The approaching nucleophiles.

Section – B

11. a) Write the difference between a conductor and an insulator in terms of band theory. b) Write the difference between a conductor and a semi conductor in terms of band theory. (a). Difference between a conductor and an insulator in terms of band theory: In a Metal the outer Orbitals of very large number of atoms overlap to form a very large number of Molecular Orbitals that are delocalised over the Metal. As a result a large number of energy levels are crowded together into bands. The highest occupied energy band is called the valence band while the lowest unoccupied energy band is called conduction band. The energy difference separating the valence band and conduction band is called band gap or energy gap.





Conduction band

Empty band

Forbidden Zone

(Large energy gap)

filled band

Valance Band Insulator (b)

Overlapping bands

Partially filled band

Metal (b)

Ener

gy

In conductors the valence band is only partially filled. Electrons may easily excited from lower energy level to higher energy level by supplying a very small amount of energy, when a voltage is applied to a metal crystal, electrons are excited to the unoccupied orbitals in the same band and more towards the positive terminal. Thus, a material with partly filled energy band is a conductor. In the of insulations, the highest occupied band is completely filled, the energy gap between the fully filled balance band and the vacant conduction band is very large and it is not possible to excite the electrons to the conduction band. So they become insulators. (b) Difference between a conductor and a semi conductor in terms of band theory: In a metal the outer orbitals of very large number of atoms overlap to form a very large number of molecular orbitrals that are delocalized over the metal. As a result a large number of energy levels are crowded together into bands. The highest occupied energy band is called the valence band while the lowest unoccupied energy band is called conduction band. The energy difference separating the valence band and conduction band is called band gap or energy gap. In conductors the valence band is only partially filled. So electrons may easily excited from lower energy level to higher energy level by supplying a very small amount of energy, when a voltage is applied to a metal crystal, electrons are excited to the unoccupied orbitals in the same band and move towards the positive terminal. Thus, a material with partly filled energy band is a conductor. In the case of semiconductors, the forbidden band i.e, the energy gap between valence band and conduction band is little. The thermal energy available at room temperature is enough to excite some electrons from the highest occupied band to the next permitted energy band. So conductivity occurs but their conductivity is in between the conductors and insulators.





Moles of Solute

Volume of solution in litre

5g

40 g mol-1

Moles of Solute

Volume of solution in litre

0.125Mol ×1000ML L-1

450ML

12. Define molarity? Calculate the molarity of a solution containing 5 gr of NaoH in 450 ml of solution. A. Molarity: - Molarity (M) is defined as number of Moles of the Solute dissolved in one litre (or one cubic decimeter) solution. Molarity (M) = Solution: - Moles of NaoH = = 0.125Mol

Volume of the solution in litres = 450Ml/1000mL L-1

Using equation M = Molarity = = 0.278M

= 0.278 Mol L-1

=0.278 Mol dm-3

13. Describe the purification of colloidal solutions by the phenomenon of dialysis with a neat diagram. A.

Dialysis is a process of removing a dissolved substance from a colloidal solution using a suitable membrane. The apparatus used for this purpose is called dialyses. A bag of suitable membrane containing the colloidal solution is suspended in a vessel containing a continous flow of water. The bag made with animal membrane, or parchment paper or cellophane sheet. Allow the molecules or ions to diffuse through it into the water and pure colloidal solution is left behind in the bag. 14. Explain briefly the extraction of aluminium from bauxite.

A. 1) Bauxite is digested with hot conc. NaoH at 523K, Al2O3 dissolves impurities are filtered off.

Al2O3 + 2NaoH + 3H2O (1) 2Na [Al (OH)4] aq

The aluminate is alkaline in nature and is neutralized by passing CO2 gas. Al2O3 is precipitated

2Na [Al (OH)4] (aq) + Co2(g) Al2O3 ∙ XH2O + 2NaoHCo3





1470K

The sodium Silicate remains in the solution and the insoluble hydrated alumina is filtered, dried and

heated to give pure Al2O3 Al2O3 ∙ XH2O Al2O3 (S) + H2O (g)

(2) The purified Al2O3 is mixed with Na3 AlF6 or CaF2 which lower the melting point of the mix and also

increases conductivity. The fused matrix is electrolyzed. Steel vessel with lining of carbon acts as cathode and graphite anode is used. The overall reaction mat be

2Al2O + 3C 4Al + 3CO2

Reactions: - At cathode Al3 + (melt) + 3e- Al

At anode (cs) + O2- (melt) Co (g) + 2e-

(cs) + 2O2- (melt) Co 2(g) + 4e-

Flow Diagram: - Bauxite Al2O3 ∙ 2H2O

digested with hot Conc. NaoH

Al2O3 dissolves – impurities eliminated

Na [Al (OH)4] Co2 bubbled Al (OH)3 ppt Pure alumina

Electrolysis of alumina mixed with molten cryolite or CaF2 at 1200K using carbon

. lining of cell – cathode carbon rods – anode.





NH3

NH3

NH3

Cl

Cl

Cl

Cocl3 . 6NH3

Ex: - 1

H3N

Pure aluminium Metal

Electrolytic cell for the extraction of auminium 15. Write the postulates of Wermer’s theory of coordination compounds and explain with two examples? A. Werner proposed two types of valencies to explain the formation and the structure of these complex compounds. (i) The primary valence: - This corresponds to the oxidation state of the central transition metal ion in the complex compounds. This primary valence is satisfied by a negative ion only in the complex compound. (ii) Secondary valence: - This is equal to the number of chemical groups bound to central metal ion. The no. of secondary valencies is called co – ordination number of the metal. (iii) These secondary valencies are directed in space which determines the shape of the complex.

CO

Co – ordination no: 6 Structure: Octahedral.

The ligand NH3 satisfies only

secondary valency.





NH3

Cl

NH3

NH3 NH3

Cl

Cl

Cocl3 . 5NH3

Ex: - 2

Ex: - 3

Cl-

NH3

NH3 NH3

Cl-

H3N

H3N

Cocl3 . 4NH3

CO


One Cl- ion satisfies both

primary and secondary valencies.

CO+3


Two Cl- ion satisfies both






Ex: - 4

Cl-

Cl-

Cl-

NH3

Polymerisation

Polychloroprene or Neoprene Chloroprene

In Werner’s representation, dotted lines represent primary Valency whereas thick lines represent secondary valency. some negative ligands, depending upon the complex, may satisfy both primary and secondary valencies, such ligands do not ionize. iv) Primary valencies are ionisable valencies, secondayy valencies are non – ionisable valencies. 16. How are the following prepared and write their uses? a) Neoprene b) Buna - S A. Synthetic rubber is any vulcanisable rubber like polymer which is capable of getting stretched to twice its length. It returns to its original shape and size when the stretching force is removed. a) Neoprene: - Neoprene is formed by the free radical polymerization of chloroprene.

2 2

1c

n C CHCH CH

2 2

cl

nCHCH C CH

l

Uses: - Neoprene is used in the manufacture of Conveyer belts gaskets and hosepipes. (b) Buna – S: - Buna – S is obtained by the copolymerization of 1, 3 – Butadiene and styrene.

nCH2= CH – CH = CH2 + NC6H5 – CH = CH2 (CH2 – CH = CH – CH2 – CH2 – CH) Buna – S

Uses: - Buna – S is used in making automobile tyres and footwear. 17. Explain the structure of primary, secondary, tertiary and quaternary proteins. A. The structure and shape of proteins can be studied at four different levels, i.e. primary, secondary, tertiary and quarternary. (i) Primary structure of Proteins: - Proteins may have one or more polypeptide chains. Each polypeptide in a protein has amino acids linked with each other in a specific sequence. It is this sequence of amino acids that is said to be the primary structure. The primary structure represents the constitution of the protein.

CO+3


Three Cl- ion satisfies both






(ii) Secondary structure of proteins: - The secondary structure of proteins refers to the shape in which a long poly peptide chain can exist. These chains are found to exist in two different types of structure (1) at helix and (2) 훽 – pleated sheet structure. These structure arise due to the regular folding of the back bone of the polypeptide chain due to hydrogen bonding between C = 0 and –NH groups of the peptide bond. (iii) Tertiary structure of proteins: - The tertiary structure of proteins represents over all folding of the polypeptide chains i.e. further folding of the secondary structure. It gives rise to two major molecular shapes viz. fibrous (thread like) and globular. (iv) Quarternary structure of proteins: - Some of the proteins are composed of two or more polypeptide chains referred to as sub – units. The special arrangement of these sub units with respect to each other is known as quarternary structure. 18. What are Cationic and anionic detergents? Give examples for each type. A. (i) Cationic detergents: - cationic detergents are quarternary ammonium salts of amines with acetates, chlorides or bromides as anions. Cationic part possesses a long hydrocarbon chain and a positive charge on nitrogen atom.

Ex: - CH3(CH2)15-N+(CH3)3Br̅ Cetyl trimethyl ammonium bromide is a popular cationic detergent. It is used in hair conditioners. (ii) Anionic detergents: - Anionic detergents are Sodium salts of sulphonated long chain alcohols or hydrocarbons. In these detergents, the anionic part of the molecule is involved in the cleasing action. Sodium salts of alkyl benzene sulphonates are an important class of anionic detergents. Ex: - Sodium dudecyl benzene sulphonate.

CH3(CH2)11 SO3- Na+

Section – C

19. a) What are electro chemical cells? How are they constructed? Explain the working of the different types of galvanic cells. b) Determine the values of Kc for the following reaction

Ni(s)+2Ag(eq)+ Ni(eq)

2+ + 2Ag(s) ∈( ) = 1.05V

A. a) The device in which chemical energy is converted into electrical energy is called galvanic cell or electro chemical cell or voltaic cell. In galvanic cell, a redox reaction is carried out in an indirect manner and the decrease in free energy during the chemical process is made to appear as electrical energy. The indirect redox reaction is such that reduction and oxidation processes are carried out in separate vessels. The working of different types of

galvanic cells can be understand by considering the Zn – CUSO4 reaction as the basis of the cell reaction.

A zinc strip is dipped in zinc sulphate solution and a copper strip is dipped in copper sulphate solution taken in separate beakers,. The two metallic strips which act as electrodes are connected by the conducting wire through a volt meter. The two solutions are joined by a U – tube known as salt bridge which





contain some electrolyte such as Kcl, KNO3 or NH4Cl. Along with gelatin or agar – agar to convert it into semi

solid paste. (i) Zinc undergoes oxidation to form zinc ions.

Zn(s) Zn 2+

(aq) + 2e- (oxidation)

(ii) The electrons liberated during oxidation are passed through the connecting wire to copper strip (iii) Copper ions move towards copper strip, pick up electrons and get reduced to copper atoms which are deposited on the copper strip. At the zinc strip oxidation of zinc atoms takes place and becomes a source of electrons acquiring negative charge. It acts as anode, since oxidation occurs at it. At the copper strip reduction of copper ions takes place and acquires positive charge. It acts as cathode since reduction takes place. The flow of electrons from zinc strip to copper strip produce electric current through the outer circuit from copper to zinc strip which is indicated by the deflection in voltmeter.

b) Ecell-

= .

log Kc

Given Ecell-

= 1.05V; R = 8.314J;

I ̅= 298K; n = 2; F = 96500

1.05 = 2.3030 ×8.314 ×298

2 ×96500 log Kc

= 0.0295 log K eq or Keq = Anti log .

.

= 3.411 × 1035 19 (a)





20. How is ammonia manufactured by Haber’s process? Explain the reactions of ammonia with a) ZnSo4(eq) b) CuSo4(eq) c) Agcl(s)

Ans. On a large scale, ammonia is manufactured by Haber’s process.

N2 (g) + 3H2 (g) ⇌ 2NH3 (g); ∆ F H° = -46.1. KJmol-1 The optimum conditions for production of ammonia are a pressure of about 2000 × 105 pa and

a temperature of 700K. Iron oxide is the catalyst with small amounts of K2O and Al2O3.

Compressed mixture of N2 and H2 in the volume ratio is heated to 700K at a pressure of 200 atm and passed over finely divided iron oxide mixed with small amounts of K2O and Al2O3. Ammonia formed is liquified and unreacted mixture of N2 and H2 again pumped into catalytic chamber. (a)With ZnSO4, Zinc hydroxide precipitate is formed,

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





108inter 2nd year chemistry e

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