13.Sr_Chem.pdf

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BOARD OF INTERMEDIATE EDUCATIONSENIOR INTER CHEMISTRY

MODEL PAPER (English Version)Time: 3 Hours Max. Marks: 60

SECTION - AI. (i) Very Short Answer Type questions.

(ii) Answer ALL questions.(iii) Each question carries TWO Marks. 10 2 = 20

1. What is "Ebullioscopic Constant"?2. Calculate the equilibrium constant of the reaction.

Cu(s) + 2 Ag+ (aq) Cu +2 (aq) + 2 Ag(s)E ocell = 0.46 V

3. Give the composition of the alloys:a) Brass and b) German Silver

4. NO is paramagnetic in gaseous state but diamagnetic in liquid and solid states.Why?

5. Why Conc. H2SO4, P4O10 and anhydrous CaCl2 can not be used to dry ammonia?

6. Why Zn+2 is diamagnetic where as Mn+2 is paramagnetic?7. What are artificial sweetening agents? Give 2 examples.8. What are tranquilizers? Give 2 examples.9. Write equations for carbylamine reaction of any one aliphatic amine and one

aromatic amine.10. What is Hell-Volhard - Zelensky reaction?

SECTION - BII. (i) Short Answer Type questions.

(ii) Answer any SIX questions.(iii) Each question carries FOUR Marks. 6 4 = 24

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www.eenadupratibha.netONLINE-15 MP-2 NEW SYLLABUS

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11. Calculate the efficiency of packing in case of a metal of body-centred cubic crystal.

12. What is relative lowering of vapour pressure? The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A non-volatile, non-electrolytesolid weighing 0.5 g when added to 39 g of benzene; vapour pressure of the solution is 0.845 bar. Calculate the molar mass of the solid substance.

13. a) Write possible chain isomers for C4H9Br.b) Predict the order of reactivity of the following compounds in SN1and SN2

reactions:

C6H5CH2Br, (C6H5)2CHBr, C6H5CH(CH3)Br, (C6H5)2 C(CH3)Br14. Outline the principles of refining metals by the following methods.

a) Zone refining b) Electrolytic refiningc) Poling d) Vapour phase refining.

15. Action of soap is due to emulsification and micelle formation. Comment.

16. Write the IUPAC names of the following compounds:a) K2[Zn(OH)4] b) [Ni(CO)4]c) K3[Cr(C2O4)3] d) [CO(NH3)4(H2O)Cl]Cl2

17. Write a brief note on amino acids.

18. a) What is PHBV? How is it useful to man?b) Classify the following as addition and condensation polymers: Terylene,

Bakelite, Poly Vinyl Chloride, Polythene.

SECTION - CIII. (i) Long Answer Type questions.

(ii) Answer any TWO questions.(iii) Each question carries EIGHT Marks. 2 8 = 16

19. a) How are XeF2 and XeF4 prepared? Give their structures.b) Explain the structures of BrF5 and IF7.

20. Discuss the effect of catalyst and temperature on the rate of reaction.21. a) Explain the acidic nature of phenols and compare with that of alcohols.

b) Explain: i) Stephen reaction ii) Etard reaction.



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ANSWERS

SECTION - A

1. What is "Ebullioscopic constant"?

A: The elevation in the boiling point of one molal solution.

Tb = Kb . m2. Calculate the equilibrium constant of the reaction.

Cu (s) + 2 Ag+ (aq) Cu +2 (aq) + 2 Ag (s)E ocell = 0.46 V

0.0591A: E ocell = log Kcn

0.0591E ocell = log Kc = 0.46 V2

0.46 2 log10 Kc = = 15.60.0591

Kc = 1015.6 = 1015 100.6 = 3.92 1015

3. Give the composition of the alloys:

a) Brass and b) German SilverA: a) Brass: 60% Cu + 40% Zn

b) German Silver: 25 to 40% Cu + 25 to 35% Zn + 40 to 50% Ni.4. NO is paramagnetic in gaseous state, but diamagnetic in liquid and solid

states. Why?

A: NO is odd number (electrons) molecule in (.. .N = . .O ..) gaseous state, hence it isparamagnetic. Whereas NO dimerises in liquid and solid states. N2O2 is havingeven number of electrons (odd + odd = even), so it is diamagnetic.

5. Why Conc. H2SO4, P4O10 and anhydrous CaCl2 can not be used to dry ammonia?

A: NH3 reacts with Conc. H2SO4 to form (NH4)2SO4 NH3 reacts with P4O10 to form (NH4)3PO4



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NH3 reacts with CaCl2 forming CaCl2 . 8 NH3

So NH3 is dried using quick lime (CaO)6. Why Zn+2 is diamagnetic where as Mn+2 is paramagnetic?

A: Zn+2 = [Ar] 4s0 3d10 all electrons are paired.Mn+2 = [Ar] 4s0 3d5 has 5 unpaired electrons.As no unpaired electrons present in Zn+2, it is diamagnetic, As 5 unpaired d electrons are present in Mn+2, it is paramagnetic.

7. What are "Artificial sweetening agents"? Give 2 examples.

A: The chemical substances not only controls the intake of calories but also gives

sweet taste are called "artificial sweetening agents".

e.g.: Sucralose, Aspartame.

8. What are tranquilizers? Give 2 examples.

A: The drugs which are used to reduce mental diseases and stress are called

tranquilizers.

e.g.: Seconal, Luminal.

9. Write equations for carbyl amine reaction of any one aliphatic amine and onearomatic amine.

A: A dirty smelling compound isocyanide or carbyl amine is formed when

aliphatic or aromatic primary amine is heated with chloroform and ethanolic

KOH.Heat

C2H5NH2 + CHCl3 + 3 KOH C2H5NC + 3 KCl + 3 H2O

HeatC6H5NH2 + CHCl3 + 3 KOH C6H5NC + 3 KCl + 3 H2O

10. What is "Hell-Volhard-Zelensky reaction"?

A: The reaction in which halogenation takes place at alpha position of a carboxylic

acid (having alpha hydrogen) when treated with a halogen (Cl2 or Br2) in presence of red phosphorus.



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H Red phosphorus

H

H3C C COOH H3C C COOH + HCl

H + Cl Cl Cl

SECTION - B

11. Calculate the efficiency of packing in case of a metal of body - centred cubiccrystal.

A: B.C.C. crystal contain 8 atoms at 8 corners and one atom at the centre of thebody.

1No.of the atoms in the unit cel = corners (8 ) + body centre(1) = 1 + 1 = 2. 8In each diagonal, 3 atoms are present.

C = AF = Length of body diagonal = r + 2r + r = 4r .......... (1)In EFDFD2 = FE2 + ED2

b2 = a2 + a2 = 2a2

From (1) & (2)3 a = 4r

4 a = r3In AFDAF2 = FD2 + DA2

c2 = b2 + a2

= 2a2 + a2

= 3a2

c = 3 a ........... (2)4

Volume of the cube = a3 = ( r)334

Volume occupied by 2 atoms = 2 r33



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Volume of 2 atomsPacking efficiency = 100

Volume of unit cell

42 r3 1003

=

4( 3r)33= 68%

12. What is relative lowering of vapour pressure? The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A non-volatile, non-electrolyte solid weighing 0.5 g when added to 39 g of benzene, vapourpressure of the solution is 0.845 bar. Calculate the molar mass of the solidsubstance.

A: Relative lowering of vapour pressure (R.L.V.P.): The ratio of lowering ofvapour pressure to vapour pressure of pure solvent (Po)

Po - PsR.L.V.P. = Po

Po - Ps W2 M1 =

Po M2 W10.850 - 0.845 0.5 78 =

0.850 M2 390.850 0.5 78

M2 = = 17039 0.00513. a) Write possible chain isomers for C4H9Br.

b) Predict the order of reactivity of the following compounds in SN1andSN2 reactions:C6H5CH2Br, (C6H5)2CHBr, C6H5CH(CH3)Br, (C6H5)2C(CH3)Br

A: a) CH3CH2CH2CH2Br n - butyl bromideCH3 CH2 CH CH3 2 - Bromobutane

CH3 Br

H3C C Br 2 - Bromo 2 - Methyl Propane

CH3



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Br H H H

b) C6H5 C CH3 > H5C6 C Br > H5C6 C Br > H5C6 C Br (SN1) C6H5 C6H5 CH3 H

CH3 H H H

H5C6 C Br < H5C6 C Br < H5C6 C Br < H5C6 C Br (SN2) C6H5 C6H5 CH3 H

14. Outline the principles of refining metals by the following methods.a) Zone refining b) Electrolytic refiningc) Poling d) Vapour phase refining

A: a) Zone refining: Impurities are more soluble in molten state than in the solidstate of metal. Mobile circular heater is fixed on the impure metal at one end.Impurities are moving along with the heater and removed at the end of the rod,by cutting it.e.g.: Germanium.

b) Electrolytic refining: In this method in pure metal is taken as anode. Thinsheet of pure metal is taken as cathode. Soluble salt of the same metal is takenas electrolyte. By passing electricity through it, pure metal gets deposited onthe cathode.e.g.: Copper.

c) Poling: Impurities present in molten metal can be removed either as gases oras slag when it is stirred with poles of green wood.e.g.: Blister copper.

d) Vapour phase refining: In this method, metal is converted into easily decomposable volatile vapour compound and it is decomposed to get puremetal on heating.

350 K 450 Ke.g: Ni + 4 CO Ni(CO)4 Ni + 4 CO

15. Action of soap is due to emulsification and micelle formation. Comment.A: Soap is considered as Sodium Sterate. It has C17H35COO-

(Sterate) and Na+ ions. Sterate ion has non polar part(C17H35) tail and polar part (COO-) head. Tail part dissolvesgrease and head part attracts water to form emulsion (oil inwater type). In concentrated soap solution of soap, micelles



Oil

Sterate micelle

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are formed due to association of many sterate ions. Due to formation of emulsion and micelles, grease of a cloth is to be removed (upon rinsing in water).

16. Write the IUPAC names of the following compounds:

a) K2[Zn(OH)4] b) [Ni(CO)4]c) K3[Cr(C2O4)3] d) [CO(NH3)4(H2O)Cl]Cl2

A: a) K2[Zn(OH)4] - Potassium tetrahydroxozincate (II)b) [Ni(CO)4] - Tetracarbonyl nickel (0)c) K3[Cr(C2O4)3] - Potassiumtrioxalato chromate (III)d) [Co(NH3)4(H2O)Cl]Cl2 - Tetra ammine aqua chloro cobalt (III) chloride.

17. Write a brief note on amino acids.

A: Compounds containing amino group (-NH2) and acid group (-COOH) arecalled amino acids.

H

The general formula of - amino acid is R C COOH.

NH2Amino acids are obtained by hydrolysis of proteins. They are amphoteric. In aqueous solution they form Zwitter ions

H (R C COO- )NH3+There are 10 non-essential amino acids (which can be synthesised, in the body)and 10 essential amino acids (supplied through food). They are classified intoacidic or basic or neutral amino acids.

18. a) What is PHBV? How is it useful to man?b) Classify the following as addition and condensation polymers: Terylene,

Bakelite, Poly Vinyl Chloride, Polythene.

A: a) PHBV: Poly - hydroxy butyrate - Co - - hydroxy valerate (PHBV) isa copolymer of 3 - hydroxy butanoic acid and 3 - hydroxy pentanoic acid. Itis a condensation polymer.It is used to make cover of the capsules, used in orthopaedic devices and in speciality packing.



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SECTION - C

19. a) How are XeF2 and XeF4 prepared? Give their structures.b) Explain the structures of BrF5 and IF7.

A: a) XeF2The reaction between Xe and F2 at 1 bar pressure and 673 K temperature givesXeF2.

1 barXe(g) + F2 (g) XeF2 (s) :673 K

Xe = 5s2 5px2 5py

2 5pz1 5d1 (1st excited state)

Xe undergoes sp3d hybridization. Due to presence of 2 bond pairs and 3 lonepairs, the shape of XeF2 is linear.

XeF4XeF4 is formed when Xe & F2 reacts together in 1 : 5 ratio at 7 bar pressureand 873 K temperature.

7 barXe + 2 F2 XeF4 873 K

Xe = 5s2 5px2 5py

1 5pz1 5d1 5d1 (2nd excited state)

Xe undergoes sp3d2 hybridization. Due to presence of 4 bond pairs and 2 lonepairs, the shape of XeF4 is square planar.

b) BrF5Br = 4s2 4p

x1 4py

1 4pz1 4d1 4d1 (2nd excited state)

Br undergoes sp3d2 hybridization. It has square pyramidalshape due to presence of 5 bond pairs & 1 lone pair.IF7I = 5s1 5p

x1 5py

1 5pz1 5d1 5d1 5d1 (3rd excited state)

Iodine undergoes sp3d3 hybridization. Due to presence of7 bond pairs in its 3rd excited state, the shape of IF7 ispentagonal bipyramidal.

20. Discuss the effect of catalyst and temperature on the rate of reaction.A: Effect of Temperature: The temperature dependence of the rate of a chemical

reaction can be accurately explained by Arrhenius equation K = A . e-Ea/RT.



F

F

Xe

:

:

F

FF

FXe

..

..

F

FF

FF

Br

..

F

F

F

F

FF

F

I

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Where A is frequency factor. If temperature is increased, kinetic energy of molecules will be increased. The number of effective collisions (molecules possess equal or greater energy than that of activation energy) increases, so therate of reaction increases. It has been found that for a chemical reaction with risein temperature by 10 , the rate constant is nearly doubled.

By taking natural logarithm of both sides of Arrhenius equationEaln K = ln A - RT

EaAt temperature T1 ln K1 = ln A - RT1EaAt temperature T2 ln K2 = ln A - RT2

Ea Ealn K2 - ln K1 = - + RT2 RT1K2 Ea 1 1ln = [ - ]K1 R T1 T2

K2 Ea 1 1 log = [ - ]K1 2.303R T1 T2 Effect of catalyst: A catalyst increase the rate of forward reaction as well as rateof backward reaction and helps in attaining equilibrium sooner. A catalyst doesnot alter G and equilibrium constant. A catalyst gives alternate path by reducing the activation energy between reactants and products. It is very clearfrom Arrhenius equation that lower the value of Ea, faster will be the rate of reaction.



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21. a) Explain the acidic nature of phenols and compare with that of alcohols.b) Explain: i) Stephen reaction ii) Etard reaction

A: a) Proton (H+) donor is acid. -OH group is electron withdrawing group and isattached to sp2 carbon of benzene ring. As electronegativity of sp2 carbon ofphenol is more, electron density decreases on oxygen. This will increase thepolarity of O - H bond and results increase in ionisation of phenols than thatof alcohols. In alkoxide ion (RO-), the negative charge is localised only onoxygen, where as in phenol, negative charge is delocalised. Due to resonance,stable phenoxide ion is formed by losing H+ ion easily.

R - O - H R - O :- + H+

+ H+

Phenoxide ion is more stable than that of alkoxide ion. Acidity of phenols ismore than that of alcohols. When electron withdrawing group like -NO2 isattached to benzene in ortho, para positions in phenol, acidic nature will increases further. From pKa data we will understand that phenol is milliontimes more acidic than that of ethanol.

b) i) Stephen reaction: Alkyl cyanides on reaction with SnCl2 & HCl givesimine, which on hydrolysis give aldehyde.

H3O+CH3CN + SnCl2 + HCl CH3CH = NH CH3 CHO



::

::

:

OH : O :-

:

: O : : O : : O : : O :

:

: O :

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ii) Etard reaction: Methyl group (-CH3) on reaction with Chromyl Chloride(CrO2Cl2) gives a complex, which on hydrolysis gives aldehyde.

CH3 + CrO2Cl2 CH(OCrOHCl2)2 CHOCS2 H3O

+

Toluene Chromium complex Benzaldehyde

A.N.S. Sankara Rao,Senior Lecturer in Chemistry.



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13.Sr_Chem.pdf