SAMPLE QUESTION PAPER CHEMISTRY (043) CLASS XII (2013-14)

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Sample paper of Chemistry. Chemistry Theory(043). Class –XII. Time : 3:00 hrs. MM : 70 ... Q.8 Arrange following in order of increasing their basic character.

SAMPLE QUESTION PAPER CHEMISTRY (043) CLASS XII (2013-14) BLUE PRINT

S .NO 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

UNIT

VSA(1) SA1(2)

SOLID STATE SOLUTIONS ELECTROCHEMISTRY CHEMICAL KINETICS SURFACE CHEMISTRY GENERAL PRINCIPLES OF EXTRACTION p-BLIOCK ELEMENTS D & fBLOCK ELEMENTS CO-ORDINATION COMPOUNDS HALOALKANES AND HALOALKANES ALCOHOLS, PHENOLS AND ETHERS ALDEHYDES, KETONE AND CARBOXYLIC ACIDS AMINES BIOMOLECULES POLYMERS CHEMISTRY IN EVERYDAY LIFE

1 1

2 1

1

1

1 1

SAII(3) LA(5) TOTAL Qs/Ch 1 2 1 1 3 1 2 1 2 1 1 1

8 5 3

2

2

4

2

2

4

2

6

2 2 2 1

4 4 3 3

30

70

1 1 1 1 1

8

4 5 5 5 4 3

3 2 1

1 1

1 1 1 1

WEIGHTAGE

20

27

15

Sample paper of Chemistry Chemistry Theory(043) Class –XII Time : 3:00 hrs MM : 70 __________________________________________________________________________ General Instructions ___________________________________________________________________________ i. All questions are compulsory ii. Marks for each question are indicated against it iii. Q.No.1 to 8 are VSA type questions of 1 marks each iv. Q.No. 9 to 18 are SA type questions of 2 marks each v. Q.No. 19 to 27 are also SA type questions of 3 marks each vi. Q.No. 28 to 30 are LA type questions of 5 marks each vii. Use log tables if necessary, use of calculator is not allowed ___________________________________________________________________________ Q.1 Which stoichiometric defect in crystal decreases density of a solid? Q.2 Express relationship b/w conductivity & molar conductivity of a cell. Q.3 Draw structure of PCl5 molecule Q.4 Write IUPAC name of H2NCH2COOH compound Q.5 Draw structure of DNA. Q.6 What do you mean by “shape –selective catalysis” of a reactions? Q.7 What are condensation polymers? Q.8 Arrange following in order of increasing their basic character C2H5NH2, (C2 H5) 2 NH, (C2H5) 3N, NH3 Q.9 What is meant by (a) Primary structure of proteins (b) Peptide linkage Q.10 Write four main difference b/w RNA & DNA Q.11 Write reactions occurring during corrosion of iron Q.12 (a) What is the unit of rate constant in 1st order reactions (b) What is “temperature coefficient” Q.13 Determine the emf of cell in which the following reaction takes place: Ni(s) + 2Ag+(0.002M) (aq) Ni2+ (0.16M)(aq) + 2Ag(s), Eo =1.05V o (1F = 96500 coulomb , E value of Ni = - o.23V and of Ag is 0.80V) Q.14 Explain the following giving one example for each (i) Fittig’s reaction (ii) Friedal Crafts alkylation of anisole Q.15 What do you mean by lanthanoid contraction what are its consequences Q.16 State reasons for (a) All P-Cl bonds in PCl5 molecules ate not equivalent (b) N2 is inert at room temperature Or (a) NH3 is stronger base than PH3 (b) SF6 is kinetically is an inert substance Q.17 Mention two important uses of each: (a) teflon (b) SBR( styrene butadiene rubber) Q.18 How would you get (a) Phenyl benzoate acid from phenol (b) Salicylic acid from phenol Q.19 What do you mean by (a) Coagulation (b) peptization (c) electrophoresis Or (a) Micelles (b) emulsions (c) lyophilic colloids Q.20 Silver metal crystallises with a face centred cubic lattice. The length of unit cell is found to be 4.077 × 10–8 cm. Calculate atomic radius and density of silver. (atomic mass of Ag = 108u, NA = 6.02 × 1023 mol–1)

Q.21

A first order gas reaction A2 +B2(g) 2A(g) + 2B(g) at the temperature 400°C has the rate Constant k = 2.0 × 10–4 s–1. What percentage of A2 B2 is decomposed on heating for 900 seconds.Also calculate the half life period.

Q.22

Describe the basic principles behind each of the following (a) Vapour phase refining of metal (b) Zone refining (c) electrolytic refining of a metal Complete the following reactions (a) MnO4- + Fe2+ (b) Cr2O7 2- + Fe2+ (c) KMnO4 heated For complex K4 [Fe(CN) 6], identify following (a) Oxidation No. of Fe (b) Hybridization of Fe (3) Name of complex (a) Explain why aniline is less basic than ammonia (b) How would you convert (i) Aniline to benzene diazonium chloride (ii) Aniline to fluorobenzene (a) Out of C6H5CH2I & C6H5CH2Cl which one is more reactive in SN2 substitutions reaction (b) Alcohols have higher boiling points why? (c) What do you mean by racemic mixture Explain following with one example each (a) Artificial sweetening agent (b) Analgesics ( c) Antipyretics (a) State Henry’s laws mention some of its applications (b) Why is freezing point of depression of 0.1M NaCl solution nearly twice that of 0.1M glucose

Q.23

Q.24 Q.25

Q.26

Q.27 Q.28

solution? or

(a) (i)State the difference b/w molarity & molality of solution (ii) what is a brine solution? (b) ) A solution of glycerol (m .m=92 g/mol) in water was prepared by dissolving some glycerol in 500g of water . this solution has a boiling point of 100.420C.what mass of glycerol was dissolved to make this solution ? kb for water =0.512kkg/mol Q.29

Q.30

(a) How will you distinguish b/w (i) Primary & secondary alcohol (ii) Acetaldehyde & acetone (b) How will you convert (i)Ethanol into iodoform (ii) Toluene into benzoic acid (iii) Acetyl chloride to ethanol or (a) How will you distinguish b/w (i) Primary & secondary amines (ii) Alcohols & carboxylic groups (b) Convert : (a) Methanol into acetic acid (b) Chlorobenzene into phenol (c) Sodium acetate into methane (a) Give reasons (i) H2O is a liquid while H2S is gas (ii) N2 is inert at room temperature (b) Complete the following reactions: (i) P4 + HNO3 (Conc) (ii) P4 + NaOH + H2O (iii) CaF2 + H2SO4 Or (a) Give reasons (i) N does not form NCl5 while P can form PCl5 (ii) Red phosphorus is inert at room temperature (b) Complete the following reactions (i) Cl2 + NaOH (Hot & Conc.) (ii) NH3 (excess) + Cl2 (iii) ( NH4) 2 SO4 + Ca(OH) 2

ANSWER KEY OF XIITH CLASS SAMPLE PAPERS 1. Schottky defect 2. Molar conductivity =conductivity * 1000/molarity 3.

1 1 1

4. 2-Aminoethanoic acid. 1 5. Correct double helix structure of DNA 6. Which depends on the pore structure and the shape of reactants and products

1

1 7. A condensation polymer is defined as a polymer that involves elimination of small molecules during its synthesis, or contains functional groups as part of it 1 8. NH3 < C2H5NH2< , (C2H5) 2 NH < (C2H5) 3N 1 9. (a) correct explanation 1

(b) A covalent bond formed by joining the carboxyl group of one amino acid to the amino group of another, with the removal of a molecule of water 1

10. Four differences

½+½+½+½

11. Oxidation reaction: Iron is oxidized (loses e-) Fe(s)

Fe2+(aq) + 2 e-

Reduction reaction: Oxygen is reduced (gains e-) O2 + 4e- + 4 H+ Fe2+ + 2OH-

Fe(OH)2

2 H2O

Fe2O3 (rust)

1+1

12. (a) s-1 1 (b) The temperature coefficient is the relative change of a physical property when the T is changed by 1 K 1 13. Formula used ½

n=2 substituting the values Ecell=0.9142 V 14. (i)

½ ½ ½

1 (ii)

1

15 Lanthanide contraction is a term used in chemistry to describe the decrease in ionic radii of the elements in the lanthanide series 1 Two consequences: 1. many of the fifth- and sixth-period transition elements show remarkable similarities in their physical and chemical properties. For example, hafnium is so similar to zirconium in atomic radius 2. It is too difficult to separate the lanthanoids due to their similar sizes ½ +½

16. a). Due to more BP-BP repulsions felt by axial than equatorial one 1 b). Requires very high bond dissociation enthalpy due to triple bond in N2 molecule 1

OR a) Because due to high electro negativity of N, electron pair is more readily available in NH3 than in PH3 as electro negativity decreases on moving down the group 1 b) SF 6 is chemically inert due to the reason that the six F atoms protect the sulphur atom from attack by the regents to such an extent that even thermodynamically most 1

17. (a) I . As a material resistant to heat Ii for coating in non- sticky cookware (b) In bubble gums 18. (a) C6H5OH + C6H5COCl → C6H5OCOC6 H5 + HCl (b)

½+½ 1 1

1 19.(a) Coagulation is the process of precipitation of colloidal partical by adding a suitable electrolyte 1 (b) peptization is the formation of colloidal sol from a fresh ppt 1

( c) Electrophoresis is the movement of colloidal particals under the influence of electric field 1 OR (a) Micelles: aggregate of partials (atom, ion, simple molecule) 1 (b) Emulsions is a colloidal sol in which dispersed phase and dispersion medium are liquids 1 ( C) lyophilic colloids are water loving colloids 1 20. Given: a = 4.077 × 10–8 cm, Z = 4, M = 108 g mol–1, NA = 6.022 × 1023 d =Z M/a3* NA       ½ substitution ½ d = 10.58 g cm–3 ½ r =? FORMULA USED ½ substitution ½ r = 1.44 × 10–8 cm ½ 21 Formula used ½ Substituting the values ½ Answer 16.5% 1 T1/2 =0.693/k ½ T1/2 = 3465 s ½ 22. (a) chemical reaction the metal is converted to a compound, which forms a vapour, which is decomposed to get pure metal 1 (b) Metals of very high purity can be obtained by this method. The impure metal rod is mounted horizontally and heated by a circular electric heater at one end in an atmosphere of inert gas to form a thin molten zone. By slowly moving the heater, the molten zone is moved from one end of the rod to the other end. Pure metal crystallises while impurities pass into the molten zone. By repeated the passes of the molten zone very high purity can be attained at one end. The other end is discarded. Ge, Si and Ga used in semi conductors are refined in this manner. 1

( c) The method is based on the process of electrolysis. The crude metal is made the anode a thin sheet of pure metal the cathode. The electrolyte is the solution of a salt of the metal. On passing electricity the metal dissolves from the anode and an equal number of metal ions of the solution gets deposited at the cathode. The impurities settle down below the anode mud (eg:refining of Cu using CuSO4 solution as electrolyte). At anode Cu ------> Cu 2+ + 2electron At cathode Cu 2+ + 2electron -------> Cu 1



23. (a) 5 Fe2+ + MnO4- + 8 H+ -> 5 Fe3+ + Mn2+ + 4 H2O (b) Cr2O72- + 6Fe2+ + 14H+ = 2Cr3+ + 6Fe3+ + 7H2O

1

1

( c) 2KMnO 4 → MnO 2 + K 2 MnO 4 + O 2 1 24. (a) Oxidation no. = +2 1 2 3 (b) d sp 1 (c ) Potassium hexacyanoiron(II) 1 25. (a) Aniline is less basic than ammonia because due to resonance , lone pair on N get involved and is not available for donation but in ammonia it is not so. Lone pair of electron

is available for donation in ammonia and is more basic. 1 (b) (i) C6H5NH2 + HNO2 + HCl → [C6H5N2]Cl + 2 H2O 1 (ii)

1

26. (a) C6H5CH2 I is more reactive in SN2 substitutions 1 (b) due to presence of intermolecular H- bonding 1 ( c) Equimolar mixture of d and l- form of a compound to have net zero specific rotation. 1 27. (a) Artificial sweetening agents that are non-nutritive in nature are used as substituents for sugar (specially in soft drinks). Examples are saccharin (500 times sweeter than sucrose) and cyclamates1 (b) A medication that reduces or eliminates pain 1 ( c) Reducing or tending to reduce fever. 1 28.(a) At a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid." 1 Applications: 1. In Scuba diving 2, in soft drinks 1+2 (b) formula used 1 Calculation of molar masses ½ Calculation of no. of moles of CCL4=0.45 & C6H6 =0.38 ½ Mole fraction of benzene= 0.457 OR (a) (i) Molarity is T dependent but molality is independent of it. (ii) o.91 NaCl % (aq)

1 1

(b) ΔTb= 0.42K

1

Formula used Wb =37.7g

29. (a) (i) Lucas regent test (ii) tollen’s reagent (b)

1 1

1 1

(i)

1 (ii)

1 (iii) CH3CHO (Pb/BaSO4)  CH3CH2OH Or (a) (i)Hinsberg test positive for primary amines and correct reactions (ii) NaHCO3 teast is given by carboxylic acids and correct reactions (b) : (a) CH3OH + CO → CH3COOH

1 1 1 1

The process involves iodomethane as an intermediate, and occurs in three steps. A catalyst, metal carbonyl, is needed for the carbonylation (step 2).[41] 1. CH3OH + HI → CH3I + H2O 2. CH3 I + CO → CH3COI 3. CH3COI + H2O → CH3COOH + HI (b)

1 ( C) CH3COONa + (NaOH+CaO)  CH4

1

30. (a) (i) due to strong H-bonding present in H2O than in H2S( no H-bonding) 1 (ii) Due to triple bond between two N-atoms,it has very high Bond dissociation enthalpy 1 (b) (i) P4 + 20 HNO3 --> 4 H3PO4 + 20 NO2 + 4 H2O 1 (ii) P4 + 4 NaOH + 2 H2O = 2 PH3 + 2 Na2HPO3 1 (iii) CaF2 + H2SO4 --> 2HF + CASO4 1 OR (a) (i) Due to non- availability of vacant d-orbital. On N, it cannot extend its valency and does not form pentahalides 1. (ii) (b) (i) Cl2(aq) + 2NaOH(aq)

NaOCl(aq) +NaCl(aq) + H2O(l)

(ii) NH3 + Cl2 → NH2Cl + HCl

1

1

NH2Cl + Cl2 → NHCl2 + HCl NHCl2 + Cl2 → NCl3 + HCl

(iii) Ca(OH)2 + (NH4)2SO4 --> CaSO4 + 2(NH4)OH

1