ARMY PUBLIC SCHOOL,LUCKNOW
CENTRALISED PRE BOARD I
CHEMISTRY(043)
MARKING SCHEME
Q.NO. ANSWERS MARKS
1. B 1
2. B 1
3. C 1
4. B 1
5. D 1
6. D 1
7. D 1
8. C 1
9. D 1
10. C 1
11. B 1
12. A 1
13. B 1
14. A 1
15. D 1
16. C 1
17. (i)conversion of propene to propanol-hydroboration oxidation reaction or any 1
other suitable method
H2O/H+ 1
(ii)CH3COCH3 + CH3MgBr→ (CH3)3COMgBr-----------------→C(CH3)3OH
18. Δ Tb(obs) =100.180C – 1000C = 0.180C ½
Δ Tb = Kb × m ½
Δ Tb(calculated)=0.512×1=0.512 ½
i=Δ Tb(observed)/Δ Tb(calculated) = 0.180/0.512 = 0.35 ½
OR
α = 0.5, n=2, α = i-1/n-1 ½
0.5= i-1/2-1 i=1.5 ½
Δ Tf = i×Kf×m ½
= 1.5×1.86×1=2.79 K
Freezing point of solution =273-2.79 =270.21K ½
19. (i)Activated complex:It is defined as unstable intermediate formed between 1
reacting molecules.It is highly unstable and readily changes into product.
(ii)Collision frequency:The number of collisions that take place per second
per unit volume of the reaction mixture is called collision frequency. 1
20. (i)Ti4+does not have unpaired electron,therefore can’t absorb energy from 1
visible region and radiate colour,V4+ has one unpaired electron,undergoes d-d
transition by absorbing high from visible region and radiate violet colour.
(ii) It is due to similarity in atomic and ionic size,there is more horizontal 1
similarity.Secondly,in transition elements incoming electron goes to inner
shell(d-orbitals),whereas in main group elements,the innermost electron goes
=to outermost shell.
21. (i)[Co(NH3)5Cl]SO4 + BaCl2(aq) → [Co(NH3)5Cl]Cl2 + BaSO4(white ppt) 1
[Co(NH3)5SO4]Cl +AgNO3(aq)→ [Co(NH3)5SO4]NO3 + AgCl(white ppt)
(ii) Cl- being a weak ligand does not allow pairing to take place and so its 1
complex is paramagnetic whereas CO being a strong ligand allows pairing to
take place and its complex is diamagnetic.
22. Log K2/K1=Ea/2.303R[T2-T1/T1T2] ½
Ea=2.303×8.314×313×293×log 4/20 ½
Ea=19.147×313×293×0.6021/20×1000 1
=52.86 kJ mol-1 1
0
23. Λ HCOOH= Λ 0H+ + Λ 0HCOO- ½
=349.6 + 54.6
= 404.2 S cm2 mol-1 ½
α = Λ m/ Λ m0 = 46.1/404.2 = 0.114 ½
α=0.114 × 100 = 11.4 % ½
Ka = cα2/ 1-α = 0.025×(0.114)2/ 1- 0.114 ½
=3.67× 10-4 mol L-1 ½
24. PA0- PA/ PA0 = xB = wB ×MA/ MB ×WA ½
1- PA/PA0 = 20×18/180×500 ½
PA/PA0= 1- 1/250 =249/250 1
PA=17.536 ×249/250 1
=17.465 mm
25. 4FeCr2O4 + 8Na2CO3 + 7O2 → 8 Na2CrO4 + 2Fe2O3 + 8 CO2 1
2 Na2CrO4 + 2H+ → Na2Cr2O7 + NaCl ½
Na2Cr2O7 + 2KCl → K2Cr2O7 + 2NaCl ½
Effect of pH:
Cr2O72- + 2OH- → 2CrO42- + H2O 1
(orange) (yellow)
(dichromate ion) (chromate ion)
26. IUPAC name-Hexacyanidoferrate(III) ; hybridisation- d2sp3 ½+½
Ambidentate ligand-Ligand which can ligate through two different atoms is
called ambidentate ligand. 1
Chelating ligand-When a di- or polydentate uses its two or more donor atoms
to bind a single metal ion,it is said to be a chelating ligand. 1
27. (i) CH3CH2CH2ONa + CH3CH2CH2I → CH3CH2CH2OCH2CH2CH3 + NaI ½+½
Sodium propoxide propoxypropane
(ii)Sodium phenoxide + C2H5I → ethoxybenzene + NaI ½+½
(iii)C(CH3)3ONa + CH3I → C(CH3)3OCH3 + NaI ½+½
Sodium tert-butoxide 2-Methoxy-2-methylpropane
OR
(i)First step involves protonation of Propan-1-ol to get protonated propan-1-ol 1
(ii) Protonated propanol reacts with another molecule of propan-1-ol to form 1
protonated ether
(iii)Protonated ether loses proton to form 1-propoxypropane 1
Note:-All the reactions involved in mechanism to be written with proper
arrow movement.
28. Dil.Ba(OH)2
(i) 2CH3COCH3 ---------------------------→C(CH3)2(OH)CH2COCH3 1
Zn(Hg)/conc.HCl
(ii) Acetophenone -----------------------→Ethyl benzene + H2O 1
Pd/BaSO4
(iii) C6H5COCl + H2---------------------------→ C6H5CHO + HCl 1
29. (a)Tert.butyl chloride has branching,minimum surface area,less van der 1
waals’forces of attraction as compared to n-butyl chloride.
(b) It is because of steric hindrance,nucleophile cannot attack easily. 1
(c) 1-Bromo-2-methyl butane is more reactive because carbocation formed 2
will rearrange to form tert.carbocation which is more stable.
OR
(i)4-Chloro-pent-2-ene 1
(ii) CH3CH(Cl)CH2CH2CH3 is chiral 1
30. (a) 4-hydroxy-3-methoxy benzaldehyde 1
(b) structure of 2-hydroxy benzenecarbaldehyde 1
OH- H+
(c)C6H5CHO + CH3CHO ------------------→C6H5CH(OH)CH2CHO-----------→
Δ 2
C6H5CH=CHCHO
OR
(c)(i) 4-Methylpent-3-en-2-one is formed. 1
Ba(OH)2 Δ
2CH3COCH3------------------→(CH3)2C(OH)CH2COCH3 -----------→
(CH3)2C=CHCOCH3
P2O5 , Δ
(ii) 2CH3COOH--------------------→ (CH3CO)2O + H2O 1
31. (a) H2-O2 cell is an example of fuel cell.Fuel cells are those cells in which 1
chemical energy of a fuel is converted into electrical energy.
Advantages: (i)Its efficiency is higher 1
(ii) it does not create pollution
(b) It is a type of secondary cell which has nickel hydroxide and metallic 1
cadmium as electrodes.
Cd(s) + 2 Ni(OH)3(s) ----------→ CdO(s) + 2Ni(OH)2(s) +H2O(l) 1
(c ) It is because NH3 escapes and products cannot be converted into reactants 1
back.
OR
(a) I=? , m= Z× I × t ½
1.2 g = 12×I×60×60 ½
I = 1.2×96500/ 12×60×60 ½
= 965/360 = 2.68 A ½
(b)NaCl(aq)---------→ Na+ + Cl-
H2O ------------→ H+ + OH-
At cathode: 2H+ + 2e- ---------→ H2(g) ½
At anode : 2Cl- - 2e- ---------→ Cl2(g) ½
It is because E0 H+/ H2 is higher than E0 Na+/Na and Cl2 is evolved at anode as 1
O2 will need overpotential.
(c ) Cr2O72- + 14H+ + 6e- ---------→ 2Cr3+ + 7H2O 1
32. (a)RX+NH3-----→RNH2----→R2NH -------→R3N------→R4N+X- 1
(Ammonolysis reaction)
Coupling reaction-Benzene diazonium salt +aniline--→p-Amino azo benzene 1
Benzene diazonium salt + phenol-------→p-hydroxy azo benzene
NOTE-Marks to be awarded for writing reactions .
(b)Hinsberg reagent is used to distinguish between primary,secondary and
tertiary amine
RNH2 + C6H5SO2Cl ---------→RNHSO2C6H5 (soluble in alkali) 1
R2N + C6H5SO2Cl---------→R2NSO2C6H5(insoluble in alkali) 1
R3N + C6H5SO2Cl--------→ No reaction 1
OR
(a)(i) Carbylamine reaction or any other suitable test.(Chemical equation ½ +1/2
should be mentioned)
(ii) Hinsberg reagent test.(Chemical Equation should be mentioned) 1/2 +1/2
(b)(i)RCONH2 + Br2 +KOH --------→RNH2 +KBr+K2CO3+H2O 1
NaNO2 + HCl(00 -50C)
(ii) Aniline --------------------------------→ benzene diazonium 1
chloride(chemical reaction to be written)
1
(iii)Phthalimide + KOH------→Pot.phthalimide+RX--------→N-alkyl
phthalimide+H2O/OH--------→ Salt of phthalic acid +primary amine
(reaction to be mentioned)
33. (a)Essential amino acids are those amino acids which must be part of our 2
diet.They are not synthesized by our body,e.g.valine,leucine,isoleucine,phenyl
alanine,etc.
Non-essential amino acids are those amino acids which are produced in our
body ,e.g.glycine,alanine,serine,cystine,etc.
(b)(i)The -CONH- bond formed by condensation reaction between amino 1
acids in proteins and polypeptides is called peptide linkage.
(ii) The sequence in which amino acids are linked with each other in 1
polypeotide chain forms primary structure.It does not change on denaturation.
(iii)The process in which secondary and tertiary structures of protein are
destroyed but the primary structure remains the same is called denaturation of 1
proteins.It is brought about by heating,freezing,change in pH,etc.
OR
(a) Deficiency of Vit.A- xerophthalmia,night blindness ½
Deficiency of vit.c-scurvy,bleeding of gums ½
Sources of vit.A-carrot,fish liver oil etc ½
Sources of vit.c-citrus fruits,amla etc ½
(b) Nucleic acids are the polymers of nucleotides containing ribose or 2
deoxyribose as sugar,heterocyclic base and phosphoric acid.
Their functions are:
(i)help in synthesis of proteins
(ii)responsible for heredity
(c ) Nucleoside is formed of sugar and base and nucleotide is formed of
sugar,base and phosphoric acid. (1)
