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CBSE-XII-2017 EXAMINATION CBSE-XII-2015 EXAMINATION

CHEMISTRY
Paper & Solution Code: A
Time: 3 Hrs. Max. Marks: 100

Q 1. Write the formulae of any two oxoacids of phosphorus.

Answer:
(i) H3 PO3 (Phosphorous acid)
(ii) H3 PO4 (Phosphoric acid)

Q 2. Which would undergo SN2 reaction faster in the following pair :

Answer:

C6 H5  CH2  CH2  Br (due to less stearic hindrence)

Q 3. Out of AlCl3 and NaCl, which is more effective in causing coagulation of a negative sol and why ?
Answer:

AICI3 is more effective due to high Positive Charge density on Al 3 as compared to Na

Q 4. Write the formula of a compound in which the element Y forms ccp lattice and atoms of X occupy 1/3 rd
of tetrahedral voids.
Answer:
The no. of atoms Y in CCP = N no. of tetrahedral voids = 2N
1
no.of X atoms  2 N 
3
2
 N
3
x : y
Formula
2
N : N
3

2 : 3
 x2 y3

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 Study Materials
NCERT Solutions for Class 6 to 12 (Math & Science)
Revision Notes for Class 6 to 12 (Math & Science)
RD Sharma Solutions for Class 6 to 12 Mathematics
RS Aggarwal Solutions for Class 6, 7 & 10 Mathematics
Important Questions for Class 6 to 12 (Math & Science)
CBSE Sample Papers for Class 9, 10 & 12 (Math &
Science)
Important Formula for Class 6 to 12 Math
CBSE Syllabus for Class 6 to 12
Lakhmir Singh Solutions for Class 9 & 10
Previous Year Question Paper
CBSE Class 12 Previous Year Question Paper
CBSE Class 10 Previous Year Question Paper
JEE Main & Advanced Question Paper
NEET Previous Year Question Paper

Vedantu Innovations Pvt. Ltd.

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 CBSE-XII-2015 EXAMINATION

Q 5. Write the IUPAC name of the given compound :

Answer:
2, 2 - Dimethylpropan-1-ol

Q 6. Why do transition elements show variable states ? How is the variability in oxidation states of d-block
different from that of the p-block elements ?
Answer:
Transition elements show variable oxidation states since in these elements there is a small energy difference
between 3d and 4s orbital therefore 3d and 4s both e. participate in the bond formation hence transtion
elements shows variable oxidation states. Oxidation State of P block elements both type that is positive and
negative but d block elements shows only positive oxidation states.

Q 7. (i) Write down the IUPAC name of the following complex :

 Pt ( NH3 )(H2O)CI2 
(ii) Write the formula for the following complex :
tris(ethane-1,2-diamine)chromium(III) chloride
Answer:
(i) Ammineaquadichloridoplatinum(II)
(ii) Cr(en)3  Cl3

Q 8. Calculate the time to deposit 1.5 g of silver at cathode when a current of 1.5A was passed through the
solution of AgNO3. (Molar mass of Ag = 108 g mol-1, 1 F = 96500 C mol-1)
Answer:

Ag   e  
 Ag ( s )
108 gm Ag required  96500 C
96500
 1.5 gm Ag required  1.5
108
 1340.2 C
Q
t
i
1340.2
t
1.5
t  893.5 sec.

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 CBSE-XII-2015 EXAMINATION

Q 9. Write the reagents used in the following reactions:

(i) C6 H5  CO  CH3 
?
 C6 H5  CH 2  CH3
(ii) CH3  COOH 
?
 CH3  COCI

OR
Arrange the following compounds in increasing order of their property as indicated :

i  CH3CHO, C6 H5CHO, HCHO

(reactivity towards nucleophilic addition reaction)
(ii) 2,4-dinitrobenzoic acid, 4-methoxybenzoic acid, 4-nitrobenzoic acid (acidic character)
Answer:
(i) Zn . Hg / HCl
(ii) PCl5
Or
(i) C6H5CHO< CH3CHO< HCHO
(ii) 4-methoxybenzoic acid < 4-nitrobenzoic acid < 2, 4-dinitrobenzoic acid

Q 10. (i) Why are aquatic species more comfortable in cold water than in warm water ?
(ii) What happens when we place the blood cell in saline water solution (hypertonic solution) ?Given reason.
Answer:
(i) aquatic species ismore comfortable in coldwater than inwarmwater it is due to themore dissolution of O 2
in cold water
(ii) Blood cell will be squeeze.
 dissolution of gases is a exothermic procees means at low temperature gases more dissolve in Water

Q 11. (i) Name themethod used for the refining of titanium.

(ii) What is the role of Zn in the extraction of silver ?
(iii) Reduction of metal oxide to metal becomes easier if the metal obtained is in liquid state. Why ?
Answer:
(i) vanArkelmethod
(ii) Zn displace silver form its salt solution.
2 Ag (CN )2   Zn 
 Zn(CN )4   2 Ag( s )
 2

(iii) Since in the liquid state  S = +ve

Q 12. (i) E0 value for the Mn3+/Mn2+ couple is positive (+ 1.5 V) whereas that of Cr 3+/Cr2+ is negative (.0.4
V). Why ?
(ii) Transitionmetals form coloured compounds.Why ?
(iii) Complete the following equation :
2MnO4  16H   5C2O42 
Answer:

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 CBSE-XII-2015 EXAMINATION

3
(i) Mn Couple is positive
Mn2
Mn3 
 Mn2 due to stable half filled (d5) electronic configuration it has reduction tendency.
Cr 2 
Cr 3 have oxidation tendency
Cr 3 is more stable than Cr 2 due to stable t23g electronic Configuration.
(ii) due to d-d transition
(iii) 2MnO  16H   5C2O42 
10CO2  8H2O  2Mn2

Q 13. (i) What type of isomerism is shown by Co( NH3 )5 ONO Cl2 ?
(ii) On the basis of crystal field theory, write the electronic configuration for d 4ion if 0  P.
(iii) Write the hybridization and shape of  Fe(CN )6  .
3

Answer:
(i) it shows linkage isomerism
Co( NH3 )5 ONOCl2 and Co( NH3 )5 NO2 Cl2
(ii) If  0 < P then
d 4  t2g3
e1g
(iii)  Fe(CN )6 
3

hybridisation 
 d 2 sp3
Shape 
 actahedral

Q 14. Predict the products of the following reactions :

(i)

(a) KMnO4 / KOH

(ii) C6 H5  CH 2  CH3 ?
(b) H 
(iii)

Answer:

(ii) C6 H5  CH2  CH3 

( a ) KMnO4
(b ) H 
C6 H5  COOH

(iii)

Q 15. Write the names and structures of the monomers of the following polymers :

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 CBSE-XII-2015 EXAMINATION

(i) Nylon-6,6
(ii) Bakelite
(iii) Polystyrene
Answer:
(i) Nylon 6,6

Monomer  Adipic acid, hexamethylenediamine

 
HOOC - (CH 2 ) 4 - COOH H 2 N - (CH 2 )6 - NH 2

(ii) Bakelite

Q 16.
(i) Which one of the following is a disaccharide : starch, maltose, fructose, glucose
(ii) What is the difference between acidic amino acid and basic amino acid ?
(iii) Write the name of the linkage joining two nucleotides.
Answer:
(i) Maltose
(ii) acidic amino acid have -COOH > -NH2 group
where as basic amino acid have -NH2 > -COOH group
(iii) Phosphodiester linkage

Q 17. Vapour pressure of water at 20°C is 17.5 mm Hg. Calculate the vapour pressure of water at 20°C
when 15 g of glucose (Molar mass = 180 g mol-1) is dissolved in 150 g of water.
Answer:

P A  17.5 mm Hg WA  150 gm
WB 15 gm M A  18 gmmol1
M B  180 gm mol 1
PA  Ps WB  M A

PA M B  WA
17.5  Ps 15  18

17.5 180  150
Ps  17.325 mm Hg

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 CBSE-XII-2015 EXAMINATION

Q 18. Examine the given defective crystal :

Answer the following questions :

(i) Is the above defect stoichiometric or non-stoichiometric ?
(ii) Write the term used for the electron occupied site.
(iii) Give an example of the compound which shows this type of defect.
Answer:
(i) non . stoichiometric
(ii) F - Center (Ferb Centre)
(iii) NaCl in presence of excess Na

Q 19. How do you convert the following :

(i) Prop-1-ene to Propan-2-ol
(ii) Bromobenzene to 2-bromoacetophenone
(iii) 2-bromobutane to But-2-ene
OR
What happens when
(i) ethyl chloride is treated with NaI in the presence of acetone,
(ii) chlorobenzene is treated with Na metal in the presence of dry other
(iii) methyl chloride is treated with KNO2 ? Write chemical equations in support of your answer.
Answer:

OR
(i) C2 H5  Cl  Na l C2 H5  l  NaCl
acetone

(ii) 2C6 H5  Cl  2Na 

dryether
C6 H5  C6 H5  2NaCl
(iii) CH3  Cl  KNO2 
CH3  ONO  KCl

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 CBSE-XII-2015 EXAMINATION

Q 20. Give reasons for the following :

(i) p-nitrophenol is more acidic than p-methylphenol.
(ii) Bond length of C . O bond in phenol is shorter than that in methanol.
 
(iii) (CH3 )3 C  Br on reaction with sodium methoxide Na OCH3 gives alkene as the main product and
an ether.
Answer:

(i) P-nitrophenol is more acidic it is due to -I, -M effect of NO2 group

(ii) due to resonance bond length of C – O bond in phenol is shorter than that is methanol.

Since 3° alkyl halide gives more preference for elemination reaction as compared to substution reaction

0
Q 21. Calculate Ecell and  r G0 for the following reaction at 25°C :
A2  B   A3  B
Given : Kc  1010 ,1F  96500 C mol 1
Answer:

G  2.303 RT log KC
G  2.303  8.314  298  log 1010
G  57058.483 J mol
 5. 7 104
G  nFEcell
G
Ecell  
nF
(5.7 104 )
Ecell 
1 96500
Ecell  5.9 101V

Q 22. Define adsorptionwith an example.Why is adsorption exothermic in nature ?Write the types of
adsorption based on the nature of forces between adsorbate and adsorbent.
Answer:
Adsorption  The accumulation of molecular species at the surface rather than in the bulk of a solid or
liquid is called adsorption.

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 CBSE-XII-2015 EXAMINATION

e.g. water molecules adsorbed by silicagel

* during the adsorption process there is a force of attraction between adsorbate and adsorbent hence energy
is released.
* Two types
(a) Physisorption (b) Chemisorption.

Q 23. Seeing the growing cases of diabets and depression among young children,Mr. Lugani, the principal
of one reputed school organized a seminor in which he invited parents and principals. They all resolved this
issue by strictly banning junk food in schools and introducing healthy snacks and drinks like soup, lassi,
milk, etc. in school canteens.
They also decided to make compulsory half an hour of daily physical activities for the students in the
morning assembly.After sixmonths,Mr. Lugani conducted the health survey inmost of the schools and
discovered a tremendous improvement in the health of the students. After reading the above passage, answer
the following questions :
(i) What are the values (at least two) displayed by Mr. Lugani ?
(ii) As a student, how can you spread awareness about this issue ?
(iii) What are antidepressant drugs ? Give an example.
(iv) Name the sweetening agent used in the preparation of sweets for a diabetic patient.
Answer:
(i) (a) He is very concious towards the health of students.
(b) He is very deceplined.
(ii) As a student i am also motivate to my friend.
(iii) Phenelzine
(iv) Saccharin (ortho-sulphobenzimide)

Q 24. For the hydrolysis of methyl acetate in aqueous solution, the following results were obtained :
t/s 0 30 60
-1
(CH3 COOCH31/mol L 0 60 0 30 0 15
(i) Show that it follows pseudo first order reaction, as the concentration of water remains constant.
(ii) Calculate the average rate of reaction between the time interval 30 to 60 seconds.
OR
(a) For a reaction A + B  P, the rate is given by
Rate = k [A]2[B]
(i) How is the rate of reaction affected if the concentration of A is doubled ?
(ii) What is the overall order of reaction if B is present in large excess ?
(b) A first order reaction takes 23.1 minutes for 50% completion. Calculate the time required for
75%completion of this reaction. (Given : log 2 = 0.301, log 3 = 0.4771, log 4 = 0.6021)
Answer:
Case 1

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 CBSE-XII-2015 EXAMINATION

2.303 a
( I ) K1  log
t ax
2.303 0.60
K1  log
30 0.30
2.303
K1  log 2
30
Case 2
2.303 a
K2  log
t ax
2.303 0.60
K2  log
60 0.15
2.303
K2   2log 2
60
2.303
K2   log 2
30
K1  K2 it means it follows Pseudo first order reaction.

R
( II ) average rate  
t
 0.15  0.30 
  
 60  30 
0.15

30
1

200
 5 103 sec

OR
(a) (I)
rate1  K A B
2

rate2  K 2 A B
2

rate2  4 K A B
2

r2  4r1

(II) Second order

(b) For 50%Completion of reaction
2.303 N
K log 0
t N
N0  initial amount of the substance
N = amount left after t lime

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 CBSE-XII-2015 EXAMINATION

2.303 100
K log
23.1 50

2.303
K log 2 _________(1)
23.1
For 75%Completion of reaction
2.303 100
K log
t 25
2.303
K log 4
t
2.303
K  2log 2 _________(2)
t
Divide equation 1 and 2
2.303
log 2
23.1 K

2.303
 2log 2 K
t
t = 46.2 minute

Q 25.
(a) Account for the following :
(i) Bond angle in NH4 is greater than that in NH3.
(ii) Reducing character decreases from SO2 to TeO2.
(iii) HClO4 is a stronger acid than HClO.
(b) Draw the structures of the following :
(i) H2S2O8
(ii) XeOF4
OR
(a) Which poisonous gas is evolved when white phosphorus is heated with conc. NaOH solution ? Write tha
chemical equation.
(b) Write the formula of first noble gas compound prepared byN. Bartlett. What inspired N. Bartlett to
prepare this compound ?
(c) Fluorine is a stronger oxidizing agent than chlorine. Why ?
(d) Write one use of chlorine gas.
(e) Complete the following equation : CaF2  H2 SO4 
Answer:
(a)
(i) in NH4 and NH3 nitrogen is SP3 hybridised but due to IP  bP replusion bond angle  in NH3 but
there is no lP in NH4 hence its bond angle is 109 28' but in NH3 it is 107°
(ii) due to inert pair effect.
(iii) due to higher oxidation state of Cl and high oxygen content in HClO 4 it is more acidic then HOCl.
in HClO4 O.S of Cl = +7
in HClO O.S of Cl = +1

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 CBSE-XII-2015 EXAMINATION

OR
 a P4  3NaOH  3H2O 
 PH3  3NaH 2 PO2
* Poisonous gas is PH3 (Phosphine)

(b) Formula  Xe PtF6

N Barlett Prepared the compound O2  PtF6 . He then realised that the first jonisation enthalpy of melecular
oxygen (1175 KJ mol 1 ) was almost identical with that of xenon (1170 KJ mol 1 ) he made efforts to prepare
same type of compound with Xe and was successful in Preparing another red colour compound Xe PtF6
(c) due to low dissociation bond enthalpy and High hydration enthalpy F2 is a strong oxidising agent.
(d) Use  In sterilising drinking water.
(e) CaF2  H2 SO4  CaSO4  2HF

Q 26. An aromatic compound .A. of molecular formula C7H7ON undergoes a series of reactions as shown
below. Write the structures of A, B, C, D and E in the following reactions :

OR
(a) Write the structures of the main products when aniline reacts with the following reagents :
(i) Br2 water
(ii) HCl
(iii) (CH3CO)2 O / pyridine
(b) Arrange the following in the increasing order of their boiling point : C2 H5 NH2 , C2 H5OH , (CH3 )3 N
(c) Give a simple chemical test to distinguish between the following pair of compounds :
(CH3 )2  NH and (CH3 )3 N
Answer:
C7 H7OH

OR

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 CBSE-XII-2015 EXAMINATION

(b) C2 H5  OH  C2 H5 NH2  (CH3 )3 N

(c) (CH3 )2 NH react with Hinsberg reagent C6 H5SO2Cl 

CH3 3 N does not react with Hinsberg reagent

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