CBSE XII 2015 - CHEMISTRY
Time allowed: 3 hours
Maximum Marks : 70
General Instructions:
(i) All questions are compulsory
(ii) Questions number 1 to 5 are very short answer questions and carry 1 mark each.
(iii) Questions number 6 to 10 are short answer questions and carry 2 marks each.
(iv) Questions number 11 to 22 are also short answer questions and carry 3 marks each.
(v) Question number 23 is a value based question and carry 4 marks.
(vi) Questions number 24 to 26 are long answer questions and carry 5 marks each.
(viii) Use log tables, if necessary. Use of calculators is not allowed.
Q.1
A.1
Out of AlCl3 and NaCl, which is more effective in causing coagulation of a negative sol and why?
[1]
AlCl3 is more effective in causing coagulation of a negative sol.
A.2
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.
[1]
The formula of the compound is X2Y3.
Q.3
A.3
Write the formulae of any two oxoacids of phosphorus.
Formula of oxoacids of Phosphorus – H3PO3 and H3PO4
[1]
Q.4
Write the IUPAC name of the given compound:
CH3
[1]
Q.2
CH3—C—CH2—OH
A.4
CH3
IUPAC name of the compound is 2, 2-dimethyl propanol.
Q.5
Which would undergo SN2 reaction faster in the following pair:
A.5
C6H5—CH2—CH2—Br and C6H5—CH—CH3
|
Br
2
C6H5CH2CH2Br will undergo SN mechanism faster.
Q.6
[1]
(i) Why are aquatic species more comfortable in cold water than in warm water?
[2]
(ii) What happens when we place the blood cell in saline water solution (hypertonic solution)? Give
reason.
A.6
(i)
(ii)
Q.7
A.7
Q.8
A.8
Aquatic species are more comfortable in cold water than in warm water because according to
Henry’s law high the value of KH at a given pressure, the lower is the solubility of the gas in
the liquid.
KH value for O2 increases with increases of temperature indicating that the solubility of gases
decreases with increase in temperature.
When the blood cell is placed in saline water solution (hypertonic solution) the blood cell will
shrink due to osmosis.
Calculate the time to deposit 1.5 g of silver at cathode when a current of 1.5 A was passed through the
solution of AgNO3. (Molar mass of AB= 108 gmol–1, 1 F = 96569 C mol–1)
[2]
Given;
I = 1.5 A
Wt. of Ag deposited = 1.5g


Ag  e  Ag
M. M = 108 g
108 g of Ag is deposited by 96500 C.
 96500

1.5 g of Ag is deposited by 

1.5  C = 1340.2 C.
 108

According to Q = IT,
Q 1340.2
T= 
I
1.5
T = 893.3 sec or 14.88 min.
Why do transition elements show variable oxidation states? How is the variability in oxidation states of
d-block different from that of the p-block elements?
[2]
Transition elements show variable oxidation states due to :
(a)
Incomplete filling of d orbitals
(b)
Comparable energy levels of (n – 1)d and ns orbitals due to which electrons from both participate
in reactivity.
Oxidation states of d-Block differ from p-block in following respects :
(a)
(b)
Q.9
A.9
Due to incomplete filling of d orbitals; the oxidation states differ from each other by unity ex–
VII ; VIII ; VIV; while for p-block oxidation states differ by unit of two, ex- Cl  1,  3,  5,  7
In p-block; higher oxidation states are not favoured by heavier members (due to inert pair effect),
while opposite is true in d-block.
eg. in group 6, Mo (VI) and W(VI) are more stable than Cr (VI).
(i) Write down the IUPAC name of the following complex:
[Pt(NH3)(H2O)Cl2]
(ii) Write the formula for the following complex:
tris(ethane-1, 2-diamine) chromium(III) chloride
(i)
The IUPAC name of the complex is :Ammineaquadichloridoplatinum (II)
(ii)
The formula is :
[Cr (en)3]Cl3
[2]
Q.10 Write the reagents used in the following reactions:
?
(i) C6 H5  CO  CH3 
 C6 H5  CH2  CH3
?
(ii) CH3  COOH 
 CH3  COCl
A.10
[2]
OR
Arrange the following compounds in increasing order of their property as indicated:
(i) CH3CHO, C6H5CHO, HCHO
(reactivity towards nucleophilic addition reaction)
(ii) 2, 4 dinitrobenzoic acid, 4-methoxybenzoic acid, 4-nitrobenzoic acid (acidic character)
(i)
C6 H5  CO  CH3  C6 H5CH2CH3
Reagent used : Zn – Hg in HCl
(ii)
CH3COOH  CH3COCl
Reagent used : SOCl2 or PCl3
or
(i)
Increasing order of reactivity towards addition reaction : C6 H5CHO  CH3CHO  HCHO
(ii)
Increasing order of acidic strength :4-methoxy benzoic acid < 4-nitro benzoic acid < 2, 4 – dinitro benzoic acid.
Q.11 Vapour pressure of water at 20oC is 17.5 mm Hg. Calculate the vapour pressure of water at 20oC when
15g of glucose (Molar mass = 180 g mol–1 is dissolved in 150 g of water.
[3]
o
A.11 PH2O  17.3mm Hg
WH2O  150g
MC6H12O6  180gmol1
WC6H12O6  15g
MH2O  18gmol1
Acc. to Relative lowering in vapour pressure,
WC6H12O6 M H2O
PHo 2O  PH2O
=

o
M C6H12O6 WH2 O
PH2O
17.3  PH2O
15 18

17.3
80 150
On solving equation,
PH2O  17.1mm Hg

Q.12 Examine the given defective crystal:
X+
Y–
X+
Y–
X+
Y–
X+
Y–
X+
Y–
+
–
+
–
X
Y
X
e
X+
Y–
X+
Y–
X+
Y–
Answer the following questions:
(i) Is the above defect stoichiometric or non-stoichiometric?
(ii) Write the term used for the electron occupied site.
[3]
A.12
(iii) Give an example of the compound which shows this type of defect.
(i)
The above defect is a non-stoichiometeric defect.
(ii)
The term used for the electron occupied site is F-centre.
(iii) Example :- NaCl, KCl etc.
Q.13 Calculate E 0cell and  r G 0 for the following reaction at 25oC
A.13
A2  B 
 A3  B
Given : Kc = 1010, 1 F = 96500 C mol–1
Given,
,
IF = 96500 c mol-1
K c  1010
n = 1.
nE ocell
log K c =
0.059
1 E ocell
log1010 
0.059
o
Ecell  0.59V
[3]
o
Go  nFEcell
= -1 × 96500 × 0.59
= -56935 J mol-1
Q.14 Define adsorption with an example. Why is adsorption exothermic in nature? Write the types of
adsorption based on the nature of forces between adsorbate and adsorbent.
[3]
A.14 The phenomenon of attracting and retaining the molecule of a substance on the surface of a liquid or
a solid resulting into a higher concentration of the substance on the surface than in the bulk is called
adsorption.
eg – Adsorption of water vapours by silica gel.
During adsorption, there is always a decrease in the residual forces of the surface, i.e, there is
decrease in surface energy which appears as heat. Therefore, it is an exothermic process.
Types of adsorption based on the nature of forces between adsorbate and adsorbent are of two types
physisorption and chemisorption.
Q.15 (i) Name the method 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 in liquid state. Why? [3]
A.15 (i)
The method for refining of Titanium is –
Van Arkel’s Method (Vapour Phase Refining)
(ii)
Zinc acts as a reducing agent for extraction of silver from its complex, [Ag(CN)2]- .
(iii) Reduction of metal oxide is easier in liquid state as entropy change is higher and
consequently  G for net redox reaction is more negative.
Q.16 (i) E 0 value for the Mn3+/Mn2+ couple is positive (+ 1.5 V) whereas that of Cr3+/Cr2+ is negative
(–0.4 V). Why?
[3]
(ii) Transition metals form coloured compounds. Why?
(iii) Complete the following equation:
A.16
2MnO4  16H  5C2O42 
(i)
E° value for Mn3+ / Mn2+ couple is positive since Mn+2 is stable due to half-filled ‘d’ orbitals
5
(3d ) and has less tendency to oxidize to Mn+3 while for Cr3+ / Cr2+ E° value is negative since
Cr+2 after oxidation becomes stable as Cr+3; configuration d3 – half filled t2g level.
(ii)
Transition metals form coloured compounds because in aquous state the d orbitals undergo
splitting into two energy levels and electrons show d-d transitions which generates colour.
(iii) Equation :
Given
2MnO4  16H   5C2O42 
2MnO4  16H  5C2O24  2Mn 2  8H2O  10CO2
Q.17 (i) What type of isomerism of shown by [Co(NH3)5ONO]Cl2?
(ii) On the basis of crystal field theory, write the electronic configuration for d4 ion if 0  P.
(iii) Write the hybridization and shape of [Fe(CN)6]3–.
(Atomic number of Fe = 26)
[3]
A.17 (i)
The complex shows :
Linkage isomerism
(ii)
According to crystal field theory, for a d4 ion, if 0  P, it is a weak field, high spin situation
so fourth electron enters one of the eg orbitals and configuration is – t2g3eg1
(iii)
 Fe  CN 6 
3
Since CN- is a strong field ligand it pairs up the electrons; : hybridisation is d2sp3 and
complex has octahedral geometry.
Q.18 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
[3]
OR
A.18
What happens when
(i) ethyl chloride is treated with Nal in the presence of acetone,
(ii) chlorobenzene is treated with Na metal in the presence of dry ether,
(iii) methyl chloride is treated with KNO2?
Write chemical equations is support of your answer.
Convert the following :
(i)
Propene to Propan-2-ol
OH
|
H2O
CH3CH  CH 2 
 CH3  CH CH3
H 2SO4
(ii)
Bromobenzene to 2-bromoacetophenone
(iii)
2-Bromobutane to But-2-ene
Br
|
alc.KOH
CH3  CH  CH 2CH3 

 CH3CH  CHCH3
or
acetone
C2 H5Cl  NaI  C2 H5I  NaCl
(i)
(ii)
(iv)
CH3Cl  KNO2  CH3  O  N  O .
Q.19 Give reasons for the following:
[3]
(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
not an ether.
A.19 (i)
p-nitrophenol is more acidic than p-methyl phenol because nitro is electro with drawing,
stablises resonance and hence enhances acidic strength while methyl is electron releasing
group decreases acidic strength.
(ii)
Bond length of C – O bond in phenol is shorter than that in methanol because in phenol due
to resonance it acquired partial double bond which has shorter bond length.
(iii) (CH)3 C – Br on reaction with sodium methoxide gives alkene as main product because it
follows   elimination reaction.
Q.20 Predict the products of the following reactions:
H2N–NHCONH2
(i) CH3–C = 0
|
CH3
(a)KMnO4 / KOH
?
(ii) C6 H5  CH2  CH3 
(b)H
COOH
(iii)
conc. HNO3/H2SO4
[3]
A. 20 (i)
(ii)
O
O
||
||
NH 2  NH  C  NH 2
CH3  C  O 
 CH3  C  N  NH  C  NH 2
|
|
CH3
CH3
 
C6 H5CH2CH3 
 C6 H5COOH
 b.H
a. KMnO1KOH
(iii)
Q.21 Write the names add structures of the monomers of the following polymers :
(i) Nylon-6, 6
(ii) Bakelite
(iii) Polystyrene
A.21 Names and structures of monomers of –
(i)
Nylon – 6, 6
(ii)
[3]
Bakelite
Phenol
&
Formaldehyde HCHO
(iii)
Polystyrene
Styrene
Q.22 (i) Which one of the following is a disaccharide:
[3]
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.
A. 22 (i)
Disaccharide – Maltose
(ii)
Amino acids are classified as acidic or basic on the basis of relative number of amino &
carboxyl groups in their molecule.
Acidic amino acids – When number of carboxyl groups are more as compared to amino
groups, the amino acid is acidic.
ex – Aspartic acid
R  CH2  COOH
Basic amino acids – When number of amino groups are more than carboxyl groups, it is
basic. ex –
(iii)
The linkage joining the two nucleotides is named as –
phosphodiester linkage.
Q.23 Seeing the growing cases of diabetes and depression among young children, Mr. Lugani, the principal of
one reputed school organized a seminar in which he invited parents and principals. They all resolved this
issue by strictly banning junk food in schools an{ 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 six months, Mr. Lugani conducted the health
survey in most of the school's 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.
[4]
A.23 (i)
The values displayed by Mr. Lugani
 Humanitarian values
 Very Responsible Behaviour
(ii)
As a student, we can spread awareness in our school campus by cultural activities, dramas,
debates, seminars, notice board posters, charts and outside the campus by campaigns.
(iii)
Anti-depressant drugs are a class of Tranquilizers (neurologically active drugs) which cure
depression, stress, mental illness. example - Equanil.
(iv)
Sweetening agent – Saccharin
Q.24 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:
[5]
(C7H7ON)A
Br2 + KOH
(C6H5NH2)
NaNO2 +HCl
273K
CHCl3 + NaOH
D
B
CuCN
C
H2O
E
OR
(a) Write the structures of the main products when aniline reacts with the following reagents:
(i) Br2 water
(ii) HCl
(iii) (CH3CO)2O / pyridine
(b) Arrange the following in the increasing order of their boiling points:
C2H5NH2
C2H5OH,
(CH3)3N
(c) Give a simple chemical test to distinguish between the following pair of compounds:
(CH3)2 – NH and (CH3)3N
A.24
or
(a)
(b)
Increasing order of boiling point :  CH3 3 N  C2H5 NH2  C2H5OH
(c)
(CH3 )2  NH and (CH3 )3 N can be distinguished using Hinsberg’s reagent test.
Q.25 For the hydrolysis of methyl acetate in aqueous solution, the following results were obtained: [5]
t/s
0
30
60
[CH3COOCH3]/mol L–1
0.60
0.30
0.15
(i) Show that if 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)
A.25
(i)
As conc. of water remain constant
r  k1 CH3COOCH3 
k1 
 K1  K  H2O
CH3COOCH3 o
2.303
log
t
CH3COOCH3 t
At t = 30°,
2.303
0.60
k1 
log
30
0.30
1
2 1
k  2.303 10 s
At t = 60,
2.303
0.60
log
60
0.15
1
2 1
k  2.303 10 s
As the value of k1 at t = 30 and t = 60 is same this confirms that it follows pseudo first order
reaction.
Average rate of reaction between the time internal 30 to 60 seconds,
C  C1
0.15  0.30 0.15
= 5 103 molL1s 1
ravg   2


t 2  t1
60  30
30
k1 
(ii)
(a)
OR
For a reaction A  B  P, the rate is given by
Rate = K  A  B
(i)
If the conc. of A is doubled, the rate of reaction is increased by four times.
(ii)
If B is present in large excess, the over all order is second order.
Given,
Order = I.
t 50%  23.1min
For I order,
0.693
t 50% 
k
0.693
k
 0.03

23.1
R n  R o  0.75R o  0.25R o
2
(b)
k
Q.26 (a)
(b)
(a)
(b)
(c)
(d)
(e)
A.26
(a)
 R o
2.303
log
t
R 
t 75% =
[R ]o
2.303
log
0.03
0.25 [R ]o
t 75% 
2.303  0.60
 46.25sec.
0.03
Account for the following:
[5]

(i) Bond angle in NH 4 is greater than that in NH3.
(ii) Reducing character decreases from SO2 to TeO2.
(iii) HClO4 is a stronger acid than HClO.
Draw the structures of the following:
(i) H2S2O8
(ii) XeOF4
OR
Which poisonous gas is evolved when white phosphorus is heated with conc. NaOH solution?
Write the chemical equation.
Write the formula of first noble gas compound prepared by N. Bartlett. What inspired N. Bartlett
to prepare this compound?
Fluorine is a stronger oxidizing agent than chlorine. Why?
Write one use of chlorine gas.
Complete the following equation:
CaF2  H2SO4 
NH3 has trigonal pyramidal structure with three bond pairs with hydrogen and one lone pair
at apex which offers greater repulsion.
While NH 4 has all bond pairs with four hydrogen; hence repulsions are lesser and bond
angle greater.
(ii)
(iii)
(b)
(i)
(ii)
(a)
(b)
(c)
(d)
(e)
Reducing character decreases from SO2 to TeO2 as down the group E – O bond
enthalpy decreases and stability of the compounds decreases.
HClO4 is a stronger acid than HClO since greater the number of oxygen atoms greater
is the pull on electrons, and greater is the ease with which hydrogen liberates as a
proton (H+).
XeOF4
OR
The gas is phosphine ie PH3
P4  3NaOH  3H 2O  PH3  3NaH 2 PO2
(Sodium hypophosphite)
The formula of the first noble gas compound prepared by N. Bartlett was - Xe PtF6 .
This was based on the realisation that the first ionisation enthalpy of molecular oxygen (1175
kJ mol-1) was almost identical with that of xenon (1170 kJ mol-1)
Fluorine is a stronger oxidising agent than chlorine due to –
(i)
higher electronegativity
(ii)
low F – F bond dissociation enthalpy
(iii)
small size and high hydration enthalpy.
Use of chlorine gas:
Manufacture of poisonous gases ex – Phosgene; tear gas.
Equation ;
CaF2  H2SO4  CaSO4  2HF