Paper 2013, Set-3 “Delhi” Class XII-Science : Chemistry, Boardstudymoz.com/files/Exam...
Transcript of Paper 2013, Set-3 “Delhi” Class XII-Science : Chemistry, Boardstudymoz.com/files/Exam...
Class XII-Science : Chemistry, BoardPaper 2013, Set-3 “Delhi”
Ge ne ral Instructio ns:Ge ne ral Instructio ns:
(i) All questions are compulsory
(ii) Question numbers 1 to 8 are very short-answer questions and carry 1 mark each.
(iii) Question numbers 9 to 18 are short-answer questions and carry 2 marks each.
(iv) Question numbers 19 to 27 are also short-answer questions and carry 3 marks.
(v) Question numbers 28 to 30 are long-answer questions and carry 5 marks each.
(vi) Use Log Tables, if necessary. Use of calculators is not allowed.
Question 1
Question 1
Questions
Q 1Q 1
What type of substances�
would make better Permanent Magnets, Ferromagnetic or Ferrimagnetic ?�
( 1)( 1)
Solution:
Ferromagnetic substance�
would make better permanent magnets because when the ferromagnetic�
substance is placed in a magnetic field, all domains get oriented in�
the direction of magnetic field and a strong magnetic effect is�
produced.
Q 2Q 2
Arrange the following in increasing order of their basic strength in aqueous solution: ( 1)( 1)
CH3.NH2, (CH3)3N, (CH3)2NH
Solution:
(CH3)3N < CH3NH2 < (CH3)2NH
Q 3Q 3
What is the composition�
of ‘Copper matte’? ( 1)( 1)
Solution:
Composition of ‘Copper�
matte’ is Cu2S and FeS.
Q 4Q 4
What is the covalency�
of nitrogen in N2O5? ( 1)( 1)
Solution:
In N2O5,�
the covalency of N is restricted to 4 due to sp2�
hybridisation of nitrogen atom involving one 2s and three 2p�
orbitals.
Q 5Q 5
What is a glycosidic�
linkage? ( 1)( 1)
Solution:
The linkage between the�
two monosaccharide units through oxygen atom accompanied by the loss�
of a water molecule is called glycosidic linkage.
Q 6Q 6
Write the IUPAC name�
of(CH3)2 CH.CH(Cl)CH3. ( 1)( 1)
Solution:
Q 7Q 7
Which compound in the�
flowing pair undergoes faster SN1 reaction? ( 1)( 1)
Solution:
The carbocation formed�
when compounds I and II undergo SN1 reaction are shown�
below:
As 3° carbocation�
is more stable than 2° carbocation, hence compound I undergoes�
faster SN1 reaction.
Q 8Q 8
Write the structure of�
p-Methylbenzaldehyde molecule. 11
Solution:
Q 9Q 9
What is the difference�
between multi-molecular and macromolecular colloids? Give one example�
of each. ( 2)( 2)
Solution:
Difference between�
Multi-molecular and macromolecular colloids: �
P o int o fP o int o f
Diffe re nceDiffe re nce
Multi-mo le cular Co llo idMulti-mo le cular Co llo id Macro mo le cular Co llo idMacro mo le cular Co llo id
Definition When a large number of atoms
or small molecules�
(having diameters of less than
1nm) of a substance combine�
together in a dispersion
medium to form aggregates
having size in�
the colloidal range, the
colloidal solutions thus formed
are�
called multimolecular colloids.
When substances which have
very high molecular�
masses are dispersed in
suitable dispersion medium, the
resulting�
colloidal solutions are known as
macromolecular colloids.
Example Gold sol, Sulphur sol. Starch, Cellulose.
Q 10Q 10
(a) Which metal in the�
first transition series (3d series) exhibits + 1 oxidation state most�
frequently and why? ( 2)( 2)
(b) Which of the�
following cations are coloured in aqueous solutions and why?
Sc3+, V3+,�
Ti4+, Mn2+
(At. Nos. Sc = 21, V =�
23, Ti = 22, Mn = 25)
Solution:
(a) Cu is the only�
metal in the first transition series (3d series) which shows +1�
oxidation state most frequently. This is because the electronic�
configuration of Cu is 3d10 4s1 and after�
losing one electron it acquires a stable 3d10�
configuration.
( b)( b) The color of�
cations is dependent on the number of unpaired electrons present in�
d-orbital. The electronic configuration of the following cations is�
as follows:
Sc (Atomic number 21) =�
3d1 4s2 and Sc3+ = 3d0�
4s0 . As d-o rbital is e mpty, it is co lo urle ss.As d-o rbital is e mpty, it is co lo urle ss.
V (Atomic number 23) =�
3d3 4s2 and V3+ = 3d2�
4s0 . As d-o rbital is having 2 unpaire d e le ctro ns, it�As d-o rbital is having 2 unpaire d e le ctro ns, it�
unde rgo e s d-d transitio n and sho ws gre e n co lo ur.unde rgo e s d-d transitio n and sho ws gre e n co lo ur.
Ti = (Atomic number 22)�
= 3d2 4s2 and Ti4+ = 3d0�
4s0 . As d-o rbital is e mpty, it is co lo urle ss.As d-o rbital is e mpty, it is co lo urle ss.
Mn = (Atomic number 25)�
= 3d5 4s2 and Mn2+ = 3d5�
4s0 . As d-o rbital is having 5 unpaire d e le ctro ns, it�As d-o rbital is having 5 unpaire d e le ctro ns, it�
sho ws pink co lo r.sho ws pink co lo r.
Q 11Q 11
What happens when ( 2)( 2)
(i) PCl5�
is heated?
(ii) H3PO3�
is heated?
Write the reactions�
involved.
Solution:
(i) PCl5 on heating gives�
PCl3 and Cl2:
(ii) H3PO3�
on heating gives orthophosphoric acid and phosphine:
Q 12Q 12
18 g of glucose,�
C6H12O6 (Molar Mass = 180 g mol−1)�
is dissolved in 1 kg of water in a sauce pan. At what temperature�
will this solution boil? ( 2)( 2)
(Kb for�
water = 0.52 K kg mol−1, boiling point of pure water�
= 373.15 K)
Solution:
w1 = weight�
of solvent (H2O) = 1 kg and w2 = weight of�
solute (C6H12O6) = 18 gm
M2 = Molar�
mass of solute (C6H12O6) = 180 g mol−1
Kb = 0.52 K�
Kg mol−1
Q 13Q 13
Explain the mechanism�
of the following reaction: ( 2)( 2)
Solution:
The mechanism of the�
reaction is given below:
Q 14Q 14
(a) Give an example of�
zone refining of metals.
(b) What is the role of�
cryolite in the metallurgy of aluminium? ( 2)( 2)
Solution:
( a)( a) Zo ne �Zo ne �
R e fining o f me tals:R e fining o f me tals: This method is used for production of�
semiconductor and other metals of very high purity like germanium,�
silicon, boron, gallium and indium.
( b)( b) R o le o f�R o le o f�
cryo lite in me tallurgy o f Aluminium: cryo lite in me tallurgy o f Aluminium: Cryolite is added to lower�
the melting point of mixture and to increase the conductivity of�
electrolyte.
Q 15Q 15
Write the dispersed�
phase and dispersion medium of the following colloidal systems: 22
(i) Smoke
(ii) Milk
O RO R
What are lyophilic and�
lyophobic colloids? Which of these sols can be easily coagulated on�
the addition of small amounts of electrolytes?
Solution:
(i) Dispersed phase in�
smoke: Solid and dispersion medium in smoke: Gas
(ii) Dispersed phase in�
milk: Liquid Fat and dispersion medium in milk: Water
O RO R
Lyo phillic co llo ids:Lyo phillic co llo ids: It is made�
up of two words; ‘Lyo’ meaning liquid and ‘Phillic’�
meaning loving, so those colloids which are attracted by the liquid�
(solvent), are called as lyophillic colloids. These are also called�
reversible sols. These are quite stable and cannot be easily�
coagulated.
Lyo pho bic co llo ids:Lyo pho bic co llo ids: It is made�
up of two words; ‘Lyo’ meaning liquid and ‘Phobic’�
meaning repelling, so those colloids which are repelled by the liquid�
( solvent), are called as lyophobic colloids. These are also called�
irreversible sols and these are unstable and can be easily coagulated�
due to lack of protecting layer around charged colloidal particles,�
they easily form cluster. Hence, they got easily coagulated on�
addition of small amount of electrolyte.
Q 16Q 16
The conductivity of�
0.20 M solution of KCl at 298 K is 0.025 S cm−1.�
Calculate its molar conductivity. ( 2)( 2)
Solution:
k (S cm−1)�
= 0.025 S cm−1 and molarity (mol L−1)�
= 0.20 M
Q 17Q 17
Account for the�
following:
(i) The C−Cl�
bond length in chlorobenzene is shorter than that in CH3 −�
Cl.
(ii) Chloroform is�
stored in closed dark brown bottles. ( 2)( 2)
Solution:
(i) This is due to�
partial double bond character of C-Cl bond (due to resonance in�
C6H5Cl).
(ii) Chloroform in the�
presence of air gets oxidised to phosgene. Phosgene is carbonyl�
chloride & is represented as COCl2. To prevent the�
formation of phosgene, chloroform is stored in dark coloured bottles.�
The reaction is represented as CHCl3 + 1/2 O2 ®�
COCl2 + HCl
Q 18Q 18
How will you convert:�
( 2)( 2)
(i) Propene to�
Propan-1-ol?
(ii) Ethanal to�
Propan-2-ol?
Solution:
(i)
(ii)
Q 19Q 19
After watching a�
programme on TV about the adverse effects of junk food and soft�
drinks on the health of school children, Sonali, a student of Class�
XII, discussed the issue with the school principal. Principal�
immediately instructed the canteen contractor to replace the fast�
food with the fibre and vitamins rich food like sprouts, salad,�
fruits etc. This decision was welcomed by the parents and the�
students. ( 3)( 3)
After reading the above�
passage, answer the following questions:
(a) What value are�
expressed by Sonali and the Principal of the school?
(b) Give two examples�
of water-soluble vitamins.
Solution:
(a) The values showed�
by Sonali are awareness regarding adverse effect of junk food and�
concern for the health of her school mates.
The value showed by the�
principal is responsible behavior in listening to Sonali’s�
views and taking prompt action in replacing junk food with healthy�
food.
(b) The water soluble�
vitamins are vitamin B-complex and vitamin C.
Q 20Q 20
How would you account�
for the following? ( 3)( 3)
(i) Transition metals�
exhibit variable oxidation states.
(ii) Zr (Z = 40) and Hf�
(Z = 72) have almost identical radii.
(iii) Transition metals�
and their compounds act as catalyst.
O RO R
Complete the following�
chemical equations:
(i) �
ii.
(iii) �
Solution:
(i) The variable�
oxidation states of transition elements are due to the participation�
of ns and (n−1)d-electrons in bonding.�
Lower oxidation state is exhibited when ns-electrons take part�
in bonding. Higher oxidation states are exhibited when (n −�
1) d-electrons take part in bonding.
(ii)�This is because the atomic radii of 4d and 5d transition elements are�nearly same. This similarity in size is consequence of lanthanide�contraction. Because of this lanthanide contraction the radii of Hf�becomes nearly equal to that of Zr.(iii)�Transition elements act as good catalyst in chemical reaction because�they can lend electrons or withdraw electrons from the reagent,�depending on the nature of the reaction. The ability of transition�metals to be in a variety of oxidation states, the ability to�interchange between the oxidation states and the ability to form�complexes with the reagents and be a good source for electrons make�transition metals good catalysts.O RO R
Complete the following�
equations:
(i)�
(ii)�
2CrO42- + 2H+ ® Cr2O7
2-�
+ H2O
(iii)�
2MnO4- + 5C2O4
2-�
+ 16H+ ® 2Mn2+ + 10CO2 + 8H2O
Q 21Q 21
(a) Which one of the�
following is a food preservative? ( 3)( 3)
Equanil, Morphine,�
Sodium benzoate
(b) Why is bithional�
added to soap?
(c) Which class of�
drugs is used in sleeping pills?
Solution:
(a) Sodium benzoate is�
used as a food preservative whereas equanil is a tranquillizer and�
morphine is a narcotic analgesic.
(b) Bithional is an�
antiseptic so it is added to soaps to reduce odours producing�
bacterial decomposition of organic matter on the skin.
(c) Tranquillizers�
relieve stress, fatigue by inducing sense of well being, so they are�
used in the making of sleeping pills.
Q 22Q 22
(a) What type of�
semiconductor is obtained when silicon is doped with boron? ( 3)( 3)
(b) What type of�
magnetism is shown in the following alignment of magnetic moments?
(c) What type of point�
defect is produced when AgCl is doped with CdCl2?
Solution:
(a) When silicon is�
doped with boron, p-type semiconductor is obtained.
(b) The magnetism shown�
in the alignment of magnetic moments is ferromagnetism.
(c) Impurity defect is�
produced when AgCl is doped with CdCl2.
Q 23Q 23
Give the structures of�
products A, B and C in the following reactions:
(i) �
(ii) �
( 3)( 3)
Solution:
(i)�
(ii)�
Q 24Q 24
Write the IUPAC names�
of the following coordination compounds: ( 3)( 3)
(i) [Cr(NH3)3Cl3]
(ii) K3[Fe(CN)6]
(iii) [CoBr2(en)2]+,�
(en = ethylenediamine)
Solution:
IUPAC Nomenclature:
(i)�Triamminetrichlorochromium (III)(ii)�Potassium hexacyanoferrate (III)(iii) Dibromidobis�
(ethane-1, 2-diammine) cobalt (III) ion
Q 25Q 25
Determine the osmotic�
pressure of a solution prepared by dissolving 2.5 × 10−2�
g of K2SO4 in 2L of water at 25°C, assuming�
that it is completely dissociated. ( 3)( 3)
(R = 0.0821 L atm K−1�
mol−1, Molar mass of K2SO4 =�
174 g mol−1)
Solution:
w2 = 2.5 �
10−2 g (Mass of K2SO4) and M2�
= 174 g mol−1 (Molar mass of K2SO4)
V = 2L, R = 0.0821 L�
atm K−1 mol−1 and T = 25°C =�
298 K
Osmotic pressure, p =�
Q 26Q 26
Calculate the emf of�
the following cell at 298 K: ( 3)( 3)
Solution:
At ano de :At ano de : Fe ®�
Fe2+ + 2e-
At catho de :At catho de : 2H+�
+ 2e- ® H2
So, total number of�
electrons (n) transferred = 2
Given: Eoce ll�
= +0.44 Volt
Temperature (T) = 298 K
Therefore, Ece ll�
= 0.44(−0.02955 × − 3) = 0.44 + 0.08865 = 0.53�
Volt.
Q 27Q 27
Write the names and�
structures of the monomers of the following polymers:
(i) Bakelite
(ii) Nylon-6
(ii) Polythene ( 5)( 5)
Solution:
(i)
(ii)
(iii)
Q 28Q 28
(a) Give reasons for�
the following:
(i) Bond enthalpy of F2�
is lower than that of Cl2.
(ii) PH3 has�
lower boiling point than NH3.
(b) Draw the structures�
of the following molecules:
(i) BrF3
(ii) (HPO3)3
(iii) XeF4 ( 5)( 5)
O RO R
(a) Account for the�
following:
(i) Helium is used in�
diving apparatus.
(ii) Fluorine does not�
exhibit positive oxidation state.
(iii) Oxygen shows�
catenation behavior less than sulphur.
(b) Draw the structures�
of the following molecules:
(i) XeF2
(ii) H2S2O8
Solution:
( a)( a)
(( i) Bond�
enthalpy of F2 is lower than that of Cl2�
because F atom is small in size and due to this the electron-electron�
repulsions between the lone pairs of F-F are very large. Thus, the�
bond dissociation energy of F2 is lower than that of Cl2.
(ii) PH3 has�
lower boiling point than NH3 because NH3 molecule�
possess intermolecular hydrogen bondings which binds them strongly�
whereas PH3 has weaker Vander Waal’s forces. Thus,�
PH3 has lower boiling point than NH3.
( b)( b) The�
structures of following molecules are as follows:
(i) BrF3,�
Bent T-shape
(ii) (HPO3)3,�
cyclic structure
(iii) XeF4,�
Square planar
O RO R
(a)(i)�Helium mixed with oxygen under pressure is given to sea-divers for�respiration. Air is not given to sea-divers because nitrogen present�in air being soluble in blood will give a painful sensation called�bends by bubbling out blood on moving from high pressure(in deep sea)�to the atmospheric pressure.(ii)�Fluorine being the most electronegative atom does not exhibit�positive oxidation state because the electrons in fluorine are�strongly attracted by the nuclear charge because of small size of�fluorine atom and therefore, removal of an electron is not possible.(iii)�Sulphur shows catenation behavior more than that of oxygen because�the oxygen atom is smaller in size as compared to sulphur, the O-O�bonds in oxygen experiences repulsions due to the lone pairs present�on oxygen atom and therefore, are weaker as compared to the S-S�bonds.(b)�The structure of following molecules are as follows:(i)�XeF2, Linear
(ii)�H2S2O8
Q 29Q 29
(a) Although phenoxide�
ion has more number of resonating structures than Carboxylate ion,�
Carboxylic acid is a stronger acid than phenol. Give two reasons.
(b) How will you bring�
about the following conversions?
(i) Propanone to�
propane
(ii) Benzoyl chloride�
to benzaldehyde
(iii) Ethanal to�
but-2-enal ( 5)( 5)
O RO R
(a) Complete the�
following reactions:
(i) �
(ii) �
(iii) �
(b) Give simple�
chemical tests to distinguish between the following pairs of�
compounds:
(i) Ethanal and�
Propanal
(ii) Benzoic acid and�
Phenol
Solution:
(a) On losing a proton, carboxylic�
acids forms carboxylate ion and phenol forms phenoxide ion as�
follows:
RCOO-
Carboxylate
ion
Phenoxide
ion
Now, the negative charge is delocalized�
in both molecules as follows:
The conjugate base of carboxylic acid�
has two resonance structures in which negative charge in delocalized�
over two oxygen atoms (since O is more electronegative than C) which�
stabilizes the carboxylate ion.
On the other hand, in phenoxide ion the�
charge is delocalized over entire molecule on the less�
electronegative atom (Carbon), thus resonance of phenoxide is not�
important in comparison to resonance in carboxylate ion.
Further, in carboxylate ion the�
negative charge is effectively delocalized over two oxygen atoms�
whereas it is less effectively delocalized over one oxygen atom and�
less electronegative carbon atom.
Thus, Phenol is less acidic than�
carboxylic acids. In other words, carboxylic acids are stronger acids�
than phenol.
( b)( b)
(i) Conversion of�
Propanone to Propane:
(ii) Conversion of�
Benzoyl chloride to benzaldehyde:
(iii) On treatment with�
dilute alkali, ethanol produces 3-hydroxybutanal gives But-2-enal on�
heating.
O RO R
( a)( a)
(i) It is an example of�
Canniz aro re actio n.Canniz aro re actio n.
(ii)�It is an example of Hell-Volhard-Zelinsky reaction.
(iii)
( b)( b)
(i) Distinguish test�
between ethanal and propanal:
Io do fo rm Te st:Io do fo rm Te st:�
Ethanal gives iodoform test.
CH3CHO +�
4NaOH + 3I2 ® CHI3 (Yellow ppt.) + HCOONa�
+ 3NaI + 3H2O
Propanal does not give�
this test.
CH3CH2CHO�
+ 4NaOH + 3I2 ® No Reaction.
(ii) Distinguish test�
between Benzoic acid and Phenol:
NaHCONaHCO 33 ��
Te st:Te st: When Benzoic acid reacts with NaHCO3, brisk�
effervescence of CO2 gas evolved.
Phenol does not give�
this test.
C6H5OH�
+ NaHCO3 ® No Reaction.
Q 30Q 30
(a) A reaction is�
second order in A and first order in B.
(i) Write the�
differential rate equation.
(ii) How is the rate�
affected on increasing the concentration of A three times?
(iii) How is the rate�
affected when the concentrations of both A and B are doubled?
(b) A first order�
reaction takes 40 minutes for 30% decomposition. Calculate t1/2�
for this reaction. (Given log 1.428 = 0.1548) ( 5)( 5)
O RO R
(a) For a first order�
reaction, show that time required for 99% completion is twice the�
time required for the completion of 90% of reaction.
(b) Rate constant ‘k’�
of a reaction varies with temperature ‘T’ according to�
the equation:
Where Ea is�
the activation energy. When a graph is plotted for �
�
a straight line with a slope of −4250 K is obtained. Calculate�
‘Ea’ for the reaction.
(R = 8.314 JK−1�
mol−1)
Solution:
( a)( a)
(i) A reaction is�
second order in A and first order in B.
Differential rate�
equation:−
(ii) On increasing the�
concentration of A three times i.e. 3A:
,�
i.e. 9 times the initial rate.
(iii) On increasing the�
concentration of A and B as 2A and 2B:
,�
i.e. 8 times the initial rate.
( b)( b)
Now, it takes 40 min�
for 30% decomposition i.e. reactant left after 40 min is 70% of its�
initial concentration.
O RO R
( a)( a) For a first�
order reaction,
Case 1:Case 1: If ‘t’�
is the time required for 99% completion then x = 99% of a
Case 2: Case 2: If ‘t’�
is the time required for 90% of completion then x = 90% of a
Therefore, the time�
required for 99% completion of 1s t order reaction is twice�
the time required for 90% completion.
( b) ( b) �
Ea ®�
Activation energy
The above equation is�
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like y = mx + c where if we plot y v/s x we get a straight line with�
slope ‘m’ and intercept ‘c’.
So, slope is equal to �