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 �

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