cbse xii chemistry 2011 solution

7/30/2019 CBSE XII Chemistry 2011 Solution

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CBSEXII

Solution to Board Paper Set 1 (Delhi)Chemistry Theory

Solution 1: This statement means that some of the physical properties ofcrystalline solids such as electrical resistance or refractive index showdifferent values when measured along different directions in the samecrystals.

Solution 2. The Molar conductivity of a solution at a given concentration isrelated to conductivity of that solution, by the following relation.

m kA

=

Where,m

is molar conductivity and k is the conductivity of the solution.

Solution 3. Electrophoresis is the phenomenon of movement of colloidalparticles under the applied electric potential.

Solution 4. XeF2 is a linear molecule and adopts the following structure:

Solution 5.

The IUPAC name of the given structure is 2, 2-Dimethylbromopropane.

Solution 6.


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Solution 7. The solubility order of the given amines is as followsC6H5NH2 < (C2H5)2NH < C2H5NH2Reason:The more extensive the H-bonding, the higher is the solubility. C2H5NH2

contains two H-atoms whereas (C2H5)2NH contains only one H-atom. ThusC2H5NH2 undergoes more extensive H-bonding than (C2H5)2NH. Hence, thesolubility in water of C2H5NH2 is more than that of (C2H5)2NH.Further, the solubility of amines decreases with increase in the molecularmass. This is because the molecular mass of amines increase with anincrease in the size of the hydrophobic part. The molecular mass of C6H5NH2is greater than that of C2H5NH2 and (C2H5)2NH. Thus, the solubility ofC6H5NH2 is less than that ofC2H5NH2 and (C2H5)2NH.

Solution 8. A polymerthat can be decomposed by bacteria is called abiodegradable polymer.For example: poly- -hydroxybutyrate-CO- -hydroxyvalerate (PHBV) is

biodegradable aliphatic polyester.

2 2

3 2 3 n

O CH CH C O CH CH C

| || | ||

CH O CH CH O

PHBV

Solution 9. The process of corrosion is a redox reaction that involvessimultaneous oxidation and reduction reactions. It can therefore be referredto as an electrochemical reaction.

In the process of corrosion, due to the presence of air and moisture,oxidation takes place at a particular spot of an object made of iron. That spotbehaves as the anode. The reaction at the anode is can be written as follows.

Anodic reaction:2Fe (s) Fe (aq) 2e+ +

Electrons released at the anodic spot move through the metallic object andgo to another spot of object. There, in the presence of H+ ions, the electronsreduce molecular oxygen. This spot behaves as the cathode. There H+ ions


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come either from H2CO3 , which are formed due to the dissolution of carbondioxide from air into water or from the dissolution of other acidic oxides fromthe atmosphere in water.The reaction corresponding at the cathode is written as follows.

Cathodic reaction: 2 2O (g) 4H (aq) 4e 2H O(l)+ + +

Thus, the overall reaction is:2

2 22Fe(s) O (g) 4H (aq) 2Fe (s) 2H O (l)+ ++ + +

Also, ferrous ions are further oxidized by atmosphere oxygen to ferric ions.These ferric ions combine with moisture, present in the surroundings, to formhydrated ferric oxide (Fe2O3, xH2O) i.e., rust.

Solution 10.

( ) ( ) ( ) ( )2 o

s aq aq s

2

ocell

1

o or cell

o 1r

Ni 2Ag Ni 2Ag , E 1.05V

The galvanic cell of the given cell reaction is depicted as :

Ni(s) |Ni || Ag | Ag(s)

Now, the s tandard cell potential is

E 1.05V

n 2

F 96500 C mol

G nFE

G 2 96500 Cmol 1.0

+ +

+ +

+ + =

=

=

=

=

=

1

1

5 V

202650 Jmol

202.65 kJmol

=

=

Solution 11. The rate expression can be defined as an expression in whichthe rate of reaction is given as the product of the molar concentration of thereactants, with each term raised to some power, which may or may not bethe stoichiometric coefficients of the reacting species in a balanced chemicalequation.The rate constant can be defined as the rate of reaction when theconcentration of each of the reactant is taken as unity.

Example: 2 22NO(g) O (g) 2NO (g)+

The rate expression for the above reaction can be written as follows:2

2Rate k [ NO] [O ]= (Experimentally determined)

Now, if the concentration of NO and O2 is taken to be unity, then the rateconstant is found to be equal to the rate of the reaction.


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Solution 12.

(i) The Shorter N O bond in 2NO

is due to the existence of resonance

in 2NO . The resonating structure can be drawn as follows.

Due to resonance in 2NO

, the two bonds are equivalent. This leads to a

decrease in bond length. Thus, the N O bond length in2

NO resembles a

double bond.

Now, the resonating structure for 3NO

can be drawn as:

As seen from the above resonating structure of 3NO , the three oxygenatoms are sharing two single bonds and one double bond. So the real N-Obond length resembles a single bond closely.

This explains the existence of shorter bond length of the N-O bond in 2NO

than in 3NO

.

(ii) The kinetic inertness of SF6 can be explained on the basis of itsstructure.


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As seen from the above structure, the six fluorine (F) atoms protect thesulphur atom from attack by the regents to such an extent that eventhermodynamically most favourable reactions like hydrolysis do not occur.

OR

(i) In gaseous and liquid state, PCl5 has a trigonal bipyramidalstructure. In this structure, the two axial P Cl bonds are longerand less stable than the three equatorial P Cl bonds. This isbecause of the greater bond pair bond pair repulsion in then axialbonds. Hence, all the bonds in PCl5 are not equivalent.

(ii) Because of stronger S-S bonds as compared to O-O bonds, sulphurhas a greater tendency for catenation than oxygen.

Solution 13.

(i) In aqueous solution, Cu+ ion undergoes oxidation to Cu2+ ion. Therelative stability of different oxidation states can be seen from theirelectrode potentials.

( )+ o

redCu (aq) e Cu (s), E 0.52V+ =

( )2+ o

redCu (aq) 2e Cu(s), E 0.34V+ =

Due to more reduction electrode potential value of Cu+, itundergoes oxidation reaction quite feasibly. Hence, Copper (I) ionis not known in aqueous solution.

(ii) The actinoids show a larger number of oxidation states because ofvery small energy gap between the 5f, 6d and 7s sub-shells.Hence all their electrons can take part in bond formation.

Solution 14. Riemer-Tiemann reaction: Riemer-Tiemann reactioninvolves the treatment of phenol with chloroform in the presence of aqueoussodium hydroxide at 340 K followed by hydrolysis of the resulting product togive 2-hydroxybenzaldehyde (salicylaldehyde). The chemical reaction can berepresented as follows.(i)


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(ii)Friedel-Crafts acetylation of anisole: Friedel-Crafts acetylation ofanisole involves the treatment of anisole with either acetyl chloride or aceticanhydride to give 2-methoxyacetophenone (as a mirror product) and 4-methoxyacetophenone (as a major product), the chemical reaction can berepresented as follows.

Solution 15.(i) Phenol on reaction with concentrated HNO3 results in the formation

of picric acid.

(ii) 2-Methyl propene can be obtained from 2-methyl propanol by thereaction of the later with alc. KOH


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Solution 16. The -form of glucose and -form of glucose can be

distinguished by the position of the hydroxyl group on the first carbon atom.In open chain -glucose, the hydroxyl group on the first carbon atom istowards the right whereas, in the closed ring - glucose, the hydroxyl groupon the first carbon atom is below the plane of the ring.On the other hand, in open chain -glucose, -glucose, the hydroxyl group

on the first carbon atom is towards the left whereas, in the closed ring -

glucose, the hydroxyl group on the first time atom is above the plane of thering.The structures of open and closed -form and - form of glucose can be

drawn as follows.

Solution 17.I. Primary structure of proteins: Each polypeptide chain in

a protein has amino acids linked with each other in a specific


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sequence. This sequence of amino acids is said to be theprimary structure of proteins.

II. Secondary structure of proteins: The secondary structureof proteins refers to the shape in which a long polypeptidecan exist. The two different secondary structure possible are -Helix structure and pleated sheet structure.

-Helix structure: In -Helix structure, a polypeptidechain forms all possible hydrogen bonds by twisting into ahelix with NH group of each amino acid residue andhydrogen bonded to >C=O of an adjacent turn of helix.

-Helix structure: In a -pleated structure, all peptidechains are stretched out of nearly maximum extensionsand then laid side by side which are held together byintermolecular hydrogen bonds.

Solution 18. (i) Uses of Bakelite:(a) It is used for making combs.(b) It is used for manufacturing electrical switches.

(ii) Uses of Nylon 6:(a) It is used for making tyre cords.(b) It is used for making fabrics and mountaineering

ropes.Solution 19.Given, silver crystallizes in fcc unit cell

aSo, r

2 2=

Where e is the radius of the silver atom and a is the edge length

Now, edge length = 400 pm = 10400 10 cm

Thus,10

10

400 10 cmr

2 1.414

141.44 10 cm

141.4pm

=

=

=

Thus, the radius of the silver atom was found to be 141.4 pm

Solution 20.(a)The plot of [N2O5] v/s t is as follows


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[N2O5] (M) Time (min) log[N2O5]

0.400 0.00 -.03979

0.289 20.0 -0.5391

0.209 40.0 -0.6798

0.151 60.0 -0.8210

0.109 80.0 -0.9625

From the plot, log [N2O5] v/s t, we obtain

( )0.70 0.60Slope40 20

0.70 0.60 0.10

20 20

=

+ = =

Also, slope of the line of the plot =k

2.303


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3 1

k 0.10

2.303 20

0.10or k 2.303 0.01151

20

or, k 1.15 10 min

=

= =

=

(b) After 100 min

2 5 o

2 5 t

1

[N O ]2.303k log

t [N O ]

After 100 min

2.303 0.400k log

t 0.098

0.1406 min

=

=

=

(c) The initial rate of reaction

2 5

3

4 1

r k[N O ]

1.15 10 0.400

4.6 10 s

=

=

=

Solution 21.(i) Production of high vacuum: Traces of air can be adsorbed by

charcoal from a vessel, evacuated by a vacuum pump to give avery high vacuum.

(ii) Heterogeneous catalysis: The gaseous reactants are adsorbedon the surface of the solid catalysts. As a result, the concentrationof the reactants increase on the surface and hence the rate of thereaction increases.

(iii) Froth floatation process: This process is used to remove ganguefrom sulphide ores. The basic principle involved in this process isadsorption.In this process, a mixture of waterpine oil is taken in tank. Theimpure powdered sulphide ore is dropped in through hopper andthe compressed air is blown in through the agitator is rotator is

rotated several times. As a result, froth is formed and the sulphideores get adsorbed in the froth. The impurities settled down and arelet out through an outlet at the bottom. The froth formed iscollected in froth collector tank. After sometime, the ore particles inthe froth collecting tank start settling gradually, which are thenused for further metallurgical operations.

OR


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(i) A micelle is an aggregate of surfactant molecules dispersed in a aliquid. A micelle in aqueous solution forms as aggregate such thatthe hydrophilic head regions are in the centre of micelle.

(ii) Peptization is the process of conversion of a precipitate into acolloidal sol by shaking it with the dispersion medium in thepresence of an electrolyte. The electrolyte used in this reaction isknown as a peptizing agent.

(iii) Desorption is the process of removing an adsorbed substance fromthe surface through which it was adsorbed.

Solution 22.(i) Vapour phase refining : Vapour phase refining is the process of

refining metal by converting it into its volatile compound and then,decomposing it to obtain a pure metal. The basic principle involvedin this process are:

(a)The metal should form a volatile compound with an availablereagent, and(b)The volatile compound should be easily decomposable so that

the metal can be easily.Nickel, zirconium, and titanium are refining using this method.

(ii) Electrolytic refiningof a metal is the process of refining impuremetals by using electricity. In this process, impure metal is madethe anode and a strip of pure metal is made the cathode. A solutionof a solution salt of the same metal is taken as the electrolyte.When an electric current is passed, metal ions from the electrolyteare deposited at the cathode as pure metal and the impure metalfrom the anode dissolves into the electrolyte in the form of ions.The impurities present in the impure metal gets collected below theanode. This is known as anode mud.

n

n

Anode : M M ne

Cathode : M ne M

+

+

+

+


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(iii) In the process of leaching, the finely divided silver is treated with

dilute solution of sodium cyanide while a current of air is

continuously passed. As a result, silver pass into the solutionforming solution dicyanoargenate(I) while the impurities remainunaffected which are filtered off.

( )2 22Ag S 4NaCN 2Na[Ag CN ] Na S

Sodium dicyanoarg enate (I)

+ +

Solution 23.

(i)2 2

2 4 4 2 25C O 2MnO 16H 2Mn 8H O 10CO + ++ + + +

(ii)Heated

4 2 4 2 22KMnO K MnO MnO O + +

(iii) 2 32 7 2 2Cr O 8H 3H S 2Cr 7H O 3S + ++ + + +

Solution 24.(i) K4[Mn(CN)6]

Name: Potassium hexacyanomanganate(II)Stereochemistry Does not show geometric or optical isomerismMagnetic behaviour Paramagnetic

(ii) [Co(NH3)5Cl] Cl2Name: Pentaamminedchloridocobalt (III) chlorideStereochemistry Does not geometric isomerism but is optically activeMagnetic behaviour Paramagnetic

(iii) K2[Ni(CN)4]Name: Potassium tetracynoinickelate (II)Stereochemistry Does not show geometric or optically isomerismMagnetic behaviour Diamagnetic

Solution 25.


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(i) Haloalkanes can easily dissolve in organic solvents of low polaritybecause the new forces of attraction set up between haloalkanesand the solvent molecules are of same strength as the forces ofattraction being broken.

(ii) A mixture of equal amounts of two enantiomers is known asracemic mixture.For example: When a 3o halide undergoes substitution with KOH,the reaction proceeds through SN1 mechanism forming the racemicmixture in which one of the products has the same configuration asa reactant, while the product has an inverted configuration.

(iii) The SN1 substitution reaction involves the formation of carbocation,which is not affected by the presence of bulky groups.Thus, C6H5CH(C6H5)Br will be more reactive towards SN1substitution reaction forming racemic mixture.

Solution 26.(a) The basicity of amines depends on the +I effects of the alkyl group.

The presence of CH3 group in alkylamine increases the electrondensity on the nitrogen atom and thus increases the basicity.Hence, alkylamine is more basic than ammoniaCH3NH2 > NH3

(b) (i)


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(ii)

Solution 27.(i) Detergents: A detergents is a surfactant or a mixture ofsurfactants having cleaning properties in dilute solution. Commonly,detergent refers to alkylbenzenesulphonates. For example: Sodiumdodecylbenzene sulphonate(ii) Food preservatives: Food preservatives are chemicals thatprevent food from spoilage due to microbial growth. Table salt, sugar,


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vegetable oil, sodium benzoate (C6H3COONa), and salts of propanoic acid aresome examples of food preservatives.(iii) Antacids: Any drug that is used to counteract the effects of excessacid in the stomach and raise the pH to an appropriate level is called anantacid.Example: Omeprazole

Solution 28. (a) Molartiy is defined as the number of moles of solutedissolved per litre of solution.

Mathematically M =3

Number of moles of solute

Volume of solution in litres (dm )

Molality of a solution is defined as the number of moles of solute

dissolved in 1000 grams of solvent.

Mathematically, m =Number of moles of the solute

Mass of solvent in kg

While molarity decreases with an increase in temperature, molalityis independent of temperature. This happens because molalityinvolves mass, which deos not change with a change intemperature, while molarity involves volume, which is temperaturedependent.

(b) Given w2 = 10.50 g

w1 = 200g

Molar mass of MgBr2 (M2) = 184 g

Using the formula,

fT =f 2

1 2

1000 k w

w M

=1000 1.86 10.50

200 184

= 19.530200 184

= 0.53

Now, Tf = To - fT

= 273 0.53 = 272.47 K

OR


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(a) Osmosis : The process of flow of solvent molecules from puresolvent to solution or from solution of lower concentration ofsolution of higher concentration through a semi permeablemembrane is called osmosis.

Osmotic pressure : The pressure required to just stop the flow of

solvent due to osmosis is called osmotic pressure () of thesolution. Yes, the osmotic pressure of a solution is colligativeproperty. The osmotic pressure is expressed as.

nRT

V =

Where, = osmotic pressure

n = number of moles of solute

V = volume of solution

T = temperature

From the equation, it is clear that osmotic pressure depends uponthe number of moles of solute n irrespective of the nature of thesolute. Hence, osmotic pressure is a colligative property.

(b) Given, Kb = 0.512 k kg mol-1

w2 = 15.00 g

w1 = 250.0 g

M2 = 58.44 g

Using the formula,

b 2b

1 2

1000 K wT

w M

=

=1000 0.512 15.00

250.0 58.44

= 7.68014.600

= 0.52

Now, Tb = To + bT

= 373 + 0.53 = 373.53 K

Ans29. (a)


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(i) Propanal (CH3CH2CHO) can be distinguished from propanone(CH3COCH3) by iodoform test.

Being a methyl ketone, propanone on treatment whith I2/NaOH

undergoes iodoform reaction to give a yellow ppt. of iodoform

CH3COCH3 + 3NaOI CHI3 + CH3COONa + 2NaOH

Propanone Iodoform

Propanal on the other hand does not give this test.

CH3CH2CHONaOI

No yellow ppt. of

Propanal Iodoform

(ii) Benzaldehyde (C6H5CHO) and acetophenone (C6H5COCH3) canbe distinguished by iodoform test.

Acetophenone, being a methyl ketone on treatment with I2/NaOHundergoes iodoform reaction to give a yellow ppt. of iodoform. Onthe other hand, benzaldehyde does not give this test.

C6H5COCH3 + 3NaOI

Acetophenone

C6H5COONa + CHI3 + 2NaOH

Iodoform

C6H5CHONaOI

No yellow ppt of iodoform

Benzaldehyde

(b)

(i)


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(ii) CH3CH2CH2CH2OH 2 2 7 2 42 4(i) K CrO /HSO

(ii) Dil.H SO CH3CH2CH2COOH

Butanol Butanoic acid

(iii)

OR

(i) Cannizaro reaction

In this reaction, the aldehydes which do not have an -hydrogen atom, undergo self oxidation and reduction


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(disproportionation) reaction on treatment with a concentratedalkali.

Example:

(ii) Decarboxylation

The decarboxylation reaction can be carried out either by usingsoda lime or by electrolysis

Using soda limeSodium salts of carboxylic acids when heated with soda lime(NaOH + CaO) in the ratio 3:1 undergoes decarboxylationreaction to yield alkanes.

R COONa NaOH CaOHeat

R H + Na2CO3

(Alkane)

Electrolytic decarboxylationElectrolysis of aqueous solutions of sodium of potassium saltsof carboxylic acids give alkanes having twice the number ofcarbon atoms present in the alkyl group of acid. This isknown as Kolbes decarboxylation.

2RCOONa 2RCOO- + 2Na+

H2O 2OH- + 2H+

At Anode:-

2H+ + 2e- H2

(b)

(i)


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(ii)

(iii)

C6H5CONH23

H O

heat

+

C6H5COOH

Benzoic acid

Ans30. (a)

(i) As we move down the group 17, the size of the atom increasesfrom fluorine to chlorine. The larger difference in the size of N and Cl

results in the weakness of strength of N Cl bond. On the other hand,the difference in size of N and F is small; consequently the N F bondis quite strong. As a result, NF3 is an exothermic compound.

(ii) Due to the small size of F atom, the three lone pair of electrons oneach F atom F F molecule repels the bond pair. As a result, F F ismost reactive of all the four common halogens.

(b)

(i) C + 2H2SO4 2SO2 + CO2 + 2H2O

Sulphur dioxide

(ii) P4 + 3NaOH + 3H2O2

,CO PH3 + 3NaH2PR2

Phosphine

(iii) Cl2 + 3F2 2ClF3


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(excess) Chlorine trifluoride

OR

(a)

(i) In a period, the electro negativity decreases in the order Cl > S >P. As a result, the loss of H+ ions decreases.

Thus, the acidic strength of the hydrides decreases in the followingorder: HCl > H2S > PH3

(ii) The tendency to form pentahalides decreases down the group 15due to inert pair effect i.e., in Bi the s-electrons remain inert and do not takepart in bonding.

(b)

(i) P4 + 10SO2Cl2 4PCl5 + 10SO2

(ii) 2XeF2 + 2H2O 2Xe + 4HF + O2

(iii) I2 + 10HNO3 2HIO3 + 10NO2 + 4H2O

(conc).

cbse xii chemistry 2011 solution

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