SOLUTIONS...3. 1 (b) Fe. 3+ ions of FeCl. 3. neutralize the charge on the colloidal particles of...

S O L U T I O N SSAMPLE QUESTION PAPER - 6Self Assessment_________________________________________

Time : 3 Hours Maximum Marks : 70

1. Ethoxy-2-methyl propane 1

2. No. of X atoms per unit cell = 2 4

23 3

× =

No. of Y atoms per unit cell = 1 Hence, the formula of the compound is

A4/3B, or A4B3 1

3. Antiferromagnetism is a property of a substance due to which their domains are oppositely oriented and cancel each other‘s magnetic moment. 1

e.g., MnO ↑¯↑↓↑↓↑↓[CBSE Marking Scheme, 2014]

4. PCl5(s) Trigonal bipyramidal

1

[CBSE Marking Scheme, 2014]

5. (a) To complete the electric circuit. ½

(b) To maintain electrical neutrality around the electrodes. ½

6. (a) 95% by volume of ethanol and 5% by volume of water. 1

(b) Negative deviation because DH becomes negative. 1

7.(i)

1

(ii)

18. m = Z it

t = =W 1.27, =M 63.5

mm

zi i = current = 2Å, t = ?

Z = 1

2 96500×

t = 1.27 2 96500

1930 sec 21.4 hr63.5 2× × = =

× 1OR

Cu2+ + 2e– ¾® Cu 63.5 Cu is deposited = 2 × 96500 C 1.27g Cu is deposited

= 2 × 96500 × 1.27/63.5 C 1

t = 2 96500 1.27

63.5 2× ×

×

= 1930 sec 19. Similarity : Lanthanoids and actinoids both

show variable valency/contraction in size/stable in + 3 oxidation state. (Any two) 1+1

Difference : Lanthanoids do not form oxo ions, actinoids form oxoions. C

HE

MIS

TRY

Osw

aal C

BSE

Cla

ss -

12, E

xam

inat

ion

Sam

ple

Que

stio

n P

aper

s

2 | Oswaal CBSE Sample Question Paper, Chemistry, Class - 12 e.g., CeO2

2+/actinoid contraction is greater than lanthanoid contraction. (Any one)

10. (i) Roasting is a process of heating the ore in the regular supply of air below its melting point. This process is used for converting sulphide ores into their respective oxide.1

(ii) Calcination is a process of heating the ore in limited absence supply of air or no supply below its melting point. This process is used for converting carbonate and hydrated oxide ores to their respective oxides. [CBSE Marking Scheme, 2014] 1

OR (i) Froth floatation process in metallurgy : A

low grade sulphide ore is concentrated by separating it with silica and other matter by froth floatation process using pine oil and frothing agents. 1

(ii) Vapour phase refining of metals : It is based on the principle that certain metals are converted to their volatile compounds while impurities are not affected during the compound formation. The compound formed decomposes on heating to give a pure metal.[CBSE Marking Scheme, 2014]

11.

−

3

3

COOCo |

COO

;

IUPAC name : Trisoxalato-

cobalt (III) ion structure ½+½

Structure :

(ii) [Cr(CO)6] : IUPAC name : Hexacarbonyl-chromium (O)

Structure : Octahderal

½+½

(iii) [Pt Cl3(C2H4)] : IUPAC name : Trichloro (ethene) Platinum (IV)

Structure : ½+½

12.

So, Pt = 0.4 – x + x + x = 0.4 + x 1

or x = Pt – 0.4

Now Ps (for SOCl2) = 0.4 – x = 0.4 – (Pt – 0.4) = 0.4 + 0.4 – Pt

= 0.8 – Pt = 0.8 – 0.7 = 0.1 ½

So, taking account of given data

K = P2.303 2.303 0.4

log logP 100 0.1

i

st

=

½

= ×=2.303 2.303 0.6021log 4

100 100 1

= 1.387 × 10–2 s–1


13. (i) Peptide linkage is the linkage that exists between the monomeric amino acids in a polypeptide chain. This linkage occurs between the —COOH group of one amino acid and the NH2 group of next amino acid. The linkage occurs by condensation of one water molecule and exists as —CONH linkage. 1

(ii) Primary structure is a specific sequence in which amino acids are linked to each other by peptide bond in a polypeptide chain. This sequence decides the function of a protein. 1

(iii) Denaturation is the disruption or loss in the biological activity of native conformation of a protein when subjected to change in pH, temperature, etc. It may be reversible or irreversible. e.g., Coagulation of lactal-bumin to form cheese.

14. (i) Aniline does not undrgo Friedel-Crafts reaction due to formation of salt with aluminium chloride (AlCl3), which is a levis acid and acts as catalyst in the reaction. Due to this, nitrogen of aniline acquires positive charge and hence acts as a strong deactivating group for further reaction. 1

Solutions | 3

(ii) Ethylamine dissolves in water due to intermolecular hydrogen bonding as shown in figure, while in aniline, due to large part, the extent of hydrogen bonding decreases considerably and hence anline is insoluble in water. 1

H H | | - - - N – H - - - O – H – - - - N – N - - - O – H - - - | | | | C2H5 H C2H5 H (iii) In aniline, the lone pair of electrons on the N-atom is delocalized due to reasonance with the

benzene ring. As a result, electron density on nitrogen decreases. However, in methylamine (CH3NH2), due to +I effect of CH3 group, the electron density increases on the N-atom. Thus, the Kb value of methylamine is higher due to which its pKb value of less than that of aniline.

[CBSE Marking Scheme, 2014] 1

15. (a) Role of NaCN in the extraction of gold form gold ore : To convert it into dissolvable cyanide complex in case of silver.

Ag2S + 4NaCN ¾® 2Na [Ag(CN)2] + Na2S 1 (b) SiO2 in the extraction of copper from copper mate : It acts as a flux and converts the impuri-

ties into slag.

FeO + SiO2 ¾® FeSiO3 1 (c) Iodine in the refining of zirconium (slag) :

Zr + I2 870 K

¾¾¾¾® ZrI4 2075 K¾¾¾¾¾¾¾¾®

Tungsten filament Zr + 2I2 pure 1

16. (a) Natural rubber is cis-polyisoprene and is obtained by 1, 4-polymerization of isoprene units. This cis-configuration at double bonds does not allow the polymer chains to come closer for effective interactions and hence the intermolecular forces are quite weak. As a result, natural rubber (cis-polyisoprene) has a randomly coiled structure and shows elasticity.

1

(b) Natural rubber has following disadvantages :

(i) It is soft and sticky at high temperatures and brittle at low temperatures. Thus, it is used at a narrow temperature range (283-335K).

(ii) It has a large absorption capacity, has low tensile strength and low resistance to abrasion.

(iii) It is not resistant to the action of organic solvents.

(iv) It is easily attacked by oxygen and other oxidising agents.

Thus to improve the properties of natural rubber, it is vulcanised by heating it with about 5% sulphur at 373-415 K. Thus obtained rubber has excellent elasticity over a large range of temperature, has low water absorption tendency, is resistant to the action of organic solvents and oxidising agents. 2

17. Total no. of voids = No. of octahedral voids + No. of tetrahderal voids No. of moles = 0.5 mol

4 | Oswaal CBSE Sample Question Paper, Chemistry, Class - 12 Q 1 mol contains 6.022 × 1023 atoms \ 0.5 mol will contain 6.022 × 1023 × 0.5 = 3.011 × 1023 = atoms \ No. of octahedral voids present in compound = 3.011 × 1023 voids 1 \ No. of tetrahedral voids = 2 × no. of octahedral voids = 2 × 3.011 × 1023 = 6.022 × 1023 voids 1 = 9.033 × 1023 voids 1

OR (a) Schottky defect since the ionic crystal of type A+B– as equal number of cations and anions are

missing from their lattice sites. 1 (b) Schottky defect decreases the density of the substance. 1 (c) Schottky defect is shown by ionic substances in which the cation and anion are of almost

similar sizes. [CBSE Marking Scheme, 2014] 118. (i) [Cr(H2O)6]Cl3 will show hydrate isomerism i.e., [CrCl(H2O)5]Cl2H2O and [CrCl2(H2O)4]Cl.2H2O 1 (ii) When D0 > P, It means pairing energy is low hence electron will remain in t2g. The e– are

paired hence it is known as low spen complex. 1

(iii) [CoF6]3–

1

19. Prevaiting temperature : Physisorption decreases with increase in temperature i.e., low temperature is favourable for physisorption. ½

For example, adsorption of gases like N2 and H2 on surface of charcoal at low temperature. Chemisorption first increases, then decreases with increase in temperature i.e., high temperature is

favourable for chemisorption. e.g., adsorption of N2 or H2 on the surface of iron catalyst at high temperature. ½

Surface area of adsorbent : Both physisorption and chemisorption increases with increase in surface area. For example, the adsorption of gases like N2 and H2 increases when the surface area of charcoal or iron is increased. ½+½

Activation energy of the process : No appreciable activation energy is needed in case of physisorption while high activation energy is needed for chemisorption. For example, for the formation of nitric oxide from nitrogen and oxygen gas some amount of activation energy is required. ½+½

OR (i) Production of Vacuum in a vessel : The adsorption of air in liquid air helps to create a high

vacuum in a vessel. This process is used in high vacuum instruments like Dewar flask for storage of liquid air liquid hydrogen. The remaining traces of air can be adsorbed by charcoal from a vessel evacuated by a vacuum pump to give a very high vacuum. 1

(ii) Heterogeneous catalysis : Adsorption of reactants on the solid surfaces of the catalyst increases the rate of reaction. There are many gaseous reactions of industrial importance which involve solid catalyst. e.g., Manufacturing of ammonia using iron as a catalyst, manufacturing of H2SO4 by contact process.

(iii) Froth floatation process in metallurgy : A low good sulphide ore is concentrated by separating

Solutions | 5if with silica and other matter by froth floatation process using pine oil etc as frothing agent.


20. C2H5NC + 2H2O H+

¾¾¾¾® C2H5NH2 HNO2¾¾¾¾® C2H5OH + N2 + H2O

“A‘‘ “B‘‘ Ethyl alcohol 3

C2H5NH2 + CHCl3 + 3KOH ¾® CH3CH2NC + 3KCl + 3H2O “B‘‘ “C‘‘

A-Ethyl isocyanide B-Ethyl amine C-ethyl isocyanide21. (i) Antihistamines are used for the treatment of allergy and hence are often known as anti-allergic

drugs. Since the allergic reactions are caused due to liberation of histamine in the body, that is why these drugs are know as antihistamines. e.g., Diphenylhydramine. 1½

(ii) Antioxidants are the the important category of compounds which prevent oxidation of food materials. e.g., Butylated hydroxyl toluene (BHT). 1½

22. Alkenes react with water in presence of acid as catalyst to form alcohols. In case of unsymmetrical alkenes, the addition reaction takes place in accordance with Markomikov‘s rule.

Mechanism : Step 1 : Protonation of alkene to form carbocation by electrophitic attack

H2O + H+ ¾® H3O

+ 1

Step 2 : Nucleophilic attack of water on carbocation. 1

Step 3 : Deprotonation to form alcohol. 1

23. (a) Chemical formula is FeCl3. 1 (b) Fe3+ ions of FeCl3 neutralize the charge on the colloidal particles of blood. This leads to

coagulation of blood. Bleeding therefore stopped. 1 (c) This phenomenon is known as coagulation or floculation. 1 (d) All house wives must keep small bag or kit in their kitchen. It must have ferric chloride kept

in small bottle or Potash alum, burnol and bandages etc. Minor accidents are very common in kitchen. The kit can be very helpful to deal with emergency.

24. (a) (i) Molarity of a substance in a solution is equal to the number of moels of the substance present in one litre of solution. 1½

Molarity (M) = No. of moles of substanceVolume of solution in L

(ii) Molal elevation constant (Kb) is also called as ebullioscopic constant. It is equal to the change or elevation in boiling point of 1 molal solution. 1½

DTb = Kb × m Hence, when molality = 1, DTb = Kb (b) Mass of urea, WB = 15 g, Molar mass of urea, MB = 60 g

As the solution is isotonic,

\ purea = pglucose

6 | Oswaal CBSE Sample Question Paper, Chemistry, Class - 12 Curea = Cglucose

Þ 1560

=

glucoseW

180

Wglucose = 1660

× 180 = 45 g

Hence, 45 g of glucose is present it 1 L of solution. 2

OR

(a) A mixture of ethanol and acetone shows positive deviation.

For a mixture of two liquids A (acetone) and B ethanol, when the attractive forces between A and B are smaller/weaker than between A-A and B-B, positive deviation from Raoult‘s law is observed. 2

(b) 10 g glucose is present in 100 g solution i.e., 90 g of water = 0.090 kg water

10 g glucose =

10180

mol = 0.0555 mol, ½

90 g H2O = 9018

= 5 moles

Molality = 0.0555 mol0.090 kg

= 0.617 m = 0.612 mol/kg

1

100 g solution = 1001.2

mL = 83.33 ml

= 0.83333 L ½

Molarity = 0.0555 mol0.083333 L

= 0.67 M 1


25. (a) (i) Aldol condensation : Aldehydes and ketones having at least one a-hydrogen undergo a reaction in the presence of dilute alkali as catalyst to form b-hydroxy aldehydes (aldol) or b-hydroxy ketones (ketol) respectively. This is known as aldol reaction.

1

(ii) Cannizaro reaction : Aldehydes which do not have a-hydrogen atom, undergo self oxidation and reduction reaction on treatment with concentrated alkali. In this reaction, one molecule of aldehyde is reduced to alcohol while another is oxidized to carboxylic acid salt. 1

Solutions | 7 (b) (i) Ethanal and Propanal : Iodoform test : Ethanal gives a positive iodoform test by giving yellow precipitate while propanal

does not give this test. CH3CHO + 3NaOI ¾® HCOONa + 2NaOH + CHI3 ¯ ½ Iodoform (yellow ppt) CH3CH2CHO + 3NaOI ¾® NO Yellow ppt. ½ (ii) Benzaldehyde and acetophenone. Iodoform test : Acetophenone is a methyl ketone. It responds to iodoform test while benzaldehyde

does not. C6H5COCH3 + 3NaOI ¾® C6H5COONa + 2NaOH + CHI3 ¯ ½ Iodoform (Yellow ppt) C6H5CHO + 3NaOI ¾® No Yellow ppt ½ (iii) Benzoic acid and ethyl benzoate. Sodium bicarbonate test. Benzoic acid gives effervescence with NaHCO3 solution while ethyl ben-

zoate does not.

C6H5COOH + NaHCO3 ¾® C6H5COONa + CO2 + H2O ½ Sodium benzoate

C6H5COOC2H5 + NaHCO3 ¾® No reaction ½OR

(a) (i) 4-chloro pentan-2-one

(ii) But-2-en-1-ol 1

(b) (i) CH3COOH Br2/P¾¾¾¾® CH2 – COOH + HBr 1

| Br a-Bromocarboxylic acid

(ii) CH3 – CHO LiAlH4¾¾¾¾® CH3 – CH2 – OH 1

Ethanol

(iii) CH3COCH3 + 4 [H] Zn-Hg¾¾¾¾®

HCl CH3CH2CH3 + H2O 1

Propane


26. (i) High enthalpies of atomization of transition elements are attributed to the involvement of (n – 1)d electrons in addition to ns electrons in the interatomic metallic bonding. 1

(ii) Transition metals exist as catalysts as they exhibit variable oxidation states. 1

(iii) Actinoid contraction from element to element is greater than lanthanoid contraction due to poor shielding effect of 5f orbitals. 1

(iv) One of the factor responsible for positive value of reduction potential is the stable oxidation state of metal. Mn2+ has more stable d5 configuration Mn3+ has less stable d4 configuration. Thus, Mn3+ ¾® Mn2+ will take place more rapidly. 1

Whereas, Cr3+ has d3 configuration and Cr2+ has d4. Thus, Cr3+ ¾® Cr2+ will not take place so rapidly as d3 is stable in Cr (t32g) Therefore, Eº value of Mn

3+/Mn2+ is much more positive then for Cr3+/Cr2+ couple. 1

(v) In Sc, the last electron enters 3d orbital. Therefore, regarded as transition element. 1

8 | Oswaal CBSE Sample Question Paper, Chemistry, Class - 12OR

(a) The radii of 4d and 5d transition elements resemble closely. 1

(b) It can form unstable intermediates with reactants which readily change into products. 1

(c) Alkaline solution of KMnO4 is known as Bayer‘s reagent. 1

(d) KMnO4 gets decomposed to K2MnO4 in presence of light 1

2KMnO4 D

¾¾¾¾® K2MnO4 + MnO2 + O2 1 (e) In zinc salts, Zn2+ does not have unpaired electrons. Therefore, there is no possibility of

undergoing d-d transitions while in Cu2+ ions, due to presence of unpaired electron, there is d-d transitions. 1

l

SAMPLE QUESTION PAPER - 7Self Assessment_________________________________________


1. Ethylenediaminetetra acetate is used and its denticity is six, EDTA binds with metal in octahedral manner by two N-atoms and 4 acetate oxygen atoms.


2. Coagulation power increases with increase used for coagulation. Thus, BaCl2 is more effective in causing coagulation. 1

Ba++ > K+

Br |3. CH3—CH—CH3 will undergo faster SN

1 reaction because it gives more stable in termedia te i . e . , CH 3— C+

H—CH 3

(2º carbocation) as compared to CH3—CH2—CH2

+ (1º carbocation).

4.


5. Lyophobic sol is solvent “repelling‘, less stable, irreversible and sol particles are not solvated while lyophillic sol is solvent “loving‘‘ very stable, reversible and sol particles are heavily solvated in them. [CBSE Marking Scheme, 2014] 1

6. (i) Specific rate or rate constant of a reaction is equal to the rate of reaction when the concentrations of all the reactants are taken as unity. ½

A + B ¾® Products Rate = K[A] [B], where k is the rate

constant, A and B are reactants when [A] = 1 mol L–1 [B] = 1 mol L–1, Rate = K. ½

(ii) Energy of activation of a reaction is the extra energy supplied to the reactants so that they can form intermediate called as activated complex and change to products after crossing the activated complex. It is denoted by Ea.


7. (i) P4 + 8SOCl2 ¾®4PCl3 + 4SO2 + 2S2Cl2 1

(ii) 3F2 (excess) + Cl2 300°C

¾¾¾¾® 2ClF3[CBSE Marking Scheme, 2014] 1

8. Every Sr2+ ion causes one cation vacancy (because two Na+ ions are replaced by one Sr2+). Therefore, introduction of 10–3 moles of SrCl2 per 100 mole of NaCl would in-troduce 10–3 mole cation vacancies in 100 mole of NaCl.

No. of vacancies per mole of NaCl = –310

100 1

= 10–5

OR

Density, d = 11.2 g cm–3 edge length a = 4 × 10–8 cm NA = 6.022 × 10

23 ½

No. of atoms per unit cell for FCC (Z) = 4

d = ××3 A

Z M

Na ½

CH

EM

ISTR

Y O

swaa

l CB

SE C

lass

-12

, Exa

min

atio

n Sa

mp

le Q

uest

ion

Pap

ers

10 | Oswaal CBSE Sample Question Paper, Chemistry, Class - 12

\ M = 3

ANZ

d a× ×

= –24 2311.2 64 10 6.022 10

4× × × ×

= 107.91 g 2


9. (a)

1

(b) Reimer Tiemann reaction :

1

10. (a) Four 1 (b) The two poisonous gases which can be

prepared from Cl2 gas are : (i) Phosgene

(ii) Mustard gas ½+½11. Nernst equation : Fe + 2H+ ® Fe2+ + H2 Concentration of Fe2+ = 0.001 MH+ = 1M

E0Fe2+ = + 0.44 V Ecell = E°cathode – Eºanode = 0 – (– 0.44) =

+ 0.44 V ½

Ecell = 20.059 (0.001)(1)

0.44 log2 (1)

− − 1

= 0.059

0.44 log 0.0012

− −

= 0.44 – 0.0295 log 10–3 ½ = 0·44 – 0·0295(–3)

= 0.44 + 0.0885 = – 0.5285 V 112. Biodegradable detergents are those detergents

which can be decomposed by micro organ-isms present in water. They have straight unbranched chains of hydrocarbons. e.g., : Sodium lauryl sulphate. 1½

Non-biodegradable detergents are those de-tergents which cannot be degraded by the micro organisms. They have highly branced hydrocarbon chains. e.g., : Sodium 4-(1, 3, 5, 7-tetramethyloctyl) benzene sulphonate.

[CBSE Marking Scheme, 2014] 1½

13. (a) Extraction metallurgy of metals : For the extraction of silver (Ag), formation of soluble cyanide complex is used. In this process, silver ore Ag2S is treated with NaCN or KCN solution in presence of air

to form soluble complex known as sodium argento cyanide. From this complex, silver metal is precipitated by adding zinc soap.

Ag2S + 4NaCN ¾® Silver sulphide

2Na[Ag(CN)2] + Na2S 1 Sodium argento cyanide 2Na[Ag(CN)2] + Zn ¾®

Na2[Zn(CN)4] + 2Ag ¯ ½ (b) Analytical chemistry : The detection of

Cu(II) ions involves the formation of a deep coloured complex, [Cu(NH3)4]

2+ on addition of ammonia solution to a solution of Cu2+ ion. Due to it formation of coordintion entries froms the basis for their detection and estimatation by classical and instrumental method of analysis.

Cu2+ + 4NH3 ¾® [Cu(NH3)4]2+ 1

Deep blue colouration. [CBSE Marking Scheme, 2014] ½

14. Essential amino acids : Those amino acids which cannot be synthesized in the body and must be obtained through diet are known as essential amino acids. e.g., : Valine, Leucine.1½

Non-essential amino acids : Those amino acids which can be synthesized in our body are known as non-essential amino acids. Eg : Glycine, Alanine.

[CBSE Marking Scheme, 2010] 1½

15. (i) Glucose has five —OH groups which form hydrogen bond with water. Due to this extensive intermolecular hydrogen bonding, glucose is soluble in water. Cyclohexane is a non-polar molecule with no —OH group, hence it does not dissolve in polar solvent water. 1

Solutions | 11 (ii) Pentacelate D-glucose does not contain

any free adlehyde group as it is a cyclic structure of glucose. It is a lytic structure of glucose in acetylation forms penta acetate of D-glucose.

(iii) Deficiency of vitamin A causes night blind-

ness and zeropthalima (hardening of cornea of eye). that is why vitamin A is essential for us. 1

16. (a) ICl is more reactive than I2 because I—Cl bond is weaker than I—I bond due to less bond dissociation energy. Hence, ICl breaks easily to form halogen atoms which readily bring about the reactions.

(b) Due to its low solubility as compared to N2 in blood, a mixture of helium and oxygen is used in diving kit used by deep sea divers. 1

(c) As radon is a radioactive element with a short half life of 3.82 days, it becomes difficult to study the chemistry of radon.1

17. (i)

1 (ii)

1

(iii) CH3CH2 Cl + 2Na + Cl CH2CH3 ether

¾¾¾¾® CH3CH2CH2CH3 + 2NaCl 1 OR

(i)

1

(ii) C2H5Cl AgNO2¾¾¾¾® C2H5NO2 + AgCl 1

Ethyl nitrate

(iii) CH3 – CH – CH2 – CH2 – CH3 alc. KOH

¾¾¾¾® CH3 – CH – CH2 – CH2 – CH3 1 | | Br OH 2 bromo pentane pentan-2-ol

[CBSE Marking Scheme, 2013]18. (i) It is difficult to prepare pure amines by

ammonolysis of alkyl halide because pri-mary amines formed react with more alkyl halide to form 2º and 3º amines. Thus, we get mixture of amines, which is hard to be separated. 1½

(ii) Electrophilic substitution in case of aromatic amines takes place more readily than in benzene because —NH2 group is electron realeasing which increases the electron density on benzene ring thus making it favourable for electrophilic attack. 1½

12 | Oswaal CBSE Sample Question Paper, Chemistry, Class - 1219. (i) Sorption is defined as the process in which

both absorption and adsorption take place simultaneously. e.g., : the dye stuffs taken up by cotton fibres. 1

(ii) Tyndall effect : When a convering and strong beam of light is passed through a colloidal solution its path becomes visible. This effect is known as Tyndall effect. Eg. The blue colour of sky is due to the scat-tering of light by colloidal dust particles present in air. 1

(iii) Electrophoresis is a phenomenon involving the migration of colloidal particles under the influence of electric field towards the oppositely charged electrode. 1

20. (a) Nickel is refined by vapour phase refining (Mond‘s process)

Ni + 4CO 330-350 K

¾¾¾¾® Ni(CO)4 Nickel carbonyl Nickel carbonyl is subjected to higher

temperature till it gives pure nickel on de-composition 1

Ni(CO)4 450-470 K

¾¾¾¾® Ni + 4CO (b) Zirconium : Van Arkel method is used for

refining of zirconium

Zr + 2I2 ¾® ZrI4 The metal oxide is decomposed on a tungesten

filament, electrically heated to 1800 K. The pure metal is deposited on the filament.

ZrI4 ¾® Zr + 2I2. 1 (c) Tin : Liquation is used for refining tin.

The melting tin is made to flow on a sloping surface. In this way, it is separated from higher melting impurities. 1

21. (i) Cr2O72– + 14H+ + 6I– ¾®

2Cr3+ + 7H2O + 3I2 1 (ii) Cr2O7

2– + 6Fe2+ + 14H+ ¾®2Cr3+ + 6Fe3+ + 7H2O 1

(iii) Cr2O72– + 8H+ + 3H2S ¾®

2Cr3+ + 7H2O + 3S 122. Mass of solution = 1000 × 1.04 = 1040 g Mass of solute = 74.5 g Mass of solvent = 1040 – 74.5 = 965.5 g To calculate molality,

m = B

B A

W 1000 74.5 10001.0375 m

M W 74.5 965.5× = × =

Assuming KCl ¾® K+ + Cl–

i = 2 DTf = in Kf = 2 × 1.0357 × 1·86 1

Freezing point of solution

= 273 K – 3.852 K = 269.148 K 123. (i) The co-worker stopped him from opening

the knob because a gaseous compound NCl3 is formed which is explosive in na-ture. 1

(ii) NH3 + 3Cl2 ¾® NCl3 + 3HCl 1 (iii) The following values were exhibited by

the worker :

(a) irresponsibility and (b) illiteracy values exhibited by co-worker are concern, case responsibility and scientific knowledge. 2

24. Let the formula of sample be (Fe2+)x(Fe

3+)y0. ½ x + y = 0.93 ...(1) ½ Total positive charge on ferrous and ferric

ions should balance the two units of negative charge on oxygen. Hence, 2x + 3y = 2 ...(2) ½

or 32

x y+ = 1 ...(3) ½

On subtracting equation (i) from equation (3) we have,

3¾2

y – y = 1 – 0.93

or 1¾2

y = 0.07

or y = 0.14 Putting the value of y in equation (1) we get

x + 0.14 = 0.93 ½

x = 0.93 – 0.14 = 0.79 ½

Fraction of Fe2+ ions present in the sample

= 0.79¾¾0.93

= 0.81 1

Metal deficiency defect is present in the sample because iron is less in amount than that required for stoichiometric composition.1

OR (i) Schottky defect : When an atom or ion

is missing from its normal site, a lattice vacancy (Schottky defect) is formed. Equal number of cations and anions are missing in ionic solids. e.g., in NaCl. 1

(ii) Frenkel defect : When an atom or ion oc-cupies interstitial position instead of usual normal position, Frenkel defect is created. e.g., AgCl. 1

(iii) Interstitials : The vacant positions in the crystal lattice are called interstitials or voids. Eg : Transition metals. 1

Solutions | 13 (iv) F-centres : Electrons trapped in anion

vacancies are called F-centres. If we heat NaCl in sodium vapours, the vacant posi-tion of anions are occupied by electrons forming F centres. 1

(v) Impurity defects : When SrCl2 or CaCl2 is

added to molten NaCl for crystallization. Some of the Na+ ions are displaced by Sr2+ or Ca2+ ions. Each Sr2+ or Ca2+ ion replaces two Na+ ions. It takes place on one of the ions and other site remains vacant. The number of bivalent cations added is equal to number of cationic vacancies. 1

25. (a) The esterification of carboxylic acids with alcohols is a kind of nulceophilic acyl substitution. Protonation of the carbonyl oxygen activates the carbonyl group towards nucleophilic addition of the alcohol. Proton transfer in the tetrahderal intermediate converts the hydroxyl group into H2O

+ group which being a better learing group, is eliminated as neutral water molecule. The protonated ester so formed finally loses a proton to give the ester. 2½

(b) The molecular masses of these compounds are in the range of to 74. Since only butan-1-ol of molecules are associated due to ex-tensive inter molecular hydrogen bonding, therefore, the boiling point of butan-1-ol would be the highest. Butanal is more polar than ethoxyethane. Therefore, the inter mo-lecular dipole-dipole attraction is stronger in the former. n-pentane molecules have only weak vander Waals forces. Hence, the increasing order of boilding points of compounds is :

CH3CH2CH2CH2CH3 < H5C2—OC2H5 < CH3CH2CH2CHO < CH3CH2CH2CH2OH

OR

(i)

(ii)

(iii)

(iv)

(v)

26. (a) ( i) IBr2– is isostructrural to XeF2.

Struture: 1

(ii) BrO3– is isostructural to XeO3.

Structure: 1


(b) (i) SF6 is kinetically inert as 6 fluorine atoms protect the S atoms from attack by regents to such an extent that even thremodynamically most favourable reactions like hydrolysis do not occur.

1

(ii) NF3 is an exothermic compound whereas NCl3 is an endothermic com-pound because in case of NF3, N-F bond strength is greater than F—F bond strength while in case of NCl3, N—Cl bond strength is lower than Cl—Cl bond strength. Thus, the formation of NF3 is spontaneous while energy

has to be supplied during formation of NCl3. 1

(iii) HCl is a stronger acid than HF though fluorine is more electronegative than chlorine because HF has high bond dissociation energy than HCl due to shorter bond length. 1

OR

(a) Consequences of lanthanoid contraction :

(i) 4d and 5d transition series have almost similar atomic radii. ½+½

(ii) It is difficult to separate lanthanoids from their mixture.

(b) (i) 8MnO4– + 3S2O3

2– + H2O ¾®8MnO2 + 6SO4

2– + 2OH– 1

(ii) Cr2O72– + 3Sn2+ + 14H+ ¾®

2Cr3+ + 3Sn4+ + 7H2O 1

(c) Electronic configuration

No. of unpaired electrons

Ti3+ = [Ar] 3d14s0 1

V3+ = [Ar] 3d24s0 2 ½+½

Fe2+ = [Ar] 3d64s0 4 ½+½

Mg2+ = [Ne] 3s0 0

Hence, Fe2+ has maximum number of unpaired electrons.

l

SAMPLE QUESTION PAPER - 8Self Assessment_________________________________________


1.

1

2. Emulsion is a collodial solution of a liquid in a liquid medium i.e., the dispersion of finely divided droplets of one liquid in another liquid. Eg : Butter


3. The aluminate in solution is neutralized by CO2 gas and hydrated Al2O3 is precipitated.


4. Crystalline solids are anisotropic in nature

because they show different values of their optical and electrical properties in different directions in the some crystal. This is due to different arrangement of their particles in different directions.


5. Dichloro Bis (ethylenediamine) platinum (IV) nitrate [CBSE Marking Scheme, 2014]1

6. (i) 2KClO3 Heat¾¾¾¾®

MnO2 2KCl + 3O2 1

(ii) 6XeF4 + 12H2O ¾® 4Xe + 2XeO3 + 24HF + 3O2


7. (i) Ethylamine and aniline : Azo test

Ethylamine is a primary aliphatic amine which gives of N2 gas with the formation of alco-hol. ½

CH3CH2NH2 + HNO2 ¾® C2H5OH + N2 ̄ + H2O

Ethylamine Ethanol Nitrogen gas

Aniline being an aromatic primary amine forms a briliant orange dye with b-naphthol in sodium hydroxide ½

(ii) Aniline and Benzyl amine :

Benzyl amine will produce N2 gas on reaction with nitrous acid along with the formation of alcohol.

C6H5CH2NH2 + HNO2 ¾® C6H5CH2OH + N2 ¯ + H2O ½

Nitrogen gas

Aniline is an aromatic amine which diazotise on reacting with HNO2. It forms benzenediazo-nium chloride which on treatment with alkaline solution of 2-naphthol gives orange dye.½ C

HE

MIS

TRY

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8. (i) H2SO3

1

(ii) N2O5 :

1

9. All at corners so × =1 8 18

Cd at face centres = × =1 6 32

Au1Cd3OR

Given, r = 125 pm

For FCC unit cell body diagonal is equal to 4 times the radius of atom

\ 4r = 2a

Þ a = 4 4 125 353.55 pm2 2

r ×= = 2

10. (a) Formalin is 40% aqueous solution of metha-nol whereas trioxane is a trimer of methanol1

(b) 2-Bromo 2-methyl but 2-en-al. 111. AB A+ + B–

Moles dissolved m mol 0 0 Moles after dissociation m (1 – a) ma ma Total moles present in solution

= m(1 – a) + ma + ma = m[1 – a + a + a] = m[1 – a] 1

Degree of ionization, a of AB = 10 0.1100

= ½

Vn‘t Hoff factor, i =

Moles of solutes actually present in solutionMoles of solute dissolved

½

= [1 ]m

m+ α

= 1 + a = 1 + 0.1 = 1.1 112. Fuel cells are the electrical cells which are

designated to convert the energy from combustion of fuels such as hydrogen, methane, methanol etc. directily into electrical energy. This cell was used for providing electrical power in Apollo space programme. 1

In this cell, hydrogen and oxygen are bub-bled through a porous carbon electrode into concentrated aqueous NaOH. Catalyst like finely divided platinum or palladium metal are incorporated into the electrodes for increasing the rate of electrode reactions. (H2—O2) fuel cell electrode reactions are :

Cathode : O2(g) + 2H2O(l) + 4e– ¾® 4OH–(aq)

Anode : 2H2(g) + 4OH–(aq) ¾®

4H2O(l) + 4e– 1

Overall reaction : 2H2(g) + O2(g) ¾® 2H2O(l) 1

13. (i) CN– ion is known but CP– ion is known, It is because P is larger is size and cannot form a stable triple bond with carbon like nitrogen. 1

(ii) NO2 has an odd electron, which makes the molecule relatively unstable. By combining with another molecule NO2, it becomes stable as after it gets dimerise, it has even number of electrons.

(iii) I2 is non-polar while ICl is a polar mol-ecule. Thus, the bond between ICl breaks easily making it more reactive. 1

OR

(i) The electron donor species are known as Lewis base. Nitrogen atom in NH3 has a lone pair of electrons which is available for donation. Thus, it acts as a Lewis base. 1

(ii) NO2 contains odd number of valence elec-trons. It behaves as a typical odd electron system. On dimerisation it gets converted to stable N2O4 molecule with even number of electrons. 1

Solutions | 17

(iii) PCl3 reacts with moist air or moisture to form HCl fumes. ½

PCl3 + 3H2O ¾® H3PO3 + 3HCl ½14. (i) Antiferility drugs : Steriods are the ac-

tive ingredients of the pil functioning as an antifertility agents. When taken, it controls the female menstrual cycle and ovulation. The birth control pill is essen-tially a mixture of synthetic estrogen and progesterone derivatives, which are more potent than the natural hormones. e.g., : Norethindrone and ethynylestradiol. 1½

(ii) Tranquilizers : These are the class of chemical compounds which are used for treatment of stress, mild and severe mental diseases. These are used to relieve stress, fa-tigue, inducing a sense of well-being. They form an essential component of sleeping pills. Tranquilizers also form an important category of so called psychotherapeutic drugs e.g., : Derivatives of barbituric acid i.e., veronal, amytal. 1½

15. (i) When glucose reacts with NH2OH, glucose monoxime is formed.

HOH2C — (CHOH)4 — CH2OH + HONH2 HOCH2—(CHOH)4—CH = NOH 1

Glucose monoxime

(ii) When glucose reacts with nitric acid, sac-charic acid is formed.

H O H 2 C — ( C H O H ) 4 — C H O HNO3/oxidation¾¾¾¾¾¾¾¾®

HOOC2—(CHOH)4—COOH 1

Saccharic acid

(iii) When glucose react with Br2 water, gluconic acid is formed.

HOH2C—(CHOH)4—CHO + Br2 ¾®HOCH2—(CHOH)4—COOH 1

Gluconic acid16. Addition polymers (Chain growth polym-

erization) : These polymers are obtained by chain growth polymerisation of monomers, containing one or more double bonds. The addition polymers formed by the polymeriza-tion of a single monomer species are known

as homopolymers. For example : Polythene

nCH2 = CH2 ¾® —(CH2—CH2—)n ½ Ethene Polythene (Homopolymer)

The polymers formed by the addition poly-merisation from two different monomers are known as copolymers. For example ; Buna-S

1

nCH2 = CH — CH = CH2 +

1, 3-Butediene

nC6H5CH = CH2 ¾® styrene ¾ (CH2—CH = CH—CH2—CH2—CH)n |

C6H5 Condensation polymers (Step growth polym-

erization) : These polymers are obtained by step growth polymerization of two different monomers containing at least two functional groups. In these polymerization reactions, the elimination of small molecules such as water, alcohol, hydrogen, chloride etc. take place. For example : Nylon 6, 6 is formed by the condensation of hexamethylene diamine with adipic acid 1

nH2N(CH2)6NH2 + nHOOC(CH2)4COOH ¾® [—NH(CH2)6NHCO(CH2)4CO] — nH2O ½

17. (i) Rate = k[A]x [B]y

From experiment 1, 0.096 =K[0.30]x [0.30]y ...(a)

From experiment 3, 0.192 =K[0.30]x [0.60]y ...(b)

Dividing (a) by (b), we get

1 12 2y

= or 2 = 2y or y = 1 1

From experiment 3, 0.192 =k[0.30]x [0.60]y ...(c)

From experiment 4, 0.768 =k[0.60]x [0.60]y ...(d)

Dividing (b) by (d), we get

1 14 2x

= or 2x = 4 or x = 2 1

(ii) Rate = K[A]2 [B]1

18. (i) Neoprene Chloroprene/2-chloro-1, 3-butadiene

CH2 = C – CH = CH2 | Cl

18 | Oswaal CBSE Sample Question Paper, Chemistry, Class - 12 (ii) Polystyrene : Styrene

(iii) Polypropene : Propene

[CBSE Marking Scheme, 2014] 119. (i) [CO(NH3)5Cl]Cl2 IUPAC name : Pentamine

chloride cobalt (III) ½

Geometry : Octahedral as coordination number is 6

Magnetic moment : n = 0, m = ( 2)n n +

\ m = 0 , Diamagnetic ½ (ii) [CrCl3(PY)3]

IUPAC name : Trichlorido trispyridine chro-mium (III) ½

Geometry : Octahedral as coordination number is 6.

Magnetic moment : n = 3, m

= ( 2) 3(3 2) 15n n + = + = ½

(iii) K4[Mn(CN)6]

IUPAC name : Potassium hexacyano magnate (II) ½

Geometry : Octahderal as coordination number is 6.

Magnetic moment :

n = 1, ( 2) 1(1 2) 3n n + = + = ½

20. (a) Mond process for refining nickel Vapour phase refining : In this method, the metal is converted into its volatile compound which is collected in another container. It is than decomposed by heating to a higher temperature to give the pure metal. The two basic requirements are : 1½

(a) The metal should form a volatile compound with some suitable reagent.

(b) The volatile compound should be easily decomposable to give back the pure metal.

Ni + 300 330-350 K¾¾¾¾® Ni(CO)4 Impure Nickel carbonyl

450-470 K¾¾¾¾® Ni + 4CO

Pure

(ii) Column chromatography : This method is based on the fact that the various compo-

nents of a mixture are adsorbed to different extents on an adsorbent.

In chromatography, components of a mixture of compounds are distributed between two phases-a stationary phase and a mobile phase in column chromatography. Stationary phase can be a solid or a liquid while the mobile phase is a gas, liquid or super critical fluid. 1½

21. (a) Thermal decomposition of ammonium dichromate

(NH4)2Cr2O7 D

¾¾¾¾® N2 + 4H2O+ Cr2O3 1

(b) Reaction of Cl2 with cold and dilute NaOH

2NaOH + Cl2 ¾® NaCl + NaOCl + H2O 1 (c) When phosphine is passed through mer-

curic chloride solution

3HgCl2 + 2PH3 ¾® Hg3P2 + 6HCl 122. (a) The precipetate adsorbs one of the ions

of the electrolyte used as peptizing agent. This causes the development of positive or negative charge on the precipitate. Ultimately, the precipitate breaks up into particles of colloidal size on shaking. 1

(b) Smoke coming out of the chimney in the factories is a colloidal solution of carbon particles in air. The smoke is led through a chamber fitted with plates having charge opposite to that of smoke particles. The carbon particles in smoke on coming in contact with the plates lose their charge and get precipitated. The particles get collected on settling. 1

(c) Colloidal gold is used for intramuscular injection. This is because colloidal medi-cines are more effective as they have large surface area and easily assimilated. 1

23. (i) These get burnt, produce CO2 which pol-lutes atmosphere and need to be replaced again and again. 1

(ii) (a) Brine solution is readily available in the form of sea water.

(b) Chlorine is obtained as by product which is very easily used. ½+½

(iii) It makes use of KCN which is very poi-sonous and can be harmful if not handled properly. 1

(iv) AgNO3 can be used for silver plating and AuCl3 can be used for gold plating. 1

Solutions | 1924. (a) (i) Clemmensen reaction : Reduction of ketones/aldehydes to alkanes using zinc amalgem

and hydrochloric acid. 1

(ii) Cannizaro reaction : Redox reaction in which two molecules of an aldehyde are reacted to produce a primary alcohol and a carboxylic acid using a hydroxide base. 1

(b) (i)

(ii)

(iii)

OR

(a) (i) Hell Volhard Zelinsky reaction : Organic reaction to convert a carboxylic acid with an a-hydrogen and a halogen, to an a-halo carboxylic acid, using a phosphorus catalyst and water.

RCH2COOH (i) X2/Red Phosphourus¾¾¾¾¾¾¾¾¾¾®

(ii) H2O R – CH – COOH

| X

(ii) Wolf Kishner reduction reaction : Organic reaction used to convert an aldehyde or ketone to an alkane using hydrazine, base and thermal conditions. 1

(b) (i) C2H5CN H2O/H

+¾¾¾¾®

partial CH3 – CH2 – CONH2

Br2/KOH¾¾¾¾® CH3 – CH2 – NH2 HNO2¾¾¾¾®

CH3 CH2OH [O] ¯ KMnO4 CH3COOH

(ii) CH3 – CH2 – CH2 – CH2OH [O]¾¾¾¾®

KMnO4 CH3CH2CH2 – COOH

(iii)



MnO4– + 8H+ + 5Fe2+ ¾®

5Fe3+ + Mn2+ + 4H2O+

25. (a) Formation of KMnO4 : It is prepared by fusing manganese dioxide (MnO2) with caustic potash (KOH) or potassium carbonate in the presence of air to give a green mass due to the formation of potassium manganate.

2MnO2 + 4KOH + O2 heat

¾¾¾¾® 2K2MnO4 + 2H2O 1 Potassium manganate (green mass) The solution is then treated with a current of

Cl/O3/CO2 to Oxidise Potassium mangnate to potassium permangnate. The solution is concentrated and the dark purple crystals of potassium permangnate separate.

2K2MnO4 + Cl2 ¾® 2KCl + 2KMnO4 1 (b) Reaction of acidified permanganate solution

with (i) SO2 2MnO4

– + 5SO2 + 2H2O ¾® 5SO42– + 2Mn2+

+ 4H+ 1 (ii) Oxalic acid 2MnO4

– + 16H+ + 5C2O42– ¾® 2Mn2+ + 8H2O

+ 10H2O 1OR

(i) Transition metals are much harder than

alkali metals as they have stronger metallic bonding. 1

(ii) Actinoid contraction is greater than lantha-noid contraction from element to element due to poorer shielding of 5f electrons in actinoids. 1

(iii) Transition elements have smaller size and higher charge. Therefore, they have high value of hydration enthalpy. 1

(iv) The transition metals (except Zn, Cd and Hg) are hard and have high melting and boiling points. This is due to presence of stronger interatomic interactions which leads to strong metallic bonds among atoms. 1

(v) As we move from left to right the 1st se-ries, as effective nuclear charge increases, it is expected that 1st ionization enthalpy should show an increasing alters the rela-tive energy of 4s and 3d orbitals. Thus, there is reorganization energy accompanying ionization enthalpy. This results into the release of exchange energy which increases as the number of electron increases in the dn configuration and also from the transfer-ence of Selectrons into orbitals. 1

26. (a) Effect of pressure on solubility of gas :

The concentration of dissolved gas is proportional to the pressure on the gas above the solution. 1 (b) p = KH x ½ p = partial pressure of the gas in vapour phase x = mole fraction of the gas in solution KH = Henry‘s law constant ½ (c)

Solutions | 21 (d) The solubility of gas is releated to mole fraction in aqueous solution. The mole fraction of the

gas in the solution is by applying Henry‘s law. Thus : X(Nitrogen) =

5

H

P (nitrogen) 0.987 bar1.29 10

K 76480 bar= = × ½

As 1 litre of water contains 55.5 mol of it, therefore if n represents number of moles of N2 in solution.

X(Nitrogen) = –5 mol 1.29 10

mol 55.5 mol 55.5n n

n= = ×

+ ½

(n is denominator is negleted as it is < < 55.5) Thus n = 1.29 × 10–5 × 55.5 mol = 7.16 × 10–4

mol ½

= –47.16 10 mol 100 mol

0.716 mol1 mol

× × = ½

OR

(a) The molecular mass obtained with the help of colligative property is sometimes different from normal molecular mass, it is called abnormal molecular mass. 1

The factors which bring abnormality are :

(i) Association : When solute particles undergo association number of parti-cles becomes less and molecular mass

determined with help of colligative property will be more. 1

(ii) Dissociation : When solute particles undergo dissociation there is an in-crease in the number of particles, thus the observed molecular weight is decreased. 1

(b) Non-ideal solutions showing positive deviation have forces of attraction less than that of between pure components. Therefore, their vapour pressure is higher. 1

For example : water and benzene, chloroform and methanol. ½

l

SAMPLE QUESTION PAPER - 9Self Assessment___________________________________


1. Isopropyl alcohol or Propan-2-ol CH3 – CH – CH3 |

OH [CBSE Marking Scheme, 2014] 1

2. The substances which are added to enhance the non-wettability of the mineral particles in froth floation process are known as col-lectors e.g : Pine oil, fatty acids, xanthates

(any one) [CBSE Marking Scheme, 2014] 1

3. Diazotisation reaction.


4. Amylase. 1

5. Linkage and coordination isomerism[CBSE Marking Scheme, 2014] 1

6. (i) The resulting solution will show nega-tive deviation. After mixing liquids X and Y, temperature will be high. 1

(ii) When we place the blood cell in wa-ter, water will flow into the cell and it would swell up due to osmosis. 1

7. Kohlrausch‘s law of independent migration of ion states that “at infinite dilution when dissociation is complete each ion makes a definite contribution to molar conductivity of the electrolyte irrespective of other ion with which it is associated i.e., each ion

contributes to the limiting molar conductiv-ity of the electrolyte. e.g.,

L°HCl = l°Cl– + l°H+ Conductivity of a solution is the conductance

of ions present in unit volume of sovution. As with dilution, number of ions present in unit volume of solution decreases, con-ductivity also decreases. 2


OR

CH3COOH is a weak electrolyte. On dilu-tion, the number of ions increases due to increase in degree of dissociation as shown below : 1

CH3COOH + H2O CH3COO– + H3O

+

CH3COONa is a strong electrolyte. On dilution, there is not much change in the number of ions. Only interionic attraction decreases. Therefore, there is a small increase in Lm.

8. (i) Zero order reaction 1

(ii) The slope of the curve is negative i.e., negative of rate constant.


9. (i) P4 + H2O ¾® No reaction 1 (ii) XeF4 + O2F2 ¾® XeF6 + O2


10. (i) Reimer-Timann reaction : Phenol reacts with CHCl3 in the presence of NaOH at 340 K followed by hydrolysis to form salicyladehyde. ½

CH

EM

ISTR

Y O

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SE C

lass

-12

, E

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Solutions | 23 (ii) Williamson synthesis : It is used for the preparation of both symmetrical ethers. It is a nucle-

ophiclic substitution reaction (SN2). ½

R – X + R O: Na+ ® R – OR' + NaX (where x = halogen) e.g., : CH3Cl + CH3CH2 — O

–Na+ ® CH3 – O – CH2 – CH3 + NaCl ½ [CBSE Marking Scheme, 2014]

11. (a) (i)

(ii)

(b) (i) CH3I will react faster in SN2 reaction. ½

(ii) CH3Cl will react faster in SN2 reaction. [CBSE Marking Scheme, 2014] ½

12. K1 = 2.15 × 10–8 L/mol. s K2 = 2.39 × 10

–7 L/mol. s T1 = 650 K T2 = 700 K R = 8.314 JK

–1 mol–1

log

2

1

kk

=

2 1

1 2

T T2.303 T T

EaR

−

½

or log 7

182.39 10

2.15 10

−

−××

=

× ×

E 700 – 6502.303 8.314 700 650

a ½

or log 1.111 × 10 = ×

E 502.303 8.314 4,55,000

a

log 11.11 = ××E 1

2.303 8.314 9100a

1.046 = ×E 1

19.147 9100a

Ea = 1.046 × 19.147 × 9100

= 182252.6342 J mol–1 = 182.25 kJ mol–1


13. (a) Zwitter ion structure of glycine is 1

O O

|| || R – CH – C – O – H R – CH – C – O–

| |

:NH2 +NH2 (Zwitter ion)

(b) The change in the specific rotation of sugar from dextro-rotatory (+) to levorotatory (–) after its hydrolysis is known as inversion of sugar. ½+½

14. Analgesic medicines are those drugs which are used as pain relievers. e.g.,: Aspirin. 1

These are of two types : (i) Non-narcotic Drugs. (ii) Narcotic Drugs. Non-narcotic drugs reduces fever/has anti-

blood clotting action and is used in prevention of heart attacks/relieves skeletal pain, etc. Eg Aspirin Paracetamol. ½

Narcotic drugs are used for the relief in post

operative pains/cardiac pain / pains of ter-minal cancer. e.g., : Morphine. ½


15. Enzymes are the proteins which performs the role of biological catalysts in the body. ½

Receptors are the membrane proteins which create a signal transduction pathway so they cause the cell to communcate. ½

Catalytic action of enzymes :

24 | Oswaal CBSE Sample Question Paper, Chemistry, Class - 12 The enzymes perform two major functions in

catalyzing the reaction :

(i) The first function is to hold the substrate for a chemical reaction through interactions such as ionic bond, hydrogen bonding, vander Waals etc. Active sites of enzymes hold the substrate molecule in a suitable position, so that it can be attacked by the reagent effectively. 1

(ii) The second function of an enzyme is to provide functional groups that will attack the substrate and carry out chemical reaction. 1

16. a = 1.496 × 10–10 m

We know, radius r = 3

4 × 1.49 × 10–10

= 0.647 × 768 × 10–10 m 1

d = 19.3 g/cm3 = 1.496 × 10–10 m

= 1.496 × 10–8 cm 1

d = 3ZM

Na o 1

M = 3

0NZ

d a× ×

= 8 3 2319.3 (1.496 10 ) 6.022 102

× × × ×

= 19.456 g/mol 117. (a) (i) CH3NHCH(CH3)2 is N-Methyl propan-

2-amine. ½ (ii) m-BrC6H4NH2 is 3-Bromobenzamine or

3-Bromoaniline ½ (b) (i) Chirality is the property of a molecule

containing a carbon attatched to four different groups whereas chiral centre is the carbon atom which is attached to four different groups. 1

(ii) Enantiomers are non-superimposable mirror images of each other. They have optical roation equal in magnitude but opposite in sign while Diastereomers are non-superimposable and non-mirror images of each other. They differ in optical rotation. 1

18. (a) This is because of electron releasing ten-dency of alkyl group. The alkyl group shows +I effect which increases the electron density at Natom in alkyl amine which makes it more basic than ammonia. 1

(b) (i) C6H5NH2 NaNO2 + HCl¾¾¾¾¾¾¾®

273 K C6H5N2

+Cl–

NaNO2 + HCl¾¾¾¾¾¾¾®273 K

C6H5NO2

(ii) C6H5NH2

NaNO2 + HCl¾¾¾¾¾¾¾®273 K

C6H5N2+Cl–

KI¾¾¾¾¾¾¾® C6H5I

19. (a) Butanamine (1º) CH3CH2CH2CH2NH2 ½

(b) Butan-2-amine (1º) CH3—CH2—CH—CH3 ½

|

NH2(c) 2-Methyl propanamine (1º)

CH3—CH—CH—NH2 ½

|

CH3(d) 2-Methyl propan-2-amine

CH3 |

(1º) CH3—C—NH2 ½

|

CH3 (e) N-Methyl propanamine (2º)

CH3—CH2—CH2—NH—CH3 ½

(f) N-Ethylethanamine (2º)

CH3—CH2—NH—CH2—CH3 ½20. a-glucose and b-glucose differ only in the ori-

entation of the hydroxyl group at C1 position. In a-glucose the OH group is on right hand side at C1 position whereas in b-glucose the OH group is on left hand side at C1 position a-glucose is the monomer unit of starch and b-gluose is the monomer unit of cellusose. 2

The term pyranose is used for carbohydrates derived from word pyran. Its chemical struc-ture includes that of a six membered ring with 5 carbon atoms and one oxygen atom. e.g., : Glucose. [CBSE Marking Scheme, 2011] 3

21. (a) (i) Novolac and Bakelite (structure) : No-volac is a linear condensation polymer of o-hydroxy methyl phenol formed by reaction between phenol and formal-dehyde. Bakelite is a cross-linked con-densation polymer of o-and p-hydroxyl methylphenol formed by the reaction between phenol and formaldehyde. 1

(ii) Buna-S is formed by addition poly-merisation. It has weak intermolecu-lar forces of attraction. Terylene is obtained by condensation between ethylene glycol and terephthalic acid and has stronger intermolecular force of attraction (dipole-dipole) 1

(b) Elastomers or rubbers have the weakest

Solutions | 25intermolecular forces of attraction followed by plastics while fibres have the strongest forces of attraction. Thus, the increasing in-termolecular forces of attraction of follows the order : Elastomer < Plastic < Fibre ½

Buna-S, Polythene, Nylon-6, 6 ½22. (i) Brown colour is formed due to evalution of

I2. Excess of Cl2 reacts with I2 to form colourles ICl3.

Cl2 + 2KI ¾® 2KCl + I2 ½ Brown

3Cl2 + I2 ¾® 2ICl3 ½ Colourless

(ii) The factors which are responsible for the abnormal behaviour of fluorine are :

(a) Smallest size ½

(b) Highest electronegativity ½

(c) High electron affinity ½

(d) Non-availability of d-orbitals ½

(iii) White phsophorus is much more reactive than red phsophorus because white phos-phorus is monomeric whereas red phso-phorus is polymeric. White phsophorus has lower bond dissociation energy than

red phsophorus. 1

OR

(i) ICl is more reactive than I2 because I—Cl bond is weaker than I—I bond due to bond dissociation energy. Hence, ICl breaks eas-ily to form helogen atoms which readily bring about the reactions.

(ii) Bond strength of S-S bond is higher than that of O—O band. The values are 226 kJ/mol and 142 kJ/mol respectively. Therefore, sulphur has a greater tendency for catena-tion than oxygen. Oxygen on the other hand forms O = O due to the smaller size of oxygen. 1

(iii) As radon is a radioactive element with a short half life of 3.82 days, therefore it becomes difficult to study the chemistry of radon. 1

23. (a) Lokesh is kind hearted and helpful person. He knows how soil caused pollution is caused by chemical industrial waste.

10FeSO4 + 2KMnO4 + 8H2SO4 ¾® K2SO4 + 2MnSO4 + 5Fe2(SO4)3 + 8H2O 1+1

(ii) 5H2C2O4 + 2KMnO4 + 3H2SO4 ¾® K2SO4 + 2MnSO4 + 8H2O + 10CO2 1

24. (a) Gatterman-Koch reaction :

1

(b) Etard reaction :

(c) Stephen reaction :

RCN + SnCl ¾® RCH = NH H3O

+¾¾¾¾® RCHO 1

(d) Wolff-Kishner reduction :

(e) Hell-Volhard-Zelinsky reaction :

Or


(a) 2HCHO conc. KOH

¾¾¾¾¾® HCOOK + CH3OH “A” “C” “B”

CH3OH Cu/578 K

¾¾¾¾¾® HCHO “B” “A” 1+1+1+1

HCOOK + HCl ¾¾¾¾¾® HCOOH + KCl “C” “D”

(b) It is due to resonance

The position of carbonyl group keeps changing. Therefore, carboxylic acids do not give reactions of aldehydes and ketones. 1

25. (a) Let A and B represent the solvent and the solute respectively. XA and XB are the mole fraction of solvent and solute respectively, PA and PA are the vapour pressure of A in the pure liquid and solution respectively.

PA µ xA ...(1) xA + xB = 1 or xA = 1 – xB ...(2) ½ From (1) and (2), we get PA = PA

0 (1 – xB) ½

or A°A

P

P = (1 – xB)

or xB = °

A A AB° °

A A

1 P P P

P Px

− −= =

Relative lowering of vapour pressure of a solu-tion is equal to mole fraction of the solute.

(b) °

A A°A

P P

P

− = ixB, where i is the Vant Hoff

factor ...(3)

In the present case, as benzoic acid completely

dimerises in benzene, i = 1¾2

Now, no. of moles of benzoic acid =

Given mass 61 1Molar mass 122 2

= =

Now, no. of moles of benzene = 50078 Mole fraction xB =

No. of moles of benzoic acidNo. of moles of benzene + No. of moles of benzoic acid

=

12

500 178 2

+

Substituting the values in equation (3) we get,

°A A

A

P PP− =

11 1 12

500 1 500 12 478 2 78 2

× = ×+ +

AP 66.6

66.6−

=

1 1500 1478 2

×+

PA = 64.17 torr 3 (ii) In the absence of dimerisation

1 – A

°A

P

P

= 1

500 178 2

+

1 – AP

66.6 =

1500 178 2

+

PA = 61.78 torr 1OR

(a) The process of flow of solvent molecules from lower concentration to higher concen-tration of solution is termed as osmosis. 1

(b) Osmotic pressure (p) is the pressure which must be applied to the solution to prevent the passage of solvent through a semiper-meable membrane. 1

(c) Osmotic pressure is a colligative property as it depends on the number of solute mol-ecule and not on their identify. For dilute solutions, it has been found experimentally that osmotic prossure is proportional to the molarity, C of the solution at a given temperature T. Thus :

p = CRT Here p is the osmotic pressure and R is the

gas constant.

p = 2 RTVn

Here, V is the volume of a solution in litres containing n2 moles of solute. If w2 grams of

Solutions | 27solute, of molar mass M2 is present in the

solution, then n2 = 2 2

2 2

RT and V=

M Mw w

π

or M2 = 2RTV

wπ

Thus, knowing the quantities w2, T, p, and V we can calculate the molar mass of the solute.3

26. (i) Due to lanthanoid contraction, Hf has similar atomic size (158 pm) to that of Zr (160 pm). 1

(ii) f-block elements show limited number of oxidation states due to large energy gap between outer f, d and s-sub shells. 1

(iii) Due to the presence of large number of unpaired electrons in their atoms, they form strong interatomic metallic bonds. Hence, transition metals have high enthalpies of at-omization. 1

(iv) The variability in oxidation states of tran-sition metals is due to incomplete filling of d-orbitals in such a way that their oxidation states differ from each other by unity. e.g., Fe2+ and Fe3+. In case of non-transition elements, the oxidation stats differ by units of two. e.g., Pb2+ and Pb4+ and this is due to ns electrons.1

(v) K2Cr2O7 + 2NaOH ¾® Orange

K2CrO4 + Na2CrO4 + H2O 1 Yellow Yellow [CBSE Marking Scheme, 2013]

OR (a) (i) Electronic configuration : Lanthanoids : All Lanthanoids have elec-

tronic configuration with 6s2 common but with

variable occupancy of 4f and 5d sub shells. Actinoids : All actinoids have the electronic configuration of 7s2 and variable occupancy of 5f and 6d subshells. 1

(ii) Oxidation states : Lanthanoids : Besides common oxidation states

of + 3, some lanthanoids exhibit + 2 and + 4 oxidation states.

Actinoids : Besides + 3, they show higher oxidation states of + 4, + 5, + 6 and + 7 also + 2 is quiterare.

(iii) Chemical reactivity : Lanthanoids : These are less reactive metals

and form oxides, sulphides, nitrides, hydrox-ides and halides etc. These also form H2 with acids. They show a lesser tendency for complex formation.

Actinoids : They are highly reactive metals especially when are in finely divided state. They form a mixture of oxide and hydride by action of boiling water. They combine with non-metals even at moderate temprature. They show a greater tendency for complex forma-tion.

(b) KMnO4 as oxidising agent in acidic me-dium :

(i) Ferrous salts to ferric salts : MnO4

– + 5Fe2+ + 8H+ ¾® Mn2+ + 5Fe3+ + 4H2O 1

(ii) H2S to sulphur :

2MnO4– + 5H2S + 6H

+ ¾® 2Mn2+ + 8H2O + 5S [CBSE Marking Scheme, 2011] 1

l

SAMPLE QUESTION PAPER - 10Self Assessment___________________________________


1. Dialysis is the process of removal of soluble impurities from sols by a semipermeable membrane. e.g., : ferric hydroxide sol.


2. 2-Methyl propan-1-al


3. Fluorine (F2) is a stronger oxidising agent than chlorine (Cl2) because the reduction potential of F2 is higher than that of Cl2. It is due to the low bond dissociation energy and high hydration energy of F– ion as compared to Cl– ion. Moreover, the tendency to accept an electron readily also makes F2 a strong oxidizing agent.


4. Metallic solid can conduct electricity in solid state while ionic solids cannot due to absence of mobile electrons.


5. Glycogen. 16. A unit cell is characterized by the following

parameters : (a) The dimensions of the unit cell along

the edges : These are represented by a, b and c. The edge may or may not be mutu-ally perpendicular. 1

(b) The angles between the edge : These are represented by a, b and g. The angle a is between b and c, b is between a and c and g is between a and b. 1

OR

d = 3 –30

A

Z × M

N 10a × ×

Z= –3 –8 2 23 –1

–14.0g cm (5 10 cm) (6.023 10 mol )

(56 16) g mol

× × ×

+

= 4 i.e., 4Fe2+ and 402– ions.


7. (i) [Cr (NH3)2 Cl2 (en)] Cl

diamminedichlorido chromium (I) chloride 1

(ii) Penta-amine nitrosocobalt (III) ion [Co(NH3)5 ONO]

2+ 18. (a) Ag+(aq) + e– ¾® Ag(s) E0 = + 0.80 V

H+(aq) + e– ¾® 1¾2

H2(g) E0 = 0.00V

On the basis of their standard redirection potential (E0) values, cathode reaction is given by the one with higher Eº values.

Thus, Ag+(aq) + e– ¾® Ag(s) reaction will be more feasible at cathode. 1

(b) Limiting molar conductivity : When the concentration approaches zero, the molar conducitivity When the concentration ap-proaches zero, the molar conducitivity is known as limiting molar conductivity.

It is represented by Lm. 19. In electrolytic refining, the impure metal to

be purified is made the anode, while a thin sheet of the pure metal is made the cathode. The electrolyte taken is a salt solution of the same metal.

When current is passed through it, metal ions from the solution goes to cathode and accept electron and gets deposited on it metallic form. In the mean time, equal amount of metal at anode lose electrons C

HE

MIS

TRY

Osw

aal C

BSE

Cla

ss -

12,

Exa

min

atio

n Sa

mp

le Q

uest

ion

Pap

er

Solutions | 29and come to the solution. The pure sheet of metal made as cathode gets thicker while the impure metal electrode gets thinner. The other impurities gets settled below anode as diode mud.

Metals like Cu, Zn etc. are purified by this method. [CBSE Marking Scheme, 2014] 2

10. (i) The deficiency of vitamin A causes night blindness while deficiency of vitamin C causes scrury, bleeding of gums and loosening of teeth. Thus, both the vita-mins are essential for us to prevent these diseases. ½+½

(ii) Nucleoside : A unit formed by the at-tachment of base to 1¢ position of sugar is known as nucleoside. 1

Nucleotide : It is phosphate ester of nucleoside made up of three parts i.e., phsophoric acid group, pentose sugar and a nitrogenous base.


1

11. (i) Animals hides are colloidal in nature.

When a hide which has positively charged

particles, is soaked in tanning which con-

tains negatively charged colloidal particles,

mutual coagulation takes place. This results

in the hardening of leather. 1

(ii) Lyophilic colloids have great affinity for

the dispersion medium i.e., dispersed phase

particles are solvated to a greater extent

in case of lyophilic colloids. Hence, lyo-

philic sols are relatively more stable than

lyophobic sols. 1

(iii) It is necessary to remove CO when am-

monia is prepared by Haber‘s process

because it acts like a poison which reduces

the activity of catalyst ion. 1

12. (i) Aspirin acts as a blood thinner. Therefore,

the chances of heart attack are reduced. 1

(ii) Artificial sweeteners have almost zero

calorific value. They do not require insulin

which is restricted quantity in diabetic per-

sons. On the other hand, if natural sweet-

ener is consumed by a diabetic person, it

will raise the level of blood sugar. 1

(iii) Soaps are decomposed by micro organ-

isms present in sewage whereas detergents

are non-decomposable as they are highly

brached. 1

13. Altthough chlorine is an electron withdrawing group, yet it is ortho-para-directing in nature in elecrtrophilic aromatic substitution because when chlonine is present in benzene ring it releases electron by resonance whereas it acts as withdrawing group only through inductive effect.

By inductive effect, chlorine atom destabilizes the intermediate carbocation formation but by resoance, chlorine atom stabilises the intermediate carbocation and effect is more at o and p-positions. 2

Resonance effect opposes inductive effect which is stronger than resonance effect.


14. (i) CH3CH2CH2COOH ½+½

Butanoic acid

(ii) CH3—CH = CH—COOH

But-2-en-1-oic acid or 2-butenoic acid ½+½

(iii) CH3—CH(CH3)—COOH

2-methyl propanoic acid

[CBSE Marking Scheme, 2012] ½+½

15. (a) Alcohols are more soluble in water than the hydrocarbons of comparable molecular masses because alcohols form hydrogen bonds with water. 1

(b) Ortho-nitrophenol is more acidic than ortho-methoxy phenol because —NO2 is electron withdrawing group hence more acidic as electron withdrawing group a-acidic character. 1

(c) Due to the presence of an alkyl group, higher electron density is found in alkox-ideion. 1

16. Edge length = 200 pm Volume of the unit cell = (200 × 10–10 cm)3 =

8 × 10–24 cm3 1

In an FCC unit cell = 23

200 4

24 10

××

= 33.3 × 10–23 g

Density = Mass of unit cell

Volume of unit cell =

–23

–24 333.3 10 g

8 10 cm

×

×


OR

(a) In schottky defect positive and negative ions are missing from their respective position leaving behind a pair of holes. 1

That is why it lowers the density of solids.

(b) Phosphorus contains five valence electrons. When silicon is doped with. Phosphorus only

four valence electrons of phosphorous are used to form four covalent bonds with four neigh-bouring groups. Then the fifth electron gets delocalised and increases the conductivity. 1

17. (i) (NH4)2Cr2O7 heat

¾¾¾¾® N2 + 4H2O + Cr2O3 1

(ii) 2NaOH + Cl2 ¾® NaCl + NaOCl + H2O 1

(iii) 2PH3 + 3HgCl2 ¾® Hg3P2 + 6HCl 1

18. (a) RCONH2 + Br2 + 4NaOH D

¾¾¾¾® R—NH2 + Na2CO3 + 2NaBr + 2H2O 1

(b) Hoffmann Bromamide reaction.

(c) CH3 – CH – CH2 – CONH2 Br/NaOH

¾¾¾¾¾® | Cl 2-Chloro butanamide CH3 – CH – CH2 – NH2 | Cl

2-Chloro propanamine ½+½

19. (a) t = 60 min A = 12

A0

K = 0A2.303

logAt

½

= 0A2.303

log60 ½A

1

K = 2.303

log 260

K = ×2.303 0.301060

½

= 0.011 min–1 1

(b) k = 2.303

loga

t a x−

or 2.303

kt

= log

aa x−

420 10 9002.303

−× ×

= log

aa x− ½

log

aa x−

= 0.7815 ½

aa x−

= 0.1547 ½

Solutions | 31

1

xa

− = 0.1547

xa

= 0.8453 = 84.53% ½

20. (i) Micelles may be defined as the cluster or aggregated particles formed by associated colloids in solution e.g., sodium lauryl sulphate.

[CH3(CH2)11SO4–Na+], sodium stearate

[CH3(CH2)16COO–Na+]. 1

(ii) Peptization is the process of conversion of precipitate into colloidal solution in presence of peptizing agent. Peptizing agent is generally an electrolyte. 1

(iii) Desorption is the process of removing an absorbed substance from a surface on which it is absorbed. Eg. animal charcoal removes colour from surface of organic compound. 1


21. (a) Plane polarised light is the light which has vibrations in one plane only. 1

(b) A substance which rotates the plane of polarized light either towards clockwise or anti-clockwise direction is called opti-cally active substance. This phenomenon is called optical activity. 1

(c) Asymmetric molecule is a molecule in which at least one carbon is linked to four different groups. 1

22. (a) Despite having an aldehyde group, glucose does not give 2, 4-DNP test becuse —CHO group is not free in glucose. It is bound to —OH groups in the cyclic structure. This indicates that glucose does not exist in straight chain structure. 1

(b) Structure : 1

(c) D represents configuration of the compound i.e., —OH group on first chiral C atom from the bottom is downwards. (+) sign

shows that it is dextrorotatory i.e., rotates the planes of polarized light to the right.1

23. (i) (a) Nuclear explosion should be carried out far away from populated areas, like in deserts, so that people are not affected by radiations.

(b) Scientists should use lead coats covering whole body while carrying out nuclear explosions. 1½

(c) Nuclear wastes should be dumped in boron steel containers and buried deep inside the earth. 2

(ii) They may cause harm to flesh/skin, cause cancer and birth defects. 2

24. (a) CuF2 is coloured due to vacant d-orbital and presence of unpaired electron. 1

(b) (i) Cr in CrO42– has an oxidation state

of +6. It can reduce its oxidation state to +3 (in Cr3+). Thus, it can act as an oxidising agent. 1

Mn in MnO42– has oxidation state +

6 and has tendency to achieve noble gas configuration by going to + 7 state (d0) so it acts as reducing agent.

(ii) Zr and Hf have identical sizes due to lanthanoid contraction. 1

(iii) Oxides of Mn in lower oxidation states are ionic and form basic solution, whereas oxides in highest oxidation state are covalent and form acedic solution in water. 1

(iv) Mn2+ has the maximum number of unpaired electrons and therefore, has the maximum paramagnetic charac-ter. 1

OR

(a) HCl get oxidized to Cl2 by KMnO4 whereas H2SO4 does not. 1

(b) (i) Metals have maximum oxidation state in oxoanious as oxide ion has high electro negativity and small size and ability to form multiple bonds help it stabalises higher states.

(ii) Ce4+ can gain an electron and therefore it acts as an oxidizing agent. 1

Ce4+ + e– ¾® Ce3+ (iii) These electrons have voids in which

32 | Oswaal CBSE Sample Question Paper, Chemistry, Class - 12small atom can fit tightly to give in-terstitial compounds.

(iv) Zn2+ does not have vacant d-orbitals

or unpaired electrons in d-orbital. Cu2+ has one unpaired electron and vacant d-orbital so d-d transition is possible in Cu2+. 1

OH

|

25. CH3CHO + CH3CHO dil NaOH¾¾¾¾¾®

Aldol CH3CH – CH2CHO –H2O/Heat¾¾¾¾¾® CH3CH = CHCHO

(A) (A) Aldol (b-hydroxy aldehyde) 2-Butenal

5

OR

(i) Compound (A) forms 2, 4-DNP derivative, therefore, it is an aldehyde or a ketone. ½

(ii) As it does not reduce Tollen‘s or Fehling‘s reagent (A) must be ketone. ½

(iii) (A) responds to Iodoform test, therefore it should be a methyl ketone. ½

(iv) The molecular formula of (A) indicates high degree of unsaturation, yet it does not decolourise bromine water or Baeyer‘s reagent hence it must be aromatic. ½

(v) Compound (B), being an oxidation product of a ketone should be a carboxylic acid. The mo-lecular formula of (B) indicates that itr should be benzoic acid and compound (A) indicates that it should be phenyl methyl ketone (acetophenone). Methyl ketones react with I2/NaOH to give a yellow ppt. of iodoform.

Solutions | 3326. Empirical formula calculation :

Element Percentage At. mass Relative number of

atoms

Dividel by least No. of

atoms

Simplestratio

C 69.77 12 69.775.81

12= 5.81 5

1.16=

5

H 11.63 1 11.6311.63

1= 11.63 10

1.16=

10

O 18.60 16 18.601.16

16= 1.16 1

1.16=

1

1 Empirical formula = C5H10O Empirical formula mass = 5 × 12 + 10 × 1 +

16 = 86

n = Molecular mass 86

1Empirical formula mass 86

= =

Molecular formula = C5H10O 1

It forms addition compound with NaHCO3, therefore it is an aldehyde or ketone.

It does not reduce Tollen‘s reagent, therefore, it is not an aldehyde.

It gives iodoform test, therefore it is methyl ketone. Hence, the compound is

CH3CO—CH2—CH2—CH3CH3

K2Cr2O7/¾¾¾¾¾¾¾¾®conc. H2SO4,D

CH3COOH + CH3CH2COOH 1

(b) (i) Because of two chlorine atoms in dichlo-roethanoic acid, —I effect is double in this compound, favouring the release of protons. That is why dichloroethanoic acid is strong acid than monochloric ethanoic acid hence it has lower pKa.1

(ii) It is because benzoate ion is stabilized due to resonance whereas CH3COO

– is destabilised due to + I inductive effect of CH3 group. 1

OR

(a) Benzoic acid is less soluble in water than acetic acid due to larger hydrocarbon part in benzoic acid. 1

(b) It is because the ketone obtained would further react with Grignard reagent to form tertiary alcohol. 1

(c) Acetyl chloride is much more reactive than acetic anhydride and hence with acetyl chloride the acylation reaction are very fast and hence difficult to control, while with acetic anhydride are slow and hence easily controlled. 1

(d) An electron withdrawing group destabilzes the carbonium ion. Thus, a decativating group like —COOH deactivates all posi-tions in the ring but it deactivates the o-and p-positions even more as compared to meta position. Thus, the attack occurs mainly at meta position. 1

(e) Formic acid has aldehydic group

H — — OH, it contains aldehyde like hydrogen and thus is readily oxidised. Therefore, it reduces Tollen‘s reagent. 1

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SOLUTIONS...3. 1 (b) Fe. 3+ ions of FeCl. 3. neutralize the charge on the colloidal particles of...

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Transcript of SOLUTIONS...3. 1 (b) Fe. 3+ ions of FeCl. 3. neutralize the charge on the colloidal particles of...