SOLVED C.B.S.E. PAPER 2019 With Class–XII · Draw structures of geometrical isomers for this...

SOLVEDPAPER*

WithMarking Scheme

C.B.S.E.2019

Class–XIIDelhi / Outside Delhi

Chemistry

*Note : This paper is solely for reference purpose only. The format has now been modified by CBSE for March 2020 examination.

Time allowed : 3 Hours Max. Marks : 70

General Instructions : (i) All questions are compulsory. (ii) Section A : Question number 1 to 5 are very short answer questions and carry 1 mark each. (iii) Section B : Question number 6 to 12 are short answer questions and carry 2 marks each. (iv) Section C : Question number 13 to 24 are also short answer questions and carry 3 marks each. (v) Section D : Question number 25 to 27 are long answer questions and carry 5 marks each. (vi) There is no overall choice. However, an internal choice has been provided in two questions of one mark, two ques-

tions of two marks, four questions of three marks and all the three questions of five marks weightage. You have to attempt only one of the choices in such questions.

(vii) Use log tables, if necessary. Use of calculator is not allowed.

Delhi Set 1 56/1/1

SECTION A

1. Out of NaCl and AgCl, which one shows Frenkel defect and why? [1] 2. Arrange the following in increasing order of boiling points. [1]

(CH3)3N, C2H5OH, C2H5NH2 3. Why are medicines more effective in colloidal state? [1]

OR What is the difference between an emulsion and a gel? 4. Define ambidient nucleophile with an example. [1] 5. What is basic structural difference between glucose and fructose? [1] OR Write products obtained after hydrolysis of lactose.

SECTION B

6. Write the balanced chemical equations for the following process : [2]

(i) XeF2 undergoes hydrolysis.

(ii) 2MnO is heated with concentrated HCl.OR

Arrange the following in order of property indicated for each set. (i) 2 2 2 2H O, H S, H Se, H Te – increasing acidic character.

(ii) HF, HCl, HBr, HI – decreasing bond enthalpy. 7. State Raoult’s law for a solution containing volatile components. Write two characteristics of the solution which

obeys Raoult’s law at all concentration. [2] 8. For a reaction, [2]

the proposed mechanism is given below :

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2 Oswaal CBSE Solved Paper - 2019, Chemistry, Class-XII


(1) ( )2 2 2H O I H O IO slow− −+ → +

(2) ( )2 2 2 2H O IO H O I O fast− −+ → + +

(i) Write rate law of the reaction. (ii) Write overall order of reaction. (iii) Out of step (1) and (2), which one is rate determining step? 9. When MnO2 is fused with KOH in presence of KNO3 as an oxidizing agent, it gives a dark green compound

(A). Compound (A) disproportionate in acidic solution to give purple compound (B). An alkaline solution of compound (B) oxidises KI to compound (C) whereas an acidified solution of compound (B) oxidises KI to (D). Identify (A), (B), (C) and (D). [2]

10. Write IUPAC name of the complex ( ) 22Pt en Cl . Draw structures of geometrical isomers for this complex. [2]

OR Using IUPAC norms write the formulae for the following : (i) Hexaamminecobalt(III) sulphate (ii) Potassium trioxalatochromate(III)

11. Out of [ ]36CoF − and ( ) 33Co en

+ , which one complex is, [2]

(i) Paramagnetic, (ii) More stable, (iii) Inner orbital complex and (iv) High spin complex

(Atomic number of Co 27= ) 12. Write structures of compound A and B in each of the following reactions: [2]

(i)

CH2CH3

KMnO4-KOH AH3O

+

B

(ii)

OH

CrO3 A BH2N-NH-CONH2

SECTION C

13. The decomposition of NH3 on platinum surface is zero order reaction. If rate constant k is 3 14 10 Ms− −× , how long will it take to reduce the initial concentration of 3NH from 0.1 M to 0 064. M . [3]

14. (i) What is the role of activated charcoal in gas mask? [3] (ii) A colloidal sol is prepared by the given method in figure. What is the charge on hydrated ferric oxide colloidal

particles formed in the test tube? How is the sol represented?FeCl3

Solution

NaOH

Solution

(iii) How does chemisorption vary with temperature? 15. An element crystallizes in fcc lattice with a cell edge of 300 pm. The density of the element is 10.8 g cm-3. Calculate

the number of atoms in 108 g of the element. [3] 16. A 4% solution w/w of sucrose (M = 342 g mol-3) in water has a freezing point of 271.15 K. Calculate the freezing

point of 5% glucose (M = 180 g mol-1) in water. (Given: Freezing point of pure water = 273.15 K) [3]


Oswaal CBSE Solved Paper - 2019, Chemistry, Class-XII 3


17. (a) Name the method of refining which is : (i) Used to obtain semiconductor of high purity, (ii) Used to obtain low boiling metal.

(b) Write chemical reactions taking place in the extraction of copper from Cu2S. [3] 18. Give the reasons for following: (i) Transition elements and their compounds acts as catalysts. (ii) E° value for (Mn+2|Mn) is negative whereas for (Cu+2|Cu) is positive. (iii) Actinoids show irregularities in their electronic configuration. [3] 19. Write the structure of monomers used for getting the following polymers : (i) Nylon-6,6 (ii) Glyptal (iii) Buna – S [3]

OR

(i) Is

CH3

CHCH2[ [

n

a homopolymer or copolymer? Give reason.

(ii) Write the monomers of the following polymer:

NHN NH – CH2

NN

NH

(iii) What is the role of Sulphur in vulcanization of rubber? 20. (i) What type of drug is used in sleeping pills? [3] (ii) What type of detergents are used in toothpastes? (iii) Why the use of alitame as artificial sweetener is not recommended?

OR Define the following terms with suitable example in each : (i) Broad-spectrum antibiotics (ii) Disinfectants (iii) Cationic detergents 21. (i) Out of (CH3)3C-Br and (CH3)3C-I, which one is more reactive towards SN1 and why? [3] (ii) Write the product formed when p-nitrochlorobenzene is heated with aqueous NaOH at 443 K followed by

acidification. (iii) Why dextro and laevo – rotatory isomers of Butan-2-ol are difficult to separate by fractional distillation?

22. An aromatic compound ‘A’ on heating with Br2 and KOH forms a compound ‘B’ of molecular formula C6H7N which on reacting with CHCl3 and alcoholic KOH produces a foul smelling compound ‘C’. Write the structure and IUPAC names of compound A, B and C. [3]

23. Complete the following reactions : [3]

(i)

CHO

NaCN/HCl

(ii) (C6H5CH2)2Cd + 2CH3COCl

(iii)

CH3 — CH — COOH

CH3—

(i)Br /Red P2 4

(ii)H O2

OR Write chemical equations for the following reactions : (i) Propanone is treated with dilute Ba(OH)2. (ii) Acetophenone is treated with Zn(Hg)/ Conc. HCl. (iii) Benzoyl chloride is hydrogenated in presence of Pd/BaSO4.




24. Differentiate between the following : [3] (i) Amylose and Amylopectine (ii) Peptide linkage and Glycosidic linkage (iii) Fibrous proteins and Globular proteins

OR Write chemical reactions to show that open structure of D-glucose contains the following : (i) Straight chain (ii) Five alcohol groups (iii) Aldehyde as carbonyl group

SECTION D

25. ocellE for the given redox reaction is 2.71 V.

Mg(s) + Cu2+(0.01 M)

Mg2+(0.001 M) + Cu(s) [5]

Calculate cellE for the reaction. Write the direction of flow of current when an external opposite potential applied is

(i) Less than 2.71 V (ii) Greater than 2.71 V

OR (a) A steady current of 2 amperes was passed through two electrolytic cells X and Y connected in series containing

electrolytes FeSO4 and ZnSO4 until 2.8 g of Fe deposited at the cathode of cell X. How long did the current flow? Calculate the mass.

(Molar mass: Fe = 56 g mol-1, Zn = 65.3 g mol-1, 1 F = 96500 C mol-1)

(b) In the plot of molar conductivity Λm vs. square root of concentration ( )c12 , following curves are obtained for two

electrolytes A and B:

Λ

Answer the following : (i) Predict the nature of electrolytes A and B.

(ii) What happens on extrapolation of Λm to concentration approaching zero for electrolytes A and B? 26. (a) How do you convert the following? [5] (i) Phenol to Anisole. (ii) Ethanol to Propan-2-ol. (b) Write mechanism for the following reaction :

C H OH CH CH H OH SO

2 5 443 2 2 22 4

K → =

(c) Why phenol undergoes electrophilic substitution more easily than benzene?OR

(a) Account for the following: (i) o-nitrophenol is more steam volatile than p-nitrophenol. (ii) t-butyl chloride on heating with sodium methoxide gives 2-methylpropene instead of t-butylmethylether. (b) Write the reaction involved in the following: (i) Reimer-Tiemann reaction. (ii) Friedal –Crafts Alkylation of Phenol. (c) Give simple chemical test to distinguish between Ethanol and Phenol.




27. (a) Give reasons for the following: [5] (i) Sulphur in vapour state shows paramagnetic behaviour. (ii) N-N bond is weaker than P-P bond. (iii) Ozone is thermodynamically less stable then oxygen. (b) Write the name of the gas released when Cu is added to : (i) Dilute HNO3 (ii) Conc. HNO3

OR (a) (i) Write the disproportionation reaction of H3PO3. (ii) Draw the structure of XeF4. (b) Account for the following: (i) Although fluorine has less negative electron gain enthalpy yet F2 is a strong oxidizing agent. (ii) Acidic character decreases from N2O3 to Bi2O3 in group 15. (c) Write a chemical reaction to test Sulphur dioxide gas. Write the chemical equation involved.

Delhi Set 2 56/1/2

SECTION A

2. Arrange the following in increasing order of base strength in gas phase : [1] (C2H5)3N,C2H5NH2,(C2H5)2NH 3. Why the conductivity of silicon increases on doping with phosphorus? [1] 5. Write the IUPAC name of the given compound :

[1]

SECTION B

8. Write two differences between an ideal solution and a non-ideal solution. [2] 10. Write IUPAC name of the complex [Cr(NH3)4Cl2]

+ . Draw structures of geometrical isomers for this complex. [2]OR

Using IUPAC norms write the formulae for the following : (i) Pentaammine nitrito-o-cobalt (III) chloride (ii) Potassium tetracyanonickelate (II)

11. Out of 3

6CoF-é ùê úë û and ( )4

3

32C OCo-é ù

ê úë û which one complex is : [2]

(i) diamagnetic,

(ii) more stable,

(iii) outer orbital complex and

(iv) low spin complex.

(Atomic number of Co 27= )

SECTION C

17. (i) Write the role of ‘CO’ in the purification of nickel. (ii) What is the role of silica in the extraction of copper? (iii) What type of metals are generally extracted by electrolytic method? [3]




18. Give reasons for the following : (i) Transition metals form alloys. (ii) Mn2O3 is basic whereas Mn2O7 is acidic. (iii) Eu2+ is a strong reducing agent. [3] 20. (i) Why bithional is added in soap? (ii) Why magnesium hydroxide is a better antacid than sodium bicarbonate? (iii) Why soaps are biodegradable whereas detergents are non-biodegradable? [3]

OR Define the following terms with a suitable example in each : (i) Antibiotics (ii) Artificial sweeteners (iii) Analgesics 21. Write the structures of main products when benzene diazonium chloride reacts with the following reagents.

(i) CuCN (ii) CH3CH2OH (iii) KI

Delhi Set 3 56/1/3

SECTION A

1. Arrange the following in decreasing order of solubility in water: [1] (CH3)3N, (CH3)2NH, CH3NH2

2. What type of stoichiometric defect is shown by ZnS and why? [1]

3. Write on the stereochemical difference between NS 1 and NS 2 reactions. [1]

SECTION B

7. State Henry’s law and write its two applications. [2] 11. Write the hybridization and magnetic character of the following complexes: [2]

(i) ( ]22 6[Fe H O) +

(ii)

( )5Fe CO

12. Write structures of main compounds A and B in each of the following reactions: [2]

(i) CH3CH2CNCH3MgBr / H3O

+

A BLiAIH4

(ii)

CH3

(i) CrO3 /(CH3CO)2O

(ii)H3O+/∆

AH2N-NH2 B

SECTION C

17. How will you convert the following? [3] (i) Impure Nickel to pure Nickel. (ii) Zinc blende to Zinc metal.

(iii) [Ag(CN)2]

- to Ag. 22. Write equations of the following reactions: [3]

(i) Acetylation of aniline (ii) Coupling reaction

(iii) Carbyl amine reaction

24. Define the following with a suitable example in each. [3] (i) Oligosaccharides




(ii) Denaturation of protein (iii) Vitamins

OR Write the reactions involved when D-glucose is treated with the following reagents:

(i) Br2 water (ii) H2N-OH

(iii) (CH3CO)2O

SOLUTIONS

Delhi Set 1 56/1/1

SECTION A

Answer 1. AgCl , Due to large difference in their size/ Due to small size of Ag+ ion. [CBSE Marking Scheme, 2019] [½ + ½]

Detailed Answer : Frenkel defect is shown by AgCl due to small size of Ag+ ion. Frenkel defect is not shown by NaCl because alkali

metal ions cannot fit into interstitial sites. [1]

Answer 2. (CH3)3N < C2H5NH2 < C2H5OH [CBSE Marking Scheme, 2019] 1

Answer 3. Due to large surface area these are easily assimilated or adsorbed. [CBSE Marking Scheme, 2019] 1

OR

Emulsion – both dispersed phase and dispersion medium are liquid Gel- Dispersed phase is liquid while dispersion medium is solid [CBSE Marking Scheme, 2019] 1

Answer 4. Nucleophiles having two nucleophilic centres. CN– /SCN– / NO2

– (any one) [CBSE Marking Scheme, 2019][½ + ½]

Detailed Answer : An ambidient nucleophile is an anionic nucleophile in which the negative charge is delocalized by resonance over

two unlike atoms. [½]

O••••

•CH2

•H3C

O••••

CH2H3C

••

[½]

Answer 5. Glucose has aldehydic group while fructose has ketonic group/ Glucose is aldose while fructose is ketose. [CBSE Marking Scheme, 2019] 1

Detailed Answer : Glucose is aldohexose that means functional group present in sugar is aldehyde. Fructose is a ketohexose that

means functional group in sugar is ketone. [1]




OR

Glucose and Galactose [CBSE Marking Scheme, 2019] 1

Detailed Answer : Lactose on hydrolysis gives glucose and galactose. [1]

SECTION B

Answer 6. (i) 2XeF2 (s) + 2H2O(l) → 2Xe (g) + 4 HF (aq) + O2(g) 1 (ii) MnO2 + 4HCl → MnCl2 + Cl2 + 2H2O [CBSE Marking Scheme, 2019] 1

Detailed Answer : (i) XeF2 is soluble in water and undergoes hydrolysis to produce Xe metal and O2 gas. [1] 2XeF2 + 2H2O → 2Xe + 4HF + O2

(ii) MnO2 reacts with concentrated HCl to produce manganese chloride, water and chlorine gas. [1] MnO2 + 4HCl → MnCl2 + H2O + 2Cl2

OR

(i) H2O < H2S < H2Se < H2Te 1 (ii) HF > HCl > HBr > HI [CBSE Marking Scheme, 2019] 1

Detailed Answer :

(i) O, S, Se and Te are all in the same group in the periodic table. As, you move down the group, the atomic radius increases. In binary acids, the H - Te bond is longest and the H - O bonds is shortest. The H - Te bond is therefore weakest and the H - O bond is strongest. Thus H2O is the weakest acid while H2Te is the strongest acid.

The order of the given compounds in increasing acidic character is: H2O < H2S < H2Se <H2Te (ii) As you move down the group, the atomic size increases. Hence, HI is the strongest acid while HF is the

weakest acid. Thus, the bond enthalpy is inversely related to the size of the halogen atoms. The order of the given compounds in decreasing bond enthalpy is: HF > HCl > HBr > HI [2]

Answer 7. For a solution of volatile liquids, the partial vapour pressure of each component of the solution is directly

proportional to its mole fraction present in solution. 1 (i) DmixH = 0 (ii) DmixV = 0 (iii) The components have nearly same intermolecular force of attraction (any two) [CBSE Marking Scheme, 2019] ½ + ½

Detailed Answer : Raoult’s Law for a solution containing volatile components - The Partial vapour pressure of each volatile

components of the solution are directly proportional to mole fraction present in solution. The ideal solutions have characteristics to obey Raoult’s law for the following conditions.

1. Theenthalpyofmixingofthepurecomponentstoformthesolutioniszero,∆mix H = 0. That means no heat is absorbed or released during the mixing of the two pure components.

2. Thevolumeofthemixingalsobezero,∆mix V = 0. That means the total volume of the solution is equal to the sum of the volume of the two components.

3. In pure components, A and B, the intermolecular attractions between solute-solute interactions and solvent-solvent interaction are almost similar to the solute-solvent interaction. [2]

Answer 8. (i) Rate = k [H2O2] [ I

–] 1 (ii) Order = 2 ½ (iii) Step 1 [CBSE Marking Scheme, 2019] ½




Detailed Answer : (i) The rate law for the reaction is :

Rate I=−

[ ]= [ ]

−d H O

dtk H O2 2

2 2

(ii) This reaction is first order with respect to both H2O2 and I- The overall order of the reaction is bimolecular, 2. The order of the reaction is determined from the slowest

step of the reaction mechanism. (iii) The first reaction is slow, so this is the rate determining step. [2]

Answer 9. A = K2MnO4 / MnO2

4–, B = KMnO4 / MnO2–

4 , C= IO3– or KIO3 , D= I2 [CBSE Marking Scheme, 2019] ½ × 4

Detailed Answer : (A) K2MnO4 green (B) KMnO4 purple (C) Mn2O3

(D) Mn2O4 [2]

Answer 10. Bis(ethan-1,2-diamine)dichloridoplatinum (II) 1

Cl

Pt

Cl

en

en

Pten en

Cl

Cl

Cis Trans

[CBSE Marking Scheme, 2019] ½ + ½

Detailed Answer : IUPAC names of the complex [Pt(en)2Cl2] is dichlorobis(ethylenediamine)platinum(II) The geometrical isomers of [Pt(en)2Cl2] is

Pt enen

Cl

Cl

Pten

Cl

Trans isomer

Cl

en

Cis isomer [2]OR

(i) [Co(NH3)6]2(SO4)3 1 (ii) K3[Cr(C2O4)3] [CBSE Marking Scheme, 2019] 1

Answer 11. (i) [CoF6]

3–

(ii) [Co(en)3]3+

(iii) [Co(en)3]3+

(iv) [CoF6]3– [CBSE Marking Scheme, 2019] ½ × 4




Answer 12.

(i) A =

COOK—

B =

COOH—

(ii) A =

O——

B =

= N—NH—C—CH2

— —O

[CBSE Marking Scheme, 2019] ½ × 4

Detailed Answer : (i) Ethyl benzene when treated with KMnO4 and KOH, undergoes oxidation to produce potassium benzoate.

This potassium benzoate when treated with an acid forms benzoic acid.

H2C

KMnO4 - KOH

CH3 O OK

- +O OH

H3O+

Benzoic acid (B)

Potassium Benzoate (A)

(ii) Cyclohexanol is oxidized by CrO3 to cyclohexanone. Cyclohexanone when treated with semicarbazide produces cyclohexanone semicarbazone.

OH

CrO3

O

H2N-NH-CONH2

CyclohexanoneSemicarbazone (B)

Cyclohexanone (A)

N - NH - CONH2

[2]

SECTION C

Answer 13.

t = [ ] [ ]R R0− t

k 1

= [ . ]0.1

X−

−0 064

4 10 3 1

= 9s. [CBSE Marking Scheme, 2019] 1

Detailed Answer : The equation for the zero order reaction is: [A]=[A0] – kt Given: [A0]=0.1 M [A]=0.064 K=4×10-3 M/s Plugging in the values in the equation, and solving the equation for t, we get : 0.064 = 0.1 – (4×10–3×t) 0.064 – 0.1 = – (4×10–3×t) – 0.036 = – (4×10–3×t)

t s=

−

− ×=

−0 036

4 109

3.

Final answer : 9 seconds. [3]




Answer 14. (i) Adsorption of toxic gases 1 (ii) Negative charge ; Fe2O3.xH2O/OH– ½ + ½ (iii) Increases with increase in temperature/ First increases then decreases. [CBSE Marking Scheme, 2019] 1

Detailed Answer :

(i) The activated charcoal adsorbs the poisonous gases rather than oxygen and provides fresh oxygen for inhaling.

(ii) When ferric chloride is added to NaOH, a negatively charged sol is formed with adsorption of OH– ions. The sol is represented as : [Fe(OH)3]OH–

(iii) Adsorption is an exothermic process. According to Le Chatelier’s principle, at low temperature forward reaction is favourable. As temperature increases enough energy is being provided for the molecules to reach the activation energy. Therefore, initially as temperature increases, chemisorption increases. As the high temperature helps in bond breaking, after certain temperature, chemisorption decreases. [3]

Answer 15.

d = Zma3N

; m = Mass of element , N = number of atoms 1

N = 108 4

10 8 27 10 24×

× × −. 1

= 1.48 × 1024 atoms 1OR

M = a dZa

3 × ×N

½

= 27 10 6 022 10 10 84

24 23× × × ×− . .

1

= 43.88 g mol–1 ½ 43.88 g mol–1 contains 6.02 × 1023 atoms

So, 108 g contains = 6 02 10 10843 88

23..

× ×

= 1.48 × 1024 atoms 1

[CBSE Marking Scheme, 2019]

Detailed Answer :

Given: cell edge length, a = 300pm = 3x10-8 cm

Mass of element 108 g=

An element crystallizes in FCC crystal lattice, so there are 4 atoms per unit cell in FCC crystal structure.

Volume of the unit cell ( ) ( )33 8 V a 3 10 cm−= = ×

23 3V 2.7 10 cm−= ×

Volume of 107g of element can be calculated using its mass and density as :

3m 107V 9.90 cm

d 10.8 = = =

Number of units in 39.90 cm volume 3

2323 3

9.90 cm3.669 10 unit cells

2.7 10 cm−= = ×

×

Each unit cell contains 4 atoms, thus, total number of atoms can be calculated as:

23 24 3.669 10 4 1.467 10 atoms= × × = × [3]




Answer 16. ∆Tf = Kf m ½

Kf =∆Tf × M2 1

2 1000×

×w

w

= 2 342 964 1000× ×

×

= 16.4 K 1 ∆Tf = Kf m’

= K

Mfw

w2

2 1

1000××

= 16 4 5 100095 180. × ×

×

= 4.8 K 1 ∆Tf = Tf° – Tf

4.8 = 273.15 – Tf

Tf = 268.35 K ½ [CBSE Marking Scheme, 2019]

Detailed Answer :

Given :

( )Sucrose solution 4% w / w=

1M 342 g mol−=

Freezing point of solution 271.15 K=

Freezing point of pure water 273.15 K=

Glucose solution 5% =

1M 180 g mol −=

To calculate :

Freezing point of 5% glucose solution. Formula :

f fT i K m∆ = × ×

moles of solute

mKg of solvent

=

mass

molesmolar mass

=

Sucrose solution is 4% (w/w) which means there is 4.0 grams of sucrose dissolved in 100g of solution. Mass of solution = mass of solute + mass of solvent 100.0 g = 4.0g + mass of solvent. Hence, mass of solvent = 100.0 – 4.0 = 96.0 g

1 kg96 g 0.096 kg

1000 g× =

Moles of solute

1

mass 4.0g moles 0.011695 moles

molar mass 342 g mol −= = =

Molality of solution :




moles of solute 0.011695 molesm 0.1218 m

kg of solvent 0.096 kg water= = =

Sucrose is a non - electrolyte, hence i 1=

fT Freezing point of solvent Freezing point of solution∆ = −

fT 273.15 K 271. 15 K 2.00 K∆ = − =

f fT i K m∆ = × ×

f2.00 K i K 0.1218= × ×

1fK 16.42 Km−=

For glucose solution,

Glucose solution is 5% (w/w) which means there is 5.0 grams of glucose dissolved in 100g of solution. Mass of solution = mass of solute + mass of solvent 100.0 g = 5.0g + mass of solvent. Hence, mass of solvent = 100.0 – 5.0 = 95.0 g

1 kg95.0 g 0.095 kg

1000 g× =

Moles of solute

1

mass 5.0g moles 0.0277 moles

molar mass 180 g mol −= = =

Molality of solution :

moles of solute 0.0277 molesm 0.2923 m

kg of solvent 0.095 kg water= = =

Glucose is a non - electrolyte, hence i 1=

f fT i K m∆ = × ×

fT 1 16.42 0.2923∆ = × ×

1fT 4.801 Km−∆ =

fT Freezing point of solvent Freezing point of solution∆ = −

4.801 273.15 K Freezing point of solution= −

Freezing point of solution 273.15 K – 4.801 K 268.35 K= =

Thus, the freezing point of the glucose solution is 268.35 K [3]

Answer 17. (a) (i) Zone refining (ii) Distillation ½ + ½ (b) 2Cu2S + 3O2 → 2Cu2O + 2SO2 1 2Cu2O + Cu2S → 6Cu + SO2 [CBSE Marking Scheme, 2019] 1

Detailed Answer :

(a) (i) Zone refining (ii) distillation

(b) Roasting of the sulfide ore: 2 2 2 22Cu S 3O 2Cu O 2SO+ → +

(If iron sulphide is present) : 2 22FeS 3O 2FeO 2SO+ → +

Removal of iron oxide as slag : ( )2 3FeO SiO FeSiO slag+ →




Reduction of coper(I) oxide : 2Cu O C Cu CO+ → +

2 2 22Cu O Cu S 6Cu SO+ → + [3]

Answer 18. (i) Due to variable oxidation state 1 (ii) Mn2+ is stable due to exactly half filled 3d5configuration/Duetohigh∆aH

0andlow∆hydH0 for Cu2+/Cu is positive. 1

(iii) Due to comparable energies of 5f , 6d and 7s orbitals. [CBSE Marking Scheme, 2019] 1

Detailed Answer :

(i) Transition metals have the ability to adsorb many other substances on their surface and activate them as a result of chemisorption. Transition metals also exhibit a variety of oxidation states, and can change oxidation states relatively easily. This makes transition metals and their compounds good catalysts.

(ii) Eo values are indicative of the stability of the oxidized form of the element. The lower the Eo value, more stable the oxidized form of the element. Mn2+ with a half filled d-subshell (d5) is stable, so Mn is easily oxidized to Mn2+, making the Eo value negative. Cu2+ with a partially filled d subshell (d9) is not stable, and is relatively easily reduced to element form. This makes its Eo value positive.

(iii) The electronic configurations of actinoids show irregularities because the energies of their 5f, 6d, and 7s orbitals are close to each other. Electrons can easily move between these subshells. [3]

Answer 19. (i) HOOC(CH2)4COOH, H2N (CH2)6 NH2

(ii) HO — CH2 — CH2 — OH,

HOOC—

—

COOH

(iii) CH2 = CH – CH = CH2,

CH = CH2

—

[CBSE Marking Scheme, 2019] 1 × 3

Detailed Answer : (i) Hexamethylenediamine: NH2(CH2)6NH2 [1] Adipic acid: HOOC(CH2)4COOH (ii) Ethylene glycol : HOH2C - CH2OH

Phthalic acid : HOOC COOH

[1]

(iii) 1,3-butadiene: CH2=CH-CH=CH2

Styrene :

CH2

[1]

OR

(i) Homopolymers , single repeating unit ½ + ½

(ii)

N

—

—H N2

NN

NH2

—NH2

, HCHO (Or names of monomers) 1

(iii) Sulphur forms cross links at the reactive sites of double bonds and thus the rubber gets stiffened / To improve the physical properties of rubber by forming cross links. [CBSE Marking Scheme, 2019] 1




Detailed Answer : (i) The given structure is homopolymer. A copolymer is formed, when two or more different types of monomer

are linked in the same polymer. A homopolymer is formed, when only one monomer is linked in the same polymer [1]

(ii)

,

[1]

(iii) In vulcanization of rubber, the sulphur acts as crosslinking agent. On vulcanization, sulphur forms cross links at the reactive sites of double bonds and thus the rubber gets stiffened. [1]

Answer 20. (i) Tranquilizers 1 (ii) Anionic detergents 1 (iii) It is difficult to control the sweetness. [CBSE Marking Scheme, 2019] 1

Detailed Answer : (i) The drug used in sleeping pills is tranquilizers. (ii) Anionic detergents are used in toothpaste. Anionic detergents are sodium salts of a sulphonated long chain

of alcohols. (iii) It is high potency sweetener, Alitame is high more stable than aspartame. It is 2000 times sweeter than sucrose.

The control of the sweetness of food is difficult while using it. [3]OR

(i) Antibiotics which kill or inhibit a wide range of Gram-positive and Gram-negative bacteria. Example- Chloramphenicol (or any other) ½ + ½ (ii) The chemicals which either kill or prevent the growth of microorganisms when applied to inanimate objects

such as floors, drainage system, instruments, etc. Example – 1% Phenol solution (or any other) ½ + ½ (iii) Cationic detergents are quarternary ammonium salts of amines with acetates, chlorides or bromides as

anions where Cationic part is involved in cleansing action. Example – Cetyltrimethylammonium bromide (Or any other) [CBSE Marking Scheme, 2019] ½ + ½

Detailed Answer : (i) Broad-spectrum antibiotics – The antibiotics that are used to kill or inhibits a wide range of Gram-positive

and Gram-negative bacteria are said to be broad-spectrum antibiotics. [1] (ii) Disinfectants – These are the chemical which either kills or prevents the growth of microorganisms.

Disinfectants are used on inanimate objects such as floors, drainage system, instruments etc. [1] (iii) Cationic detergents – These detergents are quaternary ammonium salts of amines with acetate, chlorides

or bromides as anions. Cationic part possesses a long hydrocarbon chain and a positive charge on nitrogen atoms. [1]

Answer 21. (i) (CH3)3C–I , Due to large size of iodine / better leaving group / Due to lower electronegativity. ½ + ½

(ii)

Cl

—

NO2

—

(i) NaOH, 443K

(ii) H+

OH

—

NO2

—

1 (iii) Because enantiomers have same boiling points / same physical properties. [CBSE Marking Scheme, 2019] 1

Detailed Answer : (i) Within a family, larger atoms are better nucleophiles. I- > Br- > Cl- > F-, therefore I- is better leaving group

than Br-.




(ii)

Cl OH

(i) NaOH, 443 K

(ii) H+

NO2 NO2

p-nitrocholoro obenzene 4 - nitrophenol

The presence of an electron withdrawing group (-NO2) at ortho- and para- positions increases the reactivity of haloarenes.

(iii) Butan-2-ol has four different groups attached to the second tetrahedral carbon. Therefore, it is chiral molecule and the dextro and laevo – rotatory isomers of Butan-2-ol are enantiomers. The enantiomers possess identical physical properties like melting point, boiling point, solubility, refractive index etc. They only differ with respect to the rotation of plane polarised light. Therefore, dextro and laevo – rotatory isomers of Butan-2-ol are difficult to separate by fractional distillation. [3]

Answer 22.

A =

CONH2

—

B =

NH2—

C =

NC

—

A= Benzamide, B = Aniline, C = Phenylisocyanide / Benzeneisonitrile [CBSE Marking Scheme, 2019] ½ × 6

Detailed Answer :

[3]

Answer 23. (i) C6H5 – CH(OH) – CN 1 (ii) 2 CH3COCH2C6H5 + CdCl2 1 (iii) (CH3)2 – C(Br)COOH [CBSE Marking Scheme, 2019] 1

Detailed Answer :

(i)

H

hydroxy(phenyl)acetonitrile benzaldehyde

CHONaCN / HCl

OH

N

(ii)

(C6H5CH2)2Cd + 2 CH3COCl 2

O

CH3

+ CdCl2

1-phenylpropan-2-one




(iii)

[3]

OR

(i)

2CH —CO—CH2 3 CH —C—CH CO—CH3 2 3

Ba(OH)2

—CH3

—

OHPropane

(Ketone) 1

(ii)

C—

— —

O

—CH3 Zn-Hg

HCl

CH2CH3

—

1

(iii)

C—

— —

O

—Cl H2

Pd-BaSO4

CHO—

1 [CBSE Marking Scheme, 2019]

Detailed Answer : (i)

(ii)

ethyl benzene

(iii)

O

CI

benzoyl chloride benzaldehyde

H2

Pd/BaSO4

CHO

[3]

Answer 24. (i) Amylose is water soluble component while amylopectin is water insoluble 1 (ii) Peptide linkage is – CONH– formed between two amino acids while glycosidic linkage is an oxide linkage

between two monosaccharides. 1




(iii) In fibrous protein, the polypeptide chains run parallel while in globular , the chains of polypeptides coil around to give a spherical shape (or any other correct difference.) 1


Detailed Answer : (i)

Amylose AmylopectinChemically it is straight chain polymer. Chemically it is branched chain polymer.

It is water soluble. It is insoluble in water.

It constitutes 15-20% of starch. It constitutes 80-85% of starch.

(ii)Peptide linkage Glycosidic linkage

Peptide linkage is formed between two amino acids.

Glycosidic linkage is formed between two monosaccharides.

Hydrolysis of peptide linkage forms two amino acids.

Hydrolysis of glycosidic linkage forms two monosaccharides.

It links carbon and nitrogen atoms. It links two carbon atoms through an oxygen atom.

It is found in polypeptides. It is found in polysaccharides.

It can be given as –CONH-. It can be given as –C-O-C-.

(iii)Fibrous proteins Globular proteins

When the polypeptide chains run parallel and are held together by hydrogen and disulphide bonds then fibrous proteins is formed.

When the chains of polypeptides coil around to get a spherical shape then globular proteins is formed.

Generally it is insoluble in water. Generally it is soluble in water.

They have comparatively stronger intermolecular forces of attraction.

They have weak intermolecular hydrogen bonding.

[3]OR

(i)

CH —CH —CH —CH —CH —CH3 2 2 2 2 3(CHOH)4

—CHO—

CH OH2

HI, ∆

1

(ii)

(CH—O—C—CH )3 4(CHOH)4

—CHO

—

CH OH2

Acetic anhydride

—CHO — —O

—

CH —O—C—CH2 3— —

O 1

(iii)

(CHOH)4(CHOH)4

—CHO

—

CH OH2

Br water2

—COOH

—

CH OH2 1 [CBSE Marking Scheme, 2019]

Detailed Answer : (i) On prolonged heating with HI, it forms n-hexane, suggesting that all the six carbon atoms are linked in a

straight chain.




HC

C OH

OH

OH

OH

H

OH

H

H

H

H2

C

C

C

C

O

D-glucose

HI

H3C

CH2

CH2

CH2

CH2

CH3

hexane

(ii) Acetylation of glucose with acetic anhydride gives glucose pentaacetate which confirms the presence of five –OH groups.

HC

C OH

OH

OH

OH

H

OH

H

H

H

H2

C

C

C

C

OHC

CH3

CH3

H3C

CH3

C O

O

O

OH

H

O

H

H

H

H2

C

C

C

C

C

C

C

O

O

D-glucose

Acetic anhydride

O

C

O

O

(iii) Glucose on reaction with mild oxidising agent like bromine water get oxidised to six carbons carboxylic acid (gluconic acid). This indicates that the carbonyl group is present as an aldehydic group.

H

H

H

OH

OH

O

OH

OH

HO

H

H

H

H

OH

OH

OH

OH

HO

H

Br2 water

Gluconic acidD-glucose

COOH

[3]

SECTION D

Answer 25.

Ecell = E°cell – 0 059.

n log Kc 1

= E°cell – 0 0592.

log

1010

3

2

−

−

1

= 2.71+ 0.0295 1 Ecell = 2.7395 V (i) Cu to Mg / Cathode to anode / Same direction 1 (ii) Mg to Cu / Anode to cathode / Opposite direction [CBSE Marking Scheme, 2019] 1

Detailed Answer :

2o

cell cell 2

Mg0.059E E log

n Cu

+

+

= −




[ ][ ]cell0.0010.059

E 2.71 log2 0.01

= −

( )cellE 2.71 0.0295 2.74 V= − − = [2]

(i) When external opposite applied voltage is less than 2.71, it is less than ocellE , therefore, the electrons will

flow from the anode to the cathode, and current will flow from cathode (copper electrode) to anode (magnesium electrode).

(ii) When external opposite applied potential is greater than 2.71, it is greater than ocellE , therefore, the reaction

will be reversed, and the current will flow from anode to cathode. [3]

OR

(a) m = zIt ½

2.8 = 56 22 96500

× ××

t

½

t = 4825 s = 80.417 min

mm1

2 =

EE1

2 ½

2 8.m nZ

= 562

265 3

×.

1

mZn = 3.265 g 1 (b) (i) A- strong electrolyte , B-Weak electrolyte 1 (ii) Λ0m for weak electrolytes cannot be obtained by extrapolation while Λ0m for strong electrolytes can be

obtained as intercept. [CBSE Marking Scheme, 2019]

Detailed Answer :

(a) Charge required to deposit 2.8 g Fe:

1

mass 2.8 gmol Fe 0.05 mol

molar mass 56 g.mol−= = =

2 F charge is required to discharge 1 mol of Fe2+ ions as Fe, therefore deposition of 0.05 mol Fe will need

96500 C0.05 2 0.1 F 0.1 F 9650 C

F× = = × =

The quantity of charge is related to current as Q It=

Therefore, the time needed to deposit 2.8 g Fe is : Q 9650 C

t 4825 sI 2 A

= = =

So, the current flowed through the cells for 4825 seconds.

The amount of Zn deposited in cell Y can be calculated using Faraday’s second law :

mass Zn Eq.wt Zn molar mass Zn / charge on zinc ionmass Fe Eq wt Fe molar mass Fe / charge on iron ion

= =

65.3 g / 2

mass Zn 2.8 g 3.265 g ~ 3.3 g56 g / 2

= × =

Therefore, the mass of Zn deposited in cell Y in the same time is 3.3 g. [3]

(b)

(i) Molar conductivity of strong electrolytes increases linearly as the square root of the concentration decreases; therefore, electrolyte A is a strong electrolyte. Molar conductivity of weak electrolytes increases non-linearly as square root of concentration decreases; therefore, electrolyte B is a weak electrolyte. [1]




(ii) As concentration of strong electrolyte approaches zero, the molar conductivity of the plot intercepts the

molar conductivity axis, giving the limiting value of molar conductivity 0mE .

The plot of molar conductivity of weak electrolyte tends to infinity as its concentration approaches zero; it does not intersect the molar conductivity axis. [1]

Answer 26.

(a) (i)

—OH

+ NaOH

—

ONa

CH X3

—

O—CH3

1

(ii) CH3CH2OH PCC, Heat → CH3 – CHO (i) CH MgBr (ii) H+

3 → CH3CH(OH) – CH3 1

(or any other correct method)

(b)

H—C—C—O—H + H+

—H

—

H

—H

—

H

FastH—C—C—O —H+

—H

—

H

—H

—

H

—H

½

H—C—C—O —H+

—H

—

H

—H

—

H

SlowH—C—C + H O2

+

—H

—

H

—H

—

H

—H

½

H—C —C– +

—H

—

H

—H

—

H

C = C + H+

H

—H

H

—H

——

1 (c) Due to involvement of lone pair of oxygen in delocalisation makes the benzene ring electron rich. 1 [CBSE Marking Scheme, 2019]

Detailed Answer :

(a) (i) Phenol to Anisole

OH OCH2

Phenol Anisole

–+

(ii) Ethanol to propan-2-ol

2

13

OH

1

2 OH

propan-2-o1ethanol

Ethanol (a primary alcohol) when oxidized partially with the partial oxidizing agent such as PCC (Pyridinium chlorochromate) will be converted into acetaldehyde.

If we treat acetaldehyde with Grignard reagent (CH3-Mg-X) followed by acid hydrolysis, we will get propan-2-ol which is the desired product.




+

(b)

+

(c) The –OH group attached to the benzene ring in phenol is the highly activating group. The lone pair on oxygen is delocalized into the benzene ring and makes the phenol more electron rich than simple benzene. Thus, phenol undergoes electrophilic substitution more easily than benzene. [5]

OR

(a) (i) o-Nitrophenol is steam volatile due to intramolecular hydrogen bonding while p-nitrophenol is less volatile due to intermolecular hydrogen bonding. 1

(ii) Due to the formation of stable intermediate tertiary carbocation / CH3O– being a strong base favours

elimination reaction. 1

(b) (i)

—

OH

—

O Na+–

H+

—OH

CHCl + aq NaOH3—

CHO—

CHO

1 (ii) (Award 1 mark if attempted in any way) 1 (c) Add neutral FeCl3 to both the compounds, phenol will give violet colouration while ethanol does not. 1 [CBSE Marking Scheme, 2019]

Detailed Answer : (a) (i) o-Nitrophenol forms intramolecular H bond whereas molecules of p-Nitrophenol get associated through

intermolecular H bond. During boiling, the strong intermolecular H bonding increases the boiling point but intramolecular H bonding cannot do so.

(ii)

NaOCH3

heat2-methylprop-1-eue

CI

2

3 1

As the given alkyl halide is tertiary in nature, therefore on reaction with sodium methoxide, elimination reaction takes place in place of substitution and hence, alkene is formed as a major product. Therefore, in the given reaction, 2-methyl propene is formed as a major product.

(b) (i) The Reimer-Tiemann reaction is a chemical reaction used for the o-formylation of phenols.

OH OH O

HCHCI3

3 KOH




(ii)

OHOH

O

O

CIR

R

AICI3

HCI

12

34

5

6

+

O

RO

+

Phenol

C-acylation O-acylation

(c) Phenol being more acidic than ethanol will turn the blue litmus paper red but ethanol doesn’t.

Haloform test is also used to distinguish ethanol and phenol. Ethanol when treated with NaOH and iodine,iodoform is obtained which is yellow needle shaped. But phenol do not show positive haloform test. [5]

Answer 27. (a) (i) In vapour state sulphur partly exists as S2 molecule which has two unpaired electrons like O2. 1 (ii) Due to greater interelectronic repulsion. 1 (iii) Because decomposition of ozone into oxygen results in the liberation of heat (DH is negative) and

an increase in entropy (DS is positive), resulting in large negative Gibbs energy change (DG) for its conversion into oxygen. 1

(b) (i) NO gas/ Nitric oxide 1 (ii) NO2 gas / Nitrogen dioxide [CBSE Marking Scheme, 2019] 1

Detailed Answer :

(a) (i) In gas phase, Sulphur exists mostly as diatomic molecule S2. The bonding in this molecule is similar to that of dioxygen O2. According to molecular orbital theory it has two unpaired electrons in antibonding molecular orbitals.

(ii) The N-N sigma bond length is less than the P-P sigma bond length. The shorter separation means the non-bonding electrons (lone pairs) on the nitrogen atoms repel each other strongly, making its bond weaker than the P-P bond:

(iii) The enthalpy change for decomposition of ozone has a negative value, it results in liberation of heat :

o3 22O 3O H 142 kJ / mol→ …∆ = −

The increase in number of moles of gas means the entropy change for the reaction is positive. This reinforces the

effect of negative enthalpy change, and results in a large negative value of oG∆ for the decomposition reaction, making it less stable than oxygen.

(b)

(i) Dilute HNO3: Nitric oxide, NO.

(ii) Conc. HNO3: Nitrogen dioxide, NO2. [5]

OR

(a) (i) 4H3PO3 → 3H3PO4 + PH3 1

(ii)

F F

FF

Ke

1




(b) (i) Due to small size and low bond dissociation enthalpy 1 (ii) As the size increases, electronegativity decreases / non-metallic character decreases 1 (c) 5SO2 + 2MnO–

4 + 2H2O → 5SO24– + 4H+ + 2Mn2+ [CBSE Marking Scheme, 2019] 1

Detailed Answer : (b) (i) The bond dissociation energy of F2 is quite low, and the small size of the fluoride ion means it its hydration

energy is high. These two compensate for the less negative value of electron gain energy, making F2 a strong oxidizing agent. 1

(ii) As we move down a group, the atomic size and metallic character increases while electronegativity decreases. This leads to the acidic character of the oxide decreasing from top to bottom, N2O3 to Bi2O3. 1

(c) Sulphur dioxide turns an acidified potassium dichromate solution green :

( )2 2 2 7 2 4 2 4 2 2 433SO K Cr O H SO Cr SO H O H SO+ + → + + 1

rrr

Delhi Set 2 56/1/2

SECTION A

Answer 2. C2H5 NH2 < (C2H5)2N < (C2H5)3N [CBSE Marking Scheme, 2019] 1

Detailed Answer :

Base strength is the ability to donate lone pair. Due to inductive effect of 2 5C H , the negative charge density on nitrogen atom increases. Therefore, lone pair is easily available for donation.

(C2H5)3N > (C2H5)2 NH > C2H5NH2 [1]

Answer 3. Due to formation of n-type semiconductor providing free electrons. [CBSE Marking Scheme, 2019] 1

Detailed Answer :

Silicon is a group 14 element and therefore, has four valence electrons. Phosphorous is in group 15 and has five valence electrons. When silicon is doped with phosphorous, it replaces a silicon atom in the lattice. Phosphorous uses only 4 out of its valence electrons for bond formation with the four Si atoms surrounding it. This leaves one electron in phosphorus atom. This electron is delocalized - free to move anywhere in the crystal and increases its conductivity. It makes silicon an n-type semiconductor. [1]

Answer 5. 4-chlorobenzenesulphonic acid [CBSE Marking Scheme, 2019] 1

SECTION B

Answer 8.

Ideal Non –ideal

Obeys Raoult’s law at all range of concentration Does not obey

∆mixH=O,∆mixV = O ∆mixH≠0,∆mixV≠0 (or any other difference)

[CBSE Marking Scheme, 2019] 1 + 1

Detailed Answer :

Ideal solution Non-ideal solution

Raoult’s Law Solutions which obey Raoult’s law over the entire range of concentrations.

Do not obey Raoult’s law over the entire range of concentrations.




mix HD The enthalpy of mixing of the pure components to form a solution is zero.

mix H 0D =

The enthalpy of mixing of pure components to form a solution is not zero. mix H 0D >

mix VD The volume of the mixing is also zero,

mix V 0D = .

That means the total volume of the solution is equal to the sum of the volume of the two components.

The volume of the mixing is not zero, mix V 0D ¹ .

The total volume of the solution is not equal to the sum of the volume of the two components.

Intermolecular interactions

In pure components, A and B, the intermolecular attractions between solute-solute interactions and solvent-solvent interaction are almost similar to the solute-solvent interaction.

In pure components, A and B, the intermolecular attractions between solute-solute interactions and solvent-solvent interaction are not similar to the solute-solvent interaction.

[2]

Answer 10. Tetraamminedichloridochromium(III) ion 1

Co

—

—

—

——

—

NH3

NH3 NH3

Cl

Cl

NH3

+

Co

—

—

—

——

—

NH3

NH3 NH3

Cl

Cl

+

NH3

cis trans [CBSE Marking Scheme, 2019] ½ + ½

OR

(i) [Co(NH3)5ONO]Cl2 1 (ii) K2[Ni(CN)4] [CBSE Marking Scheme, 2019] 1

Answer 11. (i) [Co(C2O4)3]

3–

(ii) [Co(C2O4)3]3–

(iii) [CoF6]3–

(iv) [Co(C2O4)3]3– [CBSE Marking Scheme, 2019] ½ × 4

SECTION C

Answer 17. (i) To produce a volatile complex, which decomposes on further heating to give pure nickel. 1 (ii) To remove impurities (FeO) by forming a slag. / Acts as a flux. 1 (iii) More reactive metals having large negative electrode potential. [CBSE Marking Scheme, 2019] 1

Detailed Answer :

(i) In Mond process for refining nickel, nickel is heated with CO to form ( )4Ni CO which is decomposed at a higher temperature to obtain the pure metal. So CO acts like a catalyst in this process.

( ) ( ) ( )Heat Heat4

Ni impure 4CO Ni CO Ni pure 4CO+ ¾¾¾® ¾¾¾® +

(ii) The ore is heated in a furnace after mixing with silica. In the furnace, iron oxide reacts with silica to form a slag of iron silicate. The copper is produced in the form of copper matte. This contains Cu2S and FeS. The copper matte then charged into silica line convertor and it converts any remaining FeS, FeO into slag and Cu2S/CuO into metallic copper.

The role of silica in copper extraction is to remove the iron oxide obtained during the process of roasting.

( )2 3FeO SiO FeSiO slag+ ®

The silica act as a flux. (iii) The metals which are highly reactive in nature are generally extracted through electrolytic process. For

example, Sodium, Aluminium, Calcium etc. [3]




Answer 18. (i) Due to comparable radii / comparable size. 1 (ii) In Mn2O3, Mn is in +3 (lower) oxidation state while in Mn2O7, Mn is in higher oxidation state (+7) 1 (iii) Because its stable oxidation state is +3. [CBSE Marking Scheme, 2019] 1

Detailed Answer : (i) Transition metals easily form alloys with other transition metals because they have almost similar size. So

they can easily replace each other in the crystal lattice. (ii) The transition metal oxides in the lower oxidation state of metals are basic in nature and in higher oxidation

state they are acidic in nature. The oxidation state of Mn in Mn2O3 is +3 and Mn2O7 has +7. Therefore, Mn2O3 is basic and Mn2O7 is acidic.

(iii) The common oxidation state of lanthanide metals is +3. Eu2+ is formed by losing the two s electrons acquires half filled (4f7) configuration. But still, they oxidize to their common +3 state. So the Eu2+ loses one electron and is oxidized to Eu3+. So, Eu2+ acts as a strong reducing agent. [3]

Answer 20. (i) To impart antiseptic properties. 1 (ii) Magnesium hydroxide is better alternatives because of being insoluble, it does not increase the pH above

neutrality. 1 (iii) Because in soaps hydrocarbon chains are not branched. [CBSE Marking Scheme, 2019] 1

Detailed Answer : (i) Biothional is generally added to soaps to impart antiseptic properties and reduces the odours produced by

the bacterial decomposition of organic matter on the skin. (ii) Antacids neutralize excess acid produced in the stomach. Magnesium hydroxide is a better antacid because

of its insolubility. Being insoluble, even an excess will not allow the pH to increase above neutral. Sodium bicarbonate being soluble and its excess can make stomach alkaline and trigger more production of acid.

(iii) Soaps are sodium and potassium salts of fatty acids. Straight chain hydrocarbons are easily degraded by microorganisms. The fatty acids used to make soaps contain straight chain hydrocarbon part, making them biodegradable. Detergents mostly contain branched chain hydrocarbon. The bacteria cannot digest branched chain compounds. Hence they are non-biodegradable. [3]

OR

(i) An antibiotic refers to a substance produced wholly or partly by chemical synthesis, which in low concentrations inhibits the growth or destroys microorganisms by intervening in their metabolic processes. E.g.,Penicillin ½ + ½

(ii) Chemicals which are sweet in taste and with low calories. Eg- Saccharin ½ + ½ (iii) Analgesics reduce or abolish pain without causing impairment of consciousness, mental confusion,

incoordination or paralysis or some other disturbances of nervous system. Eg- Aspirin (Or any other correct example) [CBSE Marking Scheme, 2019] ½ + ½

Detailed Answer : (i) Antibiotics are used as drugs to treat infections because of their low toxicity for humans and animals.

Antibiotics are chemical substances which are produced by microbes and inhibit the growth or even destroy other microorganisms. Examples – penicillin, salvarsan.

(ii) Artificial sweeteners : These are the Organic substances which have been synthesized in the lab and are known to be many times sweeter than sugar. For example, Aspartame, Saccharin, Sucralose.

(iii) Analgesics : The drugs which are used for painkilling effect on human body. For example, Narcotic analgesics are used to get relief from the pain in terminal cancer. [3]

Answer 21.

(i)

—

CN

(ii)

(iii)

—

I

[CBSE Marking Scheme, 2019] 1 × 3




Detailed Answer :

(i)

N3+ CN

CuCN

CN

(ii)

N3+ CN

CH3CH2OH

(iii)

N3+ CN

KI

I

[3]rrr

Delhi Set 3 56/1/3

SECTION A

Answer 1. CH3NH2 > (CH3)2NH > (CH3)3N [CBSE Marking Scheme, 2019] 1

Detailed Answer : The more extensive the hydrogen bonding, more is the solubility in water.

( ) ( )3 2 3 32 3CH NH CH NH CH N> >

[1]

Answer 2. Frenkel defect due to large difference in size of ions. [CBSE Marking Scheme, 2019] 1

Answer 3.

SN1 SN2

Produces racemic mixture Produces racemic mixture

[CBSE Marking Scheme, 2019] 1

Detailed Answer : An SN1 reaction proceeds with racemisation and an SN2 reaction proceeds with complete stereochemical inversion.

[1]

SECTION B

Answer 7. Henry’s law states that “the partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of

the gas (x) in the solution” 1 l To increase the solubility of CO2 in soft drinks l At high altitudes the partial pressure of oxygen is less than that at the ground level. This leads to low

concentrations of oxygen in the blood and tissues of people living at high altitudes or climbers. l Scuba divers must cope with high concentrations of dissolved gases while breathing air at high pressure

underwater. Increased pressure increases the solubility of atmospheric gases in blood. (Any two) ½ + ½ [CBSE Marking Scheme, 2019]




Detailed Answer : Statement : At a constant temperature, the solubility of a gas in a liquid is directly proportional to the pressure of

the gas. Applications:

(i) To increase the solubility of 2CO in soft drinks and soda water, the bottle is sealed under high 2CO pressure. (ii) At high altitudes, the atmospheric pressure is low. Therefore, the partial pressure of oxygen is less than that

at ground level. This leads to hypoxia as its solubility in blood decreases. [2]

Answer 11. (i) sp3d2, Paramagnetic ½ + ½ (ii) dsp3 / trigonal bipyramidal , Diamagnetic [CBSE Marking Scheme, 2019] ½ +½

Detailed Answer :

(i) In ( ) 22 6Fe H O

+ the oxidation state of Fe is 2+ . The outermost electronic configuration of Fe is 6 2 03d 4s 4p .

2H O is weak field ligand and do not cause pairing up of electrons.

xxxx xx xx xx

3d 4s 4p 4d In this complex, there are 4 unpaired electrons. Therefore, it involves sp3d2 hybridisation and is paramagnetic.

(ii) In ( )5Fe CO the oxidation state of Fe is zero. The outermost electronic configuration of Fe

is 6 2 03d 4s 4p . CO is strong field ligand, causes pairing up of 4s electrons into the 3d orbitals.

xxxx xx

3d 4s 4p

xx xx

In this complex, there is no unpaired electron. Therefore, it involves 3dsp hybridisation and is diamagnetic. [2]

Answer 12. A ⇒ CH3CH2CO – CH3 , B ⇒ CH3CH2 – CH(CH3) – OH ½ + ½ A ⇒ C6H5CHO , B ⇒ C6H5 – CH = N – NH2 ½ + ½


Detailed Answer :

(i)

(ii)

CH3

(i) CrO3 /(CH3CO)2O

(ii)H3O+

CHO

H2N- NH2

N NH2

+ H2O

B A [2]

SECTION C

Answer 17. (i) Nickel is heated in a stream of carbon monoxide forming a volatile complex named as nickel tetracarbonyl.

This complex is decomposed at higher temperature to obtain pure metal. 1 (ii) ZnS is roasted to give ZnO which is heated with reducing agent coke to give Zn 1 (iii) Complex is treated with zinc , displacement reaction occurs to give pure Ag. 1 [CBSE Marking Scheme, 2019]




Detailed Answer :

(i) The conversion of impure Nickel to pure Nickel is done by the Mond process. (ii) The conversion of Zinc blende to Zinc metal is done by fractional distillation.

(iii) The conversion of ( )2Ag CN−

to Ag is done by Leaching. [3]

Answer 22.

(i)

C H —N—H + CH —C—O—C—CH2 5 3 3

—

H

—

O

— —

O

—

C H —N—C – CH + CH2 5 3 3COOH

—

H

—

O

—

— N NCl + H —�+ –

— OHOH–

— N = N — — OH + Cl + H O–2

p-Hydroxyazobenzene (orange dye) 1

(ii)

— N NCl + H —≡+ –

— NH2

H+

— N = N — — NH + Cl + H O–22

1

(iii) R — NH2 + CHCl3 + 3KOH Heat → R — NC + 3KCl + 3H2O [CBSE Marking Scheme, 2019] 1

Detailed Answer : (i) Acetylation of aniline

NH2

(CH3CO)2O

Pyridine

HN CH3

Aniline

O

N-Phenylethanamide(Acetanilide)

(ii) Coupling reaction

Benzene diazoniumchloride

N+ NCl + H OH OH

N N OH +Cl H2O +

Phenol p-hydroxyazobenzene

(iii) Carbyl amine reaction

R - NH2 + CHCl3 + 3 KOH

HeatR - NC + 3 KCl + 3H2O [3]

Answer 24. (i) Carbohydrates that yield two to ten monosaccharide units, on hydrolysis, are called oligosaccharides.

Example- Sucrose or any other. 1 (ii) When a protein in its native form, is subjected to physical change like change in temperature or chemical

change like change in pH, the hydrogen bonds are disturbed. Due to this, globules unfold and helix get uncoiled and protein loses its biological activity. Example- Boiling of egg. 1

(iii) Organic compounds required in the diet in small amounts to perform specific biological functions for normal maintenance of optimum growth and health of the organism. Example- Vitamin A 1

(Or any other correct example) [CBSE Marking Scheme, 2019]

Detailed Answer : (i) Oligosaccharides – Carbohydrates that yield two to ten monosaccharide units, on hydrolysis are called

oligosaccharides. For example – sucrose. (ii) Denaturation of protein – When a protein in its native form is subjected to a physical change like a change

in temperature or a chemical change like a change of pH, the hydrogen bonds get disturbed. Due to this, globules unfold and helix gets uncoiled and protein loses its biological activity. This is called a denaturation of proteins. For example – curdling of milk.




(iii) Vitamins – Vitamins are organic compounds that we need in small amounts in our diet to perform specific biological functions for normal maintenance of optimum growth and health of the organism. For example – vitamin A. The deficiency of vitamin A leads to xerophthalmia. [3]

OR

(i)

(CHOH)4(CHOH)4

—CHO

—

CH OH2

Br water2

—COOH

—

CH OH2 1

(ii)

(CHOH)4(CHOH)4

—CHO

—

CH OH2

NH OH2

—CH = N—OH

—

CH OH2 1

(iii)

(CH—O—C—CH )3 4(CHOH)4

—CHO

—

CH OH2

Acetic anhydride

—CHO — —O

—

CH —O—C—CH2 3— —

O 1 [CBSE Marking Scheme, 2019]

Detailed Answer :

(i)

H

H

H

OH

OH

O

OH

OH

HO

H

H

H

H

OH

OH

OH

OH

HO

H

Br2 water

HO

Gluconic acidD-glucose

O

(ii)

(iii)

(CHOH)4

CH2CH3

OHCCH3CO2O (CH-O-CO-CH3)4

CH2-O-CO-CH3

OHC

glucose pentaacetateD-glucose [3]rrr




Outside Delhi Set 1 56/2/1

SECTION A

2. Arrange the following in decreasing order of basic character : C6H5NH2, (CH3)3N, C2H5NH2 [1] 3. What type of colloid is formed when a solid is dispersed in a liquid ? Give an example. [1] 4. Out of Chlorobenzene and Cyclohexyl chloride, which one is more reactive towards nucleophilic substitution

reaction and why ? [1]

SECTION B

7. Give reasons : (a) Cooking is faster in pressure cooker than in cooking pan. (b) Red Blood Cells (RBC) shrink when placed in saline water but swell in distilled water. [2] 10. Write IUPAC name of the complex [Co(en)2(NO2)Cl]+. What type of structural isomerism is shown by this

complex ? [2]OR

Using IUPAC norms, write the formulae for the following complexes : (a) Hexaaquachromium(III) chloride (b) Sodium trioxalatoferrate(III)

SECTION C

14. (a) Write the dispersed phase and dispersion medium of dust. (b) Why is physisorption reversible whereas chemisorption is irreversible ? (c) A colloidal sol is prepared by the method given in the figure. What is the charge on AgI colloidal particles

formed in the test tube ? How is this sol represented ?

KI Solution

AgNOsolution

3

[3] 17. Write the name and principle of the method used for refining of (a) Zinc, (b) Germanium, (c) Titanium. [3] 18. Give reasons for the following : (a) Transition metals form complex compounds. (b) Eo values for (Zn2+/Zn) and (Mn2+/Mn) are more negative than expected. (c) Actinoids show wide range of oxidation states. [3] 19. Write the structures of monomers used for getting the following polymers : (a) Nylon-6 (b) Terylene (c) Buna-N [3]

OR (a) Is [– CH2 – CH(C6H5) ]– n a homopolymer or copolymer ? Give reason. (b) Write the monomers of the following polymer :

[ O—CH—CH —C—O—CH—CH —C ]2 2– –—

CH3

— —

O

—

C H2 5

— —

O

n

(c) Write the role of benzoyl peroxide in polymerisation of ethene.





SECTION A

1. What is the basic structural difference between starch and cellulose? [1]OR

Write the products obtained after hydrolysis of DNA. 2. Arrange the following in increasing order of pKb values : [1]

6 5 2 2 6 5 3 6 5 2C H CH NH , C H NHCH , C H NH

3. What type of colloid is formed when a liquid is dispersed in a sol? Give an example. [1] 4. Out of Chlorobenzene and p-nitrochlorobenzene, which one is more reactive towards nucleophilic substitution

reaction and why? [1] 5. Out of KCl and AgCl, which one shows Schottky defect and why? [1]

OR Why does ZnO appear yellow on heating?

SECTION B

6. When 2 4FeCr O is fused with Na2CO3 in the presence of air it gives a yellow solution of compound (A). Compound (A) on acidification gives compound (B). Compound (B) on reaction with KCl forms an orange coloured compound (C). An acidified solution of compound (C) oxidises Na2SO3 to (D). Identify (A), (B), (C) and (D). [2]

7. Give reasons : (a) A decrease in temperature is observed on mixing ethanol and acetone. (b) Potassium chloride solution freezes at a lower temperature than water. [2] 8. Write balanced chemical equations for the following processes : (a) Cl2 is passed through slaked lime. (b) SO2 gas is passed through an aqueous solution of Fe(III) salt. [2]

OR (a) Write two poisonous gases prepared from chlorine gas. (b) Why does Cu2+ solution give blue colour on reaction with ammonia? 9. Write structures of main compounds A and B in each of the following reactions : [2]

(a) 5 2 4PCl H / Pd BaSO6 5C H COOH A B

-¾¾¾® ¾¾¾¾¾¾¾®

(b) ( )

( )3

3

Zn Hg /conc. HCl(i)CH MgBr3 ii H O

CH CN A B+¾¾¾¾¾¾® ¾¾¾¾¾¾¾¾®

10. Define the following terms with a suitable example of each : [2] (a) Chelate complex (b) Ambidentate ligand

OR Using IUPAC norms, write the formula for the following complexes : [2] (a) Tetraamminediaquacobalt(III) chloride (b) Dibromidobis(ethane-1,2-diamine)platinum(IV) nitrate 11. (a) Using valence bond theory, write the hybridization and magnetic character of the complex [Fe(CN)6]

4- . (Atomic no. of Fe=26)

(b) Write the electronic configuration of d6 on the basis of crystal field theory when :

(i) 0 P andD <

(ii) 0 PD >

[2]




12. Define order of reaction. Predict the order of reaction in the given graphs: [2]

(a) t1/2

[R]o

(b) t1/2

[R]o

Where [R]0 is the initial concentration of reactant and t1/2 is half life.

SECTION C

13. (a) Write the dispersed phase and dispersion medium of milk. (b) Why is adsorption exothermic in nature? (c) Write Freundlich adsorption isotherm for gases at high pressure. [3] 14. The following data were obtained for the reaction :

A 2B C+ ¾¾®

Experiment [A]/M [B]/M Initial rate of formation of C/M min-1

1 0.2 0.3 4.2 × 10–2

2 0.1 0.1 6.0 × 10–3

3 0.4 0.3 1.68 × 10–1

4 0.1 0.4 2.40 × 10–2

(a) Find the order of reaction with respect to A and B. (b) Write the rate law and overall order of reaction. (c) Calculate the rate constant (k). 15. Write the name and principle of the method used for refining of (a) Tin, (b) Copper, (c) Nickel. [3] 16. Give reasons for the following : [3] (a) Transition metals show variable oxidation states. (b) E° value for (Zn2+ /Zn) is negative while that of (Cu2+ /Cu) is positive. (c) Higher oxidation state of Mn with fluorine is +4 whereas with oxygen is +7. 17. An element X with an atomic mass of 81 u has density 10.2 g cm– 3. If the volume of unit cell is 2.7 × 10–23

cm3, identify the type of cubic unit cell. (Given : NA = 6.022 × 1023 mol–1) [3]

18. A solution containing 1.9 g per 100 mL of KCI (M = 74.5 g mol–1) is isotonic with a solution containing 3 g per 100 mL of urea (M = 60 g mol–1). Calculate the degree of dissociation of KCI solution. Assume that both the solutions have same temperature. [3]

19. Write the structures of monomers used for getting the following polymers : [3] (a) Nylon-6,6 (b) Bakelite (c) Buna -S

OR (a) Write one example each of : (i) Thermoplastic polymer (ii) Elastomers (b) Arrange the following polymers in the increasing order of their intermolecular forces : Polythene, Nylon-6,6, Buna-S (c) Which factor provides crystalline nature to a polymer like Nylon? [3] 20. Among all the isomers of molecular formula C4H9Br, identify (a) the one isomer which is optically active.




(b) the one isomer which is highly reactive towards SN2. (c) the two isomers which give same product on dehydrohalogenation with alcoholic KOH. [3] 21. (a) Pick out the odd one from the following on the basis of their medicinal properties : Equanil, Seconal, Bithional, Luminal (b) What type of detergents are used in dishwashing liquids? (c) Why is the use of aspartame limited to cold foods? [3]

OR Define the following terms with a suitable example of each : [3] (a) Antibiotics (b) Antiseptics (c) Anionic detergents 22. (a) What are the products of hydrolysis of maltose? (b) What type of bonding provides stability to a-helix structure of protein ? (c) Name the vitamin whose deficiency causes pernicious anaemia. [3]

OR Define the following terms: [3] (a) Invert sugar (b) Native protein (c) Nucleotide 23. (a) Give reasons : (i) Benzoic acid is a stronger acid than acetic acid. (ii) Methanal is more reactive towards nucleophilic addition reaction than ethanol. (b) Give a simple chemical test to distinguish between propanal and propanone. [3] 24. Complete the following reactions:

(a) CN

H2 /Ni

(b)

BrCH3

H3 PO2+ H2O

N Cl–+

2

(c) CH2 – NH2

3 Ethanolic KOH+ CHCl3

OR How do you convert the following : [3] (a) N-phenylethanamide to p-bromoaniline (b) Benzene diazonium chloride to nitrobenzene (c) Benzoic acid to aniline

SECTION D

25. (a) Give equations of the following reactions : (i) Phenol is treated with conc. HNO3. (ii) Propene is treated with B2H6 followed by H2O2/OH–. (iii) Sodium t-butoxide is treated with CH3Cl. (b) How will you distinguish between butan-1-ol and butan-2-ol? (c) Arrange the following in increasing order of acidity : [5] Phenol, Ethanol, Water

OR (a) How can you obtain Phenol from (i) Cumene, (ii) Benzene sulphonic acid, (iii) Benzene diazonium chloride? (b) Write the structure of the major product obtained from dinitration of 3-methylphenol.




(c) Write the reaction involved in Kolbe’s reaction. 26. (a) Account for the following : (i) Tendency to show –3 oxidation state decreases from N to Bi in group 15. (ii) Acidic character increases from H2O to H2Te. (iii) F2 is more reactive than CIF3, whereas CIF3 is more reactive than Cl2. (b) Draw the structure of (i) XeF2, (ii) H4P2O7. [5]

OR (a) Give one example to show the anomalous reaction of fluorine. (b) What is the structural difference between white phosphorus and red phosphorus? (c) What happens when XeF6 reacts with NaF? (d) Why is H2S a better reducing agent than H2O? (e) Arrange the following acids in the increasing order of their acidic character : HBr, Hl, HCl, HF, [5]

27. (a) The conductivity of 0.001 mol L–1 acetic acid is 4.95 × 10–5 S cm–1. Calculate the dissociation constant if o

mÙ for acetic acid is 390.5 S cm2 mol–1.

(b) Write Nernst equation for the reaction at 25°C :

( ) ( ) ( ) ( )2 32Al S 3Cu aq 2Al aq 3Cu s++ + → +

(c) What are secondary batteries? Give an example. [5]OR

(a) Represent the cell in which the following reaction takes place :

( ) ( ) ( ) ( )2 32Al s 3Ni 0.1M 2Al 0.01M 3Ni s+ ++ ¾¾® +

Calculate its emf if ocellE = 1.41V.

(b) How does molar conductivity vary with increase in concentration for strong electrolyte and weak electrolyte?

How can you obtain limiting molar conductivity ( )omÙ for weak electrolyte? [5]


SECTION A

1. Out of Chlorobenzene and Cyclohexyl chloride, which one is more reactive towards nucleophilic substitution reaction and why? [1]

2. Arrange the following in decreasing order of solubility in water :

( )2 5 2 5 2 6 5 22C H NH, C H NH , C H NH

3. What type of colloid is formed when a solid is dispersed in a gas? Give an example. [1] 5. What is the difference between amylase and amylopectin? [1]

OR Write the products obtained after hydrolysis of lactose. [1]

SECTION B

7. Give reasons : [2] (a) An increase in temperature is observed on mixing chloroform and acetone. (b) Aquatic animals are more comfortable in cold water than in warm water. 10. Define the following terms with a suitable example of each : [2] (a) Polydentate ligand (b) Homoleptic complex

OR Using IUPAC norms, write the formulae for the following complexes: [2] (a) Potassium tri(oxalate)chromate(III) (b) Hexaaquamanganese(II) sulphate




11. (a) Although both [NiCl4]2– and [Ni(CO)4] have sp3 hybridisation yet [NiCl4]

2– is paramagnetic and [Ni(CO)4] is diamagnetic. Give reason. (Atomic no. of Ni = 28)

(b) Write the electronic configuration of d5 on the basis of crystal field theory when.

(i) o PD < and

(ii) o PD > [2] 12. Write structures of main compounds A and B in each to the following reactions: [2]

(a) ( )3CH OH /dry HCl gPCC3 2CH CH OH A B¾¾¾® ¾¾¾¾¾¾¾¾®

(b) NaOI6 5 3C H COCH A B¾¾¾® +

SECTION C

14. (a) Write the dispersed phase and dispersion medium of butter. (b) Why does physisorption decrease with increase in temperature? (c) A colloidal sol is prepared by the method given in the figure. What is the charge on AgI colloidal particles

formed in the test tube? How is this sol represented? [3]

KI solution

AgNO3 solution

17. Write the principle of the following : [3] (a) Hydraulic washing (b) Chromatography (c) Froth-floatation process 18. Give reasons for the following : [3] (a) Transition metals have high enthalpies of atomization. (b) Manganese has lower melting point even though it has a higher number of unpaired electrons for bonding. (c) Ce4+ is a strong oxidizing agent. 19. Write the structures of monomers used for getting the following polymers : [3] (a) Novolac (b) Neoprene (c) Buna-S

OR (a) Write one example each of : (i) Cross- linked polymer (ii) Natural polymer (b) Arrange the following in the increasing order of their intermolecular forces: Terylene, Buna-N, Polystyrene (c) Define biodegradable polymers with an example. [3] 23. (a) Write the product when D-glucose reacts with Br2 (aq). (b) What type of bonding provides stability to a-helix structure of protein? (c) Name the vitamin whose deficiency causes pernicious anaemia. [3]

OR Define the following terms: (a) Invert sugar (b) Native protein (c) Nucleotide [3]




SOLUTIONSOutside Delhi Set 1 56/2/1

SECTION A

Answer 2. (CH3)3N > C2H5NH2 > C6H5NH2 [CBSE Marking Scheme, 2019] 1

Answer 3. Example- Paints (Or any other correct example) 1 [CBSE Marking Scheme, 2019]

Answer 4. Cyclohexyl chloride; Because of partial double bond character of C-Cl bond in Chlorobenzene / Resonance effect

/ sp3 hybridised carbon in cyclohexyl chloride whereas sp2 carbon in chlorobenzene. [CBSE Marking Scheme, 2019] 1

SECTION B

Answer 7. (a) Due to increase of pressure in cooker, boiling point of water increases. (b) RBC looses water in saline water and absorb water in distilled water due to osmosis. (Or any other correct reason) 1 + 1 [CBSE Marking Scheme, 2019]

Answer 10. Chloridobis(ethane-1,2-diamine)nitrito-N-cobalt(III) ion Linkage isomerism 1 + 1

OR (i) Cr(H2O)6]Cl3 (ii) Na3[Fe(ox)3] [CBSE Marking Scheme, 2019] 1 + 1

SECTION C

Answer 14. (i) Dispersed phase = solid ; Dispersion medium = gas 1 (ii) Due to weak van der Waal’s forces in physisorption whereas strong chemical forces in chemisorption. 1 (iii) Positively charged , AgI/Ag+ ½ + ½ [CBSE Marking Scheme, 2019]

Answer 17. (a) Distillation/ Electrolytic refining : The impure metal is evaporated to obtain the pure metal as distillate/

The more basic metal remains in the solution and the less basic ones go to the anode mud. ½ + ½ (b) Zone refining : Impurities are more soluble in the melt than in the solid state of the metal. ½ + ½ (c) Van Arkel method : The metal should form a volatile compound which decomposes at higher temperature

to pure metal. [CBSE Marking Scheme, 2019] ½ + ½

Answer 18. (i) Due to small size, high ionic charge and availability of d-orbital. 1 (ii) Due to stable 3d10 configuration in Zn2+ and 3d5 configuration in Mn2+. 1 (iii) Due to comparable energies of 5f, 6d and 7s orbitals / levels. 1 [CBSE Marking Scheme, 2019]




Detailed Answers : (a) Transition metals are able to form complex compounds due to the small size of metal, high ionic charge and

availability of vacant d-orbital. (b) E° values of (Mn2+/Mn) and (Zn2+/Zn) are more negative than expected due to the greater stability of half

filled d-subshell of Mn2+(3d5) and completely filled d-subshell of Zn2+(3d10). (c) Due to a very small energy gap between 5f, 6d and 7s subshells resulting in easier excitation of the outermost

electrons to higher energy levels.

Answer 19.

(a)

N——

—H

C=O—

CH2

H2C—

H2C—

H2C

—

CH2

—

(b)

HOH C—CH OH, HOOC—2 2 —COOH

(c) CH = CH—CH = CH , CH = CH2 2 2

—CN

1 + 1 + 1 [CBSE Marking Scheme, 2019]

Detailed Answers : (a) Caprolactam

N——

—H

C=O—

CH2

H2C—

H2C—

H2C

—

CH2

—

(b) Ethylene glycol HO — CH2 — CH2 — OH

Terephthalic acid

HOH C—CH OH, HOOC—2 2 —COOH

(c) Buta–1, 3– diene CH2 = CH — CH = CH2

Acrylonitrile CH2 = CH — C ≡ NOR

(a) Homopolymer; As the same monomer is repeated. ½ + ½

(b) CH —CH—CH —COOH, CH —CH —CH—CH —COOH2 2 2 2 2

—OH

1 (c) It acts as an initiator. 1 [CBSE Marking Scheme, 2019]

Detailed Answers : (a) It is a homopolymer because single type of monomer unit i.e., C6H5 — CH = CH2 undergoes polymerisation. (b) 3– hydroxy butanoic acid nCH3 — CH — CH2 — COOH | OH 3– hydroxy pentanoic acid nCH3 — CH2 — CH — CH — CH2 — COOH | OH (c) Benzoyl peroxide is the reagent which generates free radical formed on decomposition And for polymerization

of ethene, the source of free radical is required to initiate the chain reaction. These free radicals are produced by benzoyl chloride.





SECTION A

Answer 1. Starch is a polymer of a- glucose whereas cellulose is a polymer of a- glucose. [CBSE Marking Scheme, 2019] 1

Detailed Answer : In starch, the glucose monomers are in alpha configuration while in cellulose, the glucose monomers are in beta

configuration. [1]OR

2-deoxyribose + nitrogen containing heterocyclic base + phosphate [CBSE Marking Scheme, 2019] 1

Detailed Answer : The products formed after hydrolysis of DNA are : (i) Pentose sugar, (ii) Phosphoric acid and (iii) Nitrogenous bases. [1]

Answer 2. C6H5CH2NH2 < C6H5NHCH3 < C6H5NH2 [CBSE Marking Scheme, 2019] 1

Answer 3. Gel , Example- Cheese / butter [CBSE Marking Scheme, 2019] 1

Detailed Answer : Example : cheese, butter, gel, jellies (Any one) [1]

Answer 4. p-Nitrochlorobenzene ; Due to electron withdrawing nature of –NO2 group. [CBSE Marking Scheme, 2019] ½ + ½

Detailed Answer : The relative p-nitrochlorobenzene is more reactive towards nucleophilic susbstitution reaction due to its

electron-withdrawing inductive and resonance effects which results in the stabilised carbanion formed by the –NH2 group. [1]

Answer 5. KCl , due to comparable sizes of K+ and Cl– [CBSE Marking Scheme, 2019] ½ + ½

Detailed Answer : KCl shows Schottky defect as this defect is shown by ionic crystal compounds in which the cation and anion are

of almost similar sizes. [1]OR

On heating excess Zn2+ ions move to interstitial sites and the electrons to neighbouring interstitial sites/ because of metal excess defect due to presence of extra Zn2+ cations at interstitial sites

[CBSE Marking Scheme, 2019] 1

Detailed Answer :

Heating 22

1ZnO Zn O 2e

2+ -¾¾¾¾® + +

Because excess Zn2+ ions moves to interstitial sites and the electrons move to neighbouring voids. This electron is responsible for the colour and electrical conductivity in crystals. [1]

SECTION B

Answer 6. A = Na2CrO4, B=Na2Cr2O7, C= K2Cr2O7, D= Na2SO4 [CBSE Marking Scheme, 2019] ½ × 4 = 2




Detailed Answer :

2 4 2 3 2 2 4 2 3 24FeCr O 8Na CO 7O 8Na CrO 2Fe O 8CO+ + ¾¾® + +

[A]

2 4 2 2 7 22Na CrO 2H Na Cr O 2Na H O+ ++ ¾¾® + +

[B]

2 2 7 2 2 7Na Cr O 2KCl K Cr O 2NaCl+ ¾¾® +

[C]

( )2 2 7 2 3 2 4 2 4 2 4 2 4 23K Cr O 3Na SO 4H SO 3Na SO K SO Cr SO 4H O+ + ¾¾® + + +

[D]

[A] 2 4Na CrO : Sodium chromate, [B]Na2Cr2O7 : Sodium dichromate, [C] K2Cr2O7 : Potassium dichromate, [D]Na2SO4 : Sodium sulphate. [2]

Answer 7. (a) Ethanol-acetone interaction is weaker than pure ethanol or acetone interactions. (b) On adding KCl, vapour pressure of the solution decreases. [CBSE Marking Scheme, 2019] 1 + 1

Detailed Answer : (a) When ethanol is mixed with acetone it shows positive deviation from Raoult’s law and acetone molecules get

in between the host molecules and break some hydrogen bonds, which requires higher energy than energy released in the formation of new hydrogen bonds. This results fall in temperature.

(b) According to Raoult’s law, when a non-volatile solid is added to the solvent, its vapour pressure decreases resulting in decrease in freezing point. Also, the freezing point of water is higher than when water contains KCl. [2]

Answer 8. (a) 2Ca(OH)2 + 2Cl2 → CaCl2 + Ca(OCl)2 +2H2O 1 (b) SO2 + 2Fe3+ + 2H2O → 2Fe2+ + SO4

2– + 4H+ [CBSE Marking Scheme, 2019] 1

OR

(a) Mustard gas, tear gas, phosgene (Any two ) ½ + ½ (b) Because it forms blue coloured complex [Cu(NH3)4]

+2 (aq) or Equation [CBSE Marking Scheme, 2019] 1

Detailed Answer :

(a) Phosgene (COCl2), tear gas (CCl3NO2), mustard gas ( )4 8 2C H Cl S . (Any two) (b) Cu2+ion reacts with ammonia solution to form a deep blue coloured complex because of the property of

ammonia to form complex compounds. Ammonia acts as a lewis base due to presence of lone pair of electrons on the nitrogen atom. Hence, it is able to form coordinate bond with electron deficient molecules or a number of transition metal cations.

Cu2+(aq) + 4NH3(aq) → [Cu(NH3)4]2+

Tetra-amine copper (II) (blue colour) [2]

Answer 9. (i) A = C6H5COCl, B= C6H5CHO ½ + ½ (ii) A= CH3COCH3, B = CH3CH2CH3 [CBSE Marking Scheme, 2019] ½ + ½

Detailed Answer :

(a) 5 2 4PCl H / Pd BaSO6 5 6 5 6 5C H COOH C H COCl C H CHO

-¾¾¾® ¾¾¾¾¾¾¾®

[A] [B] Benzoyl chloride Benzaldehyde




(b)

( )3

3

(i) CH MgBr Zn Hg /conc.HCl3 3 3 3 2 3 2(ii) H O

CH CN CH C CH CH CH CH H O+→ − − → +�O

[A] [B] Acetone Propane [2]

Answer 10. (a) A complex formed by bi or polydentate ligands with metal. Example- [Co(en)3]

3+ ½ + ½ (b) A ligand which can ligate through two different donor atoms. Example-SCN– ½ + ½ (Or any other correct example)


Detailed Answer : (a) Chelate complex is the complex formed when a di – or polydentate ligand uses its two or more donor atoms

to bind a single metal atom. (b) Ligand which can ligate through two different atoms is called ambidentate ligand. [2]

OR

(i) [Co(NH3)4(H2O)2]Cl3 1 (ii) [Pt Br2(en)2](NO3)2 [CBSE Marking Scheme, 2019] 1

Answer 11. (a) d2sp3, diamagnetic ½ + ½ (b) (i) t2g

4eg2 (ii) t2g

6eg0 [CBSE Marking Scheme, 2019] ½ + ½

Detailed Answer :

(a) Oxidation number of Fe is + 2 in ( ) 46Fe CN

-é ùë û

4s 4p 4d3d

Fe2+ – 3d6

CN– being strong ligand pairs with 3d orbital of Fe2+

4s 4p3d

Six CN– ions give their six pairs of electron to the empty orbitals. The orbitals utilized for hybridization

is d2 sp3.– –

Magnetic character -Diamagnetic. [1]

Answer 12. It is defined as the sum of powers to which the concentration terms are raised in the rate law equation. 1 (i) First order (ii) zero order [CBSE Marking Scheme, 2019] ½ + ½

SECTION C

Answer 13. (i) Dispersed phase = liquid ; Dispersion medium = liquid (ii) Due to the formation of new bonds / force of attraction between adsorbate and adsorbent.

(iii) xm

= kp0 = k

[CBSE Marking Scheme, 2019] 1 × 3 = 3




Detailed Answer : (a) Dispersed phase in milk - Liquid/oil Dispersion medium in milk - Liquid/water. (b) Because adsorption leads to a decrease in the residual forces on the surface of the adsorbent. This causes a

decrease in the surface energy of the adsorbent. Therefore, adsorption is always exothermic.

(c) x 1

log log k log pm n

= +

At high pressure x

pm

µ ° [3]

}

Q

a

y

Ox

log k (Intercept)

log p

Slope =1n

log = xm

b

Answer 14. Rate = k[A]p[B]q

On solving (a) Order with respect to A = 2, B = 1 ½ + ½ (b) Rate = k [A]2[B]1 ; overall order = 3 ½ + ½ (c) Experiment 1 : 4.2 × 10−2 = k (0.2)2 (0.3); k = 3.5 1 Experment 2 : 6.0 × 10−3 = k (0.1)2 (0.1); k=6 (Full marks may be awarded for any one correct answer)


Detailed Answer : Let the order of reaction with respect to A be x and with respect to B be y.

∴ Rate of reaction = [ ] [ ]x yk A B According to details given , 4.2 × 10–2 = k[0.2]x [0.3]y

... (1) 6.0 × 10–3 = k[0.1]x [0.1]y

... (2) 1.68 × 10–1 = k[0.4]x [0.3]y

... (3) 2.40 × 10–2 = k[0.1]x [0.4]y

... (4) Dividing equation 4 by 2, we get

[ ] [ ]

[ ] [ ]

x y2

3 x y2.40 10 k 0.1 0.46.0 10 k 0.1 0.1

-

-´

=´

[ ]

[ ]

y

y0.4

40.1

=

( ) ( )1 y4 4=

y = 1 Dividing equation (1) by (3), we get

[ ] [ ]

[ ] [ ]

x y2

1 x y4.2 10 k 0.2 0.3

1.68 10 k 0.4 0.3

-

-´

=´

[ ]

[ ]

x

x0.1

0.250.2

=




( ) ( )x0.25 0.5=

( ) ( )2 x0.5 0.5= x = 2 (i) So, the rate of reaction with respect to A is 2 and with respect to B is 1. (ii) Rate law = k[A]2[B] Overall order of reaction is 3.

(iii) Rate constant, K = [ ] [ ]2

Rate

A B

= ( ) ( )

3

26.0 10

0.1 0.1

-´

2 2 16.0 mol L min- -= [3]

Answer 15. (a) Liquation : Metals having low melting points than impurities. ½ + ½ (b) Electrolytic refining : The more basic metal remains in the solution and the less basic ones go to the anode

mud. ½ + ½ (c) Mond’s process : Ni should form a volatile compound with a suitable reagent which decomposes at higher

temperature to pure Ni. [CBSE Marking Scheme, 2019] ½ + ½

Detailed Answer : (a) Liquation. Principle : This method is based on the lower melting point than the impurities and tendency of the molten

metal to flow on the sloping surfaces. (b) Electrolytic refining. Principle : This method is based on the phenomenon of electrolysis where pure metal gets deposited on

cathode when electricity is passed through the electrolyte. (c) Vapour phase refining (Mond’s process). Principle : The metal is converted into its volatile compound and collected else where which gives pure metal

when decomposed at high temperature. [3]

Answer 16. (i) Because of comparable energies of (n – 1)d and ns orbitals / Incomplete filling of d-orbital. (ii) Because of stable 3d10 configuration of Zn2+ whereas due to low hydration enthalpy and high enthalpy of

atomization of Cu2+. (iii) Due to the ability of oxygen to form multiple bonds with metal. [CBSE Marking Scheme, 2019] 1 × 3 = 3

Detailed Answer : (a) Because of availability of partially filled orbitals and comparable energies of ns and (n – 1)d orbitals. (b) E° value for (Zn2+/Zn) is negative due to stable completely filled d10 configuration in Zn2+. The positive value

of (Cu2+/Cu) accounts for its ability to liberate H2 from acids due to its high enthalpy of atomization and low hydration energy.

(c) Mn can form multiple bonds with oxygen by using 2p orbital of oxygen and 3d orbital of Mn because of which it shows highest oxidation state of +7 with fluorine, Mn cannot form multiple bonds thus shows an oxidation state of +4. [3]

Answer 17.

d = ZM

Na x a3 ½

Z = dx aaNM

3

1

= 10 2 6 022 10 2 7 10

81

23 23. . .× × × × − ½

= 2 ½ Hence, lattice is bcc. [CBSE Marking Scheme, 2019] ½




Detailed Answer : Given ; d = 10.2 g cm–3, M = 81u, V= a3 = 2.7 × 10-23cm3, NA = 6.022 × 1023 mol–1

3A

zMd

a .N=

3

Ada Nz

M=

=23 2310.2 2.7 10 6.022 10

81

-´ ´ ´ ´

= 2.04 ≈ 2 So, the given element has bcc cubic unit cell. [3]

Answer 18. p1 (urea) = p2 + (KCl) ½ C1RT = i C2RT

n11V

= i

n22V

(V1 = V2) 1

360

= i ×

1 974 5..

i = 1.96 ½

∝ = in

−−

11

= 1 96 12 1. −

− = 0.96 or 96% [CBSE Marking Scheme, 2019] 1

Detailed Answer :

KCl Ureap = p

i 0.19 0.0821 300

74.56 1´ ´ ´

´ = 0.3 0.0821 300

60 1´ ´

´ (As temperature is same)

i 0.19 0.374.5 60´

=

0.3 74.5

i 1.9660 0.19

´= =

´

KCl ionizes as : KCl K Cl+ -+

Initial conc. 1 O O Conc. at equilibrium 1 - a a a Number of effective particles after dissociation = 1 – a + a + a = 1 + a

Van’t Hoff factor = Number of moles after

Number of moles before dissociationdissociation

1

11α

α+

= = +

1.96 = 1 + a ; a = 1.96 – 1 = 0.96 a = 0.96 × 100 = 96% [3]

Answer 19. (a) HOOC(CH2)4COOH, H2N(CH2)6NH2

(b)

—

OH

, CH O2




(c) CH2 = CH – CH = CH2,

—

CH = CH2


Detailed Answer : (a) Nylon 6, 6

Hexamethylene diamine : ( )2 2 26H N CH NH

Adipic acid : ( )2 4COOH CH COOH- -

(b) Bakelite

Phenol :

OH

Formaldehyde : HCHO (c) Buna – S

1, 3-Butadiene : 2 2CH CH CH CH= - =

Styrene :

CH = CH2

[3]

OR

(a) (i) Polythene ½ + ½ (ii) Buna-S (Or any other correct example) (b) Buna-S < polythene < Nylon-6,6 1 (c) Hydrogen bonding [CBSE Marking Scheme, 2019] 1

Detailed Answer : (c) Strong intermolecular forces like hydrogen bonding and linear structure lead to close packing of polymer

chains that imparts crystalline character. [1]

Answer 20. (i) CH3CH2CH(Br)CH3 1 (ii) CH3CH2CH2CH2Br 1 (iii) (CH3)3CBr and (CH3)2CHCH2Br [CBSE Marking Scheme, 2019] ½ + ½

Detailed Answer : (a) 2–Bromobutane (b) 1– Bromobutane (c) 1–Bromo–2–methylpropane and 2–Bromo–2–methylpropane. [3]

Answer 21. (i) Bithional (ii) Non-ionic detergents (iii) Because it is unstable at cooking temperature. [CBSE Marking Scheme, 2019] 1 + 1 + 1

Detailed Answer : (a) Bithional as it is an antiseptic while others are tranquilizers. (b) Non-ionic detergents. (c) Because it is unstable at cooking temperature. [3]

OR

(a) These are chemical substances produced by micro-organisms which kill or inhibit the growth of microorganisms. Ex.- Penicillin ½ + ½




(b) These are chemical substances which kill or prevent the growth of microorganisms when applied on living tissues. Ex.- Dettol ½ + ½

(c) These are sodium salts of sulphonated long chain alcohols or hydrocarbons. / Anionic part of the molecule is involved in cleansing action. Example- sodium lauryl sulphate. ½ + ½

(Or any other one correct example) [CBSE Marking Scheme, 2019]

Detailed Answers : (a) Antibiotics are the chemical substances (prepared wholy or partially by chemical synthesis) which in low

concentration, either kill or inhibit the growth of microorganisms. E.g., penicillin. (b) Antiseptics are the chemicals which prevent the growth of microorganisms or kills them but are not harmful

to human beings E.g., Dettol. (c) Anionic detergents are sodium salts of sulphonated long chain alcohols or hydrocarbons. E.g., sodium

dodecylbenzene sulphonate. [3]

Answer 22. (a) Glucose + Glucose (b) Hydrogen bonding (c) Vitamin –B12 [CBSE Marking Scheme, 2019] 1 + 1 + 1

Detailed Answer : (a) 2 molecules of glucose. (b) Intramolecular hydrogen bonding. (c) Vitamin B12. [3]

OR

(i) Hydrolysis of sucrose brings a change of sign of rotation from dextro (+) to laevo (–) and the product is named as invert sugar.

(ii) Protein found in biological system with unique three dimensional structure and biological activity is called native protein.

(iii) A unit formed by the combination of nitrogenous base , pentose sugar and phosphate . [CBSE Marking Scheme, 2019] 1 + 1 + 1

Detailed Answers : (a) Invert sugar : Hydrolysis of sucrose brings about a change in sign of rotation from dextro (+) to laevo (–) and

the product formed is called invert sugar. (b) Native protein : Protein found in the biological system with a unique 3–dimensional structure and biological

activity. (c) Nucleotide : When nucleoside is linked to phosphoric acid at 5’–position of sugar molecule, the unit obtained

is called nucleotide. [3]

Answer 23. (a) (i) Due to greater electronegativity of sp2 hybridised carbon to which carboxyl carbon is attached / Due to

greater resonance stabilization of carboxylate ion with the benzene ring. (ii) Because carbonyl carbon of methanal is more electrophilic than that of ethanol / due to + I effect of

methyl group in ethanal, reactivity decreases. 1 + 1 (b) On heating with Tollens’ reagent / [Ag(NH3)2]

+, propanal forms silver mirror whereas propanone does not. ( or any other suitable chemical test) [CBSE Marking Scheme, 2019] 1

Detailed Answer : (a) (i) Strength of acid depends on the ease of release of H+ ions. Benzoic acid contains benzene ring which

is electron withdrawing where as acetic acid contains methyl group which is electron releasing. The benzoate ion resulted from dissociation of benzoic acid stabilized by resonance where as the acetate ion resulted from dissociation of acetic acid is not stabilized. Therefore, benzonic acid easily releases H+

ion than acetic acid.

(ii) In methanal, presence of comparatively bulky group than methanol attached to carbonyl group hinders the attack of nucleophile. Also 3CH group present in ethanol decreases the positive charge on carbonyl carbon by +I effect which is not possible in methanal. As Nu attack is favourable with more positive charge and less hindrance at carbonyl carbon, therefore methanal is more reactive than ethanol.




(b) Propanal being an aldehyde when heated with Tollen’s reagent to give silver mirror but propanone being a ketone does not.

– –

[3]

Answer 24.

(a)

CH NH2 2—

(d)

—

CH3

—Br

(c)

CH2NC—

(or any other suitable method) [CBSE Marking Scheme, 2019] 1 × 3 = 3

Detailed Answer :

(a)

CN NH2

H2 /Ni

(b)

CH3

BrBrCH3

H3 PO2+ H2O + N2 + H3 PO3 +HCl

N Cl–+

2

(c)

CH2 NCCH2 – NH2

3 Ethanolic KOH+ CHCl3 + 3KCl + 3H2O

OR

(a) (i)

H—N—C—CH3

—

— —

O

Br2

CH3COOH

H—N—C—CH3

—

— —

O

—

Br

OH or H+

NH2

——

Br 1

(ii)

N Cl2

—

NO2

—

+ HBF4

NaNO2

Fluboric

acid

N BF2 2

—

+ –

Cu, ∆

1

(iii)

COOH— ∆

–H2O

NH2

—

+ NH3

COONH2—

Br2 + KOH





Detailed Answer : (a) N-phenylethanamide to p-bromoaniline :

NHCOCH3NH – C – CH3

Br2

Br Br

NH2

CCl4

aq. NaOH

∆

||Ο

N-phenylethanamide

(b) Benzene diazonium chloride to nitrobenzene:N2BF4N2Cl NO2

NaNO2

∆

Benzene diozonium

chloride

+ HBF4 + N2 + NaBF4Cu,

Fluoroboric acid

Nitrobenzene

(c) Benzoic acid to aniline: [3]COClCOOH CONH2 NH2

SOCl2 NH3 Br2 /NaOH

Benzoic acid

Benzamide Aniline Benzoylchloride

SECTION D

Answer 25.

(a) (i)

OH

—

OH

—

Conc. HNO3

—O N2 —NO2

—

NO2 1

(ii)

CH —CH = CH + (H—HBr ) (CH —CH —CH ) B 3CH —CH —CH —OH3 2 2 3 3 2 2 3 3 2 2

H O2

3H O OH2 21 1

(iii)

CH —C—ONa + CH Cl3 3

—CH3

—

CH3

— +

CH —C—O CH3 3

—CH3

—

CH3 1 (b) On heating with NaOH / I2 , butan - 2 - ol forms yellow ppt of iodoform (CHI3) whereas butan -1-ol does

not. (Or any other test) 1 (c) Ethanol < water < Phenol [CBSE Marking Scheme, 2019] 1

Detailed Answer : (a) (i)

OH

NO2

OH OH

Phenol NO2

+ conc. HNO3

O-Nitrophenol

p-Nitrophenol

+




(ii) CH3 – CH = CH2 + B2H6 CH3 – CH2 – CH2 – BH2

CH3 – CH2 = CH2

3CH3 – CH2 – CH2 – OH + B(OH)3

(CH3 – CH2 – CH2 )2 BH(CH3 – CH2 – CH2 )3 B

3H2 O2 , OHH2 O

CH3 – CH = CH2

Trialkyl borane

Propene

Propan – 1 – ol

–

(iii)

NaO– C

CH3

CH3

CH3 C

CH3

CH3

CH3+ CH3Cl + NaClCH3O

Sodiumt-butoxide

Chloromethane

2-Methoxy 2-Methylpropane

(b) Iodoform test : Butan – 2 –ol (secondary alcohol) reacts with iodine and NaOH to give yellow precipitate of CHI3 (iodoform) whereas butan –1–ol does not react with iodine and NaOH.

3 2 3 2

OH |CH CH CH CH 4I 6NaOH+ + 2 5 3 2C H COONa CHI 5NaI 5H O

(Chloroform)

¾¾® + + +

2 2 2 3 2OH – CH CH CH CH I NaOH No reaction- - - + + ¾¾®

(c) Ethanol < Water < Phenol [5]OR

(a) (i)

CH —CH3

—

O2

—CH3

CH —C—O—OH3

—

—CH3

H+

H2O

OH

—

+ CH COCH3 3

1

(ii)

—

(i) NaOH

SO H3 OH

—

(ii) H+

1

(iii)

—

H O2

N Cl2 OH

—

+ N + HCl2Warm

+ –

1

(b)

OH

——

NO2

—

CH3

—

O2N

1




(c)

OH

—

NaOH

ONa

—

(i) CO2

(ii) H+

OH

—

—COOH


Detailed Answer : (a) (i)

CH C

CH3 CH3

CH3 OH

H+

H2O

O2

CH3 O – O – H

+ CH3 COCH3

Cumene cumene hydroperoxide Phenol Acetone

(ii) OHSO3H

(i) NaOH

(ii) H+

Benzenesulphonicacid

Phenol

(iii) OHN2ClNH2

H2O

Warm

NaNO2

+ HCl+ N2 + HCl

(b) OH

CH3

NO2

O2NOH

CH3

Dinitration

(c) Kolbe’s reaction :OHOH ONa

NaOH 400K, 4-7 atm+CO2

COONa COOH

H+

Sodium phenoxide Sodium salicylatePhenol Salicylic acid

OH

Answer 26. (a) (i) Due to increase in size and metallic character. 1 (ii) Due to decrease in bond dissociation enthalpy. 1 (iii) Due to low bond dissociation enthalpy of F-F bond than Cl-F bond whereas Cl-Cl bond has higher

bond dissociation enthalpy than Cl-F bond. 1

(b) (i)

F

F

Xe

(ii)

O

P

HO

OH

O

P

O

OH

OH

1 + 1 [CBSE Marking Scheme, 2019]




Detailed Answer : (a) (i) Due to increase in the size of the atom and the metallic character. (ii) Due to increase in size of the central atom is O < S < Se < Te, the distance between central atom and

hydrogen atom also increases. This results in increase in bond length, decreasing bond dissociation enthalpy. The bond cleavage becomes easier, increasing the acidic character from H2O to H2Te.

(iii) Interhalogen compounds are more reactive than halogen compounds. But in case of fluorine, due to small size of fluorine it has high electronegativity and low bond energy so it is more reactive than ClF3. Therefore, ClF3 is more reactive than Cl2.

(b) (i) XeF2F

Xe

F

(ii) 4 2 7H P O

P

O

OHOHO

P

O

OHOH

OR

(a) 2F2 (g) + 2H2O(l) → 4H+ (aq) + 4F+ (aq) + O2 (g)

(b) White phosphorus is discrete tetrahedral whereas red phosphorus is polymeric / or structures drawn.

(c) It forms Na+ [XeF7]– / XeFe + NaF → Na+ [XeF7]

–

(d) Due to lower bond dissociation enthalpy of H-S bond than H-O bond.

(e) HF < HCl < HBr< HI [CBSE Marking Scheme, 2019] 1 × 5

Detailed Answer : (a) It forms only one oxoacid as compared to the rest of the halogens that form a number of oxoacids.

(b)

White phosphorus : It exits as discrete to tetrahedral P4 unit.

Red phosphorus : It exits in the form of polymeric chain.

P

P

PP

P

P

PP

P

P

PP

P

P

P

(c) 6 7XeF NaF Na XeF -+ é ù+ ¾¾® ë û .

(d) Due to larger size of Sulphur and weak S-H bond in comparison to size of oxygen and O-H bond. Therefore, H2S easily donates electron in comparison to H2O.

(e) HF < HCl < HBr < HI.

Answer 27.

Λm = KC

= 4 95 10 10005 1

1

3. ××

− −

−S cm

0.001 molLcm

L = 49.5 S cm2 mol–1 1




a = ΛΛ

cm

cm° = 49 5 2 1

2 1. S cm mol

390.5 S cm mol

−

− = 0.126 1

(a) K = cα

α

2

1( )− = 0 001 0 126

1 0 126

1 2. ( . ).

mol L− ×−

= 1.8 × 10–5 mol L–1

(If K= ca2, then K= 1.6 × 10−5 mol L–1 1

(b) E(cell) = E cellO( )-

– 0 059

6

3 2

2 3.

log[ ][ ]AlC

+

+u 1

(c) Batteries which are rechargeable

Example- Lead storage, Ni-Cd batteries (Or any other one example ) ½ + ½


Detailed Answer :

5 1 3

m 1k 4.95 10 S cm 000 cmc L0.001 mol L

- -

-´ 1

Ù = = ´

= 49.5 S cm2 mol–1

cmom

49.50.127

390.5Ù

a = = =Ù

( )

( )22 35 1c 10 0.127

k 1.85 10 molL1 1 0.127

-- -a ´

= = = ´-a -

(b) [ ]

[ ]

23o

cell cell 32

RT AlE E log

nF Cu

+

+= -

(c) Secondary battery are those batteries that can be recharged after use by passing current through the electrodes in the opposite direction i.e., from the negative terminal to the positive terminal. Example, lead storage battery. [5]

OR

(a) Al(s) | Al3+(0.01M) || Ni2+ (0.1 M) | Ni(s) 1

E(cell) = E cellO( )- –

0 0596

3 2

2 3.

log[ ][ ]AN

ll

+

+ ½

E(cell) = 1.42V – 0 0596

0 010 1

2

3.

log[ . ][ . ]

1

E(cell) = 1.4298V or Ecell = 1.42V ½ (b) Λm decreases with increase in concentration for both strong & weak electrolyte Λm

0 can be obtained for weak electrolyte by applying Kohlrausch law / Λ°m = V + l°+ n– l°– 1 + 1


Detailed Answer :

(a) ( ) ( ) ( ) ( )2 32Al s 3Ni 0.1M 2Al 0.01M 3Ni s+ ++ ¾¾® +

Al(s) | Al3+ || Ni2+ |Ni(s) → Cell rotation

[ ]

[ ]

2

cell 30.0591 0.01

E E log6 0.1

= -

0.0591 11.41 log

6 0.1= -




0.05911.41 log10

6= -

0.05911.41

6= -

= 1.41 – 0.00985 = 1.40015 V (b) When the concentration of weak electrolyte becomes very low, its degree of ionization rises. This increase

leads to increase in the number of ions in the solution. Thus, the molar conductivity rises sharply of a weak electrolyte at low concentration. The molar conductivity of strong electrolyte decreases a bit with an increase in concentration. This is due to increase in interionic attraction due to higher number of ions per unit volume. On dilution, ions move apart, weakening interionic attractions and thus conductance increases.

Limiting molar conductivity for weak electrolytes is obtained by using kohlrausch law of independent migration of ions. [5]

rrr


SECTION A

Answer 1. Cyclohexyl chloride, because chlorobenzene is resistant to nucleophilic substitution reaction/ sp3 hybridised

carbon in cyclohexyl chloride whereas sp2 carbon in chlorobenzene. [CBSE Marking Scheme, 2019] ½ +½

Detailed Answer :

Cyclohexyl chloride is more reactive towards nucleophilic substitution reaction because in cyclohexyl chloride chlorine atom is attached to sp3 hybridised while in chlorobenzene it is sp2 hybridised. Thus it has higher tendency to release electrons. In chlorobenzene, due to resonance, cleavage of C-Cl bond is difficult than cyclohexyl chloride which makes it less reactive towards nucleophilic substitution. [1]

Answer 2. C2H5NH2 > (C2H5)2NH > C6H5NH2 [CBSE Marking Scheme, 2019] 1

Answer 3. Aerosol , Example- Smoke (Or any other correct example) [CBSE Marking Scheme, 2019] ½ + ½

Detailed Answer :

Aerosol is formed when a solid is dispersed in gas. Example, smokes. [1]

Answer 5. Amylose is a linear polymer of a- glucose whereas amylopectin is a branched polymer of a- glucose/ Amylose is

water soluble whereas amylopectin is water insoluble. (Or any other correct difference) 1 [CBSE Marking Scheme, 2019]

Detailed Answer :

Amylose is a long unbranched chain of a-D-glucose units held by C1-C4 glycosidic linkage whereas amylopectin is a branched chain polymer of a-D-glucose units, in which chain is formed by C1-C6 glycosidic linkage while branching occurs by C1-C6 glycosidic linkage. [1]

OR

Glucose + Galactose [CBSE Marking Scheme, 2019] 1

Detailed Answer :

β −D galactose and β −D glucose. [1]




SECTION B

Answer 7. (a) Due to stronger interaction between chloroform and acetone than pure chloroform or acetone interactions. 1 (b) Because of high solubility of oxygen gas / low KH value in cold water than in warm water. 1 [CBSE Marking Scheme, 2019]

Detailed Answer : (a) On addition of chloroform and acetone, chloroform forms strong hydrogen bonding with acetone.

C = O – – H – C

Cl

Cl

Cl

H3C

CH3

This results in release of energy due to increase in attractive forces. Hence, the dissolution is an exothermic process.

(b) At a given pressure, the solubility of oxygen in water increases with decrease in temperature. Therefore, the concentration of oxygen in sea is more in cold water and thus, the presence of more oxygen at lower temperature makes the aquatic animals more comfortable in cold water. [2]

Answer 10. (i) A ligand having several donor atoms. Example- EDTA 1 (ii) A complex in which a metal is bound to only one kind of donor groups / ligands. Example- [Co(NH3)6]

3+ 1 (or any other correct example)


Detailed Answer : (a) A ligand which contains several donor atoms in a single ligand is known as polydentate ligand. Example,

N(CH2 CH2 NH2)3. (b) Complexes in which a metal is bound to only one kind of donor groups are known as homoleptic complexes.

Example, [Co(NH3)6]3+. [2]

OR

(i) K3[Cr(C2O4)3] 1 (ii) [Mn(H2O)6]SO4 [CBSE Marking Scheme, 2019] 1

Answer 11. (a) In [NiCl4]

2–, Cl– is a weak field ligand due to which there are two unpaired electrons in 3d orbital whereas in [Ni(CN)4]

2–, CN- is a strong field ligand due to which pairing leads to no unpaired electron in 3d- orbital/ or structural representation ½ + ½

(b) (i) t2g3eg

2 (ii) t2g5eg

0 [CBSE Marking Scheme, 2019] ½ + ½

Detailed Answer : (a) In [NiCl4]

2– Ni is in +2 oxidation state and each Cl– donates a pair of electron. So, Cl– acts as a weak ligand and does not cause any forced pairing. Thus, electrons remains unpaired making it paramagnetic.

In [Ni(CO)4], Ni is in zero oxidation state and CO acts as a strong ligand causing forced pairing. Thus, no electron remains unpaired making it diamagnetic. [1]

Answer 12. (a) A = CH3CHO, B = CH3CH(OH)OCH3 ½ + ½ (b) A and B = CHI3, C6H5COONa [CBSE Marking Scheme, 2019] ½ + ½

Detailed Answer :

(a)

CH CH OH2 2PCC

CH — CHO2CH OH/dryHCl(g)3

Acetaldehyde[A]

CH — CH2

—

OCH3

—

OHHemiacetal

[B]




(b)

C H COOCH6 5 33NaOI

C H COONa + CHI + 2NaOH56 3

Sodium

benzoate

[A]

Iodoform

[B]

[2]

SECTION C

Answer 14. (i) Dispersed phase = Liquid Dispersion medium = Solid/liquid 1 (ii) Because physisorption is exothermic in nature. / With rise in temperature, desorption starts 1 (iii) Negatively charged , AgI/I– [CBSE Marking Scheme, 2019] ½ + ½

Detailed Answer : (a) Dispersed phase : Liquid Dispersion medium : Solid (b) Because it is an exothermic reaction. (c) When AgNO3 solution is added to KI solution, negatively charged sol of AgI is formed due to elective

adsorption of I- ion from the dispersion medium. It is represented as given below. AgI/I–

Negatively charged. [3]

Answer 17. (a) This is based on difference in gravities of the ore and gangue particles. 1 (b) Different compounds of a mixture are differently adsorbed on an adsorbent. 1 (c) The mineral particles become wet by oil while the gangue particles by water. 1 [CBSE Marking Scheme, 2019]

Detailed Answer : (a) Hydraulic washing : It is based on the differences in gravities of the ore and the gangue particles. (b) Chromatography : Different components of a mixture are differently adsorbed on an adsorbent. (c) Froth-floatation process : This is based upon the preferential wetting of mineral/ ore particles by oil while the

gangue particles by water. [3]

Answer 18. (i) Because of strong interatomic interactions / Strong metallic bonding between atoms. 1 (ii) Due to stable 3d5 configuration , interatomic interaction is poor between unpaired electrons. 1 (iii) Because Ce is more stable in +3 oxidation state. [CBSE Marking Scheme, 2019] 1

Detailed Answer : (a) The transition elements have high enthalpies of atomization because they have large number of unpaired

electrons in their atoms. This results in stronger interatomic interaction and stronger bonding between atoms. (b) Because of high heat of atomization and strongly bounded half-filled a orbital electrons with nucleus results

in strong interatomic interaction. (c) The formation of Ce4+ is promoted by its noble gas configuration reverting to the common +3 state. E° value

for Ce4+/Ce3+ is + 1.74 V thus readily gains an electron and acts as a strong oxidizing agent. [3]

Answer 19.

(a)

OH

—

, HCHO

(b) H C = C—CH = CH2 2

—Cl




(c)

CH = CH—CH = CH2 2 ,

CH = CH2

—


Detailed Answer : (a) Novolac :

Phenol

OH

and

Formaldehyde HCHO

(b) Neoprene :

Chloroprene 2 2

Cl |

CH C – CH CH= = (c) Buna –S : Butadiene 2 2CH CH – CH CH= =

Styrene 6 5 2C H CH CH= [3]

OR

(a) (i) Bakelite (ii) Proteins (Or any other suitable example) ½ + ½ (b) Buna-N < Polystyrene < terylene 1 (c) Polymers which are easily degraded by microorganisms Example- PHBV (Or any other suitable example) [CBSE Marking Scheme, 2019] ½ + ½

Detailed Answer : (c) Biodegradable polymers are those polymers which can be decomposed by microorganisms and do not cause

water pollution. Example, PHBV. [1]

Answer 23. (a) Gluconic acid (b) Hydrogen bonding (c) Vitamin –B12 [CBSE Marking Scheme, 2019] 1 + 1 + 1

Detailed Answer :

(a)

CHO

(CHOH)4

CH2OH

Br2

COOH

(CHOH)4

CH2OH

D - Glucose D - Gluconic acid

D-Gluconic acid is formed (b) Intramolecular hydrogen bonding. (c) Vitamin B12. [3]

OR

(i) Hydrolysis of sucrose brings a change of sign of rotation from dextro (+) to laevo (–) and the product is named as invert sugar. 1




(ii) Protein found in biological system with unique three dimensional structure and biological activity is called native protein. 1

(iii) A unit formed by the combination of nitrogenous base, pentose sugar and phosphate. 1 [CBSE Marking Scheme, 2019]

Detailed Answer : (a) Invert sugar : Hydrolysis of sucrose brings about a change in sign of rotation from dextro (+) to laevo (–) and

the product formed is called invert sugar. (b) Native protein : Protein found in the biological system with a unique 3–dimensional structure and biological

activity. (c) Nucleotide : When nucleoside is linked to phosphoric acid at 5’–position of sugar molecule, the unit obtained

is called nucleotide. [3] rrr


SOLVED C.B.S.E. PAPER 2019 With Class–XII · Draw structures of geometrical isomers for this...

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