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Chemistry | Class 12th

CBSE Board Paper 2018

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CBSE Board Paper 2018 Set - 1

General Instructions:

1. All questions are compulsory.

2. Questions number 1 to 5 are very short answer questions and carry 1 mark each.

3. Questions number 6 to 10 are short answer questions and carry 2 marks each.

4. Questions number 11 to 22 are also short answer questions and carry 3 marks each.

5. Question number 23 is a value based question and carries 4 marks.

6. Questions number 24 to 26 are long answer questions and carry 5 marks each.

7. Use log tables, if necessary. Use of calculators is not allowed.

Time allowed: 3 Hours Max Marks: 100

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1. Analysis shows that FeO has a non-stoichiometric composition with formula Fe0.95O. Give reason.

Section A

1

Section B

6. Calculate the freezing point of a solution containing 60 g of glucose (Molar mass = 180 g mol–1) in 250 g of water.

(Kf of water = 1.86 K kg mol–1)

2

2. CO(g) and H2(g) react to give different products in the presence of different catalysts. Which ability of the catalyst is shown by these reactions?

1

3. Write the coordination number and oxidation state of Platinum in the complex [Pt (en)2Cl2 ].

1

4. Out of chlorobenzene and benzyl chloride, which one gets easily hydrolysed by aqueous NaOH and why ?

1

5. Write the IUPAC name of following:

1

4

OR

7. For the reaction

2N2O5(g) → 4NO2(g) + O2(g) ,

the rate of formation of NO2(g) is 2.8×10-3Ms-1 . Calculate the rate of disappearance of N2O5 (g) .

2

8. Among the hydrides of Group-15 elements, which have the

(a) Lowest boiling point ?

(b) Maximum basic character ?

(c) Highest bond angle ?

(d) Maximum reducing character?

2

9. How do you convert the following ?

(a) Ethanal to Propanone

(b) Toluene to Benzoic acid

2

Account for the following :

(a) Aromatic carboxylic acids do not undergo Friedel-Crafts reaction.

(b) pKa value of 4-nitrobenzoic acid is lower than that of benzoic acid.

5

10. Complete and balance the following chemical equations :

(a) Fe 2+ + MnO4- + H+ →

(b) MnO4-+ H 2O + I – →

2

11. Give reasons for the following :

(a) Measurement of osmotic pressure method is preferred for the determination of molar masses of macromolecules such as proteins and polymers.

(b) Aquatic animals are more comfortable in cold water than in warm water.

(c) Elevation of the boiling point of 1 M KCl solution is nearly double than that of 1 M sugar solution.

3

12. An element ‘X’ (At. mass = 40 g mol–1) having f.c.c. structure, has unit cell edge length of 400 pm. Calculate the density of ‘X’ and the number of unit cells in 4 g of ‘X’. (NA = 6.022 × 1023 mol–1 )

3

Section C

6

13. A first order reaction is 50% completed in 40 minutes at 300 K and in 20 minutes at 320 K. Calculate the activation energy of the reaction.

(Given: log 2 = 0.3010, log 4 = 0.6021, R = 8.314 JK–1 mol–1)

3

14. What happens when

(a) a freshly prepared precipitate of Fe(OH)3 is shaken with a small amount of FeCl3 solution?

(b) persistent dialysis of a colloidal solution is carried out?

(c) an emulsion is centrifuged ?

3

15. Write the chemical reactions involved in the process of extraction of Gold. Explain the role of dilute NaCN and Zn in this process.

3

16. Give reasons:

(a) E° value for Mn3+/Mn2+ couple is much more positive than that for Fe3+/Fe2+

(b) Iron has higher enthalpy of atomization than that of copper.

(c) Sc3+ is colourless in aqueous solution whereas Ti3+ is coloured.

3

7

17. (a) Identify the chiral molecule in the following pair ?

(b) Write the structure of the product when chlorobenzene is treated with methyl chloride in the presence of sodium metal and dry ether.

(c) Write the structure of the alkene formed by dehydrohalogenation of 1-bromo-1-methylcyclohexane with alcoholic KOH.

3

18. (A), (B) and (C) are three non-cyclic functional isomers of a carbonyl compound with molecular formula C 4H 8O . Isomers (A) and (C) give positive Tollen’s test whereas isomer (B) does not give Tollen’s test but gives positive Iodoform test. Isomers (A) and (B) on reduction with Zn(Hg)/conc. HCl give the same product (D).

(a) Write the structures of (A), (B), (C) and (D).

(b) Out of (A), (B) and(C) isomers, which one is least reactive towards addition of HCN?

3

19. Write the structure of the main products in the following reactions:

3

8

20. (a) Why is bithional added to soap?

(b) What is the tincture of iodine ? Write its one use.

(c) Among the following, which one acts as a food preservative ?

Aspartame, Aspirin, Sodium Benzoate, Paracetamol

3

21. Define the following with an example of each :

(a) Polysaccharides

(b) Denatured protein

(c) Essential amino acids

3

22. (a) Write the formula of the following coordination compound : Iron(III) hexacyanoferrate (II)

(b) What type of isomerism is exhibited by the complex

(c) Write the hybridisation and number of unpaired electrons in the complex [CoF6 ]3- . (Atomic No. of Co = 27)

3

OR

(a) Write the product when D-glucose reacts with conc. HNO3

.

D-glucose + HNO3→

(b) Amino acids show amphoteric behaviour. Why?

9

23. Shyam went to a grocery shop to purchase some food items. The shopkeeper packed all the items in polythene bags and gave them to Shyam. But Shyam refused to accept the polythene bags and asked the shopkeeper to pack the items in paper bags. He informed the shopkeeper about the heavy penalty imposed by the government for using polythene bags. The shopkeeper promised that he would use paper bags in future in place of polythene bags.

Answer the following :

(a) Write the values (at least two) shown by Shyam.

(b) Write one structural difference between low-density polyethene and high-density polyethene.

(c) Why did Shyam refuse to accept the items in polythene bags ?

(d) What is a biodegradable polymer? Give one example?

4

Section D

10

24. (a) Give reasons:

(i) H3PO3 undergoes disproportionation reaction, but H3PO4 does not.

(ii) When Cl2 reacts with an excess of F2. ClF3 is formed and not FCl3.

(iii) Dioxygen is a gas while Sulphur is a solid at room temperature.

(b) Draw the structures of the following:

(i) XeF4

(ii) HCLO3

5

OR

(a) When concentrated sulphuric acid was added to n unknown salt present in a test tube a brown gas(A) was evolved. This gas intensified when copper turning were added to this test tube . On cooling, the gas (A) changed int a colourless solid(B).

(i) Identify (A) and (B).

(ii) Write the structures of (A) and (B).

(iii) Why does gas (A) change to solid on cooling?

(b) Arrange the following in the decreasing order of their reducing character:

HF, HCl, HBr, HI

(c) Complete the following reaction:

XeF4 +SbF5→

Section E

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25. (a) Write the cell reaction and calculate the e.m.f. of the following cell at 298 K.

Sn (s )|Sn 2+ (0.004 M )||H+(0.020 M )|H 2 (g )(1bar)| Pt (s)

(Given: E0 Sn+2/Sn=-0.14V )

(b) Give reasons:

(i) On the basis of E0 values, O2 gas should be liberated at anode, but it is Cl2 gas which is liberated in the electrolysis of aqueous NaCl.

(ii) The conductivity of CH3COOH decreases on dilution.

5

OR

(a) For the reaction

2AgCl(s) + H2(g) (1 atm) → 2Ag(s) + 2H+ + 2H+ (0.1M) + 2Cl- (0.1M),

ΔG° = -43600J at 35° C.

Calculate the e.m.f. of the cell.

[log 10-n = -n]

(b) Define fuel cell and write its two advantages.

26. (a) Write the reactions involved in the following :

(i) Hofmann bromamide degradation reaction

(ii) Diazotisation

(iii) Gabriel phthalamide synthesis

(b) Give reasons :

(i) (CH3)2NH is more basic than (CH3)3N in an aqueous solution.

(ii) Aromatic diazonium salts are more stable than aliphatic diazonium salts.

5

12

OR

(a)Write the structures of the main products of the following reactions:

(b) Give a simple chemical test to distinguish between Aniline and N,N-dimethylaniline.

(c) Arrange the following in the increasing order of their pKb value:

C6H5NH2, C2H5NH2, C6H5NHCH3

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1. The non-stoichiometry occurs because of the instability in the FeO molecule as undergoing the oxidation of Fe2+ to Fe3+ effectively replacing a small portion of Fe2+ with two thirds their number of Fe3+.

Thus, for every three "missing" Fe2+ ions, the crystal contains two Fe3+ ions to balance the charge. This change in composition violets the law of constant composition as the composition of the compound usually varies in a continuous manner over a narrow range. Thus FeO becomes a non-stoichiometric compound, and the formula is written as Fe(1-x)O, where x is a small number representing the deviation from the "ideal" formula.

Generally, it is written as Fe0.95O or found in the form of FeO.Fe2O3 represented as Fe3O4.

2. The characteristic is shown by the catalyst is selectivity, i.e.; the catalyst determines the products formed.

1. If CO and H2 react in the presence of Ni (nickel) as catalyst, the product is methane.

2. If CO and H2 react in the presence of Cu (copper) as catalyst the product formed is methanol.

Section A (Solutions)

Solutions (Set-1)

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3. Given Complex Compound

[Pt (en)2Cl2]

Coordination no. = denticity × number of ligand

Coordination number = 2 × 2 + 2 × 1 = 6.

The net charge of the complex is zero while there are two chloride ions, each with a negative charge. Therefore, the Pt metal center will have a (+2) oxidation state, which may also be written as Pt(II).

4. Benzyl chloride will get easily hydrolysed as the halogen is attached to 1° carbon atom.

Benzyl chloride can undergo SN1- SN1 mechanism easily to form benzyl carbonium ion. Benzyl carbonium ion found after losing the leaving group(-Cl) is very stable because of resonance hence reacts easily. It is the most stable carbocation. So the final answer is“ Benzyl chloride is more easily hydrolysed by aqueous NaOH than that of chlorobenzene. ”

5. The IUPAC name of the above compound is 3,3-dimethyl1-pentan-2-ol.

First, we choose the longest chain in that compound and naming starts from right side because OH substitute will be numbered less.

So, at 2 position, there is OH substitute and at 3rd position 2 methyl groups are substituted.

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6. We know

ΔT f= k f × m

Where ΔTf is the depression in the freezing point

Kf freezing constant for water

m is molality

The molality(m) of a given solution of glucose:

Section B (Solutions)

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m= 𝑛𝑜.𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒

𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 𝑖𝑛 𝑘𝑔

Given,

Mass of solute= 60 g

Mass of solvent = 250 g

Molecular mass of solute = 180

no. of moles of solute = 60𝑔

180𝑔 mole

mass of solvent in kg = 250𝑘𝑔

1000

so, m= 60

180 𝑚𝑜𝑙𝑒

250𝑘𝑔

1000

m= 1.33 mol kg-1

Putting the values in the equation, we get

ΔTf= 1.86 × 1.33K

ΔT f= 2.4738K

ΔT f = T f × 2.4738K

Δt(solution) = t(solvent) – ΔTf

ΔT f= Tf0 2.4738K

= 273.15 - 2.4738K

= 270.65K

7. 2N2O5(g) → 4NO2(g) + O2(g)

The rate of the reaction is given by

Rate of disappearance of N2O5 = −Δ[𝑁2𝑂5]

Δt

And, −Δ[𝑁2𝑂5]

2Δt =

Δ[NO2]

4Δt =

Δ[O2]

Δt

or

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−Δ[𝑁2𝑂5]

Δt =

1

2 Δ[NO2]

Δt

Given,

Δ[NO2]

Δt= 2.8×10-3 Ms-1

The rate of disappearance of N2O5= 2.8×10−3

2 M/sec= 1.4× 10-3

M/sec

So, the rate of disappearance of N2O5 would be half of rate of production of NO2. So the rate of disappearance of N2O5 is 1.4 ×10-3M/sec

8. The Hydrides of the group 15 elements are NH3, PH3, AsH3, SbH3, BiH3.

(a) The lowest boiling point is of PH3

(b) Maximum basic character is shown by NH3

(c) Highest bond angle is for NH3

(d) BiH3 has the maximum reducing character.

9. (a)

In the first step, Propanal reacts with Grignard reagent and water to form its alcohol and then through oxidation process in the presence of CrO3, Alcohal converts into ketone group.

(b)

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In the first step, when Toluene reacts with KMnO4-KOH in the presence of heat, toluene goes into oxidation and further after hydrolysis, It converts into Benzoic Acid.

(a) As –COOH group is present as an electron withdrawing group, and thus its electron withdrawing nature makes the aromatic ring electron deficient and secondly Friedel-Craft reaction is electrophilic substitution reaction and electrophilic substitution reaction does not prefer to occur in electron deficient ring.

(b) pKa value of 4-nitrobenzoic acid is lower than that of benzoic acid because 4- nitro benzoic acid is more acidic than the benzoic acid. The -NO2 being an electron withdrawing group withdraw electron towards itself resulting in ease of carboxylic proton release, hence increasing the stability and ka of benzoic acid is lower than that of 4-nitrobenzoic acid and thus, pka value of 4-nitrobenzoic acid is lower than that of benzoic acid.

10. (a) 5Fe + 2+ MnO4- + 8H + → 5Fe +3 + Mn +2 + 4H 2O

It is the type of redox reaction of Fe2+ with MnO4- in the acidic

medium,

Fe2+ → Fe3+ + e-(oxidation reaction)-------(i)

Mn+7 + 5e- → Mn2+ (Reduction reaction) ---------(ii)

MnO4- + 5e- + 8H+ → Mn2+ + 4H2O------(iii)

Multiplying euation (i) with 5 we get,

5Fe2+ → 5Fe3+ + 5e- --------(iv)

Adding equation (iii) & (iv), we get

OR

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5Fe + 2+ MnO4- + 8H + → 5Fe +3 + Mn +2 + 4H 2O

(b) 2 MnO4- + H 2O + I - → 2 MnO2+ 2OH - +IO3

-

It is the type of redox reaction of MnO4- with I- in the basic

medium,

Mn+7 + 3e-→ Mn(+4) (Reduction Reaction)--------(i)

I-→ I+5 + 6e- (Oxidation Reaction)-------(ii)

I-→ IO3- + 6e- + H2O-------(iii)

MnO4- +3e- + H2O → MnO2 + OH-

Multiplying above equation by 2, we get

2MnO4- + 6e- + 2H2O → 2MnO2 + 2OH- -------(iv)

Adding equation (iii) and (iv), we get

2 MnO4- + H 2O + I - → 2 MnO2+ 2OH - +IO3

-.

11. (a) The Molar masses of macromolecules like polymers and protein are measured through osmotic pressure method.

The osmotic pressure method uses molarity of solution which has large magnitude even for dilute solutions, given that polymers have poor solubility, osmotic pressure measurement is used to determine their molar mass. Macromolecules such as protein are not stable at high temperatures, and measurement of osmotic pressure are done at room temperatures, it is useful for determination of molar masses of proteins.

(b) p =Kh * χ

Where Kh = Henry's law constant.

Solubility is indirectly proportional to Henry's constant.

As temperature increases, Kh increases; hence, solubility decreases.

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Hence the availability of dissolved gas oxygen in water is at a lower temperature. Thus the aquatic animal feels more comfortable at lower temperatures than at higher temperatures.

(c) Since the elevation of boiling point is

TB = i× Kb× m

In both 1M KCl & 1M sugar. The solvent is the same, but KCl is ionic due to which it dissociates completely to K+ and Cl - ions. Hence the value can't Hoff factor is twice in 1M KCl then 1M sugar due to which elevation in boiling point is nearly double for 1M

As colligative property depends only on a number of particle and KCl as electrolyte will produce a double number of a particle of sugar.

12. Given:

Atomic mass of element(m)=40 gmol-1 length of unit cell(a)=400 pm=4× 10-8 cm; Z=4(Fcc).

Now density is given by the formula

Where Z is the number of an atom in the unit cell

M is the molecular mass

NA is the Avogadro number

V is the volume of the unit cell.

V=a3=(4× 10-8)3 cm =64× 10-24 cm3

Putting the values,

Density, D = 4×40

6.023×1023×64×10−24

Density, D=160/38.5=4.1 g cm-3.

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13. Rate constant for first order reaction is given by:

Where k is the rate constant, t is the time, Ro is the initial concentration and R is the final condition.

At 300K,K300 = 2.303

40log (

100

50)

=0.058 × log 2

=0.058 × 0.301

=0.017

At 320K,

K320 = 2.303

20log (

100

50)

=0.11× log 2

=0.11 × 0.3010=0.034

Now,

log = ( )

Where Ea is the activation energy

Section C (Solutions)

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K1 = 0.017, T1 is the temperature of 300 K

K2 = 0.034, T2 is the temperature of 320 K

Putting the value in the above equation,

K

Ea=28,805.7 j mol-1

14. (a) When freshly prepared precipitate of Fe(OH)3 is shaken with a small amount of FeCl3 solution, a positively charged sol of Fe(OH)3 is formed due to adsorption of Fe3+ ions.

(b) On persistent dialysis, a colloidal solution traces of electrolyte is completely removed leaving colloids unstable, and finally, coagulation takes place

(c) On centrifugation of emulsion, separation of the two constituent liquids will take place.

15. 4 Au(s) + 8 CN-(aq)+ 2H2O (aq)+ O2(g) → 4[Au(CN)2]-(aq) + 4OH-

2[Au(CN)2]- (aq)+ Zn (s) → [Zn(CN)4]2- (aq)+ 2Au (s)

Gold is leached with a dilute solution of NaCN in the presence of air (For O2) from which the metal is obtained later by replacement. in the extraction of gold and silver, the metal is leached with NaCN or KCN, which is an oxidation reaction, during this process Ag is formed by the replacement of Zinc.

16. (a) Mn2+ has a d5 configuration, and the extra stability of half-filled d-orbitals is comprised when another electron is taken out to give Mn3+, On the other contrary, Fe3+ attains a half-filled orbitals configuration when Fe3+ gets oxidized to Fe3+ .

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Hence, the Eo value for Mn3+/Mn2+ couple has more positive Eo value.

(b) Fe has 3d64s2 outer electronic configuration whereas Cu has 3d104s1 configuration. Now, more the number of unpaired electrons in d-orbital, more favourable are interatomic attractions and thus higher atomization enthalpies. Hence, Fe has4- unpaired d-electrons has more enthalpy of atomization than copper having no unpaired d electron.

(c) Sc3+ has 3d10 configuration whereas Ti3+ has a 3d1 configuration. As there are no electrons in d-orbitals for Sc3+ ion, there is no transition of electrons y absorption of energy and hence no emission in visible range imparting colour to the Sc3+

17. (a) The molecule (i) is chiral molecule.

Chiral molecules are those which have different groups at all 4 centres of carbon and also don’t show symmetry.

Below diagram is also above type of molecule.

Thus, Below compound is chiral molecule.

(b) Chlorobenzene reacts with methyl chloride in the presence of sodium metal and dry ether to give toluene. This reaction is known as Wurtz-Fitting reaction.

(c) In the 1- bromo -1-methylcyclohexane, all β-hydrogen atoms are equivalent . Thus dehydrohalogenation takes place, in the reaction of this compound with KOH

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18. (a) Compound A and C give tollen’s test which indicates that they are aldehydes. Compound C gives iodoform test which means it contains a carbonyl carbon so, with formula C4H80 the structure of the compound would be CH3COCH2CH3 .Now upon reduction with Zn/Hg conc. HCl, the corresponding alkanes are obtained, so reduction of B gives Butane (D, so the isomer A have to be linear chain aldehyde, giving Butane on reduction. So the last isomer possible is compound C, 2-Methyl propanal. As shown below

A is CH3 CH2 CH2CHO

B is

C is

(b) will be least reactive as the carbonyl carbon is sterically hindered and most reactive would be compound A towards the addition of HCN.

19. (i)

In the above reaction, NaBH4 reacts with ketone and ester group and due to the reduction, ketone converts into hydro-carbon and ester converts into ketone and ether. NaBH4 acts as a reducing agent here.

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In the above reaction, Styrene reacts with water in acidic medium then due to hydrolysis, it converts into Phenol derivatives of Benzene. Oxidation of styrene occurs here due to hydrolysis.

In the above reaction, Phenyl ethyl ether reacts with Hydrogen-Iodide through the SN1 mechanism.

Phenol and Ethyl Iodide are obtained as product which are present in major in the solution.

20. (a) Bithional added to soaps to impart antiseptic properties and to reduce the odour produced by bacterial decomposition of organic matter on the skin

(b) Tincture of iodine, iodine tincture, or weak iodine solution is an antiseptic. It is usually 2–7% elemental iodine, along with potassium iodide or sodium iodide, dissolved in a mixture of ethanol and water. Tincture solutions are characterized by the presence of alcohol.

(c) Sodium benzoate - is a food preservative.

A food preservative is a substance or a chemical that is added to products, pharmaceutical drugs, paints, biological samples, cosmetics, wood and many other products to prevent decomposition by microbial growth or by undesirable chemical changes.

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21. (a) Polysaccharides : These are the saccharides containing large number monosaccharides units joined together by glycosidic linkage they yield a large number of monosaccharides upon hydrolysis . eg:- starch, cellulose

(b) Denatured protein : Denatured protein are the protein when subjected to physical change like a change in temperature or chemical change like pH, the hydrogen bonds are disturbed and loses its biological activity. Since denaturation reactions are not strong enough to break the peptide bonds, the primary structure (sequence of amino acids) remains the same after a denaturation process, eg:- The coagulation of egg white on boiling.

(c) Essential amino acids : The amino acids which cannot be synthesized inside the body and must be obtained from diet taken by us are known as essential amino acids.eg:- Histidine, Valine etc.

(a)

OR

27

When D-Glucose reacts with nitric acid then D-Glucose through oxidation reaction converts into Saccharic acid as above. In this terminal aldehyde(-CHO) and alcohol(-OH) converts into Acetic acid(-COOH).

(b) As amino acid contains both acidic and basic group present into them. i.e. COOH and NH2 in the same molecule can lose a proton and amino group can accept a proton, giving rise to dipolar ion.

Hence, shows amphoteric nature.

22. (a) The molecular formula for coordination compound Iron(III) hexacyanoferrate(II)

Fe4[ Fe (CN )6]3

(b) [Co(NH3)5 Cl]SO4 will show ionisation isomerism, and the possible isomers are

[Co(NH3)5Cl]SO4 and [Co(NH3)5 SO4] Cl

(c) Electronic configuration of Co3+ ions is sp3d3 hybridized orbitals of Co3+, with six pairs of electrons from six F- ions.

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Hybridisation = sp 3d 3

Number of unpaired electron = 4

There are 4 unpaired electron in [CoF6]3+

23. (a) Shyam was aware of the environment al a responsible citizen and had moral values.

(i) Shyam was aware of the hazards of polythene use by us and was willing to share the information with shopkeeper for a good cause.

(ii) Shyam told the shopkeeper about the replacement of the polythene bags by paper bags.

(b) Low-Density Polythene is branched chain structure, whereas the high-density polyethene has a linear chain structure.

(c) Shyam refused to take the items in polythene bags as polythene is nor biodegradable neither recyclable, so the one used polythene packet go to environment and get accumulated, creating hazards to the environment.

(d) Biodegradable polymers contain functional groups similar to functional groups present in biopolymers, so they get degraded in the environment by decomposers and thus are environment-friendly.

For example ploy β-hydroxybutyrate-co-β-hydroxy valerate(PHBV).

24. (a) (i)H3PO3:Oxidation number of O=+ 3

H3PO4 : Oxidation number of P = +5

The H3PO3 contains one P-H bond in addition to P=O and P-OH bonds. These types of oxoacids tend to undergo

29

disproportionation to give orthophosphoric acid. Whereas in H3PO4 is at +5 oxidation state hence no disproportionation takes places.

(ii) When we react Cl2 with an excess of F2, ClF3 is formed and not FCl3 because fluorine can exist only in – 1 oxidation state whereas chlorine can show positive oxidation states when combined with the more electronegative atom.

In ClF3 oxidation number of Cl is + 3 which is possible due to the availability of d-orbitals.

(iii) In oxygen molecule, there is p-p overlap between two oxygen atoms forming a double bond. The intermolecular forces in oxygen are weak Vander Waal’s forces, and therefore oxygen exists as a gas, on the other hand, the S are linked by single bonds and form polyatomic complex molecules. Hence sulphur in a solid.

(b) (i) The structure of XeF4 is square planar in structure.

(ii) HCLO3 or chloric acid is

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(a) (i) The brown gas A is NO2 or nitrogen dioxide. On cooling, it dimerises to N2O4 and solidifies as a colourless solid.

(ii) The structures of A and B.

A=

B=

(iii) Compound A that is NO2 contains an odd number of molecules. On dimerization, it is converted to stable N2O4 molecule with even numbers of electrons and have better intermolecular forces to get solidified. Thus it changes to solid on cooling.

(b) Decreasing order of reducing character-

HI>HBr>HCl>HF

(c) Xef4 + SbF5→[XeF3]+ +[SbF6]-

OR

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25. (a) The Half cell reaction can be written as:

= -0.11× (log 0.004) V=0.26 V

Ecell= 0.0+ 0.0592 × (log H+)

E cell= 0.0591 × (-1.7)

Ecell= - 0.10 V

(b) (i) The standard reduction potential of water is slightly less, and therefore it has slightly more chance of getting oxidised. However, in concentrated solution of NaCl, oxidation of chloride ions is preferredthan water at the anode. The unexpected result is due higher voltage required for electrolysis due to over voltage.

The reaction at the anode:

(ii) The number of ions furnished by an electrolyte depends upon the degree of dissociation with an increase in dilution the degree of dissociation increases. As a result molar conduction of acetic and increases with dilution.

(a) ΔG 0 = -43600 J at 25oC.

Calculate the e.m.f. of the cell.

Section D (Solutions)

OR

32

[log 10–n = – n]

According to the Nernst equation:

Ecell =-0.45-(0.059) log(0.1× 0.1)

=-0.51× log(10-2)=-0.51× (-2)=1.02 V

Ecell=1.02V

(b) These are voltaic cells in which the reactants are continuously supplied to the electrodes. There are designed to convert the energy from the combustion of fuel directly into electrical energy.

Advantage : -

(1) A fuel cell works with an efficiency of 70 %

(2) These are no objectionable by-products and therefore, its a pollution free source of energy.

26. When an amide is treated with bromine in an aqueous solution of sodium hydroxide, degradation of amide take place leading to the formation of primary amines.

R – CH2 – C(0) – NH2 + 4 NaOH + Br2 →R – CH2 – NH2 + Na2 CO3 + 2NaBr + 2H2O

(ii) Diazotisation

Aromatic primary amines react with nitrous acid at low temperatures to form diazonium salts. This conversion of

33

aromatic primary amines into diazonium salts is nown as diazotization.

(iii) Gabriel phthalimide synthesis

Gabriel phthalimide synthesis is a very useful method for the preparation of aliphatic primary amines. It involves the reaction of phthalimide with ethanolic potassium hydroxide to form potassium salt of phthalimide. The product is further heated with alkyl halide, followed by alkaline hydrolysis to yield the corresponding amine.

(b) (i) (CH3)2NH is more basic than (CH3)3N in aqueos solutions because in(CH3)3N the lone pair of electrons on nitrogen atom is responsible forits basicity are quite hindered by the three methyl groups, hence are less available. Due to which it is less basic as compared to(CH3)2NH.

(ii)Aromatic diazonium salts are stable than aliphatic diazonium salts because the positive charge on nitrogen atom is stabilized by the resonance with attached phenyl group.

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(a)

(b) Aniline can be distinguished from N,N-dimethyl aniline by diazo couping reaction. Anline would react with benzene diazonium chloride to give yellow dye, where as N,N-dimethyl aniline won’t undergo this reaction.

OR

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(c) C2H5NH2 < C6H5NH2 < C6H5NHCH3

In this, C2H5NH2 has highest kb because its electron are not busy in resonance, whereas C6H5NH2 and C6H5NHCH3 has comparably low kb because the molecules of these compounds are busy in resonance.

Also, kb is reverse of pkb.

Thus, above orders are correct.

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