Category: Class 12

Chemistry

Section-A

Q.1. The analysis shows that FeO has a non-stoichiometric composition with formula Fe_0.95O. Give reason. 

Answer: Shows metal defeciency defect / It is a maxture of Fe²⁺ and Fe³⁺ / Some Fe²⁺ ions are replaced by Fe³⁺ / Some of the ferrousion get oxidised to ferric ions.

Q.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?

Answer: CO(g) and H₂(g) react in the presence of different catalysts to give different products, this shows that the action of a catalyst is highly selective in nature.

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

Answer: Coordination number: 6;

Oxidation state: +2

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

Answer: Benzyl chloride would be easily hydrolysed compared to chlorobenzene. In the given reaction conditions, hydrolysis proceeds by nucleophilic substitution mechanism and the benzyl carbonium ion formed after losing the leaving group (-Cl) is better stabilized (through resonating structures) hence reacts easily.

Q.5. Write the IUPAC name of the following: 

Answer: The IUPAC name would be 3, 3- Methyl-pentan-2-ol.

Section-B

Q.6. Calculate the freezing point of a solution containing 60 g of glucose (Molar mass = 180 g mol^-1) in 250 g of water. (K_f of water = 1.86 K kg mol^-1) 

Answer: Molality (m) of a given solution of Glucose:

m = [(60/180) g mol^-1/250 g] × 1000 = 1.33 mol kg^-1

Now, depression in freezing point is given by, ΔT_f = K_fm

Putting the given values,

 ΔT_f = K_fm = 1.86 × 1.33 = 2.5

So, the freezing point of the solution would be = 273.15 K – 2.5 K = 270.65 K.

Q.7. For the reaction 

2N₂O₅>(g) → 4NO₂(g) + O₂(g) the rate of formation of NO₂(g) is 2.8 × 10^-3 Ms^-1. 

Calculate the rate of disappearance of N₂O₅(g).

Answer: Rate of reaction for the given reaction can be given as,

Rate = 1/2 {-Δ[N₂O₅]/Δt} or {-Δ[N₂O₅]/Δt} = 1/2 {[NO₂/ Δt]}

So, the rate of disappearance of N₂O₅ would be half of the rate of production of NO₂ (given 2.8 × 10^-3Ms^-1).

So, the rate of disappearance of N₂O₅ is 1.4 × 10^-3 Ms^-1.

Q.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?

Answer : 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.

Q.9. How do you convert the following?

(a) Ethanal to Propanone

(b) Toluene to Benzoic acid

OR

Q.9. Account for the following:

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

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

Answer: (a) Conversion of ethanol to Propanone:

OR

Answer: (a) Aromatic carboxylic acids do not undergo Friedel-Crafts reaction because the carboxyl group is deactivating for electrophilic substitution reaction, secondarily, the catalyst aluminium chloride gets bonded to the carboxyl group.

(b) pK_a value of 4-Nitrobenzoic acid is lower than benzoic acid, which means 4-Nitrobenzoic acid is more acidic than benzoic acid. Being an electron-withdrawing group, the -NO₂ group withdraws electrons towards itself resulting in ease of carboxylic proton release, hence increasing the acidity.

Q.10. Complete and balance the following chemical equations: 

(a) Fe²⁺ + MnO^–₄ + H⁺ →

(b) MnO^–₄ + H₂O + I^– →

Answer: (a) 5Fe²⁺ + MnO^–₄ + 8H⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺

(b) 2MnO^–₄ + H₂O + I^– → 2MnO₂ + 2OH^– + IO^–₃.

Section-C

Q.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 boiling point of 1M KCl solution is nearly double than that of 1M sugar solution.

Answer: (a) Molar masses of macromolecules like polymers and proteins are measured through osmotic pressure method. The osmotic pressure method uses ‘molarity’ of solution (instead of molality) which has a large magnitude even for dilute solutions, given that polymers have poor solubility, osmotic pressure measurement is used for determination of their molar masses. Macromolecules such as proteins are not stable at high temperatures and because measurement of osmotic pressure is done at around room temperature, it is useful for determination of molar masses of proteins.

(b) The solubility of gases in liquids decreases on increasing the temperature. Hence, the availability of dissolved oxygen in water is more at lower temperatures hence, the aquatic animals feel more comfortable at lower temperatures than at higher temperatures.

(c) Elevation of boiling point is a colligative property and hence depends on the number of solute particles in the solution. Now, 1 M KCl would have twice the number of solute particles, as KCl dissociates into K⁺ and C1^– , compared to sugar solution (as sugar does not undergo any dissociation). So, the elevation of boiling point is nearly double for 1M KCl solution compared to 1M sugar solution.

Q.12. An element ‘X’ (At. mass = 40 g mol^-1) having f.c.c. the structure has a 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).

Answer : 

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

N_A 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=160/38.5=4.1 g cm^-3

Q.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) 

Answer: Rate constant for a first-order reaction is given by,

Q.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?

Answer: (a) When FeCl³ is added to a freshly prepared precipitate of Fe(OH)₃, a positively charged sol of hydrated ferric oxide is formed due to adsorption of Fe³⁺ ions.

(b) When persistent dialysis of the colloidal solution is carried out, traces of electrolytes present in the sol are removed almost completely leaving the colloids unstable and finally, coagulation takes place.

(c) Emulsions are centrifuged to separate them into constituent liquids.

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

Answer: Extraction of gold involves leaching the metal with a dilute solution of NaCN or KCN in the presence of air (for O₂) from which the metal is obtained later by replacement method (using Zinc).

The reactions involved are:

4Au(s) + 8CN^–(aq) + 2H₂O(aq) + O₂(g) → 4[AU(CN)₂]^– (aq) + 4OH^–(aq)

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

Q.16. Give reasons:

(a) E⁰ value for Mn³⁺/Mn²⁺ couple is much more positive than that for Fe³⁺/Fe²⁺.

(b) Iron has a higher enthalpy of atomization than that of copper.<br> (c) Sc³⁺ is colourless in aqueous solution whereas Ti³⁺ is coloured. 

Answer: (a) Mn²⁺ has a d⁵ configuration, and the extra stability of half-filled d-orbitals is compromised when another electron is taken out to give Mn³⁺, On the contrary, Fe³⁺ attains a half-filled orbital configuration when Fe²⁺ is oxidized to Fe³⁺. Hence, the E⁰ value for Mn³⁺/ Mn²⁺ couple has more positive E⁰ value.

(b) Fe has a 3d⁶4s² outer electronic configuration whereas Cu has 3d¹⁰4s¹ configuration. Now, the more the number of impaired electrons in d-orbital, more favourable are interatomic attractions and thus higher atomization enthalpies. Hence, Fe having 4 unpaired d-electrons has more enthalpy of atomization than copper having no unpaired d-electron.

(c) Sc³⁺ has a 3d⁰ configuration whereas Ti³⁺ has a 3d¹ configuration. As there are no electrons in the d orbital for Sc³⁺ ion, there is no transition of electrons by absorption of energy and hence no emission in the visible range imparting colour to the Sc³⁺ ion.

Q.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.

Answer: (a) The molecule (i) is a 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-Fittig 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.

Q.18. (A), (B) and (C) are three non-cyclic functional isomers of a carbonyl compound with molecular formula C₄H₈O. Isomers (A) and (C) give positive Tollens 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 the addition of HCN?

Answer: (a) Compound A and C give positive Tollens test which indicates that they are aldehydes. Compound C gives Iodoform test which means it contains a carbonyl group with a methyl group attached to the carbonyl carbon so, with formula C₄H₈O the structure of compound would be CH₃COCH₂CH₃ (Butanone).

Now upon reduction with Zn(Hg)/conc. HCl, the corresponding alkanes are obtained, so a reduction of B gives Butane (D), so the isomer A has to be a linear chain aldehyde (Butanal), giving Butane (compound D) on reduction. So, the last isomer possible is compound C, 2-Methyl propionaldehyde. The reactions involved are shown below with the structures of compounds:

(b) Out of the three isomers A, B and C, compound B’ (Butanone) would be least reactive towards the addition of HCl as the carbonyl carbon is sterically hindered and most reactive would be compound A (Butanal) towards the addition of HCN.

Q.19. Write the structures of the main products in the following reactions: 

Answer: (i) Sodium borohydride doesn’t reduce esters, so the product would be,

Q.20. Answer the given questions:

(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

Answer: 1. Bithional is added to soaps to impart antiseptic properties to soap.

2. Tincture of iodine is 2-3 per cent mixture of iodine in the alcohol-water mixture. It is used as an antiseptic.

3. Sodium benzoate is used as a food preservative.

Q.21. Define the following with an example of each: 

(a) Polysaccharides

(b) Denatured protein

(c) Essential amino acids

OR

Q.21. (a) Write the product when D-glucose reacts with conc. HNO₃.

(b) Amino acids show amphoteric behaviour. Why?

(c) Write one difference between α-helix and β-pleated structures of proteins.

Answer: (a) Polysaccharides: Polysaccharides are food storage materials and most commonly found carbohydrates in nature. These are the compound which is formed of a large number of monosaccharide units joined together by glycosidic linkages. Example. Starch, main storage polysaccharide of plants.

(b) Denatured protein: Proteins have a unique three-dimensional structure in their native form. If the native form of protein is subjected to any physical change (such as temperature change) or any chemical change (such as a change in pH), the hydrogen bonds are disturbed. Due to this, globules unfold and helix get uncoiled due to which protein loses its biological activity. This is called denaturation of the protein. During denaturation 2° and 3° structures of proteins are destroyed but 1° structure remains intact. Coagulation of egg white is an example of denaturation of the protein.

(c) Essential amino acids: amino acids which are not synthesized in our body and have to be obtained through diet are known as essential amino acids. Example: Tryptophan

OR

Answer: (a) D-Glucose gets oxidized to give saccharic acid, a dicarboxylic acid on reacting with nitric acid.

(b) Amino acids show amphoteric behaviour due to the presence of both acidic (carboxylic group) and basic (amino group) in the same molecule. So, in the basic medium, the carboxyl group can lose a proton and in acidic medium, the amino group can accept a proton.

(c) In α-helix structure the polypeptide chain forms all possible hydrogen bonds by twisting into a right-handed screw (helix) with the -NH group of each amino acid residue gets hydrogen-bonded to the -C = O of an adjacent turn of the helix (Intramolecular bonding), whereas in β-structure all peptide chains are stretched out to nearly maximum extension and then laid side by side which are held together by intermolecular hydrogen bonds (intermolecular bonding).

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

(b) What type of isomerism is exhibited by the complex [Co(NH₃)₅ Cl]SO₄?

(c) Write the hybridisation and number of unpaired electrons in the complex [CoF₆]_3-.

(Atomic number of Co = 27) 

Answer: (a) The molecular formula of Iron(III) α-cyanoferrate(II) is Fe₄[Fe(CN)₆]₃ 

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

(c) Electronic configuration of Co³⁺ ion is,

Electronic configuration of sp³d² hybridized (as F is a weak field ligand) orbitals of Co³⁺ , with six pairs of electrons from six F ions.

There are 4 impaired electrons in [CoF₆]₃.

Section-D

Q.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 polythene and high-density polythene.

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

(d) What is a biodegradable polymer? Give an example.

Answer: (b) Low-density polythene has a branched-chain structure, whereas the high-density polythene has a linear chain structure.

(c) Shyam refused to take the items in polythene bags as polythene is non-biodegradable neither recyclable,

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

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

Section-E

Q.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

OR

Q.24. (a) When concentrated sulphuric acid was added to an unknown salt present in a test tube a brown gas (A) was evolved. This gas intensified when copper turnings were added to this test tube. On cooling, the gas (A) changed into 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 decreasing order of their reducing character: HF, HCl, HBr, HI

(c) Complete the following reaction:

XeF₄ + SbF₅ →

Answer: (a) (i) In H₃PO₃ (orthophosphoric acid) oxidation state of phosphorus is +3 and it contains one P-H bond in addition to P = O and P-OH bonds. These types of oxoacids tend to undergo disproportionation to give orthophosphoric acid (P has +5 state) and phosphine (P has +3 state). Whereas in H₃PO₄ (orthophosphoric acid), Phosphorus is in +5 state hence no disproportionation takes place in H₃PO₄.

(ii) When Cl₂ reacts with an excess of F₂, ClF₃ is formed-and not FCl₃ because Fluorine can’t expand its valency and can show only -1 oxidation state, whereas Cl can expand its valency due to the availability of d-orbitals.

(iii) Dioxygen is a gas while sulphur is a solid at room temperature this is because sulphur have S₈ molecules and these are packed to give different crystal structure, whereas dioxygen is a diatomic molecule (O₂) and it does not have enough intermolecular attraction and thus exists in gaseous form.

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

(iii) Compound A, that is, NO₂ contains an odd number of valence electrons. It behaves as a typical odd molecule. On dimerization, it is converted to stable N₂O₄ molecule with an even number of electrons (thus colourless) 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]^–.

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

Sn(s) | Sn²⁺ (0.004 M) || H⁺> (0.020 M) | H²(g) (1 bar) | Pt(s)

(Given: E° Sn²⁺/Sn = – 0.14V)

(b) Give reasons:

(i) On the basis of E° 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.

OR

Q.25. (a) For the reaction

2AgCl(s) + H₂(g) (1 atm) → 2Ag(s) + 2H⁺ (0.1M) + 2Cl^– (0.1M), ΔG° = -43600 J at 25°C.

Calculate the e.m.f. of the cell. [log 10^-n> = -n]

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

Answer: (a) The half cell reactions can be written as;

The reaction at the anode with a lower value of Ecell is preferred and therefore, water should get oxidized to give O₂ but on account of overpotential of oxygen, Cl^– gets oxidized preferably, liberating Cl₂ gas.

(ii) The conductivity of CH₃COOH decreases on dilution because the number of ions per unit volume that carry the current in a solution decreases on dilution.

OR

Answer: 

(b) Galvanic cells that are designed to convert the energy of combustion of fuels like hydrogen, methane, methanol etc. directly into electrical energy are called fuel cells.

Advantages of fuel cells are:

• Fuel cells produce electricity with an efficiency of about 70% compared to thermal plants whose efficiency is about 40%.
• Fuel cells are pollution-free.

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

(i) Hofmann bromamide degradation reaction

(ii) Diazotisation

(iii) Gabriel phthalimide 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.

OR

Q.26. (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 pK_b values:

C₆H₅NH₂, C₂H₅NH₂, C₆H₅NHCH₃

Answer: (a) (i) Hofmann bromamide degradation reaction: Acetamide can be considered for example. In this reaction, Acetamide (CH₃CONH₂) undergoes Hofmann degradation in the presence of Bromine and NaOH to give Methanamine.

CH₃CONH₂ + Br₂ + 4NaOH → CH₃NH₂ + Na₂CO₃ + 2NaBr + 2H₂O.

(ii) Diazotisation: The conversion of primary aromatic amines into diazonium salts is known as diazotization.

(iii) Gabriel phthalimide synthesis: This reaction is used for the preparation of primary amines. Phthalimide on treatment with ethanolic potassium hydroxide forms potassium salt of phthalimide which on heating with alkyl halide followed by alkaline hydrolysis produces the corresponding primary amine.

(b) (i) (CH₃)₂ NH is more basic than (CH₃)₃N in aqueous solutions because in (CH₃)₃N the lone pair of electrons on the nitrogen atom is responsible for its 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 more stable than aliphatic diazonium salts because the positive charge on the nitrogen atom is stabilized by the resonance with an attached phenyl group.

Answer: 

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

Chemistry

Set-I

Section – A

Q.1.Out of KCl and AgCl, which one shows Schottky defect and why? 

OR

Q.1.Why does ZnO appear yellow on heating?

Answer: KCl shows schottky defect because cation & anions are of similar size. 

OR

Answer: ZnO on heating loses oxygen leaving behind their electrons at that position due to electrons it appears yellow in colour.

Q.2. Arrange the following in decreasing order of basic character:

C₆H₅NH₂, (CH₃)₃N, C₂H₅NH₂

Answer: Decreasing order of basic character:

CH₃CH₂NH₂ > (CH₃)₃N > C₂H₅NH₂

Q.3. What type of colloid is formed when a solid is dispersed in a liquid ? Give an example. 

Answer: Sols are formed when a solid is dispersed in a liquid.

Example – Paints.

Q.4. Out of Chlorobenzene and Cyclohexyl chloride, which one is more reactive towards nucleophilic substitution reaction and why?

Answer: Cyclohexyl chloride is more reactive towards nucleophilic substitution reaction because the carbon bearing the chlorine atom is deficient in electron and seeks a nucleophile. In Chlorobenzene the carbon bearing the halogen is a part of an aromatic ring and is electron-rich due to the electron density in the ring.

Q.5. What is the basic structural difference between starch and cellulose? 

OR

Q.5. Write the products obtained after hydrolysis of DNA.

Answer: Starch consists of two components- amylose and amylopectin. Amylose is a long linear chain of α-D -(+)-glucose units joined by C₁, C₄ glycosidic linkage (α-link). Amylopectin is a branched-chain polymer of α-D-glucose units, in which the chain is formed by C₁-C₄ glycosidic linkage and the branching occurs by C₁-C₆ glycosidic linkage. On the other hand, cellulose is a straight-chain polysaccharide of β-D-glucose units joined by C₁-C₄ glycosidic linkage (β-link).

OR

Answer: Hydrolysis of DNA yields a pentose sugar (β-D-2deoxyribose), phosphoric acid and nitrogen-containing heterocyclic compounds called bases (Adenine, Guanine, Cytosine and Thymine).

Section – B

Q.6. 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. 

OR

Q.6. (a) Write two poisonous gases prepared from chlorine gas.<br> (b) Why does the Cu₂₊ solution give a blue colour on reaction with ammonia?

Answer: (a) Cl₂ is passed through slaked lime to give bleaching powder [Ca(OCl)₂]

2Ca(OH)₂ + 2Cl₂ → Ca(OCl)₂ + CaCl₂ + 2H₂O

(b) When SO₂ gas is passed through a Fe(III) aqueous solution, Fe(III) is reduced to Fe(II) ion:

2Fe₃₊ + SO₂ + 2H₂O → 2Fe²⁺ + SO₂^2- + 4H⁺

OR

Answer: (a) Two poisonous gases prepared from chlorine – Phosgene (COCl₂) and tear gas (CCl₃NO₂).

(b) Nitrogen in ammonia has a lone pair of electrons, which makes it a Lewis base. It donates the electron pair and forms linkage with metal ions-

Q.7. Give reasons:

(a) Cooking is faster in a pressure cooker than in cooking pan.

(b) Red Blood Cells (RBC) shrink when placed in saline water but swell in distilled water. 

Answer: (a) Boiling points increase in increasing the pressure in case of liquids. Water used for cooking attains a higher temperature than the usual boiling temperature inside the pressure cooker due to the existing high pressure inside the pressure cooker vessel. This leads to a faster flow of water inside the vegetables or grains etc. resulting in faster cooking of food in a pressure cooker than in the cooking pan.

(b) Red blood cells shrink when placed in saline water because of exosmosis, i.e., water comes out from the cell to surrounding (more concentrated) to equate the concentration. Whereas, when placed in distilled water concentration within the cell becomes more than the surrounding, hence water comes inside and endosmosis takes place to equate the concentrations.

Q.8. Define the order of the reaction. Predict the order of reaction in the given graphs:

where [R]₀ is the initial concentration of reactant and t_1/2is a half-life.

Answer: It is defined as the sum of powers to which the concentration terms are raised in the rate law equation.

(a) In this graph, as t_1/2 is independent of initial reactant concentration, it is a first-order reaction.

(b) In this graph, as tin is directly proportional to the initial concentration of reactant hence, it is a zero-order reaction.

Q.9. When FeCr₂O₄ 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 Na₂O₃ to (D). Identify (A), (B), (C) and (D). 

Answer: 

Q.10. Write IUPAC name of the complex [Co(en)₂(NO₂)Cl]⁺. What type of structural isomerism is shown by this complex? 

OR

Q.10. Using IUPAC norms, write the formula for the following complexes:

(a) Hexaaquachromium (III) chloride

(b) Sodium trioxalatoferrate (III)

Answer: IUPAC name of [Co(en)₂(NO₂)Cl]⁺ is Chlorobis(ethane-1, 2-diamine)nitro cobalt(III).

This compound shows geometrical isomerism.

OR

Answer: (a) Hexaaquachromium(III) chloride- [Cr(H₂O)₆]Cl₃

(b) Sodium trioxalatoferrate(III)- Na₃[Fe(C₂O₄)₃]

Q.11. (a) Although both [NiCl₄]_2- and [Ni(CO)₄] have sp³ hybridisation yet [NiCl₄]⁴ 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) ∆₀ &lt; P and 

(ii) ∆₀&nbsp;&gt; P [2]

Answer: (a) [NiCl₄]^2- is a high spin complex and there are two impaired electrons with 3d8 electronic configuration of a central metal atom, hence it is paramagnetic. Whereas in [Ni(CO)₄] Ni is in zero oxidation state and contains no unpaired electrons, hence it is diamagnetic in nature.

(b) (i) Electronic configuration of d₅ when ∆o &lt; P is given as t_2g³ e_g²

(ii) Electronic configuration of d⁵ when ∆o &gt; P is given as t_2g⁵ e_g⁰.

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

Answer: 

Section – C

Q.13. The following data were obtained for the reaction:

A + 2B → C

(a) Find the order of reaction with respect to A and B.

(b) Write the rate law and overall order of the reaction.

(c) Calculate the rate constant (k).
Answer: The reaction is A + 2B → C

(a) Find the order of reaction with respect to A and B.

(b) Write the rate law and overall order of the reaction.

(c) Calculate the rate constant (k).

Answer: The reaction is A + 2B → C

(a) It can be seen that when the concentration of A is doubled keeping B constant, then the rate increases by a factor of 4 (from 4.2 × 10^2- to 1.68 × 10^2-). This indicates that the rate depends on the square of the concentration of the reactant A. Also when the concentration of reactant B is made four times, keeping the concentration of reactant A constant, the reaction rate also becomes 4 times (2.4 × 10^2- to 6.0 × 10^3-). This indicates that the rate depends on the concentration of reactant B to the first power.

(b) So, the rate equation will be:

Rate = k[A]²[B]

Overall order of reaction will be 2 + 1 = 3.

(c) Rate constant can be. calculated by putting the values given.

4.2 × 10^2- M min^-1= k (0.2)²(0.3) M

k = 0.0420.0123.5 min^-1

Q.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 of Agl colloidal particles formed in the test tube?

How is this sol represented? 

Answer: (a) In dust, the dispersed phase is solid particles and the dispersion medium is air (gas).

(b) Physisorption occurs only because of physical attractive forces such as van der Waals forces between molecules of adsorbate and adsorbent, hence that can be reversed on the application of bigger forces but chemisorption occurs due to the chemical reaction between molecules of adsorbate and adsorbent, and hence can’t be reversed.

(c) When KI solution is added to AgNO³ a positively charged sol results due to absorption of Ag⁺ ions from dispersion medium-AgI/ Ag⁺(positively charged).

Q.15. An element X with an atomic mass of 81 u has density 10.2 g cm^3-. If the volume of the unit cell is 27 × 10^-23 cm³, identify the type of cubic unit cell. (Given: NA = 6.022 × 10 mol^-1.

Answer: 

Q.16. A solution containing 19 g per 1oo mL of K (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 KCl solution. Assume that both the solutions have the same temperature. 

Answer: Two solutions having the same osmotic pressure at a given temperature are called isotonic solution. Now in the given problem, the KCl and urea solutions are given to be isotonic.

Osmotic pressure π is given by the equation

π = (n²/V)RT,       where n₂ = moles of solute,

                                              V = volume of solution in litres.

                                    Also, n₂ = w₂/M₂,

                                 where W₂ = grams of solute  and M₂ = molar mass of solute.

The other given information is

The molar mass of KCl = 74.5 g mol^-1

Weight of KCl, W₂= 1.9 g,  V = 100 mL

So, for KCl

                 n = (w₂/M₂ × V)RT

                 nRTKCl = 1.9/(74.5 × 100) = 2.55 × 10^-4

Now as the solutions are isotonic at the same temperature:

                        πRTKCl = πRTurea

Hence, substituting the values for urea:

          2.55 × 10^-4 = 3/M₂ × 100

         M₂ = 117.6

So, the experimentally determined molecular weight of urea is found to be as 117.6, so the degree of dissociation can be given as:

Osmotic pressure (TT) = Experimentally determined.

So, Urea dimerized in the given experimental solution.

Q.17. Write the name and principle of the method used for refining of (a) Zinc, (b) Germanium, (c) Titanium.

Answer: (a) Distillation is used for refining zinc. As zinc is a low boiling metal, the impure metal is evaporated and the pure metal is obtained as a distillate.

(b) Zone refining is used for refining Germanium. This method is based on the principle that the impurities are more soluble in the melt than in the solid-state of the metal.

(c) Titanium is refined by van Arkel method. This method is used for the removal of oxygen and nitrogen present as an impurity. The crude metal is heated in an evacuated vessel with iodine to obtain metal iodide, which volatilizes being covalent. Later this metal iodide is decomposed through electrical heating to obtain the pure metal.

Q.18. Give reasons for the following:

(a) Transition metals form complex compounds.

(b) E⁰ values of (Zn²⁺/Zn) and (Mn²⁺/Mn) are more negative than expected.

(c) Actinoids show a wide range of oxidation states.

Answer: (a) Transition elements have partly filled d-orbitals due to which they have variable oxidation states which enables them to bind with a variety of ligands and hence form complex compounds.
(b) Oxidation of Zn to Zn²⁺ leads to a completely filled d10 configuration in Zn²⁺, making it more stable. Also, Mn/Mn²⁺ conversion leads to a half-filled stable d⁵ configuration of Mn²⁺ ion. Hence, E⁰ value for Zn/Zn²⁺ and Mn/ Mn²⁺ conversion has negative values.

(c) Actinoids show a wide range of oxidation states due to their partially filled f-orbitals and they have comparable energies as well.

Q.19. Write the structures of monomers used for getting the following polymers:

(a) Nylon-6

(b) Terylene

(c) Buna-N 

OR

Q.19. (a) Is [CH²-CH(C⁶H⁵)]ⁿ homopolymer or copolymer? Give reason.

(b) Write the monomers of the following polymers:

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

Answer: Structures of monomers

(a) Caprolactam is monomer of Nylon-6

Q.20. (a) Pick out the odd one from the following on the basis of their medicinal properties:

Equanil, Seconal, Bithional, Luminal

(b) What types of detergents are used in dishwashing liquids?

(c) Why is the use of aspartame limited to cold foods? 

OR

Q.20. Define the following terms with a suitable example of each:

(a) Antibiotics

(b) Antiseptics

(c) Anionic detergents

Answer: (a) ‘Bithionol’ is the odd one here, as it is an antiseptic whereas others are tranquilisers.

(b) Liquid dishwashing detergents are non-ionic type.

(c) Aspartame is an artificial sweetener which is unstable at cooking temperature hence its use is limited to cold foods.

OR

Answer: (a) Antibiotics: These are the compounds (produced by microorganisms or synthetically) which either inhibit the growth of bacteria or kill bacteria. Example: Penicillin.

(b) Antiseptics: These are the chemicals used to kill or prevent the growth of microorganisms when applied to the living tissues.

Example: Soframycin.

(c) Anionic detergents: These are sodium salts of sulfonated long-chain alcohols or hydrocarbons. In these, the anionic part of the molecule is involved in the cleansing action.

Example: Sodium lauryl sulphate.

Q.21. Among all the isomers of molecular formula C₄H₈Br, 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 the same product on dehydrohalogenation with alcoholic KOH. 

Answer:

(a) 2- Bromobutane is optically active as C-2 is a chiral carbon here having all the four different groups attached to it.

(b) 1-Bromobutane being primary alkyl halide is highly reactive towards SN₂ reaction.

(c) 2-Bromo-2-methylpropane and 1-Bromo-2-methylpropane would give the same product after dehydrohalogenation.

Q.22. Complete the following reactions:

OR

Q.22. How do you convert the following:

(a) N-phenylethylamine to p-bromoaniline

(b) Benzene diazonium chloride to nitro-benzene

(c) Benzoic acid to aniline

Answer:

Answer: 

Q.23. (a) Give reasons:

(i) Benzoic acid is a stronger acid than acetic acid.

(ii) Methanal is more reactive towards nucleophilic addition reaction than ethanal.

(b) Give a simple chemical test to distinguish between propanal and propanone.

Answer (a) (i) Benzoic acid is a stronger acid than acetic acid because the benzoate anion (the conjugate base of benzoic acid) formed after loss of H⁺ is stabilized by resonance, whereas acetate ion (CH³COO^–) has no such extra stability. Hence, Benzoic acid has more tendency of losing proton compared to acetic acid hence more acidic.

(ii) Methanal is more reactive towards nucleophilic addition reaction than ethanal because in ethanal there is a methyl group attached to the carbonyl carbon (centre for nucleophilic attack) and +1 effect of the methyl group decreases the nucleophilicity of carbonyl carbon by increasing the electron density at carbonyl carbon.

(b) Propanal and propanone can be distinguished using Tollen’s reagent by silver mirror test. Propanal being an aldehyde reacts with Tollen’s reagent to give silver deposition whereas propanone being a ketone does not give the reaction.

Q.24. (a) What is the product of hydrolysis of maltose?

(b) What type of bonding provides stability to the α-helix structure of a protein?

(c) Name the vitamin whose deficiency causes pernicious anaemia. 

OR

Q.24. Define the following terms:

(a) Invert sugar

(b) Native protein

(c) Nucleotide

Answer: (a) On hydrolysis maltose gives two molecules of α-D-glucose.

(b) α-Helix structure of proteins is stabilized by hydrogen bonds between -NH group of each amino acid and -COOH group of amino acid at adjacent turn.

(c) Deficiency of Vitamin B₁₂ causes pernicious anaemia.

OR

Answer: (a) Invert sugar: It is a mixture of glucose and fructose obtained after hydrolysis of sucrose. Sucrose is dextrorotatory, but after hydrolysis gives a mixture of dextrorotatory glucose and levorotatory fructose which outweighs in magnitude and hence the whole mixture becomes levorotatory hence the mixture obtained is called invert sugar.

(b) Native protein: a Protein found in a biological system with a unique three-dimensional structure and biological activity is called a native protein.

(c) Nucleotide: They are building blocks of DNA/RNA. These consist of a pentose sugar moiety attached to a nitrogenous base at V position and a phosphoric acid molecule at 5′ position.

Example: 

Section – D

Q.25. (a) The conductivity of 0.001 mol L^-1 acetic acid is 4.95 × 10^-5 S cm^-1. Calculate the dissociation constant if ∧⁰_m for acetic acid is 390.5 S cm² mol^-1.
(b) Write Nest equation for the reaction at 25°C:

2Al(s) + 3Cu²⁺ (aq) → 2 Al³⁺ (aq) + 3Cu (s) (c)

What are secondary batteries? Give an example. 

OR

Q.25. (a) Represent the cell in which the following reaction takes place:

2Al (s) + 3 Ni²⁺ (0.1M) → 2Al³⁺ (0.01M) + 3 Ni (s)

Calculate its emf if E⁰_cell = 1.41 V.

(b) How does molar conductivity vary with increase in concentration for strong electrolytes and weak electrolytes? How can you obtain limiting molar conductivity (∧m0) for weak electrolyte?

Answer: (a) Conductivity ∧_m of a solution is given by the following equation:

∧_m = K/C

where k is the dissociation constant and c is the concentration of the solution.<br> Here, given.

Conductivity, k = 4.95 × 10^-5 S cm^-1<br> Limiting molar conductivity,<br> ∧⁰_m = 390.5 S cm² mol^-1

Concentration,

c = 0.001 mol L^-1 = 1 × 10^-3mol L^-1

Substituting the given values in above equation

Molar conductivity,

(b) Nernst equation for the given reaction can be written as

(c) A secondary battery can be recharged after use, bypassing current through it in the opposite direction so that it can be used again.

Example: The most important secondary cell is lead storage cell. It consists of a lead anode and a grid of lead packed with lead dioxide as a cathode. A 38% solution of sulphuric acid is used as an electrolyte.

OR

Answer: (a) The cell can be represented as

(b) For strong electrolytes, the molar conductivity is increased only slightly on dilution. A strong electrolyte is completely dissociated in solution and thus, furnishes all ions for conductance. However, at higher concentrations, the dissociated ions are close to each other and thus, the interionic attractions are greater. These forces retard the motion of the ions and thus, conductivity is low. With a decrease in concentration (dilution), the ions move away from each other thereby feeling less attractive forces from the counterions. This results in an increase in molar conductivity with dilution. The molar conductivity approaches a maximum limiting value at infinite dilution designated as ∧m0.

In the case of weak electrolytes as the solution of a weak electrolyte is diluted, its ionization is increased. This results in more number of ions in solution and thus, there is an increase in molar conductivity, also there is a large increase in the value of molar conductivity with dilution, especially near-infinite dilution. However, the conductance of a weak electrolyte never approaches a limiting value. Or in other words, it is not possible to find conductance at infinite dilution (zero concentration).

So, limiting molar conductivity for weak electrolytes are obtained by using Kohlrausch law, from the limiting molar conductivities of individual ions (λ0).

Kohlrausch law of independent migration of ions states that limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and the cation of the electrolyte.

∧_m⁰ = λ⁰⁺ + λ⁰

Q.26. (a) Give the equation of the following reactions:

(i) Phenol is treated with cone. 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-l-ol and butan-2-ol?

(c) Arrange the following in increasing order of acidity: Phenol, ethanol, water.

OR

Q.26. (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 the denitration of 3-methyl phenol.

(c) Write the reaction involved in Kolbe’s reaction.

Answer: (a) (i) Phenol is treated with conc. HNO3 to obtain 2,4,6-trinitrophenol picric acid.

(ii) Propene undergoes hydroboration-oxidation when treated with B₂H₆ followed by hydrogen peroxide in basic medium to give propan-1-ol.

(iii) Methyl tert-butyl ether is produced when sodium tert-butoxide is treated with methyl chloride.

(b) Butan-l-ol and Butan-2-ol can be distinguished using Lucas reagent (ZnCl²⁺HCl), where butan-2-ol would react with Lucas reagent in around 5 minutes to give a white precipitate of 2-chlorobutane, whereas butan-l-ol won’t give any reaction at room temperature.

(c) Increasing order of acidity can be given as Ethanol < water < phenol

OR

Answer: (a) (i) Phenol from cumene

(ii) Phenol from benzene sulphonic acid

(iii) Phenol from benzene diazonium chloride

(b) The combined influence of -OH and -CH₃ groups determine the position of the entering groups, also the sterically hindered positions are not substituted.

(c) In Kolbe’s reaction phenol is reacted with CO₂ in the presence of sodium hydroxide, followed by acidification, to give a carboxylic acid group on 2-position of phenol-

Q.27. (a) Account for the following:

(i) The 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 ClF₃ is more reactive than Cl₂.

(b) Draw the structure of (i) XeF₂ (ii) H₄P₂O₇. 

OR

Q.27. (a) Give one example to show the anomalous reaction of fluorine.

(b) What is the structural difference between white phosphorous and red phosphorous?

(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: HF, HCl, HBr and HI

Answer: (a) (ii) Acidic character increases from H₂O to H₂Te due to decrease in E—H bond dissociation enthalpy down the group. Thus it becomes easy to lose proton going down the group.

(iii) F₂ is more reactive than ClF₃ because of the small size of fluorine atom F—F bond, bond dissociation enthalpy is low (thus is reactive).

Whereas ClF₃ is more reactive than Cl₂ because ClF₃ is an interhalogen compound with weak Cl—F bond (compared to Cl—Cl bond) due to the difference in atomic sizes (hence ineffective overlap of orbitals).

(b) (i) Structure of XeF₂ is linear.

(ii) Structure of H₄P₂O₇.

OR

Answer: (a) Fluorine reacts with cold sodium hydroxide solution to give OF₂.

2F₂ (g) + 2NaOH (aq) → 2NaF (aq) + OF₂ (g) + H₂O(l)

(c) XeFg reacts with NaF as follows:

XeF₆ + NaF → Na⁺[XeF₇]^–

(d) Ability to reduce is judged by the ease with which an atom can donate its electrons to the species which is getting reduced. Now, the size of oxygen atom in H₂O is smaller than that of Sulphur atom in H₂S, due to which the lone pair of electrons on oxygen are more attracted by the oxygen nucleus, making it difficult to donate electrons (by oxygen compared to Sulphur, while in H₂S the influence of the nucleus is less on lone pair of electrons of sulphur and hence, it can give away its electrons, easily compared to oxygen, and thus acts as a better reducing agent.

(e) The increasing order of acidic character can be written as

HF < HCl < HBr < HI

Chemistry 

Set-II

Section – A

Q.2. Arrange the following in increasing order of pKb values:

C₆H₅CH₂NH₂, C₆H₅NHCH₃, C₆H₆NH₂ 

Answer: These can be arranged in increasing order of pKb values as follows:

C₆H₅CH₂NH₂ < C₆H₅NHCH₃ < C₆H₅NH₂

Q.3. What type of colloid is formed when a liquid is dispersed in a solid? Give an example. 

Answer: When a liquid is dispersed in a solid, a ‘gel’ is formed.

Example: Butter.

Q.4. Out of chlorobenzene and p-nitrochloro-benzene, which one is more reactive towards nucleophilic substitution reaction and why? 

Answer: p-Nitro chlorobenzene would be more reactive towards nucleophilic substitution reaction compared to chlorobenzene. In chlorobenzene the carbon bearing the halogen is a part of the aromatic ring and is electron-rich due to the electron density in the ring so it does not attract the nucleophile. The -NO₂ substitution lessens the electron density on the benzene ring due to its electron-withdrawing nature, making the electron density on ringless compared to chlorobenzene, hence p-nitro chlorobenzene attracts nucleophiles better.

Section -B

Q.7. Give reasons:

(a) A decrease in temperature is observed in mixing ethanol and acetone.

(b) Potassium chloride solution freezes at a lower temperature than water. 

Answer: (a) Upon mixing molecules of ethanol and acetone have strong intermolecular attractions due to which heat is evolved from the reaction system and hence cooling of mixture is observed.

(b) Potassium chloride solution is a solution of non-volatile solute KCl and water solution. We know that, at the freezing point of a substance, the solid phase (here ice) is in dynamic equilibrium with the liquid phase. A solution freezes when its vapour pressure equals the vapour pressure of the pure solid solvent. Now, according to Raoult’s law when a non-volatile solid is added to the solvent (in this case it is KCl), its vapour pressure decreases and now it would become equal to that of solid solvent at a lower temperature. Thus, the freezing point of the solvent decreases.

Q.10. Define the following terms with a suitable example of each:

(a) Chelate complex

(b) Ambidentate ligand

OR

Q.10. Using IUPAC norms, write the formula for the following complexes:

(a) Tetraamminedichlorocobalt (III) chloride

 (b) Dibromobis (ethane-1,2-diamine) platinum (IV) nitrate

Answer: (a) Chelate complex: Chelate complexes are coordination or complex compound consisting of a central metal atom attached to a large molecule, called a ligand, in a cyclic or ring structure. The ligands are bi or polydentate i.e., they can attach to metal atom through two or more than two binding sites. An example of a chelate ring occurs in the ethylenediamine-nickel complex.

(b) Ambidentate ligand: Ligands which can ligate (attach to the metal atom) through two different atoms is called ambidentate ligand. One example of such ligand is NO^–₂ , this can bind through both the atoms, nitrogen and oxygen.

OR

Answer: IUPAC names

Tetraamine Diaqua Cobalt(III)chloride- [Co(NH₃)₄(H₂O)₂]Cl₃

(b) Dibromobis(ethane-1,2-diamine) platinum(IV) nitrate -[PtBr₂(en)₂](NO₃)₂.

Section – C

Q.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.

Answer: (a) Dispersed phase of milk is liquid and the dispersion medium of milk is liquid.

(b) During the process of adsorption molecules of adsorbate and adsorbent come closer to form physical or chemical bonds hence getting stabilized, in this process heat is evolved leading the overall process to be exothermic.

(c) Freundlich adsorption isotherm for gases at high pressure.

It is an empirical relationship between the quantity of gas adsorbed by a unit mass of solid adsorbent and pressure at a particular temperature:         x/m = kp_1/2(n > 1)

Where x is the mass of the gas adsorbed on mass m of the adsorbent at pressure p, k and n are constants which depend on the nature of the adsorbent and the gas at a particular temperature.

The relationship is generally represented in the form of a curve x/m is plotted against the pressure. This curve always approaches saturation towards high pressure, thus indicating that at adsorption through increases with increase in pressure till a limit and at high pressures, no further adsorption is observed.

Q.15. Write the name and principle of the method used for the refining of

(a) Tin,

(b) Copper,

(c) Nickel. 

Answer: (a) Tin: It is refined through liquidation. In this method, a low melting metal like tin is made to flow on a sloping surface, where the higher melting impurities are left behind and the lower melting metal is collected at the sloping end.

(b) Copper: It is refined through electrolytic refining. Anode is made of impure copper and pure copper stripes are taken as the cathode. They are dipped in acidified solution of copper sulphate, as an electrolyte. The net result of electrolysis is the transfer of copper in pure form from the anode to the cathode and the impurities get deposited as anode mud.

(c) Nickel: It is refined through Mond’s process. In this process, Nickel is heated in a stream of carbon monoxide forming a volatile complex, nickel tetracarbonyl.

The carbonyl is subjected to a higher temperature so that it is decomposed giving the pure metal.

Q.16. Give a reason for the following:

(a) Transition metals show variable oxidation states.

(b) E⁰ value of (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. 

Answer: (a) Transition metals show variable oxidation states because their d-orbitals are incompletely filled and different arrangements of electrons are possible according to the chemical environment of metal ions. hence, the ions can occupy variable oxidation states.

(b) E° value for Zn/Zn²⁺ is negative because the conversion of Zn to Zn²⁺ gives it a completely filled d⁵ configuration and extra stability gained by Zn²⁺. Whereas conversion of Cu to Cu²⁺ does not give any extra stability, hence it has a positive E⁰ value.

(c) Mn has the highest oxidation state of +4 with fluorine but with oxygen, it is +7. This is due to the ability of oxygen to form multiple bonds with the metal ion, whereas fluorine being of small size and devoid of d-orbitals can’t form multiple bonds.

Q.19. Write the structures of monomers used for getting the following polymers:

(a) Nylon-6, 6

(b) Bakelite

(c) Buna-S 

OR

Q.19. (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?

Answer: (a) Monomers of Nylon-6,6 are adipic acid and hexamethylenediamine.

(b) Monomers of bakelite are phenol and formaldehyde:

(c) Monomers of Buna-S are 1,3-Butadiene and Styrene

OR

Answer: (a) (i) Example of thermoplastic polymer – polythene, polystyrene.

(ii) Example of elastomer – Neoprene.

(b) In increasing order of their intermolecular force, they can be arranged as:

Buna-S < Polythene < Nylon-6,6

(c) Strong intermolecular forces between the polymer molecules, such as hydrogen bonding lead to closed packed structure, thus imparting crystalline nature to the polymers.

Chemistry

Set-III

Section – A

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

Answer: Cyclohexyl chloride is more reactive towards nucleophilic substitution reaction because the carbon bearing the chlorine atom is deficient in electron and seeks a nucleophile. In Chlorobenzene the carbon bearing the halogen is a part of an aromatic ring and is electron-rich due to the electron density in the ring.

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

(C₂H₅)₂NH, C₂H₅NH₂, C₆H₅NH₂ 

Answer: Decreasing order of solubility in water is:

C₂H₅NH₂ > (C₂H₅)₂NH₂ > C₆H₅NH₂.

Q.3. What type of colloid is formed when a solid is dispersed in a gas? Give an example.

Answer: An aerosol is the type of colloid formed when solid is dispersed in gas.

Example: smoke and dust.

Q.5. What is the difference between amylose and amylopectin? 

OR

Q.5. Write the products obtained after hydrolysis of lactose.

Answer: Amylose is a long linear chain of α-D-(+)-glucose units joined by C₁-C₄ glycosidic linkage (α-link), whereas Amylopectin is a branched-chain polymer of α-D-glucose units, in which the chain is formed by C₁-C₄ glycosidic linkage and the branching occurs by C₁-C₆ glycosidic linkage.

OR

Answer: Galactose and Glucose are the products obtained after hydrolysis of lactose.

Section – B

Q.7. Give reasons: 

(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.

Answer: (a) A mixture of chloroform and acetone forms a solution with a negative deviation from Raoult’s law. This is because chloroform molecule is able to form a hydrogen bond with acetone molecule as shown by the following figure:

This decreases the escaping tendency of molecules for each component, and consequently, the vapour pressure decreases and the temperature of the solution is increased because of stability attained by the molecule by associating and releasing energy.

(b) The solubility of gases in liquids increases on decreasing temperature, hence cold water has more dissolved oxygen because of which aquatic species find themselves more comfortable in cold water as compared to hot water.

Q.10. Define the following terms with a suitable example of each:

(a) Polydentate ligand

(b) Homoleptic complex

OR

Q.10. Using IUPAC norms, write the formula for the following complexes:

(a) Potassium tri (oxalato) chromate (III)

(b) Hexaaquamanganese (II) sulphate.

Answer: (a) Polydentate ligands: Ligands with several donor atoms are called polydentate ligands. These can bond with the metal ion in a complex with the different donor atoms present in them.

Example: N(CH₂CH₂NH₂)₃.

(b) Homoleptic complex: Complexes in which a metal atom is bound to only one kind of donor groups, e.g., [Co(NH₃)₆]²⁺ are known as homoleptic complex.

OR

Answer: (a) K₃[Cr(C₂O₄)₃]

(b) [Mn (H₂O)₆] SO₄.

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

Answer:

Section -C

Q.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? 

Answer: (a) Butter is an example of ‘Gel’ type of colloid. Here the dispersed phase is liquid and dispersion medium is solid.

(b) Physisorption occurs because of physical attractive forces, like Vander Waals forces between molecules of adsorbate and adsorbent, hence that can be reversed on the application of bigger forces. Hence, when the temperature is increased, the movement of adsorbed molecules increases, resulting in disturbed attractive forces, detachment of adsorbed molecules from the adsorbent surface hence physisorption decreases.

(c) When the AgNO₃ solution is added to KI, silver iodide, AgI, is precipitated. The precipitated silver iodide adsorbs iodide ions from dispersion medium and negatively charged colloidal sol results. It can be shown as AgI/I_– (negatively charged).

Q.17. Write the principle of the following:

(a) Hydraulic washing

(b) Chromatography

(c) Froth-floatation process 

Answer: (a) Hydraulic washing: This method of concentration of ores is based on the differences in gravity of the ore and gangue particles. It is a type of gravity separation. An upward stream of running water is used to wash the powdered ore. The lighter gangue particles are washed away and the heavier ores are left behind.

(b) Chromatography: Chromatography is a physical method of separation of a mixture in which the components to be separated are distributed between two phases, stationary and mobile phase. The stationary phase may be a solid or a liquid supported on a solid or a gel. The mobile phase may be either a liquid or a gas.

(c) Froth floatation process: Froth floatation is a physicochemical method of concentrating fine minerals. This process utilizes the difference in surface properties of valuable minerals and gangue (impurity) particles. For example, removal of gangue from sulphide ores.

Q.18. Give reasons for the following:

(a) Transition metals have high enthalpies of atomization.

(b) Manganese has a lower melting point even though it has a higher number of unpaired electrons for bonding.

(c) Ce⁴⁺ is a strong oxidising agent. 

Answer: (a) Transition element has high effective nuclear charge and a large number of valence electrons ((n-1) d electrons). So, as a result of the greater number of electrons participating, very strong metallic bonds are formed. As a result of the strong inter-atomic metallic bonding, the transition metals have high enthalpies of atomization.

(b) Manganese has a lower melting point even though it has a higher number of impaired electrons for bonding. Melting point depends on the intermolecular or interatomic forces. Stronger the forces, the higher the melting point. In Mn there is half-filled 3d subshell (3d⁵ configuration) which makes it stable and hence, it does not make additional covalent bonds with nearby atoms hence, it has less melting point.

(c) Ce⁴⁺ is a strong oxidising agent because Ce⁴⁺ oxidizes others and itself gets reduced to the common and preferred 3+ oxidation state of lanthanide elements.

Q.19. Write the structures of monomers used for getting the following polymers:

(a) Novolac

(b) Neoprene

(c) Buna-S 

OR

Q.19. (a) Write on example each of

(i) Cross-linked polymer

(ii) Natural polymer

(b) Arrange the following in the increasing order of their intermolecular forces:<br> Terylene, Buna-N, Polystyrene

(c) Define biodegradable polymers with an example.

Answer: (a) Novolac is the polymer of 2-hydroxymethyl phenol which is obtained by reaction of phenol and formaldehyde.

Answer: 

(b) Increasing order of their molecular forces:

 Buna-N < Polystyrene < Terylene

(c) Biodegradable polymer: These are synthetic polymers designed so as to contain functional groups similar to ones present in biopolymers. These are thus easily degraded by environmental degradation process hence, known as Biodegradable polymers.

Example: Poly (β-hydroxybutyrate-co-β-hydroxy valerate (PHBV).

Q.23. (a) Write the product when D-glucose reacts with Br₂aq.

Answer: