Chemistry
Section-A
Q.1. The analysis shows that FeO has a non-stoichiometric composition with formula Fe0.95O. 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. (Kf 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, ΔTf = Kfm
Putting the given values,
ΔTf = Kfm = 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) pKa 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) pKa 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
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=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 pKb 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.
