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+ + SO22- + 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 t1/2is 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 t1/2 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 < P and
(ii) ∆0 > 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 < P is given as t2g3 eg2
(ii) Electronic configuration of d5 when ∆o > P is given as t2g5 eg0.
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 × 102- to 1.68 × 102-). 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 × 102- to 6.0 × 103-). 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 × 102- 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 cm3-. 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 ∧0m 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 E0cell = 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> ∧0m = 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 = kp1/2(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:
