(i) chlorine gas is passed through a hot concentrated solution of NaOH? (ii) sulphur dioxide gas is passed through an aqueous solution of a Fe (III) salt? (b) Answer the following:
(i) What is the basicity of H3PO3 and why?
(ii) Why does fluorine not play the role of a central atom in interhalogen compounds? (iii) Why do noble gases have very low boiling points?
30. (a) Illustrate the following name reactions: (i) Cannizzaro’s reaction
(ii) Clemmensen reduction
(b) How would you obtain the following: (i) But-2-enal from ethanal
(ii) Butanoic acid from butanol (iii) Benzoic acid from ethylbenzene
OR
(a) Give chemical tests to distinguish between the following: (i) Benzoic acid and ethyl benzoate
(ii) Benzaldehyde and acetophenone.
(b) Complete each synthesis by giving missing reagents or products in the following:
(i) SOCl
heat 2
¾¾¾®
(ii) C6H5CHO ¾H NCONHNH¾ ¾ ¾ ¾2 ¾ 2®
(iii) CH2 ¾¾¾ ® CHO
CBSE (All India) SET–II
Questions Uncommon to Set–I
2. What are lyophobic colloids? Give one example for them.
3. Why is it that only sulphide ores are concentrated by froth floatation process? 194 Xam idea Chemistry—XII
COOH
5. Write the IUPAC name of the following compound:
6. Draw the structure of 2, 6-Dimethylphenol.
9. Define the following terms in relation to crystalline solids: (i) Unit cell
(ii) Coordination number Give one example in each case.
12. A reaction is of second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is reduced to half? What is the unit of rate constant for such a reaction? 14. Describe the principle controlling each of the following processes:
(i) Zone refining of metals. (ii) Electrolytic refining of metals.
15. Explain giving a suitable reason for each of the following:
(i) Transition metals and their compounds are generally found to be good catalysts.
(ii) Metal-metal bonding is more frequent for the 4d and the 5d series of transition metals than that for the 3d series.
19. What mass of NaCl must be dissolved in 65.0 g of water to lower the freezing point of water by 7.50°C? The freezing point depression constant (Kf) for water is 1.86 C/m. Assume van’t Hoff factor
for NaCl is 1.87. (Molar mass of NaCl = 58.5 g).
22. Write the structures and names of all the stereoisomers of the following compounds: (i) [Co(en)3]Cl3
(ii) [Pt(NH3)2Cl2]
(iii) [Fe(NH3)4Cl2]Cl
CBSE (All India) SET–III
Questions Uncommon to Set-I and Set-II 1. Define ‘activation energy’ of a reaction. 2. What is meant by ‘reverse osmosis’?
3. What type of ores can be concentrated by magnetic separation method?
11. Differentiate between molality and molality values for a solution. What is the effect of change in temperature on molarity and molality values?
14. Describe the principle controlling each of the following processes: (i) Preparation of cast iron from pig iron.
(ii) Preparation of pure alumina (Al2O3) from bauxite ore.
CH3 H Br H H H
15. Explain giving reasons:
(i) Transition metals and their compounds generally exhibit a paramagnetic behaviour. (ii) The chemistry of actinoids is not as smooth as that of lanthanoids.
18. Write such reactions and facts about glucose which cannot be explained by open chain structure. 21. How would you account for the following:
(i) NF3 is an exothermic compounds but NCl3 is not.
(ii) The acidic strength of compounds increases in the order: PH3 < H2S < HCl
(iii) SF6 is kinetically inert.
22. Write the state of hybridization, the shape and the magnetic behaviour of the following complex entities:
(i) [Cr(NH3)4 Cl2]Cl
(ii) [Co(en)3]Cl3
(iii) K2 [Ni (CN)4]
26. Write the names and structures of the monomers of the following polymers: (i) Buna-S
(ii) Dacron (iii) Neoprene
CBSE (All India) SET–I
1. Order of a reaction may be defined as the sum of the powers of the concentration terms of the reactants in the rate law expression.
2. The catalysis in which the pore structure of the catalyst and the size of the reactant and product molecules are comparable is called shape selective catalysis.
3. The naturally occurring chemical substances present in the earth’s crust which can be obtained by mining are called minerals, while mineral from which metals can be extracted economically are called ores.
4. The steady decrease in the atomic and ionic radii (having the same charge) with increase in atomic number as we move across the series from lanthanum to lutetium is known as lanthanoid contraction. 5. 3-Bromoprop-l-ene 6. CH — CH | Cl — CH — C | | O — CH 3 2 3 7. CH3CH2OH conc. H SO 443K 2 4 ¾¾ ¾ ¾¾ CH® 2 = CH2 + H2O Ethanol Ethene 8. (C6H5)2 NH < C6H5NH2 < C6H5N(CH3)2 < CH3NH2 196 Xam idea Chemistry—XII
Solutions
Solutions
9. We can determine the atomic mass of an unknown metal by using the formula of density of its unit cell.
d (density) = Z (No. of atoms per unit cell) M (atomic mass)
a (cell 3
´
edge)´NA (Avogadro number)
By knowing d, Z, a and NA, we can calculate M, the atomic mass of metal. 10. Packing efficiency = Z volume of one atom
Volume of cubic unit cell 100 ´ ´ = Z r a ´ ´ 4 3 100 3 3 p
For a simple cubic lattice, a = 2r and Z = 1 \ Packing efficiency = 1 3 2 100 3 3 ´4 ´ pr r ( ) = p 6´100 = 52.36% = 52.4%
11. (i) Raoult’s law: It states that for any solution, the partial pressure of each volatile component in the solution is directly proportional to its mole fraction.
(ii) Henry’s law: It states that the partial pressure of a gas in vapour phase (P) is proportional to its mole fraction (x) in the solution.
12. An experimentally determined expression which relates the rate of reaction with the concentration of reactants is called rate law while the rate of reaction when concentration of each reactant is unity in a rate law expression is called rate constant.
(i) Comparing power of mole in L–1 mol s–1 and (mol L–1)1–n s–1, We get
1 = l – n Þ n = 0 i.e., zero order reaction
(ii) Again comparing power of mole in L mol–1s–1 and (mol L–1) 1–n s–1, we get –1 = 1 – n Þ n = 2, i.e., second order reaction
13. For a first order reaction
t = 2 303.
k log
[A] [A]
o
Here, k = 2.4 × 10–3 s–1, [A] = [A] 3 4[A]
o- o =
[ ]A o
4 , t = ? Substituting these values in the equation, we get
t = 2.303 2.4 10 s log [A]o [A] 3 1 O 4 ´ - - t = 2.303 2.4 10´ -3s-1 log4 = 2 303 2 4 10 3 0 6021 . . ´ - ´ . s t = 577.7 s = 578 s
14. (i) In this method, the metal is converted into its volatile compound and collected elsewhere. It is then decomposed to give pure metal.
Ti I Ti I Impure metal 2 500 K 4 +2 ¾¾¾® Ti I4 1700 K Ti 2I Pure metal 2 ¾¾¾® +
(ii) This method of concentration of ore is based upon the principle that the surface of sulphide ores is preferentially wetted by oils while that a gangue is preferentially wetted by water. 15. (i) Cr2+ is reducing as its configuration changes from d4 to d3, a more stable half filled t2g
configuration while Mn3+ is oxidising as Mn3+ to Mn2+results a more stable half filled d5
configuration.
(ii) It is due to greater number of unpaired electrons in (n–1)d and ns orbitals at the middle of the series.
16. (i) 8MnO4–(aq) + 3S2O32–(aq) + H2O (l) ¾ ®¾ 8MnO2(s) + 6SO42–(aq) + 2OH–(aq)
(ii) Cr O2 2–7 (aq) 14H (+ + aq)+6Fe2+(aq) ¾ ®¾ 2Cr3+(aq)+6Fe3+(aq)+7H O( )2 l
OR
(i) This is because Cu(I) ion is unstable in aqueous solution and undergo disproportionation. 2Cu (+ aq) ¾ ®¾ Cu2+(aq) + Cu( )s
(ii) This is because due to lanthanoid contraction the expected increase in size does not occur.
17. (i) Peptide linkage: The amide (— C | | O
— NH—) linkage between two a-amino acids formed with the loss of a water molecule is called a peptide linkage.
(ii) The six membered cyclic structure of glucose is called pyranose structure (a-orb-), in analogy with heterocylic compound pyran.
18. Structural difference between DNA and RNA
DNA RNA
1. The sugar present in DNA is 2-deoxy D-(–) ribose. 1. The sugar present in RNA is D-(–)-ribose. 2. DNA has double stranded a-helix structure. 2. RNA has single a-helix structure.
The common bases present in both DNA and RNA are adenine (A), guanine (G) and cytosine (C). 198 Xam idea Chemistry—XII
H HO H H OH CH OH2 H OH OH H O a – D – (+) – Glucopyranose
19. MB = W R T V B´ ´ ´ p ....(i)
Here, WB = 8.95 mg = 8.95 × 10–3g , R = 0.0821 L atm mol–1 K–1 T = 25°C = (25 + 273) K = 298 K , p = 0.335 torr = 0 335 760
. atm V = 35 mL = 35 × 10–3 L
Substituting these values in the equation (i), we get MB = 8.95 10 g 0.0821 L atm mol K 298K 760 0.335 atm 3 3 1 1 ´ ´ ´ ´ ´ - - - 5 10´ -3L = 14193. 3 g mol –1
20. These are of two types (i) Hydrophilic
Stability: More stable as the stability is due to charge and water envelope surrounding the sol particles.
Nature: Reversible
Examples: Starch, gum, etc. (ii) Hydrophobic
Stability: Less stable as the stability is due to charge only. Nature: Irreversible
Examples: Metal hydroxide like Fe(OH)3 and metal sulphide like As2S3. OR
(i) On passing electric current through a sol, colloidal particles start moving towards oppositely charged electrode where they lose their charge and get coagulated (electrophoresis).
(ii) Scattering of light by the colloidal particles takes place and the path of light becomes visible (Tyndall effect).
(iii) The positively charged colloidal particles of ferric hydroxide sol get coagulated by the oppositely charged Cl– ions provided by NaCl.
21. (i) This is because bond dissociation enthalpy of H—S bond is lower than that of H—O bond. (ii) This is because two bonds share a double bond in the resonance hybrid structure of NO2– while three bonds share a double bond in the resonance hybrid structure of NO3–. This is because NO2– has bond order 1.5 while NO3– has bond order 1.33.
(iii) This is due to tendency of oxygen to form multiple bonds with metal atom.
22. (i) Ambidentate ligand: A ligand which can coordinate to central metal atom through two different atoms is called ambidentate ligand. For example NO2– ion can coordinate either through nitrogen or through oxygen to a central metal atom/ion.
(ii) Denticity: The number of coordinating groups present in ligand is called the denticity of ligand. For example, bidentate ligand ethane-1, 2-diamine has two donor nitrogen atoms which can link to central metal atom.
H N— CH — CH — N H2 2 2 2
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