13345450 Class Xii Cbse 2009 Board Chemistry Solutions

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Class XII, Chemistry Board Exam 2009 Solutions Code no 56/1 1. Metallic substances are conductors in solid state as well as in molten state Ionic substances are insulators in solid state but conductors in molten state and in aqueous solution. (1) 2. The process of changing the colloidal particles in a sol into the insoluble precipitate by addition of some suitable electrolytes is known as coagulation process. (1) 3.Pyrometallurgy is the branch of metallurgy which consists of the thermal treatment of minerals and metallurgical ores and their concentrates to bring about physical and chemical transformations in the materials to enable recovery of valuable metals. (1) 4.Red phosphorus is polymeric which consists of chains of P4 tetrahedra linked together whereas white phosphorus consists of a single tetrahedron (1) 5. Hexen-3-ol

(1)

6.

(1) 7. In aqueous solutions, the increasing order of basic strengths is: (CH3)2NH > CH3NH2 > (CH3)3N > NH3 (1) Get the best brains on your side. Watch Topper TV! Log on to www.topperlearning.com Call 0120 – 4646900 or e-mail us at [email protected] 

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8. It indicate the number of carbon atoms present in both the monomers of nylon 6,6 which are hexamethylene diamine and adipic acid. (1) 9. Lead storage battery is a secondary cell.

(1/2)

Anode: Pb(s) + SO4 2- (aq) → PbSO4 (s) + 2e-

(1/2)

Cathode: PbO2(s) + SO4 2- (aq) + 4H+(aq) + 2e- → PbSO4 (s) + 2H2O(l) (1/2) Overall cell reaction: Pb(s) + PbO2(s) + 2H2SO4 (aq) → 2PbSO4 (s) + 2H2O(l)

(1/2)

Or MnO4-(aq) + 8H+ (aq) + 5e- → Mn2+ (aq) + 4H2O(l) ◦ Sn4+ (aq) + 2e- → Sn2+ (aq) E = -0.15V

◦ E = +1.51V

Cathode reduction reaction: MnO4 (aq)  8H (aq)  5e  Mn2  (aq)  4H2O (l) Anode-Oxidation reaction: Sn2+ (aq) → Sn4+ (aq) + 2eRedox equation: 2MnO4-(aq) + 16H+(aq) + 5Sn2+ (aq) → 2Mn2+ (aq) + 8H2O(l) + 5Sn4+ (aq) (1)

◦ ◦ Cell potential = E reduction - E oxidation

(1/2)

= 1.51 - (-0.15) = 1.66 V

(1/2)

10. (i) An elementary step in a reaction is a chemical reaction in which one or more of the reactants react directly to form products in a single reaction step. (1) Get the best brains on your side. Watch Topper TV! Log on to www.topperlearning.com Call 0120 – 4646900 or e-mail us at [email protected] 

3 (ii) Rate of a reaction is defined as the rate of decrease in concentration of any on of the reactants or the rate of increase in concentration of any one of the products. (1) 11.(i) In this method ,the impure metal acts as anode which undergoes oxidation and pure metal acts as cathode which undergoes reduction. The metal ions deposit on the cathode as metal. (1) (ii) Vapour phase refining is based on the principle that metal is converted to its volatile compound and collected which decomposes to give pure metal. (1) 12.

(i) 2XeF2 + 2H2O → 2Xe + 4HF + O2

(1)

(ii) 2PH3 + 3HgCl2 → Hg3P2 + 6HCl

(1)

13. (i) 2MnO4-(aq) + 5C2O42- (aq) + 16H+(aq)→ 2Mn2+ (aq) + 8H2O + 10CO2 (1) (ii) Cr2O72- (aq) + 6Fe2+(aq) + 14H+(aq)→ 2Cr3+ +6Fe3+(aq) + 7H2O

(1)

14.

(i)

will undergo S N 1 reaction faster than

is a tertiary alkyl halide and because secondary alkyl halide (1 Mark)

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is a

4

(ii)

will undergo S N 1 reaction faster than

because

is a

is a

secondary alkyl halide and primary alkyl halide.

(1)

15. (i)

HI  (1) (ii) CH3CH2CH=CH2 + HBr → CH3CH2CH-CH3 │ Br (1 Mark) 16. Four bases in DNA are Adenine ,Guanine, Thymine and Cytosine (1) Thymine is not present in RNA.

(1)

17. Vitamin A and D are fat soluble vitamins. Get the best brains on your side. Watch Topper TV! Log on to www.topperlearning.com Call 0120 – 4646900 or e-mail us at [email protected] 

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A D

Sources Carrots, butter

Disease Xerophthalmia

Exposure to sunlight, fish

(1) Rickets (1)

Thermoplastic polymers

Thermosetting polymers

Linear or slightly branched long chain molecules Can undergo softening on heating and hardening on cooling

Cross linked or heavily branched molecules (1)

18.

Molecular Structure Behaviour

Do not soften on heating (1)

Ans.19 k = 0.0051 min-1 [A0] = 0.10 M t = 3 hours = 3× 60 min k = 2.303 log [A0] t [A] 0.0051 = 2.303 log 0.1 3× 60 [A] log [A] = -1.39 [A] = 4.074 × 10-2 M

Ans.20.

(1) (1) (1)

a = 409 pm, r = ?

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21.

Ecell = 0.422 V

◦ Since E

+ Ag /Ag

>E

◦

2+ Cu /Cu

So, Ag / Ag is cathode and Cu2+/Cu is the anode.

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7 Ecell  Ereduction  EOxidation  0.80  0.34  0.46V

(1 / 2 )

Cell Re action : 2Ag  Cu  Cu2   2Ag 2

Ecell Ecell

Cu2   Ag 0.059  log  E cell  2 n  Ag  Cu 0.059 [0.1]  0.46  log 2 [Ag ]2

(1 / 2 )

2

0.422  0.46  0.029{log[0.1]  log  Ag  } log[.1]  2log  Ag   1.310 (log101 )  2log[Ag ]  1.310 1  2log[Ag ]  1.310 2.310 2  log  Ag   1.155  Ag   Antilog(1.155) 2log  Ag  

 Ag   0.0699M

(1 )

(1 )

22. (i) If an electrolyte is added to a hydrated ferric oxide sol , the colloidal (1) particles get precipitated (ii) Scattering of light takes place and the the path of the beam is (1) illuminated (iii) The colloidal particles move towards oppositely charged (1) electrodes, get discharged and precipitated 23. (i) The splitting of the degenerate d orbitals into eg and t2g orbitals due to the presence of ligands in a definite geometry is called crystal field splitting. (1) (ii) Linkage isomerism arises in coordination compounds containing ambidentate ligands in which two different atoms of the same ligand can form coordinate bond with metal ion Get the best brains on your side. Watch Topper TV! Log on to www.topperlearning.com Call 0120 – 4646900 or e-mail us at [email protected] 

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Example: Co(NH3 )5 NO2  Cl2 and Co(NH3 )5 ONO Cl2

(1)

(iii) Ligands which can bind the central metal atom through two different atoms are called ambidentate ligands (1) Example: SCN ,NCS

OR [Co(NH3)6]3+

Co3  (Ground state) 3d6 4s0 4p0

d2sp3 hybridisation of Co3+

Structural shape – Octahedral Hybrid Orbitals- d2sp3 Magnetic behaviour- Diamagnetic

Cr(NH3 )6 

3

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(1)

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Cr3  (Ground state) 3d3 4s0 4p0

d2sp3 Hybridisation of Cr3 

Structural shape – Octahedral Hybrid Orbitals- d2sp3 Magnetic behaviour- Paramagnetic

(1)

Ni(CO)4 Ni - 3d8 4s2

sp3 hybrid orbitals of Ni

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Structural shape- tetrahedral Hybid orbitals- sp3 Magnetic behaviour- Diamagnetic

(1)

24.(i) Hydrogen bonding in ethanol causes the boiling point of ethanol to be higher (1) (ii) Phenol on releasing a proton forms phenoxide win which is resonance stablised .So, phenol is more acidic than ethanol. (1) (iii) NO2 groupshas -I effect or electron withdrawing inductive effect. (1)

25 (i) Ability to show variable oxidation state and to form complexes (ii) Lanthanoid contraction (iii) 5f, 6d and 7s levels are of comparable energies. 26 (a)

(1) (b)

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(c)

(1)

27. (i) Non ionic detergents do not contain any ion in their constitution. (1) (ii) Food preservatives are substances which present spoilage of food due to microbial growth. (1) (iii) Disinfectants are chemicals which either kill or prevent the growth of microorganisms .They cannot be applied on a living tissue. (1) 28. (a) (i)Mole fraction of a component is the ratio of number of moles of the component to the total number of moles of all the components. (1) (ii) Van’t Hoff factor is the ratio of normal molar mass to the abnormal molar mass. (1) Van’t Hoff factor is the ratio of observed value of colligative property to calculated value of colligative property assuming no association or dissociation. (b) Mass of protein = 100mg = 0.1 g V= 10 mL

(1)

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(1) M = 13.98 g Molar mass of protein=13.98 g

(a) (i) (ii)

(1) Or

All the properties which depend on the number of solute particles irrespective of their nature relative to the total number of particles present in the so solution are known as colligative properties. (1) Molality of solution is the number of moles of solute present in 1 kilogram of solvent. (1)

(b)

pn2  KH x XN2 (in air)  0.78 1000  55.5mol 18 Sincenn2  nH2O

nH2O 

xn2 

(1)

nN2 nH2O

nN2  0.78 * 55.5  43.29 mol

(1)

Since 43.29 moles of N2 are present in 1000 ml of water, concentration= 43.29 molar.

(1)

Ans. 29.(a) (i) H2S2O8

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(1)

(ii)

HClO4

(1) (b) (i) N is more electronegative than P. So lone pair of electrons is concentrated more on nitrogen. (1) (ii) Oxygen has a small size as compared to sulphur. So, the lone pairs on oxygen repel each other. Thus, the O-O bond is weak whereas in case of sulphur, the lone pair do not repel to the same extent. As a result, S-S bond is stronger than O-O bond. So, S has a greater tendency for catenation than oxygen. (1) (iii) The net energy release for

1  X  (aq) is more negative for X2 (g)  2

fluorine as compared to chlorine i.e. sum of dissociation energy, electron gain enthalpy and hydration energy is more negative for fluorine. So, it a stronger oxidising agent than chlorine. (1) Or (a) (i) H2S2O7 Get the best brains on your side. Watch Topper TV! Log on to www.topperlearning.com Call 0120 – 4646900 or e-mail us at [email protected] 

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(1) (ii) HClO3

(1) (b) (i) Because of partial double bond character of (N-O) bond.

(1)

(ii) The axial bonds due to more repulsion are longer than equatorial bonds. (1) (iii) ICl is an interhalogen compound having electronegativity difference. The I-Cl bond is more reactive than I2 . (1) 30. (a) (i) Aldehydes which do not have an -hydrogen atom undergoes self oxidation and reduction on treatment with concentrated alkali. (1) (ii) Carboxylic acids having an -hydrogen are halogenated at the position on treatment with chlorine or bromine in the presence of small amount of red phosphorus to give - halocarboxylic aids. (1) (b) (i) Propanal and propanone Fehling’s test 

CH3CH3CHO  2Cu2   5OH  CH3CH2COO  Cu2O  3H2O Re d ppt Propanone being a ketone will not give this test. (1) Get the best brains on your side. Watch Topper TV! Log on to www.topperlearning.com Call 0120 – 4646900 or e-mail us at [email protected] 

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(ii) Acetophenone and benzophenone Iodoform test

Benzophenone does not give this test.

(1)

(iii) Phenol and Benzoic and

FeCl3 test

C6H5OH  FeCl3  (C6H5O )6  Fe]3   3HCl Blue violet colour Benzoic acid does not give this test.

(1)

OR (a) (i)

(1) (ii)

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(1)

(b)

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(1) CrO3 CH3  CH2  CH2  CH2  OH   CH3  CH2  CH2  COOH

C

B

dehydration CH3  CH2  CH2  CH2  OH   CH3  CH2  CH  CH2 H2O

But  1  ene (1)

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(1)