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Note : Draw diagrams and write equations wherever necessary.

Section - iNote : (i) Answer all the questions. (15 × 1 = 15) (ii) Choose the most suitable answer :1. Match the List -I and List -II correctly by using the code give below.

List - I List - II

(A) Debroglie Relation (1) Δx Δp ≥ h/4π

(B) Bohr’s quantum condition (2) En = –2π2 me4 /n2h2

(C) Energy of an electron in an atom

(3) 2πa = nλ

(D) Uncertainty Principle (4) λ = hmv

Codes: (A) (B) (C) (D)

(a) (2) (4) (1) (3)(b) (4) (3) (2) (1)(c) (4) (2) (3) (1)(d) (2) (1) (4) (3)

2. The lightest gas which is non-inflammable is(a) He (b) H2 (c) N2 (d) Ar

3. Consider the following statements;(I) d-block elements are coloured because they absorb some energy

for d-d transition.(II) K2Cr2O7 is a powerful reducing agent.(III) Oxidation state of Ni in Ni(CO)4 is zeroWhich of the above statement/s is/are correct?(a) I and III (b) I and II (c) I,II and III (d) II and III

4. Ceria is used in(a) toys (b) tracer bullets (c) gas lamp material (d) all the above

5. The time taken for 10g of initial amount of a substance to become 5g in a decay is ‘X’ hours and for 1g of initial amount to become 0.5g in the same decay is ‘y’ hours, then(a) x > y (b) x < y (c) x = y (d) x >> y

6. Rutile is _____________(a) TiO2 (b) Cu2O (c) MoS2 (d) Ru

7. For the reaction 2Cl(g) → Cl2, the signs of ΔH and ΔS respectively are(a) +,– (b) +, + (c) –, – (d) –, +

8. In the reversible reaction 2HI � ⇀�↽ �� H2+I2, Kp is(a) greater than Kc (b) less than Kc(c) equal to Kc (d) zero

9. Consider the following potential energy diagram and Identify the correct representation.

- - - - -

- - - - - -

A

B

C

BEnergy

Reaction Co-ordination

(a) AB - Activation Energy (b) BC - Threshold Energy(c) BC - Activation Energy (d) AB - Threshold Energy

GOVT. MODEL QUESTION PAPER - I12th

STD.

10. Water soluble dye is mixed with emulsion and emulsion remains colourless then the emulsion is(a) o/w (b) w/o (c) o/o (d) w/w

11. The compound that reacts fastest with Lucas reagent is(a) Butan - 2- ol (b) Butan - 1- ol(c) 2-methyl propan -1-ol (d) 2-methyl propan -2-ol

12. According to Lewis concept of acids and bases, ethers are(a) neutral (b) acidic (c) basic (d) amphoteric

13. When benzaldehyde is treated with HCN followed by hydrolysis it yields(a) Benzoic acid (b) Lactic acid(c) Maleic acid (d) Mandelic acid

14. Statement (I) : Lower members of carboxylic acids are highly soluble in water.Statement (II) : This is due to hydrogen bonding between acids and water.(a) Statement(I) is correct but Statement (II) is false(b) Statement (I) and (II) are correct and Statement (II) is the correct

explanation for Statement (I)(c) Statement(I) is false but statement (II) is Correct(d) Statement (I) and (II) are correct and Statement (II) is not a correct

explanation for Statement(I)15. Which of the following answers carbylamine reaction?

(a) ethyl amine (b) sec - butylamine(c) tert-butylamine (d) all the above

Section - iiAnswer any six questions and question number 23 is compulsory: (6 × 2 = 12)16. A triprotic oxy acid of phophorus reacts with AgNO3 to give a yellow

precipitate. Write the suitable reaction.17. Give the formula for the given complexes.

(a) Tris (ethylenediammine) cobalt(III) chloride(b) Triamminetrinitro cobalt (III)

18. What is Q value of a nuclear reaction?19. What is pseudo first order reaction? Give an example.20. State Kohlrausch’s law.21. The standard reduction potential for the reaction Sn4+ + 2e- → Sn2+ is

+ 0.15V. Calculate the free energy change of the reaction.22. Draw the cis and trans form of 2-pentene.23. Substantiate with suitable evidence that in fructose the ketone group is

adjacent to one of the terminal carbon atom.24. Mention the characteristics of dyes.

Section - iiiAnswer any six questions and question number 29 is compulsory : (6 × 3 = 18)25. The uncertainty in the position of a moving bullet of mass 10g is 10–5m.

Calculate the uncertainty in its velocity.26. Explain the factors affecting ionization energy.27. Discuss the structure of AX3 and AX7 interhalogen compound.28. Show that decrease in free energy change is equal to net work done by

the system.29. Derive an expression for Kc for decomposition of PCl5.30. Explain the cell terminologies used in Electrochemical cell.31. When organic compound C3H8O3 is acetylated with acetic arhydride it

gives C9 H14O6. How many –OH groups are there in the compound? Write the structure and reaction.

32. Distinguish between diethylether and anisole.33. Explain the mechanism of aldol condensation reaction.

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2 ___________________________________________________________ Sura’s XII Std - Govt. Model Question Papers

Section - iVAnswer all the questions : (5 × 5 = 25)34. (i) Draw the molecular orbital diagram for oxygen molecule and

calculate its bond order (3)(ii) How do electro-negativity values help to find the nature of bond.

(2)(or)

(i) Discuss the consequences of lanthanide contraction. (3)(ii) [Ni(CN)4]

2– is square planar where as [NiCl4]2– is tetrahedral why?

(2)35. (i) Explain the extraction of gold from its ore. (3)

(ii) What is philosopher’s wool? How is it prepared? (2)(or)

(i) Write notes on Schottky defect. (3)(ii) Define entropy and give its unit. (2)

36. (i) State the optimum condition to obtain maximum yield of NH3 in Haber’s process (2)

(ii) Write note on consecutive and parallel reaction with example. (3)(or)

(i) Write the preparation of colloids by condensation method. (3)(ii) What is the electrochemical equivalent of a substance when 150gm

of it is deposited by 10 ampere of current passed for 1 sec? (2)37. (i) Discuss the optical isomerism in tartaric acid. (3)

(ii) CH

CHA BHOCl Ca OH2

2

2 → →( ) . Identify A and B (2)

(or)(i) How does formaldehyde react with NH3 and CH3MgI/H2O, H+?(ii) Substantiate with suitable reason that chloroacetic acid has more

acid strength than acetic acid. (2)38. (i) An organic compound (A) with molecular formula C2H7N

dissolves in acid solution. It reacts with NaNO2/HCl to give (B) of molecular formula C2H6O. (B) reacts with acetic anhydride forming a pleasant smelling liquid (C) Identify A,B,C. Explain the reactions. (3)

(ii) Convert aniline to S- diphenylthiourea (2)(or)

(i) Give evidences to show glucose has a aldehyde group, 6 carbon linear chain, 5-OH groups. (3)

(ii) How is Nylon 66 prepared? (2)

ANSWERSSection - i

1. (b) (4) (3) (2) (1)2. (a) He3. (a) I and III4. (c) gas lamp material 5. (b) x < y 6. (a) TiO27. (d) –, +8. (c) equal to Kc9. (c) BC - Activation Energy10. (a) o/w11. (d) 2-methyl propan -2-ol12. (c) Basic13. (d) Mandelic acid14. (b) Statement (I) and (II) are correct and Statement (II) is the correct

explanation for Statement (I)15. (a) Ethyl amine

Section - ii16. H3PO4 + 3AgNO3 → Ag3PO4 + 3HNO3 Ortho phosphoric acid Silver phosphate17. (a) [Co(en)3]Cl

(b) [Co(NH3)3 (NO2)3]

18. The amount of energy absorbed or released during the nuclear reaction is called Q-value of nuclear reaction.

Qvalue = (mp – mr) 931 MeV where, mp = sum of the masses of products mr = sum of the masses of reactants.19. In a second order reaction, when one of the reactant’s concentration is in

excess (10 to 100 times) of the other reactant then the reaction follows a first order kinetics and such a reaction is called pseudo first order reaction.

Eg: Acid catalysed hydrolysis of ester CH3COOCH3 + H2O H+

→ CH3COOH + CH3OH methyl acetate Acetic acid methyl alcohol20. Kohlrausch’s law states that, “at infinite dilution wherein the ionisation

of all electrolytes is complete, each ion migrates independently and contributes a definite value to the total equivalent conductance of the electrolyte.”

21. Sn4+ + 2e– → Sn2+ Eº = 0.15V Given : n = 2 electrons F = 96495 coulombs Formula : ∆G = – nFEº Solution : ∴∆G = – 2 × 96495 × 0.15 = –28.948 Free energy = – 28.948 kJ.22.

H

CH3C2H5

H

C = C

Cis - 2 - penteneCH3

HC2H5

H

C = C

Trans - 2 - pentene

23. Fructose HCN → Addition product HI P/ → 2-Methyl Hexanoic Acid. This indicates that the ketone group is adjacent to one of the terminal carbon atom (ie) C2 - carbon atom.

24. i. A dye should have a suitable colour.ii. It should be able to fix itself or be capable of being fixed to the fabric.iii. It should be fast to light.iv. It should be resistant to the action of water, dilute acids and alkalies.

Section - iii25. Given Data :

Uncertainty in position, ∆x = 10–5 m Mass of a moving bullet = 10 g Planck’s constant = 6.626 × 10–34 kg m3 sec–1

Formula : ∆x . ∆p ≥ h

4π (or) ∆x . m ∆v ≥ h

4π

Solution : 10–5 × 10 × 10–3 × ∆v =

6 626 104 3 14

34..

××

−

∆v = 6 626 104 3 14 10 10 10

34

3 5.

.×

× × × ×

−

− −

= ×

= ×

−

− −

46 368 10880

5 27 10

26

28 1

.

. sm

Dv = 5.27 × 10–28 m s–1

26. i. Atomic size :

♦ I.E.∝ 1

Atomic size

♦ Ionisation energy decreases with increase in atomic size. ♦ The larger the size of the atom, lesser is the ionisation energy.

This is due to the fact that electrons are held quite loose (i.e) lesser energy is required for removal of electrons from larger atom than the smaller one.

ii. Magnitude of Nuclear charge : ♦ I.E ∝ Magnitude of Nuclear charge. ♦ Because of the higher nuclear charge, the electrons bound

with more force and hence high energy required to remove the electron.

♦ Hence I.E. increases.

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Sura’s XII Std - Govt. Model Question Papers ___________________________________________________________ 3

iii. Screening (or) Shielding effect :

I.E.∝ 1

Screening effect ♦ The electron to be removed is thus shielded from the nucleus

by the electrons in the inner shell. ♦ Thus, the electron in the valence shell experiences less attraction

from the nucleus. ♦ Hence I.E. will be low.

iv. Shape of orbital : ♦ As s-electrons remain closer to the nucleus than p, d and

f-electrons of the same valence shell, the ionisation energy decrease in the order as s > p > d > f.

v. Arrangement of electrons (or) Electronic Configuration : ♦ More stable the electronic arrangement, greater is the I.E. ♦ As the noble gases have the stablest electronic arrangement ns2

np6,they show maximum I.E.27. i. Type AX3:

♦ Without lone pair of electrons, it is T-shaped.

♦ Trigonal bipyramidal structure. ♦ sp3 hybridisation. Eg: ClF3

ii. Type AX7: F

F

F

F

F

F

F

I ♦ Pentagonal pyramidal. ♦ sp3d3 hybridisation Eg: IF7

28. Free energy ‘G’ : i. Free Energy is defined as G = H–TS. 'G' is a state function. ii. G - Extensive; property ∆G intensive property, when mass remains

constant between initial & final states of the system.iii. Process Spontaneous Equilibrium Non-spontaneous

∆G –ve Zero +ve iv. 'G' has a single value for the thermodynamic state of the system. v. G & ∆G values correspond to the system only. vi. ∆G = ∆H – T ∆S According to I law of thermodynamics, ∆H = ∆E + P ∆ V; ∆E = q – w; T ∆ S = q ∆G = q – w + P ∆ V – q ∆G = – w + P ∆ V29. The dissociation equilibrium of PCl5 in gaseous state is written as PCl5(g) � ⇀�↽ �� PCl3(g)+Cl2(g)

Dn(g) = 2 – 1 = 1; \Kp > Kc

PCl5 PCl3 Cl2Initial no of moles a 0 0No. of moles dissociated x - -No. of moles at equilibrium a – x x xEquilibrium concentration a x

V− x

VxV

\Kc = [ ][ ][ ]

PCl ClPCl

3 2

5

=

xV

xV

a xV

×

−

= xV

Va x

2

2 ×−

K xa x Vc =

−

2

( ) x = degree of dissociation x = Number of molesdissociated

Totalnumber of moles present initially

Kc = xx V

2

1( )− [ Initially the no. of PCl5 present = 1 mole]

= x Px RT

2

1( )− [ PV =RT]

Cl

F

F

F

30. i. Current : The flow of electrons through a wire or any conductor.ii. Electrode : A metallic rod / bar/ strip which conducts electrons into

and out of a solution.iii. Anode : The electrode at which oxidation occurs. It sends electrons

into the outer circuit. It has negative charge and is shown as (–) in cell diagrams.

iv. Cathode : The electrode at which electrons are received from the outer circuit. It has a positive charge and is known as (+) in the cell diagrams.

v. Electrolyte is the salt solution in a cell.vi. Anode compartment : The compartment of the cell in which

oxidation half-reaction occurs. It contains the anode.vii. Cathode compartment : The compartment of the cell in which

reduction half-reaction occurs. It contains the cathode.viii. Half-cell : Each half of an electrochemical cell, where oxidation

occurs and the half where reduction occurs, is called the half cell.31. An organic compound with formula C3H8O3 is glycerol. It is a trihydric alcohol

and it contains 3-OH groups. When it is acetylated, with acetic anhydride, glyceryl triacetate C9H14O6 is formed. So the compound is glycerol.

Glycerol

Acetic Anhydride

Glyceryl triacetate

CH2OH

CHOH + 3(CH3CO)2O

CH2OH

CH2O−COCH3

C HO−COCH3 + 3CH3COOH

CH2O-COCH3

32.

Aromatic ether Aliphatic ethersi. Comparatively high boiling liquid. Volatile liquid.ii. Used in perfumery. Used as anesthetic.iii. Not used as solvent. Used as a solvent.iv. Can not be used as a substitute

for petrol.Mixed with alcohol, used as a substitute for petrol.

v. On heating with HI forms phenol and CH3I only.

It forms C2H5OH an C2H5I

vi. With nitrating mixture forms nitro anisoles.

Nitration does not take place.

vii. Does not form peroxide easily. Forms peroxide in air.

33. This reaction is catalysed by base. The carbanion generated is nucleophilic in nature. Hence it can bring about nucleophilic attack on carbonyl groupStep 1 : The carbanion is formed as the α-hydrogen atom is removed as a proton by the base.

OH– –CH3 – C = O CH2 – C = O + H2O| |

H H

Step 2 : The carbanion attacks the carbonyl carbon of another unionised aldehyde molecule.

–CH3 – C = O CH3– C – O

–CH2 – CHO CH2 – CHO

alkoxide ion

H H

Step 3 : The alkoxide ion formed is protonated by water to give ‘aldol’.H

H – OH

H| – | –

CH3 – C – O CH3 – C – OH + OH||

CH2CHO CH2CHO

Aldol

Section - iV34. (i) Atomic Oxygen (Z = 8) No. of electrons = 8 Electronic configuration = 1s2 2s2 2p4

Molecular Oxygen (O2) No. of electrons = 16. Its electronic configuration is, (s1s)

2 (s*1s)

2 (s2s)2 (s*

2s)2 (s2pz)

2 (p2px)2 = (p2py)

2 (p*2px)

1 = (p*2py)

1

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4 ___________________________________________________________ Sura’s XII Std - Govt. Model Question Papers

∴ Bond order = N Nb a−

= − =2

8 42

2

Magnetic property : Para magnetic - consist unpaired electrons Nature of Bond : Double bond.

2p 2p

O O2

O

Energ

y

2s 2s

σ 2s

Atomic

orbitals

Molecular

orbitals

Atomic

orbitals

σ∗2s

σ2pz

π π2py2px

π∗2px

π∗2py

σ∗2pz

(ii)

Electronegativity Nature of A – B Bond Repre-sentation Example

XA = XBXA – XB = 0

Non-polar covalent bond A – B H – H

XA > XBXA – XB is small

Polar covalent bond Aδ– – Bδ+ Oδ– – Hδ+ in H2O

XA XBXA – XB is large

Ionic bond A– – B+ Na+ Cl–

(or)(i) Consequences :

i. According to Fajan’s rule, decrease in size of ions increase the covalent character and decreases the basic character.

La3+ > Ce3+ ...>Lu3+

ii. There is regular decrease in their ionic radii.iii. Regular decrease in their tendency to act as reducing agent. iv. 2nd and 3rd rows of d-block elements are quite close in properties.v. These elements occur together in natural minerals and are difficult

to separate.Lanthanide : i. The Lanthanides form ions in oxidation state +3 which is more

stable. It is due to the fact that the ionisation and hydration energies are such that the tripositive state is more stable in aqueous solutions and in their solid compounds. In +3 oxidation state, most of the lanthanides are more stable due to f 0f 7 and f14 electronic configuration.

ii. Ce, Pr, Nd exhibit +2 and +4 states as uncommon oxidation states, because of stable f 0, f 1 and f 14 configuration.

(ii)

[Ni (CN)42-]

xx

xx

xx

dsp hybridisation2

xx

CN– CN– CN– CN–

3d 4s 4p

[Ni (Cl)42−]

xx

xx

xx

sp hybridisation3

Cl− Cl− Cl− Cl− Cl− Cl−

3d 4s 4pxx

♦ Since [Ni(CN)4]2- is paramagnetic, it must adopt tetrahedral geometry because it is the only way it could have unpaired electrons.

♦ Since [Ni(Cl)4]2- is diamagnetic, it must adopt square planar

geometry because it is the only way it could have no unpaired electrons.

35. (i) The important ores of gold is alluvial sand (or) gravel.

Mac Arthur Forrest Cyanide process : i. Concentration : Froth flotation process. ii. Roasting : The concentrated ore is roasted in excess of air when

impurities of S, As and Te are oxidised and escape as their volatile oxides.

iii. Treatment with KCN : The roasted ore is taken in large vats is then treated with a dilute (0.5%) with KCN excess of air for 24 hours. Gold dissolves in KCN to form a soluble complex called potassium aurocyanide.

4Au + 8KCN + 2H2O + O2 → 4K[Au(CN)2] + 4KOH Pot.aurocyanide iv. Precipitation of gold : The filtrate is treated with zinc dust. Gold

being less electropositive than zinc, it gets precipitated. 2K[Au (CN)2] + Zn → K2[Zn(CN)4] + 2Au ↓ Potassium tetracyano zincate (II) Gold v. Electrolytic refining :

♦ Anode is impure gold ♦ Cathode is pure gold ♦ Electrolyte is auric chloride (AuCl3) + 10-12% HCl ♦ On passing electric current, pure gold gets deposited on

the cathode.(ii) When heated in air at 773 K, zinc burns to form a white cloud of zinc

oxide which settles to form a woolly flock called philosopher’s wool.2Zn + O2

773 K → 2ZnO(or)

(i) Schottky defect :i. This defect arises when lattice points are unoccupied. ii. The unoccupied points are called lattice vacancies. iii. Crystal remains neutral. (Missing no. of cation = Missing no.

of anion)iv. Size of anion = cation v. Eg: NaCl.

Na missing+ Cl missing–

Schottky Defects in a Crystal

(ii) i. Entropy is a measure of randomness or disorder of the molecules of a system and it is a thermodynamics state function.

ii. The entropy function ‘S’ represents the ratio of heat involved (q) to the temperature (T) of the process.

S = qT

Units of entropy : i.e. (i) calories per degree per mole. (ii) eu per mole. (iii) cgs unit of entropy = cal K–1 (or) eu. (iv) SI unit is JK–1 (or) Eu.36. (i) Haber’s Process : N2(g) + 3H2(g) � ⇀�↽ �� 2NH3(g) ∆Hºf = –22 kcal/mol Temperature : 500 - 550º C Pressure : 300 - 500 atm. Catalyst : Iron (Fe) Ratio of N2 : H2 – 1 : 3 Formation of NH3 is exothermic process.(ii) i. Consecutive reactions : The reactions in which the reactant forms an intermediate and the

intermediate forms the product in one or many subsequent reactions are called as consecutive or sequential reactions. In such reactions, the product is not formed directly from the reactant.

Eg: Saponification of a diester in the presence of an alkali.

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R'OOC – (CH2)n – COOR k1 → R'OOC – (CH2)n – COOH Reactant Intermediate k2 → HOOC – (CH2)n – COOH

product ii. Parallel reactions : The reactions in which one or more

reactants react simultaneously in two or more pathways to give two or more products are known as parallel (or) side reactions.

Eg: Bromination of bromo benzene(or)

(i) i. By exchange of solvent : If a solution of sulphur or phosphorus in alcohol is poured into water, a colloidal solution of sulphur or phosphours is obtained due to low solubility in water.

ii. By change of physical state : Colloidal solutions of certain elements such as mercury and sulphur are obtained by passing their vapour through cold water containing a stabiliser.

iii. Chemical methods : It involves chemical reaction in a medium in which the dispersed phase is sparingly soluble. Some of the methods are:

♦ Double decomposition : An arsenic sulphide sol is prepared by passing H2S gas through a cold solution of arsenious oxide.

As O H S As S H O2 3 2 2 3 23 3+ + → Excess H2S is removed by passing in a stream of H2.

♦ Oxidation : A collodial solution of sulphur is obtained by passing H2S into a solution of sulphur dioxide.

2 2 32 2 2H S SO H O S+ + → ♦ Reduction: Silver and gold sols can be obtained by treating

dilute solution of silver nitrate or gold chloride with organic reducing agent like tannic acid or HCHO. AgNO Ag3 + Tannic acid → SolAuCl Au Sol3 + Tannic acid →

♦ Hydrolysis : Collodial solutions of the hydroxides of Fe, Cr, Al, etc., can be prepared by hydrolysis of their salts. A collodial solution of ferric hydroxide is obtained by boiling a dilute solution of ferric chloride.

FeCl H O Fe OH HCl3 2 33 3+ + → ( )

(ii) Given : Amount of the substance deposited, m = 150 g current strength = 1=10 ampere Time = t = 1sec. By Faraday’s first law, Formula : m = Zlt

Solution : ∴ = =×

Z mIt

15010 137.

(i) i. Tartaric acid contains two chiral carbon atoms. ii. Three isomeric forms are possible for tartaric acid.

COOH

OHH

COOH

HHO

COOH

HHO

COOH

OHH

d, l -pair of tartaric acid

H C* OH

COOH

COOH

C*HO H

C* Chiral carbon

iii. In the d-isomer, each of the two asymmetric carbon atoms rotate the plane polarised light towards right leading to overall dextro rotation.

o-dibromobenzene

p-dibromobenzeneBr

Br

+ HBr

+ HBrBr

Br

Br

Br2

iv. In the l-isomer, each of the two asymmetric carbon atoms rotate the plane polarised light towards left leading to overall laevo rotation

v. When equal amounts of d-tartaric acid and l-tartaric acid are mixed, the resulting mixture is known as racemic mixture which becomes optically inactive due to external compensation.

vi. One asymmetric carbon atom is dextro rotatory and the other is laevo rotatory both rotating to the same extent in opposite direction.

♦ The isomer is optically inactive and is called meso isomer.vii. Meso isomer becomes optically inactive due to internal

compensation. ♦ It has a symmetric plane and the molecule becomes achiral. ♦ Meso form cannot be separated because this form is a single

substance and not a mixture. ♦ Meso form is said to be a diastereomer.

(ii) CH

CH

CH OH

CH ClHOCl Ca OH2

2

2

2

2

→−

→( )

chloro hydrin (A)

Ethylene

CH CH CaCl H O

OEthylene

2 2 2 2− + +

oxide

(B)

(or)

(i)

HCHO CH MgI CH CH OH Mg IOH

H OH

Formaldehyde

+ → +−3 3 2

Ethanol

6 4 62 3 2 6 4 2CH O NH CH N H OFormaldehyde

+ → ( ) + Hexamethylene

tetramine

(ii) Acid strength increases with increase in electronegativity of substituents.

2

OCI CH C

O H← ←

← This effect weaknens the –O–H bond there by facilitating the removal

of H+ from –OH group. This acid becomes stonger than formic acid.38. (i) (i) C H NH HNO C H OH N H O

A2 5 2 2 2 5 2 2+ → + +

( ) (B)

(ii) C H OH CH C

O

O C

O

CH C H C

O

O CH CH COOH2 5 3 3 2 5 3 3− + − − − − → − − − +� � �

( )C

C H OH CH C

O

O C

O

CH C H C

O

O CH CH COOH2 5 3 3 2 5 3 3− + − − − − → − − − +� � �

( )C

A C2H5NH2 Ethyl amine

B C2H5OH EthanolC C2H5COOCH3 Ethyl acetate

(ii) 2C6H5NH2 + 2

C H6 5NH

s-Diphenyl thio urea

C=S+H2SC6H5NH

CS

(or)(i) i. Glucose HI P CH CH CH CH CH/ → − − − −3 2 2 2 3

n hexane− This indicates that 6-carbon atoms are linked in a consecutive

unbranched chain. ii. Glucose + acetic anhydride Pyridine → Penta acetate It indicates the presence of 5–OH groups in a glucose molecule.(ii) Preparation :

♦ By condensing adipic acid with hexamethylene diammine with elimination of water molecule, Nylon 66 is obtained.

♦ Chain length depends upon the time and temperature.

n nH2N (CH2)6 NH2 + HO–C–(CH2)4–C–OH – –N–(CH2)6–N–C–(CH2)4–C –n

Hexamethylenediamine Adipic acid Nylon-66

O O H H O O

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GOVT. MODEL QUESTION PAPER - 212th

STD.Time Allowed : 2.30 hours CHEMISTRY mArks : 70

Note : Draw diagrams and write equations wherever necessary.

Section - iNote : (i) Answer all the questions: (15 × 1 = 15) (ii) Choose the most suitable answer :1. Consider the following statements.

(i) Alkali metals have highest ionization potential.(ii) f- electron has the minimum screening effect.(iii) Electron-affinity is inversely proportional to its size.Which of the following statements is/are correct?(a) I and III (b) I and II(c) I,II and III (d) II and III

2. The compound with garlic odour is(a) P2 O3 (b) P2 O5 (c) H3 PO3 (d) H3 PO4

3. Match the List-I and List- II correctly by using the code given below:List - I List - II

(A) Brass (1) Surgical instruments(B) Stellite (2) Skin Ointment(C) Lunar caustic (3) Condenser tube(D) Calamine (4) Hair dyes

Codes: (A) (B) (C) (D)

(a) (2) (4) (1) (3)(b) (3) (1) (4) (2)(c) (1) (4) (3) (2)(d) (4) (3) (2) (1)

4. The co-ordination number of Ni(II) in [Ni(CN)4]2– is

(a) 2 (b) 4 (c) 5 (d) 65. The t1/2 of a particular decay is 69.32 years. The decay constant is

(a) 100 per year (b) 10 per year(c) 0.01 per year (d) 0.1 per year

6. The network done by the system is given by(a) w – PΔV (b) w + PΔV (c) –w + PΔV (d) –w–PΔV

7. If ΔG for a reaction is negative, the change is(a) Spontaneous (b) Non-Spontaneous(c) Reversible (d) Equilibrium

8. The sum of the powers of the concentration terms that occur in the rate equation is called(a) molecularity (b) order (c) rate (d) rate constant

9. Haze is an example of(a) solid dispersed in gas (b) gas dispersed in liquid(c) gas dispersed in gas (d) solid dispersed in liquid

10. Three juice samples A,B,C have the pH value as shown. Then the correct order of their [H+] ion concentration is

JUICE(A)

pH = 1

JUICE(B)

pH = 2

JUICE(C)

pH = 3(a) A > B > C (b) A = B = C(c) A < B < C (d) A > B = C

11. When phenol is distilled with zinc dust it give(a) benzaldehyde (b) benzoic acid(c) toluene (d) benzene

12. Diethylether and methyl n-propyl ether are(a) functional isomers (b) optical isomers(c) metamers (d) tautomers

13. From which of the following, tertiary butyl alcohol is obtained by the action of methyl magnesium iodide?(a) HCHO (b) CH3 CHO (c) CH3 COCH3 (d) CO2

14. Ethane can be prepared from sodium acetate by(a) clemmenson reduction (b) hydrogenation(c) decarboxylation (d) Kolbe’s electrolysis

15. Statement (I) : Trimethylamine is less basic than dimethylamine.Statement(II) : Trimethylamine has steric over crowding of methyl group than dimethylanine.(a) Statement (I) is correct but statement (II) is false(b) Statement (I) and (II) are correct and Statement (II) is the correct

explanation of Statement (I)(c) Statement (I) is false but Statement (II) is correct.(d) Statement (I) and (II) are correct and Statement (II) is not correct

explanation of Statement(I).

Section - iiAnswer any six questions and question number 20 is compulsory: (6 × 2 = 12)16. He2 molecule is not formed, why?17. Lead pipes are not used in supplying drinking water why?18. Identify the central metal ion and ligand in the given complex [Cr (en)3]Cl3.19. State Bragg’s law.20. Consider the potential energy diagrams of reaction I and II predict which

reaction will go faster and why?

Energy Energy

Reaction coordinateReaction coordinate

(I) (II)

21. Define the following terms anode and cathode in an electrochemical cell.22. Reason out cis isomer is less stable than trans isomer.23. What is isoelectric point?24. What are analgesis? Give an example.

Section - iiiAnswer any six questions and questions number 25 is compulsory : (6 × 3 = 18)25. Arrange the following as directed.

(a) Increasing order of size: O2–, F–, Mg2+, Na+

(b) Increasing order of first ionization potential: Li, Be, B(c) Increasing order of size: Fe2+, Fe, Fe3+

26. Complete following:(i) Zn + HNO3 (dil) → (ii) CuSO4 + KCN →

[6]

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Sura’s XII Std - Govt. Model Question Papers ___________________________________________________________ 7

27. Distinguish between lanthanides and actinides.28. Explain the nature of glass.29. Write notes on ultrafiltration.30. Calculate the pH of 0.1m CH3COOH solution. Dissociation constant of

acetic acid is 1.8 × 10–5 M.31. How does ethyleneglycol react with

(a) con. H3 PO4(b) anhydrous zinc chloride.

32. Give the methods of preparation of anisole.33. Explain the mechanism of Claisen Schmidt reaction.

Section - iVAnswer all the questions : (5 × 5 = 25)34. (i) Derive de-Broglie relation. (3)

(ii) Electron affinity of noble gases are zero why? (2)(or)

(i) How are noble gases separated from air by Ramsay Raleigh method.(ii) Comparing La(OH)3 and Lu(OH)3, which is more basic and why?

(2)35. (i) Explain the extraction of silver from its ore. (3)

(ii) Account for the following “transition elements form complexes”. (2)

(or)(i) Write the postulates of Werner’s theory (3)(ii) Define Q value of a nuclear reaction.

36. (i) Write the properties of ionic crystal. (2)(ii) Explain the characteristic of entropy. (3)

(or)(i) Explain the effect of temperatures on the following equilibrium

reaction. (2)N2O4(g) � ⇀�↽ �� 2NO2(g) ΔH = + 59.0 kJ/mo1

(ii) Show that the time taken for 99.9% completion of first order reaction is 10 times its half life period. (3)

37. (i) Discuss the quinonoid theory of indicators. (3)(ii) Which of the following reaction will occur and why? (2) Zn + 2H+ (dil. H2SO4) → Zn2+ + H2 Cu + 2H+ (dil. H2 SO4) → Cu2+ + H2

(or)(i) Distinguish between racemic and meso form. (3)(ii) How will you convert (a) Chlorobenzene to phenol (b) Phenol to benzene. (3)

38. (i) Explain the reaction of lactic acid with (a) dil H2SO4 (b) PCI5 (3)(ii) Explain Gabriel pthalimide synthesis. (3)

(or)(i) Explain briefly the characteristics of rocket propellant. (3)(ii) Give any two functions of lipids in biological system. (2)

ANSWERSSection - i

1. (d) II and III2. (a) P2 O33. (b) (3) (1) (4) (2)4. (b) 45. (c) 0.01 per year6. (a) w – PΔV7. (a) Spontaneous8. (b) Order 9. (a) Solid dispersed in gas 10. (a) A > B > C11. (d) Benzene12. (c) Metamers

13. (c) CH3 COCH314. (d) Kolbe’s electrolysis15. (d) Statement (I) and (II) are correct and Statement (II) is not correct

explanation of Statement(I).

Section - ii16. i. The electronic configuration of He (Z=2) in the ground state is 1s2. ii. As each helium atom contains two electrons, there will be four electrons

in He2 molecule. Keeping in view the Aufbau principle and Pauli’s exclusion principle its electronic configuration would be as follows:

iii. He2 : (s1s)2 (s*

1s)2

Nb = 2 and Na = 2

∴ =−Bond N N order b a

2 = 2 2

2−

As the bond order of He2 come out to be zero, this molecule does not exist.

17. Lead is not atlacked by pure water in the absence of air. But water containing dissolved air has a solvent action on it due to the formation of Pb(OH)2 a poisonous substance called plumbo solvence. Hence lead pipes are not used for the supply of drinking water.2Pb + O2 + 2H2O → 2Pb(OH)2

18. [Cr (en)3]Cl3 :Central metal ion : ChromiumLigand : Ethylene diammine

19. The fundamental equation that gives a simple relation between the wavelength of the X-rays, the interplanar distance in the crystal and the angle of reflection is known as Bragg’s equation.

nλ = 2d sinθ

where n is the order of reflection λ is the wavelength of X-rays d is the interplanar distance in the crystal θ is the angle of reflection20. i. k = Ae–Ea/RT

ii. Ea (Activation Energy) = Threshold Energy – Energy of colloiding molecules

iii. Ea is a characteristic value of a reactioniv. Higher the value of Ea, slower is the rate of reactionv. Hence II will occur faster.

21. i. Anode : The electrode at which oxidation occurs. It sends electrons into the outer circuit. It has negative charge and is shown as (–) in cell diagrams.

ii. Cathode : The electrode at which electrons are received from the outer circuit. It has a positive charge and is known as (+) in the cell diagrams.

22. i. In the cis-isomer similar groups are very near to each other, Vander Waals repulsion and steric hindrance make the molecule much unstable.

ii. In the trans isomer, similar groups are diagonally opposite to each other. Hence there is no such steric interaction.

iii. Generally trans isomer is more stable than cis isomer.23. The pH at which the amino acid shows no tendency to migrate when

placed in an electric field is known as iso electronic point.24. i. Analgesics are the compounds which relieve all sorts of pains

without the loss of consciousness. ii. These are also called as pain killer or pain relievers. iii. These are effective in headaches, myalgia and arthalgia. Eg: Aspirin, Novalgin.

Section - iii25. (a) Mg2+ < Na+ < F– < O2–

(b) Li < Be < B(c) Fe3+ < Fe2+ < Fe

26. i. 4Zn + 10HNO3 (dil) → 4Zn(NO3)2 + N2O + 5H2O Nitrous oxideii. CuSO4 + 2KCN → Cu(CN)2 + K2SO4 Cupric cyanide

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27.

S.No. Lanthanides Actinides(i) Binding energies of 4f electrons

are higher.Binding energies of 5f electrons are lower.

(ii) Maximum oxidation state exhibited by lanthanides is +4. Eg: Ce+4

Due to lower binding energies they show higher oxidation states such as +4, +5 and +6 in UF6 and UO2Cl2 .

(iii) 4f electrons have greater shielding effect.

5f electrons have poor shielding effect.

(iv) Most of their ions are colourless. Most of their ions are coloured U3+ (red), U4+ (green) and UO2

2+ (yellow).

(v) They do not form complexes easily.

They have much greater tendency to form complexes.

(vi) Except promethium they are non-radioactive.

All of them are radioactive.

(vii) Their compounds are less basic. Their compounds are more basic.

28. Glasses : i. When certain liquids are cooled rapidly. ii. There is no formation of crystals at a definite temperature such as occurs

on slow cooling. iii. The viscosity of the liquid increase steadily finally a glassy substance

is formed. Characteristics : i. Hardness, rigidity and ability to withstand shearing stress.ii. Optically isotropic. iii. At high temperature glasses undergo phase transition. Glasses are

regarded as amorphous solid or super cooled liquid. iv. Vitreous state - substance exist lying between solid and liquid states.

29. Ultrafiltration : ♦ Sols pass through an ordinary filter paper. Its pores are too large to

retain the collodial particles. ♦ If the filter paper is impregnated with colloidion or a regenerated

cellulose (cellophane), the pore size is much reduced. Such a modified filter paper is called an ultrafilter.

♦ The separation of the sol particles from the liquid medium and electrolytes by filtration through an ultrafilter is called ultrafiltration.

♦ It is a slow process. The colloidal particles are left on the ultra filter in the form of slime, which is stirred into fresh medium to get back the pure sol.

♦ Ultrafiltration can be employed to separate sol particles of different sizes.

30. For weak acid [H+] = K Ca ×

= 1 8 10 0 1 1 8 105 6. . .× × = ×− −

∴ pH = –log [H+] = log (1.34 × 10–3)∴ pH = 2.87

31. (a) HO – CH2 – CH2 – O H H3PO4HO – CH2 – CH2

OHO – CH2 – CH2 – OH

∆HO – CH2 – CH2

Ethylene glycol Diethylene glycol(b) H

H C OH

H C OH

H

|– –|

– –|

ZnClH O

2

2–

H

C O H

CH

|– –

||2

H

C = O

CH

|

|3

Acetaldehyde

32. (i) Williamson’s synthesis : Anisole is prepared by heating sodium phenoxide with methyl iodide. (C6H5O

– - nucleophile)

C H ONa CH I C H O CH Nal6 5 3 6 5 3+ → − − + Anisole

(ii) When phenol is treated with diazomethane, anisole is produced.

C H OH CH N C H O CH N6 5 2 2 6 5 3 2+ → − − + ↑ Diazomethane

(iii) When phenol is methylated by treating with dimethyl sulphate in the presence of NaOH anisole is manufactured.

C H OH CH SO C H OCH CH HSO6 5 3 2 4 6 5 3 3 4+ ( ) → +NaOH

33. It reacts with aliphatic aldehydes or ketones in presence of NaOH forming α,b unsaturated aldehyde or ketone.

Step 1 : NaOH –

CH3CHO CH2CHO

NaOH H+

C6H5CH = O C6H5 CH–O– C6H5CHOH(–)CH2CHO

| |CH2CHO CH2CHO

(Nucleophilic attack)

(Carbanion)

Step 2 :

NaOH –CH3CHO CH2CHO

NaOH H+

C6H5CH = O C6H5 CH–O– C6H5CHOH(–)CH2CHO

| |CH2CHO CH2CHO

(Nucleophilic attack)

(Carbanion)

Step 3 :

Similarly,NaOH

C6H5CHO + CH3COCH3 C6H5CH = CHCOCH3

C6H5 – CH– OH|CH – H|CHO

β-elimination–H2O

C6H5 – CH||CH – CHO

Cinnamaldehyde

♦ The carbanion formed from acetaldehyde or acetone brings forth a nucleophilic attack on the carbonyl carbon of benzaldehyde.

♦ The aldol type of product undergoes b-elimination (of water). ♦ This reaction is known as Claisen Schmidt reaction.

Section - iV34. (i)

According to Planck’s quantum theory,

According to Einstein equation,

E = hv ......(1)h = Planck’s constant v = frequency of the wave,

E = mc2 ......(2)m = mass of photon, c = velocity of light.

From equations (1) & (2), hv = mc2

h c.λ

= mc2 [∴v = cλ

]

∴ λ = hmc

λ = hmv

or

λ = hp

[mv = p]

(ii) i. As noble gases possess ns2np6 configuration in their valence shells, these are stablest atoms and there are no chances for the addition of an extra electron.

ii. Thus, the electron affinities of noble gases are zero.(or)

(i) Ramsay-Raleigh’s method : i. A mixture of air and oxygen is constantly admitted into a glass

globe of about 50 litres capacity. ii. Two platinum electrodes are introduced and a discharge from a

transformer of about 6000-8000 volts is passed. iii. Nitrogen and oxygen rapidly combine to form oxides of nitrogen. iv. The oxides are dissolved out in a solution of sodium hydroxide are

continuously circulated through the flask.

N2 + O2 → 2NO

2NO + O2 → 2NO2

2NO2 + 2NaOH → NaNO3 + NaNO2 + H2O

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v. Oxygen if any, is removed by introducing alkaline pyrogallol in the globe.

(ii) i. Due to lanthanide contraction the size of La3+ ions decreases regularly with increase in atomic number.

ii. According to Fajan’s rule, decrease in size of Ln3+ ions increase the covalent character and decreases the basic character between Ln3+ & OH– ion in Ln(OH)3.

iii. Since the order of size Ln3+ ions are La3+ > Ce3+ ....>Lu3+

iv. Hence La(OH)3 is the strongest base while Lu(OH)3 is the weakest base.

35. (i) The important ores of silver are : (a) Argentite Ag2S (b) Horn silver AgCl (c) Ruby silver Ag2S.Sb2S3 Extraction of silver from argentite (Ag2S) : Mac Arthur Forrest Cyanide Process : i. Concentration : Froth flotation process. ii. Treatment with NaCN: The concentrated ore is treated with a

solution of 0.4–0.6% sodium cyanide for several hours. The ore is converted into soluble sodium argento cyanide complex.

Ag2S + 4NaCN � ⇀�↽ �� 2Na[Ag(CN)2] + Na2S Sod.argentocyanide iii. Precipitation of silver : The solution of sodium argento cyanide

is filtered and the filtrate is treated with zinc dust, silver gets precipitated.

2Na [Ag(CN)2] + Zn → Na2 [Zn(CN)4]+2Ag ↓ iv. Electrolytic refining : The impure silver is further purified by

electrolytic refining. ♦ Anode - impure silver ♦ Cathode - pure silver ♦ Electrolyte - Silver nitrate solution + 1% nitric acid. ♦ On passing electric current, pure silver gets deposited at

the cathode.(ii) Transition elements have strong tendency to form complexes because of

two reasons. They are, i. Small size and high positive charge density. ii. Presence of vacant (n–1) d orbitals which are of appropriate energy

to accept lone pair and unshared pair of electrons from the ligands for bonding with them.

(or)(i) i. Every metal atom has two types of valencies

♦ Primary valency or ionisable valency ♦ Secondary valency or non ionisable valency

ii. The primary valency corresponds to the oxidation state of the metal ion. It is always satisfied by negative ions.

iii. Secondary valency corresponds to the coordination number of the metal ion or atom. It is satisfied by either negative ions or neutral molecules.

iv. The molecules or ions that satisfy secondary valencies are called ligands.

v. The ligands which satisfy secondary valencies must project in definite directions in space. So the secondary valencies are directional in nature whereas the primary valencies are non-directional in nature.

vi. The ligands have unshared pair of electrons. These unshared pair of electrons are donated to central metal ion or atom in a compound. Such compounds are called coordination compounds.

NH3

NH3

NH3

NH3

Cl

ClCoCl

H3N

H3N

Werner’s representation Eg : [Co(NH3)6]Cl3 Cl : primary valency (dotted lines) NH3 : secondary valency (solid lines). Defects of Werner’s theory Werner’s theory describes the structures of many co-ordination

compounds successfully. However, it does not explain the magnetic and spectral properties.

(ii) The amount of energy absorbed or released during the nuclear reaction is called Q-value of nuclear reaction.

Qvalue = (mp – mr) 931 MeV where, mp = sum of the masses of products mr = sum of the masses of reactants.36. (i) i. The heat of vapourisation is high.

ii. The vapour pressure of ionic crystals at ordinary temperature are very low.

iii. Their melting and boiling points are very high.iv. They are hard and brittle.v. They are insulators in the solid state.vi. Ionic crystals are soluble in water and also in other polar solvents.vii. Ionic solids are good conductors when dissolved in water.

(ii) i. Entropy ‘S’ is a thermodynamic state function (formulated from II law of thermodynamics).

ii. The entropy depends on quantity of the substance. iii. Entropy is a measure of randomness of the molecule of a system.

Entropy increases in all spontaneous process. iv. The energy of the universe remains constant although the entropy of

the universe tends to be maximum.

v. Spontaneous Equilibrium Non-spontaneous∆S > 0 ∆S = 0 ∆S < 0

vi. Units

Cgs SI MolesCal K–1 (or) eu JK–1 (or) EU Cal deg–1 mol–1

I eu = 4.184 EU.(or)

(i) Consider the equilibrium in the gaseous state such as N2O4(g) � ⇀�↽ �� 2NO2(g)

♦ Increase in pressure, decrease the volume (i.e.) reverse reaction is favoured more amount of N2O4 is formed.

♦ When pressure is decreased forward reaction is favoured. To get more product (NO2), pressure should be low.

(ii) Given data : Time required for 99.9% completion is about 10 times its half life period.

Prove that for a first order reaction the time required for 99.9% completion is about 10 times its half life period.

Formula :

k = 2 303. log

ta

a x−; t

= 0 693.k

Solution :

t99.9 = 2 303 100

100 99 9. log

.k −

= 2 303 100

0 1. log

.k =

2 303 1000. logk

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10 __________________________________________________________ Sura’s XII Std - Govt. Model Question Papers

t99.9 = 2 303 3. ×

k

t99.9 = 6 909.

k ; t1/2 =

0 693.k

∴ =kt

0 693

1 2

.

/

tt99 9

1 2

. % = 6 909

0 693.

.kk× ;

t99.9 = 6 9090 693 1 2.. /× t

t99.9 = 10 × t1/2 Time required for 99.9% completion of the reaction = 10 t1/2 37. (i) Quinonoid Theory : i. The colour change of an acid-base indicator arises as a result of

structural change. ii. An indicator exists as an equilibrium mixture of two tautomeric

forms namely, benzenoid and quinonoid forms.

benzenoid form quinonoid form iii. The two forms possess two different colours. iv. The colour change is due to the fact that one tautomer changes

over to the other. v. In acidic medium - Benzenoid form - Colourless In Basic medium - Quinonoid form - Pink colour. Eg: phenolphthalein is tautomeric mixture of the two forms.

O||C

C

OH

OH

O

O||C

C

O–

OH

OOH–

H+

Benzenoid form, colourless,exists predominantly in acidic

medium

O||C

C

OH

O

O

–

Quinonoid form, pink,exists predominantin basic medium

(ii) Zn + 2H+ (dil. H2SO4) → Zn2+ + H2 → I Cu + 2H+ (dil. H2 SO4) → Cu2+ + H2 → II

♦ Among these reactions, only reactions I will occur. ♦ Reason : Dilute acids such as H2SO4 have no action on copper in

the absence of air or an oxidising agent. ♦ Cu dissolves in acid in the presence of air. ♦ Pure Zinc is not attacked by dilute acids. However impure zinc

reacts with diluite acid with the liberation of H2. 2Cu + 2H2SO4 + O2 (air) → 2CuSO4 + 2H2O Zn + H2SO4 → ZnSO4 + H2 ↑

(or)(i)

Racemic form Meso formi. It is a mixture that can be

separated into two optically active forms

It is a single compound and hence cannot be separated.

ii. Optically inactive due to external compensation

Optically inactive due to internal compensation.

iii. Molecules of isomers present are chiral

Molecules are achiral.

iv. Mixture conains equal amount of d-isomer and l-isomer.

It is a single compound.

v. Eg: Equal amount of (+) & (–) tartaric acid

Eg: Meso tartaric acid.

(ii) (a) Chlorobenzene to phenol :

C H Cl NaOH C H OH NaClK6 5 6 5

623300

+ → + atm

Phenol

(b) Phenol to Benzene :

C H OH Zn C H ZnOPhenol6 5 6 6+ → +

Benzeene

38. (i) (a) dil H2SO4 : When lactic acid is treated with dilute sulphuric

acid, it dissociates to give acetaldehyde and formic acid.

CH CH OH COOH CH CHO HCOOHdil H SO

Lactic3 3

2 4− − → +( ) .

aciid Acetaldehhyde Formic acid

(b) PCl5 : With PCl5 it forms lactyl chloride.

CH CH OH COOH PC CH CH COC3 5 3( ) + → − −l l

|

Cl

chloridelactyl

(ii) i. This involves the treatment of phthalimide with potassium hydroxide to form potassium salt.

ii. The salt is then heated with an alkyl halide to give N-alkyl phtalimide which in turn reacts with potassium hydroxide to form potassium phthalate salt and a pure primary amine

Phthalimide

CO

CO

NH KOH

Potassium Phthalimide

CO

CO

NK RX aqueous KOH

CO

CO

N-R

Potassium Phthalate

COOK

COOK

+ RNH2Primary amine

N-alkyl Phthalimide

Phthalimide

CO

CO

NH KOH

Potassium Phthalimide

CO

CO

NK RX aqueous KOH

CO

CO

N-R

Potassium Phthalate

COOK

COOK

+ RNH2Primary amine

N-alkyl Phthalimide

(or)(i) i. Rocket motors are used both in space vehicles and in offensive

weapons such as missiles. ii. The propulsion system in most space vehicles consists of rocket

engines powered by chemical propellants. These are called rocket propellants.

iii. Propellants are combustible compounds which on ignition undergo rapid combustion to release large quantities of hot gases. A propellant is a combination of an oxidiser and a fuel.

iv. Working of propellant : When a propellant is ignited, it burns to produce a large quantity of gases. These gases then come out through the nozzle of the rocket motor.

v. The passage of gasses through the nozzle of the rocket motor, provides the necessary thrust for the rocket to move forward according to the Newton’s third law of motion.

Eg: Hydrazine, liquid hydrogen, polyurethane, etc.(ii) i. Wax : It act as a protective agent on the surfaces of animals and

plants. ♦ Wax coating on the surface of plants and fruits protects them

from excessive loss of moisture and becoming infected with fungi and bacteria.

ii. Phospholipids : ♦ Lecithins and cephalins are important lipids. ♦ Lecithins in liver used in organisation of cell structure. ♦ Chephalins in brain used for blood coagulation.

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s.comNote : Draw diagrams and write equations wherever

necessary.Section - i

Note : (i) Answer all the questions: (15 × 1 = 15) (ii) Choose the most suitable answer from the given

four alternatives :1. H3PO3 is a powerful reducing agent because it has

(a) O-H bond (b) P-O bond (c) O- P bond (d) P-H bond2. Paramagnetism is the property of

(a) Paired electrons (b) Completely filled electronic subshells.(c) Unpaired electrons (d) Completely Vacant electronic subshells.

3. Statement (I) : The size of M3+ ions decreases as we move through the lanthanides.Statement (II): One 4f electron show perfect shielding by another in the same subshell.(a) Statement (I) is correct but Statement (II) is false.(b) Statement (I) and (II) are correct and Statement (II) is the correct

explanation of Statement (I)(c) Statement (I) is false but Statement (II) is correct(d) Statement (I) and (II) are correct and Statement (II) is not correct

explanation of Statement(I)4. The geometry of complex ion [Fe(CN)6]

4– is(a) Tetrahedral (b) Square planar(c) Octahedral (d) Triangular

5. Fill in the blank

1123

1223

01Na Mg n+ → +?

(a) ∝ (b) d (c) p (d) n6. The enthalpy of vapourization of a liquid is 30 kJmol–1 and entropy of

vapouriztion is 75 J mol–1 K–1 its boiling point is(a) 600K (b) 500K (c) 400K (d) 300K

7. In the reversible reaction 2HI � ⇀�↽ �� H2 + I2, Kp is(a) greater than Kc (b) less than Kc(c) Equal to Kc (d) Zero

8. NH4OH is a weak base because(a) it has low vapour pressure. (b) it is only partially ionized.(c) it is completely ionized. (d) it has low density.

9. Consider the following statements.(I) Order of a reaction may be zero, fractional or integral values.(II) Order of a reaction can be determined theoretically.(III) Higher order reactions are not common.Which of the above statement/s is/are not correct?(a) I and III (b) I and II (c) I, II and III (d) II and III

10. Match the List-I and List- II correctly by using the code given below:

List - I List - II(A) Haber’s process (1) Cupric chloride(B) Contact Process (2) Ferric Oxide(C) Deacon’s process (3) Finely divided iron(D) Bosch’s process (4) Platinized asbestos

GOVT. MODEL QUESTION PAPER - 312th

STD.Time Allowed : 2.30 hours CHEMISTRY mArks : 70

Codes: (A) (B) (C) (D)

(a) (3) (4) (2) (1)(b) (3) (4) (1) (2)(c) (4) (3) (1) (2)(d) (2) (1) (4) (3)

11. A compound that undergoes bromination easily is(a) Benzoic acid (b) Benzene (c) Phenol (d) Toluene

12. Diethylether can be decomposed with(a) HI (b) KMnO4 (c) NaOH (d) H2O

13. Benzophenone does not form additional product with sodium bisulphite because(a) Steric hindrance of phenyl groups(b) Phenyl groups reduce the activity(c) Phenyl groups increase the activity.(d) Both a and b

14. The oil of winter green is(a) methyl acetate (b) methyl oxalate(c) methyl salicylate (d) methyl formate

15. Which one of the following is a tertiary amine?(a) Ethyl amine (b) Dimethyl amine(c) Tert-butyl amine (d) Trimethyl amine

Section - iiAnswer any six questions and question number 21 is compulsory: (6 × 2 = 12)16. State Heisenberg Uncertainty principle.17. Calculate the electro-negativity values of fluorine on Mulliken’s scale

given that (Ionization potential ) F = 17.4 ev/atom, (Electron affinity) F = 3.62 ev/atom.

18. What is the action of heat on copper sulphate crystals?19. Write a note on the assignment of atoms per unit cell in fcc.20. What is common ion effect? Give example.21. Determine the standard emf of the cell and predict its feasibility.

Ag, Ag+ ││ H+, H2(g) l atm, PtThe standard reduction potential of Ag+, Ag is 0.80V

22. How do you distinguish the three isomers of di-substituted benzene using DPM (Dipole moment value)?

23. Why sucrose is a non reducing sugar?24. What are food preservatives? Give example.

Section - iiiAnswer any six questions and questions number 31 is compulsory: (6 × 3 = 18)25. Mention the uses of helium.26. How lanthinides are extracted from Monazite sand?27. Explain co-ordination and ionization isomerism with suitable examples.28. Derive a general relationship between Kp and Kc for a equilibrium reaction.29. Distinguish between simple and complex reaction.30. Explain electro osmosis.31. Identify (B), (C) and (D)

O|| LiAlH4 SOCl 2 alc.KOHCH 3 – C – CH 3 (A) (B) (C) (D)

Acetone

32. Give the mechanism involved in the esterification of a carboxylic acid with alcohol.

[11]

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12 __________________________________________________________ Sura’s XII Std - Govt. Model Question Papers

33. How can the following conversion be effected?(a) Nitrobenzene to anisole(b) Aniline to iodobenzene.

Section - iVAnswer all the questions : (5 × 5 = 25)34. (i) Draw the MO diagram of N2 molecule and predict its bond order.

(3)(ii) How ionization energy is affected by atomic size and nuclear charge?

(3)(or)

(i) Discuss the chemistry behind Holme’s signal. (2)(ii) Explain the extraction of zinc from its ore. (3)

35. (i) Write the common and maximum oxidation state of lanthanides.(ii) Mention the function of haemoglobin.

(or)(i) What is spallation reaction? (2)(ii) Give the uses of radio active isotopes in medicine. (3)

36. (i) Explain Bragg’s Spectrometer method. (3)(ii) State Lechatelier’s principle. (2)

(or)(i) State various statements of II law of thermodynamics. (3)(ii) The initial rate of a first order reaction is 5.2 × 10–6 mol lit–1 S–1 at

298K. When the initial concentration of reactant is 2.6 × 10–3 mol lit–1. Calculate the first order rate constant of the reaction at same temperature. (2)

37. (i) Derive Henderson equation. (3)(ii) Using IUPAC convention write the cell diagram for zinc-copper

cell. (2)(or)

(i) Describe the conformations of cyclohexanol, comment on their stability. (3)

(ii) Give the possible ether isomers for molecular formula C4H10O. (2)

38. (i) An organic compound (A) of molecular formula C6H6O gives violet colour with neutral FeCl3 (A) gives maximum of two isomers (B) and (C) when an alkaline solution of (A) is refluxed with CCl4 (A) also reacts C6H5N2Cl to give compound (D) which is a red orange dye. Identify (A),(B),(C) and (D). Explain with suitable chemical reaction. (5)

(or)(i) How is the structure of glucose elucidated. (3)(ii) What are chromophores? Give examples. (2)

ANSWERSSection - i

1. (d) P-H bond2. (c) Unpaired electrons 3. (a) Statement (1) is correct but Statement (II) is false.4. (c) Octahedral5. (c) p6. (c) 400K7. (c) Equal to Kc8. (b) it is only partially ionized9. (d) II and III10. (b) (3) (4) (1) (2)11. (c) phenol12. (a) HI13. (a) Steric hindrance of phenyl groups14. (c) methyl salicylate15. (c) Tert-butyl amine

Section - ii16. Heisenberg’s uncertainty principle states that “it is impossible to measure

simultaneously both the position and velocity of a microscopic particle with absolute accuracy or certainty”.

∆x . ∆p ≥ h4π

∆x - uncertainty in position of particle; ∆p - uncertainty in position of particle

17. Electronegativity of fluorine

=( ) + ( )

×

= + = =

IP EAF F

2 2 817 4 3 62

5 621 025 06

3 75

.. .

..

..

18. On heating, CuSO4. 5H2O loses its water of crystallization and decomposes at 720oC to give cupric oxide and sulphur trioxide.CuSO4. 5H2O 100

4 2°

− →CH O

CuSO4. H2O 230

2

°− →CH O Blue

CuSO4 720° →C

CuO + SO3↑ White Cupic oxide

19. i. A face centred cubic lattice has 4 atoms per unit cell.ii. The lattice has 8 atoms in the corners (Nc) and 6 atoms on the faces

(Nf) (one on each face)

Contribution by atoms on the corners to unit cell = Nc

888

= = 1

Contribution by atoms on the faces to unit cell = Nf

262

= =3

\ The number of atoms per unit cell in fcc

= N Nc f

8 2+ = 8

862

+ = 1 + 3 = 4

20. i. The reduction of the degree of dissociation of a salt by the addition of a common ion is called common ion effect.

Eg: Silver chloride. AgCl(s) � ⇀�↽ �� Ag+

(aq) + CI– (aq)ii. When sodium chloride is added to the solution the concentration of

Cl– ions will increases. iii. The equilibrium shown above will be shifted to the left to form more

of solid AgCl. Thus, the solubility of AgCl will decrease.21. Ag, Ag+ ││ H+, H2(g)

E°cell = E EH H Ag Ag° − °+ +, ,2

= 0 – 0.80E°cell = – 0.80This reaction is not feasible.

22. i. Ortho, meta and para isomers differ mainly in physical properties like melting point, refractive index, solubility.

ii. Their behaviour under an applied electric field differs widely. This behaviour is measured by a quantity called ‘Dipole moment’.

iii. For disubstituted benzenes the magnitude of DPM depends on the angle between the two substituents.

Eg: Dichloro benzene.

= 60° = 120o = 180°θ

Cl

ClCl

Cl

Cl

Cl θ θ

♦ Their DPM, is in the order: Ortho isomer > Meta isomer > Para isomer

♦ For para dichloro benzene DPM = 0. ♦ Hence dipole moment measurement is the best method of

distinguishing the three isomers of disubstituted benzenes23. Sucrose C12H22O11 does not contain an aldehyde group, So it does not

reduce Tollen's reagent and Fehling's solution. Hence it is said to be non-reducing sugar.

24. i. A chemical substance which prevents the spoilage of food material by destroying the food-spoiling microorganisms in it is called a food preservative.

ii. Sodium benzoate, and potassium metabisulphite are used for food preservation.

Section - iii25. i. Because of its lightness and non-inflammability helium is used to

filling balloons for meteorological observations.ii. Because of its lightness it is used in inflating aeroplane tyres.

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iii. Helium oxygen mixture is used by deep-sea divers in preference to nitrogen oxygen mixtures. This prevents “bends” when a diver comes to the surface.

iv. A mixture of oxygen and helium is used in the treatment of asthma.v. Liquid helium (b.pt 4.2K) is used as cryogenic agent for carrying out

various experiments at low temperatures.vi. It is used to produce and sustain powerful super conducting magnets

of modern NMR Spectrometers and Magnetic Resonance Imaging system (MRI) for clinical diagnosis.

26. Monazite sandHeat210°C

withH SO2 4

for several hoursGrey mud

cold water

Unreacted monazite sandSiO , TiO , ZrSiO , etc2 2 4

(recycle sand)

Filtrate containing(Ln , Th , H O ,3+ 4+ +

3

HSO ,SO ,H PO )4 4 2 4– 2– –

Neutralisedacidity

to proper(or) added HF

Precipitate of (ThPO4)(or)

Precipitate of Th

Filtrate containinglanthanide and phosphate ions

NaOH (or)oxalic acid

Lanthanide hydroxides(or) oxalates

Physical methods :i. Anhydrous F– & Cl–

Ar atmosphereCa K-

/1270 → Individual metal

ii. Trifluorides of lanthanides

Ca,Li → Pure metal.

27. i. Co-ordination isomerism: In a bimetallic complex, the distribution of ligands between the two coordination spheres can vary, giving rise to isomers called the coordination isomers. This phenomenon is called coordination isomerism.

[CoIII(NH3)6] [CrIII(CN6)] and

Haxammine hexacyano cobalt (III) chromate (III)

[CrIII(NH3)6] [CoIII(CN6)]

Haxammine hexaacyano chromium (III) cobaltate (III)

ii. Ionisation isomerism. Co-ordination compounds having the same molecular formula but forming different ions in solution are called ionisation isomers. This property is known as ionisation isomerism.

[Co(NH3)5Br]SO4 Pentaamminebromocobalt (III) sulphate Red violet isomer [Co(NH3)5SO4]Br Penta ammine sulphatocobalt (III) bromide Red isomer Red violet isomer yields sulphate ion and red isomer furnishes bromide

ion in solution.28. i. Consider a general chemical equilibrium reaction in which the

reactants and products are in gaseous phases, aA + bB + cC + ... � ⇀�↽ �� lL + mM + nN ...

then Kp = p p pp p p

L Mm

Nn

Aa

Bb

Cc

1 . . .... . ...

where p is the partial pressure of the respective gases. ii. In terms of molar concentrations of reactants and products,

Kc = [ ] .[ ] .[ ] ....[ ] .[ ] .[ ] .......

L M NA B C

m n

a b c

1

For any gaseous component ‘i’ in a mixture, its partial pressure ‘pi’ is related to its molar concentration ‘Ci’ as

Ci = pRT

i [since Pi = nV

i RT]

where nV

i = Ci = number of moles of i per litre. V = volume in litres.

Substituting the concentration terms by partial pressures,

Kc = ( / ) ( / ) ( / ) ...( / ) .( / ) .( / ) ...

p p pp p p

Ll

Mm

Nn

Aa

Bb

Cc

RT RT RTRT RT RT

= p p pp p p

Ll

Mm

Nn

Aa

Bb

Cc

l m n a b c

RT. .

( ...) ( ...)1

+ + + − + + +

Kc = K

RT

png( )∆

∴ Kp = K RTcng( )∆

∆ng = (number of moles of gaseous products) – (number of moles of gaseous reactants)

Kp = K RTcng( )∆

Case ∆ngKp – Kc

Relationship Example

1 +ve Kp > Kc 2H2O + 2Cl2 � ⇀�↽ �� 4Hcl + O2

(g) (g) (g) (g)

2 –ve Kp < Kc 3H2 + N2 � ⇀�↽ �� 2NH3

(g) (g) (g)

3 zero Kp = Kc H2 + I2 � ⇀�↽ �� 2NI (g) (g) (g)

29. S.No Simple reactions Complex reactions

i. They occur in single step. They occur in many steps.ii. Overall order values are small.

Total and pseudo order values lie between 0, 1, 2 and 3.

Overall order values are large and greater than 3. Sometimes fractional orders such as 1

213

32

, , are seen.

iii. No side reactions Many side reactions are present.iv. Products are formed directly from

the reactants.In some complex reactions, products are not formed in steps directly from the reactants.

v. Experimental rate constant values agree with the calculated values.

Experimental overall rate constant values differ from the calculated values.

30. i. In a sol, the dispersion medium carries an equal but opposite charge to that of the dispersed particles.

ii. The movement of dispersion medium under the influence of applied electric potential when the dispersed particles are prevented from moving is known as electro-osmosis.

iii. The phenomenon of electro-osmosis can be demonstrated by using a U-tube in which a plug of wet clay (a negative colloid) is fixed. The two limbs of the tube are filled with water to the same level.

iv. The platinum electrodes are immersed in water and potential applied across them

v. Water level rises on the cathode side and falls on anode side. vi. The movement of medium towards the negative electrode shows that

the charges on the medium is positive. vii. Similarly for a positively charged colloid, electro-osmosis will take

place in the reverse direction.+ –

Platinumelectrode

Deionizedwater

Wet clay(Negative)

ORIGINALSOL LEVELS

ELECTRO-OSMOSIS

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14 __________________________________________________________ Sura’s XII Std - Govt. Model Question Papers

31. CH CO CH CH CH

OH

CH CH CH

Cl

CH SOLiAlH SOCl3 3 3 3

23 3 2

4− − → − − → − − + ++ HCl

↓

−CH C3 HH CH HCl− +2

Acetone (A)

Isopropylalcohol (B)

CH CO CH CH CH

OH

CH CH CH

Cl

CH SOLiAlH SOCl3 3 3 3

23 3 2

4− − → − − → − − + ++ HCl

↓

−CH C3 HH CH HCl− +2

Isopropyl chloride (C)

Propylene (D)

alc.KOH

32. Carboxylic acid reacts with alcohols in the presence of mineral acid as catalyst and forms esters. This reaction is called esterification.CH COOH C H OH CH COOC H H OH

Acetic acid3 2 5 3 2 5 2+ → +

+

EEthanol Ethyl acetate

Mechanism of esterification :Step 1: Protonation of carboxylic acid.

3 3HCH CO OH CH CO O

+ +− − → −

H

H

Step 2: Attack by nucleophile.

2

2 5

3 2 5 3H O

H

OC HH H

CCH CO O C H OH CH3 C O H CH H

O

+

++ −

−

|− − + → − − → −

||

..

33. (a) Nitrobenzene to anisole :

NO2 NH2 OH ONa OCH

NaOH→[ ]6 HSn/HCl→ HNO3

HCl2NaNO +→ CH I3NaI−→

Nitro Benzene Aniline Phenol Sodium Phenoxide Anisole

NO2 NH2 OH ONa OCH

NaOH→[ ]6 HSn/HCl→ HNO3

HCl2NaNO +→ CH I3NaI−→

Nitro Benzene Aniline Phenol Sodium Phenoxide Anisole

(b) Aniline to Iodobenzene.

NH2 I

Aniline Benzene diazonium Choloride Iodobenzene

NaNO .dil.HCl2273 278K

Diazotisation−→

N N Cl+

−≡ −

KIΛ→

Section - iV34. (i) N (Z = 7) = 1s2 2s2 2p3

N2 = 14 electrons Electronic configuration of N2 (σ1s)

2 (σ*1s)

2 (σ2s)2 (σ*

2s)2 (π2px)

2 = (π2py)2 (π2pz)

2

∴ Bond order = N Nb a−

= − =2

8 22

3

� Nature of bond : Triple bond (N ≡ N) � Magnetic property : Diamagnetic (no unpaired electrons)

(ii) i. Atomic size :

♦ I.E.∝ 1Atomic size

♦ Ionisation energy decreases with increase in atomic size. ♦ The Larger the size of the atom, lesser is the ionisation energy.

This is due to the fact that electrons are held quite loose (i.e) lesser energy is required for removal of electrons from larger atom than the smaller one.

ii. Magnitude of Nuclear charge : ♦ I.E ∝ Magnitude of Nuclear charge. ♦ Because of the higher nuclear charge, the electrons bound

with more force and hence high energy required to remove the electron.

♦ Hence I.E. increases.(or)

(i) i. Perforated container is filled with calcium phosphide and calcium carbide.

ii. When thrown into sea, water reacts with calcium carbide and calcium phosphide to give acetylene and phosphine.

iii. Phosphine gets ignited spontaneously also ignites acetylene. iv. Thus a bright red flame is produced accompanied by huge smoke. Ca3P2 + 6H2O → 2PH3↑ + 3Ca(OH)2 CaC2 + 2H2O → C2H2 ↑ + Ca(OH)2(ii) The important ores of zinc are (a) Zinc blende, ZnS (b) Calamine, ZnCO3 (c) Zincite, ZnO The extraction of zinc from zinc blende involves the following steps. i. Concentration : By froth-floatation process. ii. Roasting : The concentrated ore is then roasted in the presence of

excess of air at about 1200K.

2ZnS + 3O2 ∆ → 2ZnO + 2SO2↑

iii. Reduction : Zinc oxide is mixed with powdered coke and heated to 1673 K in a fire clay retort, in which ZnO is reduced to zinc metal.

ZnO + C 1673K → Zn + CO

iv. Purification : Electrolytic refining, � Anode - impure zinc � Cathode - pure zinc � Electrolyte - Zinc sulphate solution + a little of dil.H2SO4 � On passing electric current, pure zinc gets deposited at the

cathode.35. (i) Consequences :

i. According to Fajan’s rule, decrease in size of ions increase the covalent character and decreases the basic character.

La3+ > Ce3+ ...>Lu3+

ii. There is regular decrease in their ionic radii.iii. Regular decrease in their tendency to act as reducing agent. iv. 2nd and 3rd rows of d-block elements are quite close in properties.v. These elements occur together in natural minerals and are difficult

to separate.(ii) i. Haemoglobin in the red blood cells carries oxygen from the lungs

to the tissues. ii. It delivers the oxygen molecule to myoglobin in the tissues. When

the oxygen has been released for cell respiration, haemoglobin loses its bright red colour and becomes purple.

iii. It then combines with the waste carbon dioxide produced by the cells and deposits in the lungs so that the gas can be exhaled.

iv. Both haemoglobin and myoglobin having the same structrue but myoglobin is a monomer and heamoglobin is a tetramer.

v. Both are iron-porphyrin complex called heme group in which iron is in +2 oxidation state.

vi. The working part of haemoglobin is a heme group containing an Fe2+ cation coordinated to four nitrogen atoms of porphyrin group and one nitrogen atom of histidine group. The sixth octahedral site is available to bind oxygen molecule.

(or)(i) These are the reactions in which high speed projectiles may chip a heavy

nucleus into several fragments. 29Cu63 + 2He4 + 400 MeV → 17Cl37 + 141H

1 +16 0n1

(ii)

No. Isotope Name Usei. 1H

3 Tritium Measure water contant of the body

ii. 6C11 Carbon - 11 Brain scan

iii. 6C14 Carbon - 14 Radio immunology

iv. 53I131 Iodine - 131 Diagnosis of damaged heart muscles

and hyper thyroidismv. 80Hg197 Mercury - 197 Kidney scanvi. 15P

32 Phosphorous - 32 Detection of eye tumour

vii. 26Fe59 Iron - 59 Diagnosis of anaemia

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viii. 27Co60 Cobalt - 60 Treatment of cancer

ix. 11Na24 Sodium - 24 Location of blood clots and circulatory disorder

x. 79Au198 Gold - 198 Curing of cancers

36. (i) i. Bragg’s spectrometer method is used for studying crystals using

X-rays. ii. X-rays is allowed to fall on the crystal mounted on a rotating table.

The rotating table is provided with scale and vernier, from which the angle of incidence θ can be measured.

iii. An arm which is rotating about the same axis as the crystal table, carries an ionisation chamber. The rays ionise the gas present inside.

iv. Due to the ionisation, current is produced and measured by electrometer, which is a direct measure of intensity of reflected beam from the crystal.

v. For different angles of incidence, the corresponding ionisation current is measured from the electrometer.

These values are plotted in the form of graph.θ 5.9º 11.85º 18.15º

Sin 0.103 0.205 0.312Ratio 1 2 3

vi. For NaCl, the maximum reflection or peaks for 100 plane occurred at θ = 5.9o, 11.85o and 18.15o.

vii. These peaks represent the first, second and third order reflections. viii. The ratio confirms the correctness of Bragg’s equation.

X -ray tube

SlitDiffracted

X-rays

Path of

Recorder

Ionisation

chamber

Turn table

Scale to measure

rotation of crystal

Crystal

Incident

X-rays

X -ray spectrometer

θ

(ii) According to Le-Chatelier’s principle, “if a system at equilibrium is subjected to a disturbance or stress, then the equilibrium shifts in the direction that tends to nullify the effect of the disturbance or stress”.

(or)(i) Statements of second law of thermodynamics : (i) Kelvin-Planck statement : “It is impossible to construct an

engine which operated in a complete cycle will absorb heat from a single body and convert it completely to work without leaving some changes in the working system”.

(ii) Clausius statement : “ It is impossible to transfer heat from a cold body to a hot body by a machine without doing some work”.

(iii) Entropy statement : “A process accompanied by increase in entropy tends to be spontaneous”.

(iv) Efficiency statement : “Efficiency of a machine can never be cent percent”.

(v) According to II law of thermodynamics : The heat efficiency of any machine is given by the value of ratio of output to input energies. Output can be in the form of any measurable energy or temperature change while input can be in the form of heat energy or fuel amount which can be converted to heat energy.

% Efficiency = outputinput

× 100

% Efficiency = T T

T1 2

1

− × 100

By II law, T2 < T1,% efficiency less than 100.

(ii) Given data : Rate = 5.2 × 10–6 mol lit–1 sec–1

Initial concentration [A] = 2.6 × 10–3 Formula : Rate = k [A]1

Solution : 5.2 × 10–6 mol lit–1 sec–1 = k × 2.6 × 10–3 lit–1

= 5 2 102 6 10

2 106 1 1

3 13. sec

.×

×= ×

− − −

− −−mol lit

mol lit

∴k = 2 × 10–3 s–1

37. (i) The pH of an acid buffer can be calculated from the dissociation constant,

Ka of the weak acid and the concentrations of the acid and the salt used. The dissociation equation of weak acid HA, may be represented as,

HA � ⇀�↽ �� H+ + A–

Ka = H AHA

+ − [ ]

[H+] = Ka HAA

[ ]

− ............ (1)

The weak acid is only slightly dissociated and its dissociation is further depressed by the addition of the salt (Na+A–) which provides A– ions. (Common ion effect).

♦ The equilibrium concentration of the unionised acid is nearly equal to the initial concentration of the acid.

♦ The equilibrium concentration [A–] is presumed to be the initial concentration of the salt added.

[H+] = KAcidSalta × [ ]

[ ] ..............(2)

[Acid] = initial concentration of the acid.[Salt] = initial concentration of the salt.Taking negative logs of both sides of the equation, (2), we get,

− = − −+log[ ] log log [ ]

[ ]H K Acid

Salta ................(3)

But, –log [H+] = pH and –log Ka = pKa ∴Equation (3) becomes,

pH = pK log [Acid][Salt]a −

................(4)

pH = pK + log [Salt]

[Acid]a

................(5)

The above equation is called Henderson-Hasselbalch equation (or) Henderson equation.

Similarly for a basic buffer, the Henderson equation is pOH = pKb + log [Salt]

[Base]

(ii) IUPAC Conventions: In 1953 IUPAC recommended the following conventions for writing cell diagrams. We will illustrate these with reference to Zinc-Copper cell.

i. A single vertical line (|) represents a phase boundary between metal electrode and ion solution (electrolyte). Thus the two half-cells in a voltaic cell are indicated as

Zn | Zn2+ Cu2+ | Cu

ANODE HALF-CELL CATHODE HALF-CELL

PhaseBoundary

It may be noted that the metal electrode in anode half-cell is on the left, while in cathode half-cell it is on the right of the metal ion.

ii. A double vertical line represents the salt bridge, porous partition or any other means of permitting ion flow while preventing the electrolyte from mixing.

iii. Anode half-cell is written on the left and cathode half-cell on the right.

iv. In the complete cell diagram, the two half-cells are separated by a double vertical line (salt bridge) in between. The zinc-copper cell can now be written as

Salt Bridge Zn | Zn2+ | | Cu2+ | Cu ANODE HALF-CELL CATHODE HALF-CELL v. The symbol for an inert electrode, like the platinum electrode is

often enclosed in a bracket. For example,

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16 __________________________________________________________ Sura’s XII Std - Govt. Model Question Papers

Mg | Mg2+ | | H+ | H2(Pt)

ANODE HALF-CELL CATHODE HALF-CELL

InertElectrode

vi. The value of emf of a cell is written on the right of the cell diagram. Thus a zinc-copper cell has emf 1.1 V and is represented as,

Zn | ZnSO4 | | CuSO4 | Cu E = + 1.1 V → Direction of electron flow If the emf acts in the opposite direction through the cell circuit it

is denoted as a negative value. Cu | CuSO4 || ZnSO4 | Zn E = – 1.1 V ← Direction of electron flow vii. The negative sign also indicates that the cell is not feasible in

the given direction and the reaction will take place in the reverse direction only. The overall cell reaction for E = –1.1 V of the daniel cell is

Cu(s) + Zn2+(aq) → Cu 2+

(aq) + Zn(s)

(or)(i) i. Cyclohexanol exist in two chair forms. They are interconvertible

and exist in equilibrium.

I IIOH

OH

E

0.7 KCal

Axial

Equatorial

Boat

11 KCals

ii. In one form (I), the -OH is axially oriented, in the form (II) the -OH is equatorially oriented.

iii. The energy of the axial conformer is little higher than the equatorial conformer.

iv. The axial -OH experiences a steric interaction with axial H present at the third position .

v. This is called 1 : 3-diaxial interaction and this decreases the stability of axial conformer.

vi. In the equilibrium mixture axial conformer is about 10% and equatorial conformer is about 90%.

(ii) Isomers in alcohols. i. CH3 – CH2 – CH2 – CH2 – OH 1-butanol ii. CH CH CH CH

OH

3 2 3− − −

| 2-butanol

iii.

CH

CH CH CH OH

3

3 2

|

|− − −

OH

2-methyl - 1 - propanol

iv.

CH

CH C OH

C

3

3

|

|− −

HH3

2-methyl - 1 - propanol

Isomers in ethers. v. CH5–CH2–O–CH2–CH3 Ethoxy ethane vi. CH3–O–CH2–CH2–CH3 1-methoxy propane CH3 | vii. CH3–CH–O–CH3 2-methoxy propane.38. (i) i. Compound A giving violet colour with neutral ferric chloride is

phenol (C6H5OH) ii. (B) and (C) isomers are formed when alkaline solution of (A) is

refluxed with CCl4.

HOOC

OHOH OH

COOH+

ClC 4

(A) (B) (C)

NaOH

iii. (A) when treated with C6H5N2Cl, forms a red orange dye (D).

HO N = N HOOH−

273K

(D)

H + Cl − N = N

HO N = N HOOH−

273K

(D)

H + Cl − N = N

vi. Compound Compound Name Formula

A Phenol C6H5OH

B o-Hybroxy benzoic acid COOH

OH

C p-Hydroxy benzoic acid

HOOC

OH

D p-Hydroxy azo benzene

N = N HO

(or)(i) i. The molecular formula of gluose is C6H12O6. ii. Glucose /P

3 2 2 2 2 3HI CH CH CH CH CH CH→ − − − − −

n-hexane

This indicates that 6 carbon atoms are linked in a consecutive unbranched chain.

iii. Glucose + H2O → Neutral solution. This indicates that the molecule does not contain a –COOH group. iv. Glucose reacts with hydroxylamine to form an oxime and adds a

molecule of HCN to give a cyanohydrin. This confirms the presence of a carbonyl group ( C = O) in glucose.

v. Glucose Br / H O(O)

2 2 → Gluconic acid.

This indicates the presence of an aldehyde –CHO group. vi. Further oxidation of gluconic acid with Conc. HNO3 gives saccharic

acid. Gluconic acid Conc. HNO

(O)3 → Saccharic acid.

This indicates the presence of primary alcohol. vii. It reduces Tollen’s reagent and Fehling’s solutions. This confirms

the presence of an aldehyde group. viii. Glucose + acetic anhydride Pyridine → Penta acetate.

This indicates the presence of 5 (–OH) groups in a glucose molecule. ix. From the above evidences, we conclude that the structure of glucose

is,HOC

| CHOH

| CHOH

| CHOH

| CHOH

|CH2OH

IUPAC NAME :2, 3, 4, 5, 6 penta hydroxy hexanal

***

*

(ii) i. An organic compound appears coloured due to the presence of certain unsaturated groups. (the groups with multiple bonds) in it.

ii. Such groups with multiple bonds are called chromophores.

-NOO

Nitro

-N=N-Azo

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