[1]

Higher Secondary Board Examination 2011 12th ClassCHEMISTRY

Instructions: (i) All questions are compulsory. (ii) Read the instructions of question paper carefully and write theiranswer.(iii) There are two section -section -A and section-B in the question paper (iv) In section -AQ.Nos.1 to 4 are objective types question ,which contain Fill up the blanks, ture/False, Match the

Columns and choose the correct answer. Each question is allotted 5marks.(v) Q.Nos.5 to 14

carry 4 marks each. (vii) Q.15 to17 carry 5 marks each.

(Section-A) (Objective Type Questions) (5 marks each)

Objective Type Questions Answer :

1.Write the correct answer from the given options :

(A) Number of atoms in body centred cubic unit cell is:

(i) 1 (ii) 2 (iii) 3 (iv) 4

(B) Reason for stability of lyphobic sol is :

(i) Brownian movement (ii) Electric charge

(iii) Tyndall effect (iv) Brownian movement and electric charge

(C) How many phases are present in a colloidal solution?

(i) 1 (ii) 2 (iii) 3 (iv) 4

(D) Which of the following is the strongest acid ?

(i) HBr (ii) HCI (iii) HF (iv) HI.

(E) Number of unpaired electrons in 2Fe+ ion is :

(i) 0 (Zero) (ii) 4 (iii) 6 (iv) 3

2.Match the pairs correctly :

(a) Bad conductor (i) Adsorption of poisonous gases

(b) Gas mask (ii) Conversion of starch into sugar

(c) Diastase (iii) An ore of zinc metal

(d) Calamine (iv) Laboratory reagent

(e) Hydrogen sulphide (v) Does not allow to flow electricity.

3. Fill up the blanks-

(a) Total .........................types of crystal system are there.

(b) Ethyl amine is .............basic than ammonia.

(c) Protein is a polymer of ..................................

(d) ............................. gave the concept of “parmanu” first.

(e) Trinitrotoluene is an .............................. substance.

4. Give answer in one word of each -

(i) Which type of solid is SiC ?

(ii) On which factor colligative properties of a solution depends.

(iii) Give the name of the reaction in which oxidation and reduction occur simultaneously.

(iv) Write the formula of half -life for the first order reaction.

(v) What is nitrating mixture?

(Section-B) (Very Short Answer Type Questions) (4 marks each)

5. Write any four differences in order of reaction and molecularity.

Or

[2]

Write four difference in rate of reaction and rate constant.

6. Draw labelled diagram of Blast furnance used in metallurgy of iron.write chemical reaction occurring in everyzone.

Or

Draw labelled digram of electrolytic cell used for electrolysis of alumina. write the chemical reactions occurringin it.

7. Explain hydrides of nitrogen family on the following points :

(i) Name and formula (ii) Basic nature

(iii) Reducing property (iv) Bond angle.

Or

Explain hydrides of oxygen family on the following points :

(i) Name and formula (ii) Thermal stability

(iii) Reducing property (iv) Acidic property

8. Explain hybridization in 5

AB and7

AB type of interhalogen compounds and draw their structure.

Or

Expain hybridization in 2

XeF and 4

XeF compounds. Draw their structures.

9. Explain ionisation isomerism and linkage isomerism in co-ordination compounds.

Or

Write IUPAC names of the following co-ordination compounds.

(i) ( )3 44Cu NH SO (ii) ( )2 36

Cr H O CI

(iii) ( )4

Ni CO (iv) [ ]2 4K HgI

10. Differentiate in primary alcohol, secondary alcohol and tertiary alcohol on the following points :

(i) oxidation; (ii) dehydrogenation.

Or

(i) What change in colour is observed in phenol in presence of oxygen? Explain with reaction.

(ii) How can we obtain phenol from benzene diazonium chloride?

11. Draw labelled diagram of laboratory method for preparation of iodoform from alcohol. Write related chemicalequations.

Or

Write the following reactions:

(i) Carbylamine reaction (ii) Friedel-crafts alkylation reaction

(iii) Wurtz reaction (iv) Frankland reaction.

12. How will you obtain the following ?

(i) Acetaldehyde form Acetyl chloride (ii) Acetone from calcium acetate

(iii) Acetic acid from ethyl acetate (iv) Ethane from acetic acid.

Or

(i) What is Tollens reagent? Write its reaction with acetaldehyde.

(ii) Write method of preparation of urotropine and its use.

13. Write functions and sources of the following biomolecules/elements :

(i) Protein, (ii) Carbohydrates, (iii) Fat, (iv) Calcium

[3]

Or

Write functions and sources of the following vitamins :

(i) Vitamin ‘A’, (ii) Vitamin ’C’, (iii) Vitamin ’D’, (iv) Vitamin ’E’

14. Explain the following medicines and give one example of each :

(i) Antacid, (ii) Sulpha drugs, (iii) Antiseptics, (iv) Sedatives.

Or

Explain in brief about the following:

(i) Nagarjuna, (ii) Charak

(Long Answer Type Question ) (5 marks each)

15. Write 5 differences in solutions having positive deviation and solution having negative deviation.

Or

What are constant boiling mixtures? Write 3 differences in Ideal solution and Non-ideal solution.

16. What are weak electrolytes? Give one example. Find out molar conductivity of LiBr aq. solution at infinite

dilution when ionic conductances of Li+ ion and Br

− ion are 2 238.7s cm mol

− and 2 178.40 s cm mol

−

respectively.

Or

What are strong electrolytes? Find out the molar conductivity of aq. solution of 2

BaCl at infinite dilution

when ionic conductances of 2Ba+ ion and Cl − ion 2 1127.30 s cm mol

− and 2 175.34 s cm mol

− respectively.

17. Write general electronic configuration of d-block elements.Explain the following properties of d-block elements:

(i) Coloured ion and coloured compound formation, (ii) Formation of alloy.

Or

Write general electronic configuration of f − block elements. Write any 2 uses of lanthanides. Write any 2

uses of actinides.

[4]

SOLUTIONSECTION -A

1. (A) ii (B) iv (C) ii (D) iv (E) ii (F)2. (A) v (B) i (C) ii (D) iii (E) iv3. (A) 7 (B) more

(C) amino acids (D) Kanada(E) strong explosive

4. (i) Crystalline solid(ii) Number of particles(iii) Redox reaction

(iv) 1/ 2

0.693t

K=

(v) 2 4 3

H SO HNO+5. Order of reaction 1. It is the number of molecules which participate

in reaction and whose concentration is changed. 2. Order of reaction is determined by experimen-

tally. 3. Fractional values are also possible. 4. Zero value is possible.

Molecularity 1. It is the total number of molecules which partici-

pate in the reaction. 2. It is a theoretical concept. 3. It is always a whole number. 4. Its value is never zero.

ORRate of reaction

1. It is expressed in terms of consumption of reac-tants or formation of product per unit time.

2. It depends on concentration of reactant at par-ticular moment.

3. It generally decreases with the progress of reac-tion.

4. Its unit of mol 1 1l cm− − .

Rate constant 1. It is proportionality constant in differential form in

rate law of rate equation. 2. It is independent of concentration of reactant. 3. It does not depend on the progress of reaction. 4. It changes according to order of reaction.6. Labelled diagram of Blast furnace :

ConicalArrangem ent

Charge

Gases ( )2,C O C O

1 500°C

1 300°C

1 000°C

6 00°C

Twaiyer Twaiyer

Heat Bricks

Air

S lag Liquidm attle

Fig. B last furnace

( )2 3 3F e O O C a C O+ +

Reactions in Blast furnace :

(i) Coke burns in presence of air of form 2

CO in

excess of CO is formed.

2 2C O CO+ →

1500

2 2C

C CO CO°+ →

(ii) CO reduces haematite ore ( )2 3Fe O to

ferrousoxide( )FeO at 600°C.

2 3 22CO Fe O FeO CO+ → +

(iii) About 750°C CO reduced FeO into Fe .

2FeO CO CaO CO+ → +

(iv) At 110°C 3

CaCO decomposes into CaO .

3 2CaCO Fe CO→ +This way the iron obtained is called spongy iron.

(v) CaO reacts with silica present in ore and formslag.

2CaO SiO+ →

3C a S iO

Calcium silicateSlag floats over molten iron from where it is re-moved through the tapping hole.

OR

If Bauxite contains 2 3

Fe O as impurity,it is puri-

fied by Baeyer’s process.Electrolysis of alumina (Hall-Heroult process) :Electrolysis of alumina is done in an iron tankfitted.

Carbon AnodeCarbonPlasterCathode

IronTank

Alum ina

Alum ina

Liquid Alum iniummetal

Fig. Electrolysis o f Alum ina

with a lining of gas carbon which acts as cath-ode. Graphite rods act as anode and a bulb isfitted in parallel as indicator for alumina content.The electrolyte is a mixture of alumina and cryo-

lite ( )3 6Na AlF . The melting point of alumina is

very high (2050°C). But in presence of cryolite

[5]

( )3 6Na AlF and fluorsper ( )2

CaF , it melt at

870°C. Then pure alumina is added to the tankwhich dissolve in cryolite mixture.Reaction : Reaction take place as below :

3 6 33Na AlF NaF AlF→ +

3

3 3AlF Al F+ −→ +

3 3 3F e F− −→ + (oxidation) at anode

33Al e Al

+ −+ → (reduction) at cathode

2 3 3 22 12 4 3Al O F AlF O+ → + ↑7. (i) Name and formula :

Name Formula

1. Nitrogen : Ammonia3

NH

: Nitric acid3

HNO

2. Phosphorous : Phosphine3

PH

(ii) Basic nature : Ammonium is a mono-basicanhydride that dissolves in water in form of

4NH OH (ammonium hydroxide).

Ammonium hydroxide is a base. Hence, it has amore basic nature. Nitric acid is also a strongmono-basic acid which ionized in water and form

of 3

H O+ and 3

NO− ions.

Hence, it has also more basic nature. Phosphenecombine with halogen acids in gaseous state andforms phosphonium. There phosphonium halidesdecompose as soon as possible and reformphosphone.

3PH HX+ →

4P H X

Phosphonium halide

[ ],X Cl Br or I=

2

4 3

H OPH X PH H X

+ −→ + +Hence it has more acidic property due to

formation of H+ and it has also more basic

nature.Therefore, it is clear hydrides of nitrogen andnitrogen family has more basic nature.

(iii) Reducing property : Ammonia gas is strongreducing agent. It reduces hot metallic oxidesinto metals.Nitric acid is a strong oxidant i.e. reducing agent,as it easily decomposes to produce nascentoxygen.

32HNO → [ ]2 22H O N O O+ +

N itro gen d i-ox id e

On passing phosphene gas into the solutions ofcopper, silver and mercury salts metal phosphidesare precipitated.

4 3 3 2 2 43 2 3CuSO PH Cu P H SO+ → +

3 3 3 33 3AgNO PH Ag P HNO+ → +Hence, it has strong reducing property.Therefore from the above reasons and explanationit is clear that hydrogen of nitrogen family havestrong reducing property.

(iv) Bond angle :Ammonia :

H

HH N

l.p.-b.p.

Bond angle is 107.3°.Nitric acid :

H

ON

120°0.96Å

1.41Å130Å

1.22Å

1.22Å114°

O

O

OR (i) Name and formula :

Name Formula

Water2

H O

Calcium hydroxide ( )2

Ca OH

Hydrogen di-sulphide2

H S

Sulphuric acid2 4

H SO

(ii) Thermal stability : Thermal stability of hydridesdecreases in class size of atoms effected to thebond power. As well as size of atoms increasesthe bond power decreases.

2 2 2 2H O H S H Se H Te> > >

Decrease Thermal stability

(iii) Reducing property : Element of the group havetendency to gain electrons. Their atoms formstable octet after gaining two electrons. They loseelectron with difficulty. They form anion but not

[6]

cation. This is because they have highelectronegativity. On moving down the groupelectronegativity decrease and metallic characterincreases.

(iv) Acidic property : 2 4

H SO is a strong and which

is ionized in water in two steps :

2 4 4H SO H HSO+ −+�

2

4 4HSO H SO+ −+�

2 4H SO is the main hydride of the oxygen family

and acidic properties of other hydrides are nearly

similar to it a extent [Except 2

H O neutral].

8. ( )5 5 5,AB IF BrF Type : These type of

compounds show 3 2sp d hybridization. Their

structure is square pyramidal in which there islone pair of electron at one place. Like in the

figure, formation of 5

IF shown.

5s 5p

1 atom in ground state

1 atom in excited state

sp d hydridiza tion3 2

5d

F

F

FF

F

Fig. sp3d2 hybridization in 5

IF

( )7 7AB IF Type : Its geometry is pentagonal

bipyramidal which is formed by 3 2sp d

hybridization. Hybridization in 7

IF molecule is

3 3sp d .

5s 5p

Iodine a tom in excited state

5d

sp d hybridization3 2

I

F

FF

F

F

F

Fig. Hybridization in ( )7 7AB IF

OR

Structure and shape of 2

XeF molecule : In 2

XeF

molecule, Xe atom is in 3

sp d hybrid state

hence its structure is trigonal bipyramidal but itsgeometry is linear. It can be understood by

valence bond theory. In ground state Xe atomhas no unpaired electron.

5s 5p 5dXenon atomground stateXenon atomexcited sta te

XeFem olecule

sp d Hybridization3 2

X e

F

F

or

X e

F

F

180°

Fig. Structure of 2

XeF molecule

In excited state one electron from 5 p orbital

jumps to vacant 5d orbital hence it contains two

unpaired electrons. Now 3sp d hybridization takes

place in it as a result, five hybridized 3sp d orbitals

are formed which are directed towards five cornersof a trigonal bipyramid. Among these three arecompletely filled while two are half-filled, i.e. itcontains unpaired electrons. Orbitals containingunpaired electrons overlap p-orbitals of fluorine

to form 2

XeF molecule in which shape of 2

XeF is

linear and three lone pairs of electrons are presentat three corners of an equilateral triangle. Bond

[7]

length of Xe and Fe is 2.0 +0.01Å and

F Xe F− − bond angle is 180°. Due to repulsionbetween three lone pairs of electrons and twobonded electron pairs, the shape of the moleculebecomes linear/

Structure and shape of 4

XeF molecule : In

xenon tetrafluoride, xenon atom is linked with fourfluorine atoms through covalent bonds. Here,8+4=12 valence electrons are present aroundcentral atom.

5s 5p 5dXenon atomground state

Xenon atomexcited sta te

XeFem olecule

sp d Hybridization3 2

F F

FF

orX e

F

90°

F

F

F

X e

F

F

XeF (linear)Xenon difluoride

2

200pm

F F

FF

XeF (square planar)Xenon tetrafluoride

4

Two electrons from 5 p orbitals in Xeget excited

and are promoted to 5d orbitals which result in

four unpaired electrons. Now, 3 2sp d hybridization

takes place which results in six hybrid orbitalsdirected towards six corners of octahedron.Among them two are completely filled and four

are half-filled. Xe is linked with four flourine atomsthrough these these half filled orbitals, byoverlaping with p − orbitals of fluorine. Due to

3 2sp d hybridization structure of molecule should

have been octahedral, but due to the presence oftwo lone pair of electrons, geometry of themolecule gets distorted and becomes squareplanar which contains two non-bonding electronpairs.

9. Ionization isomerism : When complex of equalmolecular weight and idential molecular formulaexchange their ions attached through primaryvalency by anions present in co-ordination spherethen this type of isomerims is called ionizationisomerism. Such type of molecules furnishdifferent ions in solution.

( )3 43[ ]C o N H B r S O

(I )

and ( )3 45[ ]C o N H S O B r

(II)

I and II complexes furnish 2

4SO − and Br

−

anions in solution. Solution of complex I gives

white precipitate of 4

BaSO with 2

BaCl due to

the presence of 2

4SO − ion, while solutions of II

gives yellow precipitate of AgBr with 3

AgNO

due to the presence of Br− ion.

Linkage isomerims : Linkage isomerism occurswhen different atoms of the ligand are attachedto the central metal ion. The structure obtainedare called linkage isomers. Such type of ligandsare called ambidenate ligand.

( ) ( )3 5[ ]N i N H S C N C l

(I)

( ) ( )3 3[ ]N i N H N C S C l

(II)

In structure I2Ni + is linked by thiocyanate

sulphur and in structure II it is linked by nitrogenatom.

OR

(i) ( )3 44Cu NH SO

: Tetra amine copper

( )II sulphur

(ii) ( )2 36[ ]Cr H O Cl : Hexa aqua chromium

( )III chloride

(iii) ( )4

Ni CO : Tetra carbonyl nickel ( )O

(iv) [ ]2 4K HgI : Potassium tetra iodo mercurate

( )II

[8]

10. (1) Oxidation : The oxidizing agents generally usedalcohol by oxidation of alcohols are acid

dichloromate, acid of alkaline 4

KMnO and dilute

3HNO .

(i) A primary alcohol is easily oxidized to analdehyde and then to an acid both containing thesame number of carbon atoms as the originalalcohol.

3 2C H C H O H

E thy l a lcoho l

O→3

C H C H O

A ce ta ld eh yd e

O→

3C H C O O H

Acetic acid (ii) A secondary alcohol on oxidation gives a ketone

with the same number of carbon atoms as theoriginal alcohol, ketones are oxidized with dificultybut prolonged action of oxidizing agents producecarboxylic acids containing fewer number ofcarbon atoms than the original alcohol.

C H O H

Isopropyl alcohol

3C H

3C H

O→ C O=

Acetone

3C H

3C H

[ ]4 O→

3 2 2C H C O O H C O H O+ +

Acetic acid (iii) A tertiary alcohol is resistant to oxidation in

neutral or alkaline solutions but is readly oxidizedby an acid oxidizing agent giving a mixture ofketone and acid each having lesser number ofcarbon atoms than the original alcohol.

C O H-

3C H

3C H

3C H

Tert-Butyl alcohol

O→ C O=3

C H

3C H

Acetone

[ ]3 O→ 3CH COOH HCOOH+

(2) Dehydrogenation (Action of hot reducedcopper at 300°C) : Different types of alcoholsgive different products when their vapours are

passed over Cu gauge at 300°C.Primary alcohols lose hydrogenand yield analdehyde.

3 2C H C H O H

E thy l a lcoho l300

Cu

C°→3 2

C H C H O H+

A ce ta lde h yd eSecondary alcohols lose hydrogen and yield aketone.

C H O H3C H

3C H

Isopropyl alcoho l

300

Cu

C°→ 2C O H= +

3C H

3C H

Acetone

Tertiary alcohols are not dehydrogenated but losea water molecule to give alkenes.

C O-

3C H

3C H

3C H

Tert-Butyl alcohol

300

Cu

C°→2 2

C C H H O= +3C H

3C H

2-M ethylprop-1-ene

OR (i) In the presence of air pure phenol oxidises into

quinone.

O H

2O+ →

O

QuinoneO

This quinone again combines with two moleculesof phenol by H-bond and gives pink phenoquinone.

2

O H

+

O

O

→ O H

O O H O

Phenoquione (P ink colour) (ii) Obtain phenol from benzene diazonium

chloride : When atomatic primary amine reacts

with nitrous acid ( )2NaNO HCl+ at 273-283K,

then diazonium salt is produced. This ontreatment with dilute acid or boiling with watergets decomposed and phenol is produced.

2N H

2

0 10

NaNO HCl

C

+− °→

2N C l

+

2H O

Heat→

O H

Phenol2

N HCl+ +

[9]

11. (1) Laboratory method : Chloroform is prepared inthe laboratory by the action of water and bleachingpowder on ethyl alcohol or acetone.

70°C

Na CO (solution)+Iodine C H OH

2 3

2 5

Condenser

Fig. Preparation of chloroform(Labora tory method)

Method : About 100 gm of bleaching powdermade into a paste by adding about 200ml of waterand taken in a flask fitted with a condenser. Now25ml of alcohol or acetone is added and themixture is distilled chloroform colects as a heavyliquid under water.

It is washed with dilute NaOH solution then withwater, dried over fused calcium chloride andredistilled.The available chlorine of bleaching powder actsas oxidising as well as chlorinating agent duringthe preparation of chlorofom form iodoform andacetone.

( )2 2 22CaOCl H O Ca OH Cl+ → +The chemistry involved in the conversion of alcoholand acetone into chlorofom is as shown below :From alcohol : The steps involves are :

(i) Ethyl alcohol is oxidized by chlorine toacetaldehyde.

Ethyl alcohol3 2 2

C H C H O H C l+ →Acetaldehyde

32C H C H O H C l+

(ii) Acetaldehyde reacts with chlorine to give chlorali.e. trichloroacetaldehyde.

(iii) Two moles of chloral react with one mole ofcalcium hydroxide to produce chloroform.

C a3

C H

3C H

Calcium hydroxide chloral

32C C l C H O+

Chloro form3

2C H C l C a+O O C H

O O C H

Calcium fo rmate

OR (i) Carbylamine reaction : On heating chloroform

with primary amine (e.g. animline) and alcoholic

KOH solution, phenylisocyanide or carbylamineis formed which has a very bad smell and ispoisonous.

A n iline6 5 2 3

3C H N H C H C l K O H- + + →

Phenyl isocyanide

6 5 23 3C H C K C l H O+ +

Application : Chloroform and primary amine canby tested by this reaction.

(ii) Friedel-Crafts alkylation reaction : Alkylhalides react with benzene is pressence ofanhydrous aluminium chloroide to give alkylbenzene.

+ 3

CH Cl 3Anhydrous AlCl→

3C H

M ethyl benzene(Toluene)

+ HCl

Acetylation : Acetyl chloride react with benzenein presence of anhydrous aluminium chloride togive acetophenone.

+

O

3C H C C l- - 3Anhydrous AlCl→

3C C H-

Acetophenone

O

+ HCl

(iii) Wurtz reaction : When an alkyl halide is heatedwith sodium in presence of dry ether. Then higheralkane is formed. This reaction is called Wurtzreaction.

2 2 2Heat

RX Na R R NaX+ → − +

2 5 2 52C H I Na C H I+ + →

Butane2 5 2 5

2C H C H N a I+

(iv) Frankland reaction : This reaction is likely towurtz reaction. In this reaction we use zink inplace of sodium.

[10]

2R I Zn I R R R ZnI+ + → − +

12. (i) Acetaldehyde from Acetyl chloride :

A c e ty l c h lo r ine

32C H C O C l H+ 4/Pd BaSO

Xylene→

A ce ta ld eh yd e3

C H C H O H C l+

(ii) Acetone from calcium acetate :

Calcium acetate( )3

C H C O O C a∆→

Acetone3 3 3

C H C O C H C a C O+

(iii) Acetic acid from ethyl acetate :

Ethyl acetate3 2 5 2

C H C O O C H H O+ HCl→

Acetic acid

3 2 5C H C O O H C H O H+

Ethanol (iv) Ethane from acetic acid :

Acetic acid3

4C H C O O H H+2

Zn HCl

H O

+−

→Ethane

3 3C H C H

OR (i) Ammonical silver nitrate is known as Tollen’s

reagent. It reacts with acetaldehyde liberatingsilver.

3 4 4 3AgNO NH OH AgOH NH NO+ → +

2 22AgOH Ag O H O∆→ +

Tollen'sreagents

3 2C H C H O A g O+Silver

32 A g C H C O O H+

ketones do not give this reaction. (ii) When formaldehyde is treated with ammonia,

urotropine is formed.Its chemical name s hexamethylene tetraamineor Hexamine.

36 4HCHO NH+ →Urotropine

2 6 4 2( ) 6C H N H O+

13. Protein :Functions : As a catalyst in bio-chemicalreaction, motion of muscles, storage of oxygenin muscles, food for cheek, complex food for newborn body, protective action in teeth, nails andhairs.Sources : Skin, hair, nail, wool, bones, musclesmajor source of protein are plants.Carbohydrates :Functions : They provide heat to body, energyto the body for performing various functions, helpsin the manufacture of cell membrane, and theyconstitute the largest position of plants.Sources : Wheat, yeast, egg, meat, salad,beans, cereals, egg yolk, sugarcane.Fat :Functions : It is a source of carbohydrates whichremain for a stable period, it provides energy to

body, energy is also stored in form of fat.Sources : Meat , butter, egg, fish, oil, tomatoes,milk, rice, amla, goat meat.Calcium :Functions : It’s main function is to strength andprovide essential condition to bones, teeth, ribs,joints.It also provide protection towards variousdiseases and defeciencies.Functions : Bones, meat, milk, fish, egg, oil,vegetables, green leaf vegetable, papaya, mangoetc.

OR

Vitamin Source Utility or Function

Vitamin A Egg, milk, milk products, cod liver oil, fish liver oil

For vision and growth develops resistance against diseases.

Vitamin C Green vegatables, salad, citrus fruits.

For bones, teeth and healing of wounds.

Vitamin D Sunlight, milk, egg, fish liver oil

For bones, control of metabolism of calcium and phosphorous

Vitamin E Seeds, beans, milk, meat, green leaves

in man and reproduction

(i) Antacid : There are the chemical substanceswhich neutalise the excess acid and raise thepH to an appropriate level in the stomach. Themost common ailments assicated with thestomach is acid gastritis which is caused byexcess of hydrochloric acid in the stomach. Themost commonly used antacides are weak basesuch as- magnesium hydroxide, magnesiumcarbonate, magnesium tri silcate, aluminiumhydroxide gel, sodium bicarbonate etc. Howeverthese treatments control only symptoms and notthe cause. The fore with these metal salts, thepatients cannot be treated eaisly. A mjor breakthrough in the treatment of hyperacidity tookplace with the discovery that a chemicalhistamine, stimulates the secretion of pepsin andhydrochloric acid in the stomach. Thereafter adrug cimetidine was designed to prevent theinteraction of hitamine with the reciptors presentin the stomach wall. This resulted in release oflesser amount of acid. This drug was followed bythe discovery of another best drug called ranitidine(Zintac). Recently omeprazole and land soprazolehave been used as best antacides. These preventthe formation of acid in the stomach.

(ii) Sulpha drugs : They perform the same function

[11]

as antibiotics but they are synthsized in thelaboratory only. A number of sulpha drugs areused in place of antibotics. They prevent of destroythe growth of micro-organisms e.g. bactria andfungi. They have great antibacterial andpneumococci. The obtain sulphonamide

( )2 2SO NH− group in their structure.

Sulphadiazine, sulphaguanidine, sulphapyridineare some common examples of sulpha drugs,All of them are derivatives of sulphonamide/Sulphadiazine in used to cure pneumonia, throatinfection, meningits etc. Sulphaguanidine is usedto cure bacillary dysentary. Sulpha pyridine isused to cure pneumonia whereas sulphaacetamide is used to cure urinary tract infections.Sulpha thiazole is found useful againststaphylococal infections and bubonic plaguewhereas succinyl supha-thiazole is found usefulin intenstial infections such as bacillary dysentryand cholera.

(iii) Antiseptics : They may kill the bacteria or mayonly prevent their multiplication. They do notharm, the living tissues and can be applied directlyon cuts and wounds. Some common examplesof antiseptics are Dettol (Chlroexylenon+terpenoil), Savlon, acriflavin, methylene blue,gention voilet, merourochrome, boric acid andpotassium permanganate. Iodine toc, acts as astrong anti-bacterial agent in the form of tinctureof iodine (an alcohol water solution containing 2-3% iodine). In earlier days iodoform was used asan antibaceirial.

3C H

Chloroxylenol(Dettol)

3C H

C l

O H

Bithonel

C l O H

S

O H C l

C lC l

For creating antiseptic properties in soaps,bithonel is added. It removes the foul bodyodourproduced due to the bacterial decompositioin oforganic matters.

(iv) Sedative or Anti-Histamines : These drugesare also called anti allergic drugs because theyare used to treat allergy i.e. skin rashes, asthma,inflamation of tissues and itching of hives. Sinceallergy is caused due to release of histamine inthe body, these drugs are also called anti-histamines or sedatives. These drugs are widelyused in the treatment of hay fever, sneezing nasaldischarge, itching of eyes, nose and throat, nausain pregnancy and post operative vomitting. Theantihistamine drugs which are wide used are-chloropheniramine bromopheniramine, diphenylhydrazine, citrizine, promethazine etc.

OR (i) Nagarjuna : He was a famous chemist. He was

born in ‘Baluku village’ in Raipur Distt. Heprepared Gold from less costiller metals. Themethod of preparation of gold in mentioned in theGranth ‘Rasa Hridya’ He prepared ‘ Bhasma ofmercury’ also. He was appointed as V.C. ofNalanda University. ‘Nagarjuna Bridge’ on‘Krishana’ power was built on the memory of thegreat India Chemist.

(ii) Charak : Charak wa the son of a sage whotravelled from place of place on foot to cure thesuffering masses. He was the first physician topresent the concept of digestion, dosha orhumours, namely, bile, phlegm and wind. Illnessis caused when the balance among the threedosha in human body is disturbed. He prescribedmedicinal drugs to restore the balance betweenthese dosha. He alos knew the fundamentals ofgenetics. he said that a genetic defect in a child,like lameness or blindness is not due to anydefect in his mother of father but is due to defecin the ovum and sperm of parents (It is anaccepted fact today). He also studied theanatomy of the human body and various organs.According to him total number of bones, includingteeth present in the human body is 360. Hebelieved that heart has only one cavity and it isconnected to the entire body through 13 mainchannels. He revised the encyclopedic treatisewritten by Agnivesa of which came to be knownas Charaksamhita.

15. Positive deviation : 1. The vapour pressure of solution and partial

pressure of component are more expressed ByRault law.

2. At a particular temperature composition thevapour pressure of solution is maximum andboiling point is maximum.

3. The enthalpy change is more than 0. 0H∆ > 4. The volume change of solution is more than 0. 5. If intermining of solute and soluent the force of

attraction is decreased.Negative deviation :

1. The vapour pressure of solution and partialpressure of component is less expressed by Raullaw.

2. At a particular composition the vapour pressureof solutions is minimum and boiling point ismaximum.

3. The enthalpy change is less than 0. 0H∆ < .

4. The volume change is less than 0. 0U∆ < . 5. If intermining of solute and solvent the force of

attraction is increased.OR

Constant boiling mixture : Elevation in boilingpoint or adding one mole solute to 1000 gm ofsolvent.

[12]

S.N. Ideal Solutions Non-ideal solutions

1 The solutions which obeys the Rault's law at a given temperature and pressure.

The solution which do not obey the Rault's law at any temperature and pressure2 Enthalpy change is

equal to 0. Enthalpy change is not equal to zero.

3 Volume change is equal to zero.

Volume change is not equal to zero.

0H∆ = 0V∆ ≠

0V∆ = 0V∆ ≠

16. Weak electrolytes : Weak electrolytes are notionized completely. On dilution their ionizationincreases. At infinite dilution their ionization ismaximum. On dilutiion due to increase in thenumber of ions the molar conductivity of weakelectrolytes increases to a extent the limitingmolar conductivity of weak electrolytes can notbe calculated by extrapolatiion.

To find : mλ∞ (Molar conductivity of LiBr )= ?

( Li+ ion) λ ∞

+ = 2 138.75cm mol

−

( Bir− ion) λ ∞

− = 2 178.405cm mol

−

But, according to the Kohlraush’s law :

m X Yλ λ λ∞ ∞ ∞

+ −= +

1X = and 1Y = 38.7 78.40mλ∞ = +

117.10m

λ∞ =

∴Molar conductivity of LiBr is 2 1117.10 s cm mol

−

ORStrong Electrolytes : Strong electrolytes getcompletely ionized at all concentrations hencethe number of ions donot increases on dilution.At high concentration, due to high density of theions inter ionic interaction is also high. Due tothis reason mobility of ions is low. On dilutionthe distance between the ions increases resultingan increase in the mobility of ions. In this way,the molar conductivity increases on dilution.Here,

(Molar conductivity of 2Ba

+ ion)

2 1127.305 cm molλ∞ −

+ =

(Molar conductivity of Cl − ion)

2 176.345 cm molλ∞ −

− =

To find : (Molar conductivity of 2

BaCl ) ?mλ∞ =But, according to the Kohlraush’s law :

m

λ λ λ∞ ∞ ∞+ −= +

Here, 1X = , 2Y =

127.30 2 76.34mλ∞ = + ×

2 1279.98m s cm molλ∞ −=

Hence, molar conductivity of 2

BaCl is

2 1279.98 s cm mol

−

17. The general electronic configuration of transition

elements of d-block is ( ) 1 10 0 21n d ns

− −−

(i) Coloured ion and coloured compoundformation: The colour of transitional metal ions

is due to partially filled ( )1n − d-orbitals. In

transitional metal ions which contain unpaired d-electrons, transition of electrons takes place fromone d-orbital to another d-orbital. During thistransition it absorbs some radiation of visible lightand reflects the remaining radiation in the form ofcoloured light. Thus the colour of the ion iscomplementary to the colour absorbed by it.

For example : ( ) 2

2 6[ ]Cu H O

+ion appears blue

because it absorbs the red colour of the visiblelight for electrons promotion and reflects itscomplementary blue colour.

Ions Colour Ions Colour

Blue Violet

Violet Pink

Green Yellow

4Cr

+

2Mn

+

2Fe

+

3Cr

+

3Mn

+

3Fe

+

(ii) Formation of alloy : It is the homogeneous mixtureof two or more metals or metals with non-metals.Alloys are made to confer the property of metals.Transition elements have great tendency to formalloys because these elements have similaratomic size and can mutually substitute theirpositions in their crystal lattice. Alloys arecomparaatively hard and have higher m.p. thanthe elements from which they are made.

ORThe general electronic configuration of f-block

elements ( ) ( )1 14 0 1 22 1n f n d ns

− −− − .

Uses of lanthenides : (1) Ce and Th oxides are

used to gas lamps mantle. (2) Use of 2 3Pr O and

2 4Nd O for colour glass and filters.

Uses of Actinides : (1) Thorium salt use intreatment of Cancer. (2) In atomic reactorPlutonium use as a fuel for atom bomb.