how to prepare chemistry as a subject for iit

8/22/2019 How to Prepare Chemistry as a Subject for Iit

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HOW TO PREPARE CHEMISTRY AS A SUBJECT FOR IIT-JEE

You knowIIT JEEis the most prestigious & challenging examination for class XIIth students in India. It comprises

of three subjects e.q. physics, chemistry & maths. Out of these physics & chemistry play a crucial role in the

selection process of candidate. If you see the track record of previous years you will find that in chemistry &

physics student score more marks than maths, because of the theoretical nature of these subjects Chemistry isone of the top most scoring subjects. Meticulous planning of preparation can give you a marvelous increase in

your rank in IIT-JEE. Chemistry comprises of three branches physical, organic & inorganic. There are following

tips which would be useful for you to prepare for chemistry as subject. Physical chemistry which comprises of

about 40% of the entire paper can fetch you entire marks if you concentrate on understanding, the basic concept

of each chapter. My personnel suggestion is that you should make your own notes for each concept. Each

chapter has its own shortcuts, which can be understood with the help of a professional.

Important Topics of Physical Chemistry:

* Chemical Equilibrium

* Ionic Equilibria* Electrochemistry

* Solid State

* Volumetric Analysis

* Thermochemistry/Thermodynamics

* Chemical Bonding

* Atomic Structure

Besides these there are other topics also which are less important with respect to examination. So you can

prepare your studies accordingly.

Organic chemistry requires a clinical approach. You know IIT-JEE is anengineering entrance examwhere moreemphasis is given on synthetic organic chemistry rather than theoretical organic chemistry.

In order to have solid grip over the subject first of all mechanism portion should be understood.

After that chapterwise problems should be solved. The mechanism portion of organic chemistry can be classified

as follows:

* Nomenclature

* Isomerism

* Concept of acidity & basicity

* sp3-hybridised nucleophillic substitution

* sp2

-hybridised nucleophillic substitution* Elimination reactions

* Aromatic nucleophillic & Electrophillic substitution

* Free Radical mechanism

*

* Name Reactions

Use of Above Chapters:-

* When you understand sp3-hybridised carbon you can solve the question of alkyl halide, alcohol, ether,

expoxide.

* When you understand sp2-hybridised carbon you can solve the questions of alkene, alkyne, aldehyde, ketone,
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acid & its derivatives.

* These chapters also help you to understand name reactions. These reactions are specific examples of a

particular concept.

* In general organic chemistry one should understand two concepts very well.

1. Concept of acidity & basicity.* This portion definetely fetches you 2 questions precisely questions on phenol, carboxylic acids & aniline should

be practiced

2. Isomerism: In the field of isomerism more emphasis should be given to stereoisomerism more preciesely the

optical isomerism. One should learn the formula's for calculating no. of isomers both for stereosiomerism as well

as structural isomerism.

Inorganic Chemistry Course Topics:

Inorganic chemistry preparation should be more precise than exhaustive because it contribute to about 20% of

your paper. Infact there are few topics which should be concentrated more than others.

* Qualitative Analysis :- This topic should be well prepared because it contributes to major portion of inorganic.

* Coordination Compound :- Now a days this topic is getting momentum. In every entrance examination you

find at least 2 to 3 questions related to this topic.

Beside above I would like to suggest that in IIT-JEE chemistry, the more important fact is the applicative aspect

of chemistry rather than the theoretical one. Hence an expert's advise will make you understand things in a clear

manner. If you want to understand what I mean by expert advise read my next article. e.g.

* Do you understand Le-Chatlier's principle

* in volumetric

analysis ?

* Can you solve all the problems on volumetric by single method ?

* Do you know the short cuts in Ionic Equilibrium ?

Do you understand Le-Chatlier's Principle ?

It is one of the most Important principles of our daily l ife. Le-chatlier's principle states that if a system at

equilibrium is applied by an external stress (such as pressure, temperature, volume etc.) then the system adjusts

itself in such a way so as to nullify the affect of applied external stress.

Most of the time to understand the affect of external stress gas eqn. PV=nRT is used but this equation can not

handle more than two variable at a time. In order to understand the real application. of above law you can draw a

projector equation comprising of all the external variables on one side & internal variables on the other side of the

equation. Then apply the boundry condition on it to give the final result.

Example Question .;

#. For the following reaction what happens when

SO2+1/2O2 SO3

a) at const. Temperature pressure of the system is increased.

b) at const. Temperature volume of the system is increased.

c) at const. Temperature and pressure helium gas is added to the flask.

d) at const. Temperature and volume helium gas is added to the flask.

Solution:

Let us assume that at equilibrium moles of SO2 be ;SO3 be & O2 be & nT represents the total no of moles at equilibrium & total pressure be P atm.


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SO2 + 1/2 O2 SO3

Above are two prejectors to project your answers; Let us see

Ans a : In equation (i) if temperature is constant KP is fixed pressure you are increasing & nT is fixed. Therefore

lefthand side is increasing which means right hand side should also increase

nSO3 should increase or nSO2 & nO2 should decrease forward direction

Ans b : In equation (ii) if temperature fixed KP constant

Volume increased left hand side is decreases

nSO3 is decreased or nSO2 & nO2 is increased backward direction

Ans c : In equation (i) if T is fixed KP constant & P is fixed but 'He' is added means nT isincreased left hand side is decreased nSO3 is decreased backward direction

Ans d : In equation (ii) if T & V is fixed left hand side is fixed but there is no function of nT hence

addition of 'He' has no affect on it No change.

# For the following reaction at equilibrium

A(g) + 2B(g) C(g) + 3D(g)

What happens when at constant Temperature, Pressure of the system is increased but simultaneously

some amount of C was added so as to maintain the partial pressure of A constant.

a. Forward

b. backward

c. no change

d. data insufficient

Solution.

A + 2B C + 3D

Let the moles be nA, nB , nC and nD respectively & total pressure be P atm.


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In above function on left hand side both numerator & denominator is increased

Hence unless the rate is known the answer cannot be predicted

Data insufficient

Hence option (4) is correct.

Do you know that there are only 4P4 = 24 no. of questions in volumetric analysis :

Volumetric Analysis means titration of compounds which means mixing of compound A with compound B but thatis possible ony when compound A reacts chemically with compound B. In volumetric Analysis questions can be

classified under following four categories.

1. Redox titration

2. Iodometric titration

3. Acid Base Titration

4. Precipitation method

All these are manual titrations a question can be formed only on the concept of manual titration. Since there are

only four methods therefore the total no. of questions in volumetric Analysis will be only

There is a very interesting fact that you can solve all the questions by a single method called n factor method.

Moreover this method doesnot require to remember any reaction. You can solve any question without writing the

chemical reaction because molar ratio can be calculated with the help of n factor. If you want to understand this,

read my next article on n-factor.

Diels Alder Reaction:

The reversible and thermal (4 , 2 ) cyclo addition reaction between a diene and a dienolphile (dieneoving

species usually contains an active multiple bond) in which the dienophile adds 1, 4 to the diene to produce a six

membered cyclic product (adduct) is known as Diels Alder Reaction. Here the two units the diene and the

dienophile, get joined by two bonds created from the two bonds in two units, a new bond is also formed.

The reaction may also be caused by the catalytic amount of light or lewis acids.


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1.

The equilibrium lies over to right; if the temp. is raised, the equilibrium swifts to the left and the reverse diels Alder

product are obtained.

* However the dienes may be acylic, cyclic and aromatic conjugated systems. Even non conjugated Diens of

appropriate geometry may undergo the Diels Alder. Diens noted below:-

Several types of Dienophiles may be used in the Diels Alder reaction:-

(d) Diens : -C = C - C = C-, -C = C = C- (e) other then C-C :- CO2Me-N = N - CO2Me

Tetra cyano ethene is known to be the best dienophile till to date.

The procedure for Diels Alder Reaction is very simple only the mixture of the Diene and the dienophile needs

heating with or without solvent, the common solvent is benzene.

Some Example:


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It has been found that the electron donating groups in the dienes accelerate the reaction while the electron

withdrawing group retard it, on the other hand reverse effects have been found in the cases of dienophiles - the

electron withdrawing group in them accelerate the reaction whereas the electro donating group retards the same.However the opposite electronic effects on the diene and dienophile accelerate the rate but similar deccelerate

the same. Cyclic dienes in the cisoid form react more readily then the corresponding acylic dienes. In general

with the increasing sub. in the diene the rate of the reaction decreases, tetra sub. dienes in 1, 4- position do not

reacts. In fact stearic effects have pro found influence on the reaction. Large groups on 2,3 positions make it

unreactive. The cisoid conformation of such a diene gets frozen to the transoid form because of the steric strain

developed by the large groups. Bulky cis, cis 1,4 - disub. dienes also do not react. Among the four 1,4 diphenyl

butadienes only the trans isomer gives the Diels Alder Reaction.

Since the diene and the dienophile during addition can orient in two ways, a mixture of products are obtained, of

course one of them is the major product.

The discussion on the stereochemistry involves several steps :-

(a) Only conjugated dienes in the cisoid conformation undergo the reaction. Any factor that makes the diene

assume the transoid conformation will hinder the reaction.

(b) The reaction is stereospecific with respect to the dienophiles and the addition is always cis, the cis group.

Remains cis and trans group remains trans in the product.

Q1. Explain why :-

(a) Cyclo pentadiene remains as dimer but the former can be obtained from the latter on distilling at room


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temperature at 170C - 200C.

(b) Tetraphenyl cyclopentadiene is monomeric at room termperature.

Solution:

(a) Cyclo pentadiene is highly reactive as a diene and it is moderately so as a dienophile when cyclo pentadiene

is kept at room temperature the self Diels Alder Reaction occurs and a dimer is formed. However Diels Alder

Reaction is a reversible reaction. For this reason on heating the monomer, cyclo pentadiene forms.

(b) The moecule has four bulky substituents, its dimeric form experiences high steric repulsion so dimer can't

exist at room temperature.

Q.2. Explain why the following compounds do not undergo Diels Alder reaction as dienes

(i)

(ii)

(iii)

(iv)

(v)

Solution:-

(a) Dienes in the cisoid form undergoes the Diels Alder Reaction. Transoid forms do not undergo the reaction

because this forms leads to a highly strained six member. T.S. whose energy content is very high. Compound (i)

and (iv) experience high strain in the cisoid form and they remain in the transoid form. So compounds (i) and (iv)

do not undergo. Compound (iii) is locked in the transoid form, so it also not undergo the reaction. Compound (ii)1,3- butadyne is a linear molecule Hence all the C's are in sp hybridised state. The p.A. O's on and carbons of

the compound when overlap the p.A. O's of the dienophile, a highly strained 1, 2, 3 cyclo hexatriene will be

formed the triene moleity will be linear and corresponding T.S. will also have trienyl like linear moeity with high

strain and energy. It is difficult to over come the energy Barrier of the T.S. so the reaction does not occur.

For (v) the two double bonds of the molecule are nearly at right angle to one another so in this case also the T.S.

will be strained but less strained than that of compounds (i) to (iv). Hence reaction takes slowly.

Q.3. Why does anthracene undergo the Diels Alder Reaction but naphthelene does not ?

Solution:

The energy gained by the cyclo addition reaction being less then the energy lost by the naphthalene moleculeowing to the loss of aromaticity by one of the two benzene rings, naphthelene does not take part in the Diels


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Alder. In the case of anthracene the energy gained by the reaction is more than the energy lost by the loss of

aromaticity of the middle ring and so the reaction occurs.

Problems for Practice

1. Reactivity order of the following Dienophiles :

higher the electron withdrawing effects higher will be reactivity.

2. Give the structure of the product of the Diels-Alder reaction between.

a. maleic anhydride and isoprene b. maleic anhydride and 1,1'-bicyclohexenyl(I)

c. maleic anhydride and 1-vinyl-1-cyclohexene

d. 1,3-butadiene and methyl vinyl ketone

e. 1,3-butadiene and crotonaldehyde f. 2 mol 1, 3-butadiene and dibenzalacetone

g. 1,3-butadiene and -nitrostyrene (C6H5CH=CHNO2)

h. 1,3-butaddiene and 1,4-napththaoquinone (II)

i. p-benzoquinone and 1,3-cyclohexadiene

j. p-benzoquinone and 1,1'-bicyclohexenyl (I)

k. p-benzoquinone and 2 mol 1, 3-cyclohexadiene

l. p-benzoquinone and 2 mol 1, 1'-bicyclohexenyl (I)

m. 1,3-cyclopentadiene and acrylonitrile n. 1,3-cyclohexadiene and acrolein

3. From what reactants could the following be synthesized by the Diels-Alder reaction ?

a.

b.

c.

d.

e.


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f.

g.

h.

i.

4. The following observation illustrate one aspect of the stereochemistry of the Diels-Alder

Reaction: maleic anhydride + 1, 3-butadiene A(C8H8O3)

A + H2O, heat B(C8H10O4)

B + H2, Ni C(C8H12O4), m.p. 192C

fumaryl chloride (trans-ClOCCH = CHCOCl) + 1, 3-butadiene D(C8H8O2Cl2)

D + H2O, heat E(C8H10O4)

E + H2, Ni F(C8H10O4), m.p. 215C

F can be resolved; C cannot be resolved.

Does the Diels-Alder reaction involve a syn-addition or an anti-addition ?

5. Give the stereochemical formulas of the products expected from each of the following reactions. Label meso

compounds and racemic modifications.

a. crotonaldehyde (trans-2-butenal) + 1, 3-butadiene

b. p-benzoquinone + 1,3-butadiene

c. maleic anhydride + 1, 3-butadiene, followed by cold alkaline KMnO4

d. maleic anhydride + 1, 3-butadiene, followed by hot KMnO4 C8H10O8.

6. (a)

(b) benzil (C6H5COCOC6H5)+dibenzyl ketone (C6H5CH2COCH2C5H5)+base A(C29 H20O), ''tetracyclone''

A + maleic anhydride B(C33H22O4)

B + heat CO + H2 + C(C32H20O3)

(c) A + C6H5C CH D(C37H26O)

D + heat CO + E(C36H26)

(d) hexachloro-1, 3-cyclopentadiene + CH3OH + KOH F(C7H6Cl4O2)


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F + CH2 = CH2, heat, pressure (C9H10Cl4O2)G

G + Na + t-BuOH (C9H14O2)H

H + dilute and I(C7H8O), 7-ketonobronene

7. Give the likely structures for A to E, 1,3-butadiene + propiolic acid (HC = CCOOH) A(C7H8O2)A + 1mol LiAlH4 B(C7H10O)

B + methyl chlorocarbonate (CH3OCOCl) C(C9H12O3)

C + heat (short time) toluene + D(C7H8)

D + tetracyanoethylene E(C13H8N4)

Compound D is not toluene or 1,3,5-cycloheptatriene; on standing at room temperature it is converted fairly

rapidly into toluene.

1. 20 ml 2 N KMnO4 is treated with excess MnSO4 .4H2O in presence of excess H2O. The only product which

contain Mn is MnO2. The no. of ml moles of MnO2 obtained are :

a. 20

b. 13.33c. 33.33

d. 8

2. 0.7538g sample contains an unknown amount of As2O3 (M.wt. = 197.84). The sample was treated with HCl

and reducing agent, resulting in the formation of AsCl3(g), which was distilled into a beaker of water. Following

the hydrolysis of AsCl3as AsCl3+ 2H2O HAsO2+ 3H++ 3Cl

-the amount of HAsO2was determined by titration

with 0.05264 MI2 , requiring 33.64 ml to reach the equivalence point. The redox products in the titration were

H3AsO4 and I-. What was the weight percent of As2O3in sample.

a. 23.24%

b. 24.25%c. 11.0%

d. none of these

3. 80 ml 44.8 volume H2O2is mixed with 120 ml 3M H2O2 solution. The volume strength of resulting solution is

a. 38

b. 19

c. 76

d. 29

4. An equimolar mixture (w gm) of K2CO3 and KHCO3 when treated with 0.6 M H2SO4 using methyl orange as

indicator, the end point is obtained when 60 ml of H2 SO4 has been added. The same mixture (2w gm) when

treated with x M H2 SO4 using phenolphthalein as indicator the end point is obtained when 0.6 x litre of

H2SO4 has been added. x is

a. 0.1

b. 0.2

c. 0.4

d. 0.6

5. 2 mol of gas (A) and 96 gm of SO2is mixed. The mass fraction of (A) in the mixture is 5/8. The gas (A) may be


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a. SO3

b. NO2

c. O2

d. none

6. W gm of a mixture containing Na2 CO3 and NaHCO3 in 2 : 3 mol ratio requires x ml HCl for neutralisation using

phenolphthalein as indicator. 2W gm of same mixture require y ml HCl for neutralisation using methyl orange as

indicator. The value of y is

a. 5x

b. 7 x

c. 3.5 x

d. none

7. x gm of a solute is dissolved in a liquid to give a solution of volume V ml and density d gm/ml. The molality of

the solution is (the mass of 1 molecule of solute is m gm) {N0 is Avogadro number} :a. x /(Vd - x)m

b. 1000x/m(Vd-x)N0

c. x/N0 m(Vd - x)

d. 1000x/N0 (Vd -x)

8. 20 ml x M NaOH requires V ml y N H2 SO4 for complete neutralisation. Na2 SO4 formed is oxidised by 2V ml

0.1M F2 into Na2 S2 O8 . The value of y is :

a. 0.4

b. 0.8

c. 1.6d. 0.2

9. 20 ml 0.1 N of a readucing agent reduces completely X2 into X-. Now X

-is completely oxidised into X2 O

2+by

V1 ml 0.241M KMnO4 (in H2 SO4 medium). Now X2 O2+

is completely oxidised in X3 O43+

by V2 ml 0.4M

K2 Cr2 O7 (in H2 SO4 medium). The value of V1& V2 is

a. 6, (25/9)

b. 12, (25/18)

c. 6, (25/18)

d. none of these

10. 20 ml of 0.2 M KMnO4 is reduced into Mn2 O3 during oxidation of excess of hydrogen gas. Mn2 O3 is oxidised

into K2 MnO4 by CrO3 . The reduction product of CrO3 is further reduced by a reducing agent (Mol. wt. = M. eq.

wt. = M/3)into Cr. If the wt. of Cr obtained is 312 mg then the oxidation state of Cr in the reduction product of

CrO3 is :

a. 2

b. 3

c. 4

d. none

11. x ml moles of FeCr2 O4 is treated with air (containing 20% by vol O2 ) in presence of excess Na2 CO3 . The


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products are Na2 CrO4 , Fe2 O3 and CO2

. The volume of air required measured at N.T.P. is

a. 196 x ml

b. 98 x ml

c. 28 x mol

d. 168 x mol

12. The concentration of CO2 is expected to rise 440 ppm in year 2020. What would be the pH of rainwater in the

year 2020 if molar solubility of CO2 is 0.0343 L-1

bar-1

. Ka1 of CO2 = 4.210-7

.

a. 5.63

b. 4.72

c. 5.063

d. none

13. 1.0 metric ton coal containing 2.5% sulphur is burned and SO2 produced in the combustion was dissolved inrainwater of volume equivalent to 2.0cm rain fall over 2.6km

2area. Determine pH of rainwater if Ka of H2 SO3 =

1.510-2

.

a. 4.8

b. 4.08

c. 0.48

d. none

14. 30 ml 0.3 M H2 SO3 solution reduces excess KIO3 into I2 . The no. of ml moles of hypo required to convert

50% of the I2 formed into NaI area. 1.8

b. 0.9

c. 3.6

d. none

CHEMICAL EQUILIBRIUM

1. 0.01 mol of HBr(g) are heated in a vessel. At equilibrium 10-6

mol of Br2 were found. The value of equilibriumconstant for 2HBr(g) H2 (g) + Br2 (g) is :

a. 110-2

b. 110-8

c. much greater than 110-8

d. slightly less than 110-12

2. A(g) dissociates into B(g), C(g) reversibly as A(g) B(g) + 3C(g). The molar mass of A was found to be

decreased by 20%. The degree of dissociation of A is

a. (1/5)

b. (1/24)c. (1/6)


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d. (1/12)

3. The degree of dissociation of SO3 is a under the condition that equilibrium pressure is equal to KP for

2SO3(g) 2SO2(g) + O2(g) . The degree of dissociation of SO3 under the condition that KP = x equilibrium

pressure is 3 /4. The value of isa. 5

b. 1/5

c. 2

d. none

4. 3 mol of PCl5 is heated in presence of 1 mol Ne. at equilibirum the mol fraction of Cl2 was found to be 1/3. Just

after the attainment of equilibrium volume is doubled as well as temperature is increased. Knowing the

dissociation to be endothermic the no. of moles of Cl2at new equilibrium will be

a. 3

b. 1.5c. 2

d. 2.5

5. The mol fraction of N2 in a mixture of N2 and H2 is 1/3. They react to form N2 H4(g) . At equilibrium N2 H4 was

found to be 40% by mol. The value of KP P2

for N2 + 2H2 N2 H4where P is pressure of the mixture at

equilibrium is

a. 12.5

b. 25

c. 6.25

d. None

6. Solid ammonium carbamate (NH2 CO2 NH4 ) is heated in a vessel (initially evacuated). The manometer

attached with the vessel shows maximum pressure of 6P0 . Now the solid is heated at the same temperautre in

another vessel of same capacity in presence of ammonia and CO2 in molar ratio 2:1. The manometer attached

with the vessel shows minimum pressure of 3P0 . The maximum pressure shown by the manometer will be

a. 6P0

b. 9P0

c. 32P0

d. None

7. A vessel contains SO3 , SO2 , O2 , He its molar ratio of 1:2:3:4 at equilibrium. Which of the following is correct

a. KP for 2SO3 2SO2 (g) + O2 (g) is 1.2P where P is the total pressure of the vessel.

b. is the pressure due to inert gas.

c. both

d. none

8. The degree of dissociation of SO3 at equilibrium pressure of O2 equal to P0 is . KP for : 2SO3 (g) 2SO2 (g) +

O2 (g)

a. P0 [ /(1- )]2

b. [(P02

)/(1- )]


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c. [(P03

)/2(1- )]2

d. [(2P03

)/(1-2

)]

9. N2 (g) and H2 (g) are taken in 1:3 molar ratio. At equilibrium 40% by wt. ammonia is obtained. If equilibrium

pressure is P0 atm. KP for 1/2N2 + 3/2H2NH3 is

a.

b.

c.

d. None

10. A vessel contains 1mol N2 , 2 mol NH3 and 3 mol H2 at equilibrium. x moles of H2 are removed along with

volume is doubled. At new equilibrium 17gm of ammonia is found. The value of x is

a. 5.9b. 1.9

c. 2.6

d. 4.5

11. 0.5 mol of Fe(s) is treated with steam. At equilibrium 0.3 mol of solid is found. At equilibrium pressure was

found to be 6atm at 1000K. The volume of the vessel is 8.2litres. KP for 3Fe(s) + 4H2 O(g) Fe3 O4(s) +

4H2(g)

a. 1

b. 2

c. 8d. 16

12. 2 mol of N2 H4(g) is heated to form N2 and H2 . As soon as N2 and H2 are found they react to form NH3 . Two

equilibria N2 H4(g) N2 + 2H2 ........(i) and N2 + 3H2 2NH3 ........(ii), are simultaneously established.

Upto equilibrium the degree of dissociaiton of N2 H4 was found to be 50%. Which of the following is correct at

equilibrium

a. [N2 ] + [H2 ] + 2[NH3 ] = 3[N2 H4 ]

b. [N2 ] + [H2 ] + [NH3 ] = [N2 H4 ]

c. [N2 H4 ] + [NH3 ] = [N2 ] + [H2 ]

d. None of these

13. Ammonia is introduced in a vessel (initially evacuated). Ammonia is heated, it dissociates into N2 and H2 .

The vapour density of the mixture is

a. 4

b. 4.25

c. 6.5

d. 23.5

14. The degree of dissociation of PCl5 (g) at initial pressure of P0 is same as the degree of dissociation ofSO3 into SO2 and O2 at equilibrium pressure equal to KP for 2SO3 2SO2 + O2 . The value of KP for


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PCl5 PCl3 + Cl2 is

a. 4P0 /5

b. 4P0 /3

c. 4P0

d. None

15. Which of the following may be correct

a. 2NH3 (g) N2 (g) + 3H2 (g); KP =2.4910-3

atm

b. 2SO2 (g) + O2 (g) 2SO3 . yield to SO3 is maximum at low temp., high pressure.

c. C(s) + H2 O(g) CO(g) + H2 (g); KP = 9.4210-6

atm

d. both (2) & (3)

16. 4 moles of PCl5 (g) and 2 moles of CO(g) are heated in a vessel. Two equilibria : i. PCl5(g) PCl3 (g) +

Cl2 (g) ; ii. CO(g) + Cl2 (g) COCl2 (g) are established. At equilibrium 3 moles of PCl3 and 2 moles of

Cl2 are obtained. Which of the following at equilibrium is correcta. moles of CO = 1.5

b. KC (i) x KC (ii) = moles of PCl5 + moles of CO + moles of COCl2 .

c. both

d. none

17. The degree of dissociation of COCl2 into CO and Cl2 is 0.265 at equilibrium pressure of P0 . The degree of

dissociation at equilibrium pressure of 2P0 is

a. 0.3125

b. 0.53

c. 0.18d. 0.265

18. Equimolar mixture of A2 B4(g) and AB3(g) is heated.Two equilibria A2 B4(g) A2(g) + 2B2(g) .......(i) ;

2AB3(g) A2(g) + 3B2(g) ........(ii) are simultaneously established. At equilibrium moles of A2 B4 are equal to

the sum of moles of A2 and B2 . Also the degree of dissociation of A2 B4 and AB3 is same. The value of kc(i) /

kc(ii) is

a. 320/27

b. 160/3

c. 80/3

d. none

19. At 40C in a 5 litre flask LiCl.3NH3 (s) starts to decomposes to give LiCl.NH3 (s) & NH3 . After sufficient time

the pressure of NH3 was observed to be 3atm. Now more LiCl.3NH3 (s) is added at the same temperature but no

increase in pressure of NH3 is observed even after long time. Then:

a. KP = 9

b. KP< 9

c. KP = 3

d. KP> 9

20. 200gm CaCO3 and 48gm carbon is heated in a vessel. Two equilibria CaCO3 (s) CaO(s) + CO2 (g)


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.....(i) ; CO2 (g) + C(s) 2CO(g) .......(ii) are simultaneously established. At equilibrium 1 mol of CO is found

and 50gm of CaCO3 is remained. Which of the following is correct

a. KC (1) = KC (2)

b. Total amount of solid at equilibrium = 176gm

c. both

d. none


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Organic Chemistry MCQs

ORGANIC CHEMISTRY

1.

a.

b.

c.

d. none

Ans : (1)

2.

a.

b.

c.

d. none

Ans : (1)
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3.

a.

b.

c.

d. none of these

Ans : (1)

4.

a. Tircyclo [2,2,1,02,6

] heptane

b. Bicyclo [2,2,1,02,6

] heptane

c. Spiro [2,2,1,02,6

]

d. None of these

Ans : (1)


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5.

a.

b.

c.

d.


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Ans : (3)

Because the most acidic 'H' will be absorbed by the base.

6.

a.

b.

c.

d.

Ans : (3)

7. Which of the following compound can show chiral


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a.

b.

c.

d.

Ans : (1)

8.

a.

b.

c.

d.


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Ans : (3)

9.

a.

b.

c.

d.

Ans : (1)

10.


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a.

b.

c.

d.

Ans : (3)

11.

a.

b.

c.


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d.

e. none of these

Ans : (5)

12.

a.

b.

c.

d.

e. none of these

Ans : (5)


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13.

a.

b.

c.

d. none of these

Ans : (4)


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Benzene is its one of the isomers.

14.

a.

b.

c.

d.

Ans : (3)


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15.

What is Y ?

a.

b.

c.

d.

Ans : (2)

16. Hydration of the carbonyl group : C = O

a. is favoured by low pH of the medium


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b. is facilitated by high pH of the medium

c. attains a minimum value at a moderate pH

d. reaches a maximum value at a moderate pH

Ans : (4)

17. Identify the product Y and Z in the following reaction sequence.

a.

b.

c.

d.

Ans : (2)

18. Identify A to E in the following scheme :

a.

b. CH

2

N

2

; Ph

2


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CO; EtMeCHC(OH)Ph

2

; EtMeC = CPh

2

; EtCOMe

c. EtMeC = CPh

2

; EtMeCHC((OH)Ph

2

; EtCOMe ; Ph

2

CO ; CH

2

N

2

d.

Ans : (1)

19.

a.

b.


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c.

d.

Ans : (3)

20.

a.

b.

c.

d.

Ans : (3)


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21. A compound (A) of formula C

3

H

6

Cl

2

on reaction with alkali can give (B) of formula C

3

H

6

O or (C) of formula C

3

H

4

. (B) on oxidation gave a compound of the formula C

3

H

6

O

2

. (C) with dilute H

2

SO

4

containing Hg2+

ion gave (D) of formula C

3

H

6

O, which with bromine and NaOH gave the sodium salt of C

2

H

4

O

2

. Then (A) is

a. CH

3

CH

2

CHCl


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2

b. CH

3

CCl

2

CH3

c. CH

2

ClCH

2

CH

2

Cl

d. CH

3CHClCH

2

Cl

Ans : (1)

22. If pK

a

values of alcohol, water and phenol are 17, 14 and 10 respectively then which of the following reaction is

possible :

a.

b.

c.

d. None of these

Ans : (3)


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The acidic character order is

alcohol< water < phenol

basic charecter order for conjugate base is:

C

2

H

5

O-

> OH-

> C6

H

5

O-

.

The reaction in between one strong acid and other weak base or vice versa is possible.Also note that phenols

are acidic in nature.

23.

a.

b.

c.

d. none

Ans : (1)


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24.

a.

b.

c.

d. none

Ans : (1)

25.

a.

b.

c.

d. none of these

Ans : (1, 2)


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Physical Chemstry - M C Qs

PHYSICAL CHEMISTRY

1. For the equilibrium : LiCl.3NH3(s) LiCl.NH3(s) + 2NH3 , Kp =9 atm2

.At 40C. A 5 litre vessel contains0.1mole of LiCl.NH3 . How many mole of NH3

should be added to the flask at this temperature to derive the backward reaction for completion ?

a. 0.7137 mole

b. 0.7937 mole

c. 0.7837 mole

d. none of these

Ans : (3)

3 5 = n 0.0821 313

n = 0.5837

i.e,. (a - 0.2) = 0.5837

Initial mole of NH3

= a = 0.5837 + 0.2

= 0.7837 mole

2. A weak acid HA after treatment with 12mL of 0.1M strong base BOH has a pH of 5. At the end point, the

volume of same base required is 26.6mL. Ka of acid is

a. 1.810-5

b. 8.210-6

c. 1.810-6

d. 8.210-5

Ans : (2)
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Ka

= 8.219 10-6

.

3. 0.1 millimole of CdSO4 are present in 10 mL acid solution of 0.08 N HCl. Now H2 S is passed to precipitate all

the Cd2+

ions. The pH of the solution after filtering off precipitate, boiling of H2 S and making the solution 100 mL

by adding H2 O is :

a. 2

b. 4

c. 6

d. 8

Ans : (1)

4. Assuming non change in volume, calculate the minimum mass of NaCl neccesary to dissolve 0.010mol AgCl

in 100L solution.

a. 15 kg

b. 21 kg

c. 10 kg

d. 17 kgAns : (4)


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5. 338 mL clear saturated solution of AgBrO3 requires just 30.4 mL of H2 S(g) at 23C and 748 mm Hg to

preciptate all the Ag+

ions into Ag2 S. What will be KSP of AgBrO3 ?

a. 5.2910-5

b. 5.010-3

c. 3.45106

d. 5.2910-1

Ans : (1)

6. For the following reaction what happens when at constant (T), the 4HNO3 4NO2 + 2H2 O +

O2 pressure of the flask is increased but simultaneously some amount of O2 was added so as to maintain the

partial pressure of H2 O contant

.a. forward

b. backward

c. no change

d. data in sufficient

Ans : (4)


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7. For the reaction R - X + OH-

ROH +X-, the rate is give as :Rate = 5.010

-5[R-X][OH

-] +

0.2010-5

[R-X]. What percentage of R-X react by SN 2 mechanism when [OH-] = 1.010

-2M ?

a. 96.1%b. 3.9%

c. 80%

d. 20%

Ans : (4)

8. Aniline is diazotized and the diazonium salt hydrolysed to yield phenol which is brominated to produce

C6 H2 (Br3 )OH. Calculate the mass of the final product obtained from 9.3g of aniline if the yield in the two steps

is 45% and 70%(Atomic mass of Br = 80)

a. 1.04g

b. 10.43g

c. 14.89g

d. 23.17g

Ans : (2)


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9. A sample of53 I131

as iodide ion was adminstered to a patient in a carrier consisting of 0.10 mg of stable iodide

ion. After 4 days 67.7% of initial radioactivity was detected in the thyroid gland of the patient. What mass of

stable iodide ion had migrated to thyroid gland. t1/2 for iodide ion = 8 days.

a. 95.8%b. 72.3%

c. 90.8%

d. none

Ans : (1)

Thus, iodide ion left is 0.707 part of initially injected sample, however the rate decreases only 67.7% or 0.677 in

4 days, thus. If 0.707 is left then iodide ion migrated to thyroid = 1 Thus, 0.677 is left then iodide ion migrated to

thyroid

or 95.8% of the iodide ion is migrated to gland.

10. The isotopic composition of rubidium is85

Rb - 72 percent and87

Rb -28 percent.87

Rb is weakly radioactive

and decay by - -emission with a decay constant of 1.110-11 per year. A sample of the mineral pollucite was

found to contain 450mg Rb and 0.72 mg of87

Sr. Estimate the age of mineral pollution, stating any assumption

made.

a. 5.202108

yr

b. 2105yr

c. 1102

yr

d. none

Ans : (1)


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11. In the Lindemann theory of unimolecular reactions, it is shown that the apparent rate constant for such a

reaction is where C is the concentration of the reactant, k1and are constants. Calculate the

value of C for which kapp has 90% of its limiting value at C tending to infinitely large values, given = 9 105 .

a. 10-6

mole/litre

b. 10-4

mole/litre

c. 10-5

mole/litre

d. 510-5

mole/litre

Ans : (3)

12. 1 g charcoal is placed in 100mL of 0.5 M CH3 COOH to form an adsorbed mono-layer of acetic acid

molecule and thereby the molarity of CH3 COOH reduces to 0.49. Calculate the surface area of charcoal

adsorbed by each molecule of acetic acid. Surface area of charcoal = 3.01102

m2

/g.

a. 110-9

m2

b. 510-19

m2

c. 1 10-8

m2

d. none

Ans : (2)


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13. Suppose we have a cube of 1.00 cm length. It is cut in all three directions, so as to produce eight cubes,

each 0.50 cm on edge length. Then suppose these 0.50cm cubes are each subdivided into eight cubes 0.25cm

on edge length, and so on. How many of these successive subdivisions are required before the cubes are

reduced in size to colloidal dimensions (100 nm).

a. 13 subdivisions

b. 15 subdivisions

c. 19 subdivisions

d. 17 subdivisions

Ans : (4)

We find that every division in two equal halves also reduces the size of edge length to one half.

In first subdivision 1 cm is reduced to 0.5 cm In second subdivision 0.5cm is reduced to 0.25cm

14. AB, A2 and B2 are diatomic molecules. If the bond enthalpies of A2 , AB and B2 are in the ratio 1 : 1 : 0.5 and

the ethalpy of formation of AB from A2 and B2 is -100 kJ mol-1

, what is the bond enthalpy of A2

a. 400 kJ mol-1

b. 200 kJ mol-1

c. 100 kJ mol-1

d. 300 kJ mol-1

Ans : (1)


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15. When 1 mole of crystalline NaCl is obtained from sodium and chlorine gas, 410 kJ of heat is released. The

heat of sublimation of Na metal is 108 kJ mol-1

and Cl-Cl bond enthalpy is 242 kJ mol-1

. If the ionisation energy of Na is 493.0 kJ mol-1

and the electron affinity

of chlorine is 368 kJ mol-1

, the lattice energy of NaCl is

a. 764 kJ mol-1

b. -744 kJ mol-1

c. 885 kJ mol-1

d. -885 kJ mol-1

Ans : (2)

16. The enthalpy of neutralisation of a weak acid in 1M solution with a strong base is -56.1 kJ mol-1

. If the

enthalpy of ionization of the acid is 1.5 kJ mol-1

and enthalpy of neutralisation of the strong acid with a strong base is -57.3 kJ equiv-1

, what is the % ionization

of the weak acid in molar solution (assume the acid to be monobasic) ?

a. 10

b. 15

c. 20

d. 25

Ans : (3)


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17. A mineral having the formula AB2 crystallises in a cubic close packed lattice, with A atoms occupying the

lattice points. The co-ordination number of A atoms, that of B atoms and fraction of tetrahedral voids occupied

by B atoms are respectively

a. 8,4,100%

b. 2,6,75%

c. 3,1,25%d. 6,6,50%

Ans : (1)

The structures of AB2 refers to fluorite type structure

18. In a compound XY2 O4 , oxide ions are arranged in ccp and cations X are present in octahedral voids.

Cations Y are equally distributed between octahedral and tetrahedral voids. The fraction of the octahedral voids

occupied is

a. 1/2

b. 1/4

c. 1/8d. 1/6

Ans : (1)

Four oxide ions arranged in CCP would have four octahedral voids associated with them. Out of these four voids

one is occupied by X and one by Y. So, 2 voids out of 4 are occupied. So 1/2 is the answer.

19. What is the packing fraction of the two dimensional unit cell shown in figure.

a. 0.92

b. 0.74

c. 0.5234

d. none

Ans : (1)

packing fraction

20. In the vapour phase acetic acid molecule associate to the some extent to form dimers. At 50C, the pressure

of a certain acetic acid vapours is 0.0342 atm in a 360mL flask. The vapour is condensed and neutralized with

13.8mL 0.0568M NaOH. Calculate the degree of dissociation of the dimer.

a. 0.84

b. 0.94

c. 1.72

d. none of these

Ans : (1)


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21. The concentration of CO2 is expected to rise 440 ppm in year 2020. What would be the pH of rainwater in

the year 2020 if molar solubility of CO2 is 0.0343 L-1

bar-1

. Ka1 of CO2 = 4.210-7

.

a. 5.63

b. 4.72

c. 5.063

d. none of these

Ans : (1)

22. 1.0 metric ton coal containing 2.5% sulphur is burned and SO2 produced in the combustion was dissolved in

rainwater of volume equivalent to 2.0cm rain fall over 2.6km2area. Determine pH of rainwater if Ka of H2 SO3 =

1.510-2

.

a. 4.8

b. 4.08

c. 0.48

d. none of these

Ans : (1)

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