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Indian National Chemistry Olympiad Theory 2009

Instructions for students

Write your name and roll no. at the top of the first pages of all problems.

This examination paper consists of 34 pages of problems including answer

boxes.

You have 3 hours to complete all the problems.

Request the supervisor to provide you with rough sheets for rough wor.

Use only a pen to write the answers in the answer boxes. Anything

written by a pencil will not be considered or assessment.

!ll answers must be written in the appropriate boxes. !nything written

elsewhere will not be considered for assessment.

"or calculations# you must show the main steps.

$se only a non%programmable scientific calculator.

"or ob&ective type questions' (ar ! in the correct box. )ome of the ob&ective

questions may have more than one correct choice.

*alues of fundamental constants required for calculations are provided on

page +.

! copy of the ,eriodic Table of the -lements is provided at the end of the

paper.

o not leave the examination room until you are directed to do so.

The question paper will be uploaded on the /01)- website by 4th"ebruary

+22.

HBCSE, 31stJanuary 2009 1

Homi Bhabha Centre or Science Education

!ata Institute of undamenta" #esearch

$%&% 'ura( )ar*, )an+hurd, )umbai 00 0--%

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"#ndamental Constants

!vogadro constant !5 6.2++ x 72+3mol87

-lectronic charge e 5 7.62+ x 72871

(olar gas constant R 5 9.374 : ;87mol87

5 9.374 ; ,a.dm3;87mol87

5 2.29+ 5 37.? (e*@c+

7 e* 5 7.62+ x 7287:

Rydberg constant R/5 +.7A x 72879:

(ass of electron me5 .72 x 72837g

,lancBs constant h 5 6.6+? x 72834:s

)peed of light c 5 +.9 x 729ms7

!cceleration due to gravity g 5 .9 ms+

ensity of mercury 5 73.6 x 723gm3


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eV


'rob"em 1 10 )ar+s

Hydro*en atom

The electronic ground state of a hydrogen atom contains one electron in the first orbit.If sufficient energy is provided, this electron can be promoted to higher energy levels.

The electronic energy of a hydrogen-like species (any atom/ions with nuclear charge

and one electron! can be given as

1%1 The energy in "oule of an electron in the second orbit of # atom is

.0%/ mar+

1%2 The energy re$uired to promote the ground state electron of # atom to the first

e%cited state is

.1 mar+

&hen an electron returns from a higher energy level to a lower energy level, energy is

given out in the form of 'V/Visible radiation.

1%3 alculate the wavelength of light (nm! for the electronic transition of # atom from

the first e%cited state to ground state.


&ame of Student #o"" no%

,w ere #+

+

=

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constiskwhere,+

nk=


.1 mar+

In the model of hydrogen-like atom put forward by iels ohr (0!, the electron

orbits around the central nucleus. The ohr radius of n

th

orbit of a hydrogen-likespecies is given by

1% &hat is the principal $uantum number, n

of the orbit of e01 that has the same ohr

radius as that of ground state hydrogen atom2

.1 mar+

1%/ The ratio of energy of an electron in the *round statee01 ion to that of ground state

# atom is

(3! 4 (! 5

(! (6! 7

.1 mar+

The kinetic and potential energies of an electron in the # atom are given as

1% alculate the following8

a! the kinetic energy (in eV! of an electron in the ground state of hydrogen atom

b! the potential energy (in eV! of an electron in the ground state of hydrogen

atom

HBCSE, 31stJanuary 2009

:an

+

. .

=

n

=

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.2%/ mar+s

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1% 3 gaseous e%cited hydrogen-like species with nuclear charge can emit radiations of

si% different photon energies.

a! The principal $uantum number of the e%cited state is

(3! 4 (! >

(! 5 (6! 0

.0%/ mar+

b! It was observed that when this e%cited species emits photons of energy = +.45

eV when it comes to ne%t lower energy state. alculate the nuclear charge of

the species.

.1 mar+

The least energy re$uired to remove an electron from a species is known as theioni?ation energy (I.*.! of the species. The e%perimental I.*. of #e atom is

+5.>7 eV.

1%- Total energy re$uired to remove two electrons from #e is

.1%/ mar+s


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%%

#0

#

+#

>i!

#

#>

+

0#

@#

0

ii!

1

%%

#

iii!

+#

>#

0 iv!


'rob"em 2 19 )ar+s

&itro*en containin* comounds

Aeveral naturally occurring important compounds such as alkaloids, amino acids,

proteins and peptides contain nitrogen. itrogen heterocycles in the form of

pyrimidine and purine bases, are essential features of nucleic acids, which are

responsible for storage of genetic information in an organism and for transmitting the

same to its progeny.

2%1 &hich of the following amine/s is/are resolvable2

.1 mar+

2%2 The most acidic species amongst the following is

.1 mar+



# #

4

i!

# #

4ii!

#

4iii!

(#0!+

#

4iv!

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#

-r+, #

0BB#,

B#+ #

0

#B

BuCi

ii!C

i!

B C

+ > 0an

#0

B

l #++1

Indian National Chemistry Olympiad Theory 20092%3 6raw the structures of the isolableproducts of the following reactions.

a!

b!

.2 mar+s

2% Identify Band Cin the following reaction se$uence.

.2 mar+s

2%/ &hich of the following statements is true with respect to the following pair of

compounds2

5 E

i! 5is more basic than E

ii! Eis more basic than 5

iii! 5and Eare of comparable basicity .1 mar+

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#0I

3g+B

(i!

(ii! ,

H

#0

#0#

+#

+

#0I #0#

+(#

0!

0I

1 - 3g+B #+= #

+(#

0!

0

0 1

6

#0I

3g+B

i!

(ii!

Indian National Chemistry Olympiad Theory 20093lkaloids are naturally occurring biologically active nitrogeneous compounds,

isolated from plants. Atructure determination of alkaloids involves a few general

steps, some of which are mentioned below.

(a! 7eise" method8It is used to determine the presence and number of metho%yl

groups. #ere an alkaloid is heated with #I at its boiling point (+4! when

methyl iodide is formed, which is absorbed in ethanolic 3gB0and the 3gI

formed is separated and weighed.

(b! Hoffmann ehausti(e methy"ation method8 It is used to determine the

skeleton of an alkaloid. It involves the e%haustive methylation of an amine

with #0I followed by heating the resulting $uaternary ammonium salt with

moist 3g+B when elimination takes place as represented below.

ote8

6uring this reaction, the less stable alkene is formed.

2% Identify the products D6Eand DHE obtained in the #offmann e%haustive methylation

and degradation of -methylpiperidine (! in the following se$uence. ote8 H

absorbs above +99 nm in 'V.

.2 mar+s

;apaverine (+9#+B5! is an optically inactive alkaloid. Foldschmidt and co-

workers established the structure of papaverine in 777 and their work is a classic

e%ample of the application of o%idative degradation to structure determination.


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B#

BB#

onc. #+AB

5

#0I/

aB#

Veratric acid

i.ii.

fuse withe%cessaB#

dil. acid,

i.

ii.

dil. acid

e%cess

Indian National Chemistry Olympiad Theory 20092% ;apaverine reacts with four e$uivalents of #I at +4 to give papaveroline as one of

the products. The formula of papaveroline is

i! 4#0B5 ii! 4#B5

iii! 4#5B5 iv! 4#>B5

.1 mar+

;apaverine, when o%idi?ed with hot conc.

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#0

hot

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#+

BGe

BGe

BGe

GeB

i!

#+

GeB

GeB

BGe

BGe

ii!

GeB

#+

GeB

BGe

BGe

iii!

GeB

#+

BGe

BGe

BGe

iv!


2%11 ased on the above observations, the structure of papaverine is

.2 mar+s

B%idation of papaverine with hot dilute

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i! B+ ii! AiB+


'rob"em 3 1 mar+s

Chemistry of si"icon

Ailicon is the second most abundant element (+@.+ J! in the earthEs crust after

o%ygen (5>.> J!. arbon, silicon, germanium, tin and lead constitute the group 5

of the periodic table. hemistry of silicon is distinctly different from that of carbon.

Kor e%ample, under standard conditions B+ is a gas whereas AiB+ is a covalent solid.

3%1 6raw the structures ofB+and AiB+.

.1%/ mar+s

3%2% The reason for the distinct difference in the properties of B +and AiB+is

a! carbon is more electronegative than B and in case of AiB+

o%ygen is more electronegative than silicon

b! carbon has small si?e and forms a :bond withgood overlap whereas

silicon has larger si?e hence has a poor : overlap

c! carbon has only DpE orbitals and lacks DdE orbitals whereas

silicon has DdE orbitals

d! first ioni?ation potential of carbon is higher than that of silicon

(974 k" molfor and @74 k" molfor Ai!

.1 mar+



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Indian National Chemistry Olympiad Theory 20093%3 Flass, made of AiB+, a+AiB0 and aAiB0, is attacked by hydrofluoric acid with

formation of AiK4+anion. The analogous K4

+anion does not e%ist. The reason/s

is/are

a! carbon is more electronegative than silicon

b! silicon has larger atomic si?e than carbon

c! silicon has 0d orbitals which form an sp0d+hybrid orbitals

d! carbon and fluorine have comparable atomic si?es

.2 mar+s

3% Froup 5 elements have tendency to catenate. The first three members , Ai and Fe

show significant catenation. 3rrange these elements in decreasing order of their

catenation tendency.

.0%/ mar+

3%/ The order in 3% can be e%plained on the basis of

a! atomic si?e

b! electronegativity

c! bond strength

d! non-metallic character

.1 mar+

3% In a silicon manufacturing unit, silicon is obtained by heating 99 kg of pure sand

with 5> kg of high grade coke in an electric furnace. &rite the balanced e$uation of

the reaction.

.1 mar+


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3% Ailicon can be purified by converting it into volatile Ail5 (b.p. >7 L! which is

purified by fractional distillation. Ail5 can then be converted into Ai using

molecular hydrogen. &rite the balanced e$uations for the reactions involved.

.1 mar+

Ailicon is widely used in semiconductor industry in which the purity re$uired is of the

order of part per billion. This can be achieved using M?one refining techni$ueN. The

techni$ue involves moving a heater coil across an impure Ai rod.3%- In ?one refining, silicon is purified as

a! portions of Ai rod are heated and cooled so that impurities are

differentially evaporated

b! impurities are more soluble in li$uid phase than in solid

c! impurities are less soluble in li$uid phase than in solid

d! impurities are insoluble in molten Ai and can be separated

.1 mar+

The reactivities of l5 and Ail5 are different. Kor e%ample Ail5 can be easily

hydrolysed and is prone to substitution reactions, whereas l5is inert.

3%9 The observed difference is because

a! carbon atom has smaller si?e hence substitution is not possible

b! carbon is more electronegative than silicon

c! silicon has low lying unoccupied orbitals

d! -l bond is stronger than Ai-l bond.1 mar+

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Ailicon has high affinity for o%ygen to form silicates having AiB5units. Ailicates can

have chain or cyclic structures.

3%10 6raw the structure of a cyclic silicate having structural formula of

OAi4B7Pn

. 3lso determine the value of n. O#int8 AiB5 can be

shown as unitP

. mar+s

3%11 6raw the structure of the anion present in pyro%ene (GgaAi+B4!.

.1 mar+


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Indian National Chemistry Olympiad Theory 2009Ailicones are important synthetic polymers which find e%tensive applications due to

their chemical inertness and water repelling nature. They are produced via the

following reactions.

3%12 Identify

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#B

#0

#

# #

#0

nm!. Bne mole

of hydrogen bromide reacts with :at 5>o to give ;(5#@r! and )(5#@r! in the

ratio 798+9. Identify J)

.3 mar+s


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Indian National Chemistry Olympiad Theory 2009+-3minoben?oic acid on reaction with aB+/#l gives a compound &% In

mass spectrometry, a compound is heated to a high temperature under vacuum and

irradiated with a high energy electron beam to ioni?e it to form a cation. The cation

undergoes fragmentation. The masses of the fragments are recorded in the spectra.

3lkaline solution of &is subRected to a flash discharge and a specially adapted mass

spectrometer scans the spectrum of the products at rapid intervals. Gass

spectrometric analysis of &gives peaks at masses +7, 55 and @4.

% Atructures of the fragments at masses +7, 55 and @4 and &are

Gass8

+7 55 @4 ompound &

.2%/ mar+s


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'rob"em / 11 )ar+s

Chemica" +inetics

/%1 onsider a first order reaction

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b! 3ssuming the initial concentration of

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/% )adioactive decay of an element follows first order kinetics. 3fter how many years

will +>J of the initial $uantity of radium remain, if half life of radium is 4+9 years2

.0%/ mar+

The 3vogadro number can be calculated e%perimentally. In a classic e%periment,

*rnest )utherford observed that a sample of + mg left for 70 days produced

4.70 mm0of #e gas measured at +@0 < and atm pressure.

(i! alculate the moles of #e produced in the e%periment.

.1 mar+

The number of disintegrations undergone by g of )a in a second is 5.4 99gs

ii! The number of disintegrations observed in the )utherfordEs

e%periment is

.1 mar+

iii! ++4 )a emits particles to give +5 ;b. alculate the number of #e atoms

produced in the )utherford e%periment.

.1 mar+

iv! 'sing the above data, calculate the 3vogadro number.

.0%/ mar+


)a++4

77

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urve a

urve b

urve c

urve d


'rob"em 1/ mar+s

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Gatch the curves with correspondinggases

Cur(e 6as

urve a

urve b

urve c

urve d .1%/ mar+s

%3 The van der &aals e$uation for one mol of a real gas is (; 1 a/V +!(V - b! = )T. The

correct order for the value of DaE for #+, #0, and #5gases is

i! #+U #0U #5 ii! #+ #0 #5

iii! #+U #5 U #0 iv!#+ #5 #0

.0%/ mar+

% 3 diver dives to +9 m depth of sea water (W = .90 90 kg m0! and rises to the

surface $uickly without breathing. hange of pressure (in atmosphere! in coming to

water surface from the depth is

.1 mar+

%/ The total work done ("oule! on lungs if the same e%pansion is carried out at the sea

surface at constant temperature is (3ssume volume of air in the lungs is +C under

normal conditions!.

.1%/ mar+s

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% #uman body functions best when o%ygen gas has a partial pressure of 9.+ atm.

alculate the percentage of o%ygen by volume in air carried by the diver at the sea

depth of +9 m for best body function.

.1 mar+

B% 'hysica" and chemica" e>ui"ibrium

3 phase diagram summari?es the conditions at which a substance e%ists as a solid,

li$uid or gas. *ach solid line between two phases specifies the conditions of

temperature and pressure under which the two phases can e%ist in e$uilibrium. The

phase diagram of a one component system Sis shown below. ased on the diagram,

answer the following $uestions.

% &hich phases of Scan coe%ist under atmospheric conditions2

.1 mar+

%- 'nder what conditions all the three phases of Scan co-e%ist2

.0%/ mar+


T/o.+

@+.7 pc

;

(atm!

095(Tc!+7+@

a c db e

->

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%9 Gark the correct graph showing the variation of molar Fibbs function (Fm! vs.

temperature (T! along the line a-e of the phase diagram.

.1 mar+

%10 hoose the correct option/s from the following statements for triple point of system S.

a! 3ll the three phases are in e$uilibrium

b! Golar Fibbs energy for the three phases is the same

c! Golar volume of the three phases is identical

d! Golar entropy of the three phases is the same

.1 mar+

%11 3s a result of increasein e%ternal pressure, the melting point of solid Swill

a! increase

b! decrease

c! not change

.1 mar+

%12 #ow many phases e%ist at T 095 < and ; @+.7 atmosphere2

.0%/ mar+


Fm

Fm

Fm

T T

Fm

T T

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%13 Kor a chemical reaction

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%2 The p# at half e$uivalence point may be employed to estimate the p99 g of the sample was dissolved in deionised water to make >9.9 mC solution.

This solution was titrated with 9.> G #l. The p# titration curve showed two

breaks at p# = .9 and p# = >.9.


Indicator H ran*e Co"our chan*e

Gethyl yellow +. - 5.9 red-yellow

Gethyl orange 0. Z 5.5 red-orange

;henolphthalein 7.9 - .7 colorless-red

33

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% &rite the balanced e$uation corresponding to these breaks.

.1 mar+

%/ alculate the total number of moles of acid used

a! to reach e$uivalence point at p# = .9 if the volume of acid re$uired is .4 mC.

b! to reach e$uivalence point at p# = >.9 if the volume of acid re$uired is 05.0 mC.

.1 mar+

% 6etermine the number of grams of a+B0 and a#B0 and their percentage in themi%ture.

.2%/ mar+s


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% 09 mC of 9. Ga+B0 solution was titrated with 9. G #l. (The initial p# of the

a+B0 solution is around .7!. [ualitatively sketch the p# vs V #l curve for this

titration.

.1 mar+

%- hoose the correct statement/s forthe titration curve of p# vs V#l fora+B0drawn

by you in %

i! The total volume of #l re$uired to reach the +nd end point is twice that of

the first one

ii! umber of moles of B0+is e$ual to the number of moles of #B0

Z at some

point on this curve

iii! umber of moles of #B0is e$ual to twice the number of moles of

B0Z +

at some point on this curveiv! The total volume of #l re$uired to reach the +nd end point is

half that of the first one

.2 mar+s

2.2

+.2

4.2

6.2

9.2

72.2

7+.2

74.2

2.2 72.2 +2.2 32.2 42.2 ?2.2 62.2

Volume of #l (mC!

p#

Documents

INCHO-2009-paper (1)