India 2014 — incho2014-Q.pdf

Fonte: gare di altri paesi/India/chimica/incho2014-Q.pdf · Apri PDF · apri PDF p.3 Cluster: Fisica Moderna Soluzioni (stessa cartella): · · · · · · · · · · · · · ·

Problema 1

Problem 1

17 marks Metallurgy Metallurgy is the technology of obtaining metals from their ores. Depending on the metal and the ore involved, metallurgical operations are divided into three broad divisions as hydrometallurgy, pyrometallurgy and electrometallurgy. In hydrometallurgy, the metal is extracted into an aqueous solution using oxygen and a complexing agent. Hydrometallurgy is used for noble metals like silver and gold. The process is known as leaching. Potassium cyanide is traditionally used for extraction of gold.

In presence of air, native gold dissolves in a solution containing cyanide ions (CN) 1

forming [Au(CN)2] 1 . 1.1 Write a balanced equation to represent dissolution of gold in KCN solution. (0.5 mark) 1.2 Calculate the mass of KCN required to extract gold from 100 kg of a rock sample having 2.2% gold on w/w basis.

(1 mark)

Name of Student

Roll No Indian National Chemistry Olympiad

Theory 2014 HBCSE, 1st February 2014

Silver and gold both form cyanide complexes [Ag (CN)2] 1 , [Au (CN)2] 1 with formation constants as 4.2 1020 and 1.0 1037 respectively. 1.3 Calculate the equilibrium constant for the reaction.

Au+1 (aq) + [Ag(CN)2] 1 (aq) Ag+1 (aq) + [Au(CN)2] 1 (aq)

(0.5 mark) 1.4 Using 1.3 and the information given below, calculate the equilibrium concentrations of [Au(CN)2] 1 and [Ag(CN)2] 1 when solid gold is added in excess to 0.1M [Ag(CN)2] 1 solution. (under standard condition of temperature and pressure)

1. Ag+1 (aq) + e Ag(s) E0 = 0.8V
2. Au+(aq) + e Au(s) E0 = 1.83V

(2 marks) Indian National Chemistry Olympiad

Theory 2014 HBCSE, 1st February 2014

Aqua regia is a mixture of concentrated hydrochloric acid and nitric acid in the ratio of

3:1 on volume basis. It was developed by the alchemists to dissolve gold. In the

dissolution process, gold is oxidized and converted to a chloro complex and nitric acid

gets reduced to nitrogen dioxide. 1.5 Write a balanced equation for the individual reactions.

(1 mark) In pyro-metallurgy the metals are extracted from ores and concentrates by heat treatment.

This involves conversion to metal oxide. This is followed by the reduction of the oxide to

the metal using carbon or carbon monoxide. Metals like iron, copper, zinc and

manganese are obtained by this technique. Ellingham diagram is a graph showing the variation of the standard free energy of

formation ( G ) of metal oxides with temperature. The diagram also shows lines for

formation of CO and CO2.

Ref: Shriver D.F. and Atkins P.W., Inorganic Chemistry, 2nd edition, ELBS, Oxford University Press, (1994), pg no. 278. Indian National Chemistry Olympiad

Theory 2014 HBCSE, 1st February 2014

1.6 Using the Ellingham diagram given above answer the following. Mark X in the correct

box. i) The equation of lines in the diagram is given by

a) $\Delta G=\Delta H-T\Delta S$

b) $\Delta G=\Delta H+T\Delta S$

c) $\Delta G^0=\Delta H^0-T\Delta S^0$

d) $\Delta G^0=\Delta H^0+T\Delta S^0$ ii) The statement that is true for the line C CO2 is a) the standard entropy change of the above reaction is positive

b) the standard entropy change of the above reaction is negative

c) the standard entropy change of the above reaction is zero. iii) The statement that is true for the line CO CO2 is a) the standard entropy change of the above reaction is positive

b) the standard entropy change of the above reaction is negative

c) the standard entropy change of the above reaction is zero. iv) The lowest temperature below which FeO cannot be reduced by coke is

a) $750{}^{\circ }\text{C}$

b) $1300{}^{\circ }\text{C}$

c) $1000{}^{\circ }\text{C}$ v) In a blast furnace at $2000{}^{\circ }\text{C}$ the most preferred reaction that will take place for

carbon containing species is

a) C(s) + 1⁄2 O2(g) CO(g)

b) 1⁄2C(s) + 1⁄2O2(g) 1⁄2CO2(g)

c) CO(g) + 1⁄2 O2(g) CO2 g)

vi) The preferred temperature for Mg to be used as a reducing agent for SiO2 is

a) below $2200{}^{\circ }\text{C}$

b) above $2500{}^{\circ }\text{C}$

c) not possible at any temperature

(4.5 marks) Indian National Chemistry Olympiad

Theory 2014 HBCSE, 1st February 2014

In electrometallurgy, the pure metal is obtained by electrolysis of its appropriate salt. Aluminum is produced from bauxite by electro-reduction of alumina. The process is known as Hall- Herault’s process Alumina is dissolved in a conducting medium like cryolite Na3AlF6, as it cannot conduct electricity in the molten state. The temperature of melting of cryolite is reduced by adding conducting impurities to it. In the electrolytic bath, aluminum gets deposited at the cathode and oxygen gets liberated at the anode. The liberated oxygen attacks the graphite anode and forms carbon dioxide. 1.7 Write the balanced reactions involved in the electrolysis of alumina.

(1 mark)

The melting point of cryolite (molar mass 210 g) is 1282.7K. However, by adding sodium fluoride cryolite is made to melt at 1233K. 1.8 Calculate the percentage (on w/w basis) of sodium fluoride needed to be added to reduce the melting point to 1233K assuming that it dissociates completely. The heat of fusion of cryolite in this temperature range is $114{\text{kJ mol}}^{-1}$.

(3 marks) Indian National Chemistry Olympiad

Theory 2014 HBCSE, 1st February 2014

The electrolysis is carried out with a potential difference of 4.5V between the two

electrodes and the current passing through the cell is 180kA with the current efficiency of

95%. 1.9 i) Calculate the amount of energy required for production of 100kg of aluminum.

(2 marks)

ii) Calculate the rate of production of carbon dioxide per hour.

(1.5 marks) Indian National Chemistry Olympiad Theory 2014

HBCSE, 1st February 2014

Name of Student

Roll No

Topic: Thermodynamics, Electrostatics, Conservation of Energy Metodi: First Law of Thermodynamics, Ideal Gas Law, Conservation Laws Competenze: Mathematical Modeling, Physical Reasoning Fonte: Testo (PDF) — p.3

Problema 2

Problem 2

15 Marks Energy storage devices

A. Hydrogen storage as metal hydrides Hydrogen is a renewable source of energy and considered as a fuel of the future. One of the problems of its use is storage and transportation. Research has shown that several metal hydrides act as ‘hydrogen tanks’. Large quantities of hydrogen can be absorbed on them and desorbed when needed through absorption/desorption cycles. Lithium nitride, Li3N is one of the well studied hydrogen storage material. Almost all absorbed hydrogen can be subsequently desorbed if the hydrogenated sample of Li3N is evacuated to 1 Pa and heated up to $450{}^{\circ }\text{C}$. The hydrogen uptake of Li3N is a chemical reaction in two steps. Both steps involve disproportionation of H2. 2.1 Complete the following equations by identifying A, B and C.

(1.5 marks) 2.2 Under standard conditions of temperature and pressure, calculate the maximum volume of hydrogen that can be stored in 1kg of Li3N.

(1 mark)

Li3N + H2

A

B (A) + H2

C

B Indian National Chemistry Olympiad Theory 2014

HBCSE, 1st February 2014

When a certain mass of compound B was reacted with excess amount of 0.5M HCl, a colourless gas was produced. Passing this gas over CuO(s) resulted in decrease in mass of the solid by 2.096 g. 2.3 (i) Write the balanced equation for the reaction of B with HCl (ii) Calculate the mass of B taken in grams.

(1 mark) Mg and Mg2Ni also form hydrides which can be desorbed under suitable conditions of temperature and pressure. Mg2NiH4 is formed by a technique known as hybriding combustion synthesis. MgH2 is synthesised by the reaction of H2 and Mg at high temperatures and pressure in presence of a catalyst. 2.4 In Mg2NiH4, Ni(0) atoms form a face centered cubic lattice in which the Mg+2 ions occupy interstitial positions. The unit cell length of Mg2NiH4 was found to be $6.465\times 10^{-10}\text{m}$.

A hydrogen cylinder at 200C has number of moles of hydrogen per unit volume which

is (1/100) times the number of moles of hydrogen present in unit volume of Mg2NiH4.

Calculate the pressure (in bar) in this cylinder.

(2 marks) Indian National Chemistry Olympiad Theory 2014

HBCSE, 1st February 2014

The desorption equilibrium of Mg2NiH4 is as given below Mg2Ni H4 (s) Mg2Ni (s) + 2H2 (g) The equilibrium pressure of hydrogen was found to be 7.03 bar and 1.05 bar at 606K and 533K respectively. Assume that the enthalpy values do not change with temperature and $\Delta H_f^{\circ }$ (Mg2NiH4) = $-176.0{\text{kJ mol}}^{-1}$. 2.5 Calculate the standard enthalpy change (${\text{kJ mol}}^{-1}$) for the reaction

2Mg(s) + Ni(s) Mg2Ni(s)

(1.5 marks) B. Fuel cells – Green and renewable sources of energy Fuel cells convert the chemical energy of combustion in a fuel directly into electrical energy without the losses associated with the use of thermodynamic heat engines. A group of researchers have invented a fuel cell using cane sugar (C12H 22O11). This system consists of two reaction vessels. The reaction vessel on the left hand side contains cane sugar and VO2+ ions in a strong acidic solution. Air is pumped in the right hand side vessel which also contains VO2+ ions in a strong acidic solution. The assembly contains a battery at the centre that is responsible to generate electricity. In the left hand side vessel, VO2+ is reduced to V3+ and cane sugar is oxidised to CO2.

In the right hand side vessel, VO2+ is oxidised by oxygen to form VO2 +. The fuel cell

diagram is indicated below. Indian National Chemistry Olympiad Theory 2014

HBCSE, 1st February 2014

(Ref: Excell Robert H.B and Spaziante Placido (2004), A fuel power cell powered by sugar, The Joint International Conference on “Sustainable Energy and Environment, Thailand, 1-3 Dec 2004.)

2.6 Write balanced equations for all the reactions in the (a) left side reaction vessel (b) right side reaction vessel.

(2 marks) 2.7 Calculate the volume of air that has to be pumped in at $25{}^{\circ }\text{C}$ and 101 kPa into the right side vessel, for the consumption of 10g of cane sugar in the other vessel (air contains 21 % of volume oxygen).

(1.5 marks) Indian National Chemistry Olympiad Theory 2014

HBCSE, 1st February 2014

2.8 Calculate the values of X and Y (in volts) from the given information.

(1 mark) In the battery at the centre between the reaction vessels, the following half cell reactions take place during discharge V3+ (aq)+ H2O VO2+(aq) + 2H+(aq) + e …I

VO2 +(aq) + 2H+(aq)+ e VO2+(aq) + H2O …II The standard reduction potentials are E (VO2+/ V3+) = 0.34 V and E (VO2

• / VO2+) = 1.00 V respectively. Assume that the concentration of V3+(aq) and VO2 +(aq) is equal to $2.00{\text{mol L}}^{-1}$ (obtained from the reactions taking place in the reaction vessels). 2.9 Determine [V3+], [VO2 +] and [VO+2 ] in the battery when the potential of the battery is 0.32V. The temperature of the reaction vessel is $25{}^{\circ }\text{C}$ throughout.

(2.5 marks) Indian National Chemistry Olympiad Theory 2014

HBCSE, 1st February 2014

2.10 The standard Gibbs free energy change for the complete combustion of sucrose to carbon dioxide and water is $-5684{\text{kJ mol}}^{-1}$.

Calculate the theoretical limit on the efficiency of direct conversion of chemical

energy in sugar to electricity by this method under standard conditions.

(1 mark) Indian National Chemistry Olympiad

Theory 2014 HBCSE, 1st February 2014

Topic: Thermodynamics, Kinetic Theory, Conservation of Energy Metodi: Ideal Gas Law, First Law of Thermodynamics, Conservation Laws Competenze: Mathematical Modeling, Physical Reasoning Fonte: Testo (PDF) — p.9

Problema 3

Problem 3

25 marks ALKALOIDS

Alkaloids are naturally occurring nitrogen heterocycles. Although they are poisonous, when used in very small quantities they have medicinal value as well. Alkaloid A (MF C17H23NO3) on hydrolysis with Ba(OH)2 yields two products B (MF C9H10O3) and C (C8H15NO).

3.1 Compound B does not add bromine and it dissolves in NaHCO3. Heating B strongly yields compound D (MF C9H8O2). Based on these observations which of the following statement/s is/are true for compound B.

a) B has a hydroxyl group b) B is an olefin c) B has a carboxyl group d) B is an aromatic compound

(1.5 marks) 3.2 Compound D on oxidation gives benzoic acid as one of the products. Based on this observation, the possible structure/s for compound D is/ are

(2 marks) Name of Student

Roll No Indian National Chemistry Olympiad

Theory 2014 HBCSE, 1st February 2014

3.3 Compound B has been synthesized by the following reaction sequence. Draw the structures of the intermediates E-H and the structure of B.

C6H5COCH3

E F

G H

B

(Note: Product H is also obtained on treatment of G with HBr in the presence of H2O2)

(3 marks) Structure elucidation of an alkaloid makes use of several chemical and physical methods. One of them is the Hoffman’s exhaustive methylation method for determining the nature of the nitrogen heterocylic ring. The heterocyclic ring is opened up with loss of nitrogen as an amine and formation of the less substituted olefin. The reactions involved are shown below.

(Note: A cyclic amine undergoes Hoffman’s exhaustive methylation twice to give an amine and a diene).

HCN heat reduced pressure H3O+ HBr aq.NaOH ( moles)

E

F

B

H

G Indian National Chemistry Olympiad

Theory 2014 HBCSE, 1st February 2014

3.4 Compound C does not decolorise bromine and dissolves in HCl. C on treatment with conc. H2SO4 gives J. Hoffman’s exhaustive methylation of compound J gives 1, 3, 5- cycloheptatriene as one of the products. The possible structure /s for compound C is /are

(3 marks)

3.5 Compound C on controlled oxidation gives a compound K which gives a yellow precipitate with 2,4–DNP and also reacts with two equivalents of benzaldehyde in the presence of a base to give a dibenzylidine derivative, the general structure of such derivatives is shown below.

The possible structure /s for compound C based on the above observations is /are Indian National Chemistry Olympiad

Theory 2014 HBCSE, 1st February 2014

(2 marks) 3.6 i) Write the number of chiral atoms in a and b in 3.5.

a)

b)

(1 mark) ii) For the structures a-e in 3.5 the compounds that are optically active are (Mark X

in the correct box)

(1 mark) Oxidation of C with CrO3 gives dibasic acid L. L on heating with HI at $150\text{–}300{}^{\circ }\text{C}$ gives one mole of methyl iodide. 3.7 0.580 g of L is dissolved in $100{\text{cm}}^3$ of distilled water. $10{\text{cm}}^3$ of this diluted solution required $6.2{\text{cm}}^3$ of 0.1 N KOH for neutralisation. Determine the molecular mass of L. Show all the steps in the calculation.

(1 mark)

a b c d e Indian National Chemistry Olympiad

Theory 2014 HBCSE, 1st February 2014

L 3.8 Write the molecular formula of compound L.

(1 mark)

3.9 Oxidation of L gives one of the products as M. Draw the structure of L.

(1 mark) 3.10 Draw the possible tautomeric structure/s of M.

(1 mark) Indian National Chemistry Olympiad

Theory 2014 HBCSE, 1st February 2014

3.11 Based on all the observations from 3.4 -3.9 the structure of C is

(1 mark) 3.12 Draw the possible structure for alkaloid A.

(1 mark) In plant metabolism alkaloids are formed from amino acids. Alkaloid hygrine (N) is formed in nature from amino acid Ornithine (O)

The first step in the synthesis is the pyridoxal catalysed decarboxylation of Ornithine. Indian National Chemistry Olympiad

Theory 2014 HBCSE, 1st February 2014

3.13 Identify the missing intermediate Q.

(1 mark) pyridoxal phosphate Indian National Chemistry Olympiad

Theory 2014 HBCSE, 1st February 2014

3.14 Identify S, T, U and V in the following sequence.

**Note: Methylation using SAM takes place at the more nucleophilic nitrogen. (3.5 marks) 3.15 Biosynthetic pathways are usually established by isotopic labeling of potential precursors. If ornithine is labeled with 14C at C2 and 15N at nitrogen on C2, draw the structure of the hygrine obtained indicating the labeled atoms.

(1 mark) ** Indian National Chemistry Olympiad

Theory 2014 HBCSE, 1st February 2014

Topic: Thermodynamics, Geometric Optics, Modern-Quantum Physics Metodi: First Law of Thermodynamics, Thermodynamic Cycle Analysis, Ideal Gas Law, Ray Tracing Competenze: Mathematical Modeling, Physical Reasoning Fonte: Testo (PDF) — p.15

Problema 4

Problem 4

15 marks Applications of Transition Metal Complexes The transition metals are widely distributed in earth’s crust. The elements Co, Ir belong to Group IX of the periodic table. These elements as well as their compounds have several applications in chemistry. Cobalt and its compounds are used as catalysts and in making alloys, pigments and glasses. Silica, on doping with an aqua complex of Co(II) shows light pink colour in wet condition and deep blue colour in dry condition. This colour change is associated with the structural changes in the complex of Co(II) due to partial loss of water. 4.1 Write the formula of the aqua complex of Co(II) that has pink colour.

(0.5 mark) 4.2 With the above formula, write the balanced chemical equation for the reaction

responsible for the observed colour change.

(1 mark) Mercuric tetrathiocyanatocobaltate (II) is another complex of cobalt, used as a standard for calibrating magnetic balances. Its spin only magnetic moment is 3.872 Bohr magnetons. 4.3 Write the formula of the complex.

(0.5 mark) 4.4 Based on the magnetic moment alone, can the structure of the complex in 4.3 be predicted correctly to be tetrahedral or square planar?

Yes

No

(1 mark)

Name of Student

Roll No Indian National Chemistry Olympiad

Theory 2014 HBCSE, 1st February 2014

B. An oxide of cobalt, Co3O4 can exist either (a) as a normal spinel CoII(Co2 IIIO4) where 1/8th of the tetrahedral holes are occupied by CoII ions and 1/4th of the octahedral holes are occupied by CoIII ions in a cubic lattice, or (b) as an inverted spinel where 1/4th of CoII and CoIII exchange their sites.

Pairing energy $P({\text{cm}}^{-1})$ ${\Delta }_0({\text{cm}}^{-1})$ ${\Delta }_t({\text{cm}}^{-1})$

Co3+

20680

20760 9226 Co2+

20800

9200 4090 (${\Delta }_0$ and ${\Delta }_t$ imply the octahedral and tetrahedral splitting parameters respectively for oxygen as the ligand) 4.5 On the basis of the above data, calculate the crystal field stabilization energy (CFSE) in kJ mol 1 for i) a) cobalt (III) in octahedral and b) cobalt (III) in

tetrahedral sites. ii) a) cobalt (II) in octahedral and b) cobalt (II) in

tetrahedral sites (In actual calculations of CFSE values neglect the pairing energies)

(3 marks) Indian National Chemistry Olympiad

Theory 2014 HBCSE, 1st February 2014

4.6 Calculate the difference of the CFSE values obtained in 4.5 i) and 4.5 ii).

(1 mark) 4.7 Based on your above calculations, predict whether Co3O4 exists as a normal spinel

an inverted spinel

(0.5 mark) 4.8 The number of possible isomers for the complex of the type [CoIIIAX(en)2]2+ (en = ethylene diamine) is

2

3

4

(1 mark) 4.9 Draw the possible structure(s) for the cis and the trans isomers of the above complex. (Represent en by ).

(1.5 marks)

Hydrolysis of the above complex (where X is the leaving group) proceeds through dissociation mechanism. For the trans complex, the reaction proceeds via square pyramidal intermediate but for the cis complex it proceeds via trigonal bipyramidal intermediate.

cobalt (III) in octahedral and tetrahedral sites

cobalt (II) in octahedral and tetrahedral sites Indian National Chemistry Olympiad

Theory 2014 HBCSE, 1st February 2014

4.10 For the trans form, the hydrolysis will result in

a) Only cis product

b) only trans product

c) Mixture of cis and trans product

(1 mark) 4.11 For the cis form, the hydrolysis will result in

a) Only cis product

b) only trans product

c) Mixture of cis and trans product

(1 mark) C.

The complex [IrI(CO)Cl(PPh3)2] was first synthesized by Lauri Vaska in 1961 and is known as Vaska’s complex. It is a versatile catalyst for several reactions.

The addition of H2 to [IrI(CO)Cl(PPh3)2] is cis, but the addition of CH3Cl is trans. The structures of the complex and the products are given below. These reactions go through intermediate states.

4.12 (i) Draw the structures of the intermediates for the above addition reactions.

(2 marks) (ii) State the oxidation number of Ir in the products.

a)

b)

(1 mark)

Intermediate for a

Intermediate for b Indian National Chemistry Olympiad

Theory 2014 HBCSE, 1st February 2014

Topic: Conservation of Energy, Magnetism, Modern-Quantum Physics Metodi: Energy Conservation Method, Conservation Laws, Ampère’s Law, Biot-Savart Law Competenze: Mathematical Modeling, Physical Reasoning, Diagrammatic Reasoning Fonte: Testo (PDF) — p.23

Problema 5

Problem 5

18 marks Chemistry of Carbohydrates Carbohydrates are biomolecules which perform diverse functions such as being energy sources and constituents of nucleic acids. In solution, monosaccharides, the simplest unit of carbohydrates, exist in cyclic hemiacetal form. The cyclic form is in equilibrium with small quantity of the open chain form. In a trivial system of naming (D,L-system), the carbohydrates having the configuration at the asymmetric carbon of highest number same as that of R-glyceraldehyde (shown below) are called D forms while those having opposite configuration at the same carbon are called L-forms.

Monosacharides A,B and C have the following structures.

5.1 Classify sugars A, B and C into D or L forms.

A

B

C

(1.5 marks)

Some terms and their abbreviations are given below. Term Identical Enantiomer Diastereomer Abbreviation I E D Name of Student

Roll No R-glyceraldehyde B C A Indian National Chemistry Olympiad

Theory 2014 HBCSE, 1st February 2014

5.2 i) Which is the correct term to describe the relationship of each of the following structure with A? (Mark X in the correct box)

a)

b)

c)

(3 marks) 5.3 a) Some reactions of a monosaccharide U are given below. Complete them by

drawing the structures of the products with correct stereochemistry wherever

applicable.

I E D

I E D

I E D Indian National Chemistry Olympiad

Theory 2014 HBCSE, 1st February 2014

(2.5 marks) b) Monosaccharide U react with phenylhydrazine in excess to yield a

crystalline derivative called osazone.

Another aldose V and ketose W also give the same osazone as U on reaction

with excess of phenylhydrazine. Draw the structure of V and W in the Fischer

projection formula.

(1.5 marks) Indian National Chemistry Olympiad

Theory 2014 HBCSE, 1st February 2014

1,2 – or 1,3 – diols can be protected by forming cyclic acetals with one equivalent of acetone in acidic conditions. These acetals are hydrolysed back to the original hydroxyl groups in dilute acid solutions. 5.4 Draw the structure of the product G in the following reaction.

(1.5 marks)

Compound N is a vital nutrient essential for the proper growth of the human body. It is commercially synthesized from monosaccharide H. 5.5 Draw the structure of the product I (with correct stereochemistry) in the first step of the synthesis of N.

(0.5 mark) I undergoes microbial oxidation to give J whose structure is given as follows

excess dry HCl G catalyst H I I J Indian National Chemistry Olympiad

Theory 2014 HBCSE, 1st February 2014

5.6 In the conversion of I to J, which carbon of the original H is undergoing this

oxidation? (Mark X in the correct box).

C1 C2 C3 C4 C5 C6

(1 mark) 5.7 Complete the synthesis of N by drawing structures for J’, K, L and M with correct stereochemistry.

(3.5 marks) Monsaccharides and oligosaccharides that have a free – OH group on the anomeric carbon give a positive test for Tollen’s reagent. Such sugars are called reducing sugars. If this free OH group at the anomeric carbon is replaced by an ether then the result is a glycoside. The nonsugar part of a glycoside is called an aglycone. Both glycosides and oligosaccharides can be hydrolysed by acids or enzymes into their components. J J’ gives effervesence with NaHCO

• Draw M in open chain form with correct sterochemistry L N M K

Indian National Chemistry Olympiad

Theory 2014 HBCSE, 1st February 2014

A disaccharide O (C12H22O11) is reducing in nature. On hydrolysis by enzyme emulsin (which cleaves only $\beta$ linkages) O gives H as the only product.

O undergoes the following reactions.

5.8 From the structures of Q and R draw the structure of O.

(2 marks) A glycoside S on careful hydrolysis with emulsin gives the disaccharide O and the aglycone T whose structure is shown below

5.9 Draw the structure of the glycoside S with the correct stereochemistry.

(1 mark)

emulsin O H R Q O P O T Indian National Chemistry Olympiad

Theory 2014 HBCSE 1st February 2014

Topic: Conservation of Energy, Modern-Quantum Physics, Nuclear & Particle Physics Metodi: Energy Conservation Method, Conservation Laws, Photon Energy Relation, Bohr Model & Quantization Competenze: Mathematical Modeling, Physical Reasoning, Diagrammatic Reasoning Fonte: Testo (PDF) — p.27

Problema 6

Problem 6

8 marks Chemical equilibria in quantitative analysis

A. Many metal ions can be precipitated from their aqueous solutions as sulphides using hydrogen sulphide, a weak dibasic acid. The concentration of the sulphide ions in the solution can be controlled by adjusting the pH of the solution and thus the separation of the metal ions from the mixture can be achieved.

Relevant data needed for calculations in different subparts is presented below.

Dissociation constants of H2S:

K1 = 1 10 9 and K2 = 1.3 10 14

6.1 Solid NaOH is added to 1L of 0.05M solution of H2S until the pH reached the value of

12.0. Calculate- (i) Initial pH of H2S solution before the addition of solid NaOH. (ii) Concentration of all the species at pH =12.0 (Use appropriate approximations). (iii) Total mass of solid NaOH added to the solution.

(3 marks) Salt CdS Tl2S Solubility product (Ksp) 1.0 10 27 6.0 10 22

Name of student

Roll No. Indian National Chemistry Olympiad

Theory 2014 HBCSE 1st February 2014

A sample solution contains 0.003M each of Cd+2 ions and Tl+1 ions. The mixture needs to

be separated using H2S at appropriate pH. The separation is assumed to be quantitatively

complete when the concentration of Cd+2 ions in the sample solution becomes 1 10 6 of

the initial value. 6.2 Calculate the optimal pH range in which Cd+2(aq) and Tl+1(aq) can be separated

quantitatively. (The H2S concentration in this solution is 0.10 M).

(2 marks) Indian National Chemistry Olympiad

Theory 2014 HBCSE 1st February 2014

B. Volhard method is a back titration method and involves excess addition of AgNO3 and its

back titration with potassium thiocyanate using ferric alum as the indicator. The end point

is the appearance of a reddish brown colour due to the formation of iron thiocyanate

complex. 6.3 10g of pesticide sample containing arsenic was treated appropriately to convert its arsenic

content to AsO4 3 . Then 25mL of 0.045M AgNO3 was added to precipitate AsO4 3 as

Ag3AsO4. The precipitate was filtered and the unreacted AgNO3 present in the filtrate

was then titrated with 0.05M potassium thiocyanate. The observed titre value was 5.0

mL. Calculate the percentage of arsenic as AsO4 3 in the pesticide sample.

(3 marks) 1 H hydrogen [1.007, 1.009] 1 18 3 Li lithium [6.938, 6.997] 4 Be beryllium 9.012 11 Na sodium 22.99 12 Mg magnesium [24.30, 24.31] 19 K potassium 39.10 20 Ca calcium 40.08 37 Rb rubidium 85.47 38 Sr strontium 87.62 38 Sr strontium 87.62 55 Cs caesium 132.9 55 Cs caesium 132.9 56 Ba barium 137.3 87 Fr francium 88 Ra radium 5 B boron [10.80, 10.83] 13 Al aluminium 26.98 31 Ga gallium 69.72 49 In indium 114.8 81 Tl thallium [204.3, 204.4] 6 C carbon [12.00, 12.02] 14 Si silicon [28.08, 28.09] 32 Ge germanium 72.63 50 Sn tin 118.7 82 Pb lead 207.2 7 N nitrogen [14.00, 14.01] 15 P phosphorus 30.97 33 As arsenic 74.92 51 Sb antimony 121.8 83 Bi bismuth 209.0 8 O oxygen [15.99, 16.00] 16 S sulfur [32.05, 32.08] 34 Se selenium 78.96(3) 52 Te tellurium 127.6 84 Po polonium 9 F fluorine 19.00 17 Cl chlorine [35.44, 35.46] 35 Br bromine [79.90, 79.91] 53 I iodine 126.9 85 At astatine 10 Ne neon 20.18 2 He helium 4.003 18 Ar argon 39.95 36 Kr krypton 83.80 54 Xe xenon 131.3 86 Rn radon 22 Ti titanium 47.87 22 Ti titanium 47.87 40 Zr zirconium 91.22 72 Hf hafnium 178.5 104 Rf rutherfordium

23 V vanadium 50.94 41 Nb niobium 92.91 73 Ta tantalum 180.9 105 Db dubnium 24 Cr chromium 52.00 24 Cr chromium 52.00 42 Mo molybdenum 95.96(2) 74 W tungsten 183.8 106 Sg seaborgium 25 Mn manganese 54.94 43 Tc technetium 75 Re rhenium 186.2 107 Bh bohrium 26 Fe iron 55.85 44 Ru ruthenium 101.1 76 Os osmium 190.2 108 Hs hassium 27 Co cobalt 58.93 45 Rh rhodium 102.9 77 Ir iridium 192.2 109 Mt meitnerium 28 Ni nickel 58.69 46 Pd palladium 106.4 78 Pt platinum 195.1 110 Ds darmstadtium 29 Cu copper 63.55 47 Ag silver 107.9 79 Au gold 197.0 30 Zn zinc 65.38(2) 48 Cd cadmium 112.4 80 Hg mercury 200.6 111 Rg roentgenium 112 Cn copernicium 114 Fl flerovium 116 Lv livermorium 57 La lanthanum 138.9 89 Ac actinium 58 Ce cerium 140.1 90 Th thorium 232.0 59 Pr praseodymium 140.9 91 Pa protactinium 231.0 60 Nd neodymium 144.2 92 U uranium 238.0 61 Pm promethium 93 Np neptunium 62 Sm samarium 150.4 94 Pu plutonium 63 Eu europium 152.0 95 Am americium 64 Gd gadolinium 157.3 96 Cm curium 65 Tb terbium 158.9 97 Bk berkelium 66 Dy dysprosium 162.5 98 Cf californium 67 Ho holmium 164.9 99 Es einsteinium 68 Er erbium 167.3 100 Fm fermium 69 Tm thulium 168.9 101 Md mendelevium 70 Yb ytterbium 173.1 102 No nobelium 71 Lu lutetium 175.0 103 Lr lawrencium 21 Sc scandium 44.96 39 Y yttrium 88.91 57-71

lanthanoids 89-103

actinoids atomic number

Symbol

standard atomic weight 2 13 14 15 16 17 Key: 3 4 5 6 7 8 9 10 11 12 name

Notes

• IUPAC 2011 Standard atomic weights abridged to four significant digits (Table 4 published in Pure Appl. Chem. 85, 1047-1078 (2013); http://dx.doi.org/10.1351/PAC-REP-13-03-02. The uncertainty in the last digit of the standard atomic weight value is listed in parentheses following the value. In the absence of parentheses, the uncertainty is one in that last digit. An interval in square brackets provides the lower and upper bounds of the standard atomic weight for that element. No values are listed for elements which lack isotopes with a characteristic isotopic abundance in natural

terrestrial samples. See PAC for more details.

• “Aluminum” and “cesium” are commonly used alternative spellings for “aluminium” and “caesium.”
• Claims for the discovery of all the remaining elements in the last row of the Table, namely elements with atomic numbers 113, 115, 117 and 118, and for which no assignments have yet been made, are being considered by a IUPAC and IUPAP Joint Working Party. For updates to this table, see iupac.org/reports/periodic_table/. This version is dated 1 May 2013. Copyright © 2013 IUPAC, the International Union of Pure and Applied Chemistry. IUPAC Periodic Table of the Elements International Union of Pure and Applied Chemistry

Topic: Thermodynamics Metodi: Ideal Gas Law, Conservation Laws Competenze: Mathematical Modeling, Physical Reasoning Fonte: Testo (PDF) — p.33