India 2015 — incho2015-Q.pdf

Fonte: gare di altri paesi/India/chimica/incho2015-Q.pdf · Apri PDF · apri PDF p.1 Cluster: Fisica Moderna Soluzioni (stessa cartella): · · · · · · · · · · · · · ·

Problema 1

Question No 1 2 3 4 5 6 Total Marks 17 14 26 24 21 12 114 Marks Obtained

Signature of Examiner Indian National Chemistry Olympiad Theory 2015 HBCSE, 31st January 2015

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Instructions for students Write your Name and Roll No. at the top of the first pages of all problems. This examination paper consists of 36 pages of problems including answer boxes. Total marks for INChO 2015 paper are 114. You have 3 hours to complete all the problems. Blank space for rough work has been provided at the end of the paper. Use only a pen to write the answers in the answer boxes. Anything written by a pencil will not be considered for assessment. All answers must be written in the appropriate boxes. Anything written elsewhere will not be considered for assessment. You must show the main steps in the calculations, Use only a non-programmable scientific calculator. For objective type question, mark X in the correct box. Some of the objective questions may have more than one correct answer. Values of fundamental constants required for calculations are provided on page 4. A copy of the Periodic Table of the Elements is provided at the end of the paper. Do not leave the examination room until you are directed to do so. The question paper will be uploaded on the HBCSE website by 2nd February 2015. © Homi Bhabha Centre For Science Education Tata Institute of Fundamental Research V.N. Purav Marg, Mankhurd, Mumbai 400 088. Indian National Chemistry Olympiad Theory 2015 HBCSE, 31st January 2015

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Fundamental Constants

Avogadro number

NA = 6.022 x 1023 mol–1 Electronic charge

e = 1.602 x 10–19 C Molar gas constant

R = 8.314 J K–1mol–1

= 8.314 K Pa.dm3 K–1mol–1

= 0.082 L.atm K–1mol–1 1 atomic mass unit (1u)

= 931.5 MeV/C2 1 eV

= 1.602 x 10–19 J 1 cm–1

= 11.9 x 10–3 kJ mol–1 Rydberg constant

RH = 2.179 x 10–18 J Mass of electron

me = 9.109 x 10–31 kg Plancks constant

h = 6.625 x 10–34 Js Speed of light

c = 2.998 x 108 ms 1 Acceleration due to gravity g = 9.8 ms 2 Density of mercury

= 13.6 x 103 kg m 3 Faraday constant

F = 96485 C mol–1 Temperature

0 C = 273.15 K Indian National Chemistry Olympiad

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Problem 1

17 marks Oxides of nitrogen Nitrogen is one of the most important elements on earth. It forms several oxides including nitric oxide and nitrogen dioxide. Nitric oxide is an air pollutant produced in the combustion processes in automobile engines and power plants. 1.1 Write the Lewis dot structures of nitric oxide and nitrogen dioxide.

(1mark) In the chemical industry, nitric oxide is an important intermediate produced by oxidation of ammonia by oxygen. This is the first step in the Ostwald’s process of the manufacture of nitric acid. 1.2 Write the balanced equation for the oxidation of ammonia to nitric oxide by oxygen.

(0.5 mark) 1.3 In a closed vessel, at 700 K and a pressure of 1 atm, 100 kmols of ammonia are mixed with 20% excess of air than that required for complete combustion of ammonia. (air contains 80% nitrogen and 20% oxygen on volume basis)

i) Calculate the initial number of moles of oxygen and nitrogen present in the system.

.

(1 mark)

Name of Student Indian National Chemistry Olympiad

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ii) At equilibrium, 70% yield of nitric oxide was obtained in the closed vessel.

Calculate the mole percent of ammonia and water present in the mixture at

equilibrium.

(2 marks) Some equilibria of nitrogen oxides, which can be exploited commercially are shown below. (From Questions 1.4 to 1.8, you need to refer to these equations.) A) N2 (g) + O2(g) 2NO(g) B) 2NO(g) + O2(g) 2 NO2(g) C) N2O4(g) 2NO2(g)

D) 2NO2(g) +H2O(aq) HNO2(aq) + HNO3(aq) 1.4 Using the following data, calculate the standard free energy change in kJ for reaction A. The temperature is 298.15 K. Hf of NO(g): 90.37 kJ mol 1

S of N2(g): 191.5 J mol 1K 1 S of O2(g): 205 J mol 1K 1

S of NO(g): 210.6 J mol 1K 1

(1.5 marks) Indian National Chemistry Olympiad

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1.5 At 298.15 K, the $\text{DeltaGo}$ of formation for N2O4(g), and NO2(g) are 98.28 kJ mol 1 and 51.84 kJ mol 1 respectively. Starting with 1 mole of N2O4(g) at 1.0 atm and 298.15 K, calculate % of N2O4 decomposed if the total pressure is kept constant at 1.0 atm and the temperature maintained at 298.15 K.

(2 marks) 1.6 $\text{DeltaH}$ for the reaction C is 58.03 kJ. Assuming $\text{DeltaH}$ to be temperature independent, calculate the temperature at which the fraction of N2O4 decomposed is double the value of that calculated in 1.5.(The pressure is 1 atm)

(2.5 marks) Indian National Chemistry Olympiad

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1.7 The equilibrium represented by C was studied at 40 C at a certain pressure.

The density of the gaseous mixture was 5.85 g L 1. Calculate the average molecular

weight of the gaseous mixture and the degree of dissociation of N2O4 at 40 C.

$(\text{DeltaG}$ = 1.254 kJ mol 1)

(4 marks) Indian National Chemistry Olympiad

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1.8 Nitric oxide is formed in the combustion of fuel. An internal combustion engine produces 250 ppm (250 mg L 1) of NO (w/v). 100 L of air containing the produced NO was oxidized to NO2. The NO2 formed was dissolved in 100 L of water. Calculate the pH of the resulting solution. Refer to reaction D. (Given: pKa of HNO2 = 3.25)

(2.5 marks) Indian National Chemistry Olympiad

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Topic: Thermodynamics Metodi: First Law of Thermodynamics, Ideal Gas Law, Thermodynamic Cycle Analysis Competenze: Mathematical Modeling, Physical Reasoning Fonte: Testo (PDF) — p.1

Problema 2

Problem 2

14 marks Acid Base chemistry A. 2.1 Vinegar is used in food preparations. The main ingredient of vinegar is acetic acid that gives it a pungent taste. A sample of vinegar has 5% v/v acetic acid. The density of acetic acid is 1.05 g $mL-1$. a) Calculate the molarity of acetic acid in vinegar solution.

(1 mark) b) Calculate the pH of the above vinegar sample. (Ka for acetic acid = $1.75\times 10-5$)

(1 mark) 2.2 100 mL of the above vinegar sample is diluted to 250 mL and then 25 mL of the

diluted solution is titrated against NaOH solution (0.100 M).

a) Calculate the pH of the solution when 10 mL of NaOH solution was added.

Name of Student Indian National Chemistry Olympiad

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(1.5 marks) b) Calculate the pH of the solution at the equivalence point (equivalence point is

theoretical end point of the titration).

(2 marks) 2.3 A pH meter is usually calibrated using standard buffer solution for which the pH is exactly known. A buffer solution consisting of sodium acetate and acetic acid with pH = 5 is to be used for calibration of a pH meter. How many moles of sodium acetate and acetic acid are required to prepare 250 mL of this buffer solution?

(The total concentration of acetic acid in all forms in the solution is 0. 8 M).

(2 marks) Indian National Chemistry Olympiad

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B.

Two standard methods for estimation of amino acids are described below. In method

I, a sample of valine is treated with nitrous acid and the volume of nitrogen gas

released is measured. The reaction is as follows:

(CH3)2C(NH2)COOH + HNO2 $\to$ (CH3)2C(OH)COOH + N2 + H2O --------(Method I)

In method II, valine is treated with excess of perchloric acid in acetic acid (such a

titration is called as a non aqueous titration where glacial acetic acid is used as

solvent.)

The reaction is indicated below:

CH3CH(NH2)COOH + HClO4 $\to$ CH3CH(NH3)+COOH + ClO4 $-$ ---------(Method II)

After the reaction is complete, the unreacted HClO4 is determined by titrating it with

standard solution of sodium acetate.

50.0 mL of a 0.150 M solution of HClO4 is added to a sample of valine in glacial

acetic acid. The unreacted perchloric acid requires 20 mL of 0.180 M solution of

sodium acetate. 2.4 Calculate the volume of the nitrogen released (in L) at a pressure of 102658 Pa and a

temperature of $298.15^{\circ }C$ when the same quantity of sample was used in method I and

method II.

(2 marks) Indian National Chemistry Olympiad

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Fundamentals of Analytical Chemistry By Douglas Skoog, Donald West, F. Holler, Stanley Crouch C.

A sample of an aromatic acid weighing 1.743g requires 35 mL of 0.15 M NaOH for complete neutralisation. The vapour of the same acid is found to be 83 times heavier as compared to gaseous hydrogen. 2.5 Calculate the basicity of the aromatic acid.

(1.5 marks) D.

Maleic acid is a diprotic acid. Depending on pH, maleic acid can exist in solution in

different forms. If the undissociated acid is represented as H2M, the different forms

in which it can exist in the solution are H2M, $HM-1$ and $M-2$.

Let, CT = Total concentration of acid in all the forms and the fractions (represented

as n) for different forms of maleic acid can be represented as

0 = [H2M]/ CT

1 = $[HM-]/$ CT

2 = $[M-2]/$ CT

The following figure indicates variation of fractions of different forms of maleic acid

as a function of pH. Indian National Chemistry Olympiad

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2.6 Answer the following questions using the given figure. a) The pH at which 90% of $[HM-]$ is converted to $[M2-]$ b) The pKa1 and pKa2 of maleic acid are c) The indicator that can be used for the first equivalence point

(indicate the serial number of the indicator from the given table)

(3 marks) No. Indicator pH range for change of colour No. Indicator pH range for change of colour 1 Phenol red 6.8 – 8.4 3 Bromophenol blue 3.0 – 4.6 2 Bromophenol red 5.2 – 6.8 4 m-cresol purple 1.2 – 2.8 Indian National Chemistry Olympiad Theory 2015 ©HBCSE, 31st January 2015 15

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

Problema 3

Problem 3 26 marks Organic Reaction Intermediates Many organic reactions proceed through intermediates such as carbocations, carbanions, carbon radicals, carbenes etc. The structure and stability of the intermediates are the crucial factors that determine reaction mechanism. For example, in a reaction proceeding through formation of carbocation, the stability of the carbocation determines the rate of the reaction. Hammond postulate is an important tool to discuss transition state. It states that the transition state resembles that side closer to it in free energy. 3.1 Arrange the following olefins in the correct order of rate of addition of HI. (1 mark) Rearrangements of carbocations are very common. The group that migrates is generally the one that is electron rich. Thus sometimes, a product other than the one expected is formed. Compound A on treatment with a Bronsted acid gives compound C. In the reaction, B is an intermediate formed through a rearrangement. 3.2 Identify B and C. (2 marks) Name of Student Indian National Chemistry Olympiad Theory 2015 ©HBCSE, 31st January 2015 16 : … 3.3 Compound D on reaction with limited quantity of chlorine in the presence of UV light under ambient conditions gives a mixture of products. The major product E of the reaction is (1 mark) 3.4 Use the following compounds to answer the questions. i) Aromatic compound/s as they are drawn. ii) Antiaromatic compound/s. iii) Non aromatic compound/s. iv) Non aromatic as drawn but has/have resonance structure/s that is/are aromatic. v) Non aromatic, but has/have aromatic conjugate base. vi) Non aromatic, and has pKa around ( 3.8) (5 marks) Indian National Chemistry Olympiad Theory 2015 ©HBCSE, 31st January 2015 17 Aromatic compounds are difficult to reduce. However, the Australian chemist, A. J. Birch developed a method to reduce aromatic compounds to nonconjugated dienes by treatment with Li/K/Na in liquid ammonia in the presence of an alcohol. The reaction is called Birch reduction. Thus, benzene can be reduced to 1,4-cyclohexadiene. In this reaction the metal gives an electron to the aromatic ring to form a radical anion (an intermediate which is an anion and has an unpaired electron). In this intermediate, the radical centre and anionic centre are at 1,4 positions with respect to each other. Birch (1944) carried out the reduction of 3-methylanisole (3-methylmethoxybenzene) and obtained product F which on treatment with hot dilute mineral acid gave G (C7H10O). 3.5 Draw the possible structures of F. (1.5 marks) 3.6 Draw the possible stable structures of G based on the structures of F. (2.5 marks) G that is actually obtained, is not chiral. On reaction with Br2, it gave compound H which on heating with alcoholic KOH gave I (C7H9BrO). 3.7 Identify G, H and I. (2 marks) Indian National Chemistry Olympiad Theory 2015 ©HBCSE, 31st January 2015 18 rearrangement Hint: on ozonolysis it gives tetraaldehyde derivative Na : J K L M N O alcoholic KOH (excess) Br2 one equivalent CH2 Liq NH3 E. Vogel is well known for his work on bridged annulenes. Annulenes are large ring compounds containing continuous conjugation. Compound O is one such compound which is synthesized from naphthalene (J) by the following route. Compound K adds three equivalents of bromine. Carbene (:CH2) is a very reactive intermediate. It reacts with an olefinic double bond to form cyclopropane. 3.9 Draw the missing structures in the following sequence of reactions for the synthesis of compound O. (4.5 marks) Carbonyl compounds with acidic hydrogen can exhibit tautomerism and can exist in the enol form. The conjugate base of enol is called as enolate. 3.10 i) Draw the most stable enol and the corresponding enolate of the following compound. (1 mark) enol enolate O O Indian National Chemistry Olympiad Theory 2015 ©HBCSE, 31st January 2015 19 ii) Draw the structures of the most favorable products obtained from the reactants in the following reactions (1 mark) 3.11 Aza-enolates are the nitrogen analogues of enolates. Draw the structure of the aza-enolate of the following . (1 mark) 3.12 The reactant shown in 3.11 can be prepared by the condensation of i) An aldehyde and a secondary amine ii) A ketone and a secondary amine iii) An aldehyde and a primary amine iv) A ketone and a primary amine (1 mark) A Mannich reaction is a reaction between formaldehyde, a secondary amine and a ketone, an example of which is given below OH H+ O OH C C H 3 CH2COOCH3 b) C2H5O a) .. R N . O C H3 C H3 N C H3 N H C H 3 O H H O HCl EtOH, heat HCl + + Indian National Chemistry Olympiad Theory 2015 ©HBCSE, 31st January 2015 20 3.13 Identify the structures of compounds P S in the synthesis of a local anaesthetic, Tutocaine hydrochloride (S). The first step in this synthesis involves a Mannich reaction. (2.5 marks) Q P S R i) P Q R S Indian National Chemistry Olympiad Theory 2015 ©HBCSE, 31st January 2015 21 Permanganate manganate

Topic: Conservation of Energy, Oscillations & Waves, Thermodynamics Metodi: Physical Modeling Competenze: Mathematical Modeling, Diagrammatic Reasoning Fonte: Testo (PDF) — p.14

Problema 4

Problem 4 24 marks Chemistry of Potassium Permanganate Potassium Permanganate (KMnO4); chameleon mineral or Condy’s crystals is an inorganic compound which dissolves in water to give an intensely pink or purple solution. The evaporation of this solution leaves prismatic purplish-black glistening crystals. On a small scale potassium permanganate is prepared from the disproportionation of potassium manganate in acidic medium. 4.1 Write a balanced equation for this reaction. (1mark) 4.2 Draw the Lewis dot structures of manganate and permanganate species. State which of the species is paramagnetic. Calculate the spin only magnetic moment for the paramagnetic species. (2.5 marks) 4.3 When concentrated H2SO4 is added to KMnO4 it gives a greenish oily dimanganese heptaoxide. a) Write the balanced equation for this reaction. (1mark)

• 2KMnO4 + MnO2 + 2H2O Name of Student MnO43K2MnO4 + 4H+ 2KMnO4 + MnO2 + 2H2O Indian National Chemistry Olympiad Theory 2015 ©HBCSE, 31st January 2015 22 At cathode: At anode: Overall: Mn Mn Mn MnO HMnO MnO 1.18 E 2 1.51 E 3 0.95 E 2 2.09 E 4 0.90 E 4 5 4 3 2 1

b) Draw the structure of dimanganese heptaoxide. (1 mark) 4.5 On standing, the above oxide decomposes to form manganese dioxide. Write a balanced chemical equation for the reaction. (0.5 mark) The Latimer Diagram for a series of manganese species in acidic medium (pH = 1) is given below. The emf values (E ) shown are standard reduction potentials in volts. 4.6 In acidic medium, solid MnO2 is converted to Mn2+ and MnO4 . i) Write balanced equations for the half cell reactions involved and the overall reaction. (1.5 marks) ii) Using the Latimer diagram, calculate the standard electrode potential for each half cell reaction and for the overall cell reaction. Indian National Chemistry Olympiad Theory 2015 ©HBCSE, 31st January 2015 23 (3 marks) iii) Calculate K for the overall reaction. (1mark) In an acidic medium MnO4 is a strong oxidizing agent and is thus often used in redox titration. Medical shops sell 6% (w/w) aqueous solution of H2O2 as a disinfectant. X g of this H2O2 solution was titrated using KMnO4 (0.02 M) solution in an acidic medium. The sample required 15.0 mL KMnO4 solution. 4.7 Write balanced equation for the reaction involved in the titration. Calculate the amount of H2O2 in grams that was titrated. (Show the relevant steps). (3 marks) Indian National Chemistry Olympiad Theory 2015 ©HBCSE, 31st January 2015 24 The Frost Diagram (also known as oxidation state diagram) of an element (X) is a plot of volt-equivalent (NE ) for a couple X(N)/X(0) against oxidation number (N) of the element. Such a plot can be constructed from Latimer diagram. The Frost diagram for manganese species in acidic and basic condition is given below. 4.8 Using the given diagram fill in the blanks. i) The slope of the line joining two successive points is equal to of the relevant couple. ii) The most stable oxidation state of manganese in acidic condition is . and in basic condition is iii) In basic condition, the species of manganese that will disproportionate is iv) In acidic condition, the two pairs of manganese species that will comproportionate (opposite of disproportionation) are Ref: http://classes.uleth.ca/200501/chem2810a/lecture_20.pdf MnO4

HMnO4

H3MnO4 MnO2 Mn3+ Mn2+ Mn(OH)2 Mn2O3 MnO4 3 MnO2 MnO4 2 MnO4 Indian National Chemistry Olympiad Theory 2015 ©HBCSE, 31st January 2015 25 v) In basic condition, the manganese specie/s that will act as reducing agents are/is vi) In basic condition, the weakest oxidizing agent is (4.5 marks) Pourbaix diagram of manganese is the plot of the potential (E) vs pH and indicates the conditions under which different species of manganese are stable in aqueous medium at 25 C and 1M concentration. Such diagrams are used frequently in geochemical, environmental and corrosion studies. In this diagram,

1. Horizontal line separates species related by electron transfer only.
2. Vertical lines separate species related by proton transfer.
3. Slanting lines separate species related by both proton and electron transfer. Between the dashed line (a) and (b) water is stable, whereas above line (a) it is oxidized to O2 and below line (b) it decomposes to H2. Indian National Chemistry Olympiad Theory 2015 ©HBCSE, 31st January 2015 26 4.9 Based on the Pourbaix diagram of manganese given above, answer the following questions. i) Which specie/s of manganese is predominant in a) oxygen rich lakes of pH 7 b) highly oxygen depleted lakes that are contaminated with bases (pH 10)? ii) It is observed that clear well water, that is slightly acidic (pH 6) deposits solid/s on standing in toilet bowls. a) The solid/s species is/are b) The manganese species that is found dissolved in well water while it is still underground is iii) Identify the specie/s that exist/s in ocean water at pH = 8 and E = + 0.2V to + 0.6 V. (5 marks) Mn2+ Potential E (V) (s) (a) (b) MnO2(s) (s) Mn(OH)2 Mn(OH)3

(Ref: http://commons.wikimedia.org/wiki/File:Pourbaix_diagram_for_Manganese.svg) Indian National Chemistry Olympiad Theory 2015 HBCSE, 31st January 2015

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a. b. c. : .. ..

Topic: Electrostatics Metodi: Physical Modeling, Conservation Laws Competenze: Mathematical Modeling, Diagrammatic Reasoning Fonte: Testo (PDF) — p.20

Problema 5

Problem 5

21 marks

Natural Nitrogen Compounds A variety of nitrogen compounds are found in both plant and animal kingdoms. Nitrogen is present in several classes of natural products like alkaloids, nucleic acids, vitamins etc and these compounds have physiological effects. A stereogenic center is an atom, bearing groups, such that an interchange of any two groups leads to a stereoisomer. Nitrogen containing compounds can also be chiral and exhibit optical activity. 5.1 Some compounds are given below. Label them as I. Achiral

II. Chiral; enantiomers cannot be separated III. Chiral; enantiomers can be separated.

(1.5 marks) Alkaloids are basic nitrogeneous compounds of plant origin. (+) - Muscarine is a poisonous alkaloid found in some mushrooms.

Name of Student Indian National Chemistry Olympiad Theory 2015 HBCSE, 31st January 2015

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Morphine .. 5.2 Draw the enantiomer of (+) Muscarine and give its stereodescriptors.

(2 marks) Several alkaloids occur in opium which is a narcotic drug. Morphine is one of them. It is an analgesic and is used to relieve intense pain.

5.3 How many stereogenic centres are present in Morphine? a) 3

b) 4

c) 5

d) 6

(1 mark) The Hofmann exhaustive methylation is a method used in structure determination of alkaloids. The reaction involves quarternization of the nitrogen followed by elimination to yield an alkene as one of the products. It is schematically represented below.

The diacetyl derivative of morphine is heroin, which is also a narcotic drug. Heroin is subjected to Hofmann exhaustive methylation to eliminate nitrogen completely to obtain product A. Indian National Chemistry Olympiad Theory 2015 HBCSE, 31st January 2015

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A Indole 5.4 The number of moles of methyl iodide required for complete removal of nitrogen from

heroin is

a) 1

b) 2

c) 3

d) 4

(1 mark) 5.5 Draw the structure of ‘A’.

(2 marks) ‘A’ is treated with an excess of bromine. 5.6 The number of molecules of bromine consumed by one molecule of ‘A’ is

a) 2

b) 3

c) 4

d) 5

(1 mark) Another interesting class of nitrogen containing alkaloids is the Indole alkaloids that contain the indole ring.

Indole and its derivatives are synthesized by the Fischer indole synthesis, that involves an interesting acid catalysed rearrangement of an arylhydrazone as represented below.

The following indole derivative was synthesized by the Fischer synthesis using a carbonyl compound B and a substituted phenyl hydrazine derivative C. Indian National Chemistry Olympiad Theory 2015 HBCSE, 31st January 2015

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5.7 Draw the structures for B and C.

(1.5 marks) Arene diazonium salts are important intermediates that can be directly reduced to aryl hydrazines or can be used to prepare aryl hydrazones. They are also used to prepare azo dyes, in which case they function as electrophiles. 5.8 Draw the most important resonance structures of benzene diazonium ion.

(1 mark) Arene diazonium salts couple with aliphatic compounds containing acidic carbon atom (active methylene compounds) to form azo derivatives. The initial azo compound tautomerises to the hydrazo derivative. If a given compound does not permit such a tautomerism, the compound may lose a suitable group from the coupling site to allow for the tautomerism. 5.9 Complete the following reaction

(1 mark) Indian National Chemistry Olympiad Theory 2015 HBCSE, 31st January 2015

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Compound ‘J’ is an alkaloid and a redox cofactor for the enzyme alcohol dehydrogenase and plays a critical role in cell signaling pathways.

Compound ‘I’ is a key intermediate in the synthesis of ‘J’. This synthesis path involves the formation of a hydrazone. 5.10 Complete the synthesis of compound ‘I’ by drawing the missing structures in the boxes

below.

(4.5 marks)

• C G does not give precipitate with , -DNP C H , Pd, atm NaNO , HCl lewis acid, C HCl, reflux Indian National Chemistry Olympiad Theory 2015 HBCSE, 31st January 2015

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Compound ‘K’ a precursor of ‘J’ can be obtained from I. 5.11 The most basic nitrogen in compound ‘K’ is

(0.5 mark) Natural compounds containing the imidazole ring have been found to be physiologically active. Hence efforts have been made to synthesise such compounds. One such synthetic drug is Pentostatin, which is used as an antiviral and antitumour agent. A key intermediate in the synthesis of Pentostatin is compound ‘O’ (C12H10N4O5)

5.12 Complete the synthesis of compound ‘O’ by drawing the missing structures in the boxes provided.

(4 marks) a b c d e K CO O , H O ii. CH NO /t-BuOK PhCH Cl i. SOCl L Indian National Chemistry Olympiad

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Topic: Geometric Optics, Nuclear & Particle Physics Metodi: Physical Modeling Competenze: Mathematical Modeling, Diagrammatic Reasoning Fonte: Testo (PDF) — p.26

Problema 6

Problem 6

12 marks Beer-Lambert Law A. Beer–Lambert law deals with the relationship between the extent of absorption of radiation by a species and its concentration. The law states that ‘absorbance (A) is directly proportional to concentration (expressed as mol L–1) at any given wavelength for a dilute solution. Mathematically it can be expressed as A = $ϵ$ l c, where, $ϵ$ = molar absorptivity (units = L cm–1 mol–1), l = path length in cm, A = log10 (I/ Io), where Io = intensity of the incident radiation and I = intensity of the transmitted radiation. The ratio of (I / Io) is called as transmittance (T).

Method of continuous variation is one of the standard methods used for the spectrophotometric determination of the composition of a complex between a metal M and a ligand L. In this method, the sum of the molar concentrations of the metal M and the ligand L is kept constant, but their relative ratio is varied. The following graph is obtained for one such analysis. Use the graph to answer the questions from 6.1 to 6.4. (XM = CM / (CM + CL)), where CM = concentration of the metal ion in all forms, CL = concentration of ligand in all forms).

Name of Student

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 0 0.05 0.1 0.2 0.24 0.4 0.6 0.8 1 Mole Fraction Absorbance (XM) Indian National Chemistry Olympiad

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6.1 Deduce by calculations the species that absorb when XM = 0 and XM = 1 respectively.

(1 mark) 6.2 What is the ratio of the molar absorptivities of M and L?

(2 marks) 6.3 What percentage of the incident light is transmitted through solutions when

(i) XM = 0.1 and when (ii) XL = 0.2?

(1.5 marks) 6.4 Determine the composition of the complex formed. Show your calculations.

(2 marks) Indian National Chemistry Olympiad

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B. 6.5 For practical purpose, the preferred percentage transmittance range for spectrophotometric measurement should be between 20% to 65% (as the error in the measurement in this range is minimum). In an experiment which involved the determination of absorbance for a co-ordination complex of iron ( = 12000), calculate the concentrations of the complex corresponding to the above transmittance range.

(1.5 marks)

The chelate CuQ2 2– (aq) formed by Cu2+ (aq) and the complexing agent Q2 (aq) absorbs at 480 nm. When the concentration of chelating agent Q2 (aq) is five times in excess as compared to Cu2+ (aq), the absorbance of the chelate solution depends only on molar concentration of Cu2+ (aq) and obeys Beer-Lambert law. Neither Cu2+ (aq) nor Q2 (aq) absorbs at 480 nm. A solution that contains 3.30 x 10–4 M Cu2+ and 8.60 x 10–3 M of Q2– has absorbance 0.690 at $\lambda$ = 480 nm.

Another solution that was prepared by mixing 3.30 x 10–4 M of Cu2+ and 6.500 x 10–4 M of Q2– was found to have absorbance 0.610 at the same wavelength. (cell length l = 1 cm). 6.6 From the given data, calculate the equilibrium constant (called as formation constant Kf) for the following process Cu2+ (aq) + 2Q2– (aq) CuQ2 2– (aq). Indian National Chemistry Olympiad

Theory 2015 ©HBCSE, 31st January 2015 36

(4 marks) HBCSE, 31st January 2015 Rough Page HBCSE, 31st January 2015 Rough Page HBCSE, 31st January 2015 Rough Page HBCSE, 31st January 2015 Rough Page 1 IA 11A 18 VIIIA 8A 2 IIA 2A 13 IIIA 3A 14 IVA 4A 15 VA 5A 16 VIA 6A 17 VIIA 7A 3 IIIB 3B 4 IVB 4B 5 VB 5B 6 VIB 6B 7 VIIB 7B 8 9 VIII 8 10 11 IB 1B 12 IIB 2B Periodic Table of the Elements Lanthanide Series Actinide Series © 2013 Todd Helmenstine chemistry.about.com sciencenotes.org 1 H Hydrogen 1.008 3 Li Lithium 6.941 4 Be Beryllium 9.012 11 Na Sodium 22.990 12 Mg Magnesium 24.305 19 K Potassium 39.098 20 Ca Calcium 40.078 21 Sc Scandium 44.956 22 Ti Titanium 47.88 23 V Vanadium 50.942 24 Cr Chromium 51.996 25 Mn Manganese 54.938 26 Fe Iron 55.933 27 Co Cobalt 58.933 28 Ni Nickel 58.693 29 Cu Copper 63.546 30 Zn Zinc 65.39 31 Ga Gallium 69.732 13 Al Aluminum 26.982 5 B Boron 10.811 32 Ge Germanium 72.61 14 Si Silicon 28.086 6 C Carbon 12.011 33 As Arsenic 74.922 15 P Phosphorus 30.974 7 N Nitrogen 14.007 34 Se Selenium 78.09 16 S Sulfur 32.066 8 O Oxygen 15.999 35 Br Bromine 79.904 17 Cl Chlorine 35.453 9 F Fluorine 18.998 36 Kr Krypton 84.80 18 Ar Argon 39.948 10 Ne Neon 20.180 2 He Helium 4.003 37 Rb Rubidium 84.468 38 Sr Strontium 87.62 39 Y Yttrium 88.906 40 Zr Zirconium 91.224 41 Nb Niobium 92.906 42 Mo Molybdenum 95.94 43 Tc Technetium 98.907 44 Ru Ruthenium 101.07 45 Rh Rhodium 102.906 46 Pd Palladium 106.42 47 Ag Silver 107.868 48 Cd Cadmium 112.411 49 In Indium 114.818 50 Sn Tin 118.71 51 Sb Antimony 121.760 52 Te Tellurium 127.6 53 I Iodine 126.904 54 Xe Xenon 131.29 55 Cs Cesium 132.905 56 Ba Barium 137.327 57-71 72 Hf Hafnium 178.49 73 Ta Tantalum 180.948 74 W Tungsten 183.85 75 Re Rhenium 186.207 76 Os Osmium 190.23 77 Ir Iridium 192.22 78 Pt Platinum 195.08 79 Au Gold 196.967 80 Hg Mercury 200.59 81 Tl Thallium 204.383 82 Pb Lead 207.2 83 Bi Bismuth 208.980 84 Po Polonium [208.982] 85 At Astatine 209.987 86 Rn Radon 222.018 87 Fr Francium 223.020 88 Ra Radium 226.025 89-103 57 La Lanthanum 138.906 58 Ce Cerium 140.115 59 Pr Praseodymium 140.908 60 Nd Neodymium 144.24 61 Pm Promethium 144.913 62 Sm Samarium 150.36 63 Eu Europium 151.966 64 Gd Gadolinium 157.25 65 Tb Terbium 158.925 66 Dy Dysprosium 162.50 67 Ho Holmium 164.930 68 Er Erbium 167.26 69 Tm Thulium 168.934 70 Yb Ytterbium 173.04 71 Lu Lutetium 174.967 89 Ac Actinium 227.028 90 Th Thorium 232.038 91 Pa Protactinium 231.036 92 U Uranium 238.029 93 Np Neptunium 237.048 94 Pu Plutonium 244.064 95 Am Americium 243.061 96 Cm Curium 247.070 97 Bk Berkelium 247.070 98 Cf Californium 251.080 99 Es Einsteinium [254] 100 Fm Fermium 257.095 101 Md Mendelevium 258.1 102 No Nobelium 259.101 103 Lr Lawrencium [262] 104 Rf Rutherfordium [261] 105 Db Dubnium [262] 106 Sg Seaborgium [266] 107 Bh Bohrium [264] 108 Hs Hassium [269] 109 Mt Meitnerium [268] 110 Ds Darmstadtium [269] 111 Rg Roentgenium [272] 112 Cn Copernicium [277] 113 Uut Ununtrium unknown 114 Fl Flerovium [289] 115 Uup Ununpentium unknown 116 Lv Livermorium [298] 117 Uus Ununseptium unknown 118 Uuo Ununoctium unknown

Topic: Wave Optics Metodi: Experimental Data Analysis, Graph Linearization Competenze: Experimental Data Analysis, Graph Linearization, Mathematical Modeling Fonte: Testo (PDF) — p.32