selected topics for csir net & gate … a science student, ... happiness and prosperity in our...

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Selected Topics for CSIR NET & GATE chemistry 2017

Prepared by

V. Aditya Vardhan

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AACihem stry

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SELECTED TOPICS FOR CSIR NET & GATE CHEMISTRY(NEW EDITION 2017)

Study material by V. Aditya vardhan(AdiChemistry / AdiChemAdi)

Released on: 01st, May 2017Last updated on: 10th, June 2017

SAMPLE COPYOnly few pages are shown randomly for evaluation purpose

GO TO NEXT PAGE FOR THE LIST OF TOPICS COVERED

SALIENT FEATURES* Important selected topics for CSIR NET & GATE as well as other equivalent exams like SET/

SLET, TIFR etc. Also useful for UG & PG exams.* Written in a simple langualge but with very effective tone and helps you in improving your

score.* Most of the topics are provided with illustrated notes as well as worksheets/practice problems.Note: The length of topics may range from large to small. Some topics are provided with either notes or

question bank only. Missing parts will be appended in future updates.* You can take printouts of this e-book, if you wish to do so, provided you should have purchased

this book from the Author, V. Aditya vardhan.* Online access (read only) to keys to unsolved problems will be given only to those who

purchased the book from the author.* This book is going to be updated again and again with new content. So keep watching.

What is new?It is a completey reorganized study material. New topics and new questions were added.

Copyright & DisclaimerAll rights reserved. No part of this online publication may be reproduced, stored in a retrieval system, or transmitted, in any form

or by any means, without the prior written permission from the author.Being a science student, you should use your discretion while using the information given in this book. While every care has been

exercised in compiling and publishing the data contained in these pages, the author accepts no responsibility for errors or omissionsto the information or any damage caused by reading this book. Author cannot be held liable for typographical errors or otherinformation. The information is not guaranteed to be accurate since the information comes from other sources and therefore may bewholly unreliable.

VISIThttp://www.adichemistry.com/common/htmlfiles/csir-gate-chemistry.html

TO FIND MORE INFORMATION ON OTHER CSIR STUDY MATERIAL & ONLINE COACHING

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(read only)

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ii

# Topic Namefrom page

to page

1 Bioinorganic chemistry: * Distribution of elements - Bulk & Trace metals ; non metals - biological functions * Porphyrins * Oxygen transport * Hemoglobin & Myoglobin (Updated again) * Hemerythrin * Hemocyanin * Electron transport * Cytochromes - Cytochrome C oxidase - Cytochrome P450 * Iron sulfur proteins - Ferridoxins - Reiske protein * Rubredoxin * Photosystems * PSII - OEC * PSI * Metalloenzymes * Carboxy peptidase * Carboxy anhydrase * Liver alcohol dehydrogenase * Superoxide dismutase * Nitrogen fixation * Copper proteins * Siderophores * Ionophores * Ferritin, Apoferritin & Tranferrin * Application of metals in medicine * Practice questions (level 1) * Advanced questions (level 2)

1 36

2 Analytical chemistry: * Chromatography * Basic concepts & terminology * Theory of Band broadening and Van deemter equation * Gas Chromatography * HPLC * Practice questions * Advanced questions * Spectrochemical methods * Beer-Lambert’s law * Practice questions * Advanced questions * Electroanalytical methods * Polarography* Amperometry * Cyclic Voltammetry * Stripping Voltammetry * Potentiometry * Practice questions * Advanced questions * Gravimetry * Iodometry * Practice questions

37 82

3 Nuclear chemistry: Terminology, Concepts, Formulae * Stability of nucleus * Mass defect * Binding energy * magic numbers * Radioactive disintegration series * Solved questions * Practice

83 97

4 Solid state (explained through problems) 98 1195 Pericyclic reactions: 1) Cycloaddition reactions 2) Electrocyclic

reactions 3) Sigmatropic rearrangements 4) Group transfer reactions 5) Cheletropic reactions 6) Dyotropic rearrangements ; * Frontier Molecular Orbital (FMO) Theory * Woodward Hoffmann rules *

120 142

6 Mossbauer Spectroscopy: Introduction; Natural line width; Doppler effect; Recoiling effect; 57 Fe Mossbauer spectroscopy; Isomeric shift; Electric quadrupole splitting; Magnetic interactions; selection rules, Chemical applications; Practice problems

143 155

PART-1LIST OF TOPICS

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7 Thermodynamics (through problems) 156 177

8

Electron spin resonance (ESR) Spectroscopy: Introduction; Spin states of electron, Presentation of spectrum, Characteristics of 'g', Hyperfine splitting, Selection rules, EPR transitions in metal complexes & organic molecules, Practice problems

178 189

9 Electronic Spectroscopy of Transition metal complexes: Selection rules (Laporte and spin); LMCT, MLCT, calculation of CFSE , Term symbols for ground state, Russel Saunder's coupling are explained through solved problems.

190 207

1 Aldol reaction 1 112 Jahn Teller Theorem 12 223 Kohlrausch law 23 374 Wittig reaction 38 465 Hard Soft Acid Base theory (HSAB) 47 516 Miller indices 52 607 Beckmann reaction 61 658 Wilkinson catalyst 66 749 Rate laws 75 94

10 Birch reduction 95 10211 Spinels 103 10712 van der Waals equation 108 12313 SeO2 124 12914 18E Rule 130 14315 Phase rule 144 14916 Hammett equation 150 15117 Trans Effect 152 15618 le Chatelier's Principle 157 16819 Favoroskii reaction 169 17320 Silicates 174 17821 Jablonskii diagram 179 18122 Inductive effect 182 18723 Boranes & Wades rules 188 19924 Adsorption isotherms - Langmuir theorem 200 20825 Corey-Chaykovsky reaction 209 21726 f-block 218 23227 Band Theory 233 24128 DIBAL 242 24929 Catalysis using organometallics-Hydroformylation-Wacker oxidation 250 263

PART-2In the following topics, explanations are followed by practice questions.

MORE TOPICS WILL BE ADDED FROM TIME TO TIME

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A careful balance between Science and Silence brings peace, happiness and prosperity in our life.

HEMOGLOBIN AND MYOGLOBINHemoglobin (denoted as Hb) and Myoglobin (Mb) are dioxygen (O2) binding metalloproteins containing

an iron porphyrin system, heme. Both of them contain Fe(II) ion. Hemoglobin is present in Red BloodCells (RBC) and helps in transport of dioxygen from lungs to tissues. Whereas, myoglobin stores dioxygenand is present in muscles.

STRUCTUREHemoglobin contains four heme units each embedded in a globular protein sub-unit. There are two

types of protein sub-units i.e., a and ß. Myoglobin contains only one heme unit surrounded by a globularprotein, containing seven -helical and six non helical segments, made up of 153 amino acids .

Note: Heme moieties are shown in green color in above diagram. Fe(II) ion is shown in red color.

Heme is a porphyrin ring system made up of four pyrrole rings with an Fe(II) ion coordinated to nitrogensof pyrrole rings.

Globin part prevents irreversible oxidation of Fe(II) ion by providing hydrophobic environment. It enhancesthe selectivity for O2 binding. In hemoglobin, the tetramer arrangement allows for co-operativity by makingit more efficient in binding to dioxygen.

Hemoglobin and Myoglobin exist in two forms i.e., 1) deoxy form: No oxygen is bound to iron.2) oxy form: dioxygen is bound to iron.

In deoxy-hemoglobin, four of the coordinated sites of iron are occupied by nitrogens of porphyrin ring.The fifth site is occupied by Histidine residue (called proximal histidine) of globin. The sixth position isoccupied by weakly bonded water molecule. Hence some authors tend to report Fe(II) ion in deoxy formas pentacoordinated. Deoxy-hemoglobin is said to be in T-state (tense).

On the opposite side of the proximal histidine, there is one more histidine group (called distal histidine)placed near the iron ion. It forces the binding of dioxygen in "end on bent" confirmation.

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Note: The bent confirmation discourages the binding of CO to heme iron. Otherwise, CO may haveeven more affinity with the iron ion. It is observed that CO binds to hemoglobin 200X stronger thandioxygen but binds 20,000X stronger with unprotected heme.

Hemoglobin coordinated to dioxygen is called oxy-hemoglobin. It is also referred to as R-state(relaxed). In oxy-hemoglobin the sixth coordinated position of iron is occupied by dioxygen in "end onbent" geometry.

Hemoglobin is sometimes referred to as dimer of dimers since it contains two alpha and twobeta subunits.

FUNCTIONING OF HEMOGLOBIN - A BIOINORGANIC PERSPECTIVEIn deoxy-Hemoglobin, the porphyring ring is dome shaped. The Fe(II) is in high spin state and is

paramagnetic. Its size is 0.78 Ao and is positioned above the plane of the porphyrin ring. However, in oxy-Hemoglobin, the size of iron ion is reduced to 0.61 Ao and can fit into the cavity of

planar porphyrin ring and hence moves into the cavity of porphyrin ring with concomitant dragging ofproximal histidine that triggers the conformational changes in other globin subunits and thus by openingother heme sites. As a result, the binding capacity of other heme irons with dioxygen is enhanced. This isbest example for co-operativity through allostery.

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Adichemistry.comFeII

N

NH

N N

N N

CH3CH3

CH3 CH3O

-OC-H2C-H2C CH2-CH2-COO

-

N

NH

FeIII

Deoxy HbDomed porphyrin

High spin Fe(II) - out of ring

Oxy HbPlanar porphyrin

Fe ion moves into the ringBent O2

OO

NH

N

H-bond

NH

N

N N

N N

CH3CH3

CH3 CH3

O-OC-H2C-H2C CH2-CH2-COO

-

The nature of Fe in oxy-Hemoglobin or in oxy-Myoglobin is controversial. According to old Pauling model, there is a low spin Fe(II) ion that is bound to singlet O2 in oxy-Hb. Both

are diamagnetic. However, according to Weiss model, there is Fe(III) ion bound to superoxide radical anion (O2

-).Though both are paramagnetic, a strong paramagnetic coupling between them ensues diamagnetic behavior. This model is supported by the O-O stretching frequency at 1105 cm-1 in resonance Raman spectrumthat is consistent with the fact that O2 is in superoxide form. This deems to be more accurate and modernexplanation.

More information..............* Bisphosphoglycerate (BPG) reduces the affinity of Hb with oxygen. It binds at the centre of cavity

between two beta-Hb units and stabilizes the T-state. It is observed that at high altitudes (on moun-tains), the amount of BPG increases in the blood. This won’t affect the affinity of Hb with oxygen inlungs where the pO2 is very high but decreases the affinity in tissues and helps in release of dioxygen bystabilising the T-state of Hb.

* Bohr effect: The affinity of Hemoglobin with dioxygen decreases with decrease in the pH ofblood. Hence oxygen is released more efficiently in the tissues where CO2 concentration is high. Theresponse of hemoglobin to changes in pH is called the Bohr effect.

* Increase in the temperature decreases the affinity with O2.* The percent saturation with O2 curve of Hb against pO2 is sigmoidal whereas that of Mb is hyper-

bolic. This shows greater affinity of Mb with dioxygen even at low pO2 values.* Deoxyhemoglobin is the form of hemoglobin without the bound oxygen.The oxyhemoglobin has

significantly lower absorption (660 nm) than deoxyhemoglobin (940 nm). This difference is used formeasurement of the amount of oxygen in patient's blood by pulse oximeter.

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Mb - Hyperbolic curve

Hb - Sigmoidal curve

% s

atur

atio

n w

ith O

2

pO2

Why are all the oxygen carriers that contain iron and porphyrins are found inside the cells?The inside cell environment is reducing and sustains Fe(II), whereas outside the cell the O2 concentration

is high and hence the probability of the oxidation of Fe(II) ions to Fe(III) increases.

What prevents synthetic (simple) iron porphyrins from functioning as O2 carriers?In the naturally occuring porphyrins, like Hemoglobin, there is a globular protein around the heme

groups. I t prevents the irreversible oxidation of Fe(II) to Fe(III) and solvolysis of Fe(II)-O2 complex byproviding hydrophobic environment around iron. It stops the formation of Fe-O2-Fe dimer.

However in synthetic porphyrins, protein part is absent. As a result, Fe(II) is oxidized to Fe(III) and Fe-porphyrins easily dimerize to Fe-O2-Fe and then Fe-O-Fe (a -oxo dimer) and hence cannot function asoxygen carriers.

The steps, involved in conversion of free heme in aqueous solutions to hematin, a -oxo dimer, whenexposed to dioxygen, are summarized below.

Fe2+ + O2 Fe

2+O

O

Fe2+

OO Fe

2+ Fe3+

Fe3+

OO

Fe3+

Fe3+

OO

Fe4+

O Fe4+

O

Fe4+

O Fe2++ Fe

3+O

Fe3+

Binding of dioxygen molecule

Formation of peroxo complex

Formation of ferryl complexes in which iron is in +4 formal oxidation state

Hematin formation

+

Note: This -oxodiiron(III) moiety has a distinctive fingerprint that has made it easy to identify thismotif in proteins regardless of the geometry, type and number of ligands.

Magnetic susceptibility = 1.5 to 2.0 Bohr magnetons, indicating one unpaired electron. It is due to

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strong antiferromagnetic coupling of iron centres through oxygen. (Actually the value should be more)Asymmetric Fe-O stretching frequency at 730 - 880 cm-1

What is the "Cooperative effect" in hemoglobin?Coordination of one O2 leads to conformational changes in the protein chain leading to facilitated

coordination of O2 by the other 3 sub-units in hemoglobin.

Why does carbon monoxide binds strongly to iron(II) porphyrin complexes but not iron(III)porphyrins. Compare the infrared frequency of CO in this complex to that of free CO.Ans:- The Fe(II)-CO bond is strong because there are strong d -p * backbonding interactions whichstrengthen the Fe-C bond. This is much less important for Fe(III)-CO because of higher positive chargeon the metal.

The IR frequency of free CO is higher than that of CO in a porphyrin complex because backbondingadds electron density to the * antibonding orbital of the CO bond, decreasing its strength.

PRACTICE QUESTIONSFROM

BIOINORGANIC CHEMISTRY

Keys/Solutions along with extra problems (updates) are available at http://www.adichemistry.com/csir-net/chemistry/key.html

1) Complexes of which of the following metals are used in the treatment of rheumatoid arthritis:1. Gold 2. Ruthenium 3. Iron 4. Copper

2) In biological systems, the metal ion involved in the dioxygen transport besides Fe isa) Co b) Zn c) Mg d) Cu

3) In photosynthesis, the predominant metal present in the reaction centre of photosystem II is:(a) Zn (b) Cu (c) Mn (d) Fe (GATE 2009)

4) Zn in carbonic anhydrase is coordinated by three histidine and one water molecule. The reaction of CO2with this enzyme is an example of

(a) electrophilic addition (b) electron transfer(c) nucleophilic addition (d) electrophilic substitution.

5) The metals present in nitrogenase are(a) Fe and Mg (b) Mo and K (c) Mo and Fe (d) Fe and K

6) The transition metal present in vitamin B12 is_______

7) The trivalent ion of lanthanoid element which is used as NMR contrasting agent is1) Gadolinium 2) Technetium 3) Cerium 4) Lutetium

8) Match the following1) Li+ A) Ulcer treatment2) Bi3+ B) Eczema3) Sb3+ C) Anemia4) Fe2+ D) Depression

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9) The enzyme which removes C-terminal amino acid from a peptide is1) Carbonic anhydrase 2) Carboxy peptidase 3) Zymase 4) All

10) Which of the following metal is stored by metallothionein in biological systems?1) Fe 2) Cu 3) Zn 4) Co

11) The metals present in superoxide dismutase are: (CSIR NET DEC 2011)a) Zn & Fe 2) Cu & Fe 3) Mo & Cu 4) Cu & Zn

12) In human body, by mass, the most abundant nonmetal is ____ and most abundant metal is ____.

13) The copper containing non heme respiratory protein is1) Cytochrome-c 2) Hemerythrin 3) Hemocyanin 4) Myoglobin

14) The oxidation state of iron stored in ferritin is1) +2 2) +1 3) +3 4) 0

15) The enzyme which catalyzes the conversion of carbondioxide to bicarbonate is1) Carboxy peptidase 2) Superoxide dismutase3) Carboxy anhydrase 4) Hydrogenase

16) The number of methine bridges present in porphyrin is _______ and in corrin is _____ .

17) The ionophore used in the cattle feed to improve the permeability of Na+ and H+ ions is1) Crown ethers 2) Monensin 3) Ionomycin 4) Nonactin

18) Match the followingA) Silver sulfadiazine 1) Anti copper agentB) Selenium sulfide 2) Skin fungal infectionsC) tetrathiomolybdate 3) RadiodiagnosisD) Cardiolyte 4) Seborrheic dermatitis

19) The function of Na+ and K+ in biological systems is1) To maintain osmotic balance in the cells 2) Help in active & passive transport3) As charge carriers 4) All

20) Fluoride is present in the teeth in the form of1) CaF2 2) Na3AlF6 3) Ca5(PO4)3F 4) CaF2.CaCO3

21) Which of the following is a channel forming ionophore?1) Valinomycin 2) Salinomycin 3) Gramicidin 4) All

22) The porphin system in chlorophyll is called as1) porphyrin 2) chlorin 3) corrin 4) heme

23) The metal present in both superoxide dismutase and hemocyanin isa) Zn b) Fe c) Cu d) Ni

24) The metal ion present in urease is____ .

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25) The ionophore valinomycin is highly selective for:a) K + b) Na+ c) Mg2+ d) Ca2+

26) The metal ions that have highest mobility in biological media are:a) Mg(II) & Ca(II) b) Zn(II) & Fe(II) c) Na(I) & K(I) d) Ni(II) & Cu(II)

27) Cisplatin is:a) diamagnetic b) paramagnetic c) ferromagnetic d) anti-ferromagnetic

28) Hemerythrin belongs to the group of:a) non-heme iron protein b) heme-iron proteinc) binuclear copper protein d) non heme non iron protein

29) Which of the following is correct concerning the differences between hemoglobin and myoglobin?A) Hemoglobin is a monomeric protein and myoglobin is a tetrameric protein.B) Hemoglobin exhibits a hyperbolic O2 saturation curve while myoglobin exhibits a sigmoid shaped

curve.C) Hemoglobin exhibits cooperative binding of O2 while myoglobin does not.

BEER-LAMBERT LAW

BEER-LAMBERT LAW: The actual absorbance, A, of a sample is dependent on the pathlength l (incm) and the concentration c (molar conc.) of the species.

Mathematically: A = .c.l* is a is a measure of the amount of light absorbed per unit concentration at a given wavelength.* is an intrinsic property of the compound;Limitations:* Beer-Lambert law is a limiting law. It is only valid for dilute solutions. At higher concentrations, the

electrostatic interactions between neighbouring molecules distort the energy levels which cause deviationfrom the law.

* Deviations also occur when the molecules undergo chemical change in solutions (this may occur dueto change in concentration or pH).

eg., Yellow chromate and orange dichromate are in equilibrium with each other in aqueous solution.The equilibrium shifts to the right if pH is decreased (when acid is added) and shifts to the left if pH isincreased (when base is added).

2- + 2-

4 2 7 22 CrO (aq) + 2 H (aq) Cr O (aq) + H O (l)

yellow orange

Hence the absorbance for this solution depends on the pH .* There should be no scattering of light.* There should be no fluorescence or phosphorescence

INTERESTINGGuess the molar absorbtivity of Cu2+ ions in an aqueous solution of CuSO4. Whether 20 or 100,000

L mol-1 cm-1?

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You may prefer higher value as the copper sulphate solutions you have seen are usually a beautifulbright blue in colour. However, its molar absorbtivity value is 20 L.mol-1.cm-1! The bright blue colour isseen because the concentration of the solution is very high.

Now consider the beta-carotene which is responsible for the colour of carrots. It is found at exceedinglylow concentrations and still imparts beautiful orange color as its molar absorbtivity is 100,000 L.mol-1.cm-

1.

SOLVED QUESTIONS44) The unit of molar absorptivity is:

1. L mol–1 cm–1 2. L–1 mol cm–1 3. L mol cm–1 4. L mol cm

Explanation:Beer-Lambert law A = .c.lWhere A = absorbance

c = molar concentrationl = pathlength (in cm)

= molar extinction coefficient or molar absorptiviy = Acl

(note: is pronounced as epsilon)Hence the units are usually in M-1cm-1 or L mol-1cm-1 ( ‘A’ has no units: see below)

As 1 liter = 1000 cm3, the unit may be sometimes cm2 mol-1, especially in old literature.

Additional informationAbsorbance (A) = log10(1/T)

= -log10 T = -log10 (I /Io)

Where T = transmittance = I /IoI = intensity of the transmitted lightIo = intensity of the initial incident light

* Remember transmittance (T) is inversely related to absorbance (A)

Other equations to remember:

10100A = log ( )%T = 2 - log10(%T)

44.1) What is the concentration of a solution whose absorbance is 0.21 when placed in a tube ofpath length 10 cm ( max = 245 nm and max = 31,500 M-1cm-1)

Ans:- -7A 0.21c= = = 6.67 x 1031500 x 10

Ml

44.2) If the visible spectrum of [Ti(OH2)6]3+ has an absorbance maximum at 0.9 at 510 nm and

the spectrum was recorded on a 0.20 M solution in a 1.0 cm cell, what would be the value of 510 ?

Ans:- 510 = 4.5 cm-1M-1

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44.3) Transferrin is an iron-transport protein found in blood. Desferrioxamine B is an iron chelatorused to treat patients suffering from iron overload. Both are colourless in the absence of iron.The following data was obtained in aqueous solution using a 1.00 cm path length cell for bothspecies when fully complexed with iron:

----------------------------------------------------------------------------------

max / nm max / L mol–1 cm–1

iron-transferrin iron-desferrioxamine B-----------------------------------------------------------------------------------

428 3540 2730 470 4170 2290

-----------------------------------------------------------------------------------Calculate the absorbance at 470 nm of a solution that is 11.0 M in iron-transferrin and 69.5

M in iron-desferrioxamine B, assuming both are fully complexed and that the cell path lengthis 1.00 cm.

Ans:- M = 10-6 MFor iron transferrin, A = 4.6 x 10-2

For iron-desferrioxamine B, A = 1.6 x 10-1 = 16 x 10-2

Total = 4.6 x 10-2 + 16 x 10-2 = 2.06 x 10-1

44.4) The percentage transmittance of a transition metal complex at 360 nm and at 250C is 25%for a 6 x 10-4 mol L -1 solution in a 1 cm cell. The molar absorption coefficient in the unit of L mol-1 cm-

1 is(a) -31.0 x10 (b) 31.0 x10 (c) 32.0 x10 (d) 41.0 x10

Ans:- 10100A = log ( )%T = 2 - log10(%T)

= 2 - log25= 2 - 2log5= 2 - (2 x 0.6990) 0.6

3 -1 -1-4

A 0.6 = = = 1 x 10 L mol cm6 x 10 x 1cl

44.5) A solution of a compound shows a transmittance of 0.800 in a 1.0 cm cell at 525 nm. Thetransmittance in a 5.0 cm cell at the same wavelength would be

(a) 0.160 (b) 0.654 (c) 0.240 (d) 0.317

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Ans:- 10 11

2 10 2

1 1

2 2

10 11

2 10 2

10

10 2-1

2

2

log log

log = loglog 0.8001or =

5 log

-logT 5 x -log0.8 = 5 x -(3log2 + log10 ) =5 x(-0.9 +1) = 0.5

or T =antilog(- 0.5) = antil

TAA TA lButA l

Tll T

THence

og (1.5) 0.316

44.6) A solution containing 5 ppm of KMnO4 (F.W = 158) has a transmittance of 0.360 measuredin a 1cm cell at 500nm. The molar absorptivity of KMnO4 in L mol-1 cm-1 is

(a) 1.1 x 104 (b) 1.4 x 104 (c) 1.9 x 104 (d) 2.7 x 104

Ans:- 5 ppm = = = 5 grams of KMnO4 is present in 106 grams (=106 mL) of water.

-5 -16

w 1000 5 1000Hence the conc. = molarity(M) = x = x = 3.16 x 10 mol LFW V(in mL) 158 10

Now 4 -1 -1-5 -5

A -logT -log0.36 0.444 = = = = 1.4 x 10 L mol cmc 3.16 x 10 x 1 3.16 x 10 x 1cl l

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MILLER INDICES

What are Miller indices?Miller indices are a notation for describing directions and labelling planes in lattices and crystals.

Miller indices for directions in crystalsA direction is expressed in terms of its ratio of unit vectors in the form [uvw] where u, v and w are

integers. A family of crystallographically equivalent directions is expressed as <uvw>.

Miller indices for crystal planesA Miller indices for a plane are expressed as (hkl), where h, k and l are integers. A family of

crystallographically equivalent planes is expressed as {hkl}.The orientation of a surface or a crystal plane may be defined by considering how the plane intersects

the main crystallographic axes of the solid.

How to assign Miller indices?The following procedure is used to assign the Miller indices for crystal planes. To simplify the concept,

cubic crystal systems (with equal unit cell dimensions) are considered.

y

x

z

ab

c

Cubic systema=b=c

Step-1:Identify the intercepts of the plane along the crystallographic axes (x, y & z), in terms of unit cell

dimensions.Step-2:

Specify the intercepts in fractional coordinates by dividing by the respective cell-dimension i.e. a, b &c.Step-3:

Take the reciprocals of fractional coordinates.Step-4:

Clear fractions either by multiplying or dividing through by a common number in order to simplify them.Step-5:

Reduce to lowest terms wherever possible.

Finally, the numbers obtained are written in the brackets without using commas. The negative number isdenoted by a bar above it. The numbers are separated by gaps if they have two digits.

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ILLUSTRATIONSExample-1

For the plane shown below, the intercepts on x, y and z axes are a, & respectively..Remember that if any plane is parallel to an axis, the intercept along that axis is .

y

x

z

Step-1:Intercepts: a, ,

Step-2:Fractional coordinates: a/a, /a, /a = 1, , Note that for cubic systems a = b = c

Step-3:Reciprocals: 1/1, / , / = 1, 0, 0

Step-4 and 5 are not required here.Therefore, the Miller indices for this plane are: (100)

Note: The reason for why should we take reciprocals of intercepts is clear now. This operation is necessaryto eliminate values.

Example-2For the plane shown below, the intercepts on x, y and z axes are , a & a respectively..

y

x

z

Step-1:Intercepts: , a, a

Step-2:Fractional coordinates: /a, a/a, a/a = , 1, 1

Step-3:Reciprocals: / , 1/1, 1/1 = 0, 1, 1

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Therefore, the Miller indices for this plane are: (011)

Example-3For the plane shown below, the intercepts on x, y and z axes are a, a & a respectively.

y

x

z

Step-1:Intercepts: a, a, a

Step-2:Fractional coordinates: a/a, a/a, a/a =

Step-3:Reciprocals:


Example-4For the plane shown below, the intercepts on x, y and z axes are a/2, a/2 & a respectively.

y

x

z

a/2 a/2

a

Step-1:Intercepts: a/2, a/2, a

Step-2:Fractional coordinates: a/2a, a/2a, a/a =

Step-3:Reciprocals:


Example-5For the plane shown below, the intercepts on x, y and z axes are a, -a & respectively..

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x

z

a

-a

Step-1:Intercepts: a, -a,

Step-2:Fractional coordinates: a/a, -a/a, /a =

Step-3:Reciprocals:

Therefore, the Miller indices for this plane are: (1 10)

Example-6For the plane shown below, the intercepts on x, y and z axes are , 0 & respectively..

y

x

z

Step-1:Intercepts: , 0,

Step-2:Fractional coordinates: /a, 0/a, /a = , 0,

Step-3:Reciprocals: , 0

Therefore, the Miller indices for this plane would be: 0)

STOP! INFINITIES ARE NOT ALLOWED.However, infinities are not allowed in miller indices. This is the problem with planes along the axes. To

avoid this, we have to translate the plane by a unit distance along the axis or axes with non zero index/indices.

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x

z

The plane is translated by unit distance ‘a’ along the y-axis. Now the Miller index can be computed asfollows:

Step-1:Intercepts: , a,

Step-2:Fractional coordinates: /a, a/a, /a = , 1,

Step-3:Reciprocals: , 0

Therefore, the Miller indices for this plane are:

Some Generalisations* The plane is parallel to that axis for which the Miller index is zero.* The smaller a Miller index, the more nearly parallel the plane is to the axis.* The larger a Miller index, the more nearly perpendicular a plane is to that axis.* Multiplying or dividing Miller indices by a constant has no effect on the orientation of the plane* Miller indices are almost always small.

symmetry-equivalent planes A set of equivalent planes related by symmetry are indicated by using "curly" brackets. For example,

the planes: (100), (010) and (001) as shown below are symmetrically equivalent and can be representedas: {100}.

y

x

z

(100)

(001)

(010)

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COREY-CHAYKOVSKY REACTION

PRACTICE PROBLEMS (MCQ)

1) Choose the correct statement about Corey-Chaykovsky reaction.1) The sulfur ylides are stabilized when electron donating groups are attached to it.2) Dimethylsulfoxonium methylide is less stable than dimethylsulfonium methylide.3) With Michael acceptors, cyclopropanation is major route when unstable ylides are used.4) With Michael acceptors, epoxidation is major route when unstable ylides are used.

2) The major product in the following reaction is:

CHO KOH, DMSO

SPh2 BF4

1) 2)

3) 4)

CHO O

CHO O

3) The major product expected in the following reaction is:

Me3S+=O I- , NaH

DMSO

OO

1) 2)

3) 4)

OO

OO

OO

OO


OH

NPh Me3S

+=O I- , NaH

DMSO

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1) 2)

3) 4)

OH

NPh

OH

NPh

O

NHPh

O

NHPh

5) (IISc 2009)

a) b) c) d)CHO

HEtOOC

H

? major productCHO+SCOOEt

C6H6

CHO

HH

EtOOC O

HH

EtOOCO H

H

EtOOC

6) The expected product in the following reaction is:

O

SCH3

1) MeI

2) t-BuOK

1) 2) 3) 4)

O

S(CH3)2

O

SCH3

O

SCH3

O

SCH3I


Me3S+ I-

DMSO, NaH

OH

O

1) 2) 3) 4)OH

O

OH

O

OH

O

OH

O

8) What is the major product formed in the following reaction?

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O

O

O

O

BnO DMSO/THF

(H3C)2S CH2-

O

1) 2)

3) 4)

O

O

OO

BnO

OO

O

BnO

O

O

O

OBnO

O

O

O

OO

BnO

9) What is the major product in the following Corey-Chaykovskii reaction?

O

OMeMeO

(H3C)2S CH2-

O

THF, 0 oC

1) 2) 3) 4)

OMeMeO

OO

OMeMeO

O

OMeMeO

O

OMeMeO


MeO

MeOMeO

MeOOMe

O

OH Me3SO+ I-, NaH

DMSO, RT

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