C1-Allens Made Chemistry Theory {PART-1}

Compounds derived from hydrocarbons by replacement of one or more H-atoms by corresponding no. ofhalogen atoms are known as halogen derivatives.Classification :On the basis of nature of hydrocarbon from which they are obtained, hydrocarbon derivatives can be classifiedas :Halogen DerivativesAlkyl halidesAlkenyl halideAlkynyl halideAryl halideMono halidesDi-halidesTri-halidesTetra halidesPrimary halide Secondary halideTertiary halide(a) Alkyl halides : Halogen derivative of alkanes.(b) Alkenyl halides : Halogen derivative of alkenes.(c) Alkynyl halides : Halogen derivative of alkynes(d) Aryl halides : Halogen derivative of arenes (aromatic)Alkyl halides : They are further classified on the basis of halogen atoms introduced in the molecule. e.g.( i )Mono halides: These involves replacement of one H-atom by halogen atom.General formula C Hn2n+1XExample : CH Cl3Methyl chloride (Chloro methane)CH CH Br Ethyl bromide (Bromo ethane)32(ii)Dihalides : Replacement of two H-atom by halogen atoms.General formula C H Xn2n2Example : CH X22Methylene dihalideCHX 2CHX 2Ethylene dihalideVicinal dihalide orCH 3CHX 2Ethylidene dihalidegeminal dihalideor(iii)Trihalides : Replacement of three H-atoms by halogen atoms. General formula - C Hn2n-1 3X .Example : CHX Trihalo methane or haloform3(i v)Tetra halide and Perhalo compounds : Replacement of 4 H-atoms by halogen atoms(in CH 4 CCl ) 4 Tetrahalides.When all the H-atoms from an alkane are replaced by halogen atoms, then the compounds are called asperhalo compound - General formula C Hn2n–2 4X (tetra halide).CH4CX (Per halo methane)4 C H 26C X (Per halo ethane)26 ALKYL HALIDE AND ARYL HALIDE

MONO HALIDES : These are classified on the basis of nature of C-atom carrying the halogen atom .(A)Primary halide or 1 alkyl halides : Halogen atom attached with a primary or 1 C-atom.00Example : CH —X3Halo methane or methylhalideCH —CH —X32Halo ethane or ethyl halideCH —CH —CH —X3221-Halo propane or n-propyl halide(B)Secondary or 2 alkyl halides : Halogen atom linked with 2 C-atom.00Example :CH CH CH 33X2-halo propaneorIso propyl halideCH CH CH3CH 3X22-halo butaneorSec. butyl halide(C)Tertiary halide or 3 alkyl halide : halogen atom linked with 3 C-atom.00Example :RC XRR(tert.alkyl halide)Isomerism : Alkyl halides shows position and chain isomerism -Example : C H Cl37CHCHCHCl 322CHCHCH 33Cl Position isomersExample : C H Cl(a)49CH CH CH CH Cl32221 - chloro butane(b)CH 3CH CHCl2CH 31-chloro-2-methyl propane(c)CH CH CH32CH 3Cl2 - chloro butane(d)CH 3C ClCH 3CH 32-chloro-2-methyl propanea, b  Chain isomersa, c  Position isomersa, d  Chain and position isomersb,c  Chain and position iosmersb, d  Position isomersc, d  Chain isomersReactivity order : The order of reactivity of alkyl halides is -RI > RBr > RCl > RFBond energy values : C–I ( 57. 4), C–Br (65.9), C–Cl (78.5) and C–F (105.4) K.cal/moleC–I bond is most reactive because lower energy is required to break the bond. On the basis of nature of alkylgroup the reactivity order of alkyl halide is -tert > Sec. > PrimarySince alkyl groups, are electron repelling or electron releasing, larger no. of alkyl groups on C-atom of C–Xgreater is the electron density on C-atom hence ease in release of X atom as –X ion (+ I effect of alkyl group)CH 3CXCH 3CH 3CH 3C + XCH 3CH 3+–

Alkyl halides are generally more reactive than the corresponding alkanes due to the presence of polar covalent(X C) bond. So alkyl halides ( R–X) undergo nucleophilic substitution reaction.The centre for attacking Nu  is Catom +Genreal Method of Preparation of Monohalides :1 .By direct halogenation of alkanes :R—H + Cl 2U.V.light R—Cl + HCl(excess)2 .By the addition of H—X on alkenes :R—CH CHR + HXRCH —CHXR2CH2CH + HX2CH —CH X32CH –CH 3CH + HX2CH 3CH CH3XIsopropyl halide3 .By Alcohols :( a )By the action of hydrogen halides :Example : R—CH —OH 22H –XZnClRCH —X2Mechanism :R CH2OH ..H +(H—X)R CH2O H H–HO2R CH2 X R CH2X(unstable)(Product)In this reaction intermidiate carbocation is formed so rearrangement (–H shifting or 3–CHshifting) can takeeplace.ZnCl act as dehydrating agent and absorbs H O from the reaction so good yield of halide is obtained. Also it22generates H from HCl.+HCl + ZnCl23ZnCl + HReactivity order for alcohol :Reactivity stability of intermediate carbocation , so reactivity order : Tert. alc. > Sec. alc. > Pri. alc.Reactivity order of H—X is :HI > HBr > HClHI is maximum reactive so it reacts readily with 1°, 2° and 3° alcohols.R—OH + HIR—I + H O2HCl and also 1° alcohol are less reactive so ZnCl or some amount of H SO is needed to increase the224reactivity.Example : CH —CH —OH + HCl322 ZnClCH3 —CH —Cl2At normal condition :CH —CH —OH + HCl 32× (no reaction)Note : HCl + ZnCl is called as lucas reagent, alchol gives turbidity with lucas reagent.2Reactivity towards lucas reagent (difference in 1°, 2° and 3° alcohol).1° alcohol2° alcohol3° alcoholTime toin 30 min.in 5 min.in 1 min.give turbidity

( b )By the action of phosphorus halides (1N Smechanism) :R—OH + PCl5R—Cl + POCl + HCl33R—OH + PCl33RCl + H PO33PBr and PI are less stable, thus for bromides and Iodides, ( P + Br ) Or ( P + I ) mixture is used.3322( c )By reaction with thionyl chloride - (Darzen's procedure) (i2NN S and S mechanism) :R—OH + SOCl2Pyridine(1 mole)R—Cl + SO + HCl2One mole One moleBecause of less stability of SOBr and SOI , R—Br and RI does not obtained by this method.224 .Borodine – Hunsdicker's reaction : R—COOAg + X24 CClR—X + CO + AgX2Silver salt of(Cl or Br )2 2a fatty acid5 .By halide exchange :R–Cl or R—Br + KIAcetoneR–I + KCl or KBr (Conant finkelstein reaction)2CH Cl + HgF 322CH –F + HgCl (Swart reaction)32R– I and R—F can be prepared by this method only.6 .By reaction of alkanes with sulphuryl chloride (SO Cl ) :22R—H + SO Cl 2 2lightorganic peroxideR—Cl + HCl + SO2Physical Properties :(a)The lower members CH F, CH Cl, CH Br , C H Cl and C H F are gases at room temp.3332525CH I and members upto C are colourless sweet smelling liquids.318(b)Higher B.P. than parent alkanes.Decreasing order of B.P. is :R – I > R—Br > R—Cl > R—Famong isomeric R—X decreasing order of B.P. is : Primary > Secondary > tertiary(c)R—F and R—Cl lighter than waterR—Br and R—I heavier than waterDecreasing order of density is : R—I > R—Br > R—Cl > R—F(d)R—X are polar co-valent compounds but insoluble in water because they can not formH–bonds. They dissolve in organic solvents.(e)R—X burns with a green flame due to interaction of X with Cu wire.(Beilstein test)(f)The stability order is :R—F > R—Cl > R—Br > R—IR—I is least stable and darken in light due to photodecomposition.2R—I h R—R + I2Chemical Properties :A .Nucleophilic substitution reaction (S ) : Due to electronegativity difference the NC X bond is highlypolarised bond. X C

Thus the C-atom of the X C bond becomes centre to attack by a nucleophile (Nu) .X ion from R—X molecule is substituted by a Nu.  i.e.S reaction are the most common reactions in R—X.NR—X + Nu R—Nu + X These may be takes place by two ways -(a) 1N Smechanism(b) 2N S mechanismReactivity order is :R1C XR3R2>R1CH XR2>R CH12XS N 1S and SN N SN122(secondary)(secondary)(Tertiary)Mechenism of 12NNSand S:1N S Mechanism : SN stands for uni molecular nucleophilic substitution. The mechanism involves two steps.1Consider the hydrolysis of tert. butyl bromide with aqueous NaOH.Step 1: The alkyl halide ionises to give a planar corbonium ion. The corbonium ion is planar because thecentral positively charged carbon is sp hybridized.2Br CRRSlowC RRRR+ Brt–alkyl bromidePlanarStep-2 : The nucleophile can attack the planar carbonium ion from either side to give the product.C RRRFastOH–CRRROH + OHCRRRt–alkyl alcoholt–alkyl alcohol(i)Ionisation is the rate determining step because it is the slow step. In other words, the rate at whichalcohol is formed should depend upon the concentration of tertiary alkyl halide alone.Rate = K[R C—Br]3It is obvious that the reaction follows first order kinetics, therefore reaction is called 1N S.(ii)The reactivity order for 1N S reaction  stability of carbocations formed by halides.reactivity order of halides (1N S) varies as follows :Benzyl halide > Allylhalide > 3°halide > 2° halide > 1° halide > methyl halide.(iii)Remember that in case alkyl halide is optically active, SN reactions lead to racemisation.1

2N S mechanism : 2N S stands for bimolecular nucleophilic substitution. In this type of nucleophilic substitutionreaction, there occurs bond making and bond breaking simultaneously.Br CHHSlowCHHHTransition stateOHHHOBrHO C H + BrHWalden inversion –HH(i)Reactivities of alkyl halides in 2N S substitution is governed by steric factors. The bulkier the group, thatless reactive it will be.(ii)Reactivity order of alkyl halide varies as follows :CH X > 1°halide > 2°halide > 3° halide3(iii)The order of reactivity among 1° alkyl halides is :CH X > C H X > C H X etc.32537Remember that in case alkyl halide is optically active, SN reactions lead to Walden inversion.2(iv)Thus in short 3° alkyl halides react by 1N S, 1° by 2N S and 2° by either or both of them SN and SN12depend upon the nature of the alkyl halide and the reagent.(v)For a given alkyl group the order of reactivity is - (for 1N S and 2N Sboth) : RI > RBr > RCl > RF(vi)In addition to substitution reaction alkyl halide also undergo elimination reactions to form alkene withthe removal of a molecule of hydrogen halide (dehydrohalogenation). In dehydrohalogenation, hydrogenand halogen atoms are eliminated from two adjacent carbon atoms, the reaction also known aselimination may proceed by E & E mechanism (analogous to 121N S and 2N S mechanism).The order of elimination reaction is :3° halides > 2° halides > 1° halides(vii)In general 3° halides tend to react by elimination; 1° halides by substitution and 2° halides by either orboth of the reactions.B .Elimination Reactions (ER) : Alkyl halides also undergo ER in the presence of base as Nu (Loss of H—Xand formation of = bond)OR + CH2CH 2ClHROH+ CH —CH=CH + Cl23RCH 31 .Nucleophilic substitution reaction ( S ) :N HOH(Boil)X by — OHKOH aq. moist Ag O2KSH alc. NaSRR' COOAg X by — OHX by — OHX by — SHX by SRX by(R'COO)(a)(b)(c)(d)(e)(f)R—XR—OH (alcohol) + HXR—OH (alcohol) + KXR—OH (alcohol) + AgXR—SH + KXRSR' + NaXR' COOR + AgX(Alkyl alkanoate) esterAlkane thiol(Mercapto Gp.)(Mercaptane)Replacement ofProductThioether

(g)Reaction with KCN and AgCN :R—X + K CNIonic bondAlc.R—CN + KX + RNCcyanide (major) R — X + Ag CN ..covalent bondAlc.R — NC + Ag—X + RCNisocyanide (major)(h)Reaction with KNO and AgNO22 R — X + KNO2+Ionic bondAlc.R — O—NO + KX + RNO2(Alkyl nitrites) major R — X + AgNO2covalent bondAlc.R — N+ Ag—X + RONONitroalkane (major)OO(i)Reaction with NaOR' (Sodium alkoxide) : (williamson synthesis reaction)R—X + NaOR' R—OR' + NaXExample : (i)CH —CH —Cl + NaOCH 323CH —CH —O—CH (major)323 (ii)CH CH CH + NaOCH333ClCH CH CH + CH—CH=CH33O CH3 (major)32(minor)(iii) CH C Cl + NaOCH33CH 3CH C CH + NaCl + CHOH323CH 3 (Alkene) (elimination is more)CH 3(more reactive)(major)(j)Reaction with NH :3 Example : (i)32R — XNHR — NHH — X(ii)32R — XNHR — NHH — X(excess)(iii)R — X32R — XR— XNHR — NHR — NH— RR N RRRN R XRR  R X(iv)CH —CH —Cl+ NH323CH —CH —NH + HCl322(v)CH CH CH + NH 33 3ClCH CH CH + HCl33NH (or elimination product)2 (vi)CH C Cl + NH33CH 3CH 3(more reactive)CH C + NHCl34CH 3 (Elimination is more)CH 2

(k)Reaction with CHC Na :R—X + CHC Na R—CCH + NaXIFCH C X + CH3CH 3CH C3 (Elimination is more)CH 3CNa –+CH + NaX + CH2CHCH 32 .Elimination Reaction : (Dehydrohalogenation) Alkyl halides undergo - elimination on treatment withKOH (alc.)22R — C H — CH — XKOH (alc.)R—CHCH + HX2CH —CH —CH —CH —Br + KOH(alc.) 3222CH —CH —CH32CH + HBr2CH CH CH CH3H3BrAlc.KOH CH —CH3CH—CH + CH —CH —CH332CH + HBr22 - butene (80%)1 - butene (20%)Competition between substitution and elemination reactions :E - Reaction : 1° < 2° < 3°1E - Reaction : 1° < 2° < 3°2S - Reaction : 1° < 2° < 3°NS - Reaction : 1° > 2° > 3°NReactivity order of alkyl halides :12Alkyl halides22N S/E12N S/E1° Alkyl halidesMainly give substitution unlessCan not undergo SN /E11sterically hindered alkyl halideor sterically hindered base inwhich case elimination is favoured2° Alkyl halidesBoth substitution and elemination,Both substitution and eleminationstronger base/ bulkier base or highhigh temp. greater of percentagetemp. greater percentage of elimination. of elimination3° Alkyl halidesMainly eliminationBoth substitution and elimination, hightemp greater percentage of elimination.(i)22N S/ E is favoured by high conc. of good neucleophile or strong base. (CH O , HO )3Rate of Reaction (Substrate) (Reagent)(ii)11N S/ E is favoured by low conc. of poor neucleophile or weak base (CH OH, H O)32(iii)If an alkyl halide, undergoes 21NNS/ S – Reaction then 2N S will be favoured by high conc. of gooddneucleophile (negetively charged) in presence of polor aprotic solvent . where as 1N S– reaction isfavoured by low conc. of poor neucleophile (neutral) is presence of polar protic solvent.

Polar protic solvent : H O, CH OH, HCOOH.23Polar aprotic solvent : DMSO, CH CN, C H –O–C H32525Example : CH CH CH —Br + CH O32233CH OH CH CH CH —OCH + CH CH32233CH2majorminor2N S E2Example :CHOH3CH CH + CH CH 32OCH3CH CH CH 32CH 3Br + CHO3CH 3C CH 32CH 3minorSN 2majorE2Example :CHOH3C CH CH 32CH 3Br + CHO3C CH CH CH 3CH 3majorE1CH 33Example : CH CH CH CH Br + 3222C ONa CH 3CH 3CH 3CH—CHCH=CH322(E ) mech.2CHCHO32CHCHOH32CHCOO2CHCOOH3CH CH CH 33BrCH 3CH CH2(major)CH CH CH 33OO CCH 3(major)Example :CH C OCH 3 3CH 3 CH 3 CH C 3 CH 2 CH 3 (major) CH C OH 3 CH 3 CH 3 CH C Br 3 CH 3CH 3CNNH 2CH Cconc.CH O3conc.HOCH OH3H O2Via S mech.N 1Via E mech.2Example :Saytzeff rule : According to this rule \"The dehydrohalogenation of an alkyl halide results preferentially in theproduction of more alkylated alkene i.e. more stable alkene.

3 .Wurtz Reaction : When a mixture of different alkyl halides, (R - X) and (R - X) is used a mixture of alkane12is formed -R —X + 2Na + X—R 12EtherR —R + R —R + R —R + NaX121122If 'Zn' is used in place of 'Na' the reaction is called Frankland's Reaction.4 .Halogenation of alkyl halides :CH —Cl + Cl 32light (–HCl)or.U.V.CH Cl 222–HClClCHCl3 2–HClClCCl4SimilarlyC H Cl 252 ClC H Cl 2422 ClC H Cl 2332 Cl(excess)C Cl265 .Reduction : R - X are reduced to corresponding alkane.R—X + 2HRe ducing agentR—H + HX(i)CH CH Br + H322Ni CH CH + HBr33 (ii)CH CH CH Br + 2H 322Zn / HClCH - CH - CH + HBr323(iii)C H I + HI25Re d.PC H + I2626 .Formation of Organometalic compounds :(i)R—X + Mg dry ether RMgX (Grignard reagent)(ii)2C H Br + 2Zn 25dry ether(C H ) Zn + ZnBr25 22(iii)2C 2H Br + Hg(Na) 5(C H ) Hg25  2NaBrSodium Amalgam(iv)4C H Cl + 4Na / Pb 25(C H ) Pb + 4NaCl + 3Pb25 4Sodium lead Alloy Tetra ethyl lead (used as antiknocking agent)7 .Friedel - Crafts reaction :+ CHCl 3AlCl (anhyd)3+ HCl BenzeneTolueneCH 38 .Action of heat :(i)CH —CH —CH —Cl 322At or below 300 C(Re arrangement)CH CH CH33 Cl(ii)CH CH CH32 lHAbove300° CCHCH CH + HI (Elimination)32PropyleneUses :(a) As alkylation agent (Wurtz reaction)(b) As synthetic reagent(c) Lower members used as anaesthetic agent, refrigerent or solvent.

DIHALIDESGeneral formula C H X .Two H - atom of alkanes, replaced by two halogen atoms to form dihalides.n2n 2 Dihalides are classified as :(a) Gem dihalide : The term Gem is derived from geminal means - same position.Two similar halogen atoms are attached to same C - atomExample : CH CHX32ethylidene dihalide(1, 1 - dihalo ethane)CH 3CH 3CHX 2Isopropylidene dihalide(2, 2 - dihalo propane)( b )Vic dihalides : Vic term from - Vicinal means adjacent C - atomsTwo halogen atoms are attached on adjacent carbon atom.Example :CH CH22 X XH H CCH 2 X X CH HVic and Gem dihalides are position isomers.ethylene dihalidePropylene dihalide(1,2-dihaloethane)(1,2-dihalopropane)( c ) dihalides : Halogen atoms are attached with terminal C - atom. They are separated by 3 or moreC - atom . They are also known as polymethylene halides.Ex.CH CH22 XXCH CH22 Tetramethylene dichloride(1,4-dichloro butane)General Method of Preparation :(a )Gem dihalides :(i)By the reaction of PCl on carbonyl compound.5CH C H + PCl35OCH C H + POCl3ClCl3AcetaldehydeEthylidene chlorideCH C CH + PBr335OCH C CH + POBr3BrBr33Acetone2, 2 - dibromo propane(ii)By addition of halogen acids on alkynes :CHCH + HBr CH2CHBr HBrCH CHBr32 Vinyl bromide 1, 1 - dibromo ethaneCH C CH + HBr3CH C CH 3BrBr3CH C CH 3Br2HBrPropyne2, 2 - dibromo propane

( b )Vic-dihalides :(i)By the addition of halogens to alkenes :CH 2+ Br 2CH 2CHBr 2CHBr21, 2 - dibromo ethaneCH CH CH + Br322CH CH CH3 Br2 Br1, 2 - dibromo propane(ii)By the action of PCl on glycols :5CHOH2CHOH2 + 2PCl5CHCl2CHCl2+ 2POCl + 2HCl3( c )  dihalides :(i)CH2CH—CH Br + HBr 2Br—CH —CH —CH Br2221, 3 - dibromo propane(ii) + Cl 2h CH CH22ClCH 2Cl(ii) + Cl 2h CH CH22ClCH 2ClCH 2(iii) + Cl 2h Cl+ HClNote : Herehas less tendency to open the ring.Physical Properties :(i)Lower members are colourless, oily liquids with sweet smell. Higher members are solid.(ii)The reactivity of gemdihalides is lesser than vicinal or mono halides. [Reason - in presence of onehalogen atom (Strong attracting –I effect) the other halogen atom can not be so easily replaced.(iii)These are heavier than water.Chemical Properties :( i )Action of KOH(alc.) (Dehydrohalogenation)CHX2CHX2CHX2CH3 or(i) alc. KOH(ii) NaNH2CHCH(ii)Action of KOH(aq.) (Hydrolysis) It is a distinction test for gem and vic dihalides.

(a)CH 2+ 2KOH(aq.)ClCH 2ClCH 2+ 2KCl OHCH 2OHGlycolVic - dihalide.1, 2 - ethane diol.(b)CHCHXCHY32Vic-dihalide.KOH(aq.)CHCH 2OH OH CH 3propane - 1, 2 - diolGem. CH 3CHX2KOH(aq.)CH 3CHOCHCXCH323KOH(aq.)CHCOCH 33(Ketone)and(iii)Reaction with KCN : Gem and Vic dihalide gives different productsVic. :CHCl2CHCl2H O/H2++ 2KCN— 2KClCH 2CNCH 2CNCH 2COOHCH 2COOH—HO2CH 2COCH 2COOSuccinic acidSuccinic anhydrideGem. :CH 3CH— 2KClClCl2KCNCH 3CHCNCNH O/H2+CH 3CHCOOHCOOH —CO2CH 3CH COOH2Propionic acid(i)– CN group on acid hydrolysis gives - COOH(ii)Two – COOH group on one C – atom on heating always loose CO to form monocarboxylic acid.2(iii)Two – COOH group on vic. C – atom on heating loose H O to form anhydride.2(i v)Dehalogenation :Vic.CHBr 2CHBr2Same Carbon Product+ ZnCHOH3HeatCH 2+ ZnBr 2CH 2Ge m.CH —CH CH Br + Zn3223CH OH CH —CH —CH3 2CH—CH —CH + ZnBr232 (Twice Carbon Product)CH 2CHBr2CHBr2CH OH3HC2CH 2CH 2+ Zn+ ZnBr2- dihalide. cyclopropaneGRIGNARD REAGENTGeneral Method of Preparation :Grignard reagents are prepared in the laboratory by the action of alkyl halides on magnesium metal in thepresence of dry ether.R – X + Mg dryetherRMgX(Grignard reagent)(Alkyl magensium halide)

The ease of formation of Grignard reagent is in the order RI > RBr > RClEther is used to dissolved the Grignard reagent by coordination.Chemical Reactions : (i) O (ii) H O23+R—OH H CO/H O23RCH—OH2CH —CH /H O223R—CH—CH—OH 1° Alcohol22ORCHO/H O3R CHOH2HCOOEt/H O3R CHOH2RCOR/H O2R C—OH3RCOOEt/H O2R C—OH3 HCOOEtRCHO RCOR  RCN/H O2RCOR RCOOEtRCOOH CO + H O22R—H HOH or ROH orNH or Ph — OH3or R—NH or RNH — R2or CHCH or Ph — NH2R—R R—XNo reactionR N3R—CH—CH CH Alkene2ClCH —CH CH222R—NH 2Cl—NH2R—CN Cl—CNR—X X 2RCOOEtEsterClCOOEtRMgX++++ 1° Alcohol 1° Alcohol 1° Alcohol 2° Alcohol 3° Alcohol 3° AlcoholaldehydeKetoneKetoneAcidsAlkaneAlkane1° AmineCyanidesAlkyl halideARYL HALIDEIf halogen atom is directly attached to the benzene ring, then compound is called as Haloarene.Example :ClClCH 3ClCH 3Cl(Chlorobenzene) (2–Chlorotoluene) (2,4–Dichlorotoluene)General Methods of Preparation :Halogenation of Benzene :Cl+ Cl2AlCl3+ HCl

From Phenol :Cl+ PCl5+ POCl + HCl3OH(minor)3C H OH + POCl 653 (C H ) PO + 3HCl65 34(major)Sandmeyer's reaction :NCl2Cl+ CuClHCl+ N2NCl2Br+ CuBrHBr+ N2Reaction with KI :NCl2I+ KI+ N + KCl2Balz-schiemann's reaction :NCl2NBF24+ HBF4F+ BF + N32 Hunsdicker reaction :COOAgCl+ Cl2+ CO + AgCl2Raschig Process :Cl+ 2HCl + O2+ 2HO22Cu Cl222Chemical Properties :(i)Chlorobenzene is essentially inert to aqueus sodium hydroxide at room temperature.(ii)Aryl halide are very less reactive than alkyl halides in nucleophilic substitution reactions.(iii)The carbon-halogen bonds of aryl halides are too strong & aryl cations are too high in energy to permitaryl halides to ionize readily 1N S-type process.(iv)The optimal transition state geometry required for 2N S process cannot be achieved.(v)Nucleophilic attack from the side opposite the carbon-halogen bond is blocked by the aromatic ring.

The Elimination-Addition Mechanism of Nucleophilic Aromatic Substitution (Benzyne) :(i) Very stong base such as sodium or potassium amide react with aryl halide, even those without electronwithdrawing substituents to give products corresponding to nucleophilic substitution of halide by the base.ClNH 2NH 2KNH, NH23–33°CChlorobenzyne-1- C14Aniline - 1- C14Aniline-2- C14+MechanismStep-1 : Elimination stage ; Amide ion is a very strong base and brings about the dehydrohalogenation ofchlorobenzene by abstracting a proton from the carbon adjacent to the one that bears the leaving group. Theproduct of this step is an unstable intermediate called benzyne.HHHClH NH2H..........ChlorobenzeneHHHHBenzyne+ :NH + :Cl:3....–Step-2 : Beginning of addition phase ; Amide ion acts as a nucleophile and adds to one of the carbons of thetriple bond. The product of this step is a carbanion.HHHHHHHHBenzyneAryl anion:NH2NH 2......Step-3 : Completion of addition phase ; The aryl anion abstracts a proton from the ammonia used as thesolvent in the reaction.HHHHHHHHAryl anionAnilineNH 2NH 2....H–NH2 ..H+ NH2–

NOTE :The sp orbital in the plane of the ring in benzyne are not properly aligned for good overlap thus 2bonding is weaker than alkyne.(i)ClCH 3KNH2NH 3O-chloro toluene 3-MethylbenzyneCH 3NH 2NH 2KNH2NH 3o-Methylanilinem-MethylanilineCH 3CH 3+(ii)ClNH 2HN2+KNH2KNH2NH 3NH 3p-Chlorotoluenep-Methylaniline4-methylbenzynem-MethylanilineCH 3CH 3CH 3CH 3(iii)NH 2NH 3++KNH2KNH2NH 3NH 33-Methylbenzyne4-Methylbenzynep-Methylanilineo-Methylanilineo-Methylanilinep-MethylanilineCH 3CH 3CH 3CH 3CH 3CH 3NH 2NH 2ClCH 3KNH2NH 3Nucleophilic Aromatic Substitution by the addition Elimination Mechanism :(i) The generally accepted mechanism for nucleophilic aromatic substitution in nitro-substituted aryl halides.NO 2NO 2++NaOCH3NaClp-Chloronitrobenzenep-NitroanisoleClOCH3CHOH385°C

(ii)An ortho - nitro group exert a comparable rate-enhancing effect, m-chloronitrobenzyne while muchmore reactive than chlorobenzyne itself, is thousand of times less reactive than either o-orp-chloronitrobenzene.(iii)The effect of o- & p-nitro substituents is cummulative, as the rate data for substitution with methoxideion in a series of nitro-substituted chlorobenzene derivative demostrate increasing rate of reaction as :Chlorobenzene1.01-Chloro-4-nitrobenzene7 × 1010ClClNO 2Relative rate :1-Chloro-2, 4-dinitrobenzene2.4 × 10152-Chloro-1, 3, 5-trinitrobenzene(too fast to measure)ClClNO 2NO 2NO 2NO 2ON2(iv)In contrast to nucleophilic substitution in alkyl halides, where alkyl fluorides are exceedingly unreactive,aryl fluorides undergo nucleophilic substitution readily when the ring bears an o-or a p-nitro group.p-FluoronitrobenzenePotassiummethoxidePotassiumfluoridep-Nitroanisole(93%)FOCH3NO 2NO 2++KOCH3KFCHOH385°C(v)Indeed, the order of leaving group reactivity in nucleophilic aromatic substitution is the opposite of thatseen in alphatic substitution.(vi)Fluoride is the best reactive leaving group in nucleophilic aromatic substitution, iodide the least reactive.NO 2XRelative reactivity towards sodium methoxide in methanol (50°C)X =F312Cl1.0Br0.8I0.4(vii)Kinetic studies of many of the reactions described in teh section have demostrated that they follow asecond-order rate law.Rate = k[aryl halide] [nucleophile](viii)Second order kinetics is usually interpreated in terms of a bimolecular rate determining step.p-FluoronitrobenzeneSodium methoxideSodium Fluoridep-NitroanisoleFOCH3NO 2NO 2++NaOCH3NaF

Mechanism :Step -1 - Addition stage. The nucleophile, in this case methoxide ion, adds to the carbon atom that bears theleaving group to give a cyclohexadienyl anion intermediate.p-FluoronitrobenzeneCyclohexadienyl anion intermediateMethoxide ion:F::F: :OCHNO 2NO 2+:OCH3HHHHHHHH........slow3....Step -2 - Elimination stage. Loss of halide form the cyclohexadienyl intermediate restores the aromaticity ofthe ring and gives the product of nucleophilic aromatic substitution.p-NitroanisoleCyclohexadienyl anion intermediate:OCH3:F: :OCHNO 2NO 2HHHHHHHH.... ..Fast3..+F –......Fluoride ion..– ..( a )Dow process :OH+ Aq.NaOH+ NaClCl300°CHigh pressure(via benzyne mechanism)Presence of deactivating group in ortho and para position makes the nucleophilic substitution easier.Reactivity Order : (Towards nucleophilic substatitution)NO 2ClNO 2>NO 2ClClClNO 2NO 2>>NO 2( b )NH 2+ KNHStrong base2+ KBrBrliq. NH3low temperature( c )CN+ CuCN+ CuClClPyridine, 300°Chigh pressure

Fitting reaction :Cl + 2Na + Cl+ 2NaCldry ether(Diphenyl)Wurtz fitting reaction :Cl + 2Na + ClCHCHCH223dry etherCHCHCH + 2NaCl223Formation of aryl megnisium halide :Cl + MgMgClTHFReduction :XNi—Al/NaOHWith Chloral :CCl CHO +3Cl C—CH3HSO, 24HClHCl–HO2ClCl p,p–Dichloro diphenyl trichloroethaneDDT (insecticide)Electrophilic Substitution Reaction :( i )ClFeCl3ClCl+ Cl2Cl+ClCl is o– and p– directing group.(ii)ClHSO24NO 2Cl+ HNO3Cl+NO 2(iii)ClSOH3Cl+ HSO24Cl+SOH3(i v)Friedal Craft Reaction :ClCH 3Cl+ CHCl3Cl+CH 3AlCl3

SOLVED EXAMPLESEx .1Which of the following undergoes Hydrolysis most easily :(A) Cl(B) NO 2Cl(C) NO 2ClNO 2(D) NO 2ClNO 2NO 2Ans. (D)Sol.If there is more m-directing group then there will be more nuclephilic substitution reaction.Ex .2 The product in the following reaction is :Ph – Cl + Fe / Br 2Product(A) o– bromo-chloro benzene(B) p– bromo-chloro benzene(C) (A) and (B) both(D) 2, 4, 6-tribromo chloro benzeneAns. (C)Sol.Since – Cl group is deactivating and o/p directing group so only o– and p– products are formed.Ex .3The most reactive towards SN is :1(A) PhCH Cl2(B) Ph–Cl(C) CH CHCl(CH )33(D) p–NO —Ph—CH —Cl22Ans. (A)Sol.1N S the intermediate carbocation is formed. C H —CH Cl 652652 C H CH is maximum stable due to resonance.Ex .4Which of the following is used as insecticide :(1) D.D.T.(2) Chloritone(3) Chloropicrin(4) (A) and (C) both Ans. (D)



CATHODE RAYS (Discovery of e )-In 1859, Julius plucker started the study of conduction of electricity through gases at low pressure in adischarge tube. When a high voltage of the order 10, 000 volts or more was impressed across the electrodes,some sort of invisible rays moved from the –ve electrode to the +ve electrode. Since the –ve electrode isreferred to as cathode, these rays were called cathode rays.Properties of Cathode rays(1)They travel in straight lines away from cathode with very high velocity ranging from 10 to 10 m/sec.79 (2)A shadow of metallic object placed in the path is cast on the wall opposite to the cathode.(3)They produce a green glow when strick the glass wall matter. Light is emitted when they strike thezinc-sulphide screen.(4)When a small pin wheel 10.0is placed in their path, the blades of the wheel are set in motion. Thus thecathode rays consist of material particles which have mass and velocity.(5)They are deflected by the electric and magnetic fields. When the rays are passed between two electricallycharged plates, these are deflected towards the positively charged plate. It shows that cathode rayscarry -ve charge. These particles carrying negative charge were called negatrons by Thomson.ATOMIC STRUCTUREAnodeCathode+-vaccum pump(Invisible rays)(Cathode rays)Gas at low pressure (10 atm)- 4V > 10,000 voltsProduction of a shadow of the solid object by cathode raysRotation of light paddle wheel by cathode rays

The name negatron was changed to 'electron' by Stoney(6)They produce heat energy when they collide with the matter. It shows that cathode rays posses Kineticenergy which is converted into heat energy when stopped by matter.(7)These rays affect the photographic plate.(8)Cathode rays can penetrate the thin foil of solid materials.(9)Cathode rays can ionize the gases through which they pass.(10)The nature of cathode rays is independent of(a) The nature of cathode and(b) The gas in discharge tube.MEASUREMENT OF e/m FOR ELECTRON :In 1897, J.J. Thomson determined the e/m value (charge/mass) of the electron by studying the deflectionof cathode rays in electric & magnetic fields.The value of e/m has been found to be –1.7588 10 coulomb/g.8 By performing a series of experiments, Thomson proved that whatever gas be taken in the dischargetube and whatever be the material of the electrodes the value of e/m is always the same.Electrons are thus common universal constituents of all atoms.DETERMINATION OF THE CHARGE ON AN ELECTRON :The absolute value of the charge on an e was- measured by R.A. Milikan in 1909 by the Milikan'soil drop experiment.The apparatus used by him is shown in fig.An oil droplet falls through a hole in the upperplate. The air between the plates is thenexposed to X-rays which eject electrons fromair molecules. Some of these e are captured-by the oil droplet and it acquires a negativecharge.The metal plates were given an electric charge, and as the electric field between the plates wasincreased, it was possible to make some of the drops travel upwards at the same speed as they werepreviously falling.Deflection of cathode rays towards positive plate of the electric field+ + ++ + +- - - - --Oil dropstelescope toobserve speedof oil dropsX-raysChargableplatesSpray gun(atomizer)

By measuring the speed, and knowing things like the strength of the field and the density of the oil, radius of oildrops, Milikan was able to calculate the magnitude of the charge on the oil drops. He found that the smallestcharge to be found on them was approximately 1.59 10–19 C. This was recognised as the charge on an e .-The modern value is 1.602 10–19 C.MASS OF THE ELECTRON :Mass of the e can be calculate from the value of e/m and the value of e-m = ee /m. 1602 10. 17588 10198 = 9.1096 10–28 g or = 9.1096 10–31 kgThis is termed as the rest mass of the electron i.e. mass of the electron when moving with low speed. The massof a moving e may be calculate by applying the following formula.-Mass of moving e =-rest mass of ev /c12b gWhere v is the velocity of the e and c is the velocity of light.-Whenv = c  mass of e = -v > c  mass of e = imaginary-POSITIVE RAYS - (DISCOVERY OF PROTON) :The first experiment that lead to the discovery of the +ve particle was conducted by 'Goldstein'.He used a perforated cathode in the modified cathode ray tube.It was observed that when a high potential difference was applied b/w the electrodes, not only cathoderays were produced but also a new type of rays were produced simultaneously from anode movingtowards cathode and passed through the holes or canals of the cathode. These rays were termed canalrays since these passed through the canals of the cathode.These were also named anode rays as theseoriginated from anode.When the properties of these rays were studied by Thomson, he observed that these rays consisted ofpositively charged particles and named them as positive rays.The following characteristics of the positive rays we recognised :(i)The rays travel in straight lines and cast a shadow of the object placed in their path.(ii)Like cathode rays, these rays also rotate the wheel placed in their path and also have heatingeffect. Thus, the rays passess K.E. i.e. mass particles are present.Production of Anode rays or Positive rays

(iii) The rays are deflected by electric and magnetic fields towards the negatively charged plate showingthereby that these rays carry +ve charge.(iv) The rays produce flashes of light on ZnS screen(v)These rays can pass through thin metal foil.(vi) These rays can produce ionisation in gases.(vii) Positive particles in these rays have e/m value much smaller than that of e . For a small value of-e/m, it is definite that positive particles possess high mass.(viii) e/m value is dependent on the nature of the gas taken in the discharge tube, i.e. +ve particles aredifferent in different gases.Accurate measurements of the charge and the mass of the particles in the discharge tube containinghydrogen, the lightest of all gases, were made by J.J. Thomson in 1906. These particles were foundto have the e/m value as +9.579 10 coulomb/g. This was the maximum value of e/m4observed for any +ve particle.It was thus assumed that the positive particle given by the hydrogen represents a fundamental particleof +ve charge. This particle was named proton by Rutherford in 1911. Its charge was found to beequal in magnitude but opposite in sign to that of electron.Thus charge on proton = + 1.602 10 columb i.e. one unit +ve charge-19The mass of the proton, thus can be calculated.Mass of the proton = ee /m. 1602 10. 9 579 10194 = 1.672 10–24 g = 1.672  10–27 kgMass of proton in amu = 1672 10166 102424.. = 1.00757 amu.NEUTRONIn 1920, Rutherford suggested that in an atom, there must be present at least a third type of fundamentalparticles which should be electrically neutral and posses mass nearly equal to that of proton. He proposed thename for such fundamental particles as neutron.In 1932, chadwick bombarded beryllium with a stream of -particles. He observed that penetrating radiations wereproduced which were not affected by electric & magnetic fields. These radiations consisted of neutral particles, whichwere called neutrons. The nuclear reaction can be shown as+-particleBe-atomCarbon atomNeutronCharge = +2Atomic No. = 4Atomic No. = 6Charge = 0Mass = 4 amuMass = 9 amuMass = 12 amuMass = 1 amu[ He +4294Be 126C +10n]Thus a neutron is a sub atomic particle which has a mass 1.675 10–24 g approximately 1amu, or nearly equalto the mass of proton or hydrogen atom and carrying no electrical charge.The e/m value of a neutron is zero.

ATOMIC STRUCTURE :Atom is actually made of 3 fundamental particles :1 .Electron2 .Proton3 .NeutronFundamentalDiscovered By Charge Masschargem ass particle (Specific Charge)ElectronJ.J.Thomson–1.6× 10–19coloumb9.1× 10–31kg1.76× 10 C/g8 (e or )– –4.8× 10–10 esu9.1× 10–28g–1 Unit0.000548 amuProtonGoldstein+1.6× 10–19coloumb1.672× 10–27kg(P)+4.8× 10–10esu1.672× 10–24g(Ionized H1.00757 amu9.58× 10 C/g4 atom, H )++1 UnitNeutronJames Chadwick1.675× 10–27kg( n )01 01.675× 10–24g 01.00893 amuIMPORTANT POINT :1 .esu = electrostatic unit (1 cb = 3 × 10 esu)9 amu = atomic masss unit1 amu = 1.6 × 10–24 g = 1.6 × 10–27 kg2 .Order of Mass–e m< m < mpnOrder of Specific Charge–pene< e/m< e/mm mass of protonmass of electron –pem= 1837mINTRODUCTION :AtomATom (Greek word)NotDivisibleNot divisible (According to Dalton)Atom is a Greek wordand its meaning Indivisible i.e. an ultimate particles which cannot be further subdivided.John Dalton (1803 - 1808) considered that \" all matter was composed of small particle called atom.

ACCORDING TO DALTON'S THEORY :(1)Atom is the smallest indivisible part of matter which takes part in chemical reaction.(2)Atom is neither created nor destroyed.(3)Representation of atom : X .ZAWhere : A  Mass number, Z Atomic number, X Symbol of atom.Mass Number :It is represented by capital A. The sum of number of neutrons and protons is called the mass number.of the element. It is also known as number of nucleons because neutron & proton are present innucleus.A = number of protons + number of neutronsNote :It is always a whole number.Atomic Number :It is represented by Z. The number of protons present in the Nucleus is called atomic number of anelement. It is also known as nuclear charge.For neutral atom : Number of proton = Number of electronFor charged atom : Number of e = Z – (charge on atom)–Z= number of protons onlyFor Eg:17Cl35n = 18p = 17e = 17Two different elements can not have the same Atomic NumberNumber of Neutrons=Mass number – Atomic number=A – Z=(p + n) – p=nMethod for Analysis of atomic weighteg.6C 12P +  6Weight of Proton = 6 × 1.00750n 0 6Weight of Neutron = 6 × 1.00850e – 6Weight of Electron = 6 × 0.000549Weight of C atom = 12.011 a.m.u.Mass no. of C atom = 12 [P and n]Note :Mass no. of atom is always a whole no. but atomic weight may be in decimal.Q .If no. of protons in X is 16. then no. of e in X will be––2–+2(1) 14(2) 16(3) 18(4) NoneSol. No. of proton in X is = 16–2 No. of electron in X+2 is = 14

Q .In C atom if mass of e is doubled and mass of proton is halved, then calculate the percentage change in mass12–no. of C .12Sol. 6C 12P + 3e – 12e –P +n°666A  121236A  9% change = 312× 100= 25%Q .Assuming that atomic weight of C is 150 unit from atomic table, then according to this assumption, the12weight of O will be :-16Sol.12 amu = 1501 amu = 1501216 amu = 15012 × 16 = 200 UnitAtomic Weight : The atomic weight of an element is the average of weights of all the isotopes of thatelement.An element have three isotopes y , y and y and their isotopic weights are w , w , w and their123123percentage/possibility/probability/ratio of occurance in nature are x , x , x respectively then the123average atomic weight of element is –ave. wt =112233123w xw xw xxxxEx.Cl35Cl37Probability ratio7 5 %2 5 %3 :1353 37 13 1  = 1424 = 35.5Q .35Br79 : 35Br81 1 : 179 1 81 11 1  = 1602 = 80Q .An element have three isotopes and their isotopic weight are 11, 12 , 13 unit and their percentage of occurancein nature is 85, 10, 5 respectively then calculate the average atomic weight of element.Sol. Average Atomic weight=11 85 12 10 13 585 10 5 = 935 12065100  Average wt. = 1120100 = 11.2

Q .Average atomic weight of an element M is 51.7. If two isotopes of M, M , M are present then calculate the5052percentage of occurance of M in nature.50Sol.M 50M 52x +1x = 100%2x = (100 – x )21wt = w xw xxx1 12 21251.7 =5052 1212 xxxx51.7 = 50x52 1001001111x )xx ) ((5170 = 50 x + 5200 – 52x115170 = – 2x + 520012x = 301x = 151M = 15%50M = 85%52Q .Calculate the precentage of Deuterium in heavy water.Sol.D O2( H ) O122164 + 16 = 20 (Moleculer weight)420 × 100Ans= 20%Isotopes : Given by SoddyThey are the atoms of a given element which have the same atomic number (Z) but different mass number (A)i.e. They have same Nuclear charge (Z) but different number of Neutrons (A–Z).For Eg.117Cl3517Cl37n = 18n = 20e = 17e = 17p = 17p = 17Isotopes have same chemical property but different physical property.Isotopes do not have the same value of emNumber of electronmass because mass varies.(No. of electron are same but mass varies).For Eg.1(ProteiumDeuteriumTritium)1H 11H 21H 3e = 1e = 1e = 1p = 1p = 1p = 1n = 0n = 1n = 2e/m1/11/21/31H is the only normal hydrogen which have n = 0 i.e. no nuetrons1Deuterium is also called as heavy hydrogen. It represent by DEg. 26C 126C 136C 14e = 6e = 6e = 6p = 6p = 6p = 6n = 6n = 7n = 8

Isobars : Given by AstonThey are the atoms of different element which have the same mass number (A) but different Atomic number (Z)i.e They have different number of Electron, Protons & Neutrons But sum of number of neutrons & Protons i.e.number of nucleons remains same.For Eg.11H 32He3p = 1p = 2e = 1 e = 2 n = 2n = 1p + n = 3p + n = 3Isobars do not have the same chemical & physical propertyThey do not have the same value of e/mFor Eg.219 K4020 Ca40p = 19p = 20n = 21 n +p = 40n = 20 n +p = 40e = 19e = 2019 + 21 = 4020 + 20 = 40 n + p = 40Number of Nucleons sameIsodiaphers :They are the atoms of different element which have the same difference of the number of Neutrons & protons.For Eg1.5B116 C 13p = 5p = 6n = 6 n – p =1n = 7 n – p =1e = 5e = 6For Eg 2.7 N 159 F 19p = 7p = 9n = 8 n – p =1n = 10 n – p =1e = 7e = 9Isotones/ Isoneutronic species / Isotonic :They are the atoms of different element which have the same number of neutrons.For Eg. 1.1H 32He4p = 1p = 2n = 2n = 2e = 1e = 2For Eg. 2.19 K 3920 Ca40e = 19e = 20p = 19p = 20n = 20n = 20Isosters :They are the molecules which have the same number of atoms & electrons.For Eg. 1CO2N O2Atoms= 1 + 2Atoms= 2 + 1= 3= 3Electrons= 6 + 8 2×Electrons = 7 2 + 8×= 22 e–= 22e–

For Eg. 2CaOKFAtoms22Electrons20 + 819 + 928 e–28 e–For Eg. 3OF2HClOAtoms= 33Electrons= 8 + 181 + 17 + 8= 26 e–17 + 926 e–Isoelectronic Species :They are the atoms, molecules or ions which have the same number of electrons.For Eg. 1Cl–ArElectron18 e–18 e–For Eg. 2H O2NH3e = 2 + 8e = 7 + 310 e–10 e–For Eg. 3BF3SO2e = 5 +9 3×16 + 8 2×5 + 2716 + 1632 e–32 e–Nuclear Isomer :Nuclear isomers (isomeric nuclei) are the atoms with the same atomic number and same mass number butwith different radioactive properties.Example of nuclear isomers isUranium–X (half–life 1.4 min) andUranium–Z (half–life 6.7 hours)The reason for nuclear isomerism is the different energy states of the two isomeric nuclei.Other examples are3069Zn3069Zn(T1/2 = 13.8 hr)(T1/2 = 57 min)3580Br3580Br(T1/2 = 4.4 hour)(T1/2 = 18 min)QUESTIONS BASED ON NUCLEAR STRUCTUREEx.If the mass of neutrons is doubled & mass of electron is halved then find out the atomic mass of6C12 and the percent by which it is increased.Sol.Step-16C 12e = 6p = 6 = 6 amu  n = 6 = 6 amu= 12 amuIf the mass of neutrons is doubled and mass of e is halved then.–n = 12 amu p = 6 amu =18 amuImp. Note : mass of e is negligible, so it is not considered in calculation of atomic mass.–Step-2% Increment =Finalmass - Initial mass× 100Initial mass=18 121001250%

Ex.If mass of neutron is doubled, mass of proton is halved and mass of electron is doubled then find out thechange in At. wt of C6121.Remain same2.Increased by 25%3.Increased by 37.5%4.None of themSol.Step-16C 12e = 6p = 6 = 12 amun = 6If mass of neutron is doubled, mass of proton is halved and mass of electron is doubled,then new atomicmass will be :n = 12 amu p = 3 amu= 15amuStep-2% Increment =Final massInitial mass100Initial mass =15 1210012 25%THOMSON'S MODEL OF ATOM [1904]Thomson was the first to propose a detailed model of theatom.Thomson proposed that an atom consists of a uniform sphereof positive charge in which the electrons are present at someplaces.This model of atom is known as 'Plum-Pudding model'.DRAWBACKS :An important drawback of this model is that the mass of the atoms is considered to be evenly spreadover that atom.It is a static model. It does not reflect the movement of electron.RUTHERFORD's - SCATTERING EXPERIMENT-scattering experimentRuther ford observed that :(i)Most of the -particles (nearly 99.9%) went straight without suffering any deflection.(ii)A few of them got deflected through small angles.-----e -sphere of+ve charge----- - ---ZnS screenCircularfluorescentscreenSlit system[lead plate]Thin gold foil (.00004 cm)Most of– particlesstrike here- RaySource [Ra] of-rays= [ He ]24 +2[doubly ionised He particle]

(iii)A very few -particles (about onein 20,000) did not pass through thefoil at all but suffered largedeflections (more than 90°) or evencome back in the direction fromwhich they have come i.e. adeflection of 180°.Following conclusions were drawn from the above observations :(1)Since most of the -particle went straight through the metal foil undeflected, it means that there mustbe very large empty space within the atom.(2)Since few of the -particles were deflected from their original path through moderate angles; it wasconcluded that whole of the +ve charge is concentrated and the space occupied by this positive chargeis very small in the atom.Whenever -particles come closer to this point, they suffer a force of repulsion and deviate from theirpaths.The positively charged heavy mass which occupies only a small volume in an atom is called nucleus.It is supposed to be present at the centre of the atom.(3)A very few of the -particles suffered strong deflections on even returned on their path indicating that thenucleus is rigid and -particles recoil due to direct collision with the heavy positively charged mass.(4)The relation between number of deflected particles and deflection angle isµ =41sin2  [ increases µ decreases]where µ = deflected particles = deflection angleAs atomic number increases, the number of protons increases which increases the repulstion and sodeflection angle increases.APPLICATIONS OF RUTHERFORD MODELOn the basis of scattering experiments, Rutherford proposed the model of an atom, which is known as nuclearatomic model. According to this model -(i)An atom consists of a heavy positively charged nucleus where all the protons are present.(ii)The volume of the nucleus is very small and is only a minute fraction of the total volume of the atom.Nucleus has a radius of the order of 10 cm and the atom has a radius of the order of 10 cm-13-8rrradius of theatomradius of thenucleusAN 1010813 = 10 , r = 10 r5A5NThus radius (size) of the atom is 10 times the radius of the nucleus.5The radius of a nucleus is proportional to the cube root of the no. of nucleons within it.R A1/3 R = R A01/3 cmWhereR = 1.33 010 (a constant) and, A = mass number (p + n)-13R = radius of the nucleus.R = 1.33 × 10–13 A1/3 cm

(iii)There is an empty space around the nucleus called extra nuclear part. In this part electrons arepresent. The no. of electrons in an atom is always equal to no. of protons present in the nucleus. Asthe nuclear part of atom is responsible for the mass of the atom, the extra nuclear part is responsiblefor its volume. The volume of the atom is about 10 times the volume of the nucleus.15vol of theatomvol of thenucleusrrAN..FH GIK e JFH GIKJ e 4343101010338313315jj(iv)Electrons revolve round the nucleus in closed orbits with high speeds.This model was similar to the solar system, the nucleus representing the sun and revolving electrons asplanets.Drawbacks of Rutherford model :(1)This theory could not explain the stability of an atom. According toMaxwell electron loses it's energy continuously in the form ofelectromagnetic radiations. As a result of this, the e should loss-energy at every turn and move closer and closer to the nucleusfollowing a spiral path. The ultimate result will be that it will fall intothe nucleus, thereby making the atom unstable.(2)If the electrons loss energy continuously, the observed spectrum should be continuous but the actualobserved spectrum consists of well defined lines of definite frequencies (discontinuous). Hence, the lossof energy by electron is not continuous in an atom.Electromagnetic waves (EM waves) or Radiant Energy/Electromagnetic radiation :It is the energy transmitted from one body to another in the form of waves and these waves travel inthe space with the same speed as light ( 3 10 m/s) and these waves are known as Electromagnetic×8 waves or radiant energy.The radiant Energy do not need any medium for propogation.Ex : Radio waves, micro waves, Infra red rays, visible rays, ultraviolet rays, x–rays, gama rays and cosmic rays.+Nucleuse -Cosmicray's–raysx–raysUltravioletVisibleInfra redMicrowaveRadiowavesVIBGYORVioletIndigoBlueGreenYellowOrangeRed38004300450049005500590065007600(In A°)10–40.010.1150380076006 × 1063 × 1093 × 1014In Å

A wave is characterized by following six characterstics.The upper most point of the wave is called crest and thelower most point is called trough.Some of the terms employed in dealing with the waves aredescribed below.1 .Wavelength (Lambda) :It is defined as the distance between two nearest crest or nearest trough.It is measured in terms of a A (Angstrom), pm (Picometre), nm (nanometer), cm(centimetre),°m (metre)1Å = 10–10 m,1 Pm = 10–12 m,1nm = 10 m,–91cm = 10 m–22 .Frequency ( ) (nu)Frequency of a wave is defined as the number of waves which pass through a point in 1 sec.It is measured in terms of Hertz (Hz ), sec–1 , or cycle per second (cps)1 Hertz = 1 sec–1 = 1 cps.3 .Time period (T) : Time taken by a wave to pass through one point.1 T  sec.4 .Velocity  (c)Velocity of a wave is defined as distance covered by a wave in 1 sec.C = = C/Since C is constants 1/i.e. frequency is inversely proportional to 5 .Wave number   ( ) ( nu bar)  It is the reciprocal of the wave length that is number of waves presentin 1cm1 m = 100 cm1 1100cmm (1cm–1 = 100 m )-1It is measured in terms of cm , m–1–1 etc,6 .Amplitude  (a)The amplitude of a wave is defined as the height of crust or depth of trough.CC  1  QUESTIONS BASED ON EM WAVESEx.The vividh Bharti station of All India Radio broadcast on a frequency of 1368 Kilo Hertz. Calculate the wavelength of the Electromagnetic waves emited by the transmitter.Sol.As we know velocity of light (C)C = 3 × 10 m/sec.8Given (frequency)= 1368 kHzCrestCrestTroughTroughDirection of propogationaa

= 1368 × 10 Hz3= 1368 × 10 sec3–1  C81313 10 m sec1368 10 sec  = 219.3 mEx.Calculate  in cm and of yellow radiations have wavelength of 5800 Å–1Sol.As we known 1 15800 Å  815800 10 cm { 1Å = 10 cm}–8=8105800 cm–1 = 17241.4 cm–1=c =3 10×10 cm sec–1× 1.7 10 cm×4 –1=3 1.7 10××14=5.1 10×14 sec-1Ex.A particular radiostation broadcast at a frequency of 1120 Kilo Hertz another radio station broadcast at afrequency of 98.7 mega Hertz. What are the wave length of radiations from each station.Sol.Station Ist8131C3 10 m sec1120 10 sec  = 267.86 mStation IInd8161C3 10 m sec98.7 10 sec = 3.0395 mEx.How long would it take a radio wave of frequency 6 10 sec to travel from mars to the earth, a distance×3–1of 8 × 10 km ?7 Sol.Distance to be travelled from mars to earth= 8 × 10 km7= 8 × 10 m10Velocity of EM waves= 3 × 10 m/sec8Time= Dis tan ceVelocity10818 10 m3 10 m / sec = 2.66 × 10 sec.2Ex.What will be the frequency of photon of wavelength 2225 Å traveling in vacuum ?Sol.Velocity of light in vacuum = 3 × 10 m sec8–1Wavelength = 2225 × 10–10 meterFrequency = VelocityWavelength = 8103 10 meter / sec2225 10meter  = 51300010 sec2225 = 1.349 × 10 sec15–1

PLANCK'S QUANTUM THEORYAccording to planck's quantum theory :1.The radiant energy emitted or absorbed by a body not continuously but discontinuously in the formof small discrete packets of energy and these packets are called quantum.2.In case of light, the smallest packet of energy is called as 'photon' but in general case the smallestpacket of energy called as quantum.3.The energy of each quantum is directly proportional to frequency of the radiation i.e.E E = horE = hcc    Proportionality constant or Plank's constant (h)h = 6.626 10×–37 kJ sec.or6.626 10×–34 J sec(1erg = 10 J)–7or6.626 10×–27 erg sec.4.Total amount of energy transmitted from one body to another will be some integral multiple ofenergy of a quantum.E = nhWhere n is an integer andn = number of quantumhcEhhc  Ex.Calculate the energy of a photon of sodium light of wave length 5.862 10 ×–16 m in Joules.Sol.5.886 10×–16 mc = 3 10 m sec×8–1E = nhornhc{n = 1}E = hcE =341161 6.6 10Jules 3 10 m sec5.862 10m  =106.6 3105.862Joules = 3.38 10×–10 Joules.Ex.Calculate the frequency & energy of a photon of wave length 4000 ÅSol.(a)Calculation of frequency := 4000 Å= 4000 × 10–10 mC 873 10 m / sec4 10 m  =0.75 × 10 sec15–1=7.5 × 10 sec14–1(b)Calculation of energy :E=h =6.626 × 10–34 Joule × 7.5 × 10 sec14–1=4.96 × 10–19 Joule

Ex.Calculate the and frequency of a photon having an energy of 2 electron voltSol.1ev = 1.602 × 10–19 J2ev = 3.204 × 10–19 J = E(a)Calculation of wavelength ( ) :hcEor hcE  = 3481196.626 10Js 3 10 m sec3.204 10J = 6.204 × 10 m–7(b)Calculation of frequency ( ) :c = 8173 10 m sec6.204 10 m= 0.48 × 10 sec15–1= 4.8 × 10 sec14–1Ex.Which has a higher energy ?(a)A photon of violet light with wave length 4000 Åo r(b)A photon of red light with wave length 7000 ÅSol.(a)Violet light :Eviolet = hc= 3481106.626 10J sec 3 10 m sec4000 10m = 4.97 × 10–19 Joule(b)Red light :Ered = hc= 3481106.626 10J sec 3 10 m sec7000 10m = 2.8 × 10–19 JouleSo,Eviolet > EredEx.How many photons of lights having a wave length of 5000 Å are necessary to provide 1 Joule of energy.Sol.E =nhcE nhc =1034811Joule 5000 10m6.626 10Joule sec 3 10 m sec =2.5 × 10 photons18

Ex.Calculate the energy associated with the photon passing through vacuum with wavelength 9900 Å.Sol.For vacuum, velocity of photon = 3 × 10 m/sec8h =6.6 × 10–34 Joule sec = 9900 × 10–10 meterE = h = h  c = 348106.6 10J.sec 3 10 m sec6600 10m = 1619.8 109900 = 2 × 10–19 JouleBOHR'S ATOMIC MODELSome Important formulae :Coulombic force =12 2kq qrCentrifugal force = 2mvrAngular momentum = mvrIt is a quantum mechanical model. This model was based on quantum theory of radiation and Classical lawof physics.The important postulates on which Bohr's Model is based are the following :1 Postulate :stAtoms has a nucleus where all protons and neutrons are present.The size of nucleus is very small and it is present at the centre of the atom.2nd Postulate :Negatively charged electron are revolving around the nucleus in the same way as the planets are revolvingaround the sun.The path of electron is circular.The attraction force (Coulombic or electrostatic force) between nucleus and electron is equal to the centrifugalforce on electron.i.e.Attraction force towards nucleus = centrifugal force away from nucleus.3 Postulate :rdElectrons can revolve only in those orbits whose angular momentum (mvr) is integral multiple of h2  .i.e.mvr = nh2 n = Whole numberWhereh = Plank's constant, = ConstantAngular momentum can have values such as h2 , 2h2 , 3h2 , 4h2 , 5h2  ........but can not have frac-tional values such as 1.5h2 , 1.2h2 , .5h2 .......

4 Postulate :thThe orbits in which electron can revolve are known as stationary Orbits because in these orbits energy ofelectron is always constant.5 Postulate :thEach stationary orbit is associated with definite amount of energy therefore these orbits are also called asenergy levels and are numbered as 1, 2, 3, 4, 5, .... or K, L, M, N, O, ..... from the nucleus outwards.6 PostulatethThe emission or absorbtion of energy in the form of photon can only occur when electron jumps from onestationary state to another & it is E = Efinal state – Einitial stateEnergy is absorbed when electron jumps from inner to outer orbit and is emitted when electron moves fromouter to inner orbit. Radii of various orbits of hydrogen atom :Consider, an electron of mass 'm' and charge 'e' revolving around a nucleus of charge Ze (where, Z = atomicnumber and e is the charge of the proton) with a tangential velocity v. r is the radius of the orbit in whichelectron is revolving.By Coulomb's law, the electrostatic force of attraction between the moving electron and nucleus is Coulombicforce = 22KZerK =  014 (where  is permittivity of free space)vm radiusK = 9 × 10 Nm C92–2In C.G.S. units, value of K = 1 dyne cm (esu)2–2The centrifugal force acting on the electron is 2mvrSince the electrostatic force balance the centrifugal force, for the stable electron orbit. 222mvKZerr.......(1)(or)v = 22KZemr.......(2)According to Bohr's postulate of angular momentum quantization, we havemvr =  nh2v = nh2 mrShell 5Shell 4Shell 3Shell 2Shell 1Shell KShell LShell MShell NShell ONucleus

v = 22222 2 n h4m r.......(3)Equating (2) and (3)22222 2KZen hmr4m rSolving for r we get r = 2222n h4mKZewhere n = 1, 2, 3, ......, Hence, only certain orbits whose radii are given by the above equation are available for the electron. Thegreater the value of n, i.e., farther the energy level from the nucleus the greater is the radius.The radius of the smallest orbit (n = 1) for hydrogen atom (Z = 1) is r .0r =02222n h4me K =  2234223119916.626 1043.149 101.6 109 10= 5.29 × 10–11 m = 0.529 ÅRadius of n orbit for an atom with atomic number Z is simply written asthr = 0.529 × n2nÅZCALCULATION OF ENERGY OF AN ELECTRON :The total energy (E) of the electron is the sum of kinetic energy and potential energy.Kinetic energy of the electron = ½ mv2Potential energy = columbic force.dr = 22KZe.drr = 2KZerTotal energy = 1/2 mv – 22KZer......(4)From equation (1) we know that 222mvKZerr½ mv = 222KZerSubstituting this in equation (4)Total energy (E) =  22KZeKZe2rr = 2KZe2rSubstituting for r, gives usE =  2242222mZ e Kn hwhere n = 1, 2, 3, .....This expression shows that only certain energies are allowed to the electron. Since this energy expressionconsist of so many fundamental constant, we are giving you the following simplified expressions.E = –21.8 × 10–12 × 22Zn erg per atom = –21.8 × 10–19 × 22Zn J per atom = –13.6 × 22Zn eV per atom(1 eV = 3.83 × 10–23 Kcal)1 eV = 1.602 × 10–12 erg(1 eV = 1.602 × 10–19 J)[E = –313.6 × 22Zn Kcal/mole (1 cal = 4.18 J)]

The energies are negative since the energy of the electron in the atom is less than the energy of a freeelectron, i.e. the electron is at infinite distance from the nucleus which is taken as zero. The lowest energylevel of the atom corresponds to n = 1, and as the quantum number increases, E becomes less negative.When n = , E = 0, which corresponds to an ionized atom, i.e. the electron and nucleus are infinitelyseparated.H H + e (ionization)+– Calculation of velocity :We know thatmvr =  nh; v = 2hn2 mrBy substituting for r we are gettingv = 22 KZenhwhere excepting n and Z all are constantsv = 2.18 × 10 8Zncm/sec.QUESTIONS BASED ON BOHR'S MODELEx.Calculate the radius of 1 ,2 ,3 ,4stndrdth Bohr's Orbit of hydrogen.Sol.Radius of Bohr's orbitr = 0.529 × 2nZ(a)Radius of 1 orbit :str = 0.529 × 211 = 0.529 Å(b)Radius of 2 orbit :ndr = 0.529 × 221 = 0.529 × 4= 2.116 Å(c)Radius of 3 orbit :rdr = 0.529 × 231 = 0.529 × 9= 4.761 Å(d)Radius of 4 orbit :thr = 0.529 × 241= 0.529 × 16= 8.464 ÅEx.Calculate the radius ratio of 3 & 5 orbit of Herdth +r = 0.529 ×2nZÅAt. Number of He = 2Sol.r3 = 0.529 × 232= 0.529 ×92r = 0.529 5×2(5)2= 0.529 ×252

Therefore   352230.529r2r50.529235r9r25 r : r = 9 : 253 5 Ex.Calculate the radius ratio of 2nd orbit of hydrogen and 3 orbit of Lird +2Sol. Atomic number of H = 1Atomic number of Li = 32 orbit radius of hydrogennd(r ) = 0.529 2 H× 2213 orbit radius of Lird+2(r )Li = 0.529 3+2× 233  222 H23 Li20.529r1r30.5293 = 43 2 H r : r 23 Li = 4 : 3Ex.The ratio of the radius of two Bohr's orbit of Li+2 is 1:9. what Would be their nomenclature.1 .K & L2. L & M3. K & M4. K & NSol.2xx2yyn 0.529r13r9n0.52932x2ynn19xyn1n3  = K ShellM ShellEx.Calculate the radius of 2nd excited state of Li .+2Sol. 2 excited state, means e is present in 3 shell so,nd–rdr3= 0.529 ×3 33 = 0.529 3 Å×= 1.587 ÅEx.Calculate the radius ratio of 2nd excited state of H & 1 excited state of Li .st +2Sol. 2 excited state, means e is present in 3 shell of hydrogennd–rdr = 0.529 3× 231 = 0.529 9×1 excited state, means e exist in 2 shell of List–nd+2r = 0.529 2 × 223 = 0.529 ×43  23 H2 Li90.529r14r0.5293 = ndradius of 2 excited state of hydrogenst+ 2radius of 1 excited state of Li  23 H2 Lir27r4

Ex.Calculate the energy of Li+2 atom for 2nd excited state.Sol.E = –13.6 ×22ZnZ = 3 and e exist in 2 excited state, means e present in 3 shell i.e. n = 3–nd–rdE = –13.6 ×  2233 = – 13.6 eV/atomEx.Calculate the ratio of energies of He for 1 & 2+ Stnd excited state .Sol. (He ) 1 Excited state+ st:(He ) 2+2nd Excited statei.e. (He )2 shell+nd:(He )3 shell+rd–13.6 ×  2222:– 13.6 ×  222344:4914:199: 4Ex.If the P.E. of an electron is –6.8 eV in hydrogen atom then find out K.E., E of orbit where electron exist &radius of orbit.Sol. 1.P.E. = –2K.E.–6.8 = –2K.E.6.82 = K.E.K.E. = 3.4 eV2.E. = – K E. = – 3.4 eV3.Orbit = 2ndE = – 13.6 ×22Zn3.4 = – 13.6 ×221nn = 213.63.4  = 4 i.e. n = 24.r = 0.529 ×2nZÅr = 0.529 × 221Å= 0.529 4Å×= 2.16 ÅEx.The ionization energy for the hydrogen atom is 13.6 ev then calculate the required energy in ev to excite itfrom the ground state to 1 excited state.st Sol. Ionization energy= 13.6 eV

i.e.1 energy statest= – 13.6 eVEnergy of 1 excited statesti.e.2 orbitnd= –3.4 eVso,E – E21= – 3.4 + 13.6 = 10.2 eVEx.If the total energy of an electron is –1.51 ev in hydrogen atom then find out K.E, P.E, orbit radius and velocityof the electron in that orbit.Sol. GivenE = –1.5 eV(i)E = – KEK.E = – E{ Z = 1} = 1.51 eV(ii)PE = – 2 1.51× = – 3.02 eV(iii)Orbit = 3rdE = – 13.6 ×22Zn ev  – 1.51 = – 13.6 ×221n213.6n9 1.51 n = 3(iv)r = 0.529 ×3 31  = 0.529 9 = 4.761 Å×v = 2.188 10×8 × 13cm/sec = 82.188 103 = 0.729 10 cm / sec×8Ex.Calculate the velocity of an electron placed in the 3 orbit of the Li ion. Also calculate the number of revolutionsrd2+per second that it makes around the nucleus.Sol. Radius of 2 orbit = r x nd12(n)Z= 0.529 × 10 × –8 233 = 1.587 × 10 cm–8Velocity of electron in 2 orbit,ndv = 2.18 × 10 8Zn cm/sec = 82.18 1033  = 2.18 × 10 cm/sec8No. of revolutions/sec = 12 r / v= v2 r=882.18 10 cm / sec2 3.14 1.587 10 cm = 0.2187 × 1016 = 2.187 × 10 rev/sec15SPECTRUMElectromagnetic spectrum or EM spectrum :The arrangement obtained by arranging various types of EM waves in ordes of their increasing frequency ordecreasing wave length is called as EM SPECTRUM

RWlow high MWIRVisible RaysU.VX-rayCosmiclow Ehigh Elonger Shorter Spectrum :When a radiation is passed through a spectroscope (Prism) for the dispersion of the radiation, the pattern(photograph) obtained on the screen (photographic plate) is called as spectrum of the given radiationClassification of Spectrum(1) Emission(2) Absorption(a) Continuous(b) line(c) band(a) line(b) band( 1 )Emissions spectrum :When the radiation emitted from incandescence source (eg. from the candle, sun, tubelight, burner, bulb, or bypassing electric discharge through a gas at low pressure, by heating some substance at high temp) is passeddirectly through the prism and then received on the screen then the obtained spectrum is called as emissionspectrum.( a )Emission continuous spectrum or continuous spectrum :When a narrow beam of white light is passed through a prism, it is dispersed into 7 colours from violetto Red.( b )Emission line spectrum :When an atomic gas is raised to incandescence source or subjected to electrical excitation, it firstabsorbs energy & then gives it out as radiation. On examining these radiation through a spectroscopea spectrum is obtained which have well defined lines,each corresponding to a definite wave length &these lines are separated from each other by dark space. This type of Emission spectrum is called asEmission line spectrum.Screen U V regionGB IVY regionORVisibleNarrow beam of white lightInfrared regionScreenLine SpectrumAtomic gas}Slit systemIncandescencesourceStopped aftera shortperiod

Special Note :1.No two elements will have identical line spectrum since no two elements have identical energy leveltherefore the line spectrum of the elements are described as finger prints differing from each other likethe finger prints of the human beings.2.Since line spectrum is obtained by the emission of energy through the atoms of the element thereforeline spectrum is also called as atomic spectrum.( c )Emission band spectrum :If molecular form of the gas is used, it first absorbs energy for not only electron transition but forrotational, vibrational and translational then emits radiations.On examining these radiations through a spectroscope a spectrum is obtained on the screen, whichare group of closely packed lines called Bands, therefore this type of Emission spectrum is called asemission band spectrum. Bands are separated from each other by dark space.Note :Since band spectrum are caused by molecules therefore band spectrum are also called as molecularspectrum.( 2 )Absorption spectrum When white light is first passed through a solution or vapours of chemical substance or gas and then analyzedby spectroscope, it is observed that some dark lines are obtained in otherwise continuous spectrum.This type of spectrum is called as Absorption spectrum.If white light is passed through atomic gas then the obtained spectrum is called as Absorption linespectrum.If white light is passed through molecular gas then the obtained spectrum is called as Absorption bandspectrum.Hydrogen line spectrum or Hydrogen spectrum :When an electric excitation is applied on atomic hydrogen gas at Low pressure,a bluish light is emitted. whena ray of this light is passed through a prism, a spectrum of several isolated sharp line is obtained.The wave-length of various lines show that spectrum lines lie in Visible, Ultraviolet and Infra red region. These lines aregrouped into different series.Mg}Band SpectrumScreenColouredColouredColouredColouredIncandescencesourceStopped aftera shortperiodScreenGasColouredColouredColouredColouredIncandescencesource

SeriesDiscovered by regions n2n1Number of lineslymanlymanU.V. regionn = 2,3,4 ... / n =12 1n – 12 BalmerBalmerVisible regionn = 3,4,5 ... / n =22 1n – 22 PaschenPaschenInfra red (I.R.)n = 4,5,6 ... / n =32 1n – 32 BracketBracketI.R. regionn = 5,6,7 ... / n =42 1n – 42 PfundPfundI.R. regionn = 6,7,8 ... / n =52 1n – 52 HumpheryHumpheryfar I.R. regionn = 7,8,9 ... / n =62 1n – 62 QUESTIONS BASED ON SPECTRUMEx.In a hydrogen spectrum if electron moves from 7 to 1 orbit by transition in multi steps then find out the totalnumber of lines in the spectrum.Sol. Lyman=(n – 1) = 7 – 1 = 61Balmer=(n – 2) = 7 – 2 = 52Paschen=(n – 3) = 7 – 3 = 42Bracket=(n – 4) = 7 – 4 = 32Pfund=(n – 5) = 7 – 5 = 22Humphrey=(n – 6) = 7 – 6 = 12Total=21Total number of lines can be calculated as follows :Total number of lines =  2121n - nn - n+ 17 - 1 6 + 142=== 21222Ex.In a hydrogen spectrum if electron moves from 6 to 2th nd by transition in multi steps then find out the numberof lines in spectrumSol. Total number of line= 4 + 3 + 2 + 1 + 0= 10Total number of lines=  2121nnnn12  = 6 2 4 124 5102 213456 7QMLKI.R. regionBracketInfra Red regionorPaschen seriesVisible regionorBalmer seriesUltra violet region or Lyman seriesI. R. regionPfundPONFar I.R. regionHumphery

Ex.In a hydrogen spectrum if electron moves from 6 to 3 orbit by transition in multi steps then find out theth rd following steps :(a)Total number of lines in spectrum(b)Total number of lines in U.V. region(c)Total number of lines in visible region(d)Total number of lines in IR regionSol. (a)Calculation of total number of lines :=  2121nnnn12  =  636312 = 3 42 = 6(b)Calculation of number lines present in U.V. region.e moves from 6 to 3 orbit in multisteps. –thrdFor U.V. region, e should be comes into 1 shell. So the number of lines in U.V. region zero.–st(c)Calculation of total number of lines in visible region.For visible region, e should be comes into 2 shell, so the number of lines in visible region zero.–nd(d)Calculation of total number of lines in I.R. region.In I.R. region, Paschen, Bracket and Pfund series are present.Number of lines in Paschen series=n2 – 3=6 – 3=3Number of lines in Bracket series=n – 42=6 – 4=2Number of lines in Pfund series=n – 52=6 – 5=1So total number of lines=3 + 2 + 1 = 6Ex.In Balmer series of H atom/spectrum, which electronic transitions represents 3 line ?rd Sol. In Balmer series3 to 2rd nd1line4th to 2nd2 line5 to 2th nd3 lineInfinite to 2ndLast line or limiting lineSo, Ans is5 to 2th nd line3 linerdEx.In H atom if e moves, from n orbit to 1 orbit by transition in multi steps, if there are total number of lines– th st in spectrum are 10 then find out the value of n.Sol. Total number of lines =  2121nnnn12 So,10 =  212nnn2 n – n – 20 = 0 222n – 5n + 4n – 20 = 0 222 2 n (n –5) + 4 (n – 5) = 02 2 2 (n + 4) (n – 5) = 02 2 n = 52 Ex.Calculate the wavelength of 1st line of Balmer series in Hydrogen spectrum.Sol. For first line of Balmer seriesn = 2, n = 31 2