EG_Chemistry-12_(22-07-2022) a

31. (i) No geometrical isomer is possible (ii) NH3 + Cl + NH3 (ii) Two isomers (facial and meridional) are there: en NH3 Co NH3 NH3 Co en Cl NH3 Cl NH3 Cl NH3 Pt Cr Cl Cl cis trans Cl NH3 Cl Cl (cis-form also shows optical isomerism) Cl Facial NH3 (iii) Cl + Cl + Meridional NH3 32. There are three possible geometrical isomers: NH3 Br NH3 py NH3 Cl en Co en Co en Pt Pt Pt en NH3 cis trans py Cl Cl Br Br py (cis-form also shows optical isomer- All these isomers will show optical isomerism. ism) 37. (i) Diamminechloridonitrito-N-platinum (II). 33. (i) Sodium hexafluoridoaluminate(III) (ii) Potassium trioxalatochromate(III). (ii) Tetracarbonylnickel (0) (iii) Dichloridobis (ethane-1, 2-diamine) cobalt 34. (i) Na2[Cr(en)(CH3COO)4] (ii) [CoCl(en)2NO2]+ (III) chloride. (iii) [v(acac)2O] (iv) Pentaamminecarbonatocobalt(III) chloride. 35. Unidentate ligand: A molecule or an ion having (v) Mercury tetrathiocyanato-S-cobaltate (III) only one donor atom to form a coordinate bond 38. (i) [Cr(CO)5P(C6H5)3] (ii) Fe4[Fe(CN)6]3 with the central metal atom or ion is called a (iii) [Pt(NH3)4(Cl2)][PtCl4] (iv) [Co(Cl)(en)2(NO2)]+ unidentate ligand, e.g. H2O, NH3. (v) [Ni(CO)4] Bidentate ligand: A molecule or an ion having 39. (i) [Zn(OH)4]2– (ii) K2[PdCl4] two donor atoms that can form two coordinate (iii) [Pt(NH3)2Cl2] (iv) K2[Ni(CN)4] (v) [Co(NH3)5(ONO)]2+ bonds with the central metal atom is called a (vi) [Co(NH3)6]2(SO4)3 (vii) K3[Cr(C2O4)3] (viii) [Pt(NH3)6]4+ CH2NH2 (ix) [Cu(Br)4]2– (x) [Co(NH3)5(NO2)]2+ didentate or bidentate ligand, e.g. and 40. (i) Hexaamminecobalt(III) chloride COO− CH2NH2 COO− (ii) Tetraamminechloridonitrito-N-cobalt (III) Ambidentate ligand: A molecule or an ion chloride having two donor atoms but can only form one (iii) Hexaamminenickel(II) chloride coordinate bond at a particular time with the (iv) D i a m m i n e c h l o r i d o ( m e t h a n a m i n e ) central metal atom or ion is called ambidentate platinum(II) chloride ligand, e.g. CN– or NC– and NO2− or O—N—O–. (v) Hexaaquamanganese(II) ion (vi) Tris(ethane-1,2-diamine)cobalt(III) ion 36. (i) Cl + Cl + (vii) Hexaaquatitanium(III) ion Cl (viii) Tetrachloridonickelate(II) en Co en Co en (ix) Tetracarbonylnickel (0) en Cl 41. (i) x + (–1) = +2 or x = +3 cis trans (ii) x + 4 (–1) = –2 or x = +2 (cis-form also shows optical (iii) x + 3 (–1) = 0 or x = +3 isomerism). (iv) x + 2 (–1) = +1 or x = +3 (v) x + 6(–1) = –3 or x = +3 150 Chemistry-12

Topic 2. Bonding in Coordination Compounds Werner’s Theory empty orbitals, i.e. the number of empty orbitals is equal to the coordination number. Metal exhibits two types of valencies in the formation of complexes. • These empty orbitals hybridised before These are primary valencies and secondary valencies. participation in bonding and the nature of hybridisation depends upon the nature of metal • Primary valencies corresponds to the oxidation and on the nature of approaching ligand. number (ON) of the metal and are satisfied by the anions. These are ionisable and non-directional. Inner orbital complexes or outer orbital complexes • Secondary valencies corresponds to the When outer d-orbitals i.e. of nth shell are used in coordination number (CN) of the metal atom bonding, the complexes are called outer orbital and are satisfied by the ligands. These are non- complexes. They are formed due to weak field ligands ionisable and directional. Hence, geometry is or high spin ligands and hybridisation is sp3d2. They decided by these valencies. have octahedral shape. Valence Bond Theory (VBT) When d-orbitals of (n – 1) shell are used, these are known as inner orbital complex, they are formed • According to this theory, when a complex is due to strong field ligands or low spin ligands and formed, the metal ion/atom provides empty hybridisation is d2sp3. They are also octahedral in orbitals to the surrounding ligands. shape. • Coordination number shows the number of such Complexes with coordination number 6 • Inner orbital complex, [Co(NH3)6]3+ Orbitals of Co3+ ion 3d 4s 4p NH3 is a strong field ligand, so it causes pairing of electrons. d2sp3-hybridised orbitals of Co3+ [Co(NH3)6]3+ 3d 4s 4p (Low spin complex) ×× ×× ×× ×× ×× ×× six pairs of electrons from six NH3 molecules (d2sp3-hybridisation) All electrons are paired, therefore complex will be diamagnetic in nature. [tChoeFre6f]o3–re • Outer orbital complex, F– is weak field ligand, cannot cause pairing of electrons. sp3d2-hybridised orbitals of Co3+ 3d 4s 4p 4d [CoF6]3– 3d ×× ×× ×× ×× ×× ×× 4d (High spin complex) 4s 4p six pairs of electrons from six F– ligands (sp3d2-hybridisation) Complex has unpaired electrons, therefore, it will be paramagnetic in nature. Complexes with coordination number 4 Coordination Compounds 151

S. No. Inner orbital complexes Outer orbital complexes (a) Strong field or low spin ligands Weak field or high spin ligands (b) Hybridisation is dsp2 (where, one orbital is of 3d, one Hybridisation is sp3 (where one orbital of 4s and two orbitals are of 4p for first series orbital is of 4s and three of 4p) transition metal.) (c) Square planar shape Tetrahedral shape • Inner orbital complex, [Ni(CN)4]2– Orbitals of Ni2+ ion 3d 4s 4p CN– is a strong field ligand, therefore causes pairing of electrons. dsp2-hybridisation 3d 4s 4p [Ni(CN)4]2– ×× ×× ×× ×× (Low spin complex) dsp2 Four pa4irCs Nof–eglreocutrposns from All electrons are paired, so the complex will be diamagnetic in nature. • Outer orbital complex, [CoCl4]– Orbitals of Ni2+ ion 3d 4s 4p CN– is a weak field ligand, therefore it will not cause pairing of electrons. 3 hybridisation sp 3d 4s 4p [CoCl4]– 3d ×× ×× ×× ×× (High spin complex) 4s 4p Four pairs of electrons from Cl– Since complex has unpaired electrons, so it will be paramagnetic in nature. Limitations of VBT negative field of ligands, the five degenerate d-orbitals • This theory could not explain the magnetic splits up. Splitting of d-orbitals depends on the nature data quantitatively and could not predict the of the crystal field. exact existence of inner orbital and outer orbital complexes. The energy difference between splitted levels (orbitals) • It involves a number of assumptions. is denoted by D and is called crystal field splitting • It does not explain change of magnetic moment energy. with temperature. By using spectroscopic data for a number of coordination • It does not explain colour of the complexes. compounds, having the same metal ions but different ligand, the crystal field splitting for each ligand can be Crystal Field Theory (CFT) calculated. A series in which ligands are arranged in order of increasing magnitude of crystal field splitting In this theory, ligands are treated as point charges in is called spectrochemical series. case of anions and dipoles in case of neutral molecules. Spectrochemical series: The five d-orbitals have same energy when surrounded by spherically symmetrical field but on approaching I– < Br– < SCN– < Cl– < S2– < F– <<OCHN––<<CC2OO42. − <isHt2hOe < NCS– < EDTA4– < NH3 < en I 152 Chemistry-12

weakest ligand, CO is the strongest ligand because it axes. So, in such complexes d-orbital splitting is can form s as well as p-bond. inverted. Crystal Field Splitting in Octahedral Complexes • The crystal field splitting energy is less as • iTnhbreetew de-eonrbtihtaelcso, oi.red.idnxayt,edayzxaens datdazxnaarnegolreie4n5t°eodf compared to the octahedral field splitting. [ ]Dt =4 the approaching ligand and are called t2g-orbitals. 9 D0 • The other two d-orbitals, ai.xee. sdax2l–oy2nagntdhdezd2 oirrebcittiaolns • Orbital splitting energies are so low that pairing are oriented towards the of electrons are not possible so they mostly of the approaching ligand and are called eg- form high spin complexes with large number of orbitals. unpaired electrons. • Energy separation be tween ts2ugbascnrdipetg‘ os’ eistfoorf Energy dxy , dyz , dxz orbitals is denoted by Do (where t2 octahedral). It is also called crystal field splitting 2 ∆t 5 energy in octahedral complexes. ∆t • Energy odfueeg set of orbitals > energy of t2g set of 3 ∆t orbitals to repulsion with the ligands. 5 • The energy of eg orbitals will increase by (3/5) Do e and t2g will decrease by (2/5) Do. dx2 – y2, dz2 • If Do < P, (where P = energy required for d orbitals Average enery Splitting of d-orbitals e– pairing in an orbital so as to overcome free ion of the d orbitals in in tetrahedral spherical crystal crystal field repulsion) the fourth electron enters one of field Lthigeaengdos rfboritwalhsicghivDion<gPtharee ckonnofwignuarsatwioenakt2f3igeelgd1. Colour in Coordination Compounds ligands and they form high spin complexes. • The crystal field theory attributes the colour of the coordination compounds to d–d transition of • If Do > P, it becomes more energetically favourable the electrons, i.e. electrons can jump from t2g level to higher eg level. for the fourth et42lgecet0gr.oLnigtoanodccsuwphyiacht2pg roordbuitcaelswthitihs • In the absence of ligands, crystal field splitting configuration effect are known as strong field ligands and form does not occur and hence the substance is low spin complexes with least number of unpaired colourless. In the absence of unpaired electrons, electrons. d-d transition do not take place and complex is Energy dx2–y2, dz2 eg colourless. • The colour of the complex is complimentary to 3/5∆0 that which is absorbed. Barycentre ∆0 [[[CTTiiu((HH(H222OOO))66)4]]43]2+++i–s–vcBoiloluoleuetrilnienscscoololouurr, etc. [Zn(H2O)4]4+ is colourless Metal 2/5∆0 Limitations of CFT d-orbitals dxy,dxz, dyz t2g Average enery Splitting of d orbitals dx2–y2, dz2 dxy,dxz,dyz of the d orbitals in in octahedral Free metal ion spherical crystal crystal field • It does not consider the formation of covalent field bonding in complexes. Crystal Field Splitting in Tetrahedral Complexes • In tetrahedral complexes, four ligands may be • It is also unable to account satisfactorily for imagined to occupy the alternate corners of the the relative strengths of ligands, e.g. it does not explain why H2O is stronger ligand than OH–. cube and the metal ion at the center of the cube. • It does not account for the partly covalent nature • Energy of t2 set of orbitals > Energy of e set of of metal-metal bonds. orbitals as ligands are approaching between the ExERCISE 5.2 Multiple Choice Questions (MCQs) (1 Mark) 2. Predict the geometry and magnetic behaviour of [N(iiC) ld4s]2p–2. and paramagnetic 1. How many moles of AgCl will get precipitated (ii) dsp2 and diamagnetic (iii) sp3 and paramagnetic when an excess of AgNO3 solution is added to one (iv) sp3 and diamagnetic molar aqueous solution of [CrCl(H2O)5]Cl2? (i) 1 (ii) 2 (iii) 3 (iv) 5 Coordination Compounds 153

3. The colour of the coordination compounds depends 8. Assertion: n[Cart(uHre2O. )6]Cl2 and [Fe(H2O)6]Cl2 are on the crystal field splitting. What will be the correct reducing in order of absorption of wavelength of light in the visible region, for the complexes, [Co(NH3)6]3+, [Co(CN)6]3–, Reason: Unpaired electrons are present in their [Co(H2O)6]3+ d-orbitals. (i) [Co(CN)6]3– > [Co(NH3)6]3+ > [Co(H2O)6]3+ 9. mAsosmeerntticoonrr:e[sFpeon(CdiNn)g6t]o3–twioonunshpaoiwresdmelaecgtnroentisc. (ii) [Co(NH3)6]3+ > [Co(H2O)6]3+ > [Co(CN)6]3– Reason: Because it has dsp3 type hybridisation. (iii) [Co(H2O)6]3+ > [Co(NH3)6]3+ > [Co(CN)6]3– 10. Assertion: The number of unpaired electrons present in [CuCl2]– complex is zero. (iv) Co(CN)6]3– > [Co(NH3)6]3+ > [Co(H2O)6]3+ Reason: The complex is linear in solid state with 4. WocofnhAdegnuNc0tO.i1v3im,ty0o.ol2fCsmooCloull3t(ioNofnHAw3g)i5Clilslctoarrrereaetsoepbdotnwadiinttheode.xcTehses sp hybridization. Banotdhsa[Nmie(CmNa)g4]n2–etaicndbe[hNaivCiol4u]2r–. 11. Assertion: have same shape (i) 1:3 electrolyte (ii) 1:2 electrolyte Reason: Both are square planar and diamagnetic. (iii) 1:1 electrolyte (iv) 3:1 electrolyte Very Short Answer Type Questions (1 Mark) 5. AoWfghtNheOne 3c1,om3mmoplloeClxroCifslA:3◊g6CHl2aOreisotbrteaaitneeddw. Tithheefxocremssuloaf 12. What is meant by crystal field splitting energy? [AI 2013, 2011(C)] (i) [CrCl3(H2O)3] ◊ 3H2O 13. What is meant by spectrochemical series? (ii) [CrCl2(H2O)4]Cl ◊ 2H2O [Foreign 2010] (iii) [CrCl(H2O)5]Cl2 ◊ H2O 14. Contrast the magnetic behaviour of the complex entities [Fe(CN)6]4– and [FeF6]3– (Fe = 26). 6. (iv) [Cr(H2O)6]Cl3 15. How will you differentiate between the following When one mole onfitCroaCtel3s.5oNluHti3onw,a2s treated with excess of silver mol of AgCl two isomers: [CoBr(NH3)5]SO4 and was precipitated. The formula of the compound [Co(SO4)(NH3)5]Br? is: (Sample Paper 2019-20) Short Answer Type Questions-I (2 Marks) 16. What kind of hybrid orbitals are associated with: (i) [Co(NH3)5Cl2]Cl (i) Ni-atom in [Ni(CN)4]2– (ii) [Co(NH3)5Cl]Cl2 (ii) Fe-atom in [Fe(H2O)6]2+ (iii) [Co(NH3)4Cl2] (NH3)Cl (iii) Co-atom in [Co(NH3)6]3+ (iv) [Co(NH3)3Cl3] (NH3)2 (iv) Cu-atom in [Cu(NH3)4]2+? 7. The absorption maxima of several octahedral 17. The magnetic moment of [MnCl4]2– is 5.92 B.M. complex ions are as follows: Explain giving reason. (Sample Paper 2019-20) (Atomic No. of Mn = 25) [NCERT Exemplar] 18. Account for the phenomena Co(II) is stable in S.No. Compound λmax nm aqueous solution but in the presence of a strong 1. [Co(NH3)6]3+ 475 2. [Co(CN)6]3– 310 ligand, it get oxidised to Co(III). 3. [Co(H2O)6]3+ 490 19. Explain why nearly all the tetrahedral complexes are regarded as high spin complexes. The crystal field splitting is maximum for: (i) [Co(H2O)6]3+ (ii) [Co(CN)6]3– 20. Using valence bond theory, predict the geometry (iii) [Co(NH3)6]3+ and magnetic behaviour of (iv) All the complex ions have the same (i) [Co(NH3)6]3+ splitting 0 (ii) [Ni(NH3)6]2+ (Atomic No. of Co = 27, Ni = 28) Assertion-Reason Type Questions (1 Mark) 21. Describe the state of hybridisation, shape and Note: In the following questions a statement magnetic nature of the complexes: of assertion followed by a statement of reason (i) [Cr(H2O)2(C2O4)2]– is given. Choose the correct answer out of the (ii) [Co(NH3)2 (en)2]3+ following choices. (Atomic No. of Cr = 24, Co = 27) (i) Assertion and reason both are true, reason [Foreign 2010] is the correct explanation of assertion. 22. Using valence bond approach, predict the shape (ii) Assertion and reason both are true but reason and magnetic behaviour of: is not the correct explanation of assertion. (i) [Fe(CN)6]3– ion (ii) [Cr(CO)6] (iii) Assertion is true but reason is false. (Atomic No. of Cr = 24, Fe = 26) (iv) Assertion is false, reason is true. 154 Chemistry-12

23. wAdiceacamokualngynt epftoaircr.athmeafgonlleotwicin(wgA:hto[iFmleei(cC[NFNeo).(6Co]3fN–F)ies6]=4o–n26liys) (i) Crystal field splitting (ii) Linkage isomerism 24. Predict the number of unpaired electrons present in (iii) Ambidentate ligand [AI 2009] the square planar [Pt(CN)4]2–(Aiotnom. ic No. of Pt = 78) 36. Using valence bond theory explain the geometry and magnetic behaviour of pentacarbonyl iron. 37. lTohwescpoimnpcloemxp(ale)x[Ceso.(Heno)w3]3m+ aanndy (ubn)p[Mainre(CdNel)e6c]3t–roanres 25. What is the coordination entity formed when excess of aqueous KCN is added to an aqueous solution of copper sulphate? Why is it that no precipitate of are present in each complex. (Co = 27, Mn = 25) a[CquoFeo6u]3s– copper sulphide is obtained when H2S gas is passed 38. Among the two complexes: (a) and (b) through this solution? [NCERT] [wChoi(leeno)3t]h3+eronaepappepaersarbsluyeel.loGwivien solution 26. What is crystal field splitting energy? How does the the colour of the dm-aorgbniittauldseinofaDn0 decide the actual configuration of complex and explain your choice. octahedral field in a coordination 39. aaDtneodpmicNitcit(ohCrebOsitt)ar4u.lscAtoulfsroethsgeoifvt[reCaonth(sNietHiho3ny)6b]m3r+ie,dt[iNaslia(itCniNoen)a4c]o2h–f entity? [Delhi 2011 C][NCERT] c[Foleo(uCrNs )i6n]4d– ialuntde s[Foleu(tHio2nOs).6]W2+hay?re 27. of different case. (Atomic No. of Co = 27, Ni = 28) (Atomic No. of Fe = 26) [Foreign 2012][NCERT] 40. Explain on the basis of valence bond theory that [dgNeiaoim(mCaeNgtr)n4y]e2ta–inchdawsihtsiiqlseupa[aNrreiaCpmll4aa]n2g–anireotsnitcr.huacstutreetraanhdeditraisl 28. A coordination compound has the formula 41. dWiahmya[gNnieCtilc4],2e–viesnptahroaumghabgontehtiacrewtheitlreahNeid(CraOl?)4 is pCroeCcli3p.i4tNatHes3.cIhtldoroiedseniootnlaibsesrialvteerNcHhl3o, rhidowe.eWverritiet coEorxbmpitlpaalilencxowmwhhpylee[rCxe.oa(sN[HN3i)(6N]3+Hf3o)r6m]2+s the IUPAC name of the complex. 42. an inner orbital forms an outer 29. d[CiaomFa6]g3n–eitsic.pWarhaym? agnetic but [Co(CN)6]3– is OR Using valence bond theory, predict the hybridisation and magnetic charact[eCrBofS[ECo2F026]23–] (At. No. of Co = 27) 43. Explain using Crystal Field theory, why hex- aaquamanganese(II) ion contains five unpaired Short Answer Type Questions-II (3 Marks) electrons while hexacyanidomanganese(II) ion 30. Determine the hybridisation, geometry and contains only one unpaired electron. magnetic moment of the complexes: (i) [Co(NH3)6]3+ 44. Aqueous copper sulphate solution (blue in (ii) [Cr(CN)6]3– [AI 2009] colour) gives (i) a green precipitate with aqueous (Atomic No. of Co = 27, Cr = 24) potassium fluoride and (ii) a bright green solution 31. Amongst the following ions, which one has the with aqueous potassium chloride. Explain the highest magnetic moment and why? (Atomic No. experimental results. [NCERT] Cr = 24, Fe = 26, Zn = 30) [NCERT] 45. What is spectrochemical series? Explain the (i) [Cr(H2O)6]3+ (ii) [Fe(H2O)6]2+ difference between a weak field ligand and strong (iii) [Zn(H2O)6]2+ 32. What will be the correct order for the wavelength field ligand. [Foreign, AI 2010][NCERT] of absorption in the visible region for the following? 46. [Cr(NH3)6]3+ is paramagnetic while [Ni(CN)4]2– is diamagnetic. Explain why. [NCERT] [NCERT] 33. Wi[tNh it(hHe2Ohe)6l]p2+o,f[cNriy(sNtaHl3f)i6e]2ld+, t[hNeio(NryO, p2)r6e]4d–ict how 47. [ANsio(CluNti)o4]n2–oifs[Ncoil(oHu2rOle)s6s]2.+Eisxpglraeienn.[NbuCtEaRsTol]ution of many number of unpaired electrons are there in 48. (i) How many chloride ions will be precipitated when the coordination complex CoCl3.5NH3 ([AFeto(CmNic)6N]4o–.aonfdF[eFies(H262O) )6]2+ complexes. is treated with silver nitrate? 34. Draw crystal field energy level diagram and (ii) Assign the reason for non-ionic nature of the complex, CoCl3.3NH3. predict the number of unpaired electrons for the following complexes. [Atomic number of Fe Long Answer Type Questions (5 Marks) (26), Ni (28), Cr (24)] 49. A metal complex having composition Cr(NH3)4Cl2Br [[NFeiCCll44]]2__((TTeettrraahheeddrraal)l) has been isolated in two forms (A) and (B). The (i) (ii) form (A) when reacted with AgNO3 gives a white ppt. which is readily soluble is dilute NH3, while (iii) [Cr(CN)6]3+ (Octahedral) form B gives a pale yellow ppt. which is soluble in 35. Provide a suitable example and explain each the following: conc. NH3. Write the formula of (A) and (B) and Coordination Compounds 155

predict the hybridisation of chromium in each. (i) [CoF4]2– (ii) [Cr(H2O)2(C2O4)2]1– Also calculate the magnetic moment values. (iii) [Ni(CO)4] 50. For the complex [Fe(en)2Cl2]Cl where, [en = Atomic number: Co = 27, Cr = 24, Ni = 28 ethylenediamine], Identify: [Delhi 2009][DSB 2009] (i) The oxidation number of Fe 52. Write the name, structure and the magnetic behaviour of each of the following complexes: (ii) The hybrid orbitals and shape of the complex. 53. (Eii(xiip)) l[[aNPinti((NCthOHe)34b)]2oCnld(NinOg2i)n] (ii) [Co(NH3)4Cl2] Cl (iii) The magnetic behaviour of complex. coordination compounds in terms of Werner’s postulates. [NCERT] (iv) Whether the complex shows isomerism isomer. 54. On the basis of the following observations made with aqueous solutions, give secondary (v) Name the complex (At No. of Fe = 26) valencies to the metal atoms/ions in the following complexes: 51. Compare the following complexes with respect to their shape, magnetic behaviour and the hybrid orbitals involved: Formula Moles of AgCl precipitated per mole of the compounds with excess AgNO3 (i) PdCl2.4NH3 2 (ii) NiCl2.6H2O 2 (iii) PtCl4.2HCl 0 (iv) CoCl3.4NH3 1 (v) PtCl2.2NH3 0 Answers 5.2 1. (ii) 2. (iii) 3. (iii) 4. (ii) 16. (i) CN– being a strong ligand cause pairing of 5. (iv) 6. (ii) 7. (ii) 8. (ii) electrons. 9. (iv) 10. (iii) 11. (iv) ×× ×× ×× ×× 12. The difference in energy between the two sets of 3d 4s 4p d-orbitals (i.e.between eg and t2g set of orbitals) after crystal field splitting, in the presence of a 2 hybridisation dsp ligand is called crystal field splitting energy. (ii) pHa2iOribneginofg a weak field ligand does not cause electrons. 13. When ligands are arranged in the increas- ing order of their crystal field splitting energy, the ×× ×× ×× ×× ×× ×× series thus formed is termed as spectrochemical 3d 4s 4p 4d series. 32 hybridisation 14. [Fe(CN)6]4–: sp d ×× ×× ×× ×× ×× ×× (iii) NofHel3ebcetirnognas.strong field ligand causes pairing 3d 4s 4p ×× ×× ×× ×× ×× ×× 3d 4s 4p 23 d sp It is diamagnetic, m = 0.00 BM 23 hybridisation [FeF6]3–: d sp (iv) uNnHp3aibreeidnegleactsrtornontog field ligand shift the 4p from 3d. ×× ×× ×× ×× ×× ×× ×× ×× ×× ×× 3d 4s 4p 4d 32 3d 4s 4p sp d 2 hybridisation It is paramagnetic, m = 5.92 BM dsp 15. The isomer, [CoBr(NH3)5]SO4, can give white 17. Icnau[MsenCpal4i]r2i–n: Cg lo– fbedi-negleacwtreoanksfioefldMling.anHdecnacnen, oitt precipitate with barium chloride solution while contains 5 unpaired electrons results in high value the isomer [Co(SO4) (NH3)5]Br gives light yellow precipitate with AgNO3 solution. of magnetic moment. 156 Chemistry-12

3d 4s 4p 23. [Fe(CN)6]3+ ×× ×× ×× ×× ×× ×× 3d 4s 4p [MnCl4]2– ×× ×× ×× ×× Magnetic moment, m = 5(5 + 2) = 5.92 B.M. 23 18. Co(II) has the configuration of 3d7, i.e. it has 3 singly occupied orbitals. d sp Water being a weak ligand is unable to force the Complex is paramagnetic, d2sp3 hybridisation (one electrons to get paired up. In the presence of a e– is unpaired) strong field ligand, two electrons get paired up [Fe(CN)6]4– ×× ×× ×× ×× ×× ×× and third one is shifted to the higher energy shell where it can be easily lost and is thus oxidised to 3d 4s 4p form Co(III) ion. 19. Because none of the orbitals is pointing directly 23 along the approaching ligands in tetrahedral arrangement and there are only four ligands d sp instead of six (octahedral geometry), therefore Complex is diamagnetic, d2sp3 hybridisation (all crystal field splitting of tetrahedral complexes the e–s are paired up) is 4/9 of that in case of octahedral complexes. 24. For square planar shape, Pt2+ (d8) ion undergoes Consequently, crystal splitting energy for dsp2-hybridisation. Hence, two unpaired electrons tetrahedral complexes is almost always smaller in 5d orbital get paired up to make one d-orbital than the spin pairing energy, P and thus nearly empty to facilitate dsp2-hybridisation. Thus, no all the tetrahedral complexes are high spin unpaired electron is present in [Pt(CN)4]2–. complexes. 25. KCN when treated with copper sulphate forms cupric cyanide which decomposes to give cuprous 20. (i) ×× ×× ×× ×× ×× ×× cyanide and cyanogen gas. Cuprous cyanide dissolves 3d 4s 4p in excess of KCN forming a soluble complex, K3Cu(CN)4. 23 ×× ×× [CuSO4 + 2KCN → Cu(CN)2 + K2SO4] × 2 2Cu(CN)2 → Cu2(CN)2 + (CN)2 d sp 4d Cu2(CN)2 + 6KCN → 2K3[Cu(CN)4] Octahedral, diamagnetic 2CuSO4 + 10KCN → 2K3[Cu(CN)4] (ii) ×× ×× ×× ×× + 2K2SO4 +(CN)2 3d 4s 4p Since the complex is very stable, so there are no cupric ions present in the solution.Hence no 32 precipitation reaction occurs when H2S gas is passed through the solution. sp d 21. (i) Octahedral, paramagnetic 26. When ligands approaches a transition metal ion, ×× ×× ×× ×× ×× ×× d-orbitals splits up into two sets of orbitals, one 3d 4s 4p twewnietohrsgeylot(weogfe).orTrebhnieteadrligsffyiesr(etcn2agcl)leeaodnf cdernyoestrthgayel rbfieewtlwdit ehsepnhliittghthienesger 23 d sp Hybridisation is d2sp3. Shape is Octahedral, and it is paramagnetic. energy (Do) for octahedral field. Under the influence of weak field ligands, the (ii) ×× ×× ×× ×× ×× ×× energy difference, Dloowbeerttwheaennthte2gpaanirdineggesneetrgoyf 3d 4s 4p orbitals is relatively 23 and hence the distribution of electrons in t2g and eg set of orbitals occur according to the Hund’s rule, d sp 22. (i) Hybridisation is d2sp3. Shape: Octahedral, i.e. electrons can be accommodated in the manner and is diamagnetic. ×× ×× ×× ×× ×× ×× as shown below: 3d 4s 4p t21,g2, 3 → e4g, 5 → t26,g7, 8 → e9g, 10 When ligands are strong, the energy difference 23 bheigthweeretnhatn2gthaenpdaeirgi snegtesnoefrgoyrbaintdaltsheisdirsetrlaibtuivteiolny of electrons in both sets does not obey Hund’s rule. d sp The distribution occurs according to the following manner: It is octahedral and weakly paramagnetic (ii) ×× ×× ×× ×× ×× ×× 3d 4s 4p 23 t21,g2, 3, 4, 5, 6 → e7g, 8, 9, 10 d sp It is octahedral and diamagnetic Coordination Compounds 157

27. In both the complexes, iron is in +2 oxidation [Co(NH3)6]3+ ×× ×× ×× ×× ×× ×× state, i.e. it has 3d6 configuration. Ilniga[Fned(,CsNo )t6h]4e– 3d 4s 4p complex ion, CN– is a strong field unpaired electrons are paired up and Fe2+ ion undergoes d2sp3-hybridisation, forming an inner- 23 hybridisation d sp orbital, octahedral complex having no unpaired Geometry: Octahedral, three unpaired electrons. electrons are there, hence magnetic moment Ounnptahireedotehleercthroannsd,inH32Od is a weak ligand and has value, orbital are not paired up under its influence. Fe2+ ion thus undergoes m = 3(3 + 2) = 3.87 B.M sp3d2-hybridisation giving rise to outer-orbital complex with four unpaired electrons. 31. All the given three complexes have H2O as a ligand which is a weak field ligand and does not cause Because of the difference in the number of unpaired electrons, both the complexes show pairing of electrons. In [Cr(H2O)6]3+, Cr3+(d3) ion different colours. has 3 unpaired electrons. In [Fe(H2O)6]2+, Fe2+(d6) ion has 4 unpaired electrons. In [Zn(H2O)6]2+, 28. cTohme pcloemx pilsexteitsra[Camo(NmHin3e)4dCicl2h]Clolr.idNoa-mcoebaolft(tIhIiIs) Zn2+(d10) ion has no unpaired electron. Thus, chloride nInot[CcaouFs6e]3p–,aFir–inisg [Fe(H2O)6]2+ has the highest magnetic moment. 29. a weak field ligand which does 32. In all the three complexes, the central metal ion of electrons i.e. four d electrons remains unpaired, hence it is paramagnetic. is the same, hence the increasing field strength of [CoF6]3– the ligands from the spectrochemical series which ×× ×× ×× ×× ×× ×× follows the order is: 3d 4s 4p 4d H2O < NH3 < NO2– will affect the effective absorption of light from 32 the visible region. sp d In [Co (cCaNn )c6a]3u–,s CN– is a strong field ligand Hence energies absorbed for the excitation of wh ich e pairing of electrons. There are no unpaired electrons present, hence it is electrons will be in the order: diamagnetic. [Ni(H2O)6]2+ < [Ni(NH3)6]2+ < [Ni(NO2)6]4– [Co(CN)6]3– ×× ×× ×× ×× As E = hc , so the wavelength absorbed will be ×× ×× λ in opposite order. [Ni(NO2)6]4– < [Ni(NH3)6]2+ < 3d 4s 4p 23 hybridisation [Ni(H2O)6]2+ 33. CN– is a strong field ligand, so D0 is very large. d sp 30. (i) The oxidation state of cobalt in the complex Due to it, the electrons prefer to remain in the is +3. The electronic configuration of Co3+ ion is lower set of orbitals and they get paired up. Fe2+ – 2, 8, 8, 6 3d 4s 4p ∆o is large eNleHc3trboenisn.g a strong ligand causes paring of e and there are g d6 no unpaired [Co(NH3)6]3+ ×× ×× ×× ×× ×× ×× electrons. Hence it is diamagnetic. 3d 4s 4p 23 hybridisation t2g H2O being a weak field ligand has small D0 value. d sp As a result, the electrons tries to occupy both set of orbitals first singly and then pairing will occur. It has octahedral geometry, there is no unpaired electron, hence zero magnetic e ∆o is small, moment. g 4 unpaired (ii) The oxidation state of chromium in the d6 electrons. complex is +3. The electronic configuration of Cr3+ ion is Hence it is paramagnetic 3d 4s 4p t2g Chemistry-12 158

34. (i) Cl– is a weak field ligand. Thus the complex 37. (i) (en) is a strong field ligand. Thus under its is a high spin complex. All the five orbitals influence 3d6 electrons gets ipsaniroedunuppaiinretd62g are singly occupied. e0g arrangement, i.e. there electrons in this complex. (ii) Cl– is a weak field ligand. Thus it forms a high spin complex. Co3+ Ni2+ (3d8) 3d ×× ×× ×× ×× ×× ×× [Co(en)3]3+ 3d 4s 4p 23 hybridisation d sp (ii) CN– is a strong field ligand. Thus under its influence the 3d4 electrons of Mn3+ ion get arrange themselves in t2g orbitals in the manner as shown: (iii) (CN)– is a strong field ligand. Hence the Hence, there are two unpaired electrons. electrons will try occupy the orbitals of lower energy. 38. Weak field ligands does not interact with the (Cr3+ = 3d3) d-orbital electrons of the metal ion that much effectively and so they produce only small crystal 35. (i) The splitting of d orbitals under the influence of approaching ligands is called crystal field field splitting of d-electron energy states. The splitting. For example, d4 configuration in the opposite is true in case of strong field ligands. dpr4ecsoennficgeuorfawtioenakinfitehledplirgeasnendciesotf23gaesg1twrohniglefitehlde Because F– is a weak field ligand, splitting ligand is t24g eg0. between the levels is lesser in case (a) than in (ii) Linkage isomers are the ones which have case (b). Therefore, for d-d transition in (a) it ambidentate ligand. They have the same will absorb light of lower energy or of larger molecular formula but they differ in the linkage of ligand atom to the central metal wavelength than (b). Consequently, it will display atom, e.g. a complementary colour of shorter wavelength. [Co(NH3)5 NO2] Cl2, [Co(NH3)5 ONO]Cl2 Blue light has shorter wavelength than yellow (iii) A unidentate ligand which coordinates to light. So [CoF6]3+ is blue and [Co(en)3]3+ is yellow in colour. the central atom through more than one coordinating site but only one site at a time. 39. NH3 being a strong ligand causes pairing of e.g. NO2– or ONO–, SCN– or NCS–. electrons. 36. Consider [Fe(CO)5] ×× ×× ×× ×× ×× ×× 3d 4s 4p 23 hybridisation d sp NH3 3+ H3N NH3 Fe [Co(NH3)6]3+ Co (octahedral) NH3 3d 4s 4p H3N [Fe(CO)5] ×× ×× ×× ×× ×× NH3 3d 4s 4p CN– being a strong ligand causes pairing of 3 hybridisation electrons. dsp ×× ×× ×× ×× 3d 4s 4p Therefore, Fe( CsOin)c5ehnaos trigonal bipyramidal geometry and unpaired electron is present, so it is diamagnetic in nature. 2 hybridisation dsp Coordination Compounds 159

2– aInnd[NCilC– li4s]2a–,wtehaekolxigidaantdi.oTnhsutas,teit of nickel is +2 does not cause [Ni(CN)4]2– NC CN pairing of electrons. Ni (square planar) NC ×× ×× ×× ×× CN Ni(0) is d10 system i.e., all e–s are paired. 3d 4s 4p ×× ×× ×× ×× 3 hybridisation 3d 4s 4p sp Ni(Cl)42– –paramagnetic 3 hybridisation 42. lIinga[Cndo(wNhHic3h)6]p3u+,shthees tNhHe 33 dlieglaencdtroisnsatsotgroetnpgafiireeldd up, leaving two d-orbitals empty which then get sp involved in d2sp3-hybridisation, thus forming an inner orbital complex. CO Ni(CO)4 OC Ni Co3+ 3d 4s 4p ×× ×× ×× ×× ×× ×× (tetrahedral) CO NH3 NH3 NH3 NH3 NH3 NH3 23 hybridisation d sp CO fIitinehl[daNslii(gaNaHcnod3n),6f]ip2ga+u,irrNainitgiiosnofinde8l+.e2cNtrHoox3nidstahdtoiouoegnshsntaoattsetorcaocnnudgr 40. c[Nomi(pCleNx)4i]o2n–:. The oxidation state of Ni is +2 in the Its electronic configuration is [Ar] 3d8. CN– being a strong field ligand cause pairing of electrons. It undergoes dsp2-hybridisation to as pairing will create only one vacant d-orbital. form the complex. Thus, sp3d2 hybridisation is involved to form an outer orbital complex. No unpaired electron is present, hence it is Ni2+ ×× ×× ×× ×× ×× ×× diamagnetic in nature. 3d 4s 4p 4d ×× ×× ×× ×× 3d 4s 4p 32 sp d 43. Mn is in +2 oxidation state and it has the electronic Scoon, ftihguerdaitsitornibouft3iodn5.oHf 2eOlecistraonwseiask field ligand. 2 hybridisation t23g e2g, i.e. all dsp [cNomiCpll4e]x2.–:ItTsheeleocxtirdoantiiconcosntfaigteuroaftNioinisis+[2Ari]n3tdh8e. the electrons remains unpaired. Cl– being a weak field ligand does not cause pairing ↑ ↑ eg of electrons, it undergoes sp3-hybridisation to form ↑ ↑ ↑ t2 g CN– being a strong ligand pushes the electrons the complex. to get paired up and thus the distribution of ×× ×× ×× ×× electrons is t25g e0g, i.e. only one unpaired electron 3d 4s 4p is present. 3 hybridisation — — eg sp Two unpaired electrons are present in the complex ↑↓ ↑↓ ↑ t2g ion, hence it is paramagnetic in nature. 44. Copper sulphate is a coordination compound having four water molecules coordinated with 41. CInONbie(CinOg)a4, the oxidation state of nickel is zero and Cu2+ [Cu(H2O)4]2+ strong field ligand forces the unpaired blue ion, i.e. it has ions. This imparts colour to it. electrons of 3d-orbitals to get paired up. (i) When potassium fluoride solution is added, Ni(Z=28) (ground state) H(st2rOonmgeorlelicgualneds) are repl ace d by F– io ns forming the green ppt. of [[CCuu(FH4]22O– )i4o]n2+s.+ 4F– → Ni(CO)4 3d 4s 4p [CuF4] 2– + 4H2O ×× ×× ×× ×× Green ppt. 3d 4s 4p 3 hybridisation or CuSO4 + 4KF → K2[CuF4] + K2SO4 (ii) When potassium chloride is added, H2O sp 160 Chemistry-12

molecules are replaced by Cl– ions that forms are no unpaired electrons, i.e. d-d transition is t[Cheug(Hre2eOn)4s]o2l+u+ti4oCnlc–ons→istingG[Croeuefn[CsCollu4u]tCi2o–nl4+]2–4iHon2Os. not possible. Hence the complex is colourless. ×× ×× ×× ×× ×× 3d 4s 4p or CuSO4 + 4KCl → K2[CuCl4] + K2SO4 3 hybridisation 45. For any given metal cation, the magnitude of dsp crystal field splitting energy depends upon the 48. (i) Two chloride ions will get precipitated because the complex has coordination nature of ligands. When ligands are arranged in number 6 and ammonia molecule cannot acts the order of magnitude of crystal field splitting energy, the arrangement so obtained is called as primary valence as it can be satisfied only spectrochemical series. by anions. Thus, 5NH3 molecules and 1 Cl ion I– < Br– < Cl– < NO3– < H2O < < NO2– < CN– < CO act as ligands while two chloride ions satisfy the ionisable valence. The ligands which causes only a small degree of (ii) The coordination number of cobalt is 6. The three NH3 molecules and three Cl– ions are crystal field splitting are termed as weak field present inside the coordination sphere and thus will not get ionised when dissolved to ligands while those which causes a large crystal form solution. Hence, the complex is non- ionic in nature. field splitting are called strong field ligands. Ligands arranged before H2O are considered as 49. The coordination compound Cr(NH3)4Cl2Br has weak field ligands while ligands after H2O are two isomers. Since, coordination number of Cr is considered as strong field ligands. six in both the forms, thus the two forms may be represented in the following ways: 46. (i) The oxidation state of Cr in the complex ion is +3. Its electronic configuration is [Ar] 3d3. Cr [Cr (NH3)4 ClBr] Cl and [Cr (NH3)4 Cl2] Br undergoes d2sp3-hybridisation while forming (A) (B) the complex ion. [Cr(NH3)6]3+ ×× ×× ×× ×× ×× ×× [Cr (NH3)4 ClBr] Cl + AgNO3 → (A) 3d 4s 4p 23 hybridisation [Cr (NH3)4 ClBr] NO3 + AgCl . White ppt d sp There are three unpaired electrons present, AgCl + 2NH4OH → [Ag (NH3)2 Cl] + 2H2O hence it is paramagnetic in nature. Soluble in dil NH3 The oxidation state of Ni in the complex ion [Cr (NH3)4 Cl2] Br + AgNO3 → is +2. Its electronic configuration is [Ar] 3d8. (B) [Cr (NH3)4 Cl2] NO3 + AgBr . It undergoes dsp2-hybridisation to form the Pale yellow ppt. complex. CN– being a strong field ligand AgBr + 2NH4OH → [Ag (NH3)2 Br] + 2H2 O causes the unpaired electrons to get paired Soluble in conc. NH3 up. No unpaired electron is present, so the The hybridisation state of chromium in both the complex is diamagnetic in nature. complexes is d2sp3, i.e. chromium is in trivalent [Ni(CN4)]2– ×× ×× ×× ×× state (Cr3+) with three unpaired electrons in 3d 4s 4p orbital. 2 hybridisation ×× ×× ×× ×× ×× ×× 3d 4s 4p dsp 47. His2Oanisouatwere-aokrbfiietaldl lciogmanpdleaxn. dTthhiussc[oNmi(pHle2xO)h6]a2s+ 23 hybridisation two unpaired electrons. During d-d transition, d sp complementary green light is emitted. Due to the presence of three unpaired electrons, Magnetic moment: ×× ×× ×× ×× 3d 4s 4p m = n(n + 2) = 3 (3 + 2) 3 hybridisation = 15 = 3.873 BM sp 50. (i) x + (0) + (–2) + (–1) = 0 x = +3 CN– is a strong field ligand and it causes the Thus oxidation state of Fe, x = + 3 unpaired electrons to get paired up. The central (ii) Ethylenediamine(en) is a strong field ligand metal atom undergoes dsp2-hybridisation and a square planar complex is formed in which there causes pairing of electrons. Fe undergoes Coordination Compounds 161

d2sp3 hybridisation. Thus the shape of Magnetic behaviour – Diamagnetic complex is octahedral 5d 6s 6p ×× ×× ×× ×× ×× ×× × × ×× 3d 4s 4p × × ×× 23 2 d sp (ii) [Co(NH3)4Cl2]Cl dsp (iii) One unpaired electron is present, hence the Tetraamminedichloridocobalt(III) chloride. Structure – Octahedral complex is paramagnetic. Magnetic behaviour – Diamagnetic (iv) The complex has two geometrical isomers as 3d 4s 4p shown: ×× × ××× ×× × ××× cis isomer also exhibits optical isomerism d2sp3 (iii) [Ni(CO)4] Tetracarbonylnickel(0) Structure – Tetrahedral Magnetic nature – Diamagnetic 3d 4s 4p × ××× × ××× (v) Dichloridobis(ethane-1,2-diamine) iron(III) chloride sp3 51. (i) [CoF4]2– 53. Werner’s theory explains the bonding in 3d 4s 4p coordination compounds as per the following Co2+(d7) postulates: ×× ×× ×× ×× (i) In coordination compounds the metal atoms/ [CoF4]2– ions exhibit two types of valencies, namely, primary and secondary valencies. Shape –Tetrahedral Hybridisation – sp3 (ii) Primary valencies are ionisable and are satisfied by negatively charged ions. Three unpaired electrons are present– (iii) Secondary valencies are non-ionisable and Paramagnetic. are satisfied by negatively charged ions (ii) [Cr(H2O)2(C2O4)2]1– Cr3+(3d3) or neutral molecules. It is equal to the coordination number of central metal atom/ 3d 4s 4p ion and it is a fixed number. (iv) The ions/neutral molecules that satisfies the secondary valencies of metal atoms/ions [Cr(H2O)2(C2O4)2]1– in coordination compounds have particular ×× ×× ×× ×× ×× ×× spatial arrangements corresponding to different coordination numbers. Shape – Octahedral Hybridisation – d2sp3 For eg., In [CoCl(NH 3c)a5]nCbl2e- a coordination compound, the bonding explained as per Three unpaired electrons are present– the figure depicted below: Paramagnetic Cl (iii) Ni(CO)4 4s 4p H3N NH3 3d Ni(3d8) H3N Co NH3 Ni(CO)4 × ××× Cl Cl × ××× NH3 Shape – Tetrahedral Here,......represents primary valency Hybridisation – sp3 —— represents secondary valency. 54. (i) Formula of the complex: No unpaired electrons are present– [Pd(NH3)4]Cl2 Diamagnetic [Pd(NH3)4]Cl2 Water→ [Pd(NH3)4]2+ + 2Cl– 52. (i) [Pt(NH3)2 Cl(NO2)] Diamminechloridonitrito-N-platinum(II) Structure – Square planar. 162 Chemistry-12

2Cl– + 2AgNO3 → 2AgCl + 2NO3− on treatment with AgNO3 solution. Secondary valency = 6 which is satisfied by White six Cl– ions. precipitate Secondary valency = 4 which is satisfied by (iv) Formula of complex: 4 ammonia molecules. [Co(NH3)4Cl2]Cl Cl Water→ (ii) Formula of the complex: [Co(NH3)4 Cl2] [Co(NH3)4Cl2]+ + Cl– [Ni(H2O)6]Cl2 [Ni(H2O)6]Cl2Water→ [Ni(H2O)6]2+ + 2Cl– AgNO3 + Cl– → AgCl + NO3− 2Cl– + 2AgNO3 → 2AgCl + 2NO3− White precipitate White precipitate Secondary valency of complex = 6 which is Cl– Secondary valency of complex = 6, which is satisfied by four NH3 molecules and two satisfied by six H2O molecules. ions. (iii) Formula of the complex: H2(Pt(Cl))6] (v) Formula of complex: [Pt(NH3)2Cl2] H2[PtCl6] Water→ 2H+ + [PtCl6]2– It will not produce free Cl– ions in the The solution does not contain any free Cl– ions. Hence it will not give residue of AgCl, solution. Secondary valency = 4 which is satisfied by two NH3 molecules and two Cl– ions. Topic 3. Stability Importance of Coordination Compounds & Organometallic Compounds Stability of coordination compounds • Greater the value of stability constant, lower the value of dissociation constant, more stable will The stability of complex in solution refers to the degree be the complex. of association between the two species involved in the • Polydentate ligands form more stable complexes than monodentate ligands. state of equilibrium. It is expressed by the stability • Stability of complex decreases with increase constant. The instability constant or the dissociation in temperature because their formation is an exothermic process. constant of compounds is defined as the reciprocal of Applications of Coordination Compounds the formation or stability constant. • They are used in many qualitative and quantitative e.g., M+ + Lx– (ML)1 – x); K1 = [(ML)1−x ] analysis (Analysis of cations and anions). [M+ ] [Lx− ] • Hardness of water is estimated by simple titration .......................................................... with Na2 EDTA. .......................................................... • Purification of metals can be achieved through formation and subsequent decomposition of their (MLn)1– (n–1)x + Lx– (MLn)1–nx; Kn = coordination compounds e.g. Au and Ag. [(MLn )1−nx ] • They have great importance in biological systems, [(MLn−1 )1−(n−1)x ] [Lx− ] e.g. chorophyll in photosynthesis, haemoglobin as oxygen carrier in blood. M+ + nLx– (MLn)(1–nx); bn = • They are used as catalyst for many industrial [(MLn )(1−nx) ] pisroucessesdes,aes.g.hWomilkoingseonne’socuastaclyastta(Rlyhs(Pt Pihn3)3tChle] [M+ ] [Lx− ]n hydrogenation of alkenes. Zeigler-Natta catalyst. [TiCl4 + (C2H5)3Al] acts as a heterogeneous aaHnsedbrenb,=nKiKs1,1o×Kve2K,r2a…l…l Ks×tnabaKirlneitystceopnwsitsaents.taTbhieliytyarceonrsetlaantetds catalyst in the polymerisation of ethylene are a The factors on which stability of the complex depends: few examples. • Charge on the central metal atom: As the • In medicinal chemistry, there is a growing magnitude of charge on metal atom increases, interest of chelating therapy, e.g. cis-platin is used as anticancer agent. stability of the complex increases. • Nature of metal ion: The stability order is: 3d < 4d < 5d transition metal series. • Basic nature of ligands: Strong field ligands form stable complexes. Coordination Compounds 163

Organometallic Compounds R – Mg – X; etc. They contain one or more metal-carbon bond in their • s and p bonded compounds: Metal carbonyls molecules. They are of the following types: are their examples. Metal-carbon bond of metal • Sigma (s) bonded compounds: Metal-carbon carbonyls have both s and p-bond character. bond is sigma bond, e.g. (C2H5)4 Pb; Zn(C2H5)2; They have CO molecule as ligand, e.g. CO CO O CO CO OC C CO CO Mn Mn CO OC Co Co CO OC C CO CO CO CO CO [Mn2(CO)10] O [Co2(CO)8] Bonding in metal carbonyls CO CO CO Fe CO CO CO CO CO CO Cr Fe(CO)5 Ni Trigonal CO CO CO bipyramidal CO CO π π∗ Octahedral CO π [Cr(CO)6] Tetrahedral Mσ [Ni(CO)4] s bond is formed by donation of lone pair of electron C≡O from CO to vacant d-orbitals of transition metal, p-bond is formed by back donation of pair of electrons π from transition metal to vacant anti-bonding molecular Synergic bonding orbital of CO. This is called as synergic bonding in carbonyl complex. ExERCISE 5.3 Multiple Choice Questions (MCQs) (1 Mark) (iii) [Fe(C2O4)3]3– (iv) [FeCl6]3– 1. The stepwise and overall stability constant, (for n = 4) are related as : bn 4. Which of the following complexes formed by Cu2+ K2 K3 K1 K2 ions is most stable? K2K4 K3K4 (i) b4 = (ii) b4 = (i) Cu2+ + 4NH3  [Cu(NH3)4]2–, log K = 11.6 (ii) Cu2+ + 4CN–  [Cu(CN)4]2–, log K = 27.3 (iii) b4 = K1K2K3K4 (iv) b4 = K1K2K3K4 (iii) Cu2+ + 2en  [Cu(en)2]2+, log K = 15.4 2. Dissociation constant of coordination compounds (iv) Cu2+ + 4H2O  [Cu(H2O)4]2+, log K = 8.9 is related to its formation constant as 5. The anti pernicious anaemia factor which is a (i) Both are same coordination compound of Cobalt is: (ii) Reciprocal to each other (i) Cyanocobalamine (iii) Dissociation constant is half of formation (ii) Haemoglobin constant (iv) Dissociation constant is double of formation (iii) Desferrioxime B constant. (iv) Carbonic anhydrase 3. Amongst the following the most stable complex is: 6. The complex used in the treatment of cancer is (i) [Pt(NH3)4]2+ (i) [Fe(H2O)6]3+ (ii) [Fe(NH3)6]3+ 164 Chemistry-12

(ii) cis–[PtCl2(NH3)2] Very Short Answer Type Questions (1 Mark) (iii) trans–[PtCl2(NH3)2] 11. Name the ion which is estimated by using (iv) [Pt(NH3)3Cl]Cl Assertion-Reason Type Questions dimethyl glyoxime. (1 Mark) 12. What does stability constant indicate? In the following questions, a statement of assertion 13. Give names of two complexes which are used in followed by a statement of reason is given. Choose medicines. the correct answer out of the following choices on 14. Which compound is used to estimate the hardness of water volumetrically? the basis of the above passage. (i) Assertion and reason both are correct 15. Name an organometallic compound that is used as a homogeneous catalyst. statements and the reason is correct explanation of assertion. Short Answer Type Questions-I (2 Marks) (ii) Assertion and reason both are correct 16. Name two coordination compounds which are statements but reason is not correct biologically important. explanation for assertion. (iii) Assertion is correct statement but reason is 17. Name two analytical uses of coordination compounds. wrong statement. (iv) Assertion is wrong statement but reason is 18. Calculate the overall complex dissociation equilibrium constant for [Cu(NH3)4]2+ ion, given correct statement. that b4 for the complex is 2.1 × 1013. 7. Assertion: Toxic metal ions are removed by the 19. What is meant by chelate effect? Give an example. chelating ligands. [AI 2015][NCERT] Reason: Chelate complexes tend to be more stable. Short Answer Type Questions-II (3 Marks) 8. Assertion: K3[Fe(CN6)] is more stable than K4[Fe(CN)6] 20. Name the central metal atom present in Reason: Charge density on Fe3+ ion in K3[Fe(CN)6] haemoglobin, chlorophyll and vitamin B12. is more than Fe2+ion in K4[Fe(CN)6] 21. Write the structure of pentacarbonyl iron (0) and 9. Assertion: Strong field ligands form stable hexacarbonylchromium (0). complexes. 22. Explain the following: (i) Low spin octahedral complexes of nickel are Reason: Lower the value of dissociation constant, more stable is the complex. unknown. 10. Assertion: In metal carbonyl, lone pair of (ii) p complexes are known for transition metals electrons are donated by carbonyl group and only. transition metal to each other. (iii) CO is strong ligand than NH3 for most of the Reason: This types of bonding is called synergic metals. [AISB 2009] bonding. Answers 5.3 1. (iii) 2. (ii) 3. (iii) 4. (ii) 16. (i) Haemoglobin: It is an iron based complex, the red pigment in the blood that transports 5. (i) 6. (ii) 7. (i) 8. (i) 11. Ni2+ ion oxygen to all the parts of the body. 9. (ii) 10. (ii) (ii) Chlorophyll: It is a magnesium based 12. Stability constant indicates whether coordination complex, a green pigment that acts as a complex is stable or not in a solution. Higher its catalyst for the synthesis of carbohydrates value, more will be the stability of complex. in green plants. 13. (i) EDTA is used in lead poisoning. 17. (i) In quantitative analysis of a salt, DMG (ii) cis-platin, [Pt(NH3)2Cl2] is used in the (dimethylglyoxime) is used for the estimation treatment of cancer. of Ni2+ ions. 14. EDTA is used to test the hardness of water. (ii) In gravimetric estimation of gold, trans dichlorobis(ethylene diamine) copper(III) 15. Wilkinson catalyst (Ph3P)3RhCl is an example that can be used as a homogeneous catalyst in chloride is used for the estimation. hydrogenation of alkenes. Coordination Compounds 165

18. Overall complex dissociation equilibrium constant CO CO is equal to the reciprocal of the overall stability OC CO CO 21. Fe CO constant. Cr OC Here, overall dissociation constant CO CO CO = 1 1 = 4.7 × 10–14 β4 = 2.1 × 1013 Fe(CO)5 CO 19. FeSO4 solution on mixing with (NH4)2SO4 Trigonal bipyramidal [Cr(CO)6 ] Octahedral solution in 1 : 1 molar ratio forms a double salt, 22. (i) Ni(Z = 28) in its atomic or ionic form FfuerSnOis4h(NFHe24+)i2oSnOs4i.n6Hso2lOut(ioMno. hHrensacelt,)thwehsioclhutciaonn gives the test of Fe2+ ions. cannot afford 2 vacant 3d orbitals. Hence, d2sp3 hybridisation or low spin octahedral CuSO4 solution on mixing with aqueous ammonia complexes of nickel are unknown. solution in 1 : 4 molar ratio forms a coordination (ii) Because transition metals have vacant compound [Cu(NH3)4] SO4, which does not produce d-orbitals in their atoms or ions into which Cu2+ ions. Hence, it does not give the test for Cu2+ electron pairs can be donated by the ligands ions. containing p electrons. Thus dp–pp bonding is possible in them. 20. Molecule Central Metal Atom (iii) CO forms a s bond by donating an electron Chlorophyll Mg pair to metal but it also form a back bond by accepting a lone pair of d electrons from Haemoglobin Fe metal to its vacant p* orbital i.e. a p-bond is also formed. Though such back bonding is not vitamin B12 Co feasible in case of NH3 hence, it is a weaker ligand than CO. Case based questions The following questions are multiple choice 1. Case Study questions. Choose the most appropriate Coordination compounds of aspartic acid were answer: synthesized in basic and acidic media, with metal ligand M : L stoichiometric ratio 1:2. The complexes (i) According to the study, the complexes were characterized using infrared, electronic exhibited and magnetic susceptibility measurements, and (a) narrow spectrum of activity mass spectrometry. Antimicrobial activity of the (b) broad spectrum of activity compounds was determined against three Gram- (c) no activity positive and three Gram-negative bacteria and (d) activity depends on temperature and one fungus. The results obtained indicated pressure. that the availability of donor atoms used for coordination was a function of the pH of the (ii) The activity of some complexes are higher solution in which the reaction was carried out. when prepared in basic medium as compared This resulted in varying geometrical structures to that of in acidic medium because for the complexes. The complexes exhibited (a) they have enhanced lipophilicity due to a broad spectrum of activity. In some cases more chelate rings. complexes synthesized in basic medium exhibited (b) they have enhanced lipophobicity due to better activity compared to their counterpart less chelate rings. complexes obtained in acidic medium. This was (c) they have enhanced lipophobicity due to attributed to their enhanced lipophilicity as a more chelate rings. result of the increased number of chelate rings. (d) they have enhanced lipophilicity due to (Reference: T. O. Aiyelabola, D.A. Isabirye, E.O. less chelate rings. Akinkunmi, O.A. Ogunkunle, and I.A.O Ojo, Synthesis, Characterization, and Antimicrobial (iii) Which of the following is the correct Activities of Coordination Compounds of Aspartic application of complexes? Acid. Journal of Chemistry, Volume 2016, Article (a) EDTA used in lead poisoning treatment ID 7317015, 8 pages) (b) Cis-Platin used in cancer treatment (c) Wilkinson catalyst used in hydrogenation of alkenes (d) All of these 166 Chemistry-12

(iv) Which of the following is a chelating agent? metals – and in particular ruthenium(II/III) (a) S2– (b) HCN2O– complexes, due to their well known interaction (c) en (d) with DNA. OR What results the varying geometrical Ruthenium-based complexes have been widely structures of the complexes in the given studied due to their ability to strongly bind nucleic study? acids and proteins, ligand exchange kinetics (a) Isomerism shown by the complexes similar to those of their platinum counterparts, (b) Change in temperature, change their the prevalence of two main oxidation states (II structure and III) and the iron-mimicking property when (c) Solvent competes with the aspartic acid bound to biological molecules. In addition, both as ligand the commonly accessible oxidation states of (d) Availability of donor atoms is a function ruthenium are octahedral and relatively inert, of pH of the solution. the synthetic chemistry is very well established 2. Case Study (including stereochemical control). A consequence of widespread use of antimicrobials b[Rauct(eprhieanl )s3t]r2+apinrso.veHsotwoebveeirn,atchteiveinatgraoidnustctailolnthoef methyl group substituents on the phenanthroline there has been the emergence of drug-resistant ligands dramatically increased the activity against all bacteria – especially Gram positive populations of microorganisms. There is clearly bacteria and Mycobacterium tuberculosis – indicating the importance of lipophilicity on a need for the development of new classes of antibacterial activity. It was also demonstrated that the bacteria did not easily develop resistance antimicrobials, rather than drugs based upon to this class of compound. analogues of known scaffolds. Due to the success of the platinum anticancer agents, there has been considerable interest in the development of therapeutic agents based upon other transition (Reference: Fangfei Li, J. Grant Collins and F. (ii) Assertion: Ruthenium complexes shows Richard Keene, 2015, Ruthenium complexes as iron-mimicking property when bound to antimicrobial agents, Chemical Society Reviews, biological molecules. 44(8), 2529-2542) Reason: Ruthenium and iron belongs to In these questions, a statement of assertion same period and shows similar properties. followed by a statement of reason is given. Choose the correct answer out of the (iii) Aefsfescetrivtieoang:ai[nRsut(bpahcetne)r3i]a2.+ complex proves following choices. Reason: Phenanthroline (phen) acts as (a) Assertion and reason both are correct bidentate ligand in this complex. statements and reason is correct explanation (iv) Assertion: When methyl groups are for assertion. substituted on (phen) ligand in ruthenium complex then its activity increases against (b) Assertion and reason both are correct Gram positive bacteria. statements but reason is not correct Reason: Addition of methyl groups explanation for assertion. increases lipophilicity of the complex. OR (c) Assertion is correct statement but reason is Assertion: Common oxidation state shown wrong statement. by Ru in these complexes are +2 and +3. Reason: Ruthenium complexes are similar (d) Assertion is wrong statement but reason is to those of their platinum counterparts. correct statement. Answers (i) Assertion: New classes of antimicrobials are needed instead of analogous drugs based 1. (i) (b) (ii) (a) (iii) (d) (iv) (c) OR (d) upon same skeletal structure. 2. (i) (a) (ii) (c) (iii) (d) (iv) (a) OR (b) Reason: Many microorganisms are now become resistant to these drugs. Coordination Compounds 167

Analogy based questions 2. Complete the following analogy. 1. Which of the following analogies is correct Tetrahedral : A :: Square planar : B [INnni(eCrNo)r4b]2it–a:l:cBom: [pMlenxC::l4B]2–: (i) Ethylene diammine : Bidentate :: C2O42− : (i) A : outer orbital Unidentate (ii) A : (ii) [Co(NH3)6]3+ : Homoleptic complex :: complex [Co(NH3)4Cl2] : Heteroleptic complex (iii) A : sp3 hybridisation :: B : dsp2 hybridisation (iv) A : Strong field ligands : : B : weak field (iii) Geometrical isomess : : Cis-trans :: Optical ligands isomers : fac-mer 1. (ii) Answers (iv) [Ti(H2O)6]3+ : Colourless : : [Ti(H2O6)4+] : violet Colour 2. (iii) Multiple type question 1. Match the complex ions given in Column I with the colours given in Column II and assign the correct code : Column I (Complex ion) Column II (Colour) A. [Co(NH3)6]3+ 1. violet B. [Ti(H2O)6]3+ 2. Green C. [Ni(H2O)6]2+ 3. Pale blue D. [Ni(H2O)4(en)]2+ (aq) 4. Yellowish orange Code: 5. Blue (i) A (1) B (2) C (4) D (5) (ii) A (4) B (3) C (2) D (1) (iii) A (3) B (2) C (4) D (1) (iv) A (4) B (1) C (2) D (3) Answers 1. (ii) Quick revision notes through two different atoms but only one at a • Transition metals form large number of complex time are called ambidentate ligands. e.g. CN–, compounds in which the central metal atom NO2–, SCN– etc. are bound to a number of anions or neutral • Polydentate ligands forming a ring/cyclic molecules. These compounds are also known as coordination compounds as these involve structure around the central metal ion are called coordinate bonds. chelating ligands, e.g. ethane-1,2-diamine (en), • The negatively charged ions/neutral molecules oxalate ion (C2O42–). having lone pair that surrounds the metal in • Chelating ligands provide stability to the complex a particular spatial arrangement are called and this stabilisation is called chelate effect. e.g. ligands. [Co(en)3]3+ is more stable than [Co(NH3)6]3+. • The number of coordinate bonds formed by a • Ligands are categorised on the basis of number ligand around the central metal ion is referred of donor sites/atoms as unidentate (one donor to as the denticity of the ligand, e.g. EDTA4– has atoms), bidentate (two donor sites), polydentate denticity equal to 6. (more than two donor atoms) ligands. • The number of ligand donor atoms to which the • Also the monodentate ligands which can link 168 Chemistry-12

metal is directly bonded is called coordination configuration will have higher stability for a number. complex. • Oxidation number is the charge on the central EAN = Atomic Number – (charge on the metal ion, if all the ligands are removed along metal ion) + 2 × coordination number of with the shared electron pairs from the central metal metal atom. The IUPAC naming of the coordination (or complex) compounds can be exemplified as: • An ion/atom of a metal having EAN (effective atomic number) equal to the nearest noble gas Central metal atom + 3 + x – 3(2) = 0 Þ x = + 3 (oxidation number of metal) Complex: K3[Cr(C2O4)3] Name: Potassium trioxalatochromate(III) Ligand Coordination number (Oxidation number) = 3 × 2 = 6 (a oxalate is a bidentate ligand) Positively charged ion satisfying charge on the complex (counter ions) • Two principal types of isomerism in coordination • vBT, however fails to explain the optical compounds are known as properties and quantitative aspects of magnetic (i) Sterioisomerism (ii) Structural isomerism behaviour of coordination compounds. ã Sterioisomerism is further divided as • Crystal Field theory explains the quantitative Geometrical isomerism (cis-trans isomers and fac-mer isomers) and optical isomerism aspects of magnetic behaviour, spectral [dextro (d)-laevo (l)]. properties and stability of complex in a more ã Structural isomerism is further divided as: reasonable way on the basis of effect of crystal Œ Linkage isomerism: An ambidentate ligand bind through different atoms in field (provided by different ligands, assumed different isomers. as point charges) in splitting the degeneracy of Œ Coordination isomerism: Interchange of ligands between cationic and anionic d-orbital energies of the central metal atom/ion. entity of a complex. • However, too many assumptions and assuming Œ Ionisation isomerism: Interchange of ligand and counter ion in a complex. central metal atom/ion as point charge and Œ Solvate isomerism: Whether or not a anionic ligands exerting greatest splitting effect solvent molecule is directly bonded to the metal ion. posed some difficulties in explaining many • Werner gave the theory for the formation of theoretical aspects of coordination compounds. coordination compounds. • Spectrochemical series is the series of ligands in • In Werner’s theory the central metal atom is said to exhibit two types of valencies; namely, increasing field strength: primary and secondary valency. In modern chemistry, these valencies are named as ionisable I– < Br– < SCN– < Cl– < S2– < <Fe– n<<OCHN– –<<CC2OO42– < (primary) and non-ionisable (secondary) H2O < NCS– < EDTA4– < NH3 valencies. Ionisable valency is non directional • In metal carbonyls, bond between CO and metal and non-ionisable valency is directional. Primary valencies is the number of negative ions needed is strong due to synergic effect. Synergic effect to satisfy the charge on metal whereas secondary valencies are same as coordination number of is due to donation of electron pair by metal to metal. p* orbital of CO to form p-bond in addition of s • valence Bond Theory (vBT), explains the formation, magnetic nature and geometrical bond. shapes of coordination compounds by overcoming the shortcomings of Werner’s theory. • In general, the stability of coordination compounds is measured in terms of stepwise stability constants (K) and overall stability constant, b (beta) and dissociation constant. • Higher the value of stability constant and lower the value of dissociation constant, more stable will be the complex. • Moreover, the important applications of coordination compounds in various biological, metallurgical, analytical and medicinal chemistry, made the studies of coordination compounds in modern inorganic chemistry quite important. Coordination Compounds 169

Important reaCtIons 1. [Co(NH3)6]Cl3(aq) + 3AgNO3(aq) [Co(NH3)6] (NO3)3 + 3AgCl Ø (Yellow) White ppt Hexamminecobalt (III) chloride 2. [Co(NH3)5Cl]Cl2(aq) + 2AgNO3(aq) [Co(NH3)5Cl] (NO3)2 + 2AgCl Ø (Purple) White ppt Pentaamminechloridocobalt (III) chloride 3. [Co(NH3)4Cl2]Cl(aq) + AgNO3(aq) [Co(NH3)4Cl2] NO3 + AgCl Ø (Green) White ppt Tetramminodichloridocobalt (III) chloride 4. K4[Fe(CN)6](aq) 4K+ + [Fe(CN)6]4– Potassium hexacyanidoferrate(II) 5. FeSO4.(NH4)2 SO4.6H2O (aq) Fe2+ + 2SO42– + 2NH4+ + 6H2O Mohr’s salt 6. KCl.MgCl2.6H2O K+ + Mg2+ + 3Cl– + 6H2O Carnalite 7. K2SO4.Al2(SO4)3.24H2O 2K+ + 2Al3+ + 4SO42– + 24H2O Potash alum 8. [Ni(H2O)6]2+ + en(aq) [Ni(H2O)4(en)]2+ + 2H2O Hexaaquanickel(II) Pale blue Tetraaqua(ethane-1,2-diamine) 9. [Ni(H2O)4en]2+ + en(aq) nickel(II) [Ni(H2O)2 (en)2]2+(aq) + 2H2O Blue/purple Diaquabis(ethane-1,2-diamine) nickel(II) 10. [Ni(H2O)2 (en)2]2+ + en(aq) [Ni(en)3]2+ violet + 2H2O Tris(ethane-1,2-diamine) nickel(II) 11. Cu2+(aq) [Cu(NH3)4]2+ + 4NH3 Blue Tetraamminecopper(II) Common errors errors CorreCtIons (i) There is confusion among students while Practice for writing nomenclature and oxidation state of ligands. mentioning the name of a complex and oxidation state of central metal ion. It should always be noted that electron from the (ii) Students do not understand electronic valence shell will be lost first. configuration of ions. (iii) Students do not understand the concept of pairing In the presence of strong ligands like CO, CN–, of electrons. NenHer3gcyr.ysTtahlufsi,eldpasiprliintgtinogf is higher than pairing electrons will occur (iv) Students commit mistakes in drawing isomers because it is energetically favourable. of octahedral complexes with bidentate ligands such as en, ox. Draw octahedral structure first and then mark the ligands. (v) Students write wrong name of central metal ion in anionic complexes In anionic complexes, the name of metal ion ends with ‘ate’ e.g., Fe becomes ferrate, Cr becomes chromate Al becomes aluminate, Zn becomes zincate. 170 Chemistry-12

revIsIon Chart Coordination compounds are those addition molecular compounds which retain their identity in solid state as well as in dissolved state. For illustration, Coordination Sphere K4[Fe(CN)6] Ligand The entity showing central metal (The e–donating atom surrounded by ligands in Central Metal atom species to Fe, present around it in a i.e. the metal atom surrounded by CN– ions particular spatial arrangement square brackets. Coordination Number Counter Ion It represents the no. of ligand It balances the charge on the units present around central metal coordination sphere. atom/ion. Applications: Coordination are widely used in industries, scientific research, technology and in medicines. Coordination CoMPounds Ligands may be unidentate (i.e. one donor site), Arises due to difference in Arises due to different bidentate (i.e. two donor sites) or polydentate (i.e. structure of coordination spatial arrangement of more than two donor sites) compounds ligands Bidentate and polydentate ligands causes cyclisation around the central metal atom while forming the Isomerism coordinate bond. This leads to a ring structure, which is more state and the effect is known as chelate effect. Structural isomerism Stereoisomerism noMenClature: Coordination compounds/complexes Linkage isomerism Geometrical Optical are named as per IUPAC rules. Coordination isomerism isomerism isomerism Examples illustrating the rules followed is as follows: Ionisation isomerism Solvate isomerism Complex Name [CoCl2(en)2] Dichlorobis(ethane-1,2-diamine) cis-trans Cl cobalt(III) chloride form K3[Fe(C2O4)3] Potassium trioxalatoferrate(II) fac-mer form Homoleptic complexes are the ones in Only exhibited which the central metal atom is linked by one type of donor by chiral atom(s) of ligands. molecules Heteroleptic complexes are the ones in which central metal Laevorotatory atom is linked by more than one type of donor atom of Dextrorotatory ligands. Coordination Compounds 171

revIsIon Chart Stability of metal carbonyl complexes are higher due to synergic effect i.e., formation of M–C p bond by donation of d-electron pair of metal to p* orbital of C, in addition of s bond by donation of electrons from C to metal. • Expressed by stepwise stability constant (K) and overall stability constant (b) • bn = K1 × K2 .... Kn • ‘K’ depends upon–Nature and charge of metal atom, basic nature of ligands. Stability of coordination compounds Bonding in Coordination CoMPounds Werner’s Theory Valence Band Theory Crystal Field Theory Metals exhibit two types of • Empty d-orbitals of the • It considers the bonding valencies: between metal and ligand as central metal atom undergoes purely electrostatic. (a) Primary valency: hybridisation, depending • All the d-orbitals are degenerate in symmetric spherical field. • corresponding to oxidation upon the type of metal and no. of central metal atom; • However as the ligands approaching ligands and it (considered as point charges approaches the metal atom, • satisfied by anions, ionisable accept electrons from these the dx2–y2 and dz2-orbitals, and non-directional. which point directly in the line ligands to form coordinate of ligands raises higher up in energy to form eg set of orbitals (b) Secondary valency: bond. while dxy, dyz and dxz orbitals raises to lower energy, t2g-level. • corresponding to • When d-orbitals of (n–1)th shell • In case of tetrahedral field of coordination no. of central is used, inner orbital complex ligand, the d-orbitals crystal metal atom; is formed and if it is of nth shell field splitting is inverted and the crystal field splitting • satisfied by anionic or then outer orbital complex is energy is also smaller, i.e. neutral ligands; non- Dt = 4/9D0 ionisable and directional formed. • However, this theory fails to explain the colour of the complexes and quantitative interpretation of magnetic data. 172 Chemistry-12

Chapter trend—Based on past Years’ CBse exams ➣ It has been observed from this chapter that the weightage of the topics ‘Terminology, Nomenclature and Isomerism in Coordination Compounds’ and ‘Bonding in Coordination Compounds’ are maximum. Hence, these are the most important topics. ➣ From this chapter, generally 1 mark questions were asked from the topic ‘Terminology, Nomenclature and Isomerism in Coordination Compounds’. ➣ Most of the 2 and 3 marks questions asked from this chapter belong to the topic ‘Bonding in Coordination Compounds’. Questions For practice Very Short Answer Type Questions (1 Mark) 1. Write IUPAC name of the complex [CoCl2(en)2]+ . [CBSE 2017 (C)] 2. Write IUPAC name of the isomer of the following complex: [Co(NH3)5Cl] SO4 3. Write the formula for the following: Diamminechloridonitrito-N-platinum(II) 4. Write the formula for the following: Tetraammineaquachloridocobalt(III) nitrate [CBSE 2017] 5. Using IUPAC norms write the formulae for the following: (i) Sodium dicyanidoaurate(I) (ii) Tetraamminochloridonitrito-N-platinum(Iv) sulphate [CBSE 2017] 6. What type of isomerism is shown by the complex [Co(NH3]5 SCN]2+? 7. Arrange the following complex ions in increasing order of crystal field splitting en[eNrgCyE(RDT0) :E[xCermCpl6la]3r–], [Cr(CN)6)]3–, [Cr(NH3)6]3+ Short Answer Type Questions-I (2 Marks) 8. (i) Why is [NiCl4]2+ paramagnetic while [Ni(CN)4]2– is diamagnetic? [Atomic number of Ni = 28] (ii) Why are low spin tetrahedral complexes rarely observed? [CBSE 2017] 9. Write the hybridisation and shape of the following complex: (i) [FeF6]3– (ii) [Ni(CO)4] [Atomic number of Fe = 26, Ni = 28] 10. When a coordination compound CrCl3.6H2O is mixed with AgNO3, 2 moles of AgCl are precipitated per mole of the compound. Write (i) Structural formula of the complex. (ii) IUPAC name of the complex. [CBSE 2016] 11. (i) On the basis of crystal field theory, write the electronic configuration for d4 ion if D0 < P. (ii) Why are low spin tetrahedral complexes rarely observed? [CBSE 2015] Short Answer Type Questions-II (3 Marks) 12. gFeoormtheetrcicoaml pisleoxmieorno[fFteh(eenco)2mCpl2l]e+x, write the hybridisation type and magnetic behaviour. Draw one of the ion which is optically active. [Atomic number of Fe = 26] [CBSE 2017 (C)] 13. When a coordination compound CrCl3.6H2O is mixed with AgNO3 solution, 3 moles of AgCl are precipitated per mole of compound. Write (i) Structural formula of the complex (ii) IUPAC name of the complex (iii) Magnetic and spin property of complex. [CBSE 2017 (C)] 14. Explain how s[Ptatt(eNsHo3f)p2 lCatli2n] uamndin[Ptht(eNseHc3o)m6]Cpol4unwdilsl. differ in their electrolytic conductance. Give the hybridisation Coordination Compounds 173

assIgnment Time: 45 Minutes M.M.: 25 Multiple Choice Questions (MCQs) (1 Mark) 1. The charge (x) on the complex ion, [Co(NH3)2Cl4]x is : (i) +1 (ii) +3 (iii) –3 (iv) –1 2. The spin only magnetic moment of [MnBr4]2– is 5.9 BM. Predict the geometry of the complex. (i) sp3 (ii) dsp2 (iii) d2sp (iv) sp 3. Dissociation constant of coordination compounds is related to its formation constant as (i) Both are same (ii) Reciprocal to each other (iii) Dissociation constant is half of formation constant (iv) Dissociation constant is double of formation constant. Assertion-Reason Type Questions (1 Mark) In the following questions, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices on the basis of the above passage. (i) Assertion and reason both are correct statements and the reason is correct explanation of assertion. (ii) Assertion and reason both are correct statements but reason is not correct explanation for assertion. (iii) Assertion is correct statement but reason is wrong statement. (iv) Assertion is wrong statement but reason is correct statement. 4. Assertion: K3[Fe(CN6)] is more stable than K4[Fe(CN)6] Reason: Charge density on Fe3+ ion in K3[Fe(CN)6] is more than Fe2+ion in K4[Fe(CN)6] 5. Assertion: Complexes of MX6 and MX5L type (X and L are unidentate) do not show geometrical isomerism. Reason : Geometrical isomerism is not shown by complexes of coordination number 6. Very Short Answer Type Questions (1 Mark) 6. How many geometrical isomers are possible for the complex, Ni(CO)4? 7. What does ‘en’ represent? How many coordinate sites does it have? 8. Draw the structure of hexacarbonylchromium(0) Short Answer Type Questions-I (2 Marks) 9. Out of the following two coordination entities, which is chiral (optically active)? (i) cis-[CrCl2(ox)2]3– (ii) trans-[CrCl2(ox)2]3– 10. Out of CN– and CO which ligand form more stable complex with metal and why? [CBSE 2017] 11. Which of the following is more stable complex and why? [CBSE 2014] [Co(NH3)6]3+ and [Co(en)3]3+ Short Answer Type Questions-II (3 Marks) [Ti(H2O)6]3+. 12. Draw a crystal field splitting level diagram for the 3d orbital of titanium in the complex: Indicate what is meant by crystal field spitting and explain why [Ti(H2O)6]3+ is coloured? 13. How would you account for the following: N[Fie(C(CON)4)6p]3o–sisseswseeasktleytrpaahreadmraalgngeeotimc ewthryilew[hFilee(C[NNi)(6C]4N– i)s4]2d–iaismsaqgunaerteicp.lanar. (i) (ii) (iii) [Ni(CN)4]2– is diamagnetic while [NiCN4]2– is paramagnetic. Long Answer Type Questions (5 Marks) 14. Write down the IUPAC names for each of the following complexes and indicate the oxidation state, electronic configuration and coordination number. Also give the shape and magnetic moment of the complexes: (i) K[Cr(H2O)2(C2O4)2].3H2O (ii) [Co(NH3)5Cl].Cl2 (iii) [CrCl3(py)3] (iv) Cs[FeCl4] (v) K4[Mn(CN)6] Answers 1. (iv) 2. (i) 3. (ii) 4. (i) 5. (iii) 174 Chemistry-12

Topics covered 6 Haloalkanes and Haloarenes 6.1 Haloalkanes 6.2 Haloarenes 6.3 Polyhalogen Compounds C hapter map 175

Topic 1. Haloalkanes • Allylic: Carbon containing X have a carbon- carbon double bond just next to it e.g., Classification of Haloalkanes and Haloarenes Based on number of halogen atoms: CH2 = CH—CH2X, X • Monohaloalkane or monohaloarene: Compounds • Benzylic: Carbon containing X have an aromatic with one halogen atom only e.g., C2H5X, X (where, X is hologen atom like F, Cl, Br ring just next to it e.g., CH2X or I) Based on sp2 hybridised C—X bond: • Dihaloalkane or dihaloarene: Compounds with two halogen atom e.g., CH2X—CH2X, • Vinylic: Halogen atom bonded to sp2-hybridised X, X, X X carbon of a carbon-carbon double bond e.g., CH2 XX (para) = CH—X, X (ortho) (meta) • Polyhaloalkane or polyhaloarene: Compounds • Aryl: Halogen atom bonded to sp2-hybridised with three, four or more halogen atoms e.g., carbon of aromatic ring e.g., X CH2X—CH2X—CH2X, X XX Nomenclature • Common name of haloalkanes is derived by Based on sp3 hybridised C—X bond: naming the alkyl group followed by halide. • Primary (1°): Carbon containing X bonded with For aryl halides common and IUPAC name only one alkyl group e.g., CcoHn3t—aiCnHin2g—XX. is haloarenes. For dihalogen derivatives of • Secondary (2°): Carbon bonded haloarenes the prefix, o-, m- and p- are used in with only two alkyl groups e.g., common names. CH3—CHX—CH3, X • IUPAC name of haloalkanes is halosubstituted hydrocarbons. For dihalogen derivatives of • Tertiary (3°): Carbon containing X bonded with haloarenes, the numerals 1, 2; 1, 3 and 1, 4 are C|H3 used in IUPAC system. three alkyl groups e.g., CH3 — C|—CH3 X Formula Common name IUPAC name CH3Cl Methyl chloride Chloromethane CH3CH2CH2Cl 1-Chloroethane CH3–C|H–CH3 n-Propyl chloride 2-Chloropropane Isopropyl chloride Cl CH3–C|H–CH2–CH3 sec-Butyl chloride 2-Chlorobutane Cl CH3–C|H–CH2Cl Isobutyl chloride 1-Chloro-2-methylpropane CH3 Tert. butyl chloride 2-Chloro-2-methylpropane |CH3 Neopentyl chloride 1-Chloro-2,2-dimethylpropane CH3–C|–Cl CH3 |CH3 CH3–C|–CH2Cl CH3 176 Chemistry-12

Br Bromobenzene Bromobenzene m-Dibromobenzene 1,3-Dibromobenzene Br Br p-Dibromobenzene 1,4-Dibromobenzene Br o-Dibromobenzene 1,2-Dibromobenzene Br Sym-Tribromobenzene 1,3,5-Tribromobenzene Br o-Bromotoluene 1-Bromo-2-methylbenzene Br Benzyl bromide Bromophenylmethane Br Br Br Br CH3 CH2Br Preparation of haloalkanes Haloalkanes Free radical halogenation; HX, anhydrous. HX (Markovnikov Cl2/UV light ZnCl2 addition) or or or heat PX3 or HBr in the presence of peroxide Red P/X2 where X2 = Br2, I2 (Anti Markovnikov addition) or SOCl2 or PCl5 Alcohol (R-OH) Alkenes Hydrocarbon Halogen exchange method: goes to least substituted carbon atom, eg. R – X + NaI Æ R – I + NaX (Finkelstein reaction) CH3—CH = CH2 + HCl æÆ CH3—CHCl—CH3. R–Br + AgF Æ R – F + AgBr • Peroxide Effect: When HBr is added to (Swarts reaction) unsymmetrical alkenes in presence of peroxide, • Markovnikov’s Rule: When a polar compound is added to unsymmetrical alkene, negative part anti-Markovnikov’s addition takes place, eg. CH3CH = CH2 + HBr Peroxide CH3CH2CH2Br. haloalKanes and haloarenes 177

Physical Properties of Haloalkanes of the parent hydrocarbon of comparable • Nature of C–X bond in alkyl halides: X, i.e. molecular masses. halogen atom is more electronegative than carbon. ã Amongst Haloalkanes, the following trends So, the C–X bond is polar with C-atom possessing are observed: a partial positive charge and X possessing a Œ For the same alkyl group, the boiling point partial negative charge. of alkyl chlorides, bromides and iodides • Solubility: Although haloalkanes are polar in follows the order as: RI > RBr > RCl > RF, nature, yet they are practically very only slightly where R is an alkyl group. This order is soluble in water. This arises because haloalkanes followed because with increase in the size are not able to form that much strong bond of the halogen atom, the magnitude of van with water as in water-water molecules and in der Waal’s forces will increase. between haloalkane molecules. Œ In general, the boiling point of chloro, bromo and iodo derivatives of alkanes • Density: increases with increase in the number of ã Simple fluoro and chloroalkanes are lighter halogen atoms. than water while bromides, iodides and Œ For the same halogen atom, the boiling polychloro derivatives are heavier than point of haloalkanes increases with water: increase in the size or bulkiness of the alkyl group. ã R – I > R – Br > R–Cl; CCl4 > CHCl3 > CH2Cl2 Œ For isomeric alkyl halides, the boiling is the order of densities. point decreases with more branching in the hydrocarbon chain. This is because ã With increase in number of carbon atoms, branching decreases the surface area the density of molecules will increase. which decreases the magnitude of van der Waal’s forces of attraction consequently, ã The density of halogen ions increases in the boiling point of the branched chain the order: Fluoride < Chloride < Bromide < compounds is comparatively lesser than Iodide. the corresponding linear chain compounds. • Melting and Boiling points: Chemical Properties of Haloalkanes ã Molecules of organic halogen compounds • Nucleophilic substitution reaction: are generally polar. Due to their polarity R – X + aq. KOH ——Æ R – OH + KX as well as higher molecular mass than the R – X + NH3 ——Æ R – NH2 + HX parent hydrocarbon, the intermolecular R – X + KCN ——Æ R – CN + KX forces of attraction (dipole–dipole and van R – X + AgCN ——Æ R – NC + AgX der Waals) between the molecules are stronger in halogen derivatives than in case of parent alkanes. As a result, melting and boiling point of chlorides, bromides and iodides are considerably higher than those Mechanism: SN1 SN2 First order reaction: Rate = k[RX] Second order reaction: Rate = k[RX][Nu–] Racemic mixture is obtained Inversion of configuration takes place Two step reaction One step reaction Order: CH3X < 1° < 2° < 3° Order: CH3X > 1° > 2° > 3° Carried out in polar protic solvents like H2O, R-OH, Carried out in polar aprotic solvents like DMSO, etc. acetone, etc. R • Sn2: • SN1: Step I: R3CX l R ⊕ + X– R Nu– + R RR + X– Step II: RRX Nu X Nu R R RR b (Inversion of Ra R R configuration) ⊕ + Nu– → + • Primary allylic and benzylic hstaalbidileissafativoonu,rws hSiNl1e RR R Nu Nu R reactions, due to resonance R R (a) (± Racemic (b) tertiary alkyl halides predominantly favours SN2 mixture) reaction due to the stability of carbocation. 178 Chemistry-12

• Nucleophilic elimination reaction: Dehydro- greater number of alkyl groups attached to halogenation (b-elimination): doubly bonded carbon atoms. When an haloalkane having b-hydrogen atom • Reaction with metals: is heated with alcoholic solution of potassium Reaction with magnesium: Grignard’s reagent is hydroxide, there is elimination of hydrogen atom formed: from the b-carbon atom and a halogen atom from R – X + Mg dry ether RMgX the a-carbon atom. As a result, an alkene is formed as the product. Wurtz reaction: R – X + 2Na + X – R ——Æ R – R + 2 NaX • Saytzeff rule: The preferred product in dehydrohalogenation is the one which has Wurtz-Fittig reaction: R-X + 2 Na + X-Ar ——Æ R-Ar + 2 NaX ExErCisE 6.1 Multiple Choice Questions (MCQs) (1 Mark) 5. Select the name of the product formed by 1. Identify A, B, C and D: treatment of isobutylene with HBr in presence of peroxide. [CBSE Sample Paper 2020-21} (i) Isobutyl bromide (ii) tert – Butyl bromide (iii) Propane and methylbromide (iv) Isobutene 6. Which isomer of C4H9Cl is capable of showing optical isomerism? (i) CH3CH2CH2CH2Cl (ii) CH3CH2CHCH3 (i) A = C2H4, B = C2H5OH, C = C2H5NC, (iii) Cl Cl D = C2H5CN (iv) (CH3)3C–Br (ii) A = C2H5OH, B = C2H4, C = C2H5CN, 7. Name the product that will form when sec-butyl D = C2H5NC bromide is subjected to dehydrohalogenation. (iii) A = C2H4, B = C2H5OH, C = C2H5CN, (i) 2-Butene (ii) 1-Butene D = C2H5NC (iii) 2-Methylpropane (iv) 2,3-Dibromobutane (iv) A = C2H5OH, B = C2H4, C = C2H5NC, D = C2H5CN 8. Name the reagent used to convert methyl bromide into propyne. 2. Which isomer of the molecule with molecular (i) Propylsodium, CH3CH2CH2Na formula C4H9Br will have the lowest boiling point? (ii) Prop-1-en-1-ylsodium, CH3CH=CHNa Br (iii) Sodium acetalide, HC≡ CNa (i) CH3CH2CHCH3 (ii) Br (iv) But-1-yn-1-ylsodium, CH3CH2C ≡ CNa 9. Which of the following alcohols will yield the (iii) (CH3)3C–Br (iv) CH3CH2CH2CH2Br corresponding alkyl chloride on reaction with 3. 2-Bromo-2-methylpropane is allowed to react with concentrated HCl at room temperature? alcoholic KOH solution. The major product formed (NCERT Exemplar) is [CBSE 2022] (i) CH3CH2 CH2 OH (i) (CH3)3 C – OH (ii) (CH3)3 C – OC2H5 (ii) CH3CH2 −− CH −− OH (iii) (CH3)3 C – O–K+ (iv) (CH3)2 C = CH2 | CH3 (iii) 4. Select the name of the product formed by CH3CH2 −− CH −− CH2OH | treatment of isobutylene with HBr. CH3 (i) Isobutyl bromide C| H3 C −−OH (ii) tert – Butyl bromide (iv) CH3CH2 −− | (iii) Propane and methylbromide (iv) Isobutene CH3 haloalKanes and haloarenes 179

10. Which reagent will you use for the following 18. Out of CH3 CH CH2Cl and CH3 CH2 CH CH3, reaction? (NCERT Exemplar) moreCrHea3ctive Cl CH3CH2CH2CH3 which is towards SN1 reaction and CH3CH2CH2CH2Cl + CH3CH2CHClCH3 why? (i) Cl2/UV light (ii) NaCl + H2SO4 [AI 2017] (iii) Cl2 gas in dark 19. Among the isomers of pentane y(iCe5lHds12)a, which on photochemical chlorination single (iv) Cl2 gas in the presence of iron in dark monochloride. [AI 2017] X 11. Which of the following is an example of vic- X dihalide? (NCERT Exemplar) , which is an example of 20. Out of and (i) Dichloromethane (ii) 1,2-dichloroethane allylic halide? [CBSE 2017] (iii) Ethylidene chloride (iv) Allyl chloride 21. Which would undergo SN2 reaction faster Assertion Reason Type Questions (1 Mark) among the following pair: C6H5–CH2–CH2–Br and In the following questions a statement of assertion C6H5–CH(Br)–CH3. [AI 2015] followed by a statement of reason is given. Choose 22. Give IUPAC name of the following compound the correct answer out of the following choices. CH2 C—CH2Br. [Delhi 2012] (i) Assertion and reason both are correct CH3 23. Arrange the following in the order of decreasing statements and reason is correct explanation boiling points: for Assertion. (i) n-pentyl chloride (ii) Assertion and reason both are correct statements but reason is not correct (ii) isopentyl chloride explanation for Assertion. (iii) neopentyl chloride (iii) Assertion is correct statement but reason is 24. Under what conditions, 2-methylpropene can be converted into isobutyl bromide using hydrogen wrong statement. bromide as the reagent? (iv) Assertion is wrong statement but reason is 25. Arrange the following in the order of increasing density: correct statement. n-propyl chloride, n-propyl bromide, n-propyl 12. Assertion: The order of decreasing boiling point iodide. of different isomers of pentyl chloride is neopentyl chloride>isopentyl chloride>n-pentyl chloride. Reason: Boiling point decreases with branching. Short Answer Type Questions-I (2 Marks) 13. Assertion: The order of increasing density is 26. What is chirality of a compound? Give one n-propyl chloride <n-propyl bromide <n-propyl example. iodide. 27. Answer the following: Reason: Density of halogens increases as chlorine (i) Which of the following compounds can be < bromine < iodine. most easily hydrolysed by KOH and why? 14. Assertion: Thionyl chloride is preferred over phosphorus chlorides (tri and penta) for the CH3CHClCH2CH3 or CH3CH2CH2Cl (ii) W h i c h a m o n g t h e m u n d e r g o e s S N2 preparation of alkyl chlorides from alcohols. substitution reaction faster and why? Reason: Phosphorus chlorides give pure alkyl I Cl halides. or 15. Assertion: tert-Butyl bromide undergoes Wurtz reaction to give 2, 2, 3, 3-tetramethylbutane. [AI 2012] 28. Explain why vinyl halide is less reactive while Reason: In Wurtz reaction, alkyl halides react with sodium in dry ether to give hydrocarbon allyl halide is more reactive than alkyl halides? containing double the number of carbon atoms 29. Mention all the possible monochloro structural isomers that can be formed by free radical present in the halide. (NCERT Exemplar) monochlorination of (CH3)2CHCH2CH3. 16. Assertion: Dextro and laevo-isomers of 2-butanol 30. In the following pairs of halogen compounds, are difficult to separate from fractional distillation. which would undergo SN2 reaction faster? Reason: Both the isomers have very less (i) difference in their boiling point. Very Short Answer Type Questions (1 Mark) (ii) 17. Write the structure of 1-Bromo-4-chlorobut-2-ene. [Delhi 2009][NCERT] [Delhi 2017] 180 Chemistry-12

31. Why is sulphuric acid not used during the reaction CH3 CH CH2CH3 of alcohols with KI? [NCERT] (iii) (a) | (b) CH3CH2CH2CH2Br Br [NCERT] 32. Write the structures of different dihalogen derivatives of propane. [NCERT] 43. Among the isomeric alkanes of molecular formula 33. Arrange each set of compounds in the order of cCh5Hlor12i,niadteionntifyyietlhdes: one which on photochemical increasing boiling points. (i)Bromomethane, Bromoform, Chloromethane, (i) A single monochloride. Dibromomethane (ii) Three isomeric monochlorides. (ii)1 - C h l o r o p r o p a n e , I s o p r o p y l c h l o r i d e , 1-Chlorobutane [NCERT] (iii) Four isomeric monochlorides. 34. Among the following pairs of halogen compounds, [Delhi 2017] [NCERT] which compound undergoes faster SN1 reaction? 44. Which alkyl halide from the following pairs Cl Cl would you expect to react more rapidly by an SN2 mechanism? Explain your answer. (i) and (i) CH3CH2CH2Br or CH3—CH—CH2—CH3 Cl Br (ii) CH3 and Cl (ii) CH3CH2CHCH3 or H3C—C—Br [AI 2009][NCERT] 35. Write the isomers of the compound having Br CH3 molecular formula C4H9Br. [NCERT] (iii) CH3–CHCH2CH2Br or CH3CH2CHCH2Br 36. What are ambident nucleophiles? Explain with an example. [Foreign 2012, AI 2013][NCERT] CH3 CH3 37. Which compound among each of the following pairs will react faster in SN2 reaction with —OH? [AI 2011] [NCERT] (i) CH3Br or CH3I (ii) (CH3)3CCl or CH3Cl [Delhi 2014; AI 2010 (C), 2008][NCERT] 45. Write the structures of the following organic halogen compounds: 38. Write the mechanism for the following reaction: (i) 2-Chloro-3-methylpentane n-BuBr + KCN EtOH—H2O nBuCN (ii) p-Bromochlorobenzene [Delhi 2011 (C)][NCERT] (iii) 1-Chloro-4-ethylcyclohexane (iv) 2-(2-Chlorophenyl)-1-iodooctane 39. Carry out the following conversions: (v) 2-Bromobutane (i) 2-Methyl-1-propene to 2-chloro-2-methylpro- pane (vi) 4-tert-butyl-3-iodoheptane (ii) tert-butylbromide to isobutyl bromide (vii) 1-Bromo-4-sec-butyl-2-methylbenzene 40. The treatment of alkyl chlorides with aqueous (viii) 1, 4-Dibromobut-2-ene KOH leads to the formation of alcohols but in the [Delhi, AI 2011 (C)] [NCERT] presence of alcoholic KOH, alkenes are the major 46. Which one of the following has the highest dipole moment? product. Explain. [NCERT] [AI 2011] (i) CH2Cl2 (ii) CHCl3 41. Arrange the following haloalkanes in the order of increasing boiling point: (iii) CCl4 [NCERT] (i) (CH3)3C–Br (ii) (CH3)3C–Cl 47. A hydrocarbon C5H10 does not react with chlorine in dark but it gives a single monochloro compound, (iii) CH3CH2CH2CH2Br (iv) CH3CH(Br)CH2CH3 C5H9Cl in the presence of bright sunlight. Identify the hydrocarbon. [NCERT] Short Answer Type Questions-II (3 Marks) 42. Identify chiral molecules (Wedge and Dash 48. Write the equations for the preparation of 1-iodobutane from (i) 1-butanol, (ii) 1-chlorobutane representations) among the following pair of and (iii) but-1-ene. [NCERT] compounds. 49. Predict all the alkenes that will be formed by HH (i) (a) H3C (b) H3C dehydrohalogenation of the following halides with Br OH sodium ethoxide in ethanol and identify the major HO Br Br alkene: (ii) (a) H CH3 (i) 1-Bromo-1-methylcyclohexane H3C OH (b) H3C (ii) 2-Chloro-2-methylbutane H3C H (iii) 2,2,3-Trimethyl-3-bromopentane [NCERT] haloalKanes and haloarenes 181

50. Write the structures of the major organic product (iii) 1-Bromobutane, 1-Bromo-2, obtained in each of the following reactions: 2-dimethylpropane, (i) CH3CH2CH2Cl + NaI Acetone 1-Bromo-2-methylbutane, Heat 1-Bromo-3-methylbutane (ii) (CH3)3C–Br + KOH Ethanol Heat [Delhi 2017] [AI 2011] [NCERT] (iii) CH3CH(Br)CH2CH3 + NaOH water 52. Primary alkyl halide cCom4Hp9oBurnd(A(B) )r.eCaocmtspowuinthd (iv) CH3CH2Br + KCN aq. ethanol alcoholic KOH to give a (v) C6H5ONa + C2H5Cl (B) when reacted with HBr gives (C) which is an isomer of (D). When (A) is treated with sodium (vi) CH3CH2CH2OH + SOCl2 peroxide metal it gives compound (D), C8H18 which is (vii) CH3CH2CH==CH2 + HBr different from the compound formed when n-butyl bromide was treated with sodium. Give the (viii) CH3CH==C(CH3)2 + HBr [Delhi 2008 (C)] [NCERT] structural formulae of (A), (B), (C) and (D) and give the equations involved in all the reactions. 51. Arrange the compounds of each set in order of Long Answer Type Questions (5 Marks) reactivity towards SN2 displacement: (i) 2-Bromo-2-methylbutane, 1-Bromopentane, 53. Draw the structures of all the eight possible structural isomers having the molecular formula 2-Bromopentane ICU5HPA11BCrs. yNstaemme each isomer according to the (ii) 1-Bromo-3-methylbutane, and classify them as primary, 2-Bromo-2-methylbutane, secondary and tertiary bromides. [NCERT] 3-Bromo-2-methylbutane 54. Write IUPAC names of the following: 55. Draw the structures of the following compounds: [NCERT] [AI 2008, 09, 10, 11] (i) 2-Chloro-3-methylpentane (v) Propene to propyne (ii) 1-Chloro-4-ethylcyclohexane (vi) Ethanol to ethyl fluoride (iii) 4-tert.-butyl-3-iodoheptane (iv) 1, 4-Dibromobut-2-ene (vii) Bromomethane to propanone (v) 1-Bromo-4-sec.-butyl-2-methylbenzene. (viii) But-1-ene to but-2-ene [NCERT] 56. Identify A, B, C, D, E, R and R¢ in the following (ix) 1-Chlorobutane to n-octane reactions: (x) Benzene to biphenyl [NCERT] (i) (ii) 58. How the following conversions can be carried out? (i) Propene to propan-1-ol (iii) 57. How the following conversions can be carried out? (ii) Ethanol to but-1-yne (i) Ethanol to but-1-yne (iii) 1-Bromopropane to 2-Bromopropane (ii) Ethane to bromoethene (iii) Propene to 1-nitropropane (iv) Ethanol to propanenitrile (iv) Toluene to benzyl alcohol (v) 2-chlorobutane to 3,4-dimethylhexane 59. How the following conversions can be carried out? (i) Ethyl chloride to propanoic acid (ii) But-1-ene to n-butyl iodide (iii) 2-chloropropane to propan-1-ol (iv) Chloroethane to butane (v) 2-Bromopropane to 1-bromopropane 60. What happens when: (i) n-butyl chloride is treated with alcoholic KOH, 182 Chemistry-12

(ii) Bromobenzene is treated with Mg in the (v) Methyl bromide is treated with sodium in the presence of dry ether, presence of dry ether, (iii) Chlorobenzene is subjected to hydrolysis, (vi) Methyl chloride is treated with KCN? (iv) Ethyl chloride is treated with aq. KOH, [AI 2016 (C)] [NCERT] Answers 6.1 1. (i) 22. 3-Bromo-2-methylpropene. 2. (iii) Explanation: Boiling point decreases with The isomer with highest boiling point: 23. BCoHil3iCnHg 2pCoHin2tCdHe2cBrer.ases with branching thus, the branching. 3. (iv) order is: n-pentyl chloride > isopentyl chloride > 4. (ii) Explanation: It follows Markovnikov’s rule. 5. (i) neopentyl chloride H 24. In the presence of hydrogen peroxide, 2-methyl propene can be converted to form isobutyl 6. (ii) cEhxiprlaalncaatriboonn: )CH3CH2— C*— CH3 (*signifies bromide by treating it with HBr, as follows: Cl CH3 — CH = CH2 + HBr Peroxide → | 7. (i) Explanation: According to Saytzeff rule, CH3 highly substituted akene is favoured. CH3 — CH = CHBr 8. (iii) | 9. (iv) Explanation: Order of reactivity of alcohol is CH3 3° > 2° > 1°. Thus 3° alcohol reacts at room 25. n-propyl chloride < n-propyl bromide < n-propyl temperature while 1° and 2° needs catalyst iodide or higher temperature. 26. Compounds that will form non-superimposable 10. (i) mirror images are called as chiral compounds and 11. (ii) this property of non-superimposability is called 12. (iv) Explanation: Correct order is n-pentyl chirality. A carbon atom with 4 different atom(s) or groups of atom(s) shows chirality. For example, chloride > isopentyl chloride > neopentyl HH chloride 13. (i) Br C* I and H3C *C CH2CH3 14. (iii) Explanation: Thionyl chloride is preferred Cl OH because the other two products are escapable gases. Hence, the reaction gives pure alkyl Bromochloroiodomethane Butan-2-ol halides. 27. (i) 22-°Cahlkloyrlohbaultiadnee. ,Iti.ree. aCdHily3CuHnCdelCrgHo2eCsHio3niissaa- 15. (i) 16. (i) 17. BCrHC3H—2—CHC—H=C=HC2H——CCHH3 ,2Cl 18. tion to form a stable 2° carbocation, while 1-chlorobutane is 1° alkyl halide which forms a less stable 1° carbocation. Hence, Cl 2-chlorobutane undergoes hydrolysis more because 2° carbocation is more stable readily. than 1°. H CH3 CH3CHClCH2CH3 Ionisation CH3—C+—CH2CH3 –Cl– 19. CH3—C—CH3 + Cl2 hv 2-Chlorobutane 2° Carbocation (Stable) CH3 CH3 H CH3—C—CH2Cl + HCl OH– CH3—C—CH2CH3 OH x CH3 (ii) I– ion because of its bigger sBiuzteana-2n-odl lower 20. is example of allylic halide. electronegativity is a better leaving group than Cl– ion and C–I bond has lower bond 21. rCe6aHct5i—onCbHec2CauHs2eBtrhewreilils urensdoenragnoceesstfaabsitleisratSioNn2 dissociation energy than C–Cl bond, there- fore l reacts faster than of carbocation in case of C6H5CH(Br)CH3 so it action. Cl in a SN2 substitution re- prefers SN1 reaction. haloalKanes and haloarenes 183

28. In vinyl halides, C-X bond is stable due to the turn prevents the reaction between alcohol and resonance phenomenon (as in chlorobenzene). HI for the formation of an alkyl iodide. Hence, X cannot be replaced by any other atom, KI + H2SO4 —Æ KHSO4 + HI; 2HI + H2SO4 —Æ I2 + 2H2O + SO2 ion or molecule. Therefore vinyl halides is less To remove this difficulty, a non-oxidising acid such reactive than alkyl halides. as H3PO4 is used instead of H2SO4. H2C CH Cl – CH C+l CH3 CH2 OH + KI + H3PO4 Heat Ethanol H2C In allyl chloride, SN1 reaction is easier, since allyl CH3 CH2 I + KH2PO4 + H2O carbonium ion formed after the removal of Cl– ion Iodoethane is stablised by resonance. 32. (i) CH2—CH—CH3 1, 2-Dichloropropane + CH2 CH—CH2Cl CH2 CH— CH2 + Cl– + Cl Cl CH2— CH CH2 (ii) CH2—CH2—CH2 1, 3-Dichloropropane 29. 2-Methylbutane (isopentane) has four different 1, 1-Dichloropropane Cl Cl type of hydrogen atoms and replacement of any (iii) CH2CH2CHCl2 of these hydrogen atoms with chlorine atom will Cl give four isomers as shown below: (i) CH3—CH—CH2CH2Cl CH3 (iv) CH3—C—CH3 2, 2-Dichloropropane 1-Chloro-3-methylbutane Cl (ii) CH3—CH—CH—CH3 33. (i) The boiling point of these compounds increases with increase in their molecular masses. CH3 Cl chloromethane < bromomethane < dibromomethane < bromoform 2-Chloro-3-methylbutane (ii) The boiling point decreases as the branching Cl in the hydrocarbon chain increases. Isopropyl chloride < 1-Chloropropane < (iii) CH3—C—CH2CH3 1-Chlorobutane CH3 Cl 2-Chloro-2-methylbutane 34. (i) is more reactive towards SN1 reaction (iv) CH2—CH—CH2CH3 and its substitution is faster because it forms Cl CH3 1-Chloro-2-methylbutane 30. (i) being a primary halide a 3° carbocation intermediate which is more stable than 2° carbocation. + undergoes SN2 reaction faster than +> , which is a secondary halide. 3° (more stable) 2° (less stable) (ii) , undergoes faster SN1 Since iodide ion is a better leaving group, Cl because of its large size, therefore it will be (ii) released at a faster rate in the presence of the incoming nucleophile as compared to the reaction than Cl because of chloro-derivative. formation of a more stable 2° carbocation 31. H2SO4 is an oxidising agent. It oxidises HI produced during the reaction to form I2 and in intermediate, i.e. + 184 Chemistry-12 > + 2° (more stable) 1° (less stable)

35. Isomers of C4H9Br and their common and IUPAC 39. (i) CH3—C CH2 HCl names are as follows: (Markovnikov's CH3CH2CH2CH2Br ; CH3CH2CHBrCH3 addition) n-Butyl bromide sec.-butyl bromide CH3 Cl (1-Bromobutane) (2-Bromobutane) 2-Methylprop-1-ene (1° Bromide) (2° Bromide) CH3—C—CH3 CH3 CH3 CH3 2-Chloro-2-methylpropane CH3—CH—CH2Br ; CH3—C—Br CH3 CH3 Isobutyl bromide CH3 (ii) CH3—C—CH3 KOH(alc.) CH3—C—CH2 (1-Bromo-2-methylpropane) Heat tert.-butyl bromide 2-Methylpropene (1° Bromide) Br (major product) (2-Bromo-2-methylpropane) tert. butyl bromide (3° Bromide) (2-Bromo-2-methylpropane) CH3 36. The nucleophiles which have the ability to HBr CH3—CH—CH2Br attack through two different sites are called Peroxide ambident nucleophiles. An example is cyanide Isobutyl bromide group (CN–) because it can attack either (1-Bromo-2-methylpropane) through C or N atom because of the following 40. In aqueous solution, KOH is almost completely resonance structures that give rise to cyanide ionised to furnish OH– ions. OH– ion being a strong (R–CN) or isocyanide (RNC). nucleophile results into a substitution reaction with alkyl halides and leads to the formation of C– N C N– alcohols. At the same time in aqueous solution, the OH– ions are extensively hydrated (solvated). 37. (i) CH3I reacts faster than CH3Br in SN2 reaction Hydration of OH– ions further reduces the basic with –OH as I– ion is a better leaving group character of OH– ions and it therefore fails to abstract a proton (H+) from the b-carbon atom of than Br– ion. This is because of the large size of I– ion as compared to Br– ion. alkyl chloride to form an alkene. (ii) CH3Cl reacts faster than (CH3)3CCl, because In contrast, an alcoholic solution of KOH contains there is more steric hindrance in (CH3)3CCl alkoxide (RO–) ions which being a stronger base that arises due to the presence of three bulky- than OH– ions preferentially eliminates a molecule CH3 groups around the C-atom. of HCl from an alkyl chloride to form an alkene. 38. KCN is predominantly ionic and furnishes cyanide 41. (CH3)3C–Cl < (CH3)3C–Br < ions in the solution. CN– acts as a nucleophile, CH3CH(Br)CH2CH3 < CH3CH2CH2CH2Br which is an ambident nucleophile: 42. (i) C– N C N– Therefore, it can attack the carbon atom of C— (ii) Br bond in n-BuBr either through C or N atom. However, since C—C bond is more stable than (iii) C—N bond; thus the attack preferably occurs through carbon atom and n-butyl cyanide is CH3 formed as follows: 43. (i) H3C—C—CH3 KCN EtOH —H2O→ K+ + C− N CH3 H7C3 SN2 2, 2-Dimethylpropane C– N + C — Br Pathway All the hydrogen atoms are equivalent having HH four 1° carbon atoms and replacement of any of these H-atom will yield the same product. C3H7 H7C3 haloalKanes and haloarenes 185 NC ·····C····· Br NC — C + Br– HH HH

(ii) CaHP3eCnbtHan2Cec H2Cb H2Ca H3 (vi) C1 H3C2 H2C3 H—C4 H—C5 H2C6 H2C7H3 I C(CH3)3 Replacement of a, b and c hydrogen atoms will give three isomeric monochlorides. a bc d (iii) CH3 — CH — CH2— CH3 CH3 (vii) 2-Methylbutane (viii) CH2CH==CHCH2 Replacement of a, b, c and d hydrogen atoms Br Br yield four isomeric monochlorides. 46. (i) Cl 44. If the leaving group is same in different isomers H of a particular molecular formula, then reactivity of the isomers towards SN2 mechanism will C decrease with the increase in steric hindrance in the hydrocarbon chain. Cl H (i) CH3CH2CH2CH2Br is a 1° alkyl halide and µ = 1.62 D therefore it reacts faster than CH3CH2CH(Br) CH3 which is a 2° alkyl halide, having CIn—CCHl2Cbol2n, tdhseirsesrueilntafnortcdeidpobley moment of two comparatively more steric hindrance. the resultant (ii) CH3CH2CH(Br)CH3 is a 2° alkyl halide and dipole moment of two C—H bonds and therefore it reacts faster than (CH3)3CBr therefore it has the highest dipole moment which is 3° alkyl halide, having comparatively value (m = 1.62 D) than CHCl3 and CCl4. more steric hindrance than a 2° alkyl halide. (iii) CH3CHCH2CH2Br (I) and Cl Cl CH3 (ii) C CH3CH2CHCH2Br (II) Cl H CH3 µ = 1.03 D both are 1° alkyl halides. But in (II) compound dInipColHe Cmlo3,mtehnetrseissuoltpapnotseodf two C—Cl bond the CH3 group is at the C-atom which is by the resultant more closer to Br (which exerts more steric dipole moments of C—Cl and C—H bonds. hindrance) to the attacking nucleophile at the Since, the latter resultant dipole moment C-atom as compared to compound (I) having is smaller than the former, therefore the CH3 group at C-atom, which is away from the molecule on the whole has been found to have Br atom. a dipole moment (m = 1.03 D). Thus, (I) will react faster than (II). 45. (i) CH3—CH—CH—CH2—CH3 Cl Cl (iii) C Cl Cl µ = 0.0 D zCeCrol4 is a symmetrical molecule and it has dipole moment. 47. (i) The hydrocarbon with molecular formula C5H10 can either be an alkene or a cycloalkane. Cl CH3 (ii) Since, the hydrocarbon does not react with Ccaln2 ninotthbee dark, therefore it indicates that it an alkene. (iii) The h ydrocarbon wil l rseinagctlewmitohnoCchll2oirno bright sunlight to give a (ii) (iii) cdoemrivpaotuinved,,Ci.5eH. c9yCcll.oTpheenrteafnoreehiat vaisnygmamll eittsritceanl (iv) H-atoms identical. Cl (v) CH3—CH—CH2—CH3 + Cl2 Sunlight + HCl Br Cyclopentane Monochloro 186 Chemistry-12 (C5H10) Cyclopentane (C5H9Cl)

48. (i) CH3CH2CH2CH2OH I2, P CH3 Butan-1-ol CH3CH2CH2CH2I (ii) CH3 C CH3 + KOH Ethanol, heat (Dehydrohalo- 1-lodobutane nenation) NaI Br Acetone, heat (ii) CH3CH2CH2CH2Cl 2-Bromo-2-methylpropane 1-Chlorobutane CH3CH2CH2CH2I CH3 (iii) CH3CH2CH==CH2 + HBr Peroxide (iii) CH3 CH3 C CH2 + KBr + H2O But-1-ene 2-Methylpropene KI Water (CH3OH) CH CH2CH3 + NaOH (Hydrolysis) CH3CH2CH2CH2Br CH3CH2CH2CH2I Br 1-Bromobutane 2-Bromobutane CH3 CH CH2CH3 + NaBr 49. (i) 1-Bromo-1-methylcyclohexane has two OH equivalent b-hydrogen atoms, hence only one type of alkene is formed. Butan-2-ol (iv) CH3CH2Br + KCN Ethanol (aq.) (Substitution Bromoethane reaction) CH3CH2CN + KBr Propanenitrile (v) C6H5O– Na+ + C2H5Cl Williamson's synthesis Sodium phenoxide CH3 C6H5 O C2H5 + NaCl (ii) H3C C CH2 CH3 Alc. KOH or Ethylphenyl ether C2H5ONa/C2H5OH (Phenetole) (vi) CH3CH2CH2OH + SOCl2 Pyridine (Substitution Cl Propan-1-ol reaction) 2-Chloro-2-methylbutane CH3CH2CH2Cl + HCl + SO2 CH3 CH3 1-Chloropropane H3C C CH CH3 + H2C C CH2CH3 (vii) CH3CH2CH CH2 + HBr But-1-ene 2-Methylbut-2-ene(A) 2-Methylbut-1-ene(B) Peroxide CH3CH2CH2CH2Br (Major product) (Minor product) (Anti-Markovnikov 1-Bromobutane addition) Since, the alkene (A) is more substituted, Markovnikov's thus according to Saytzeff’s rule it is will be (viii) CH3CH C CH3 + HBr addition more stable and it will be the major product. CH3 reaction C| H3 C| H3 2-Methylbut-2-ene Br (iii) H3C— C — C — CH2CH3 CC22HH55OONHa→ CH3CH2 C CH3 | | CH3 Br 2,2,3-Trimethyl-3-bromopentane CH3 2-Bromo-2-methylbutane CH3 CH3 51. sTtheericrehainctdivraitnyceoffaScNt2orr.eMaoctrieotnhdeespteernicdshiunpdornantchee, CH3 C C CHCH3 + lesser will be the reactivity. Therefore the CH3 reactivity of different alkyl halides towards SN2 reaction follows the order: 1° > 2° > 3°. 3, 4, 4-Trimethylpent-2-ene (Major product, more stable) (i) CH3CH2CH2CH2CH2Br > CH3 CH2 1-Bromopentane (1°) CH3 C C CH2CH3 CH3 — CH —CH2CH2CH3 > | Br CH3 2-Bromopentane (2°) 2-Ethyl-3, 3-dimethylbut-1-ene C| H3 CH3 — C| —CH2CH3 (Minor product, less stable) Br 50. (i) CH3CH2CH2Cl + NaI acetone heat 1-Chlorobutane CH3CH2CH2I + NaCl 2-Bromo-2-methylbutane (3°) 1-Iodopropane haloalKanes and haloarenes 187

(ii) CH3 — CH— CH2CH2Br > 53. (i) CH3CH2CH2CH2CH2Br | CH3 1-Bromopentane (primary) 1-Bromo-3-methylbutane (1°) (ii) CH3—C* H—CH2CH2CH3 CH3 — CH— CH— CH3 > Br | | CH3 Br 2-Bromopentane (secondary) 2-Bromo-3-methylbutane (2°) (iii) CH3CH2CH—CH2CH3 C| H3 Br CH3 — C| —CH2CH3 3-Bromopentane (secondary) Br (iv) CH3CH2*CH—CH2Br 2-Bromo-2-methylbutane CH3 (3°) 1-Bromo-2-methylbutane (primary) (iii) CH3CH2CH2CH2Br > 1-Bromobutane (1°) (v) CH3—CH—CH2CH2Br CH3 — CH— CH2CH2Br > | CH3 CH3 1-Bromo-3-methylbutane 1-Bromo-3-methylbutane (primary) (1° with oβn-peo—siCtiHon3) group at CH3 CH3 —CH2 — CH— CH2Br > (vi) CH3CH2—C—CH3 | CH3 Br 1-Bromo-2-methylbutane (1° with αon-peo—siCtiHon3 )group 2-Bromo-2-methylbutane (tertiary) at (vii) CH3—CH—*CH—CH3 C| H3 CH3 Br CH3 — C| —CH2CH3 Br 2-Bromo-3-methylbutane (secondary) 2-dimethylpropane CH3 (1° witahttαw-op—osCitHio3n)groups (viii) CH3—C—CH2Br 52. There are be only two possible primary alkyl CH3 bromides having the molecular formula, C4H9Br. There are: 1-Bromo-2,2-dimethylpropane (primary) (A) must be isobutyl bromide because when it is Among them, optically active isomers are (ii), (iv) treated with sodium, it forms a compound different and (vii). than the one produced by n-butylbromide which 54. (i) 4-Bromopent-2-ene when reacts with sodium will form n-octane. (ii) 3-Bromo-2-methylbut-1-ene (iii) 4-Bromo-3-methylpent-2-ene CH3—CH—CH2Br KOH(alc.) CH3—C CH2 (iv) 1-Bromo-2-methylbut-2-ene (v) 1-Bromobut–2-ene CH3 CH3 (vi) 1-Bromo-2-methylprop-2-ene (A) (B) 55. (i) Isobutyl bromide 2-Methylpropene (ii) Na HBr CH3CHCH2CH2CHCH3 Br CH3 CH3 CH3—C—CH3 (D) CH3 2, 5-Dimethylhexane (C) tert-Butyl bromide [Isomer of (A)] 188 Chemistry-12

(iii) CH3 7 57. (i) (ii) 1 H3C—C—CH3 26 (iii) (iv) (v) 4 35 I Br CH3 CH3—CH—C2H5 56. (i) A = MgBr B= Cyclohexyl mag. bromide Cyclohexane (iv) (v) (ii) R—Br = CH3 — CH — CH3 Here, (vi) (vii) Br (viii) R = CH3 — CH — CH3 and (ix) Isopropyl group C = CH3 — CH — MgBr CH3 Isopropyl magnesium bromide (iii) R¢—X=(CH3)3C—X or R¢=(CH3)3C— Tertiary butyl group and D = (CH3)3C—MgX, E = (CH3)3C—H 2-Methylpropane Tert. butyl magnesium halide It should however be noted that in general tert alkyl halides does not undergo Wurtz reaction. It undergoes dehydrohalogenation reaction to form an alkene under the normal reaction conditions, as shown below: CH3 CH3 H3C—C = CH2 Na/ether H3C—C—X Mg (–HX) 2-Methylpropene CH3 Tertiary butyl halide CH3 CH3 H3C—C—MgX H2O H3C—C—H + Mg(X)OH CH3 CH3 (D) (E) haloalKanes and haloarenes 189

(x) (ii) (iii) 58. (i) (iv) (ii) (v) 60. (i) (iii) (ii) Br MgBr (iv) (v) + Mg Dry ether 59. (i) Bromobenzene Phenylmagnesium bromide (iii) Under Normal conditions, chlorobenzene is highly stable towards nucleophilic substitution reactions. However, at 623 K and 300 atm in aqueous NaOH it hydrolysis to phenol. Cl OH + NaOH(aq.) 623 K 300 atm Phenol (iv) (v) (vi) CH3—Cl + KCN Heat CH3CN + KCl (Substitution Methyl Ethane- reaction) nitrile Chloride Topic 2. Haloarenes In this reaction, halogen atom is substituted by replacing hydrogen atom of benzene ring. Here, Preparation of Haloarenes halogen atom hdaoleosanlkoat nseu.bFsotirtuthteisofnor—mCatHio3na,ltkhyel • By electrophilic substitution reaction: group to form CH3 Fe CH3 CH3 reaction would be as follows: dark + + X2 CH3 CH2X XX + X2 + HX 383 K o-Halotoluene p-Halotoluene sunlight 190 Chemistry-12

• Sandmeyer’s reaction: N2+ X – • Dow’s process: At 623 K, 300 atm aq. NaOH hydrolyse chlorobenzene. NH3 Cl OH NaNO2 + HX 273–278 K For X = Cl, Br Benzene diazonium (i) NaOH, 623 K, 300 atm halide (ii) H + N2+ X – Cu2X2 X • When an electron withdrawing group is present + N2 at ortho- and para- positions the reactivity of haloarenes increases. Aryl halide For X = I Cl NO2 (i) NaOH, 443 K HO NO2 I N2+ X – KI, warm (ii) + H NO2 NO2 • Balz-Schiemann reaction: – Cl NO2 Warm N2+ 4 H2O N2+ X – BF F HO NO2 + HBF4 Physical Properties of Haloarenes NO2 NO2 Nature of C–X Bond in Aryl halides Cl The C–X bond is polarised with sp2-hybridised C-atom • Electrophilic substitution: having partial positive charge attached to halogen ã Halogenation: atom having a partial negative charge. Melting and boiling points of haloarenes: Cl Cl • Melting and boiling points of haloarenes are nearly + Cl2 Anhyd. FeCl3 Cl the same as those of alkyl halides containing the same number of carbon atoms. + • The boiling points of monohalogen derivatives of Cl 1, 2-Dichlorobenzene benzene are in the order: (Minor) iodo > bromo > chloro > fluoro 1, 4-Dichlorobenzene ã For the same halogen atom, the melting and (Major) boiling point increases as the size of the aryl group increases. ã Nitration: ã The melting point of para isomer is quite higher than that of ortho or meta isomers. Cl Cl Cl This is due to the fact that para isomer has symmetrical structure and therefore its HNO3 NO2 molecules fits better in the crystal lattice. As conc. H2SO4 a result, intermolecular forces of attraction + are stronger and therefore greater energy is required to break its lattice and it melts at a 1-Chloro-2-nitrobenzene NO2 higher temperature. (Minor) 1-Chloro-4-nitrobenzene (Major) ã Sulphonation: Cl Cl Cl SO3H conc. H2SO4 ∆ + Chemical Properties of Haloarenes 2-Chlorobenzene SO3H • Nucleophilic substitution: Due to resonance sulphonic acid effect in haloarenes, C—X bond acquires double (Minor) 4-Chlorobenzene bond character which makes it difficult to cleave resulting haloarenes less reactive towards ã Friedel-Crafts alkylation: sulphonic acid nucleophilic sustitution. Also, due to this effect haloarenes are o and p–directing wrt to the halogen (Major) atom in electrophilic substitution reactions. This resonance stabilisation is explained as follows: X X+ X+ X+ – – – haloalKanes and haloarenes 191

ã Friedel-Crafts acylation: ã Fittig reaction: Cl 2 + 2Na Ether + 2NaCl Diphenyl ã Wurtz-Fittig reaction: XR + 2Na + RX Ether + 2NaX ExErCisE 6.2 Multiple Choice Questions (MCQs) (1 Mark) 5. Toluene reacts with a halogen in the presence 1. Choose the name of the product formed when of iron (III) chloride giving ortho and para halo benzene is reacted with excess of chlorine gas in the presence of sunlight. compounds. The reaction is (i) Benzyl chloride (NCERT Exemplar) (ii) Cyclohexyl chloride (i) Electrophilic elimination reaction (iii) para-Dichlorobenzene (ii) Electrophilic substitution reaction (iv) Benzene hexachloride (iii) Free radical addition reaction 2. Name the method used for the preparation of haloarenes from diazonium salts as shown in the (iv) Nucleophilic substitution reaction following reaction: 6. Which of the following is halogen exchange Ar − N2+Cl− CuX/HX→ Ar − X (where, X = Cl or Br) reaction? (NCERT Exemplar) (i) Balz-Schiemann reaction (i) RX + NaI RI + NaX (ii) Dow’s process (ii) C C + HX CC (iii) Fittig reaction (iii) R OH + HX ZnCl2 HX R X + H2O (iv) Sandmeyer’s reaction (iv) 3. Name the reagent used for producing fluorobenzene through benzene diazonium chloride salt. 7. Arrange the following compounds in the (i) Hydrogen fluoride (ii) Fluoroboric acid increasing order of their densities. (NCERT Exemplar) (iii) Copper difluoride (iv) Copper fluoride Cl 4. Identify the compound Y in the following reaction. (NCERT Exemplar) Cl (a) (b) (i) (ii) Cl Br Cl (c) (d) Cl Cl (iii) (iv) Cl (i) (a) < (b) < (c) < (d) Cl Cl (ii) (a) < (c) < (d) < (b) (iii) (d) < (c) < (b) < (a) Chemistry-12 (iv) (b) < (d) < (c) < (a) 192

8. Chlorobenzene is formed by reaction of chlorine 12. Arrange the following compounds in increasing with benzene in the presence of AbelCnzl3e.nWe hriinchg of the following species attacks the in order of rate of reaction towards nucleophilic this reaction? (NCERT Exemplar) substitution. (NCERT Exemplar) (i) Cl– (ii) Cl+ Cl Cl (iii) AlCl3 (iv) [AlCl4]– NO2 9. What is ‘A’ in the following reaction? (a) (b) (NCERT Exemplar) CH2 CH CH2 Cl + HCl A (c) CH2 CH CH2 NO2 (i) (a) < (b) < (c) (ii) (c) < (b) < (a) (i) Cl (iii) (a) < (c) < (b) (iv) (c) < (a) < (b) 13. Arrange the following compounds in increasing order of rate of reaction towards nucleophilic CH2 CH2 CH2 Cl substitution. (NCERT Exemplar) (ii) Cl Cl (a) (b) CH3 CH2 CH CH3 Cl (iii) Cl (c) Cl CH3 (ii) (a) < (c) < (b) (i) (a) < (b) < (c) CH CH2 CH3 (iv) (iii) (c) < (b) < (a) (iv) (b) < (c) < (a) 14. Arrange the following compounds in increasing order of rate of reaction towards nucleophilic 10. The reaction of toluene with chlorine in the substitution. (NCERT Exemplar) presence of iron and in the absence of light yields Cl Cl .................... . (NCERT Exemplar) CH2Cl CH3 (a) (b) Cl (i) (ii) Cl NO2 NO2 O2N (iii) H3C Cl (c) (iv) Mixture of (ii) and (iii) 11. Reaction of C6H5CH2Br with aqueous sodium hydroxide follows. (NCERT Exemplar) NO2 (i) SN1 mechanism (i) (c) < (b) < (a) (ii) (b) < (c) < (a) (ii) SN2 mechanism (iii) Any of the above two depending upon the (iii) (a) < (c) < (b) (iv) (a) < (b) < (c) temperature of reaction 15. Arrange the following compounds in increasing order of rate of reaction towards nucleophilic (iv) Saytzeff rule substitution. (NCERT Exemplar) haloalKanes and haloarenes 193

Cl Cl 19. Assertion : Presence of a nitro group at ortho (b) or para position increases the reactivity of CH3 haloarenes towards nucleophilic substitution. (a) Reason: Nitro group, being an electron Cl withdrawing group decreases the electron density over the benzene ring. (NCERT Exemplar) (c) 20. Assertion: Nitration of chlorobenzene leads to CH3 the formation of m-nitrochlorobenzene Reason: —NO2 group is a m-directing group. (NCERT Exemplar) 21. Assertion: Inmonohaloarenes,furtherelectrophilic substitution occurs at ortho and para positions. CH3 Reason: Halogen atom is a ring deactivator. (i) (a) < (b) < (c) (ii) (b) < (a) < (c) (NCERT Exemplar) (iii) (c) < (b) < (a) (iv) (a) < (c) < (b) 22. Assertion : Aryl iodides can be prepared by reaction of arenes with iodine in the presence of 16. Which is the correct increasing order of boiling an oxidising agent. points of the following compounds? Reason : Oxidising agent o(xNiCdiEsResTIE2 xienmtoplHarI). 1-Bromoethane, 1-Bromopropane, 1-Bromobutane, Bromobenzene (NCERT Exemplar) 23. Assertion: It is difficult to replace chlorine by (i) Bromobenzene –OH in chlorobenzene in comparison to that in < 1-Bromobutane < 1-Bromopropane chloroethane. < 1-Bromoethane Reason: Chlorine-carbon (C—Cl) bond in (ii) Bromobenzene < 1-Bromoethane chlorobenzene has a partial double bond character < 1-Bromopropane < 1-Bromobutane due to resonance. (NCERT Exemplar) (iii) 1-Bromopropane < 1-Bromobutane Very Short Answer Type Questions (1 Mark) 24. Write the name of the product formed when < 1-Bromoethane < Bromobenzene benzene is reacted with excess of chlorine gas in (iv) 1-Bromoethane < 1-Bromopropane the presence of sunlight. 25. Give an example of Wurtz-Fittig reaction. < 1-Bromobutane < Bromobenzene 26. Explain the method used for the preparation of haloarenes from diazonium salts. Assertion Reason Type Questions (1 Mark) In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. 27. Name the reagent used for producing fluorobenzene (i) Assertion and reason both are correct through benzene diazonium chloride salt. statements and reason is correct explanation 28. Name the reagents required for one step conversion for Assertion. of chlorobenzene into toluene. (ii) Assertion and reason both are correct Short Answer Type Questions-I (2 Marks) 29. What happens when chlorine is passed through statements but reason is not correct boiling toluene in the presence of sunlight? explanation for Assertion. 30. (i) Arrange the following in the increasing (iii) Assertion is correct statement but reason is order of reactivity towards nucleophilic wrong statement. substitution reactions: Chlorobenzene; 4-ni- (iv) Assertion is wrong statement but reason is trochlorobenzene; 2,4-dinitrochlorobenzene; correct statement. 2, 4, 6-trinitrochlorobenzene. 17. Assertion: The reagents required for one step conversion of chlorobenzene into toluene are (ii) Arrange the following in the order of increasing reactivity towards sulphonation CH3Cl, Na, dry ether. reaction when treated with fuming sulphuric Reason: The one step conversion of chlorobenzene acid: Benzene, Toluene, Chlorobenzene, into toluene is the Fittig reaction. Methoxybenzene. 18. Assertion: 4-Nitrochlorobenzene requires higher temperature to replace chloro group 31. Out isofmCo6rHe 5eCaHsi2lCy lhyadnrdolCys6Hed5CbHy CaqlCue6Hou5s, which one KOH? with hydroxyl group in Dow’s process while [NCERT] 2,4,6-trinitrochlorobenzene requires only warm Short Answer Type Questions-II (3 Marks) water. 32. Explain why haloarenes are comparatively Reason: When an electron withdrawing group is less reactive towards nucleophilic substitution present at ortho- and para- positions the reactivity of haloarenes increases. reactions when compared with haloalkanes? [AI 2009 (C), Delhi 2008] 194 Chemistry-12

OR Long Answer Type Questions (5 Marks) Haloalkanes undergo nucleophilic substitution 41. Even though chlorine is an electron withdrawing group, yet it is ortho-para-directing during reactions preferably, while haloarenes undergoes electrophilic aromatic substitution reactions. electrophilic substitution reactions. Explain. Why? [NCERT] [AI, Delhi 2012] 33. Account for the following: 42. Draw the structures of the major monohalo (i) Benzylic halides are highly reactive towards SN1 reaction. products in each of the following reactions: (ii) KC6OHH5CsHolCultCioHn3rcaatnhebre easily hydrolysed with (i) than C6H5CH2Cl. 34. Write the products formed in the following reactions: (ii) (i) H + HBr H H (ii) CH3CH2CH==CH2 + HCl (iii) (iii) CH2—C CH2 + HBr Peroxide H (iv) 35. Predict the order of reactivity of the following co(im) pTohuendfosutroiwsoamrdesrSicNb1raonmdobSuN2tarneeasctions: (ii) tBCe6rHtah5nCedHCI2BU6HrP,5CCA(6CCHH5nC3a)H(mC(C6eH6sH5)5oB)fBr.rt,hCe6Hf o5Cl lHo(wCiHn3g) (v) CH3CH2Br + NaI 36. Wri (vi) compounds: (i) CH3CH(Cl)CH(Br)CH3 (ii) CCHlCFH2C2CB∫r∫CClCFH2Br (iii) [AI 2014, 2009] [NCERT] (iv) (CCHC3lC3)(3pC-CCllC6H4)2CH(Br)CH3 43. Name the following halides according to the (v) IUPAC system and classify them as alkyl, allyl, (vi) (CH3)3C—CH==CClC6H4I –p [NCERT] benzyl (primary, secondary, tertiary), vinyl or aryl Explain why: 37. halides: (i) The dipole moment of chlorobenzene is lower (i) (CH3)2 CHCH(Cl)CH3 (ii) CH3CH2CH(CH3)CH(C2H5)Cl than that of cyclohexyl chloride? (ii) Alkyl halides, are immiscible with water (iii) CH3CH2C(CH3)2CH2I even though they are polar? (iv) (CH3)3CCH2CH(Br)C6H5 (iii) Grignard reagents are always prepared (v) CH3CH(CH3)CH(Br)CH3 under anhydrous conditions? (vi) CH3C(C2H5)2CH2Br [NCERT] [AI 2011, 2010 (C)] (vii) CH3C(Cl)(C2H5)CH2CH3 38. p-Dichlorobenzene has higher melting point and solubility than those of the corresponding o-and (viii) CH3CH==C(Cl)CH2CH(CH3)2 m-isomers. Discuss. [AI 2009, 2008] [NCERT] 39. Compound (A), having an empirical formula, (ix) CH3CH==CHC(Br)(CH3)2 (x) p-ClC6H4CH2CH(CH3)2 tCo7Hfo8rmis chlorinated in the presence of sunlight (xi) m-ClCH2C6H4CH2C(CH3)2 product (B), which on hydrolysis gives a (xii) o-BrC6H4CH(CH3)CH2CH3 [NCERT] compound (C). compound (C) gives a positive test with Tollen’s reagent. Identify A, B and C. 44. How will you carry out the following conversions? 40. Name the products formed in the following reactions: (i) Toluene to Benzyl alcohol (i) Nitration of chlorobenzene (ii) Benzene to 4-bromonitrobenzene (ii) Sulphonation of bromobenzene (iii) Benzyl alcohol to 2-phenylethanoic acid (iii) Chlorination of chlorobenzene (iv) Aniline to Chlorobenzene (iv) Friedel-Crafts acylation of chlorobenzene with (v) Chlorobenzene to p-nitrophenol acetyl chloride [AI 2014, Foreign 2011] haloalKanes and haloarenes 195

45. Complete the following chemical reactions: (iv) (i) (v) (vi) (ii) (iii) (vii) Answers 6.2 1. (iv) 2. (iv) 8. (ii) Explanation: Cl – Cl + AlCl3 → [AlCl4]– + Cl+ 3. (ii) Explanation: (Attacks as electrophile)  BF4−  + X− + HBF4 → Ar + 2 → Ar − F 9. (iii) Explanation: In this reaction addition of HCl takes place on doubly bonded carbon atom Ar N2 N in accordance with Markovnikov’s rule i.e., addition of negative species will take place on 4. (i) Explanation: The Cl–, Br– and CN– the carbon which has lesser number of hydrogen. nucleophiles can easily be introduced in the benzene ring of diazonium salts in the 10. (iv) Explanation: presence of Cu(I) ion. CuCl / HCl ArCl + N2 ArN+ 2X– CuBr / HBr ArBr + N2 CuCN / KCN ArCN + N2 11. (i) Explanation: Greater the stability of 5. (ii) Explanation: Aryl chlorides and bromides carbocation, greater will be its ease of can be easily prepared by electrophilic sub- stitution of arenes with chlorine and bromine respectively in the presence of Lewis acid catalysts like iron or iron(III) chloride. formation from alkyl halide and faster will be the rate of reaction. In case of alkyl vhearlyidefass, t30baelckayulshealoifdethseunhdigehrgsotSaNb1ilriteyacotfio3n0 carbocations. For the same reasons, allylic and benzylic halides show high reactivity towards the 6. (i) Explanation: Alkyl iodides are often SN1 reaction. The carbocation thus formed prepared by the reaction of alkyl chlorides/ gets stabilised through resonance as shown bromides with NaI in dry acetone. This reaction is known as Finkelstein reaction. below: ⊕ ⊕ NaCl or NaBr thus formed is precipitated in dry acetone. It facilitates the forward CH2 CH2 reaction according to Le Chatelier’s Principle. CH2 CH2 7. (i) Explanation: Density is related to molecular ⊕⊕ mass. Higher the mass, higher will be the density of the molecule. 196 Chemistry-12

12. (iii) Explanation: The presence of an electron 27. Fluoroboric acid withdrawing group (—NO2) at ortho- and para- 28. CH3Cl, Na, Dry ether; positions increases the reactivity of haloarenes. CH3 Cl The presence of nitro group at ortho- and para- positions withdraws the electron density from + 2Na + CH3Cl Dry ether + 2NaCl the benzene ring and thus facilitates the attack of the nucleophile on haloarene. The carbanion (Wurtz-Fittig reaction) thus formed is stabilised through resonance. 29. Benzyl chloride is the product formed (C6H5CH2Cl). The negative charge appeared at ortho- and para- positions with respect to the halogen substituent is stabilised by —NO2 group. 13. (iv) Explanation: The presence of electron releasing group at ortho-and para-positions 30. (i) Chlorobenzene < 4-nitrochlorobenzene decreases the reactivity of haloarenes. < 2,4-dinitrochlorobenzene < Because of the possible repulsion, it is less 2,4,6-trinitrochlorobenzene likely for the electron rich nucleophile to (ii) Benzene < Toluene < Chlorobenzene < approach electron rich arenes. Methoxybenzene 14. (iv) Explanation: More the electron withdrawing 31. Since, the carbocation C6 H5 CG+HC6 H5 (2°) is group is attached greater will be the rate of more stable than the carbocation C6 H5 CG+H2 (1°), therefore, C6H5CH(Cl)C6H5 is more readily reaction. hydrolysed than C6H5CH2Cl under SN1 reaction conditions. 15. (iii) Explanation: More the electron releasing group is attached lesser will be the rate of reaction. C6 H5 CG +H2 + Cl– 16. (iv) Explanation: The attractions get stronger C6H5CH2Cl Ionisation as the molecules get bigger in size and have C6H5CHClC6H5 Ionisation C6 H5 CG +HC6 H5 more electrons. + Cl– 17. (iii) Explanation: Conversion of chlorobenzene In aryl halides, big sized aryl group causes steric to toluene is the Wurtz-Fittig reaction. hindrance for incoming nucleophile thus, SN2 18. (i) 19. (i) pathway is not possible here. 20. (iv) Explanation: Halogen atom besides being slightly deactivating is ortho and para- 32. The C-X bond in haloalkanes is more polar and thus carbon atom is more electron deficient than directing. in case of haloarenes. Hence, haloalkanes undergo 21. (ii) Explanation: Halogen atom besides being slightly deactivating is o, p-directing; nucleophilic substitution reactions. While haloarenes therefore, further substitution occurs at shows resonance effect in which halogen atom ortho- and para-positions with respect to the releases electrons to benzene ring causing higher halogen atom. The o, p-directing influence of electron density at ortho and para positions where halogen atom can be easily understood if we electrophilic substitution reaction takes place. consider the resonating structures. Haloarenes are less reactive towards nucleophilic 22. (iii) Explanation: Reactions with iodine are substitution reactions as compared to haloalkanes because of sp2 hybridisation of carbon atom reversible in nature and require the involved in C-X bond which is more electronegative, presence of an oxidising agent (HNO3, HIO4) to oxidise the HI formed during iodination. instability of phenyl carbocation and inter- 23. (i) Explanation: It is difficult to replace chlorine electronic repulsions. by —OH in chlorobenzene in comparison to 33. (i) Benzyl halides undergo ionisation very readily to form benzyl carbocation which is that in chloroethane. stabilised by resonance. Therefore benzylic 24. C6H6Cl6 (BHC), i.e. Benzene hexachloride 25. C6H5Cl + 2Na + ClCH3 ——Æ halides are rIeonai–csXat–tiivone+towCa6Hrd5Cs HSN21 reactions. C6H5CH2X C6H5CH3 (Toluene) + 2NaCl 26. Diazonium salt on treatment with Cu2Cl2 or Benzyl halide stabilised by resonance Cu2Br2 form haloarenes (Sandmeyer’s reaction) as: Nu:– C6H5CH2Nu SN1 substitution Ar–N+2 Cl– CuX/HX Ar–X product haloalKanes and haloarenes 197

(ii) Both C6H5CHClCH3 and C6H5CH2Cl are C6H5C(CH3)(C6H5)Br < C6H5CH(C6H5)Br benzylic halides and they undergo hydrolysis < C6H5CH(CH3)Br < C6H5CH2Br (SN2 tHhorwouegvheCr6aHiqn5uCSeHNo1u2CsrleKaIcOotniHiosantb,iyonSNC16Hm5eCc+hHa2n+isCml–. reactions) 1° carbocation For SN1 reaction, reactivity order of halides is (less stable) 3° > 2° > 1°. For the two secondary bromides, since the carbocation intermediate obtained While, Ionisation from C6H5CH(C6H5)Br is more stable than the one obtained from C6H5CH(CH3)Br due C6H5CHClCH3 C6H5–CH–C+H3 + Cl– to resonance stabilisation by two phenyl 2° carbocation groups. Therefore, C6H5CH(C6H5)Br is more (more stable) reactive than C6H5CH(CH3)Br towards SN1 reactions. A phenyl group being much Since 2° carbocation is more stable than bulkier than methyl group therefore, C6H5C (CH3) (C6H5)Br is least reactive followed 1° carbocation, th erefo rreeaCd6iHly5wCiHthCKlCOHH3 by C6H5CH(C6H5)Br and C6H5CH(CH3) undergo hydrolysis more Br, while C6H5CH2Br is the most reactive towards SN2 reactions. than C6H5CH2Cl. 36. (i) 2-Bromo-3-chlorobutane 34. (i) (ii) 1-Bromo-1-chloro-1,2, 2-trifluoroethane (iii) 1-Bromo-4-chlorobut-2-yne (iv) 1, 1, 1, 2, 3, 3, 3-heptachloro-2- (ii) (trichloromethyl)propane (ii) (v) 2-Bromo-3, 3-bis(4¢-chlorophenyl) butane (vi) 1-Chloro-1-(4¢-iodophenyl)-3, 3-dimethylbut-1-ene sp2–hybrid carbon 37. (i) δ+ Clδ– (Less negative charge) 35. (i) CH3CH2CH2CH2Br < (CH3)2CHCH2Br < Chlorobenzene CH3CH2CH(Br)CH3 < (CH3)3C—Br sp2–hybrid carbon (SN1 reaction) and CH3CH2CH2CH2Br > (CH3)2CH—CH2Br δ+ Clδ– (More negative charge) > CH3CH2CH(Br)CH3 > (CH3)3C—Br (SN2 reaction) Cyclohexyl chloride In chlorobenzene, the carbon of C—Cl bond is sp2-hybridised while the carbon of C—Cl bond in cyclohexyl chloride is sp3-hybridised. For the two primary bromides, since the carbocation intermediate obtained from The polarity of C—Cl bond is more in case (CH3)2CHCH2Br is more stable than the one of cyclohexyl chloride because the carbon bobectaaiunseedoffrgormeatCeHr 3eCleHct2rCoHn2CdoHn2oBrri.nTduhcistivies atom involved in cyclohexyl chloride has lower s-character and hence it is less e(tCoffHecC3t)H2C3HCofHCH2C(2CHBHr2C3r)He2Ca2BHctr—s fgianrsotueSprN.1as Therefore, electronegative as compared to the carbon compared reactions. atom of chlorobenzene. On the account of it, the dipole moment of chlorobenzene is less CH3CH2CH(Br)CH3 is a secondary bromide than that of cyclohexyl chloride. fwrHoehlealniocltcweieos,n(tCtshh.HeeT3)ahr3bCeeovBverreresaeocisrtdiovreaidrteyirtseinrsftioinlSalcorNeyw2 terhdbeeraoincmsttieiSodrnNei1sc. (ii) Alkyl halides being polar molecules, has their molecules held together by dipole- dipole attractions. On the other hand, hindrance factor around electrophilic carbon the molecules of aHlk2yOl are held together by H-bonds. When halides are put into atom comes into play. water, then new forces of attraction between (ii) C6H5C(CH3)(C6H5)Br > C6H5CH(C6H5)Br water and alkyl halide molecules so arises > C6H5CH(CH3)Br > C6H5CH2Br (SN1 reactions), and are weaker than the forces of attraction that already exist between RX—RX molecules 198 Chemistry-12

and H2O—H2O molecules. This means that Electrophilic substitution takes place at o- and there is almost no scope for the association p- position due to +R effect inspite of –I effect of Cl. of R—X and water molecules. Hence, alkyl halides are immiscible with water. [Resonance effect of Cl stabilising the intermediate carbocation] (iii) Grignard regents (R—MgX) are highly The inductive effect is stronger than resonance reactive compounds and should always and thus causes net deactivation due to net electron withdrawl. The resonance effect tends to be prepared under anhydrous conditions. oppose the inductive effect for the attack at o-and p-positions and hence makes the deactivation less This is because they react readily with the for o-and p-attack. Reactivity is thus controlled by the stronger inductive effect and orientation moisture present in the apparatus. is controlled by resonance effect. 42. (i) RMgX + HOH Æ R—H + Mg(OH)X (ii) Alkyl magnesium halide (iii) (iv) 38. The melting point of the p-isomer is always comparatively higher than that of o- or m-isomers. This is because of the more symmetrical structure of p-isomer and close packing of its molecules in the crystal lattice. Therefore, it has stronger intermolecular forces of attraction than o- and m-isomers and hence greater energy is required to break their crystal lattice to melt or dissolve than the corresponding o- and m-isomers. In other words, the melting point of p-isomer is higher and has lower solubility than the corresponding o- and m-isomers. CH3 CHCl2 CHO 39. +2Cl2 H3O+ Sunlight Toluene Benzyl dichloride Benzaldehyde (A) (B) (C) 40. These reactions are electrophilic substitution reactions where among the ortho- and para- products, the major product is para-isomer. (i) o-and p-nitrochlorobenzene (ii) o-and p-bromobenzene sulphonic acid (iii) o-and p-dichlorobenzene (iv) o-and p-chloroacetophenone 41. Chlorine withdraws electrons through inductive effect and releases electrons through resonance. Through inductive effect, chlorine destabilise the intermediate carbocation formed during the electrophilic substitution reaction. Through resonance, chlorine tends to stabilise the carbocation and the effect is more pronounced at o- and p- positions. (v) (vi) haloalKanes and haloarenes 199