CHEM1102 2014-J-2 June 2014 Marks Compounds of d-block elements are frequently paramagnetic. Using the box notation • 2 to represent atomic orbitals, account for this property in compounds of Co2+. 2+ 7 Co has a 3d configuration: ↑↓ ↑↓ ↑ ↑ ↑ Co2+ is a d7 system, so must have at least 1 unpaired electron. Consequently it must be paramagnetic. CHEM1102 2014-J-5 June 2014 Marks • Name the complex [CoCl2(en)2]. en = ethylenediamine = NH2CH2CH2NH2 4 dichloridobis(ethylenediamine)cobalt(II) Draw all possible isomers of this complex. THE REMAINDER OF THIS PAGE IS FOR ROUGH WORKING ONLY. CHEM1102 2014-N-7 November 2014 Marks • The structure below represents the active site in carbonic anhydrase, which features a 7 Zn2+ ion bonded to 3 histidine residues and a water molecule. The pKa of uncoordinated water is 15.7, but the pKa of the water ligand in carbonic anhydrase is around 7. Suggest an explanation for this large change. The high charge on the Zn2+ ion draws electron density out of the O–H bonds in the water molecule. This weakens the O–H so the H+ is more likely to leave. The water in carbonic anhydrase is therefore more acidic, as shown by the large decrease in pKa. When studying zinc-containing metalloenzymes, chemists often replace Zn2+ with Co2+. Using the box notation to represent atomic orbitals, work out how many unpaired electrons are present in the Zn2+ and Co2+ ions. 2+ 10 Zn , 3d ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ 2+ 7 Co , 3d ↑↓ ↑↓ ↑ ↑ ↑ Zn2+ has 0 unpaired d electrons, Co2+ has 3 unpaired d electrons. Co2+ is therefore paramagnetic and will be attracted by a magnetic field. Suggest why it is useful to replace Zn2+ with Co2+ when studying the nature of the active site in carbonic anhydrase. The ions have similar radii so the properties of natural carbonic anhydrase and the version with cobalt replacing zinc should have similar biological properties. The unpaired electrons on Co2+ however mean that it is paramagnetic and the magnetism can be used to study the active site. Suggest two differences in the chemistry of Zn2+ and Co2+ ions that may affect the reactivity of the cobalt-containing enzyme. Zinc only forms +2 ions but cobalt forms +2 and +3. The cobalt-containing enzyme may be susceptible to oxidation. Zn2+ tends to form 4-coordinate tetrahedral complexes but Co2+ is slightly larger and often forms 6-coordinate octahedral complexes. The metal ion may change its coordination by bonding extra ligands. CHEM1102 2013-J-2 June 2013 Marks • Use the information already provided to complete the following table. 2– 8 (ox = oxalate = C2O4 ) n n n Formula [CrCl2(NH3)4] [Fe(ox)3] [ZnCl2(NH3)2] Oxidation state of transition +III +III +II metal ion Number of d-electrons in the transition 3 5 10 metal ion Number of unpaired d-electrons in 3 5 0 the transition metal ion Charge of complex 1+ 3- 0 (i.e. n) Is the metal atom Yes Yes No paramagnetic? The complex [PtCl2(NH3)2] has two isomers, while its zinc analogue (in the table) exists in only one form. Using diagrams where appropriate, explain why this is so. The Pt compound has square planar geometry and hence 2 isomers, where the Cl groups are either opposite each other (trans) or next to each other (cis). The Zn compound has tetrahedral geometry and hence only one structure exists CHEM1102 2013-N-3 November 2013 Marks • Complete the following table. 6 Formula Geometry of complex Ligand donor atom(s) 2– [Zn(OH)4] tetrahedral O [CoCl(NH3)5]SO4 octahedral Cl and N K4[Fe(CN)6] octahedral C – [Ag(CN)2] linear C Select any complex ion from the above table and state whether it is paramagnetic, diamagnetic or neither. Explain your reasoning. Zn2+ is d10 system. No unpaired electrons, therefore diamagnetic. ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ Co3+ is d6 system. 2 paired electrons and 4 unpaired, therefore paramagnetic. ↑↓ ↑ ↑ ↑ ↑ Fe2+ is d6 system. 2 paired electrons and 4 unpaired, therefore paramagnetic. ↑↓ ↑ ↑ ↑ ↑ Ag+ is d10 system. No unpaired electrons, therefore diamagnetic. ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ CHEM1102 2013-N-5 November 2013 Marks Which of the following complexes is/are chiral? Explain your reasoning. Use • 3 diagrams where necessary. – acetylacetonate ion = CH3COCHCOCH3 ethylenediamine = NH2CH2CH2NH2 tris(acetylacetonato)chromium(III) is chiral – mirror images are non- superimposable trans-bis(ethylenediamine)difluoridochromium(III) ion is not chiral – it is identical to its mirror image. acetylacetonatobis(ethylenediamine)chromium(III) ion is chiral – mirror images are non-superimposable CHEM1102 2013-N-6 November 2013 –1 –19 Marks • What is the solubility of Cu(OH)2 in mol L ? Ksp (Cu(OH)2) is 1.6 × 10 at 25 °C. 7 The dissolution reaction and associated solubility product are: 2+ - 2+ - 2 Cu(OH)2(s) Cu (aq) + 2OH (aq) Ksp = [Cu (aq)][OH (aq)] If x mol dissolve in one litre, [Cu2+(aq)] = x M and [OH-(aq)] = 2x. Hence: 2 3 –19 Ksp = (x)(2x) = 4x = 1.6 × 10 x = 3.4 × 10–7 M Answer: 3.4 × 10–7 M 2+ 13 The overall formation constant for [Cu(NH3)4] is 1.0 × 10 . Write the equation for the reaction of Cu2+ ions with excess ammonia solution. 2+ 2+ Cu (aq) + 4NH3(aq) [Cu(NH3)4] (aq) Calculate the value of the equilibrium constant for the following reaction. 2+ – Cu(OH)2(s) + 4NH3(aq) [Cu(NH3)4] (aq) + 2OH (aq) This reaction can be considered to occur via (i) Cu(OH)2 dissolving followed by (ii) the Cu2+(aq) ions that form being complexed by ammonia. 2+ For the formation of [Cu(NH3)4] , the equilibrium constant is: �! [��(���)� ] 13 Kstab = �! � = 1.0 × 10 [�� ][���] For the reaction of Cu(OH)2(s) with NH3(aq), the equilibrium constant is: �! ! � [��(���)� ][�� �� ] K = � [���] To obtain K, Ksp is multiplied by Kstab: K = Ksp × Kstab �! �! ! � 2+ - 2 [��(���)� ] [��(���)� ][�� �� ] = [Cu (aq)][OH (aq)] × �! � = � [�� ][���] [���] = (1.6 × 10–19) × (1.0 × 1013) = 1.6 × 10–6 Answer: 1.6 × 10–6 CHEM1102 2013-N-6 November 2013 Would you expect Cu(OH)2(s) to dissolve in 1 M NH3 solution? Briefly explain your answer. No. Equilibrium constant K is very small so the reaction lies heavily in favour of reactants. CHEM1102 2012-J-2 June 2012 2 • Compounds of d-block elements are frequently paramagnetic. Using the box notation to represent atomic orbitals, account for this property in compounds of Cu2+. 2+ 9 Cu is d ↑↓ ↑↓ ↑↓ ↑↓ ↑ is paramagnetic 4 • Provide a systematic name for the complex [NiBrCl(en)] and draw both of its possible structures. (en = NH2CH2CH2NH2 = ethylenediamine = ethane-1,2-diamine) Both of the following names are acceptable: • bromidochlorido(ethylenediamine)nickel(II) • bromidochlorido(ethane-1,2-diamine)nickel(II) Is either complex chiral? Explain your reasoning. No. Both structures are superimposable on (i.e. identical to) their mirror images. CHEM1102 2012-J-3 June 2012 Marks Complete the following table. (ox = oxalate = C O 2–) • 2 4 6 Formula Na[FeCl4] [CrCN(NH3)5]Br2 K3[VO2(ox)2]⋅3H2O Oxidation state of +III +III +V transition metal ion Coordination number of 4 6 6 transition metal ion Number of d-electrons in 5 3 0 the transition metal ion Species + 2+ + formed upon Na (aq), [CrCN(NH3)5] (aq), K (aq), – – 3– dissolving in [FeCl4] (aq) Br (aq) [VO2(ox)2] (aq) water THE REMAINDER OF THIS PAGE IS FOR ROUGH WORKING ONLY. CHEM1102 2012-N-5 November 2012 Marks The following three complex ions can all exhibit isomerism. Name the type of • 9 isomerism involved in each case and draw the structures of the isomeric pairs. 2– ox = oxalate = C2O4 + [CrCl2(NH3)4] Geometrical isomerism 3– [Fe(ox)3] Optical isomerism 3+ [Co(NH3)3(OH2)3] Geometrical isomerism 3 • Give the systematic name of each of the following compounds. en = ethylenediamine = 1,2-diaminoethane = NH2CH2CH2NH2 Cs2[PtF6] caesium hexafluoridoplatinate(IV) [Co(en)2(NH3)2]Br3 diamminebis(ethylenediamine)cobalt(III) bromide CHEM1102 2010-J-4 June 2010 Marks Complete the following table. NCS– = isothiocyanate ion • 5 bipy = 2,2'-bipyridine = (C5H4N)2 = N N Formula K2[Zn(CN)4] [Co(bipy)(NH3)4]Cl3 [Co(bipy)2(NCS)2] Oxidation state of +2 or II +3 or III +2 or II transition metal ion Coordination number of 4 6 6 transition metal ion Number of d-electrons in 10 6 7 the transition metal ion Coordination geometry tetrahedral octahedral octahedral of the complex ion List all the 2 × N (from bipy) 4 × N (from bipy) ligand donor 4 × C 4 × N (from NH ) 2 × N (from NCS-) atoms 3 CHEM1102 2010-N-2 November 2010 • Titanium has three common oxidation states, +II, +III and +IV. Using the box 2 notation to represent atomic orbitals, predict whether compounds of Ti2+, Ti3+ and Ti4+ would be paramagnetic or diamagnetic. Ti is in group 4: it has 4 valence electrons. Ti2+ therefore has (4 – 2) = 2 remaining: it has a d2 configuration. Ti3+ therefore has (4 – 3) = 1 remaining: it has a d1 configuration. Ti4+ therefore has (4 – 4) = 0 remaining: it has a d0 configuration. These electrons are arranged in the five d orbitals to minimise the repulsion between them. This is achieved by keeping the maximum number possible unpaired. Ti2+ Ti3+ Ti4+ Ti2+ and Ti3+ have unpaired electrons and are paramagnetic. Ti4+ has no unpaired electrons and is diamagnetic. • Provide a systematic name for the complex trans-[NiBr2(en)2] and draw its structure. 4 Is this complex chiral? Explain your reasoning. en = ethylenediamine = ethane-1,2-diamine trans-dibromidobis(ethylenediamine)nickel(II) or trans-dibromidobis(ethane-1,2-diamine)nickel(II) It is not chiral as it is superimposable on (i.e.