CHEM J-2 June 2014

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1 CHEM J-2 June 2014 Compounds of d-block elements are frequently paramagnetic. Using the box notation to represent atomic orbitals, account for this property in compounds of Co Co 2+ has a 3d 7 configuration: Co 2+ is a d 7 system, so must have at least 1 unpaired electron. Consequently it must be paramagnetic.

2 CHEM J-5 June 2014 ame the complex [CoCl 2 (en) 2 ]. en = ethylenediamine = H 2 CH 2 CH 2 H 2 4 dichloridobis(ethylenediamine)cobalt(ii) Draw all possible isomers of this complex. THE REMAIDER OF THIS PAGE IS FOR ROUGH WORKIG OLY.

3 CHEM ovember 2014 The structure below represents the active site in carbonic anhydrase, which features a Zn 2+ ion bonded to 3 histidine residues and a water molecule. 7 The pk a of uncoordinated water is 15.7, but the pk a of the water ligand in carbonic anhydrase is around 7. Suggest an explanation for this large change. The high charge on the Zn 2+ 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 pk a. When studying zinc-containing metalloenzymes, chemists often replace Zn 2+ with Co 2+. Using the box notation to represent atomic orbitals, work out how many unpaired electrons are present in the Zn 2+ and Co 2+ ions. Zn 2+, 3d 10 Co 2+, 3d 7 Zn 2+ has 0 unpaired d electrons, Co 2+ has 3 unpaired d electrons. Co 2+ is therefore paramagnetic and will be attracted by a magnetic field. Suggest why it is useful to replace Zn 2+ with Co 2+ 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 Co 2+ however mean that it is paramagnetic and the magnetism can be used to study the active site. Suggest two differences in the chemistry of Zn 2+ and Co 2+ 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. Zn 2+ tends to form 4-coordinate tetrahedral complexes but Co 2+ is slightly larger and often forms 6-coordinate octahedral complexes. The may change its coordination by bonding extra ligands.

4 CHEM J-2 June 2013 Use the information already provided to complete the following table. (ox = oxalate = C 2 O 4 2 ) 8 Formula [CrCl 2 (H 3 ) 4 ] n [Fe(ox) 3 ] n [ZnCl 2 (H 3 ) 2 ] n Oxidation state of +III +III +II umber of d-electrons in the umber of unpaired d-electrons in the Charge of complex (i.e. n) Is the metal atom paramagnetic? Yes Yes o The complex [PtCl 2 (H 3 ) 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

5 CHEM ovember 2013 Complete the following table. Formula Geometry of complex Ligand donor atom(s) 6 [Zn(OH) 4 ] 2 tetrahedral O [CoCl(H 3 ) 5 ]SO 4 octahedral Cl and K 4 [Fe(C) 6 ] octahedral C [Ag(C) 2 ] linear C Select any complex ion from the above table and state whether it is paramagnetic, diamagnetic or neither. Explain your reasoning. Zn 2+ is d 10 system. o unpaired electrons, therefore diamagnetic. Co 3+ is d 6 system. 2 paired electrons and 4 unpaired, therefore paramagnetic. Fe 2+ is d 6 system. 2 paired electrons and 4 unpaired, therefore paramagnetic. Ag + is d 10 system. o unpaired electrons, therefore diamagnetic.

6 CHEM ovember 2013 Which of the following complexes is/are chiral? Explain your reasoning. Use diagrams where necessary. acetylacetonate ion = CH 3 COCHCOCH 3 3 ethylenediamine = H 2 CH 2 CH 2 H 2 tris(acetylacetonato)chromium(iii) is chiral mirror images are nonsuperimposable 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

7 CHEM ovember 2013 What is the solubility of Cu(OH) 2 in mol L 1? K sp (Cu(OH) 2 ) is at 25 C. 7 The dissolution reaction and associated solubility product are: Cu(OH) 2 (s) Cu 2+ (aq) + 2OH - (aq) K sp = [Cu 2+ (aq)][oh - (aq)] 2 If x mol dissolve in one litre, [Cu 2+ (aq)] = x M and [OH - (aq)] = 2x. Hence: K sp = (x)(2x) 2 = 4x 3 = x = M Answer: M The overall formation constant for [Cu(H 3 ) 4 ] 2+ is Write the equation for the reaction of Cu 2+ ions with excess ammonia solution. Cu 2+ (aq) + 4H 3 (aq) [Cu(H 3 ) 4 ] 2+ (aq) Calculate the value of the equilibrium constant for the following reaction. Cu(OH) 2 (s) + 4H 3 (aq) [Cu(H 3 ) 4 ] 2+ (aq) + 2OH (aq) This reaction can be considered to occur via (i) Cu(OH) 2 dissolving followed by (ii) the Cu 2+ (aq) ions that form being complexed by ammonia. For the formation of [Cu(H 3 ) 4 ] 2+, the equilibrium constant is: K stab = [Cu(H 2! 3) 4 ] [Cu 2! ][H 3 ] 4 = For the reaction of Cu(OH) 2 (s) with H 3 (aq), the equilibrium constant is: K = [Cu(H 3) 4 2! ][OH! aq ] 2 [H 3 ] 4 To obtain K, K sp is multiplied by K stab : K = K sp K stab = [Cu 2+ (aq)][oh - (aq)] 2 [Cu(H 2! 3) 4 ] [Cu 2! ][H 3 ] 4 = [Cu(H 2! 3) 4 ][OH! aq ] 2 [H 3 ] 4 = ( ) ( ) = Answer:

8 CHEM ovember 2013 Would you expect Cu(OH) 2 (s) to dissolve in 1 M H 3 solution? Briefly explain your answer. o. Equilibrium constant K is very small so the reaction lies heavily in favour of reactants.

9 CHEM J-2 June 2012 Compounds of d-block elements are frequently paramagnetic. Using the box notation to represent atomic orbitals, account for this property in compounds of Cu Cu 2+ is d 9 is paramagnetic Provide a systematic name for the complex [ibrcl(en)] and draw both of its possible structures. (en = H 2 CH 2 CH 2 H 2 = ethylenediamine = ethane-1,2-diamine) 4 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. o. Both structures are superimposable on (i.e. identical to) their mirror images.

10 CHEM J-3 June 2012 Complete the following table. (ox = oxalate = C 2 O 4 2 ) 6 Formula a[fecl 4 ] [CrC(H 3 ) 5 ]Br 2 K 3 [VO 2 (ox) 2 ] 3H 2 O Oxidation state of Coordination number of umber of d-electrons in the +III +III +V Species formed upon dissolving in water a + (aq), [FeCl 4 ] (aq) [CrC(H 3 ) 5 ] 2+ (aq), Br (aq) K + (aq), [VO 2 (ox) 2 ] 3 (aq) THE REMAIDER OF THIS PAGE IS FOR ROUGH WORKIG OLY.

11 CHEM ovember 2012 The following three complex ions can all exhibit isomerism. ame the type of isomerism involved in each case and draw the structures of the isomeric pairs. ox = oxalate = C 2 O [CrCl 2 (H 3 ) 4 ] + Geometrical isomerism [Fe(ox) 3 ] 3 Optical isomerism [Co(H 3 ) 3 (OH 2 ) 3 ] 3+ Geometrical isomerism Give the systematic name of each of the following compounds. en = ethylenediamine = 1,2-diaminoethane = H 2 CH 2 CH 2 H 2 3 Cs 2 [PtF 6 ] caesium hexafluoridoplatinate(iv) [Co(en) 2 (H 3 ) 2 ]Br 3 diamminebis(ethylenediamine)cobalt(iii) bromide

12 CHEM J-4 June 2010 Complete the following table. CS = isothiocyanate ion 5 bipy = 2,2'-bipyridine = (C 5 H 4 ) 2 = Formula K 2 [Zn(C) 4 ] [Co(bipy)(H 3 ) 4 ]Cl 3 [Co(bipy) 2 (CS) 2 ] Oxidation state of Coordination number of umber of d-electrons in the Coordination geometry of the complex ion List all the ligand donor atoms +2 or II +3 or III +2 or II tetrahedral octahedral octahedral 4 C 2 (from bipy) 4 (from H 3 ) 4 (from bipy) 2 (from CS - )

13 CHEM ovember 2010 Titanium has three common oxidation states, +II, +III and +IV. Using the box notation to represent atomic orbitals, predict whether compounds of Ti 2+, Ti 3+ and Ti 4+ would be paramagnetic or diamagnetic. 2 Ti is in group 4: it has 4 valence electrons. Ti 2+ therefore has (4 2) = 2 remaining: it has a d 2 configuration. Ti 3+ therefore has (4 3) = 1 remaining: it has a d 1 configuration. Ti 4+ therefore has (4 4) = 0 remaining: it has a d 0 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. Ti 2+ Ti 3+ Ti 4+ Ti 2+ and Ti 3+ have unpaired electrons and are paramagnetic. Ti 4+ has no unpaired electrons and is diamagnetic. Provide a systematic name for the complex trans-[ibr 2 (en) 2 ] and draw its structure. Is this complex chiral? Explain your reasoning. en = ethylenediamine = ethane-1,2-diamine 4 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. identical to) its mirror image.

14 CHEM ovember 2010 Complete the following table. (EDTA = ethylenediaminetetraacetate) Formula [i(h 3 ) 6 ](O 3 ) 2 trans-[ptcl 2 (H 3 ) 2 ] a[fe(edta)] 5 Oxidation state of Coordination number of umber of d-electrons in the Coordination geometry of the complex ion List all the ligand donor atoms +II +II +III octahedral square planar octahedral Cl,, O

15 CHEM J-2 June 2009 Explain in terms of their electronic configurations and ionisation energies why the halogens (Group 17) are powerful oxidising agents. 2 Oxidising agents are themselves reduced (i.e. they gain electrons). The electronic configuration of the halogens (Group 17) is np 5. They are small atoms (atomic size decreases across a period as the nuclear charge increases). In each period, the halogen is the element with the highest number of protons in the nucleus that also has an incomplete shell. As a result, they will readily gain a single electron to form the X ion. Similarly, the high nuclear charge and small size means that they have high ionisation energies. Hence halogens are poor reducing agents.

16 CHEM J-4 June 2009 Complete the following table. (en = ethylenediamine = H 2 CH 2 CH 2 H 2 ) Formula K 2 [CoCl 4 ] a 3 [FeBr(C) 5 ] [Zn(en) 2 (O 3 ) 2 ] 9 Oxidation state of Coordination number of umber of d-electrons in the +2 (II) +3 (III) +2 (II) (Co is in Group 9 so Co 2+ has 9 2= 7) 5 (Fe is in Group 8 so Fe 3+ has 8 3 = 5) 6 (4 from en and 2 O from O 3 - ) 10 (Zn is in Group 12 Zn 2+ has 12 2= 10) Charge of the complex ion [CoCl 4 ] 2- [FeBr(C) 5 ] 3- [Zn(en) 2 (O 3 ) 2 ] Geometry of the complex ion List all the ligand donor atoms tetrahedral octahedral octahedral 4 Cl 1 Br and 5 C 4 and 2 O

17 CHEM ovember 2009 Explain why compounds of d-block elements are frequently paramagnetic. Use examples in your answer. 2 Paramagnetism is the property of a compound to be attracted by an external magnetic field. It is a characteristic of any compound with unpaired electrons. d-block elements have from 1 to 10 electrons in the d-orbitals. When forming compounds, some of these may be lost to give paramagnetic species. Species with odd numbers of electrons must be paramagnetic, species with even numbers of d-electrons, may or may not be paramagnetic. eg Cu 2+, d 9 Zn 2+, d 10 must be paramagnetic must be diamagnetic Fe 2+, d 6 is paramagnetic Provide a systematic name for cis-[co(en) 2 Cl 2 ]Cl. Is this complex chiral? Explain your reasoning by drawing the structure of the complex. en = H 2 CH 2 CH 2 H 2 = ethane-1,2-diamine = ethylenediamine 3 cis-dichloridobis(ethylenediamine)cobalt(iii) chloride or cis-dichloridobis(ethane-1,2-diamine)cobalt(iii) chloride The complex is chiral as it is not superimposable on its mirror image. Co Cl Cl + Cl Cl Co +

18 CHEM ovember 2009 Derive expressions for the equilibrium constants for the complexation of Pb 2+ (K 1 ) and of Ca 2+ (K 2 ) by EDTA 4. 3 Pb 2+ + EDTA 4 [Pb(EDTA)] 2 K 1 = Ca 2+ + EDTA 4 [Ca(EDTA)] 2 K 2 = Briefly explain why the chelating agent, EDTA, is administered as [Ca(EDTA)] 2 to treat lead poisoning and determine which of K 1 or K 2 must be greater for the therapy to be effective. K 1 must be greater than K 2 for the therapy to be effective. [Ca 2+ ] is much greater than [Pb 2+ ] in the body, so need K 1 > K 2 to form the Pb complex. If EDTA is not administered as the Ca complex, it will strip Ca 2+ from the body.

19 CHEM J-2 June 2008 Compounds of d-block elements are frequently paramagnetic. Using the box notation to represent atomic orbitals, explain why most Fe 2+ and Fe 3+ compounds are paramagnetic. 2 Iron is in Group 8 and thus has 8 valence electrons. Fe 2+ thus has (8 2) = 6 electrons and Fe 3+ has (8 3) = 5 electrons. These electrons occupy the 3d orbitals and do so to minimize the number of electron pairs, and thus minimize electron / electron repulsion. There are five 3d orbitals and thus the configurations are: Fe 2+ 3d 6 Fe 3+ 3d 5 c c With 4 and 5 unpaired electrons respectively, both Fe 2+ and Fe 3+ are paramagnetic. THE REMAIDER OF THIS PAGE IS FOR ROUGH WORKIG OLY.

20 CHEM J-4 June 2008 Complete the following table. (en = ethylenediamine = H 2 CH 2 CH 2 H 2 ) Formula (H 4 ) 2 [CoCl 4 ] [Cr(H 3 ) 5 (H 2 O)]Cl 3 cis-[ptcl 2 (en) 2 ] 9 Oxidation state of Coordination number of umber of d-electrons in the +2 (II) +3 (III) +2 (II) (Co is in Group 9 so Co 2+ has 9 2= 7) 3 (Cr is in Group 6 so Co 2+ has 6 3= 3) 6 (2 Cl and 4 from 2en) 8 (Pt is in Group 10 so Pt 2+ has 10 2= 8) Charge of the complex ion [CoCl 4 ] 2- [Cr(H 3 ) 5 (OH 2 )] 3+ [PtCl 2 (en)] Geometry of the complex ion List all the ligand donor atoms tetrahedral octahedral octahedral 4 Cl - 5 and 1 O 2 Cl - and 4

21 CHEM ovember 2008 Calculate the equilibrium constant for the following reaction. 3 AgI(s) + 2C (aq) [Ag(C) 2 ] (aq) + I (aq) Data: K stab of [Ag(C) 2 ] = ; K sp of AgI = The equations for the dissolution of AgI and the stability constant of the complex [Ag(C) 2 ] - are, respectively: AgI(s) Ag + (aq) + I - (aq) K sp = [Ag + (aq)][i - (aq)] Ag + (aq) + 2C - (aq) [Ag(C) 2 ] - (aq) K stab = 2 + [[Ag(C) ] (aq)] [Ag (aq)][c (aq)] Addition of these reactions gives the required reaction and so the equilibrium constant for the reaction is the product of the individual equilibrium constants: Hence, AgI(s) + 2C - (aq) [Ag(C) 2 ] - (aq) + I - (aq) K = K sp K stab = [Ag + (aq)][i - [[Ag(C) 2 ] (aq)] (aq)] + 2 [Ag (aq)][c (aq)] = 2 [[Ag(C) ] (aq)][i (aq)] [C (aq)] K = K sp K stab = ( ) ( ) = Answer:

22 CHEM ovember 2008 Which of the cations, [Fe(OH 2 ) 6 ] 3+ and [Fe(OH 2 ) 6 ] 2+, has the larger pk a? Briefly explain why. 2 Solutions containing both cations are acidic due to the equilibria: [Fe(OH 2 ) 6 ] 2+ (aq) [Fe(OH 2 ) 6 ] 3+ (aq) [Fe(OH 2 ) 5 (OH)] + (aq) + H + (aq) [Fe(OH 2 ) 5 (OH)] 2+ (aq) + H + (aq) The Fe 2+ and Fe 3+ cations polarize the Fe-OH 2 bond, pulling electron density from the oxygen. The oxygen, in turn, pulls electron density away towards itself in the O-H bond, causing H + to be produced. Fe 3+ has a higher charge and is smaller than Fe 2+ : it is more polarizing and so [Fe(OH 2 ) 6 ] 3+ is more acidic. The equilibrium is further to the right for [Fe(OH 2 ) 6 ] 3+ than it is for [Fe(OH 2 ) 6 ] 2+ so K a ([Fe(OH 2 ) 6 ] 3+ ) > K a ([Fe(OH 2 ) 6 ] 2+ ). As pk a = -log 10 K a, a higher K a means a lower pk a. Thus, [Fe(OH 2 ) 6 ] 3+ is the more acidic of the two cations and has the lower pk a value. Consider the compound [CrCl(OH 2 ) 4 (CS)]Cl 2H 2 O. The complex ion is [CrCl(OH 2 ) 4 (CS)] +. This contains Cr 3+ bonded to Cl -, 4H 2 O and one CS -. For each ligand, the donor atom is listed first. 3 What is the oxidation state of the? +3 or (III) What is the coordination number of the? 6 How many d-electrons in the? List all the ligand donor atoms. Cl -, 4 O and Cr is in group 6 so Cr(III) is d 3 Consider the complexes cis-[ptcl 2 (H 3 ) 2 ] and trans-[ptcl 2 (H 3 ) 2 ]. Draw the structures of the two isomers, clearly illustrating the stereochemistry. 3 The platinum is bonded to four ligands (2 Cl - and 2 H 3 ). With four ligands, two geometries are possible tetrahedral and square planar. Of these, only square planar gives rise to cis and trans-isomers and so the complexes must be square planar: Cl H 3 Cl Pt H 3 Cl Pt Cl H 3 H 3 cis-diamminedichloridoplatinum(ii) trans-diamminedichloridoplatinum(ii)

23 CHEM ovember 2008 Briefly suggest why cis-[ptcl 2 (H 3 ) 2 ] is an effective anti-cancer drug, but trans-[ptcl 2 (H 3 ) 2 ] is not. cis-[ptcl 2 (H 3 ) 2 ] is believed to act by binding the cis-pt(h 3 ) 2 group to two nearby nitrogen atoms on the bases of a strand of DA. This can only be achieved by the cis isomer the trans form has 180 between the vacant sites of the Pt(H 3 ) 2 group and is not able to bind in this way. The coordination of the platinum to the DA causes a kink in the α-helix of the DA and this prevents its replication.

24 CHEM J-2 June 2007 Compounds of d-block elements are frequently paramagnetic. Using the box notation to represent atomic orbitals, account for this property in compounds of V Paramagnetism is associated with the presence of unpaired electron spins. As vanadium is in group 5, V 3+ has two electrons in its 3d orbitals. These electrons occupy separate orbitals with the same spin to reduce the repulsion between them: 3d V 3+ thus has two unpaired electrons and is paramagnetic.

25 CHEM J-3 June 2007 Complete the following table. (en = ethylenediamine = H 2 CH 2 CH 2 H 2 ) Formula K 3 [Fe(C) 6 ] [Cu(H 3 ) 4 (H 2 O) 2 ](O 3 ) 2 cis-[crcl 2 (en) 2 ]Cl 3 Oxidation state of Coordination number of umber of d-electrons in the Species formed upon dissolving in water III or +3 II or +2 III or +3 6 (6 C - ) 6 (4 H H 2 O) 6 (2 Cl H 2 CH 2 CH 2 H 2 ) K + [Cu(H 3 ) 4 (H 2 O) 2 ] 2+ [Fe(C) 6 ] O 3 cis-[crcl 2 (en) 2 ] + Cl -

26 CHEM ovember 2007 What is a chelate ligand? 4 A chelate is a ligand with more than one donor atom that can bond to the same. Draw all possible isomers of [CoCl 2 (en) 2 ]. en = ethylenediamine = H 2 CH 2 CH 2 H 2 cis and trans geometric isomers are possible. The cis-isomer can exist as nonsuperimposable mirror images (enantiomers). Cl Co Cl Cl Co Cl Cl Cl Co trans- cis- cis- Explain briefly why the [Fe(H 2 O) 6 ] 3+ cation has a K a of M, whilst the [Fe(H 2 O) 6 ] 2+ cation has a K a of M. 2 The Fe 3+ ion has a higher charge and is smaller than the Fe 2+ ion it has a higher charge density. The higher charge density withdraws electron density from the oxygen and leading to a more polarised O H bonds that are more easily broken.

27 CHEM ovember 2007 Hemochromatosis or iron overload is a potentially fatal disorder in which excess iron is deposited in the bodily organs as insoluble hydrated iron(iii) oxide. It can be treated by administration of desferioxamine B (Desferal), a natural substance isolated from fungi. O H O H 3 Desferal H 3 O OH HO Desferal is taken over 8-12 hour periods up to six times per week. A value of log K = 30.6 is associated with the following equilibrium: O HO O Fe 3+ + LH 3 + FeL + + 3H + where LH 3 + = Desferal Briefly describe the chemical basis for the use of Desferal in iron overload therapy. The solubility of Fe 2 O 3 is very small - the equibrium for the reaction below lies far to the left: Fe 2 O 3 (s) + excess H 2 O 2Fe 3+ (aq) + 6OH (aq) Complexation of Fe 3+ ions with Desferal is very favourable the equilibrium for the reaction below lies far to the right (as K for this reaction is ): Fe 3+ + LH 3 + FeL + + 3H + where LH 3 + = Desferal The Desferal complexes all free Fe 3+ (aq) ions, so more Fe 2 O 3 must dissolve to reestablish the first equilibrium (le Chatelier's principle). Eventually all the Fe 2 O 3 will dissolve.

28 CHEM J-5 June 2006 Consider the compound with formula [CoCl 2 (H 3 ) 4 ]Br 2H 2 O. 3 Write the formula of the complex ion. [CoCl 2 (H 3 ) 4 ] + Write the symbols of the ligand donor atoms. Cl, What is the d electron configuration of the in this complex? Co 3+ : 3d 6

29 CHEM ovember 2006 Compounds of d-block elements are frequently paramagnetic. Using the box notation to represent atomic orbitals, account for this property in compounds of Cu As Cu is in group 11, it has 11 valence electrons. Cu 2+ therefore has (11 2) = 9. These occupy the five 3d orbitals: 4s 3d There is an unpaired electron and so Cu 2+ is paramagnetic. Complete the following table. 6 Formula Oxidation state of metal Coordination number of metal umber of d- electrons in the metal Species formed upon dissolving in water a 2 [CoCl 4 ] a +, [CoCl 4 ] 2- [i(h 3 ) 5 (H 2 O)]SO [i(h 3 ) 5 (H 2 O)] 2+, SO 4 2- [Cr(en) 3 ]Br [Cr(en) 3 ] 3+, Br - en = ethylenediamine = H 2 CH 2 CH 2 H 2

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