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Intro Inorganic Chem Lab

by: Dr. Drew Flatley

Intro Inorganic Chem Lab CHEM 242

Marketplace > University of Massachusetts > Chemistry > CHEM 242 > Intro Inorganic Chem Lab
Dr. Drew Flatley
GPA 3.96

Patricia Bianconi

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Patricia Bianconi
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This 43 page Class Notes was uploaded by Dr. Drew Flatley on Friday October 30, 2015. The Class Notes belongs to CHEM 242 at University of Massachusetts taught by Patricia Bianconi in Fall. Since its upload, it has received 25 views. For similar materials see /class/232349/chem-242-university-of-massachusetts in Chemistry at University of Massachusetts.

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Date Created: 10/30/15
Experiment 7 Thermochromism and Ionic Conductivity in Solid State CqugI4 Introduction A great variety of structure types are derived from lling cations into the voids of a face centered cubic FCC anion lattice One simple example is the zinc blende structure of ZnS where one half of the tetrahedral FCC voids are lled Figure 1 far left The ordered tetragonal roomtemperature structure of CqugI4 Figure 1 far right involves a more complex arrangement of cations in the tetrahedral voids and closely resembles a doubled zinc blende unit cell Figure 1 center left Cu and Hg2 ions are arranged in separate layers both sandwiched between closepacked layers of 139 ions The stability of this complicated order of the cations is tenuous When solid CqugI4 is heated the cations begin to move freely and a degree of randomness will be incorporated into the manner in which the Cu and Hg2 ions f111 tetrahedral voids As a result CqugI4 undergoes a structural phase transition to a disordered cubic structure of onehalf the volume of the tetragonal phase Figure 1 center right At high temperatures the Cu and Hg2 tetrahedral sites are indistinguishable to xrays because Xray diffraction measures the cell contents averaged over all unit cells within a crystal and on average each of the four available sites has a 50 chance of being occupied by a Cu ion a 25 chance ofbeing occupied by a Hg ion and a 25 chance ofbeing vacant giving the high temperature phase the same high symmetry cubic unit cell as ZnS The phase change is accompanied by a color change thermochromism and a marked decrease in electrical resistance ie an marked increase in electrical conductivity The thermochromic color change of CqugI4 is due to a small decrease in the semiconducting band gap 21 to 19 eV with the temperatureinduced change in structure ZnS Cqugl4 Figure 1 Left A single and doubled unit cell of zinc blende structure of ZnS Right The closely related disordered cubic hightemperature HT and ordered tetragonal lowtemperature LT structures of CqugI4 Despite the fact that four different tetrahedral voids in the cubic HT cell can host Cur and Hg2 ions only three of these sites actually contain an atom in a typical cell Ionic Conduction Unlike most electrical conductors in which electrons move in response to an applied voltage ionic conductors such as CqugI4 and the related solid AnggI4 can transport a current due to the ability of their ions to move in response to an applied voltage Ionic conductors are the solid analogues of electrolyte solutions which also conduct electricity via the motion of ions Ionic conductors are generally broken into two classes 7 those which conduct cations and those which conduct anions The latter class is of crucial importance to fuel cell applications which demand a material capable of conducting oxygen in the form of 0239 ions through a solid barrier Above its transition temperature CqugI4 exhibits ionic conductivity Fiveeighths of the tetrahedral holes and all of the octahedral holes formed by the iodide ions are vacant and these open sites provide a possible mechanism for the small copper cations to move through the crystal carrying charge It is easiest for a copper cation to jump between tetrahedral holes by moving to an octahedral hole and then to the new tetrahedral hole rather than jumping directly between tetrahedral holes Such substantial ionic conductivity is rare in ionic solids Determination of the Resistivity and Ionic Conductivity of CqugI4 An ohmmeter measures the resistance R of a material through the relation Voltage I X R where I the current in units of amperes A and where the units of resistance are Ohms S2 de ned as voltsampere However the resistance in Ohms is not independent of sample geometry The resistance increases linearly with the length of a sample and decreases proportionally to the cross sectional area of the sample You will therefore convert the measured resistance of your CqugI4 t0 the resistivity p in units of Ohmscm The resistivity of a sample is de ned to be independent of sample geometry through the formula below area pRx length The conductivity 0 of a sample is then defined as the inverse of its resistivity l a i p The units of a material39s conductivity are reported in Siemens Scm where S 182 You will therefore determine the ionic conductivity of the hightemperature structure of CqugI4 by l measuring its resistance with an ohmmeter while it is in the hightemperature phase 2 converting the hightemperature resistance to the geometryindependent resistivity and 3 calculating the hightemperature phase39s ionic conductivity Syntheses Despite having a solid state rather than molecular structure CqugI4 is readily prepared through a solution synthesis as are most nanoparticles CopperI tetraiodomercurateH CqugI4 is prepared by combining c0pperI iodide with mercuryII iodide CopperI iodide is formed by reacting copperII sulfate with potassium iodide in which the iodide ion reduces CuII to CuI thereby forming solid CuI 2Cu2 aq 4I39aq S 2CuIs I aq 1 Note that this reaction follows the HSAB rule that you studied in Experiment 1 the most stable oxidation state of the Cun ions in water is manipulated to be the softer Cu1 by the addition of the soft base I39 The E0 of this reaction halfreactions are derived from Table l in Experiment 1 is also favorable 2Cu2 21 2e39 3 2CuI 086V 2139 r 12 2e39 054v 2Cu2 4139 r 2CuI 12 aq 032v However in the presence of excess iodide ion the iodine undergoes further reaction forming triiodide ions 12 aq I39 S 1339 aq 2 Thus the true net ionic equation for the formation of copperI iodide is shown in Equation 3 2Cu2 aq 5139 aq S 2CuI s 1339 aq 3 Since the reaction is an equilibrium triiodide ions can act as oxidizing agents taking the Cu1 back to Cu2 the reverse of Equations 2 and 1 so the triiodide must be removed from solution Sodium sulfite is used to reduce the triiodide ion back to iodide Equation 4 1339aq 303239 aq 3H20 r 3139 aq SO4239aq 2H30aq 4 The solid copper I iodide can then be separated from the reaction mixture by carefully removing the excess solution of supernatant liquid A mercuryII iodide precipitate is synthesized by an anion metathesis meaning quotexchangequot reaction in which mercuric nitrate is combined with potassium iodide The nitrate anion is exchanged for the iodide The insolubility of HgIz helps to pull the equilibrium toward the right of Equation 5 Hg aq 2139 aq 3 Hglz S 5 Finally CqugI4 is prepared by adding the solid copperI iodide with heating to the mixture containing the mercuryII iodide precipitate 2 CuI s HgIz s gt CqugI4 s 6 Experimental Procedure SAFE T Y N 0T ES The mercurycontaining compounds are all quite toxic Avoid creating or breathing dust Use gloves when working with the mercury compounds and avoid eye contact Make extra efforts to ensure that all mercury waste ends up in the proper container Keep your work surfaces clean at all times so that mercury waste is not inadvertently transferred Synthesis of CqugI4 Synthesis of CuI Add 25 mL of 05 M CuSO4 solution 30 mL of 1M KI solution and five drops of 6M acetic acid to 25 mL of deionized water in a 100 mL beaker while stirring A precipitate of Cul will form Add to this precipitate with continuous very vigorous stirring a solution of0 10g Na2803 dissolved in 5 mL of water Allow the precipitate of Cul to stand for ve to ten minutes and then pour off as much as possible of the supernatant solution without losing the precipitate It may be better to slowly pipette off the supernatantit forms atiny layer on top of the CuI Synthesis of HgIz in situ In situ is Latin for in its original place In synthesis this means that an intermediate compound is not isolated before using it in a further reaction Combine 125 mL of 005 M HgN03z 15 mL of 1M KI and 50 mL of deionized water in a 200 mL beaker while stirring Note your observations Synthesis of CqugI4 Transfer the CuI precipitate into the suspension of Hglz using a stream of deionized water to wash all the CuI from the beaker Make sure to transfer all the CuIdon t lose too much precipitate Heat the mixed suspension of CuI and Hglz almost to boiling for about 20 minutes on a stirring hot plate A single brightred brickred or darkbrown solid should form Record your observations of initial colors etc If the CqugI4 appears green or light pink it should be resynthesized Filter the solution using a vacuum aspirator while it is still hot and wash the precipitate of CqugI4 with small portions of acetone Dry the solid as much as possible on the lter using the aspirator and then put on gloves collect the solid and spread it on a piece of filter paper to air dry for 10 minutes Determination of the Transition Temperature of CqugI4 The transition temperature of CqugI4 is the temperature at which it switches from the roomtemperature phase to the hightemperature conductive phase This temperature lies between 40 and 900 C for CqugI4 and can be determined using a meltingpoint apparatus mel temp by observing the thermochromism of CqugI4 Wearing gloves pack a small amount of dry CqugI4 into a piece of capillary tubing that has one end sealed and immerse the capillary in a meltemp apparatus Heat the capillary and record the temperature at which a color change takes place as the sample is heated This is the thermochromism the change in color upon heating to the phase transition that CqugI4 displays The color change is thus an indicator of the phasetransition temperature Record the color of the hightemperature phase and the phase transition temperature Allow the capillary to cool observe and record its roomtemperature color and then repeat the transition temperature measurement on the exact same sample Determine and record in your notebook whether the sample can be thermally cycled or whether the rst phase transition is irreversible Cnmpalixnnn lu Emmi Cundmtili u am Twn Plum 1wath Wem glwves cmyynursnhd cmiLgmp mg minim m wmchglaS mch prwmi Wimp pipu CmmbmakaglaS capillary le inhalf andmsen asmmpm nfl gauge mppr 3 in mm mm n 922 mg Figme mm The m Sl39lmlldbe ppm mandipr mg ppnmipm capillame mp sums CmHglmnhlat 12m m cm a mammals smka lube Thenmsen mimmgipm prepppuwm man up npnendm39 uz glaS capum mi gemlypllsh he cmiLgi dawnsn m n is packedbelween he m Emmi m 21mm The cud1g m be mgnnd suntan wiihbmh ms airshmdd mpg mppi beiween he ends imp wimsand he CmILgh Aim mg endsn he ms munnnlbe in mm m lube shnmbe packzd wiihasnhdcyhndememILgh wimp mcnnm wi h he ends nfbmhcnpprwues 18gauge Copper Wire Cqugl4 Glass Cap aryTube anwip up all mus prcmixgi andbmkznglas Emmiin imprpm mi msppsi imp gims mi mils mm mm rnzicurywnsle Enmmr Allmucurywasbzs must be segpgmimayrmm mherwasbzs 5n m heycanbe mated pmpiy wip in mud pr ynmcaplllarg wm apparmls wnhapipu m1 Alm wash ynurhands hnmughly m ynmberlch measm mi maid in 1mm in cm nfynur 53mph prcmixgi mg mg nlb pmvided Reunide mKIEId amEMni l39lz capl ary lbe whlchis wnmnnn he cap lanz plashl minim Take ynmcaplllarg wireCm gh sampb p in mm Wham m m 11m guns mi nhmmems Cannecnhz m ends imp Emmi wne m up nhmmemelenmdes mi pm in capillary at on the bench Your TA will adjust the ohmmeter to the correct scale Read and record the electrical resistance R of your CqugI4 sample at room temperature Your TA will now heat the capillary with the heat gun Read and record the electrical resistance R of your sample above the phase transition temperature The ohmmeter reading may be jumpy so record several R values and use either 1 the one that seems to be most stable or 2 an average of the values that occur in a reasonable range In the Data Analysis section of your lab report you will use this experimentallydetermined resistance value to calculate the dimensionindependent resistivity p and conductivity 0 of your sample Be sure to dispose of all the CqugI4 in the waste container provided Data Analysis 1 6 points Record your value of the transition temperature of CqugI4 and the colors of the lowtemperature and hightemperature phases 2 6 points Record the electrical resistance R of your CqugI4 sample at room temperature and the values of R that you got at high temperature If you got several values of R at high temperature state which one you will use in the remaining calculations judging by either 1 the one that seems to be most stable or 2 an average of the values that occur in a reasonable range 3 7 points Calculate the dimensionless resistivity p of CqugI4 in the hightemperature phase Show all your calculations including those for the crosssectional area of your sample 4 6 points Calculate the conductivity of CqugI4 in the hightemperature phase Discussion 1 8 points Discuss whether the CqugI4 phase transition can be thermally cycled or whether the first phase transition is irreversible and how you determined this Why or why not can this phase transition be thermally cycled 2 7 points Compare the value of the resistance R of CqugI4 in the room temperature phase and the high temperature phase What are the likely resistivity and conductivity characteristics of CqugI4 in the low temperature phase 3 7 points Discuss the magnitude of the conductivity of CqugI4 in the hightemperature phase and the reason for its difference from that of the lowtemperature phase Questions 1 6 points When you made CqugI4 you filtered it while the suspension was still hot What impurity or impurities were you trying to avoid by ltering while hot 2 6 points CqugI4 is an ionic conductor which means that charge is carried by the net movement of ions rather than by the motion of electrons through a stationary lattice Give two reasons for why the cations are mobile in this compound but the anions are stationary 3 6 points Do you expect the Cu or Hg ions to be more mobile Explain your answer Experiment 1 HardSoft Acids and Bases Altering the CquCu2 Equilibrium with Nitrogen Oxygen or Halide Ligands Pre Lab Assignments and Notes 1 Calculate the correct mass of stalen needed MW 2683 gmol see molecular structure on page 3 to form a 11 complex with CuClz using 2 mL of a 003 M copperll chloride solution 2 Write out the procedure you will use in Part G to determine the relative stabilities of CuII and Cul when ligated by salen State what you expect to observe when you carry out the procedure You need not calculate the quantities of the reagents you ll use Note The abbreviation quotenquot refers to ethylenediamine a ligand with two nitrogen donor sites bridged by an ethane moiety HzNCHZCHzNHZ Both nitrogens bind to metals as shown A H2N M IH2 H2N NH2 ethylenediamine Experimental Procedure SAFE T Y N 0T ES Aqueous ammonia ammonium hydroxide gives off corrosive vapors Handle this reagent in the hood Ethylenediamine may give off irritating vapors handle in hood H Cl aqueous ammonia and 30 H 2 02 are extremely corrosive H 202 can react violently with organic solvents and substances such as clothes hair paper and dust Wipe up all spills immediately with water If skin or eyes are exposed instantly ush the affected area with copious amounts of water for fifteen minutes H zsalen is a skin irritant if exposed wash thoroughly with water Complexation of CopperII by Nitrogen Donors A Add about lmL of 003 M copperII chloride solution a small test tube and in the hood add 1 mL of 6 M aqueous ammonia Record your observations Color Color change Precipitation Gas evolution Other changes Or no changes thoroughly and completely B Add about lmL of 003M copperII chloride solution to a small test tubes and in the hood add 1 mL of 25 aqueous ethylenediamine to the other As always in this week39s experiment record observations Reduction of CopperII to CopperI by Halide Ligands C Four 10 mL of 003M copperII chloride solution into a small Erlenmeyer ask Add 1 g of ammonium chloride to the solution and stir or swirl to dissolve Note any changes Then heat the solution to boiling on a Bunsen burner to expel oxygen and add an excess of clean copper wire 3 of heavy gauge wire is more than enough or about 01 g of clean copper tumings and continue heating Note what happens as this slow heterogeneous reaction proceeds and record the results Note39 CuH exhibits colors in the visible spectrum CuI does not exhibit visible colors and Cu0 is a red metal D Take a small test tube and add 2 mL of 003 M copperII chloride solution Add an equal amount of 01 M potassium iodide solution Note what happens and record the results as always record observations thoroughly and completely Oxidation of CopperI to CopperII By Amine Ligands E Divide the material both solution and precipitate in the test tube from procedure step D into two equal parts It should be a suspension of CuI In the hood add about 1 mL of 6 M aqueous ammonia to one part and carefully record your observations This reaction may proceed differently than reaction A Note any differences or similarities and explain them in the discussion section of your lab report F In the hood add about 1 mL of 25 ethylenediamine to the other part saved from D Again carefully record your observations G Synthesize the ligand bissalicylideneethylenediamine or salen or quotstalenquotsee Figure l by taking 3 mL of 95 ethanol in a test tube that also has a small magnetic stirring bar Heat the ethanol to boiling in an oil bath DO NOT use a Bunsen burner while stirring on a magnetic stirrerhotplate Immediately with continued heating and stirring add 030 mL 28 mmol of salicylaldehyde and then 010 mL 14 mmol of ethylenediamine 98 Heat and stir the solution to boiling about three minutes and then cool in an icewater bath to precipitate the yellow stalen Filter your product wash it with drops of icecold ethanol and dry it as much as possible on the filter by vacuum Then collect the crystals and spread them on a large piece of filter paper not the waxy paper used to weigh out reagents to airdry Weigh your product and determine and record your per cent yield 7 N N 0 7 H2N NH2 2 dOH gt OH HO 2H2O 5 sal HZSalen Cu2 CM N Ecui O O Cusalen Figure 1 Synthesis of stalen and formation of the Cusalen complex Develop an experimental scheme to test the equilibria of the oxidation states of copper when ligated by the hard donor ligand salen For example 1 you could show that CuIIsalen is not easily reduced to copperI by copper metal Equation 1 below and thus that CuII is the more stable oxidation state when ligated by salen Alternatively 2 CuIsalen should spontaneously disproportionate to Cu0 and CuII Equation 2 demonstrating that the equilibrium lies towards CuIICu0 the more stable oxidation states with this harddonor ligand CuHsalen0 Cu0 2 CuIsalen39 1 2 CuIsalen S CuHsalen0 Cu0 salenz39 2 Show your scheme to a TA for approval before attempting your procedure Carry out your scheme and record your results Table 1 Reduction Potentials Involving Copper and Copper Compounds Couple EO value 1 Cu2 1 e CuI 086V 2 Cu2 C139 e CuCl 054v 3 12 2e39 3 2139 054v 4 011 aq e39 r Cus 052v 5 Cu2aq 2e39 3 Cus 037v 6 CuCl e39 3 Cu C1quot 014v 7 CuNH342 2e39 3 Cu 4NH3 0 12v 8 Cu2aq e Cu aq 0 15v 9 Cu1 e39 3 Cu 1 0 19v 10 Cuenzz 2e39 3 Cu 2en 050V en ethylenediamine Data Analysis 1 6 points Write the equation for each reaction you did in procedures A through G both charge and mass balance should be satis ed Then identify from the observations recorded in your notebook the products you obtained from each reaction and whether they concur with the expected products A and B are ligandcomplexing reactions Expected chemical products are A CuNH342 and B Cuenz2 C and D are reduction reactions so halfreactions must added be arrive at the balanced equations Table 1 above Halfreactions 2 and 6 are relevant to procedure C and 1 and 3 are relevant to procedure D Expected chemical products are C CuCl and D CuI E and F are oxidation reactions Halfreactions 7 and 9 are relevant to procedure E and 9 and 10 to procedure F Expected chemical products are E CuNH342 and F Cuenz2 2 4 points Using Table 1 write out individual halfreactions their balanced overall redox reactions and the overall E0 potential that describe the chemical reactions in procedures A through F Halfreactions 5 and 7 are relevant to procedure A Halfreactions 5 and 10 are relevant to procedure B Use your answers to Data Analysis topic 1 above for halfreactions and balanced equations relevant to procedures C F In each case the halfreactions should sum to the net balanced redox equation with no e39 left over 3 5 points Using Table I calculate the E0 for the preparation of CuI from CuII and Cu0 in the presence of chloride ion halfreactions 2 and 6 use your answers to procedure C from the above topics Calculate the E0 for the same reaction in the absence of chloride ion halfreactions 4 and 8 4 5 points Use the result from the previous problem to calculate the free energy change AGO for the preparation of CuI from copperII and Cu0 in the presence of chloride ion Then calculate the AG0 for the same reaction in the absence of chloride ion Discussion 1 6 points State what the HSAB theory predicts and discuss whether each of your obtained reaction products concurs with the HSAB predictions See the diagram depicting hard and soft acids and bases located at the end of this EP 2 5 points Discuss whether the E0 values obtained for procedures A through F correctly predict your reaction product results In any case did your results disagree with the predictions of the reactions overall E0 potential If so explain why that might be 3 6 points Discuss whether the E0 values for each of procedures A through F concur with what the HSAB theory predicts 4 4 points Discuss what the experimental scheme you developed in procedure G concerning the equilibria of the oxidation states of copper when ligated by salen proved in terms of the HSAB theory 5 6 points Compare the electrochemical potentials and the AG0 values for the preparation of CuI in the presence and absence of the chloride ion Which preparation is more thermodynamically favorable and why What is the reason for the difference in thermodynamic favorability of the two preparations 6 3 points In what ways do our experimental conditions vary from those used to calculate the standard E0 potentials Do the actual electrochemical potential values of our experimental conditions vary a great deal from the standard E0 potentials Use the identity of the products you obtained in procedures A through F to arrive at and justify your answer 10 points Conclusions What do your results the HSAB theory and calculated electrochemical potentials show about how we can manipulate the concentrations of different oxidation states of transition metals Questions 1 2 points Why is CuI colorless while CuII tends to be colored green or blue Think about electron configurations 2 3 points Copper 1 deactivators used in gasoline include a ligand analogous to salen with 12 diaminopropane in place of en This ligand prevents the copperI catalyzed formation of gum in the gasoline Explain why this works in terms of your results Hard amp Soft Acids amp Bases Table 81 from Wullsbeq Principls a Dscn39p 39w Inorganic Chemistryquot u m N M it an Rb an Sc I 1 Ln A11 2 u m u and Mid v Cr 5 LGG M M mm 139 w u 7351 n N Lu LII r u u u F J u 2 can mum Saul 7 n N A as s m nuu n 20 19 I um I um r2 0 u m Ir n In H 39n n m n J12l m 254 20 Hum Hy 1110 ms Bariumgnu Pm Sm Gd n Dy Ha a Tm I39I no 22 us 114 1 Nu Pu Am Cm Bk c a ma m 136 12 I u u u u u I Yb N u Lecture Experiment 7 Thermochromism and Ionic Conductivity in Solid State CqugI4 This experiment studies the properties that can be obtained in what is called inorganic solid state chemistry Solid state compounds are in nite non molecular solids There are no separate molecules of CqugI4 present in the conductive thermochromic solid we will study in this experiment but instead in nite arrays of atoms packed together in different arrangements The subscripts in the formula CqugI4 simply give the ratios of the atoms in the solid You are used to molecular chemistry either in solution or in molecular species that just happen to be solids at room temperature The difference between such compounds and non molecular ones is that the packing structure or the ph of the infinite solid is what determines its properties The most common example of this is diamond and graphite Both are infinite solids with exactly the same formula Cn Diamond s phase is a cubic one in which each carbon forms four bonds to every other Graphite is hexagonal an in nite twodimensional array of stacked fused rings in which the carbons are doublebonded The properties of these two phases of carbon could not be more different because of their differences in structure Diamond is the hardest material known and the best electrical and thermal insulator Graphite is one of the softest of materials and is a good electrical and thermal conductor While in molecular solids the identity and placement of the functional groups the structure of the molecule determines its properties in solid state chemistry it is the phase of the nonmolecular solid that does so The objectives of this experiment slide 1 is to synthesize one phase of CqugI4 at room temperature then convert it to a different phase using high temperature The difference between the two phases colors and electrical conductivities caused by the phase change will then be observed Structural Descriptions of Solids The simplest structural description of an in nite solid is that it consists of close packed layers slide 2 In this description all the closepacked atoms are the same the closepacked atoms in CqugI4 are the iodide ions Every atom is coordinated by six other atoms in the same layer and three more atoms are placed in the layer above in the depressions made by the first layer There is room for only half the depressions to be lled The third layer can now be placed in two ways First all the closepacked atoms can be placed directly above the atoms in the first layers slide 2 This leads to hexagonal symmetry and therefore to hexagonal closepacking Second the third layer of atoms can be placed above the unfilled depressions of the first layer This leads to cubic symmetry and therefore to cubic close packing In each kind of close packing the space filled by the closepacked atoms is 74 This is the most efficient way to pack any spheres of the same size in space In both packings the 26 of unfilled space is organized into what are called sites voids vacancies or holes slide three Octahedral Oh sites are formed when six closepacked atoms in two different layers come into contact slides 3 and 4 These sites can hold another type of atom of a radius up to 41 of the radius of the closepacked atoms There is one Oh site for every close packed atom Tetrahedral Td sites are formed when four closepacked atoms in two different layers come into contact slides 3 and 4 These sites can hold another type of atom of a radius up to 23 of the radius of the closepacked atoms There are two Td sites for every close packed atom The organization of the vacancies in solids allows the formation of organized solid state phases such as the roomtemperature phase of CqugI4 that we will study As an example some crystals may form a cubic array with the anions closepacked because they are larger and the cations because they are smaller sitting in the Oh or Td sites Slide 5 shows sodium chloride in which the cubic closepacked array of Cl39 ions has a Na ion occupying each Oh site Its unit cell shown on slide 5 is defined as the smallest repeating volume element in the structure The translation of the unit cell in three dimensions therefore generates the entire structure For NaCl the unit cell is facecentered because one Cl39 ion lies on every face of the cube that makes up the cell slide 5 There are many shapes of unit cells some can be rectangular or hexagonal or shaped like other polygons Similarly the zinc sul de ZnS unit cell is found to have a facecentered cubic array of 8392 ions with Zn2 sitting in half the available Td sites slide 6 The doubled ZnS unit cell is analogous to the ordered roomtemperature cell of CqugI4 slide 6 The unit cell of this phase of CqugI4 consists of a closepacked array of 139 ions with Cu and Hg 2 cations sitting in the various sites in a tetragonal unit cell The unit cell is shaped like a rectangle with square ends slide 6 CqugI4 Structures We can see from slide 6 that two Cu1 and one Hg2 ions are distributed in three of the eight available Td sites in the roomtemperature structure of CqugI4 Each cation is coordinated by four 139 ions This structure is ordered because the di erent cations occur in layers first a layer of I39 then one of Cu then I39 then one of Hg2 etc The structure occurs at room temperature and lower It is an organized structure because these particular cations occur at these particular organized sites But note that this structure is defective one Hg2 ion is missing from each of its layers This is because the compound s stoichiometry demands only one Hg2 per every four I39 But leaving this Td site empty is like pulling a brick out of a wall it generates a large structural defect This phase is more unstable and therefore more reactive because of the defect Therefore at higher temperatures the compound can change its phase The 139 ions collapse to a facecentered cubic unit cell slide 6 The Cu1 and Hg2 ions are still in Td sites but their placements are not organized but randomly scattered A stronger structure is formed Three out of four Td sites are filled with either a Cu1 or a Hg2 or are empty but each unit cell can have these ions placed in di erent sites Each unit cell can be different This is the hightemperature disordered phase Now that we have had a structure change we may expect quite different properties of the two phases as we see with diamond and graphite Differences Between the Ordered and Disordered Phases A color change arises when the lowtemperature ordered phase is converted to the high temperature disordered phase This change in color with temperature is called thermochromism It arises from a change in energy between lowlying lled electron sites and highlying empty electron sites in the two phases These are called semiconducting bands and are analogous to lowlying lled and highlying empty molecular orbitals in molecular compounds The energy needed to promote an electron from the lled band to the empty one changes from 21 eV for the room temperature phase to 19 eV for the high temperature phase Thus the color change that is seen between the two phases This color change will allow you to determine the phasechange temperature of CqugI4 using a melting point apparatus The electrical conductivity of the phases of CqugI4 also changes as the high temperature phase is capable of ionic conductance slide 7 This is electrical conduction generated not by the movement of electrons through the solid but by the movement of ions which are also charged particles Ions in some phases can hop from one vacant site to another thus moving across the solid and carrying electrical charge through it When a voltage is applied the charged ions move through the in nite structure and transport a current CqugI4 conducts cations in the disordered phase Many of the Td and all of the Oh holes are empty and the small cations can hop from one end of the solid to the other Ions are much bigger and harder to move than are electrons so there are not very many good ionic conductors known In the hightemperature phase it is easier for the cations to move because the iodide closepacked lattice is more stable yet many vacant sites still remain Resistance Resistivity and Conductivity What you will measure experimentally is the resistance of the two phases of CqugI4 you will then be able to calculate their resistivity and conductivity The resistance is measured slide 8 by making a cylindrical sample of CqugI4 and applying voltage to it The resistance for both phases in ohms is then read off of an ohmmeter However resistance is not independent of sample geometry It must be converted to the dimensionless resistivity p slide 9 a measure of resistance that is not dependent on the sample s geometry The conductivity of both phases slide 10 can then be calculated In conclusion an infinite nonmolecular solid CqugI4 will be synthesized in two different phases and the difference in physical properties that different phase structure has on the same nonmolecular compound will be determined Exp enmem 4 Smdying Lhe Spectmchemiczl Series Crystal Fields nf cyan lnnuduc nn Caaxdnnanm campamds af mhmmems ne a mhg y calmed The calm esuns rm ahsmphm uf 19 n spenm wavelznghs afvmble 19 ssnemeawmn eeeuane kamth Mann me 1 mbxlals Thus these 4 kmsuans gve many kmauanm em mus men ennnaensne calm eg cabaltblue The 1 mbxlals af 5 metal mn n an Dashednl cysts eld mmmdedby an acuhz 39al may ufhgnnds ne spmm mg marge set and alawex emxgyt gt xgme 1 Thsxs z axes The mdgy exence between me We andlwwex emxgylevelsxs deagmzd s A pmnnunseaae enm a man We 9888 n y Figure 111mhtals spmbynn actshednl ayml sea The aegee af snnmng uf me 1 anqu and hence une magnum uf Ac depends an several facmrs mun ma xmpanmt ne me ennge an me new and me aenmy uf une ngnnn Understanding the trends in ligandfield splitting is simplified considerably by considering a series of complexes with the same metal in a given oxidation state the only major variable in this case is the ligand identity From a large number of studies it is known that ligands can be arranged in a sequence according to their ability to cause d orbital splitting This series is known as the spectrochemical series halides lt NC 39 lt OHquot lt oxalate lt H20 lt ECS39 lt pyridine lt NH3 lt en lt E02quot lt 2Nquot lt 20 Underlined atom is the one coordinated to the metal Note the general trend shown here halide ligands are weakerfield ligands than sulfur donors which are weaker field than oxygen donor ligands which are weakerfield than nitrogen donors which are weakerfield than multiplybonded ligands The stronger the interaction of the ligand with the metal s dorbitals the stronger the field strength of that ligand The magnitude of ligand field strength increases by a factor of about two as one moves from halide to CN39 in the spectrochemical series The structure of the oxalate ion C2042 is shown below Note the resonance forms of this compound 00 00 Figure 2 Structure of the oxalate ion The objective of this experiment is to quantify A0 for a series of CrIII complexes by electronic absorption spectroscopy CrIII compounds are d3 and their electronic spectral characteristics are reasonably easy to interpret Because of electronic selection rules see lecture the absorption band corresponding to the energy of the crystal eld strength Ao will be the one at the longest wavelength lowest energy in the spectrum and it should be more intense than any other nearby transition Ordering the octahedral CrIII compounds from longest to shortest wavelength will place the ligands in order of increasing crystal eld strength as A DC lAt7 and will allow you to build your own spectrochemical series In mixedligand complexes the Rule of Average Environments states that the observed value of A0 in such complexes is the weighted average of A0 for each of the homoleptic single type of ligand complexes The rst equation below is general the second equation is for the specific example of CrHzO4Clz If the Ads of CrHzO4Clz and CrHzO53 are known then by rearranging the equation you can solve for the A0 of CrC153 You can use this A0 to nd the A0 of other unknown complexes AOMAan 16 nA0MA6 onMB 1 A0CrH204C12 16 4A0CrH2063 2A0CIC16339 2 Syntheses In this experiment the bidentate ligand acetylacetonate acac39 will be generated via the deprotonation of acetylacetone acacH by ammonia The ammonia is generated by hydrolysis of urea Figure 3 subsequently ammonia acts as a base to deprotonate acacH Note the resonance forms of acac O k H20 gt 2NH3 C02 H2N NH2 O O H o o o 0 H30 C CH3 H3O c CH3 H3O c CH3 H2 H H Figure 3 Top Hydrolysis of urea Bottom Deprotonation of acetylacetone forms the bidentate ligand acetylacetonate acac Systematic name is 24pentanedionate Experimental Procedure You will work in pairs in this lab to prepare two compounds Cracac3 and Cren3C13 2H20 One of the pair will synthesize one compound and the other partner will prepare the other compound The procedure for the syntheses for both compounds must be written in your prelab You will then share experimental and spectroscopic data for all the compounds not just for the two that you prepare Another compound CrHzO5N033 3HzO will be provided to you you will spectroscopically analyze both this complex and your starting material CrC13 6HzO this compound is actually CrClzHzO4Cl 2HzO Finally the electronic spectrum of CrNH35Cl C12 will be provided on the class web page for your use in the Data Analysis section of your lab report How to Heat a Reaction with a Sand BathMagnetic Stirrer In this experiment both syntheses require that you heat your reaction mixtures to boiling or to quotre uxquot in a sand bath placed on top of a magnetic stirring plate Getting the reaction mixture to stir well while heating can be problematic so use the following procedure Place an empty beaker containing a stir bar on the magnetic stirrer and turn on the stirrer Check to see that the bar can stir quickly and well If it cannot use a di erent magnetic stirrer Place your sand bath on top of the magnetic stirrer and bury your ask containing the stir bar and the solution to be heated deep into the sand Turn on the magnetic stirrer very slowly and try to get the stir bar to stir vigorously There is often a problem in accomplishing this because the stir bar is too far away from the magnetic stirrer to respond well to it To correct this take sand out of your sand bath place it in a beaker and RETURN IT to the sand bath when you are finished and move your ask down towards the bottom of the bath until it almost touches the metal bottom of the sand bath The closer the bottom of the ask comes to the bottom of the sand bath the better the stir bar will stir Do not however let the bottom of your ask come into direct contact with the metal of the sand bath the ask may melt Again turn on the magnetic stirrer very slowly and keep removing sand and moving the ask down into the bath until the bar is stirring vigorously and well Now heap the extra sand back into the bath underneath and around the shoulders of the ask to help transfer heat and encourage quick heating of the solution When your reaction mixture is vigorously stirring and the extra sand has been replaced around the shoulders of the ask turn on the sand bath and heat your reaction mixture Make sure if possible that your reaction mixture is stirring vigorously throughout the heating period If the stirring stops repeat the above procedure until you can get efficient stirring again A Cracac3 the preparation of T139is24 pentanedionatechr0miumIII Dissolve 260 mg of CrC13 6HzO in 40 mL of distilled water in a 25 mL roundbottomed ask and then add 1 g of urea to the ask Measure out 08 mL of acetylacetone using the lmL graduated pipette provided and add the acetylacetone to the ask also Next place a re ux condenser on the ask after first applying a thin layer of stopcock grease to the male ground glass joint of the re ux condenser Re ux your reaction with stirring for one hour using a sand bath As the urea releases ammonia and the solution becomes basic deep maroon crystals begin to form After one hour cool the ask thoroughly in an ice bath Very thorough cooling is needed If no crystals form carefully make a small scratch on the bottom of the ask using a glass rod or metal spatula and cool the ask again Collect the crystals by suction ltration and wash them with three 03 mL portions of distilled water Dry the crystals as much as possible on the lter using the aspirator and then collect them and spread them on a piece of lter paper to air dry Determine the percentage yield and transfer the Cracac3 to a labeled vial B Cren3C132H20 the preparation of TI39isethylenediaminechr0miumIII Weigh out 100 mg of mossy zinc into a 25 mL roundbottomed ask Remove the surface layer of ZnO and generate a clean and active Zn0 surface by washing the Zn with HCl immediately prior to use To accomplish this add enough 6M HCl to cover the mossy zinc swirl the ask brie y the mossy zinc may completely dissolve upon prolonged contact with HCl and pipette off the acid Then wash the mossy zinc three times with 10 mL portions of methanol to remove as much acid as possible Weigh out 266 mg of CrC13 6H20 and add it and 1 mL of methanol to the roundbottomed ask In the hood add 1 mL of ethylenediamine Next place a re ux condenser on the ask after rst applying a thin layer of stopcock grease to the male ground glass joint of the re ux condenser Re ux your reaction with stirring for one hour using a sand bath Take care not to turn up the sand bath power too high methanol re uxes at a lower temperature 65 0C than water 100 0C Cool the purple solution in an ice bath Collect the yellow crystalline product it will appear purple because it is wet with the solution by suction filtration using a Hirsch funnel Remove any unreacted zinc with tweezers Wash the filtered product with 05 mL portions of 10 ethylenediamine in methanol until the purple solution is gone the product is yellow and the washings are colorless Follow this with a 05 mL rinse with ether to rinse away the methanol and help dry the product Dry the crystals as much as possible on the filter using the aspirator and then collect them and spread them on a piece of lter paper to air dry Determine the percentage yield and transfer the Cren3C13 2HzO to a labeled vial C Spectroscopy of the CrIII Complexes Prepare aqueous solutions of Cren3C13 2HzO and CrHzO5N033 3HzO by dissolving approximately 5 mg of each complex in about 5 ml of distilled water Because the analysis of A0 the objective of this experiment requires only that the longestwavelength km of each complex be identi ed you do not need to know the exact concentration of the solution as you do when calculating extinction coef cients using Beer39s law Also prepare a similar ethanol solution of Cracac3 Make a similar solution of CrC13 6HzO in water but note that this complex consisting of CrClzHzO4 ions in water slowly substitutes to the hexaaquo species CrHzO53 so it should be analyzed immediately after its preparation Do not prepare this solution until your name is called for use of the UVvis instrument At that point quickly add alreadyprepared portions of CrC13 6HzO and water to a cuvette mix the solution a few times by drawing it up in a Pasteur pipette and then acquire the spectrum of this solution rst Obtain the electronic absorption spectrum of each complex Record the wavelengths and absorbance values for all absorbances for each sample in your notebook Determine and record the longest wavelength of the absorbance peaks for each complex in units of nanometers This is the wavelength you will use to calculate the Ads of the complexes in the Data Analysis section of your lab report Data Analysis 1 3 points Draw the crystal eld energy levels and electron occupancies for octahedral Cr3 ions Indicate the energy gap that corresponds to the transition that is being investigated in this lab 2 10 points Prepare a table of your data including columns for max in nm and A0 in cm39l eV and kJmol a Convert the wavelengths which correspond to A0 into wavenumbers cm39l using the following relationship A0 lA nm 1 x 107 cm391 b Other energy units for A0 may be obtained using the following conversion factors lcm391 124 x104 eV 001196 kJmol c Arrange the entries in order of increasing gap energy Show your complete set of calculations for one of the complexes For the rest of the complexes just record the calculated values in the table 3 12 points List the five ligands in order of increasing ligand field strength Use the Rule of Average Environments to calculate the true ligand field strengths of two of the ligands to be included in the list above Show your complete set of calculations Discussion 1 10 points Discuss the order of the ligand field strengths you obtained comparing them with the spectrochemical series given in the introduction Does the order of ligands obtained by this experiment correspond to the established order of the spectrochemical series Explain any deviations 2 5 points List the five ligands in order of the increasing strength of their interaction with Cr and give the reasoning behind your answer 3 5 points The ligand acac39 is not included in the spectrochemical series given in the Introduction Examining the chemical structure of acac39 and of those of related ligands shown in the spectrochemical series explain where you would expect to find acac39 in the spectrochemical series Does your calculated ligand eld strength for acac39 agree with this prediction Conclusions 10 points Discuss what knowledge has been gained by this experiment Discuss the importance to inorganic chemistry of being able to ascertain the crystal eld energy gap of complexes and the ligand eld strength of ligands What information can be obtained or what advantages in other experiments can be gained about inorganic complexes using these two quantities Questions 1 5 points What is a signi cant difference that is seen between the the UVVisible spectrum of Cracac3 and the spectra of the other complexes What is the reason for this difference 2 5 points Highspin MnII and FeIII complexes are much less intensely colored than those of CrIII Why are they so weakly colored Lecture Experiment 3 Separation of the Oxidation States of Vanadium The objectives of this experiment as stated on slide 1 of the Powerpoint lecture slides is l to make complexes of the multiple oxidation states of vanadium and 2 to separate those complexes via ion exchange chromatography Why should we want to access multiple oxidation states of an inorganic element The transition metals characteristic chemistry arises from their multiple oxidation states No other group of elements displays so many stable oxidation states and no other group can change easily from one oxidation state to another as transition metals can This is because the dorbitals are the valence outermost orbitals of transition metals There are relatively many dorbitals they are closely spaced in energy and electrons within them in the various oxidation states can adopt a variety of stable con gurations rendering that particular oxidation state stable Other groups of elements main groups lanthanides and actinides do not have dorbitals as their valence orbitals and cannot undergo this chemistry The reactivity of transition metals39 di erent oxidation states is m differentnMnOz Mn4 for example is a stable compound used as a solid support while KMnO4 Mn is such a rapid oxidizing agent that it can even be explosive with organic solvents The very different reactivity of different oxidation states makes it important to choose correct one to obtain the reaction or reactivity that you desire Correct oxidation states are vital for all kinds of catalysis both biochemical and industrial the Pd 2 Pd0 oxidation states for example are very widely used in organic synthesis especially in the synthesis of highlyfunctionalized drugs and pharmaceuticals plastics and specialty organic chemicals Accessing the correct oxidation state of metals is also crucial in bioinorganic chemistry For example small clusters composed of iron and sulfur clusters Fe4S4 in cubic shapes act as electron shuttles in hundreds of enzymes Each iron is initially in oxidation state 3 Each iron can then accept an electron taking it to Fe2 The electrons are then passed along to enzyme sites where they can participate in reactions leaving the irons as Fe3 again ready to pick up more electrons and shuttle them into the reaction site by changing between the 3 and 2 oxidation states Humans and mammals use a suite of over 100 ironbased enzymes such as hemoglobin and cytochrome c to bind transport and use 02 Every enzyme depends on maintaining speci c oxidation states of iron and of other metals within the enzymes Each metal must be in the exactly correct oxidation state for the enzyme to perform its particular biochemical transformation of 02 However a class of sea creatures known as ascidians or sea squirts actually use the multiple accessible oxidation states of vanadium in an analogous suite of enzymes for the same purpose of binding transporting and using 02 Again having the vanadium metal centers in the correct oxidation states for the particular biochemical transformation of 02 by a particular enzyme is crucial In these mammalian and ascidian enzyme suites multiple oxidation states of the respective transition metals are accessed and used to bind O2 carry it to cells react it with compounds and transform it into usuallyunstable oxygencontaining ions The metals39 different oxidation states allow for the enzymes vital abilities to transform O2 to vital unusual oxidation states including H20 0 in 2 oxidation state H202 O in 1 oxidation state the peroxide ion and 0239 O in 12 oxidation state the superoxide ion All these unusual oxidation states of oxygen are essential to life and can only be generated by reaction of oxygenspecies with transition metals in variable oxidation states So the reactions of transition metals in different oxidationsstate complexes could be models for inorganic Oztransport systems Therefore an important part of inorganic chemistry is to learn to synthesize different oxidation states of the transition metals then to separate them to isolate the particular one we want That is objective of this lab Vanadium is in Group 5 in periodic table slide 2 and is a firstrow transition element It has five valence electrons this is a consequence of its being in Group 5 can therefore lose up to five electrons and therefore has a maximum oxidation state of V Lower oxidation states can be obtained and are stable down to V Other states even negative oxidation states are accessible but easily react with air and water So these states are not relevant to bioinorganic chemistry which takes place in air and water and we will not attempt to access them in this lab They can be synthesized by the use of airsensitive techniques The vanadium complexes of different oxidation states we will use in this lab are aquo complexes slide 3 VH206 VH2063 VH2050 V0H03239 quot0quot denotes the oxo ligand 02 which is doublybonded to the vanadium with a VO structure VOH03239 therefore has tetrahedral geometry while the other complexes are octahedral Note that the different oxidation states from 5 to 2 are all present To calculate these oxidation states use the following procedures VOH03239 one negative charge from OH39 ligand six from three 0392 ligands total charge on complex is 2 therefore V oxidation state is 5 VHzO5O2 two negative charges from O392 ligand none from neutral water ligands Total charge on complex is 2 so V oxidation state is 4 VHzO53 VHzO5z no negative charge from neutral water ligands total charges on complexes are 3 and 2 therefore V oxidation states are 3 and 2 respectively Remember that oxidations states are not necessarily equal to charge on complex There are shorthand notations for the aquo complexes of V which we will use given on slide 4 Note also that V0339 is also called the quotvanadatequot ion and VO2 is called the quotvanadylquot ion To synthesize different aquo complexes in different oxidation states we will start with highest oxidation state VO339 5 and use sequential reductions to access the other oxidation states The reducing agents we will use are shown on slide 5 HCl is the reducing agent that will take V5 to V and Zn amalgam will be use to reduce V4 to V3 and V quotZn amalgamquot is a quotsolutionquot of solid Zn in liquid mercury The mercury is said to quotamalgamatequot the zinc These sort of mercury amalgams can be made with many metals for example sodium or cobalt The different metals form solutions or solids when quotdissolvedquot depending on the ratios of metal and mercury used Amalgams exist as small even atomic particles of metal in this case zinc dispersed in mercury This gives the reactive metal in this case zinc very large surface area making it an excellent reducing agent As we will see zinc reacts very quickly from the amalgam to reduce the various vanadium complexes The chemistry of reduction is Zn0 lt gt Zn2 2 e39 There is no change in the mercury in any reaction the mercury is a carrier only It will not appear in any reactions that you write or redox potentials that you calculate in this experiment The standard redox potentials E0 for converting one oxidation state of V to another are given on slide 6 In this experiment we are again far from using standard conditions but the standard potentials work closely enough for our purposes For these syntheses two reducing reactions are needed for VT5 to V and for VT4 to V3 and V These are shown on slide 7 The reducing agent in the V5 to VT4 synthesis is actually chloride ionit goes from the l to the zero oxidation state DO NOT breathe the fumes of this reaction or take it out of the hood It will be evolving C12 gas which is extremely corrosive For other two reductions Zn from the amalgam is the reducing agent Because this reagent reacts very quickly it is sometimes dif cult to obtain the V3 complex as all the V4 may be rapidly reduced to V This may happen in first reduction as well with HCl reducing all the VT5 complex to V4 We use HCl and ZnHg in this experiment however because they are not airsensitive reagents ie they do not react with water or dioxygen The reagents that more slowly do this reduction and with which different oxidation states can be quantitatively synthesized are airsensitive But in a research lab using airsensitive techniques one can obtain the desired oxidation states quantitatively and selectively Once the vanadium complexes in different oxidation states are synthesized they must be separated This will be done using ion exchange chromatography slide 8 This is a very convenient very efficient separation method for many kinds of ions so we use it here But note that compounds must be ionic for this separation method to work The column is packed with an ion exchange resin which consists of an insoluble polymer base quotresinquot means an insoluble very hard tough polymer in bead form The top surface of the resin is functionalized with bound SOg39 sulfonate groups see slide each with a counter cation Resins are usually supplied in what is called quotHformquot mean the countercations of the sulfonates are H In first step of our separations we prepare the column by passing acid over it ensuring that resin in completely in the Hform Solution of the ions to be separated are then added In high concentrations they will replace H as the bound counter cations and H ions will be removed from the sulfonate groups while the vanadium complexes are bound to them The highest charged metal cation complex will be the most tightly bound as it can bind to and bridge many sulfonate anions The next highest positively charged complex will be the next most tightly bound and etc So differently charged cations can be separated by strong or weak binding to the column Now more His added in the form of HCl to the column In high concentrations equilibrium conditions will cause the least weakly bound vanadium complex the lowest positively charged complex to be displaced from the sulfonates and that complex will elute from the column This displacement of one cation by another by manipulation of the relative concentrations of the two species is called quotcompetitive bindingquot At first there is not a high enough concentration of H to break up the binding of highest charged vanadium complex cations so these remain bound on the column Therefore the various vanadium ions are separated the least highly charged complex comes off the column and can be collected while the most highly charged complex remains bound on the column We then increase the concentration of H to the point where the next most highly charged ion elutes and can be selectively collected Then the next and the next etc until the acid concentration is so high that most tightly bound vanadium cation is displaced and comes off the column We will use this method to separate for example V which will come off column first 3 Since it has a lower pos1t1ve charge from V What happens to anionic complexes such as VOH03239 They alway elute first before any cation because they are repulsed by the negativelycharged sulfonate groups and are therefore not attracted at all to the column Anions will elute very quickly as explained in this experiment s EP The Experimental Procedures shown on slide 8 are discussed at great length in this experiment s EP so read it thoroughly These are the bases and procedures for separating all kinds of ions The concentrations of three of the vanadium complex ions can be found by titration with an oxidizing agent KMnO4 potassium permanganate slide 9 Permanganate ion is a good oxidizing agent because manganese is in the 7 oxidation state its highest and therefore its least stable There is a driving force to remove electrons from other species and go to lower oxidation states As seen in slide 9 a stable oxidation state for manganese is Mn2 and our vanadium complexes will reduce permanganate to this oxidation state Using the half reactions shown in the EP for vanadium you will write balanced redox reactions showing Mn7 oxidizing V4 to V3 and V3 to VH Write these as separate reactions V4 to V3 to V not as one overall step from say V4 to V You will calculate the number of moles of permanganate used in the titration from the number of drops used and the concentration of permanganate given in the EP This is equal to the number of moles of vanadium complex present in the correct ratio to manganese as given in the balanced redox reaction In conclusion this lab is an example of how one can synthesis separate and characterize the many valuable oxidation states of transition metals an important goal in inorganic chemistry Lecture Experiment 5 Measuring Metal Magnetism The objectives of this experiment as stated on slide 1 of the Powerpoint lecture slides are l to measure the magnetic susceptibility of metal complexes 2 to determine the magnetic moment of each metal 3 to determine the electron con guration whether highspin or low spin of each metal and 4 to compare these results to the ligands crystal eld strength Recall slide 2 how to draw both the lowspin and highspin configuration for any metal in any oxidation state In the highspin con guration the electrons are rst dispersed into both the lowlying and highlying orbitals then any remaining are placed in the lowlying orbitals In the lowspin con guration the electrons are rst dispersed only in the lowlying orbitals lling the highlying orbitals only when the lowlying ones are full The magnetic susceptibilty slide 3 of a solid x is the degree of its attraction or repulsion to or from an external magnetic eld It can be measured in many ways and units as the molar magnetic susceptibility XM which we will use the paramagentic magnetic susceptibility xpm and others The magnitude of x depends on 1 the magnetic moment generated on each metal atom by electon spin 2 what is known as spinorbit coupling a factor which increases a metal s magnetic moment and 3 the alignment of the individual magnetic moments in a solid xpm or xfmo or others The magnetic moment slide 4 generated on each metal atom by electron spin is the magnetism arising from the spin of unpaired electrons on that metal It is known as Meg meaning the effective magnetic moment as without also measuring the alignment of the individual magnetic moments we cannot determine it exactly We will just refer to it as the magnetic moment of each metal atom The magnetic moment is proportional to the number of unpaired electrons on a metal It is given by the equation Mz g S S l where S is the total electron spin the number of unpaired electrons divided by 2 g is called the Lande factor or the gyromagnetic ratio It is a constant the magnetic spin contributed by one free electron Its value for a free electron is 200000 to many decimal places However when the electron is not free but located on an atom the gfactor changes slightly A carboncentered electron has a gfactor of 20000001 a titaniumcentered electron has a gfactor of l900000 and other electrons centered on other metals have slightly different g factors We will use the approximation that for all our complexes g is approximately equal to 2 So if we substitute that S number of unpaired electrons divided by 2 and that g is approximately equal to 2 we will come up with the equation that we will use for calculating the magnetic moment of our complexes MEf 2 quotI101 2 S 0 means spinonly which will be explained below The next component of x to be considered is spinorbit coupling slide 5 This actually contributes to the magnetic moment ueff and therefore contributes to x indirectly Spinorbit coupling can occur when magnetism is generated in a complex by the metal s unpaired electron spins coupling with the valence orbitals angular momentum given by the orbital s magnetic quantum number m This will increase the value of the magnetic moment and so increase the value of the compound s magnetic susceptibility In many transition metal complexes however the spinorbit coupling is quenched by the binding of ligands to the d orbitals The contribution of the spinorbit coupling to the magnetic moment of our complexes therefore approaches zero and we will ignore it or as is said use the spinonly magnetic moment and not the magnetic moment that takes account of any spinorbit coupling The last factor to be considered in measuring the value of the magnetic susceptibility is the ordering or alignment of the individual magnetic moments within the solid slide 6 The magnetic moments may align in different ways all of which may affect the magnetic susceptibility As we see from slide 6 the magnetic moments of the individual metal atoms may be randomly aligned giving rise to paramagetism in which a factor of xpm is added to the magnetic susceptibility Paramagnetic substances are attracted to external magnetic fields xpm may be weak to moderate it is the resultant of all the randomly dispersed individual magnetic moment vectors In our powdered complexes to be used in this lab xpm is considered to be weak and will we ignore its effect on the overall molar magnetic susceptibility Ferromagnetism is achieved when all the magnetic moments are aligned giving rise to a very strong resultant and a very large xfem This is the type of magnetism seen in iron and in some other magnetic metals Antiferromagnetism is achieved when the individual magnetic moments are nearly all aligned antiparallel and thus nearly cancel each other out This contributes to a very low magnetic susceptibility Although on slide 6 the alignment of magnetic moments are drawn as pure phases this does not always occur in bulk solids They may have regions of ferromagnetism and of paramagnetism or especially common regions of paramagentism and antiferromagnetism within the same actual solid When a metal in a complex has no unpaired electrons slide 7 it is said to be diamagnetic Since there is no contribution to magnetism provided by electron spin S 0 the magnetic moment is also zero The magnetic susceptibility also closely approaches zero as it is so strongly dependant on magnetic moment But the paired electrons give rise to a very slight repulsion to magnetic fields causing a very small xdia to arise This diamagnetic susceptibility is too small to be measured by our experimental apparatus and we will take the magnetic susceptibility of diamagnetic complexes to be zero It can be exploited however in many applications slide 7 and it the basis for such machines as highspeed trains Magnetic Susceptibility vs Temperature Magnetic susceptibility is very sensitive to changes in temperature The individual magnetic moments may become more ordered at low temperature or less ordered at hgih tempertaure causing xpm to change dramatically Or an alignment change may occur such as from ferromagnetic alignment to anitferromagnetic alignment Regions of the solid which are at one alignment may grow larger or change to another alignment altogether Note that ueff at lower temperatures may become different from the ueffsospinorbit coupling may arise or different magnetic interactions may appear as discussed above Many of the complexes that we will use have the formula MSO4 tzO slide 9 Many of these are actually aquo complexes such as FeSO4397HzO which is actually FeHzO5SO439HzO Whether coordinated directly to water or perhaps to the sulfate ion all the donor atoms in these complexes are oxygen This will become important when you compare the highspinlowspin properties of these compounds to their ligand field strengths In conclusion we will determine the magnetism of various transition metal complexes with which many properties of practical value can be found For example from knowing the magnetism of complexes we can determine their strengths as magnets for use in electronc applications number of unpaired electrons oxidation states ligand eld strengths and magnetic coupling of metals in such complexes as organic catalysts and bioinorganic reaction centers


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