Answer Key, Problem Set 11

Chemistry 122 Mines, Spring 2018 Answer Key, Problem Set 11 NOTE: Tro always writes the formulas of complex ions in brackets, even when they are not part of a coordination compound (i.e., even if no counterions are present). This is fine, but I (and some others) often omit the brackets (but not the overall charge!!) when just writing the formula of the complex ion. Thus, in this key, when just complex ion formulas are written, you may see them with brackets or without. But just know that in the formula of a coordination compound, the brackets are (always) shown. Coordination Compounds (General Ideas) 1. Identify the Lewis acid and the Lewis base in the reaction of oxalate ions with Fe to give [Fe(ox)3] 3-. In the process, define Lewis acid and Lewis base and contrast this with the Brønsted-Lowry definitions. Answer: Fe is the Lewis acid and oxalate ion is the Lewis base. A Lewis acid is an electronpair acceptor (and so it is usually "electron deficient", as a metal cation most certainly is), and a Lewis base is an electron-pair donor, as are most anions (although many Lewis bases are (neutral) molecules with lone pairs as well). In any complex ion, the metal cation is the Lewis acid and the ligands are all Lewis bases. 2. State the i) coordination number and ii) # and type of donor atoms in each of the following complexes. Note: The definition in Tro for coordination number is historical and is technically incorrect in modern usage since it does not consider the possibility of polydentate ligands. The most straightforward definition of coordination number is the number of donor atoms making coordinate covalent bonds to the metal cation (This is consistent with the Wikipedia entry on Coordination complex : https://en.wikipedia.org/wiki/coordination_complex, but it not as easily discernable from the Wikipedia article entitled Coordination number. One must always be careful with Wikipedia or any other online sources!) The coordination number is determined by considering both the number of ligands bound AND how many atoms each ligand uses to bind to the metal ion. In this problem, only metal complexes are shown (i.e., the charge on the complex is shown, and there are no counterions, and thus no need for brackets. However, contrast this to problem 6 below.) (a) AgCl2 - : C.N. = 2 (two monodentate chloride ligands); 2 Cl donor atoms (one from each Cl - ) (b) [Cr(H2O)5F] 2+ : C.N. = 6 (five monodentate water ligands and one monodentate fluoride ligand; 5 O donor atoms (one from each H 2 O) and 1 F donor atom (from F - ) (c) [Co(NCS)4] 2- : C.N. = 4 (four monodentate thiocyanate ligands*); 4 N donor atoms (one from each NCS - ) Note: We know here that N from NCS - is bound and not S because the N is written farthest to the left. I.e., if S were the donor atom, the complex would have been written [Co(SCN)4] 2- (d) [Zr(ox)(CN) 4 ] 2- : C.N. = 6 (one bidentate oxalate ligand and four monodentate cyanide ligands); 2 O donor atoms (from the one oxalate ligand) and 4 C donor atoms (one from each CN - ) (e) Co(en)(NH 3 ) 3 (ONO) + : C.N. = 6 (one bidentate ethylene diamine ligand, three monodentate ammonia ligands and one monodentate ONO - [nitrite*] ligand); 5 N donor atoms (two from the en ligand and one from each NH 3 ) and 1 O donor atom (from the ONO - ) ligand. Note: We know here that O from ONO - is bound and not N because the O is written farthest to the left. I.e., if N were the donor atom, the complex would have been written Co(NH3)3(NO2)3. PS11-1

(f) [Fe(EDTA)] - : C.N. = 6 (one hexadendate EDTA ligand); 2 N donor atoms and 4 O donor atoms (from one EDTA ligand) *Although you don t need to know this for my course, when the same ligand binds via a different donor atom, the name of the ligand changes. E.g., NCS - is called isothiocyanato and SCN - is called thiocyanato. And ONO - is called nitrito whereas NO2 - is called nitro (see p. 1137-8 and Fig. 25.7 in Tro). 3. What is the oxidation state of the metal (i.e., charge on the metal cation) in each of the complexes in the prior problem? Strategy: 1) Recognize that the overall charge on the complex (ion) must equal the sum of the charges of the metal cation and (all of) the ligands. 2) Let x = the charge on the metal cation. 3) Sum up the charges on all of the ligands. Then set up an equation expressing the relationship in (1) above, and solve for x. (sum of ligand charge + x = overall charge) Note: Steps (2) and (3) need not be shown explicitly to earn credit on this type of problem since some students may choose to do this arithmetic in their head. I am showing the algebraic method here to show how I am getting these answers. Answers: (a) +1 Two Cl - 's sum to -2. So (-2 + x) = -1 (overall charge) x = +1 (b) +3 One F - and 5 neutral H 2 O's add up to -1. So (-1 + x) = 2+ (overall charge) x = +3 (c) +2 Four NCS - 's equal -4. So (-4 + x) = -2 (overall charge) x = +2 (d) +4 One ox equal -2 and four CN - s = -4. So total sum of ligand charges = -6. So (-6 + x) = -2 (overall charge on complex) x = +4 (e) +2 One neutral en and three neutral NH 3 's and one NO - 2 sum up to -1. So (-1 + x) = +1 x = +2 (f) +3 One EDTA is -4. So (-4 + x) = -1 x = +3 4. i) State the coordination number of the transition metal cation in each of the following. ii) Write out the balanced equation showing each of the following compounds dissolving in water (and dissociating into a complex ion and counter ion). Be sure to include all coefficients and charges! [Hint: It may be helpful to do part (ii) before part (i).] Strategy: 1) In coordination compound formulas, look for brackets. The metal complex is within the brackets. Anything outside the brackets represents counterions (i.e., not part of the complex; just part of the coordination compound). 2) Recognize the formula of any counterions, and realize that the subscript of the counterion (not any subscript that may be a part of the formula for the counterion) indicates how many counterions there are per formula unit. This subscript will become the coefficient on the right side of the equation here. 3) To determine the charge on the complex ion, multiply the charge on the counter ion by the number of counterions per formula unit (subscript). Since the formula unit is overall neutral, and each of the compounds in this problem only has one complex ion (i.e., there PS11-2

are no subscripts after the closed bracket), the charge on the complex ion is the opposite of the sum of the charges of all of the counter ions. You should now be able to write the balanced equation. (Make the coordination compound a solid on the left and put the dissolved ions on the right, with (aq) state designations.) 4) Determine the coordination number as in Problem 2 above. Answers: (a) [Ni(NH3)6](NO2)2: [Ni(NH 3 ) 6 ](NO 2 ) 2 (s) Ni(NH 3 ) 6 2+ (aq) + 2 NO 2 - (aq) 6 NH 3 ligands bound; each binds with only one atom (N) C.N. = 6 (the NO 2 - ions are just counterions here, not ligands) (b) [Cu(NH3)4]SO4: [Cu(NH 3 ) 4 ]SO 4 (s) Cu(NH 3 ) 4 2+ (aq) + SO 4 2- (aq) 4 NH 3 ligands bound; each binds with only one atom (N) C.N. = 4 (sulfate is just a counterion) (c) [Cr(en)3]Br3: [Cr(en) 3 ]Br 3 (s) Cr(en) 3 (aq) + 3 Br - (aq) 3 en (ethylenediamine) ligands; each is bidentate (binds with two distinct N atoms) C.N. = 3 x 2 = 6 (bromide is just a counterion here) (d) K2[Ni(CN)4]: K 2 [Ni(CN) 4 ] (s) 2 K + (aq) + Ni(CN) 4 2- (aq) 4 CN - ligands; each is monodentate (binds through C) C.N. = 4 (K + is a counterion [though cations can never be ligands anyway]) 5. Give the oxidation state (i.e., charge on) the metal cation in the coordination compounds below. Also state the coordination number and coordination geometry of the complex ion, and make a sketch of it. Na is +1 [Au(CN)2] is -1. Each CN is -1, thus x + 2(-1) = -1 x = +1 [NC CN] SO4 2- is -2 [Co(NH3)5Br] is +2. Each NH3 is +0, Br is -1, thus x + 5(0) +1(-1) = +2 x = +3 (sketch on next page ) PS11-3

C2O4 2- is ox 2- en is bidentate! Thus although there are three ligands, the C.N. is 4. NH4 + is +1 [PtCl2(ox)2] is -2. Each Cl - is -1, ox (C2O4) is -2, thus x + 2(-1) + 2(-2) = -2 x = +4 Color in Coordination Compounds (Including Splitting Energy, Generally) 6. (a) 425 nm, 625 nm, 550 nm correspond to which colors of light? (choose from the following): red, blue, green Answer: 625 nm is in the red region. 550 nm is in the green region 425 nm is technically violet, but of the choices given, it is clearly closest to blue The precise edge between blue and violet is fuzzy at best. Tro lists the border at about 430 nm, while different Wikipedia articles cite 436 nm and 450 nm. The bottom line is that anything lower than about 490 is basically blue in trichromatic color theory since only the blue receptor can be stimulated to any significant degree by violet light). The ranges for B, G, and R are roughly as follows: B: 400 490 nm (490 is really cyanish, at the border between B and G) G: 490 560 nm (again 490 is really cyanish, at the border) R: ~610 750 (or 800) nm. About halfway between G and R is Y (at around 570-590 nm) Orange is around 590 610 nm, although Tro gives it a much wider range in his color wheel. PS11-4

(b) Is orange light of longer or shorter wavelength than yellow light? Answer: longer. Orange is closer to red (by ROYGBIV), and Red is the long wavelength, low photon energy end of the visible spectrum. (c) Which photons have a greater energy, those of light with = 500 nm or those with = 600 nm? Answer: 500 nm photons have greater energy. Shorter wavelengths of light correspond to higher frequencies and thus higher energy per photon (Ephoton = h = hc/ ). (d) If complex A absorbs light maximally at a higher wavelength than complex B (each with only one absorption band), which complex has the greater? Explain using the words gap, orbital(s), absorption and photon in your answer. Answer: Complex B. The complex that absorbs light with a higher wavelength (or really greater or longer would have been a better word here) is absorbing photons that have lower energy (because Ephoton is inversely proportional to wavelength; see part (c) above). If they are absorbing (experiencing absorption of) lower-energy photons, that means the electron in the species is absorbing a smaller amount of energy as it is getting excited. Thus is it not going up in energy as much, which is just another way of saying that the E, or energy gap between the orbitals of interest, is smaller. 7. Given that the color in emeralds and rubies is due to Cr (in both) even though emeralds are green and rubies are red, and given the basic shapes of the spectra of each (shown in class; two bands, one in the blue region and one at lower energy), which gemstone is providing the Cr with the stronger crystal field environment (i.e., which has the larger )? Answer: Ruby. Both metal complex ions have two absorption bands in the visible region, one in the blue and one at lower energy (see below). The difference in color basically comes from the position of the lower-energy absorption band: in ruby, which is red, the band must NOT be absorbing much red light. The band must be in the green region (so that overall, the two bands are basically absorbing the blue and green light to yield a red color). The emerald on the other hand appears green. That means that the complex must be absorbing some of the red light and letting some of the green light through. Thus in emerald, the lower-energy band must be moved toward the red end of the spectrum compared to the ruby s band. SINCE RED LIGHT IS COMPOSED OF LESS ENERGETIC PHOTONS THAN GREEN LIGHT, THE EMERALD IS THE ONE WITH THE SMALLER SPLITTING, AND THE RUBY MUST HAVE THE LARGER SPLITTING. Since the metal ion is the same in both stones, the lattice in the ruby must be providing the stronger crystal-field environment! PS11-5

Notice that this band (both bands actually) has shifted left compared to its position in the spectrum of emerald. That shift to higher energy is what causes more green and less red to be absorbed in ruby, causing ruby to appear red. (Note that not ALL the green and blue light need be absorbed for something to appear red! It s just that red must be the predominant color reaching our eyes! Actually, it is clear from the emerald spectrum that not all of the RED light need be absorbed for something to appear green! Apparently just enough red light is absorbed in emerald so that green wins out. Different shades of green are a result of slightly different amounts of the other colors getting through to our eyes.) 8. (From 21.51, Zumdahl, 7/e). Consider the following series of complexes of chromium: Complex I appears purple, Complex II appears yellow, and Complex III appears cyan (bluish-green). (a) Predict the predominant color(s) of light absorbed by each complex. (b) Based on the spectrochemical series (p. 1145 in Tro; please read the paragraph above the actual series listing!), rank the following three ligands, NH 3, H 2 O, and Cl - from strongest to weakest in terms of the tendency to make E (or ) larger or smaller. (c) Based on all the information above (including your answers to (a) and (b), if one of the complexes is Cr(H 2 O) 6, one is Cr(NH 3 ) 6, and one is Cr(H 2 O) 4 Cl 2+, determine which one is Complex I, which is Complex II, and which is Complex III). You must provide reasoning! Answers: (a) (c) Complex I (purple): green is predominantly absorbed; is Cr(H 2 O) 6 Complex II (yellow): blue is predominantly absorbed; is Cr(NH 3 ) 6 Complex III (cyan): red is predominantly absorbed; + is Cr(H 2 O) 4 Cl 2 (b) NH 3 is the strongest-field ligand (makes biggest gaps) H 2 O is the second strongest Cl - is the weakest (makes the smallest gaps) Explanations: (a) The color that a complex appears is a composite of the colors that are not getting absorbed (they are the colors that are making it to your eyes). If you use trichromatic color theory, then consider R, G, and B as the primary colors: I (appear purple): Since purple R + B, those colors must be getting transmitted, and so green must be getting absorbed. II (appears yellow): Since yellow R + G, those colors must be getting transmitted, and so blue must be getting absorbed. PS11-6

III (appears cyan): Since cyan B + G, those colors must be getting transmitted, and so red must be getting absorbed. (b) The spectrochemical series indicates that NH 3 is a stronger-field ligand than H 2 O, and H 2 O is stronger than Cl -. (c) Based on the answer in (b), the splitting energy ( ) should be largest in Cr(NH 3 ) 6, smallest in Cr(H 2 O) 4 Cl 2+, and in between in Cr(H 2 O) 6 : largest : Cr(NH 3 ) 6 > Cr(H 2 O) 6 > Cr(H 2 O) 4 Cl 2 + smallest Now consider the colors (and relative wavelengths) of light being absorbed by each complex: smallest : blue (II) < green (I) < red (III) longest Since E photon hc/, we can now order the complexes in order of E photon absorbed, which directly corresponds to, the splitting energy: largest E photon (& : blue (II) > green (I) > red (III) smallest E photon (& Since blue light has the greatest energy per photon, its absorption must occur in the complex with the largest (i.e., Cr(NH 3 ) 6 ). Red light has the smallest energy per photon, and thus its absorption occurs in the complex with the smallest (i.e., Cr(H 2 O) 4 Cl 2+ ). And green light, with the middle energy per photon, must be being absorbed in the complex with the medium (i.e., Cr(H 2 O) 6 ). 9. Write the electron configuration for each of the following atoms and ions (use the noble gas abbreviation): (a) Mn: [Ar] 4s 2 3d 5 (Note: some texts might write [Ar] 3d 5 4s 2 ; the order doesn t matter as long as the correct number of electrons in each sublevel are shown) (b) Fe: [Ar] 4s 2 3d 6 (c) Fe : [Ar] 3d 5 (or [Ar] 4s 0 3d 5 for emphasis) Note: There were 8 electrons after the Ar configuration. The ion results from the removal of 3 e - s from the neutral atom, hence 5 e - s (after Ar) remain. But since in cations, the 3d is lower in energy than the 4s, all 5 e - s end up in the d sublevel. (d) Ag + : [Kr] 4d 10 (11 e - s after Kr, take away 1 e - (to make a +1 cation) and you re left with 10. All end up in the d sublevel (4d is lower than 5s in TM cations). (e) Rh : [Kr] 4d 6 (9 e - s after Kr, take away 3 e - s but 4d lower than 5s, so all in 4d) (f) Mn 6+ : [Ar] 3d 1 (7 e - s after Ar, take away 6 e - s.but 3d lower than 4s so into 3d) NOTE: Again, remember that once electrons are removed from a first-row transition metal atom to form a cation, the 3d orbitals drop down in energy more than do the 4s, and so all the electrons end up in the d subshell. The same is true for the 4d and the 5s, and the 5d and the 6s, etc. Namely, the nd sublevel becomes lower in energy than the (n + 1)s sublevel in a transition metal cation. PS11-7

10. Predict the number of unpaired electrons for each of the following: NOTE 1: To get the number of unpaired electrons, you first need to get the number of d electrons by writing (or thinking of) the electron configuration (as in Problem 9 above). Then you put them into an orbital diagram. NOTE 2: In THIS problem, the ions are all "free" (i.e., NOT bonded with any ligands), so all five d orbitals are degenerate (i.e., they have the same energy). When ligands are present, the d orbital sublevel "splits" into groups or pairs of non-degenerate orbitals (see problems on "crystal field" theory such as #11 and #12 below. 11. Draw the d-orbital splitting diagram, including the proper number and placement of electrons, for the following octahedral complex ions (make sure you show how you determine the number of electrons in each case!). State the number of unpaired electrons in each. (a) ZrCl 6 4- (b) OsCl 6 2- (low-spin) (c) MnCl 6 4- (high-spin) Commentary: **NOTE: high-spin or low-spin is only a useful distinction if there are two possibilities for the way electrons can be arranged. This doesn t occur, for example, if the configuration is d 1 because there would be one unpaired electron regardless of whether or not the splitting was larger than P (see below) or smaller than it (we don t have to pair any electrons up in this situation, so it makes no difference). (Spin) Pairing energy (P) refers to the increase in energy that results when an electron is placed in an orbital that already has an electron in it rather than one that is empty. That is, it is the energy price paid for pairing two electrons in the same orbital. This energy increase occurs because when two electrons are in the same orbital, they spend their time (on average) in the same region of space and therefore are (on average) closer to each other than would be the case if they were in different orbitals. Since electrons repel one another, putting them in the same region of space means more repulsion and hence a higher (potential) energy. NOTE: THE PAIRING ENERGY IS BASICALLY A CONSTANT; IT DOESN T VARY MUCH BETWEEN ORBITALS WITHIN A COMPLEX OR BETWEEN METAL COMPLEXES. If the pairing energy is larger than the difference in energy between the lower-energy d orbitals and the higher energy ones (i.e., if P is larger than the crystal field splitting energy, ), then it will cost more energy to pair up an electron with one in a lower-energy d orbital than to place the electron in the empty higher energy d orbital, and so the electron will end up in the higher energy orbital (and be unpaired). If the pairing energy is smaller than, then it will cost less energy to just pair up in the lower energy d orbital, and so the electron will end up paired. Thus when the P > the total of the spins of the d electrons will be greater than if P < and so the P > case is called high spin and the P < case is called low spin. Now that I m thinking about it, although I framed the discussion above in terms of whether or not P is greater than or less than ", I actually tend to think of it more often in terms of whether or not is greater than or less than P. Mathematically it doesn t matter which way you compare them, but conceptually this makes sense to me because as noted above, it is the P that is basically constant IT IS THE VALUE THAT CHANGES BETWEEN COMPLEXES. So I ll rewrite the above logic again here. If < P, then the energy gap up to the next orbital is smaller than the energy needed to pair, and so the electron just goes up to the higher energy orbital (they don t pair up in the lower energy one). Conversely, if > P, then the energy gap up to the next orbital is larger than the energy needed to pair, and so the electrons just go ahead and pair up in the lower energy orbital. In either case, the system adopts the configuration that is the lowest in energy overall! PS11-8

Answers: (a) ZrCl 6 4-6 Cl - s -6. Overall charge is 4, so the charge on Zr must be +2. Zr has four more electrons than Kr. Subtract 2 for the +2 charge to get two electrons past Kr in the Zr 2+ cation. These go into the d sublevel. Thus, the configuration for Zr 2+ is just [Kr] 4d 2 (see Problem #9 above) Octahedral splitting pattern is: 2 unpaired electrons regardless of the magnitude of the splitting energy, so it is not designated as high spin or low spin (Note, however, that this ion is paramagnetic, since it has two unpaired e - s) (b) OsCl 6 2- (low spin) 6 Cl - s -6 Os must be +4 since overall charge is 2. Os has 22 electrons past Xe. But 14 of these go into the 4f sublevel! I may not have made this point as clearly as needed [omitted from Self-Study worksheet]. So really there are 8 electrons worthy of consideration here. Subtract away 4 to make a +4 cation. That leaves 4 electrons, all in the d sublevel: [Xe] 4f 14 5d 4 (If it is low spin, that means that the splitting E must be fairly large so I ve tried to draw it that way.) From the diagram, there are 2 unpaired e - s Low-spin means that electrons pair up before filling the higher-energy d-orbital sublevel (occurs when > P) (c) MnCl 6 4- (high spin) 6 Cl - s -6 Mn must be +2. Mn has 7 electrons past Ar. Subtracting 2 for the +2 charge, leaves 5: Mn 2+ : [Ar] 3d 5 5 unpaired electrons High-spin means that electrons don t pair up until all the d orbitals have an electon in them (occurs when < P) ** Note that the ligand in all three cases is Cl -, generally considered to be a fairly weak-field ligand, but it sometimes results in a strong field case. This is because the metal ion and oxidation state affect the splitting too. A high oxidation state (and being lower in the periodic table) makes the splitting bigger, so that is why even with Cl - you can get a low-spin (strong-field) case.** 12. If Co(ox)3 3- is observed to be diamagnetic, is ox acting as a strong-field or weak-field ligand in this complex ion? Is the complex high-spin or low-spin? Is the pairing energy greater than or less than the crystal-field splitting energy,? Answers: ox acts as a strong-field ligand; complex is low-spin; P < PS11-9

Explanation: Diamagnetic means no unpaired electrons. Ox has a charge of 2- per ligand, so 3 ox s means 6 contribution to the charge. Thus, the Co must be +3 since the overall charge is 3. Co: 9 electrons past Ar. Subtract 3 to make a +3 cation. 6 electrons in the d orbitals. (Co : 3d 6 ). There are two possibilities: P low spin strong field high spin weak field If the complex is diamagnetic, that means that it must be low spin (actually, no spin here!), and so the ox ligand is acting as a strong-field ligand here. If it were acting as a weak field ligand, then there would be 4 unpaired electrons and the complex would be paramagnetic. THIS IS HOW WE CAN TELL WHETHER SOMETHING IS HIGH-SPIN OR LOW SPIN EXPERIMENTALLY (we measure its paramagnetism using a Gouy balance or other similar device which measures the force of attraction of a compound to an externally imposed magnetic field). Since the electrons have paired up, the pairing energy must be less than the splitting energy ( > P is the same as P <! Note comment above in the answer to NT3 regarding the two different ways to look at it). 13. Although Cl - is a weak field ligand and CN - is a strong-field ligand, [CrCl6] 3- and [Cr(CN)6] 3- exhibit approximately the same amount of paramagnetism. Explain. The degree of paramagnetism is predominantly determined by the number of unpaired electrons. Since Cr has only three unpaired electrons, neither case will involve pairing any of the electrons, and so the orbital diagrams will be the same in both cases. Regardless of the value of the splitting energy, in each case, there will be 3 unpaired electrons and a similar paramagnetism. weak field case with three d electrons strong field case with three d electrons Note that for an octahedral complex, this situation will occur in the following other configurations: d 1, d 2, d 8, d 9, and d 10. Write out a few and see for yourself! PS11-10

14. What does it mean if two species (or compounds) are isomers? Give examples of a pair of coordination isomers and a pair of linkage isomers. Are these considered structural isomers or stereoisomers? Give a definition for stereoisomer, state the two subclasses of stereoisomer, and sketch a pair of stereoisomers (any kind; can be from text). Two chemical species (or compounds) are isomers if they have the same chemical formula but at least one different property. That is, they are not the same exact chemical species (or compound). Coordination isomers are coordination compounds that have the same overall formula, but whose complex ions do not have the same exact composition (there is a different coordination sphere around the TM cation). They arise when an anion that acts as a ligand in one compound is a counter anion in the other and vice versa (or in the case in which the anion and cation are both complex ions, at least one ligand is swapped between the two complex ions see Q16.). An example of a pair of coordination isomers would be: [Ru(NH 3 ) 5 Cl]Br and [Ru(NH 3 ) 5 Br]Cl. The complex ion in the first compound is Ru(NH 3 ) 5 Cl + and that in the second one is Ru(NH 3 ) 5 Br +. NOTE: To create (on paper) a coordination isomer from a given coordination compound [having a counter anion], you must swap at least one ionic ligand with a counter anion. Linkage isomers are coordination compounds (or complex ions) in which all the same ligands are bound to the same metal ion, but one or more of the ligands bind(s) (coordinates) via a different atom. One example is [Fe(NH 3 ) 5 SCN]Cl (S is bound to the metal ion) and [Fe(NH 3 ) 5 NCS]Cl (N is bound to the metal ion). One could also say that the complex ions Fe(NH 3 ) 5 SCN - and Fe(NH 3 ) 5 NCS - are linkage isomers. Both of the above types of isomers are considered to be structural isomers since they differ in the type of bonds that are made between metal and ligand (the connectivity pattern differs). Stereoisomers are compounds (or complex ions) that have the same types of bonds (and same number of these bonds), but the arrangement of these bonds in space differs in some way. In geometric isomers, the difference is in the spatial arrangement (relative positions are not the same); in optical isomers (enantiomers), the difference is in the handedness of the arrangement (i.e., they are mirror images of one another [that are not identical]). Ex. Geometric isomers: H 3 N H 3 N Pt cis Cl Cl Cl H 3 N Pt trans NH 3 Cl Ex. Optical isomers (Enantiomers): H 2 O O OH 2 Co O N N N N OH 2 Co O OH 2 O PS11-11

15. Consider the following structures, all of which have the formula [Cr(NH3)2Cl4] - (a) Identify each structure as either cis or trans. (1) and (4) are cis (two NH 3 's are at 90 to one another [same side]; (2) and (3) are trans (two NH 3 's are at 180 to one another [across] (b) Which structures are identical, and which are different? (1) and (4) are identical, and (2) and (3) are identical. If you just rotate the structures, you can see that (1) is superimposable with (4) and so is (2) with (3). (c) Do any of these structures have an optical isomer? Explain. No. The mirror images of each of these structures are superimposable with (not different than) the original structure. One can also conclude this (possibly more easily) by looking for a plane of symmetry (a plane that cuts the structure in half so that each side is a mirror image of the other). Recall that a structure that has a plane of symmetry cannot be chiral; its mirror image will be superimposable (see PowerPoint slides). Each of these structures has a plane of symmetry, and so none has an optical isomer. 16. Consider the following complexes, each of which contains at least two ethylenediamine (en) ligands. (a) Which complexes are chiral (i.e., have an optical isomer), and which are achiral (i.e., have no optical isomer [because the mirror image is identical/superimposable])? Answers: (1) No plane of symmetry chiral (Note: If the en ligands were separated (i.e., two NH 3 's, then there WOULD be a plane of symmetry and the structure would be achiral--see Problem 10!! But the "connectedness" makes the mirror images nonsuperimposable and thus gives this structure its chirality.) (2) The plane with all the N atoms (and the Cr) is a symmetry plane achiral. PS11-12

(3) Chiral. Same idea as in (i); if this were [Cr(NH 3 ) 6 ] 2+, then it would be achiral, but the bidentate nature of the en ligands breaks up the symmetry and makes the structure chiral. (4) Chiral. This structure is analogous to structure (1). (b) Draw the optical isomer (enantiomer) of each of the chiral complexes. (Hint: Just draw the mirror image! If the structure is, in fact, chiral, then the optical isomer is the mirror image [and it should not be identical to/superimposable upon the original structure].) (c) Which, if any, of the chiral complexes are optical isomers (enantiomers) of one another? Answer: (1) and (4) are enantiomers. This is easier to see if you rotate structure (1) 90 around one of the axes so that the Cl that is "below" the Cr ends up in the same horizontal plane as the other Cl (like the two Cl's are in structure (4)). Perhaps I can draw this for you later when I have access to the correct software at work. If I don t have time, just ask me about this in the review session. PS11-13

17. Tro, 25.32 (formulas only). Write the formulas and names for the coordination isomers of [Co(en) 3 ][Cr(ox) 3 ] NOTE 1: NOTE 2: NOTE 3: It does not matter which ligand is put first in the formula (en or ox). I chose to put en first in all of them just for consistency. Technically, the cation is always written first in an ionic compound, so I ve done that below. I reminded you of this in the problem itself. Because of note 2, it makes sense to analyze the charges on the two metal cations before proceeding. From the original complexes, the charges on Co and Cr must both be +3. This is the only way to get the cation and anion to have the same charge in absolute value: Co(en)3 and Cr(ox)3 3-. (Recall that en is neutral and ox is a -2 anion.) Answers (with explanation): Ignoring the charge issue for now, and focusing on the swapping issue, there are three other ways to distribute the en s and ox s in order to keep octahedral complexes for both (there is no way to get 4 bidentate ligands around these cations, so you can t consider the 2 & 4 kind of pairs) swap either one, two, or all three ligands: [Co(en)2(ox)][Cr(en)(ox)2] (left ion s charge is +3 + -2 +1; right ion s charge is +3 + 2(-2) -1) [Co(en)(ox)2][Cr(en)2(ox)] (left ion s charge is +3 + 2(-2) -1; right ion s charge is +3 + -2 +1) [Co(ox)3][Cr(en)3] (left ion s charge is +3 + 3(-2) -3; right ion s charge is +3 + 0 +3) Because of the cation listed first restriction, the proper formulas are actually: [Co(en)2(ox)][Cr(en)(ox)2] (kept the original order) [Cr(en)2(ox)][Co(en)(ox)2] (flipped the order) [Cr(en)3][Co(ox)3] (flipped the order) 18. Tro, 25.34 Which complexes exhibit geometric isomerism? (a) [Co(H 2 O) 2 (ox) 2 ] - (b) [Co(en) 3 ] (c) [Co(H 2 O) 2 (NH 3 ) 2 (ox)] + (d) [Ni(NH 3 ) 2 (en)] 2+ (e) [Ni(CO) 2 Cl 2 Answers: (a), (c), and (e) do. (b) and (d) do not. Reasoning: (a) With two bidentate and two monodentate ligands (octahedral arrangement), one can have cistrans isomerism, because the monodentate ligands can be either at 180 or 90 degrees to one another. (Analogous to the two structures shown in the answer to Problem 7(d) above) In (b), there is no pair of monodentate ligands to put in different positions, and thus no geometric isomerism possible. (c) This combination (ML2L 2(bidentate), octahedral geometry) can have cis-trans (or other geometric isomer possibilities) because the L ligands (or L ligands) can be either at 180 or 90 to one another. This is similar to Co(NH3)4Cl2 + (just pretend that two of the NH3 s were connected as an en ligand) see Figure 24.8(b) in Tro. (e) In square planar, if there are two pairs of different monodentate ligands, cis-trans isomerism will be possible (analogous to Pt(NH3)2Cl2 see Figure 24.8(a) in Tro). In (d), the presence of the small bidentate en ligand forces the two NH3 ligands to be at 90 to one another. Since they cannot be at 180 there is no trans option possible. PS11-14