{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW_6_7_AnswerKey

HW_6_7_AnswerKey - 19.1 19.2 19.3 Configuration number of...

Info iconThis preview shows pages 1–8. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 8
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 19.1 19.2 19.3 Configuration, number of unpaired electrons, and LFSE? (a) [C0(NH3)5]3+? Since the NH; ligands are neutral, the cobalt ion in this octahedral complex is Co“, which is a d6 metal ion. Ammonia is in the middle of the spectrochemical series but, since the cobalt ion has a 3+ charge, this is a strong field complex and hence is low spin, with S = 0 and no unpaired electrons (the configuration is tzgs). The LFSE is 6(0.4Ao) a 2.4A0. Note that this is the largest possible value of LFSE for an octahedral complex. (h) [Fe{0H2)5}7'+? The iron ion in this octahedral complex, which contains only neutral water molecules as ligands, is Fe“, which is a Li‘s-metal ion. Since water is lower in the spectrochemical series than NH; (i.e., it is a weaker field ligand than Nth) and since the charge on the metal ion is only 2+, this is a weak field complex and hence is high spin, with S = 2 and four unpaired electrons (the configuration is t2 4.9 2). The LFSE is s 4(0.4A0) .. 2(O.6Ao) = 0.4%. Compare this small value to the large value for the low spin complex in part (a) above. (c) [Fe(CN)6]3'? The iron ion in this octahedral complex, which contains six negatively charged CN‘ ion ligands, is Fe“, which is a ds—metal ion. Cyanide ion is a very strong-field ligand, so this is a strong-field complex and hence is low spin, with S = 1/2 and one unpaired electron. The configuration is Eggs and the LFSE lS 2.0A0. (d) [Cr(NH3)5}3+? The complex contains six neutral NH; ligands, so chromium is Cr“, a a’3 metal ion. The configuration is 13;, and so there are three unpaired electrons and S = 3/2. Note that, for octahedral complexes, only (fl—d7 metal ions have the possibility of being either high spin or low spin. For [Cr(Nl-l3)5]3+, the LFSE = 3(0.4A0) = 1.20.0. (For rid—d3, dB, and d9 metal ions in octahedral complexes, only one spin state is possible.) (e) [W(CO)5]? Carbon monoxide (i.e., the carbonyl ligand) is neutral, so this is a complex of W(0). The W atom in this octahdral complex is d6. Since CO is such a strong field ligand (it is even higher in the spectrochemical series than CN‘), W(CO)5 is a strong-field complex and hence is low spin, with no unpaired electrons (the configuration is :25). The LFSE = ammo) = 2.4a, (l) Tetrahedral [FeCl4}2‘? The iron ion in this complex, which contains four negatively charged Cl" ion ligands, is Fe“, which is a d6 metal ion. All tetrahedral complexes are high spin, since AT is much smaller than A0 (AT = (4/9)A0, if the metal ion, the ligands, and the metal—ligand distances are kept constant), so for this complex S = 2 and there are four unpaired electrons. The configuration is 23:3. The LFSE is 3(0.6AT) — (g) Tetrahedral [Ni(C0)4]? The neutral C0 ligands require that the metal center in this complex is Ni", which is a Lilo—metal atom. Regardless of geometry, complexes of rim-metal atoms or ions will never have any unpaired electrons and will always have LFSE = O, and this complex is no exception. What factors account for the ligand field strength of different ligands? It is clear that n-acidity cannot be a requirement for a position high in the spectrochemical series, since H‘ is a very strong field ligand but is not a “it-acid (it has no low energy acceptor orbitals of local n—symmetry). However, ligands that are very strong o-bases will increase the energy of the eg orbitals in an octahedral complex relative to the t2, orbitals. Thus there are two ways for a complex to develop a large value of AD, by possessing ligands that are alt-acids or by possessing ligands that are strong o—bases (of course some ligands, like CN‘, exhibit both n—acidity and moderately strong o—basicity). A class of ligands that are also very high in the spectrochemical series are alkyl anions, R" (e.g., CHg‘). These are not it-acids but, like H“, are very strong bases. Estimate the spin—only contribution to the magnetic moment? The formula for the Spin-only moment is ugo = [(N)(N + 2))”. Therefore, the spin-only contributions are: 122 Part 2: The Eiements and Their Compounds 19.4 19.5 19.6 complex N “30 = [(NJ(N+ 2n“ icolNHslsi3+ 0 0 [Fe(OH2)5]2+ 4 4.9 [Fe(CN)5]3‘ 1 1.7 remand“ 3 3.9 {W(CO)s] 0 0 {FeCldz' 4 4.9 (Ni(CO)4] 0 0 Assign the colors pink, yellow, and blue to cobaltfll) complexes? The colors of metal complexes are frequently caused by ligand—field transitions involving electron promotion from one subset of d orbitals to another (e.g., from tzg to eg for octahedral complexes or from c to t; for tetrahedral complexes). Of the three complexes given, the lowest energy transition probably occurs for [CoCl.;]2', because it is tetrahedral (AT = (4f9)(Ag)) and because Cl' is a weak-field ligand. This complex is blue, since a solution of it will absorb low energy red light and reflect the complement of red, which is blue. Of the two complexes that are left, [Co(NH3)5]2+ probably has a higher energy transition than [C0(0H2)5]2+, since NH; is a stronger-field ligand than H20 (see Table 19.1). The compiex [Co(NI-l_a,)5]2+ is yellow because oniy a small amount of visible light, at the blue end of the spectrum, is absorbed by a solution of this complex. By default, you should conclude that [Co(0H,),]“ is pink. Larger LFSE? (a) The chromium is expected to have a larger crystal field stabilization because of the t2g3 egl configuration [(0.4 x 3) — (0.6 x 1) = 0.6 A0] compared to the Manganese with an electronic configuration t2g3e22 [(0.4 X 3) —~ (0.6 x 2) = 0]. (b) Although Fe3+ and Mn2+ are iso—electronic, the higher charge on the Fe ion leads to higher stabilization energy. (c) Since water is a weak—field ligand compared to ON", the electronic configurations of these two complexes differ. For the [Fe(CN)5]3" the configuration will be t2g5 e30 [(0.4 X 5) fl 2A0] while the corresponding configuration for the aquo complex Wilt be t233 egZ [(0.4 X 3) — (0.6 X 2) = 0] . Hence [Fe(CN)5]3' will have higher stabilization. (d) The LF SE increases down the group and hence the ruthenium complex will have higher stabilization. (e) In general, tetrahedral complexes form high-spin complexes. Fe2+ with an electronic configuration of eg2 t2g3 will have smaller stabilization [(2 X 0.6) - (2 x 0.4) = 0.2 A0] compared to Co2+ with a configuration of e34 t2g3 [(4 X 0.6) — (3 X 0.4) = 1.2 AU]. Comment on the lattice enthalpies for CaO, TiO, V0, Milo, FeO, C00, and NiO? As in the answer to Exercise 819.3, there are two factors that lead to the values given in this question and plotted beiow: decreasing ionic radius from left to right across the at block, leading to a general increase in AHL from CaO to MC, and LFSE, which varies in a more complicated way for high-spin metal ions in an octahedral environment, increasing from n"J to d3, then decreasing from d3 to d5, then increasing from d5 to :13, then decreasing again from d8 to dm. The straight line through the black squares is the trend expected for the first factor, the decrease in ionic radius (the last black square is not a data point, but simply the extrapolation of the line between AHL values for CaO and M110, both of which have LFSE = 0). The deviations of AH}, values for ’l‘iO, V0, FeO, C00, and NiO from the straight line are a manifestation of the second factor, the nonzero values of LFSE for Tim", V“, Fe“, Co“, and Ni“. You will find in Chapter 23 that TiO and V0 have considerable metal—metal bonding, and this factor also contributes to their stability. pl; 0 CD CD 3800 3600 Lattice Enthalpy, kI/rnol 3400 Ca Ti V MnFeCoNi 19.7 Diamagnetic and paramagnetic Ni (consider that HClO4 is a very strong macrocyclic ligand, and two C104“ (II) complexes? Perchiorate, C104“, is a very weakly basic anion Breasted acid). Therefore, in the compound containing MOI), the neutral anions, there is probably a four—coordinate square-planar Ni(II) (d8) i S C 12+ N r N C S 19.8 Predict the structure of [Cr(OH2)6]2+? The main consequence of the Jahn-Teller theorem is that a molecule or ion if the distortion removes the degeneracy. The high—spin d4 com configuration tggaeg', which is orbitally degenerate since the single eg orbital. Therefore, by the Jahn-Teller theorem, the complex sh distortion, whereby two trans metal—ligand bonds are elongated and the other four are shortened, removes the degeneracy. This is the most common distortion observed for octahedral complexes of high—spin d4, low—spin d7, and d9 metal ions, all of which possess es degeneracies and exhibit measurable Jahn-Teller distortions. The predicted structure of the [Cr(OH2)5]2+ ion, with the elongation of the two trans Cr—O bonds shown greatly exaggerated, is shown at the right. plex {Cr(0Hz)5]2+ has the electron can be in either the dz; or the dx2_ yz 0H2 T|2+ 19.9 Explain the asymmetry in the eg <- tzg visible transition for Ti3+ (aq)? It is suggested that you explain this observation using the Jahn-Teller theorem. The ground state of Ti(0H;)53+, which is a a" complex, is not one 124 Part 2: The Elements and Their Compounds 19.10 19.11 of the configurations that usually leads to an observable Jahn—Teller distortion (the three main cases are listed in the answer to Exercise 19.8, above). However, the electronic excited state of '[‘i(0H2)53+ has the configuration @5033], and so the excited state of this complex possesses an 22 degeneracy. Therefore, the “single" electronic transition is really the superposition of two transitions, one from an on ground—state ion to an 01, excited—state ion, and a iower energy transition from an 01, ground-state ion to a lower energy distorted excited—state ion (probably D4h). Since these two transitions have slightly different energies, the unresolved superimposed bands result in an asymmetric absorption peak. 0H2 ——| 3+ H20 “‘w Ti V 0H2 OH2 Russell-Saunders term symbols? (3.) L = 0, S = 5/2? You should remember that a term, denoted by a capital ietter, is reiated to L in the same way that an orbital, denoted by a iowercase letter, is related to /: Ifl=0 1 2 3 4 5 6 orbital=s p d f g h 1' IfL=0 1 2 3 4 5 6 term=S P D F G H I The multiplicity of the term, which is always given as a ieft superscript, can always be determined by using the formuia multiplicity = ZS + l: \ ifS=O 1/2 1 3/2 2 5/2 multiplicity 2S + 1 = l 2 3 4 5 6 The terms and multiplicities listed above are not a complete list to cover all possibilities for all atoms and ions, but they will cover all possible d" configurations. As far as the situation L = O, S = 5/2 is concerned, the term symbol is 6S. In addition to answering this question, you should try to decide which a“ configurations can give rise to the term. In this case, the only a" configuration that can give rise to an r’S term is d5 (e.g., a gas~phase Mn2+ or Fe3+ion). (b) L = 3, S = 3/2? According to the reiations shown above, this set of angular momentum quantum numbers is described by the term symbol “F. The diagram below, which is another way to depict a microstate, shows how the situation L = 3, S = 3/2 can arise from a d3 configuration (e.g., a gas-phase Cr3+ ion). it can also arise from d5 and d7 configurations. L=3,S =3/2 L=2,S =1/2 IIIII IIIII m,=2 1 0—1—2 m,=2 1 0—1—2 (c) L $ 2, S = 1/2? This set of quantum numbers is described by the term symbol 2D. The microstate diagram above shows how the situation L ="«~ 2, S = 1/2 can arise from a a” configuration (e.g., a gas-phase Ti3+ ion). It can also arise from d3, d5, :17, and d9 configurations. (d) L = 1, S = 1? This set of quantum numbers is described by the term symbol 3P. It can arise from d2, d4, 5/6, and d3 configurations. Identify the ground term? (a) 3F, 3P, 1P, 1G? By definition, the ground term has the lowest energy of all of the terms. Recall Hund’s two rules: (1) the term with the greatest multiplicity lies lowest in energy; (2) for a given multiplicity, the greater the value of L of a term, the lower the energy. Therefore, the ground term in this case will be a triplet, not a singlet (rule 1). Of the two triplet terms, 3F lies lower in energy than 3F: L = 3 for 3F, L = 1 for 3P (rule 2). Therefore, the ground term is 3F. (c) 6s, 4G, 4P, 2I? The ground term is 68 because thi 'Il‘he Russell-Saunders terms for the following con 5 term has a higher multiplicity than the other terms. 19.12 y the ground term? (a) 4 s ? You can approach this exercise in the way described in Section 19.3 for th 2 possible microstates for ount for these, you can cross out (1+, 1‘), (—1+, —1‘), and one microetate from each of the other three rows under M; = 0. That [eaves ten microstates to be accounted for. 3 , 1S that arise fiom the 3p2 r two terms (see Hund’s 19.13 Identification of ground terms? species configuration L S gonad term 13* [He]232 o 0 3 Na [Neps' 0 1/2 23 T12 [Ar]de 3 1 3F Ag [Kr13d‘0 o o 's 513 + 20 = 10,642 curl and 153 = 12,920 cm" if you can determine that B = (12,920 cm'f)/(15) a 861.33 cm" and c= 3167.7 cm“. ' 3P A + 73 7B 153 1D A — SB +2C *33 +2C SB +2C 3F A ~— 88 —8B 0 Relative Energies 126 Part 2: The Elements and Their Compounds 19.15 19.16 19.17 19.18 19.19 19.20 a" configurations and ground terms? (a) Low-spin [Rh(NH3)6]3+? There are a number of ways to determine the integer n for a given til-block metal. One straightforward procedure is as follows. Count the number of elements from the lefi side of the periodic table to the metal in question. This will be the number of of electrons for a metal atom in a complex (note that an isolated gas-phase metal atom may have an smd” configuration, but the same neutral metal atom in a complex will have a d’" configuration). Then subtract the positive charge on the metal ion from this number, leaving the integer n. For example, Rh is the ninth element in period 5, so Rh0 in a complex has a d9 configuration, Rh+ has a d8 configuration, and so on. Using this procedure, the Rh3+ ion in the octahedral complex [Rh(NH3)5 3+ has a d5 configuration (9 —~ 3 $ 6). According to the d6 Tanabe—Sugano diagram (see Resource Section 6), the ground term for a tow-spin :2; metal ion is 1 A1,. (b) {Ti(H20)5]3+? Titanium is the fourth element in period 4, so Ti° in a complex has a d4 configuration. Therefore, the Ti3+ ion .in the octahedral [Tia-120k]3+ ion has a of1 configuration (4 ~— 3 = l). A correlation diagram for d‘ metal ions is shown in Figure 19.23 (Resource Section 6 does not include the a“ Tanabe— Sugano diagram). According to this diagram, the ground term for a to; metal ion is 2'ng. (c) High-spin [Fem20)5]3+? Iron is the eighth element in period 4, so Fe" in a complex (such as Fe(CO)5) has a a'8 configuration. Therefore, the Fe3+ ion in the octahedral [Fed-1510);]3+ ion has a d5 configuration (8 — 3 = 5). According to the d5 Tanabe-Sugano diagram, the ground term for a high-spin 132:2; metal ion is fiAtg. Estimate An and B for: (a) [Ni(l-I;0)5]2+? According to the dgTanabe-Sugano diagram (Resource Section 6), the absorptions at 8500 cm“, 13,800 cm”, and 25,300 cm‘1 correspond to the following spin—allowed transitions, respectively: 3n, <— 3A,,, 31“,, <— 3A,,, and 3n, <~ 3%. The ratios 1330013500 = 1.5 and 25,300/8500 = 3.0 can be used to estimate Au/B as ll, Using this value of AoiB and the fact that E/B = AOIB for the lowest-energy transition, A0 = 8500 cm"1 and B is 770 cm“. Note that B for a gas-phase Ni2+ ion is 1080 cm4 (see Table 19.6). The fact that B for the complex is only ~70% of the free ion value is an example of the nephalauxetic effect (Section 19.4). I (b) [Ni(NH3)61“? The absorptions for this complex are at 10,750 omr‘, 17,500 em”, and 28,200 cm”‘. The ratios in this case are 17,500! 10,750 = 1.6 and 28,200l10,750 = 2.6, and lead to Alt/B z 15. Thus, Au : 10,750 ear‘ and s a 720 cm”. It is sensible that s for [NiCNl-igsf’" is smaller than s for [Ni(H20)5]2+, since NH, is higher in the nephalauxetic series than is H20. Ground term and lowest energy transition for a paramagnetic octahedral Fe(II) complex? A d6 octahedral Fe(ll) complex can be either high spin (tzg‘egz, S 5 2) or low spin (tlgfi, S = 0). if the complex has a large paramagnetic susceptibility, it must be high spin, since a low spin complex would be diamagnetic. According to the d6 Tanabe-Sugano diagram (Resource Section 6), the ground term for the high-spin case (i.e., to the left of the discontinuity) is STZE. The only other quintet term is SEE, so the only spin-allowed transition is 5E3 4'— stg. The spectrum of [C0(NH3)5]3+? If this ti6 complex were high spin, the only spin-allowed transition possible would be 5E,g <— STZE (refer once again to the d6 Tanabe—Sugano diagram). On the other hand, if it were low spin, several spin-allowed transitions are possible, including ‘Ttg <—- ‘Atg, ‘Tpg <-— ‘Atg, [EB <—- Agg, etc. The presence of two moderate—intensity bands in the visible/near-UV spectrum of [CoCNI-Ig)5]3+ suggests that it is low spin. The first two transitions listed above correspond to these two bands. The very weak band in the red corresponds to a spin-forbidden transuion such as 3ng (— lAtg. Why is [Ferls‘ colorless, whereas [CoFEIT is colored? The d5 Fe“ ion in the octahedrai hexafluoro complex must be high spin. According to the d5 Tanabe—Sugano diagram (Resource Section 6), a high-spin complex has no higher energy terms of the same multiplicity as the GA”; ground term. Therefore, since no spin—allowed transitions are possible, the complex is expected to be colorless (i.e., only very weak 5 in- forbidden transitions are possible). if this Fe(lIl) complex were low spin, spin-allowed transitions such as ET“; (— ”T23, ZAZQ <— leg, etc. would render the complex colored. The at“ C0“ ion in [CoF5]3' is also high spin, but in this case a single spin—allowed transition, SEg (— stg, makes the complex colored and gives it a one-band spectrum. Explain why the nephalauxetic effect for CN‘ is larger than for NH3? These two ligands are quite different with respect to the types of bonds they form with metal ions. Ammonia and cyanide ion are both cr— bases, but cyanide is also a n-acid. This difference means that NH3 can form molecular orbitals only with the metal cg orbitals, while CN' can form molecular orbitals with the metal 6,, and tog orbitals. The formation of molecular orbitals is the way that ligands “expand the clouds” of the metal at orbitals. Chapter 19: d—Metal Complexes: Electronic Structure and Spectra 127 19.2] The origins of transitions for a complex of Co(III) with ammine and chloro ligands? Let’s start with the intense band at relatively high energy with emax = 2 X 10“ M*1 charge-transfer transrtion, smce it is too = 60 and 80 M‘1 crn‘i are probably spin-allowed mplex is not strictly octahedral, the ligand field bands are still not “' is most likely a spin-forbidden ligand field transition. 19.22 Describe the transitions of Fe3+ impurities in bottle glass? The Fe“ they were low spin, several spin-allowed ligand field transitions woul viewed through the wall of the bottle (see the Tanabe 6). Therefore, the Fe3+ ions are high spin, octahedral high-spin d5 metal ion is fiAlg, only observed when lookin transitions. ions in question are (2’5 metal ions. If d give the glass a color even when -Sugano diagram for d5 metal ions in Resource Section and as such have no spin-allowed transitions (the ground state of an cited states). The faint green color, which is 19.23 The origins of transitions for [Cr(H20).;]3+ and C1041“? The blue [Cr(l-l20)6]3+ is caused by spin-allowed but Laporte- molar'absorption coefficient, a, which is a manifestati -green color of the Cr3+ ions in forbidden ligand field transitions. The relatively low on of the Laporte-forbidden nature of the transitions, is e Odeation state of chromium in dichromate dianion is Cr(Vl), which is a“. Therefore, no ligand field who " srtions are possible. The intense yellow color is due to LMCT transitions (i.e., electron transfer from the oxide ion ligands to the Cr transfer transitions are intense bec .. 2 l Cr H20“ i VOHZ O“‘?CI\O 0H2 O 19.24 The orbitals of [CoCl(NH3)5] ...
View Full Document

{[ snackBarMessage ]}