ADVANCED INORGANIC CHEMISTRY (CHE 510) EXAMINATION II NOVEMBER 20, Name: Key

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1 ADVACED IORGAIC CEMISTRY (CE 510) EXAMIATIO II OVEMBER 20, 2006 ame: Key SOW ALL OF YOUR WORK so that you may get some partial credit. The last few pages of this exam booklet contain tables that you may need to complete this exam. Please answer questions I TE AREA PROVIDED, TE BACK OF A EXAM PAGE OR O TE CLEARLY LABELED SPARE SEET. o credit will be given for work written on the tables pages. 1

2 1a. (8 pts) Using a group theoretical approach, construct a MO energy level diagram for tetrahedral 4 +. Include appropriate symmetry labels and discuss the logic used as well as any assumptions made in the construction of the MO diagram. What is the bond order predicted for tetrahedral 4 + on the basis of your MO energy level diagram? First, identify valence orbitals of the atoms involved in the bonding: the valence orbitals of the nitrogen atom are 2s, 2p x, 2p y, and 2p z orbitals while each hydrogen atom has a 1s orbital. ext, figure out the symmetry of the orbitals involved. Thus, we make SALC s of the hydrogen 1s orbitals T d E 8 C 3 3 C 2 6 S 4 6 σ d Γ 1s Which reduces to Γ 1s = A 1 + T 2 Looking in the character table, we see that for nitrogen: 2s orbital has a 1 symmetry 2p x, 2p y, and 2p z orbitals have t 2 symmetry Recognizing that nitrogen is more electronegative that hydrogen, we can suspect that the hydrogen SALC s will have the higher energy and that the a 1 SALC is slightly lower than the t 2 SALC s because of the nodal surfaces that exist although the atoms are not formally bonded. Thus, we get the diagram below: 2

3 t 2 2a 1 t 2 t 2 a 1 1t 2 a 1 1a 1 B.O. = ½ (8) = 4 1b. (8 pts) Using a group theoretical approach, construct a MO energy level diagram for square planar 4 +. Include appropriate symmetry labels and discuss the logic used as well as any assumptions made in the construction of the MO diagram. What is the bond order predicted for square planar 4 + on the basis of your MO energy level diagram? First, we find the symmetry point group for square planar + 4 : z y + x D 4h 3

4 ext, we generate SALC s of the hydrogen atom 1s orbitals; z y x D 4h E 2 C 4 C 2 2 C 2 ' 2 C 2 '' i 2 S 4 σ h 2 σ v 2 σ d Γ 1s We find that Γ 1s = A 1g + B 1g + E u after applying our reduction formula. Looking in the D 4h character table, we see that for nitrogen: 2s orbital has a 1g symmetry 2p x and 2p y have E u symmetry 2p z orbital has a 2u symmetry Recognizing that nitrogen is more electronegative that hydrogen, we get the diagram below: e u 2a 1g b 1g b 1g + e u a 2u + e u a 2u a 1g a 1g 1e u 1a 1g B.O = ½ (# of bonding electrons -# of antibonding electrons) = ½ (6-0) = 3. ote that electrons in a 2u orbital are nonbonding. 4

5 1c. (6 pts) Construct (draw) a Walsh diagram correlating the bonding molecular orbitals of tetrahedral 4 + with those of square planar 4 +. Explain the reasoning behind your diagram. Our Walsh diagram is shown below. The nodeless a 1 - a 1g orbital does not influence shape of the molecule since it is cylindrically symmetrical. As we can see the t 2 set of molecular orbitals in tetrahedral geometry splits into the a 2u and e u set in the square planar structure. The a 2u MO is primarily 3p z in character and the overlap with the hydrogen SALC s decreases to essentially zero, producing a large rise in its energy. Though overlap between the e u (p x and p y ) orbitals and the hydrogen orbitals increase slighty, it s is not enough to compensate this increase in energy of the a 2u orbital because the P X and p y orbitals were already in good overlap in tetrahedral geometry. Thus, when a molecule possesses a total of eight electrons, tetrahedral structure will be preferred. T d D 4h a 2u orbital energy t 2 eu a 1 a 1g 109 Angle ( ) 90 See Yoshizawa et al. Chem. Phys. 2001, 271, for recent discussion of Walsh diagrams of A 4 molecules. 5

6 2a. (10 pts) Using group theory, construct a MO energy level diagram showing sigma bonding only for [Ti(CO) 6 ] 2-. Include appropriate symmetry labels for all orbitals and discuss the logic used as well as any assumptions made in constructing your MO diagram. Draw the MO diagram using the template labeled [Ti(CO) 6 ] 2- (below). First, identify valence orbitals of the atoms involved in the bonding. The valence orbitals of the titanium atom are the five 3d-, 4s-, and three 4p orbitals. The CO ligands form sigma bonds though their 3σ g OMO (we showed this in class). ext, we figure out the symmetry of the orbitals involved. Thus, we make SALC s of the CO donor sigma orbitals:!!!!!! O h E 8 C 3 6 C 2 6 C 4 3C 4 2 i 6 S 4 8 S 6 3 σ h 6 σ d Γ Co Which reduces to Γ CO = A 1g + E g + T 1u Looking in the character table, we see that for titanium: 4s orbital has a 1g symmetry 3dz 2 and 3dx 2 -y 2 have e g symmetry 3dxy, 3dxz, and 3dyz have t 2g symmetry 4p x,4p y, and 4p z orbitals have t iu symmetry Recognizing that the CO donor electrons must be at lower energy than the titanium acceptor orbitals (CO carbon is more electronegative that titanium), we get the diagram on the next page. We see that the t 1u orbitals are lower in energy than the e g, this is because the e g orbitals point directly at the ligands and are raised in energy due to repulsive interaction. 6

7 Ti 2- [Ti(CO) 6 ] 2-6CO 2t 1u 2a 1g 2e g CO!* orbitals (not involved in " bonding) t 1u E a 1g e g t 2g t 2g t 1u a 1g e g 1t 1u 1e g 1a 1g 7

8 2b. (5 pts) [Ti(CO) 6 ] 2- displays an IR-active CO stretch (ν CO ) at 1748 cm -1 while [Cr(CO) 6 ] displays an IR-active CO stretch (ν CO ) at 2000 cm -1. Use the template below to draw a MO energy level diagram for [Cr(CO) 6 ] that is consistent with its higher CO stretching frequency. Cr [Cr(CO) 6 ] 6CO 2t 1u 2a 1g 2e g CO!* orbitals (not involved in " bonding) E t 1u a 1g e g t 2g t 2g t 1u a 1g e g 1t 1u 1e g 1a 1g 8

9 2c. (5pts) Rationalize the trend in ν CO for [Ti(CO) 6 ] 2- and [Cr(CO) 6 ] on the basis of your MO diagrams and keeping in mind the electroneutrality principle. Both Ti 2- and Cr are d 6 metal centers. The IR data indicate that greater metal to CO ligand back-bonding interaction occurs in [Ti(CO) 6 ] 2- than [Cr(CO) 6 ]. This can be explained by the fact that titanium is more electropositive than Cr and has a -2 charge in [Ti(CO) 6 ] 2-, which means more electron density needs to be delocalized onto the CO ligands to maintain an essentially neutral metal center. Thus, titanium s valence orbitals and hence [Ti(CO) 6 ] 2- s molecular orbitals reside at higher energy than Cr s valence orbitals and hence [Cr(CO) 6 ] s molecular orbitals. As a result, t 2g orbitals of [Ti(CO) 6 ] 2- are closer in energy to CO π* orbitals, leading to better overlap and greater (metal d CO π*) backbonding interaction. 2d. (6 pts) The photoelectron spectrum of gas phase [Mo(CO) 6 ] is shown below. Use the spectrum to account for the energies of the molecular orbitals of the octahedral complex. int: the ionization energy of CO itself is around 14 ev. The OMOs of [Mo(CO) 6 ] are the three t 2g orbitals largely confined to the Mo atom (see MO diagram for related [Cr(CO) 6 ] on previous page for reference ) and their energy can be ascribed to that of the peak with the lowest ionization energy (close to 8 ev). The group of ionization energies around 14eV is 9

10 probably due to the Mo-CO σ bonding orbitals, as well as bonding orbitals in CO since the ionization energy of CO itself is around 14 ev. 3a. (8 pts) ame the following compounds according to IUPAC rules: (a) [Co 2 (en)( 3 ) 2 ] Diamminedichloroethylenediaminecobalt(II) (b) [Co( 3 )( 3 ) 5 ]SO 4 Pentaammineazidocobalt(III) sulfate (c) [Ag( 3 ) 2 ]PF 6 Diamminesilver(I) hexaflurophosphate (d) K 3 [Fe(C) 6 ] Potassium hexacyanoferrate(iii) 3b. (10 pts) Draw all of the possible isomers for [Co 2 (en)( 3 ) 2 ]. 3 Co 3 3 Co 3 Co 3 3 Co

11 4. (10 pts) The important structural role of Zn 2+ in biological systems can be attributed to its electronic preference for octahedral over tetrahedral geometry. True or False? Explain your reasoning False. Zn 2+ has d 10 electron configuration. The LFSE is zero for both octahderal and tetrahedral geometries (below). Thus, octahedral site stabilization energy is zero hence there is no electronic preference for octahedral geometry over tetrahedral geometry. The geometry adopted by Zn 2+ generally depends on the steric requirements of the ligands and thermochemical considerations. T d O h! O! t d LSFE = 0 LSFE = 0 11

12 5. (15 pts) Bearing in mind the Jahn-Teller theorem, rank the following compounds in terms of their degree of deviation from idealized octahedral structure: [Cr(C) 6 ] 4-, [Cu(O 2 ) 6 ] 2+, and [Cr(O 2 ) 6 ] 3+. Explain your reasoning and indicate what type of distortion can be expected (use drawings as required). [Cr(O 2 ) 6 ] 3+ < [Cr(C) 6 ] 4- < [Cu(O 2 ) 6 ] 2+ Cr 2+ = d 4 ion, Cr 3+ = d 3 ion, and Cu 2+ = d 9 ion. 2 O is a weak field ligand while C - is a strong field ligand. ence, [Cr(C) 6 ] 4- is a low spin complex while only one electron configuration is possible in the case of [Cu(O 2 ) 6 ] 2+ or [Cr(O 2 ) 6 ] 3+. While a d 3 ion is not subject to Jahn-Teller distortion since no stabilization is gained, a Jahn-Teller (J-T) distortion, which lowers the symmetry, removes orbital degeneracy, and leads to a more stable complex, is expected for both low spin d 4 - (axial compression) and d 9 electron configurations (axial compression or elongation). J-T distortion results in greater splitting of the e g orbitals than the t 2g orbitals, hence deviation from idealized octahedral is greater for [Cu(O 2 ) 6 ] 2+ than [Cr(C) 6 ] 4- (see your textbook for splitting diagrams) 12

13 6. (12 pts) Explain the differences in values of the ligand field splittings ( = o or T ) for the cobalt complexes below: Complex (cm -1 ) [Co( 3 ) 6 ] 3+ 22,900 [Co( 2 O) 6 ] 3+ 18,200 [Co( 3 ) 6 ] 2+ 10,200 [Co( 3 ) 4 ] 2+ 5,900 3 exerts a stronger ligand field than 2 O hence is greater for [Co( 3 ) 6 ] 3+ than [Co( 2 O) 6 ] 3+, both of which contain Co 3+ ions. While [Co( 3 ) 6 ] 3+ and [Co( 3 ) 6 ] 2+ are both hexaammine complexes, an increase in ionic charge (from Co 2+ to Co 3+ ) will draw the ligands more closely in and thereby increase electrostatic repulsion. Consequently, for [Co( 3 ) 6 ] 3+ is greater than for [Co( 3 ) 6 ] 2+. early, 3 is not a sufficiently stronger ligand field than 2 O to overcome the effect of the increased charge in [Co( 2 O) 6 ] 3+ versus in [Co( 3 ) 6 ] 2+. The greater the number of ligands, the greater the perturbation of the d orbitals. Thus, six coordinate complexes have greater values of than the tetrahedral complex, [Co( 3 ) 4 ] 2+ (remember that t = 4/9 o ). 13

14 7. (12 pts) For each of the following pair of complexes, identify the one that has the larger ligand field stabilization energy (LFSE). Explain your reasoning and where possible, show your work. (a) [Mn(O 2 ) 6 ] 2+ or [Fe(O 2 ) 6 ] 3+ Both Mn 2+ and Fe 3+ are isoelectronic (d 5 ) ions. Since 2 O is a weak field ligand, both [Mn(O 2 ) 6 ] 2+ and [Fe(O 2 ) 6 ] 3+ are high spin complexes. ence both have LFSE = 0. (b) [Cr(O 2 ) 6 ] 2+ or [Mn(O 2 ) 6 ] 2+ Both are high-spin complexes. While Mn 2+ is a d 5 ion and hence LFSE = 0, Cr 2+ is a d 4 ion and hence [Cr(O 2 ) 6 ] 2+ has t 2g 3 e g 1 configuration and LFSE = [(-0.4 x 3) + (0.6 x 1)] o = -0.6 o. Thus, [Cr(O 2 ) 6 ] 2+ has the larger LFSE. (c) [Ru(C) 6 ] 3- or [Fe(C) 6 ] 3- Both Ru 3+ and Fe 3+ are d 6 ions that belong to the same group. Both complexes are low spin and hence have t 2g 6 electron configuration. owever, LFSE increases down the group (as d orbital size increases, allowing for better overlap) and hence the ruthenium complex will have the higher LFSE 14