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Problems Practice for Final Exam 1a.) Which molecular orbital would be lower in energy? MO #1, or MO #2? Circle your answer. (3 pts) MO #1 MO #2 b.) Why? Briefly explain your answer above (in one or two sentences) (7 Electron density is accumulated between two nuclei and benefits from more Coulombic attractive stabilization force with MO #2. With MO #1, electrons Are forced to occupy a smaller region of space pointing away from the second nucleus -- a situation which is less stable than the isolated atoms Multiple Choice. 2. The orbital shown below is a (5 pts). a. ) u* b.) g c.) g* e.) g* f .) u g.) u* 3. BF3 is more stable than the separated B and F atoms because (5 pts). a.) Boron is more electronegative than fluorine. b.) Molecular orbitals are always more stable than atomic orbitals. c.) The bonding electron pairs are attracted to, and stabilized by two nuclei as opposed to one. d.) Hund s rule states that it should be. e.) The Pauli exclusion principle requires this. 4. The point group of the molecule shown below is: (15 pts). Cl C H H H b.) C2v e.) D2d f.) C1 c.) D3d g.) D3h d.) C3v h.) D3h 7. Draw a picture of two of the symmetry elements that describe the symmetry of the molecule below, and illustrate what each operation does to the molecule using the numbering scheme below. CH3 1 6 5 7 8 3 4 CH3 H3C H3C 2 2 3 CH3 h 1 6 5 7 8 3 4 CH3 C2 C2 4 8 7 5 6 H3 C 1 h 1 6 CH3 H3 C 2 7 5 4 8 3 H3 C 2 5. The ruthenium complex shown below belongs to the point group D2h . In this problem, you will use symmetry arguments to show that the px , py , and pz orbitals of this complex have different energies. z y OH2 NH3 Ru Cl OH2 Cl H3N x (a) First, use the D2h character table below to determine how the pz orbital of the ruthenium complex transforms under D2h symmetry, ie, assign the pz orbital a symmetry label by determining the character (1, 0, or 1) associated with each symmetry operation of the D2h group. (10 pts) Operation on the pz orbital by the following symmetry elements has the following effect: E C2(z) C2(y) C2(x) (xy) (xz) (yz) 1 Pz 1 1 1 1 1 1 Thus, the pz orbital has b1u symmetry. z y OH2 NH3 Ru Cl OH2 Cl H3N x Operation on the py orbital by the following symmetry elements has the following effect: E C2(z) C2(y) C2(x) (xy) (xz) (yz) Py 1 1 1 1 1 1 1 Thus, the py orbital has b2u symmetry. Operation on the px orbital by the following symmetry elements has the following effect: E C2(z) C2(y) C2(x) (xy) (xz) (yz) Px 1 1 1 1 1 1 1 Thus, the px orbital has b3u symmetry. 6. Multiple choice. (circle the correct answer. Note: there be may more than one answer) (4 pts) a.) The following metal ion(s) would possess an S= 1/2 spin state in a weak field ( donor) octahedral metal complex i. Fe3+ ii. Ni2+ iii. Cu2+ iv. Co2+ Fe 3+ Ni 2+ Cu 2+ Co2+ S= 3/2 S= 5/2 S= 1 S= 1/2 9. Multiple choice. (circle the correct answer. Note: there may be more than one answer) .) donor complexes of d3 ions tend to be more reactive than d3 acceptor complexes because: i. the LFSE is more negative for the donor complex ii. the LFSE is less negative for the donor complex iii. the donor complex has more filled antibonding orbitals than the acceptor complex iv. the donor complex has fewer filled antibonding orbitals than the acceptor complex 7. Comment on the bond lengthes in the complex shown below (NH3= neutra (a.) What might be the cause for these inequivalent bond lengthes? Jahn Teller distortion= elongation along the z-axis NH3 2.05 5 1.9 2+ NH3 H3 N 1.95 5 1.9 Cr 1.95 NH3 H3N 2.05 NH3 7(b.) What do these bond lengths imply about the spin state of this complex? In order for there to be a driving force for distortion, there has to be an odd number of electrons in this upper set of two orbtials. With a d4 ion, the only way this can occur is if the complex is high spin. + 1/2 1 1/2 1 + 2/3 2 yz NH3 z 2 xy xz 1/3 2 +2 H3N H3N Cr NH3 NH3 d4 elongated NH3 8a. Calculate the LFSE for [Fe(H2O)6]2+ given the following data: 10Dq= P= + 6 Dq 4 Dq high spin case 16,000 cm 1 18,000 cm 1 + 6 Dq 4 Dq low spin case LFSE(high-spin case)= 4( 4Dq) + 2(+6Dq)= 4Dq Dq = 16,000/10 cm 1 = 1,600 cm 1 LFSE(high-spin case)= 4Dq = 6,400 cm 1 (10 pts) 8b. How would this LFSE value (determined in 8a.) change if the complex were to change spin states? + 6 Dq 4 Dq low spin case vs. LFSE(low-spin case)= 6( 4Dq) + 2P= 24Dq + 2P Dq = 16,000/10 cm 1 = 1,600 cm 1 LFSE(high-spin case)= 24Dq + 2P = = 38,400 cm 1 + 36,000 cm 1 = LFSE(high-spin case)= 2,400 cm 1 Thus, this molecule would prefer to be high-spin. 10. Use symmetry arguments to show that the pz orbital of BF3 is of the appropriate symmetry to overlap with LGO (5) shown in the Figure below. What kind of bond would this form? Why are there so many LGO's? Assume that the z-axis coincides with the C3 axis, and the y-axis lies along one of the B F bonds. z peration on the pz orbital by the following symmetry elements has the follow ffect: E C3 C2 h S3 v Pz 1 1 1 1 1 1 Thus, the pz orbital has a2" symmetry. LGO(5) transforms in the same manne ossesses a2" symmetry as well. Overlap between LGO(5) and pz would be i haracter.

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