Fourth Hour Exam (Solutions)

Fourth Hour Exam 5.111 Write your name and your TA's name below. Do not open the exam until the start of the exam is announced. The exam is closed notes and closed book. 1. Read each part of each problem carefully and thoroughly. 2. Show your work. Indicate units. Use correct significant figures. 3. Make your dots on Lewis structures clearly visible. 4. If you don’t understand what the problem is requesting, raise your hand and a proctor will come to your desk. 5. Physical constants, formulas and a periodic table are given on the last page. You may detach this page once the exam has started. TRANSITION METALS 1. (32 points) 2. (10 points) CHEMICAL KINETICS 3. (16 points) NUCLEAR KINETICS 4. (10 points) OXIDATION REDUCTION 5. (10 points) 6. (10 points) 7. (12 points) Total (100 points) Name Answer Key TA 1 1. TRANSITION METALS (32 points total) (a) (3 points) Calculate the d-count for Fe3+ 5 (b) (11 points) For Fe3+, (i) in the appropriate places below, draw crystal field splitting diagrams with electrons to show orbital occupancies in both weak and strong octahedral fields. Label the diagrams (ii) with the names of the d-orbitals, and (iii) with the appropriate orbital sets eg and t2g designators.



Weak Field Octahedral Diagram Strong Field Octahedral Diagram c) (6 points) Fill in the blanks below based on your diagrams in part (b). Answer for Weak Field Diagram (i)
(ii)
(iii)
system is high spin
# unpaired electrons is 5
3 2 dn electron configuration t2g eg Answer for Strong Field Diagram system is low spin # of unpaired electrons is 1 5 dn electron configuration t 2g (d) (3 points) Calculate the crystal field stabilization energy (CFSE) for high spin Fe3+. Do not include pairing energy. 0 (e) (3 points) Calculate the crystal field stabilization energy (CFSE) for low spin Fe3+. Do not include pairing energy. -2Δo 2 (f) (6 points) Calculate the octahedral crystal field splitting energy in kJ/mol for an Fe3+ complex that absorbs light most intensely at 700. nm. Show your work. Δo = 171 kJ/mol 2. TRANSITION METALS (10 points total)
(a) (6 points) Draw d-orbitals, dz2 and dyz on top of the diagrams below. Z Z L- L- Mn+ Mn+ Y Y X X L- L-
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(b) (4 points) Predict the relative energies of just these two d-orbitals for the linear molecule drawn along the Z-axis. Explain your reasoning. dz2 is much more destabilized than dyz. Ligands right along z will repel dz2 orbitals since these orbitals are on axis.
dyz is 45 ̊ off axis, so repulsion is less. 3 3. CHEMICAL KINETICS (16 points total) The following data were obtained for the reaction O2 (g) + 2NO (g) → 2NO2 (g). Initial concentrations, mol•L-1 Experiment [O2]0 [NO]0 Initial rates, mol•L-1•s-1 1 1.10 x 10-2 1.30 x 10-2 3.21 x 10-3 2 2.20 x 10-2 1.30 x 10-2 6.40 x 10-3 3 1.10 x 10-2 2.60 x 10-2 12.8 x 10-3 (a) (3 points) Determine the order of the reaction with respect to [O2]. No need to show work. First order (b) (3 points) Determine the order of the reaction with respect to [NO]. No need to show work. Second order (c) (3 points) Write the rate law for the overall reaction. No need to show work. Rate = k[O2][NO]2 (d) (3 points) Determine the order of the overall reaction. No need to show work. 3 (e) (4 points) Calculate the rate constant k (the value and the units). Show your work. 𝑘 = 𝟏𝟕𝟑𝟎 𝑴-𝟐𝒔-𝟏 4
4. NUCLEAR CHEMISTRY (10 points) The activity of a strontium-90 source is 3.0 x 1014 Bq and its half-life is 28.1 years. Calculate the activity in Bq after 75.0 years have passed. Show your work. A=4.7x1013 Bq 5. OXIDATION REDUCTION (10 points)
(a) (6 points) Balance in BASIC solution the following skeletal equation by using oxidation and reduction half-reactions: Pb(OH)42- (aq) + ClO- (aq) → PbO2 (s) + Cl- (aq) Pb(OH)42- + ClO- PbO2 + H2O + Cl- + 2OH- (b) (4 points) The oxidizing agent is Cl+ or ClO- and the reducing agent is Pb2+ or Pb(OH)42- 5
6. OXIDATION-REDUCTION (10 points) For the following reagents under standard conditions: Au (s), Cl2 (g), Pb (s), Sn (s), Ni (s), Cd (s), Zn (s), Au+ (aq), Cl- (aq), Pb2+ (aq), Sn2+ (aq), Ni2+ (aq), Zn2+ (aq) Standard Reduction Potentials at 25°C Half-Reactions Au+ (aq)+ e- ⇒ Au (s) E°(volts) 1.69 Cl2 (g) + 2e- → 2Cl- (aq) 2H+ (aq) + 2e- ⇒ H2 (g) 1.36 Pb2+(aq) + 2 e- ⇒ Pb (s) Sn2+(aq) + 2 e- ⇒ Sn (s) -0.13 Ni2+(aq) + 2 e- ⇒ Ni (s) Cd2+(aq) + 2 e- ⇒ Cd (s) -0.26 Zn2+(aq) + 2e- ⇒ Zn (s) 0 -0.15 -0.40 -0.76 (a) (3 points) State which reagent is the strongest oxidizing agent. "+

(b) (3 points) State which reagent is the strongest reducing agent. Zn (s) (c) (4 points) State which reagent(s) will reduce Pb2+ (aq) while leaving Cd2+ (aq) unreacted. Sn(s) Ni (s) 6 7. OXIDATION-REDUCTION (12 points) A galvanic cell is constructed using the following half-reactions Half-Reactions Pb2+(aq) + 2 e- ⇒ Pb (s) Cr3+ (aq)+ e- ⇒ Cr2+ (aq) E°(volts) at 25°C - 0.13 - 0.42 Calculate the initial voltage generated by the cell at 25 ̊C if the initial concentration of Pb2+ (aq) is 0.15 M, Cr2+(aq) is 0.20 M, and Cr3+ (aq) is 0.0030 M. Show your work. E = 0.37 V 7