Practice Exam 3 Key PDF

1. (3pts) Classify each of the following orbital interactions as σ-bonding, σ-antibonding, π-bonding, π-antibonding, or non- bonding. Provide your responses in the box below each orbital pair. σ-Antibonding π-Bonding σ-Bonding...

1. (3pts) Classify each of the following orbital interactions as σ-bonding, σ-antibonding, π-bonding, π-antibonding, or non- bonding. Provide your responses in the box below each orbital pair. σ-Antibonding π-Bonding σ-Bonding Nonbonding π-Antibonding π-Bonding 2. (3pts) Aqueous permanganate will react with aqueous oxalate ([C2O4]2-) to produce the aqueous manganese (II) ion and gaseous carbon dioxide under acidic conditions. Provide the balanced chemical equation below. 𝑀𝑛𝑂4− 𝑎𝑞 + 𝐶2 𝑂4 −2 𝑎𝑞 → 𝐶𝑂2 𝑔 + 𝑀𝑛+2 𝑎𝑞 𝐶2 𝑂4 −2 𝑎𝑞 → 2𝐶𝑂2 𝑔 + 2𝑒 − x5 8𝐻 + 𝑎𝑞 + 5𝑒 − + 𝑀𝑛𝑂4− 𝑎𝑞 → 𝑀𝑛+2 𝑎𝑞 + 4𝐻2 𝑂(𝑙) x2 16𝐻 + 𝑎𝑞 + 2𝑀𝑛𝑂4− 𝑎𝑞 + 5 𝐶2 𝑂4 −2 𝑎𝑞 → 10𝐶𝑂2 𝑔 + 2𝑀𝑛+2 𝑎𝑞 + 8𝐻2 𝑂(𝑙) 3. (2pts) Which of the following solutions (i – iv) has the greatest concentration of ions? Circle your answer and show your work for the answer you selected. (i) 1.25 M Ca3(PO4)2 𝑖𝑜𝑛𝑠 ≈ 0𝑀 𝐴𝑙 𝑁𝑂3 → 𝐴𝑙 +3 𝑎𝑞 + 3𝑁𝑂3− (𝑎𝑞) 3 (ii) 2.00 M MgSO4 𝑖𝑜𝑛𝑠 = 4.00𝑀 1.50 𝑀 1.50 𝑀 4.50 𝑀 (iii) 1.50 M Al(NO3)3 𝑖𝑜𝑛𝑠 = 6.00𝑀 (iv) 2.50 M NaCl 𝑖𝑜𝑛𝑠 = 5.00𝑀 4. (2pts) Decide whether the below statements regarding crystal field theory are true or false. Circle your answer for each. (i) Decreasing the oxidation state of the metal center leads to a greater crystal True False field splitting due to an increase in the metal – ligand bonding strength. (ii) In comparing a series of ligand binding atoms within the same period, decreasing the electronegativity of a ligand’s binding atom increases its ability to True False donate electrons density to the metal center, which leads to an increase in the crystal field splitting. (iii) An increase in crystal field splitting decreases the probability that a metal True False center will reside in a high-spin electron configuration. (iv) The crystal field model assumes that the 𝑑𝑧 2 and 𝑑𝑥 2−𝑦2 orbitals participate in bonding because their lobes collectively accommodate an octahedral True False arrangement of ligands, while the remaining d-orbitals are non-bonding. 5. (4pts) Consider the set of coordination complexes (1 – 4) below when answering the following questions. No explanations are required for this part and ligand lone pairs have been omitted for clarity. (i) Which complex is most likely to reside in a low-spin configuration? Complex 1 (ii) Which complex would have greatest crystal field splitting? Complex 3 (iii) Which complex is the most likely to absorb the lowest energy visible light (e.g., red)? Complex 2 (iv) If complex 4 most strongly absorbs light with a wavelength of 600 nm, then what color does it appear? Blue 6. (2pts) Three separate 100.0 L containers are each charged with 1.00 mol of a gas at 200 K. Containers A, B, and C are filled with He, O2, and NH3, respectively. Use this information to answer the following. (i) Assuming ideal behavior, which container has the greatest pressure of gas? Container A Container B Container C All the same (ii) Assuming ideal behavior, which container has the greatest density of gas? Container A Container B Container C All the same (iii) Assuming real behavior, which container has the lowest pressure of gas? Container A Container B Container C All the same (iv) Assuming real behavior, which container has the greatest volume of available space? Container A Container B Container C All the same 7. (5pts) Identify the products for each reaction and classify each as precipitation, acid/base, or redox. Balance all reactions as necessary. (i) 2𝑁𝐻4 𝐶𝑙 𝑎𝑞 + 𝑆𝑟 𝑂𝐻 2 𝑎𝑞 → 2𝐻2 𝑂 𝑙 + 𝑆𝑟𝐶𝑙2 (𝑎𝑞) Acid/base (ii) 𝐹𝑒 𝑁𝑂3 3 𝑎𝑞 + 𝐾3 𝑃𝑂4 𝑎𝑞 → 𝐹𝑒𝑃𝑂4 𝑠 + 3𝐾𝑁𝑂3 (𝑎𝑞) Precipitation 9 (iii) 𝐶3 𝐻8 𝑂 𝑔 + 𝑂 𝑔 → 3𝐶𝑂2 𝑔 + 4𝐻2 𝑂(𝑔) Redox 2 2 (iv) 𝐹𝑒𝐶𝑙6 −3 𝑎𝑞 + 3 𝐻2 𝑁 𝐶𝐻2 2 𝑁𝐻2 𝑎𝑞 → 6𝐶𝑙− 𝑎𝑞 + 𝐹𝑒 𝑒𝑛 +3 (𝑎𝑞) Acid/base 3 (v) 𝑁𝐻3 𝑎𝑞 + 𝐵𝐹3 𝑎𝑞 → 𝐻3 𝑁 − 𝐵𝐹3 (𝑎𝑞) Acid/base 8. (3pts) Circle the compound from each set that you predict to have the highest boiling point. (i) (ii) (iii) 9. (3pts) Assign oxidation numbers to the central atom for each of the following molecules. +1 +3 +2 10. (2pts) Shown below are structures of two common organic liquids, 1,4-dioxane (a) and glycerol (b). Please identify which will have the greater surface tension, viscosity, adhesion to glass, and will be more miscible with hexane. Surface tension: (b) Viscosity: (b) Adhesion to glass: (b) Miscibility with hexane: (a) 11. A new technology has been developed which enables the separation of ideal gases with mass differences as small as 1 AMU. Analysis of a Xenon (Xe) sample obtained from an extraterrestrial planet’s atmosphere reveals that it is composed of exactly two Xe isotopes. The original sample occupies a 1.00 L vessel at STP, and upon separation into two separate 1.00 L vessels (Vessels A and B), the pressures are found to be 569.8 torr and 189.9 torr, respectively. a) (4pts) If 4.315 g and 1.461 g of gas occupy vessels A and B, respectively, then determine the molar mass and identity of each isotope. m = 1.461 g Vessel A m = 4.315 g 1𝑎𝑡𝑚 V = 1.00 L V = 1.00 L 𝑃 = 569.8 𝑡𝑜𝑟𝑟 ∙ = 0.7497 𝑎𝑡𝑚 760𝑡𝑜𝑟𝑟 T = 273 K T = 273 K 𝑃𝑉 0.7497𝑎𝑡𝑚 1.00𝐿 P = 569.8 torr P = 189.9 torr 𝑛= = = 0.03347 𝑚𝑜𝑙 𝑅𝑇 𝐿𝑎𝑡𝑚 0.08206 273𝐾 Vessel A Vessel B 𝑚𝑜𝑙𝐾 4.315𝑔 129 𝑀𝑤 = = 128.9𝑔/𝑚𝑜𝑙 54𝑋𝑒 0.03347𝑚𝑜𝑙 Vessel B 1𝑎𝑡𝑚 𝑃𝑉 0.2499𝑎𝑡𝑚 1.00𝐿 𝑃 = 189.9𝑡𝑜𝑟𝑟 ∙ = 0.2499 𝑎𝑡𝑚 𝑛= = = 0.01115 𝑚𝑜𝑙 760𝑡𝑜𝑟𝑟 𝑅𝑇 𝐿𝑎𝑡𝑚 0.08206 273𝐾 𝑚𝑜𝑙𝐾 1.461𝑔 131 𝑀𝑤 = = 131.0𝑔/𝑚𝑜𝑙 54𝑋𝑒 0.01115𝑚𝑜𝑙 b) (2pts) Calculate the average molar mass of the extraterrestrial Xe sample. 0.03347 0.01115 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑀𝑤 = χ129 129 𝑔/𝑚𝑜𝑙 + χ131 131 𝑔/𝑚𝑜𝑙 = 129 𝑔/𝑚𝑜𝑙 + 131 𝑔/𝑚𝑜𝑙 0.03347 + 0.01115 0.03347 + 0.01115 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑀𝑤 = 129.5 𝑔/𝑚𝑜𝑙 c) (6pts) On the planet the sample was obtained from, it has been determined that the atmospheric pressure is 91.9 atm, and it is 50% by mass Xe and 50% by mass isotopically pure 84 36𝐾𝑟. Determine the partial pressure of each Xe isotope on the planet. Assume 100 g sample: 1𝑚𝑜𝑙 1𝑚𝑜𝑙 𝑋𝑒: 50𝑔 ∙ = 0.3861 𝑚𝑜𝑙 𝐾𝑟: 50𝑔 ∙ = 0.5952 𝑚𝑜𝑙 𝑛𝑡𝑜𝑡𝑎𝑙 = 0.9813 𝑚𝑜𝑙 129.5𝑔 84𝑔 129 131 54𝑋𝑒: 0.75 ∙ 0.3861 𝑚𝑜𝑙 = 0.2896 𝑚𝑜𝑙 54𝑋𝑒: 0.25 ∙ 0.3861 𝑚𝑜𝑙 = 0.09653 𝑚𝑜𝑙 0.2896 𝑚𝑜𝑙 𝑃129 = χ129 𝑃𝑡𝑜𝑡𝑎𝑙 = 91.9 𝑎𝑡𝑚 = 27.1 𝑎𝑡𝑚 0.9813 𝑚𝑜𝑙 0.09653 𝑚𝑜𝑙 𝑃131 = χ131 𝑃𝑡𝑜𝑡𝑎𝑙 = 91.9 𝑎𝑡𝑚 = 9.040 𝑎𝑡𝑚 0.9813 𝑚𝑜𝑙 12. You have been hired by Cirri Industries to study the biosynthesis of tetrahydrocannabinol (THC), the primary psychoactive component of cannabis. In your studies, you have isolated 1.5000 g of an intermediate produced in the late stages of the plant’s natural THC synthesis. Using combustion analysis on 1.0000 g of the sample, you find that 2.6858 g of CO2 and 0.7998 g of H2O are produced. Mass spectrometry reveals that oxygen is the only other element present in the compound. A titration of the remaining 0.5000 g dissolved in 1.00 L water reveals that the compound is a triprotic acid (3 acidic protons), and that 41.61 mL of 0.100 M NaOH are required for neutralization. (Hint: Represent the general formula of the acid as H3A). a) (6pts) Determine the Empirical Formula of the compound. 1 𝑚𝑜𝑙 1 𝑚𝑜𝑙 𝐶 12.01 𝑔 2.6858𝑔 𝐶𝑂2 ∙ ∙ = 0.06103 𝑚𝑜𝑙 𝐶 ∙ = 0.7329 𝑔 𝐶 44.01 𝑔 1 𝑚𝑜𝑙 𝐶𝑂2 1 𝑚𝑜𝑙 1𝑚𝑜𝑙 2 𝑚𝑜𝑙 𝐻 1.008 𝑔 0.7998𝑔 𝐻2 𝑂 ∙ ∙ = 0.08877 𝑚𝑜𝑙 𝐻 ∙ = 0.08948 𝑔 𝐻 18.02𝑔 1𝑚𝑜𝑙 𝐻2 𝑂 1 𝑚𝑜𝑙 𝑀𝑎𝑠𝑠 𝑂 = 1.000𝑔 − 0.7329𝑔 − 0.08948𝑔 = 0.1776𝑔 1𝑚𝑜𝑙 𝑀𝑜𝑙𝑒𝑠 𝑂 = 0.1776𝑔 ∙ = 0.0111 𝑚𝑜𝑙 16.00𝑔 0.06103 𝑚𝑜𝑙 𝐶 0.08877 𝑚𝑜𝑙 𝐻 0.0111 𝑚𝑜𝑙 𝑂 5.5𝐶 8𝐻 1𝑂 𝐶11 𝐻16 𝑂2 0.0111 𝑚𝑜𝑙 b) (2pts) Determine the Molecular Formula of the organic molecule. 𝐻3 𝐴 𝑎𝑞 + 3𝑁𝑎𝑂𝐻 𝑎𝑞 → 3𝐻2 𝑂 𝑙 + 𝑁𝑎3 𝐴 (𝑎𝑞) 0.100 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻 1 𝑚𝑜𝑙 𝐻3 𝐴 0.04161 𝐿 𝑁𝑎𝑂𝐻 ∙ ∙ = 0.001387 𝑚𝑜𝑙 𝐻3 𝐴 𝐿 3 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻 0.500𝑔 𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 = = 360.5 𝑔/𝑚𝑜𝑙 0.001387𝑚𝑜𝑙 𝐸𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠(𝐶11 𝐻16 𝑂2 ) = 180.1 𝑔/𝑚𝑜𝑙 𝑀𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑀𝑤 360.5 = =2 Multiply Empirical formula by 2: 𝐶22 𝐻32 𝑂4 𝐸𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝑀𝑤 180.1 13. Lead(II) oxide is soluble in acidic solutions and produces a soluble product (A) in nitric acid. Solutions of A can react with potassium iodide to yield a precipitate of brilliant yellow crystals of product B. a) (4pts) Write a single set of molecular and net ionic equations for this series of reactions. Identify products A and B and name them. Reaction 1: 𝑃𝑏𝑂 𝑠 + 2𝐻𝑁𝑂3 𝑎𝑞 → 𝐻2 𝑂 𝑙 + 𝑃𝑏 𝑁𝑂3 2 𝑎𝑞 Reaction 2: 𝑃𝑏 𝑁𝑂3 2 𝑎𝑞 + 2𝐾𝐼 𝑎𝑞 → 𝑃𝑏𝐼2 𝑠 + 2𝐾𝑁𝑂3 𝑎𝑞 Net molecular 𝑃𝑏𝑂 𝑠 + 2𝐻𝑁𝑂3 𝑎𝑞 + 2𝐾𝐼(𝑎𝑞) → 𝐻2 𝑂 𝑙 + 𝑃𝑏𝐼2 𝑠 + 2𝐾𝑁𝑂3 𝑎𝑞 reaction: Complete 𝑃𝑏𝑂 𝑠 + 2𝐻 + 𝑎𝑞 + 2𝑁𝑂3− 𝑎𝑞 + 2𝐾 + 𝑎𝑞 + 2𝐼 − 𝑎𝑞 → 𝐻2 𝑂 𝑙 + 𝑃𝑏𝐼2 𝑠 + 2𝐾 + 𝑎𝑞 + 2𝑁𝑂3− (𝑎𝑞) ionic: Net ionic: 𝑃𝑏𝑂 𝑠 + 2𝐻+ (𝑎𝑞) + 2𝐼 − (𝑎𝑞) → 𝐻2 𝑂 𝑙 + 𝑃𝑏𝐼2 𝑠 𝐴 = 𝐿𝑒𝑎𝑑 𝐼𝐼 𝑛𝑖𝑡𝑟𝑎𝑡𝑒 𝑎𝑛𝑑 𝐵 = 𝐿𝑒𝑎𝑑 𝐼𝐼 𝑖𝑜𝑑𝑖𝑑𝑒 b) (4pts) You add 15.6 g of lead (II) oxide and 23.2 g of potassium iodide to 1000 mL of a 0.100 M solution of nitric acid. Identify the limiting reagent and the amount of product B that will form. 𝑃𝑏𝑂 𝑠 + 2𝐻𝑁𝑂3 𝑎𝑞 → 𝐻2 𝑂 𝑙 + 𝑃𝑏 𝑁𝑂3 2 𝑎𝑞 15.6𝑔 0.100𝑀 1.000 𝐿 Limiting Reagent Reaction 1 to find limiting reactant of whole reaction: 1𝑚𝑜𝑙 1 𝑚𝑜𝑙 𝑃𝑏 𝑁𝑂3 2 𝑃𝑏𝑂: 15.6𝑔 ∙ ∙ = 0.06989 𝑚𝑜𝑙 223.2𝑔 1 𝑚𝑜𝑙 𝑃𝑏𝑂 0.100 𝑚𝑜𝑙 1 𝑚𝑜𝑙 𝑃𝑏 𝑁𝑂3 2 𝐻𝑁𝑂3 : 1.000𝐿 ∙ ∙ = 0.0500 𝑚𝑜𝑙 Nitric acid limits the entire reaction! 𝐿 2 𝑚𝑜𝑙 𝐻𝑁𝑂3 𝑃𝑏 𝑁𝑂3 2𝑎𝑞 + 2𝐾𝐼 𝑎𝑞 → 𝑃𝑏𝐼2 𝑠 + 2𝐾𝑁𝑂3 𝑎𝑞 0.0500 𝑚𝑜𝑙 23.2 𝑔 Limiting Reagent Reaction 2 to find yield: 1𝑚𝑜𝑙 1 𝑚𝑜𝑙 𝑃𝑏𝐼2 𝐾𝐼: 23.2 𝑔 ∙ ∙ = 0.06988 𝑚𝑜𝑙 166 𝑔 2 𝑚𝑜𝑙 𝐾𝐼 1 𝑚𝑜𝑙 𝑃𝑏𝐼2 𝑃𝑏 𝑁𝑂3 2 : 0.0500 𝑚𝑜𝑙 ∙ = 0.0500 𝑚𝑜𝑙 Lead (II) nitrate is total consumed in reaction 2 1 𝑚𝑜𝑙 𝑃𝑏 𝑁𝑂3 2 𝑃𝑏𝐼2 𝑦𝑖𝑒𝑙𝑑: 0.0500 𝑚𝑜𝑙 = 23.1 𝑔 c) (4pts) Determine the concentrations of lead (II), potassium ion, and hydronium that will be present in solution at the end of this reaction. 𝐻+ = 0 𝑀 (𝑐𝑜𝑛𝑠𝑢𝑚𝑒𝑑 𝑖𝑛 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 1) 𝑃𝑏+2 = 0 𝑀 (𝑐𝑜𝑛𝑠𝑢𝑚𝑒𝑑 𝑖𝑛 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 2) 23.2𝑔 𝐾𝐼 1𝑚𝑜𝑙 1 𝑚𝑜𝑙 𝐾 + 𝐾+ = ∙ ∙ = 0.14 𝑀 1𝐿 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 166 𝑔 𝐾𝐼 1 𝑚𝑜𝑙 𝐾𝐼 14. (a) (8pts) Consider the diatomic BF molecule. Draw a MO diagram of BF using each atom’s valence orbitals. Be sure to include i) an energy axis, ii) appropriate labels for all orbitals, iii) appropriate relative energies for all atomic and molecular orbitals, and iv) the BF bond order. Compare this to the best Lewis structure for BF based on octet considerations. Do they agree with each other? MO Theory BF 𝝈∗𝟐𝒑 𝝅∗𝟐𝒑 2p 𝝈𝟐𝒑 2p 𝟔−𝟎 Energy 𝑩𝒐𝒏𝒅 𝒐𝒓𝒅𝒆𝒓 = =𝟑 𝟐 𝝅𝟐𝒑 𝝈∗𝟐𝒔 Octet-abiding Lewis: 2s 𝝈𝟐𝒔 Bond order = 3 2s This Lewis structure is consistent with MO! AOs of B MOs of BF AOs of F (b) (3pts) Advanced theoretical and experimental studies agree that the BF bond order is 1.5 and the boron has a high partial negative charge! Is MO consistent with the experiment? If not, what do you think this may indicate? MO could agree with the partial negative charge of B being negative since the HOMO would be heavily weighted toward th B atom, but it does not capture the bond order very well. This suggests that sp-mixing may introduce complex bonding effects beyond just switching the energetic ordering of σp and π. (c) Given below are the structures of the HOMO and LUMO for BeH2. Use the molecular orbitals to answer the following questions regarding reactivity. (i) (4pts) If BeH2 reacts with hydrochloric acid in the aqueous phase, do you expect BeH2 to function as a Brønsted-Lowry acid or base? Which atom would function as the acid or base? BeH2 would function as the Lewis base. H, based on HOMO, is the electron donor (ii) (2pts) In a follow-up to (i), sketch the orbital interaction between BeH2 and HCl that supports your answer. (d) Shown below is the MO diagram for the hypothetical compound KrH4. Use this diagram to answer the questions that follow. (i) (4pts) Sketch the structures of σ2 and σ3*. Ensure your sketches unambiguously account for the electronegative effect. Do not blend your orbitals, label the positions of the corresponding atoms within your MO, and include appropriate phases. σ2 : σ3*: (ii) (2pts) Place the appropriate number of electrons onto the MO diagram that would be consistent with KrH4’s electron count. Which orbital on the MO diagram would correspond to the HOMO? The LUMO? 𝐻𝑂𝑀𝑂 = σ∗3 𝐿𝑈𝑀𝑂 = σ∗4 (iii) (2pts) Based on the above, do you expect KrH4 to more likely function as an acid or a base? Why? A base – the HOMO is an antibonding orbital weighted toward H. Since electron density would build up on H, it would be an electron donor rather than an electron acceptor. By having the HOMO react, electrons would be lost from an antibonding orbital to make the resulting product compound more stable. 15. Briefly explain the experimental results below. All explanations must be restricted to 1 – 2 sentences, and you may include a picture to support your answers. (a) (2pts) While almost all Group 1 metal ionic compounds are water soluble, Li3PO4 is highly insoluble. Of the Group 1 metals, Li+ is the smallest. The lattice energy must be too high to permit dissolution. (b) (2pts) Group 1 metal oxides do not dissociate in water, but they do dissolve to create a basic solution. Explain how this would occur. They react with water to produce hydroxide. 𝐾2 𝑂 𝑠 + 𝐻2 𝑂 𝑙 → 2𝐾 + 𝑎𝑞 + 2𝑂𝐻− 𝑎𝑞 (c) (2pts) NaClO4 is known to be a stronger oxidizing agent than NaClO3. Why? The oxidation number of Cl in NaClO4 is +7, while the ON of Cl in NaClO3 is +5. The Cl in the more oxidized salt is more willing to accept electrons (more strongly attracted to them) due to the higher ON. (d) (2pts) At temperature of 300 K and pressures below 200 atm, the volume of almost all gases is overestimated by the ideal gas law. Why? This indicates the gas is more compressed than expected, and this must be due to IMFs attracting the gas particles together. (e) (2pts) It is not possible for an ideal gas to undergo a phase transition. Why? Particles in an ideal gas exert no attractive forces on one another, indicating that condensation is not possible. Attractive forces are required to stabilize the liquid or solid phases! (f) (2pts) Fluorosulfuric acid (FSO3H) is a stronger acid than sulfuric acid. Why? F helps to tug electrons away from the H in the O—H bond more readily than O (F is more EN). This allows for a greater partial positive charge build-up on H, giving rise to the higher acidity. (g) (2pts) In Lecture 1 of this semester, we discussed how replacing a glutamic acid amino acid on the surface of hemoglobin with valine (this is a mutation; side chains below) leads to hemoglobin proteins sticking to each other, causing sickle cell anemia. Why does this mutation lead to sticking between the proteins? The glutamic acid side chain is polar and can interact strongly with water, while valine’s side chain is non-polar and hydrophobic (water repelling). This will reduce hemoglobin’s water solubility and increase the sticking affinity between hemoglobin proteins through dispersion forces.