Chem Spring, 2017 Hour Exam III May 8, 2017

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1 Page 1 of 8 Chem 70 - Spring, 2017 Hour Exam III May 8, (2 points; 4 points each) Fill in the blanks. In some cases, answer choices are shown in bold italics. a. From your recall of the appropriate Orgel diagram, indicate the number of bands expected in the visible spectra of the following complexes. (Disregard any possible bands resulting from Jahn-Teller effects or spin-forbidden transitions.) [Ti(H 2 O) 6 ] + (d 1, O h ) 1 [Cr(H 2 O) 6 ] 2+ (d 4, O h ) 1 b. Circle the ground state term for each of the following complexes: [Ti(H 2 O) 6 ] + (d 1, O h ) 2 E g 2 T 1g 2 A 2g 2 T 1g [Cr(H 2 O) 6 ] 2+ (d 4, O h ) 5 E g 5 T 2g 5 A 2g T 1g c. Indicate whether the following complexes in their ground states show strong, weak, or no Jahn- Teller distortion. [Ti(H 2 O) 6 ] + (d 1, O h ) weak [Cr(H 2 O) 6 ] 2+ (d 4, O h ) strong d. [Cr(C 2 O 4 ) ] (d, ~O h ) shows visible absorption bands at 17,400 cm 1 and 2,600 cm 1, with a third band occurring well into the ultraviolet. What is the value (cm 1 ) of o for this complex, and what is the state-to-state notation (e.g., 2 E g 6 2 T 2g ) for the transition that corresponds to o? o = 17,400 cm 1 transition 4 A 2g 6 4 T 2g e. [Ti(NCS) 6 ] shows an asymmetric, slightly split visible absorption band at 18,400 cm 1. Suggest a reason for the splitting of this band. The band arises from the 2 T 2g 6 2 E g transition. While the ground state has only weak Jahn- Teller distortion, the doubly degenerate 2 E g state has a strong distortion, giving rise to two possible transition energies. f. In the reaction, [Pt(CO)Cl ]! + NH 6 [Pt(CO)(NH )Cl 2 ], which isomer, cis or trans, will be favored for the product? Answer: trans

2 Page 2 of 8 g. Consider the redox reaction, [Co(NH ) 6 ] + + [Cr(H 2 O) 6 ] 2+ 6 [Co(NH ) 6 ] 2+ + [Cr(H 2 O) 6 ] + which occurs rapidly in aqueous medium. Is this redox reaction likely to be an inner-sphere or an outer-sphere mechanism? Explain briefly. An outer sphere mechanism is more likely. Neither complex has a ligand that is capable of serving as a bridge to make a linked intermediate complex, so there is no capability to have an inner sphere mechanism. The fact that the redox reaction occurs rapidly is also consistent with an outer sphere mechanism. h. Consider the following equilibrium: [Co(NH ) 6 ] + + 6H O + º [Co(H 2 O) 6 ] + + 6NH 4 + In this reaction [Co(NH ) 6 ] + is said to be unstable but inert. Briefly explain what these terms mean with respect to this equilibrium. Stable and unstable are thermodynamic terms. That [Co(NH ) 6 ] + is unstable in this reaction implies that the equilibrium, once established, lies to the right (i.e., K > 1). Labile and inert are kinetic terms. That [Co(NH ) 6 ] + is inert to aquation in this reaction implies that the equilibrium is established slowly. This suggests that the ammine complex will persist in acid medium despite having a favorable equilibrium constant for replacement by ligand H 2 O.

3 Page of 8 2. (12 points) Consider the Tanabe-Sugano diagram for d 4 octahedral complexes shown below. List the same-spin state-to-state transitions that are expected for both d 4 high-spin and d 4 low-spin ML 6 complexes, using the usual state-to-state notation (i.e., ground state 6 excited state). d 4 high-spin d 4 low-spin 5 E g 6 5 T 2g T 1g 6 E g T 1g 6 T 2g T 1g 6 A 1g T 1g 6 A 2g What factors might alter the number of observed bands in the visible spectrum from the number of possible transitions you have identified for each case? Both the ground state of the high-spin case and the first excited state of the low spin case have strong Jahn-Teller distortions, which could cause band splitting. For the low-spin case, the close spacing of all the excited states could result in incomplete resolution, making fewer bands apparent.

4 Page 4 of 8. (40 points; 4 points each) In the spaces below the given information, write balanced chemical equations for ten (10) of the following. Be sure to cross out the eight (8) you are not answering. a. Na + 2NH (l) Fe + NH (l) Fe + 2Na + 2NH (l) 2NaNH 2 + H NH 2 (l) b. CaC 2 + H 2 O CaC 2 + 2H 2 O C 2 H 2 + Ca(OH) 2 c. CH 2 (CO 2 H) 2 + P 4 O 10 CH 2 (CO 2 H) 2 + P 4 O 10 2C O 2 + 4H PO 4 d. CaCO CaCO CaO + CO 2 e. Ca(OH) 2 (aq) + SO 2 (g) 6 Ca(OH) 2 (aq) + SO 2 (g) 6 CaSO (s) + H 2 O(l) f. BCl + H 2 O BCl + H 2 O B(OH) + HCl g. The Hall-Héroult process 1000 oc Al 2 O K 4Al + O 2 AlF 6 /elect. h. NaBH 4 + BF ether ether NaBH 4 + 4BF 2B 2 H 6 + NaBF 4 i. Mg burning in a CO 2 atmosphere 2Mg + CO 2 2MgO + C (More equations on next page)

5 Page 5 of 8 ether j. LiH + Al 2 Cl 6 ether 8LiH + Al 2 Cl 6 2LiAlH 4 + 6LiCl k. With moderate heating, NH 4 NO NH 4 NO N 2 O + 2H 2 O l. HSCN + MnO 2 4HSCN + MnO 2 (SCN) 2 + Mn(SCN) 2 + 2H 2 O m. Br 2 + OH Br 2 + 2OH Br + OBr + H 2 O n. Industrial synthesis of hydrochloric acid NaCl(s) + H 2 SO 4 (aq) NaHSO 4 (aq) + HCl(g) o. Xe + F 2 hν Xe + F hν 2 XeF 2 p. SO 2 (aq) + S(s) SO 2 (aq) + S(s) S 2 O 2 (aq) q. Raschig synthesis 2NH + OCl r. Synthesis of F 2 (g) glue N 2 H 4 + Cl + H 2 O 2KHF 2 electrolysis o H 2 + F 2 + 2KF

6 Page 6 of 8 4. (16 points; 8 points each) Based on your knowledge of structure, reactivity, and bonding of the main group elements, offer brief chemical explanations for two (2) of the following four observations. Be sure to cross out the two that you are not answering. a. The only known noble gas compounds that are stable at room temperature are formed with xenon, and those are only formed with fluorine and/or oxygen. Explain. The ionization energies of the lighter noble gases are too high to be compensated by stable bond formation. Although Kr forms some fluoride compounds at very low temperature, even KrF 2 has an endothermic H o f, whereas it is exothermic for XeF 2, which forms spontaneously when Xe and F 2 are irradiated with u.v. light. Only very electronegative O and F are capable of making bonds with Xe that are polar enough to be strong, and therefore justify the energy input of ionization. Rn has a slightly lower ionization energy and might be capable of forming compounds, but the bonds would be weaker and might not compensats for the ionization energy requirement. In any event, radon's instability ( 222 Rn t ½ =.825 days) would limit the longevity of any compound formed. b. In our survey of the main group elements, we have noted some substances that pose particularly serious hazards. Briefly describe the hazards posed by the following substances and the circumstances under which the hazard is most acute in each case. Be specific! Beryllium compounds Except for minerals, such as beryl, all beryllium compounds are extremely toxic, owing to the acidity of the Be 2+ (aq) ion. The hazrd is particularly acute from inhaling beryllium compound dusts, causing beryllicosis, which may become apparent years after chance exposure. Perchloric acid HClO 4, particularly when concentrated, is a strong oxidant that causes fires or explosions in contact with organic or other oxidizable materials. The hazard is most acute when concentrated to HClO 2 O, which is a shock sensitive explosive. Hydrofluoric acid Although HF is a weak acid, it causes severe burns, which may not be immediately detected. Prolonged exposure, particularly to minute quantities in breathed air, can result in osteoporosis. White phosphorous The white form of P 4 can be stored under what, but it spontaneously reacts with O 2 in the air in a very exothermic reaction that produces P 4 O 10. The hazard is most acute with finely divided phosphorous, which will spontaneously flame in air. Hence, it has been used in incendiary bombs.

7 Page 7 of 8 c. The basis of organic chemistry is carbon's ability to catenate without limit. Why doesn't silicon show similar behavior? What is the more common bond type for Si, in lieu of homonuclear catenation? The C-C bond strength is comparable to common heteronuclear bonds (e.g., C-H, C-O, C-Cl), which allows building of limitless chains that are stable relative to heteronuclear bonds. Moreover, carbon forms effective pπ-pπ bonds, resulting in double and triple bonds and π- delocalization. Larger silicon forms weaker homonuclear bonds than carbon, and the Si-Si bond is significantly inferior to heteronuclear bonds. The Si-O bond, which may be bridging, is more robust than the S-Si bond, and that is the ubiquitous bond type in silicon chemistry. d. In our survey of the main group elements, we have seen that diagonally related elements in periods 2 and often show similar chemistries. What is the underlying cause of similar chemistry among diagonally related pairs of elements? Give at least three specific examples of uniquely similar chemistry among diagonally related elements. The main cause of unique chemistry among diagonally related pairs of elements is similarity of charge density. First elements, in general, tend to have more covalent character, owing to high charge density (Fajan's rules). Although larger, the diagonally related elements have higher charges, giving them similar charge density. Examples: Li & Mg tend to form useful organometallic compounds Be & Al form bridged species with either 2c-2e [e.g., BeCl 2 (s) & Al 2 Cl 6 ] or c-2e bonds [e.g., BeH 2 (s) & Al 2 (CH ) 6]. B & Si form borates and silicates with M-O bridging bonds, which facilitates their mixing in borosilicate glass. [Many other examples can be cited.]

8 Page 8 of 8 Bonus Questions (4 points each) a. For substitution reactions on a particular transition metal complex, why would the observation that rate constants vary with the identity of the entering ligand be evidence for an associative mechanism (A or I a )? The rate law expressions for I a and I d are too similar (often first order) to form a basis for distinguishing between them. However, if the rate constants for different entering ligands are quite different, that suggests that the formation of an associative complex (effectively increasing the coordination number) is a rate determining step, and the mechanism is I a. The rate constants would increase with the ability of the entering ligand to form an effective bond either within (A) or in contact with (I a ) the coordinate sphere of the original complex. b. The yellow?prussate of soda", Na 4 [Fe(CN) 6 ], has been added to table salt as an anticaking agent. Why have there been no apparent toxic effects, even though this complex contains cyano ligands? [Fe(CN) 6 ] 4! is a low-spin d 6 configuration ( 1 A 1g ), with the maximum CFSE of!2.4 o (ignoring the P pairing energy). Having a nondegenerate ground state, the complex is immune to Jahn- Teller distortion and can be perfect O h. The low-spin CFSE, strong Fe!CN bonds, and absence of Jahn-Teller distortion make this a stable and inert complex. Thus, there is no danger of dissociation of toxic CN!.

Chem Spring, 2018 Hour Exam III - Part 2 May 7, 2018

Chem Spring, 2018 Hour Exam III - Part 2 May 7, 2018 Chem 370 - Spring, 2018 Hour Exam III - Part 2 May 7, 2018 Page 1 of 5 3. (40 points; 4 points each) In the spaces below the given information, write balanced chemical equations for ten (10) of the following.