Chemistry HW 5 - Kim Eugene – Homework 5 – Due Oct 2...

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Unformatted text preview: Kim, Eugene – Homework 5 – Due: Oct 2 2007, midnight – Inst: McCord I 3−  I −→   Ca3 I2 ¦¦  ¤      −    I −→ CaI       Ca+ +   ¤  −→ 1. Mg2+ , O2− correct 2. Mg2+ , Se2−   + 3−  §§ I −→  ¨¨ © I © 3. Ca + + I +  © © + Ca © + Ca I  Ca 2+  ¤ 2. Ca + Ca + Ca + 2+ −→ Ca3 I ¥¥ Ca + Ca + Ca + 2+ 3−  + ¥¥ + ¥ ¤¥ 001 (part 1 of 1) 10 points Which of the following pairs of ions would have the greatest coulombic attraction in a solid compound? + ¥¥ This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 1 4. None of these + 2− $ ¤$ $ # +I $ $$ # # " ¤" " # ##  ¤  " " " "" (( −→ − 00 + 11 '' 0 ¤0 00 I I 11 ' 00 & & && )) )) % ¤% % & && ) ¤) )) + I 4 ¤4 8 8 8 I + −→ CaI −→ A A A A BB 2− F ¤F FF F F E ¤E E E D C Ca + Ca + I F EE + −→ CaI −→ F @ ¤@ @ + I BB A 10. Ca + Ca + 3− I BB 7 Ca3+ + B ¤B 7 8 8 003 (part 1 of 1) 10 points Use electron-dot notation to demonstrate the formation of ionic compounds involving the elements Ca and I. 99 8 I 2− 99 55 9. Ca + 2+ 9 ¤9 55 Ca Explanation: E is aluminum; since it has three valence electrons it forms a +3 ion isoelectronic with Ne. −→ CaI2 correct −→ 5. E3 (SO4 ) 6. E3 (SO4 )2 Ca2 I3 − 99 5 ¤5 55 I I 66 33 + 66 22 3 8. Ca + 6 6 44 − −→ −→ 66 3 3 33 I 4 4. E(SO4 )2 + + I 44 Ca 2+ I − 11 + 1 ¤1 ' ¤' − I '' I + ( + ( ( Ca 3+ I (( 6. Ca + 7. Ca + 3. E2 (SO4 )3 correct −→ Ca2 I3 1. E2 (SO4 ) 2. E(SO4 )3 −→ 2− $  # ¤#  ¤   +I I $ I + !   2− +I  +Ca 3+  Ca 3+ I ! 002 (part 1 of 1) 10 points An element E has the electronic configuration [Ne] 3s2 3p1 . Write the formula of its compound with sulfate. 5. Ca + Ca + ! ¤! Explanation: 1 E ∝ , so the ion with the shortest radius r will have the greatest coulombic attraction. !! 3. Mg2+ , S2− −→ Ca2 I Explanation: £ £ ¤£ I −→ Ca gives up two electrons to form the Ca2+ GG £ £ ¢ ¡ 1. Ca + Ca + Ca + Kim, Eugene – Homework 5 – Due: Oct 2 2007, midnight – Inst: McCord H ¤H H I ¤I II II I HH − H II the I acquires an electron to form H cation, and R R TT + I TT SS I − − TT S S SS SS RR + −→ T T Q ¤Q R QQ Ca 2+ I RR + I 2. 0 3. 2 anion. Two iodine anions combine with one calcium cation to form an electrically neutral compound: QQ Q PP Ca + 2 −→ CaI2 004 (part 1 of 1) 10 points Which of the compounds below has bonds with the most covalent character? 1. LiCl 4. 16 5. 7 Explanation: Valence electrons are electrons that are found in the outermost energy levels of an atom. For Ar, Atomic number : 18 Electronic configuration : [Ne] 3s2 3p6 Shell Structure : 2 − 8 − 8 Therefore, Ar has 8 valence electrons. 2. MgCl2 3. BeCl2 correct 007 (part 1 of 1) 10 points The P2− anion has how many total electrons and how many valence electrons? 4. NaCl 1. 16; 7 5. CaCl2 Explanation: ∆EN is min for BeCl2 . 005 (part 1 of 1) 10 points The lattice enthalpy of calcium bromide is the energy change for the reaction 2. 17; 8 3. 15; 7 4. 18; 8 5. 15; 5 1. Ca(s) + Br2 ( ) → CaBr2 (s) 6. 16; 6 2. CaBr2 (s) → Ca2+ (g) + 2 Br− (g) correct 7. 15; 6 3. CaBr2 (s) → Ca(g) + 2 Br(g) 8. 17; 7 correct 4. CaBr2 (s) → Ca(g) + Br2 (g) 9. 17; 6 5. Ca(g) + 2 Br(g) → CaBr2 (g) 10. 16; 5 Explanation: 006 (part 1 of 1) 10 points How many valence electrons are in a Ar atom? 1. 8 correct Explanation: P2− has gained 2 electrons to give it a total of 17. In its valence, it has 3s2 3p5 , giving it a total of 7 valence electrons. 008 (part 1 of 1) 10 points What total number of valence electrons Kim, Eugene – Homework 5 Due: Oct 2 2007, midnight – Inst: McCord ‚‚ ƒƒ …… OO † † ˆˆ …… ‡‡ OO ˆˆ ‡‡ ‡‡ ‰ OO ‰ Explanation: N = 8 × 2 = 16 A = 6 × 2 = 12 S = N − A = 16 − 12 = 4 There are 4 shared electrons. The remaining electrons are placed around the oxygen atoms such that each has an octet. The Lewis dot formula for oxygen (O2 ) is O ‘‘ ‘ ‘ O ‘‘ U  ` C g g hh p qp i i r r ss u Wu t t ww v v ’ ’’ ““ 2. C H H C •• 3. 010 (part 1 of 1) 10 points H C C H •• e fe Explanation: The metal loses all of its outer (valence) electrons; the non-metal gains sufficient electrons to form an octet. The ratio of the negative and positive ions is chosen so there is no overall charge. C •• b d ’ f’ H 1. ”” b cc d correct ’ X a Wa ` 2− ’ X YY  011 (part 1 of 1) 10 points Which of the following is the correct Lewis formula for acetylene (ethyne, C2 H2 )? ’ V WV U S …… S ††…… 6. Ca2+ , 3− † † 5. 3 Ca2+ , 2 O †† 4. None is appropriate because calcium sulfide is a covalent compound. „„ S „„ „„ „„ − O „„ S 7. ƒ ƒ − ƒ ƒ ,2 3. Ca+ , 2− S ƒƒ Which of the following is the best representation of the compound calcium sulfide? 2. Ca ‚‚ 8. 2+ ‚‚ 009 (part 1 of 1) 10 points OO 6. 5. ‚‚ Explanation: Chlorine has 7 valence electrons and oxygen 6. The overall −1 charge indicates that there is 1 extra electron. The total number of valence electrons is 1 × 7 e− (from Cl) + 3 × 6 e− (from O) +1 e− (from −1 charge) = 26 e− 1. 2 Ca+ , OO 4. 5. 26 correct ‚‚ €€ 4. 32   €€ 3. OO € 3. 24 € 2.   xx 2. 28 correct O  O yy xx 1. yy xx 1. 30 Which of the following is the correct Lewis formula for oxygen (O2 )? yy should appear in the dot formula for the chlorate ion ClO− ? 3 3 H correct Kim, Eugene – Homework 5 – Due: Oct 2 2007, midnight – Inst: McCord 4. two; two – – —– –– C – C H 4. –– 5. four; four – – H 4 – – –– ˜ ˜ f˜ ˜ —˜ C 6. zero; eight ˜˜ ˜˜ 7. four; three ˜ ˜ 8. none; two ™™ ™ C ™ 9. two; six correct ™™ ™ ™ O O mm k k d d f f 2. PH3 correct g g gg 3. H2 S C 4. SiH4 dd ff e fe h h e e e fe g g h H 5. NaCl h hh ii C i i fi C i H ii i fi i i H i i i i ii Explanation: The Lewis formula for acetylene (ethyne, C2 H2 ) is pp 10. O uu C ↔ 1. C2 H4 C H 9. C H O ss H O 013 (part 1 of 1) 10 points Which of the following contains exactly one unshared pair of valence electrons? C H 8. l —l H O j fj 7. C n on ™ ™™ Explanation: v —v ™ H q fq 6. C t —t ˜ ˜˜ H r ˜ ˜ H r ˜ ˜ f˜ 5. C ˜ H Explanation: The PH3 molecule contains exactly 1 unshared pair of valence electrons: ·· H H P H 2. one; three You can check this structure for correctness by verifying that each atom has the correct number of electrons around it (8 for most elements, 2 for hydrogen) and that the structure shows the correct total number of valence electrons (calculated by adding up the available valence electrons from each atom). The C2 H4 molecule contains no unshared pairs; SiH4 contains no unshared pairs; NaCl (an ionic compound) contains 4 unshared pairs; and H2 S contains 2 unshared pairs. 3. three; six 014 (part 1 of 1) 10 points H C C H 012 (part 1 of 1) 10 points How many unshared electrons and bonding electrons exist around the central atom in ozone, O3 ? 1. one; six Kim, Eugene – Homework 5 – Due: Oct 2 2007, midnight – Inst: McCord Œ Œ F Cl Œ Œ  F Cl F Cl   ‘ ‘ ŽŽ  ‘ ‘ ŽŽ 2. 0   3.  F Cl 5. ’’ F Cl 6. ” F Cl ”” ” ”” F Cl is the most plausible • • • F Cl • • • — — — F Cl — — — ™™ F Cl ™™ –– ˜˜ ˜˜ Explanation: The Lewis formula for chlorine fluoride (ClF) is F Cl šš › œ› šš šš 017 (part 1 of 1) 10 points Which of the following is the correct Lewis formula for boron trichloride (BCl3 )? | w { zz { žž  f B ŸŸ   f ŸŸ 7. three Cl Cl f B Cl Cl ‰ Cl ¡¡ Cl ¡¡ „„ ‡ B Cl ¡ ‹‹ ‡ ¡¡ ¡ ¡ 4. ¡ ‰ …… †† ƒ ƒ ˆ ˆ 016 (part 1 of 1) 10 points Which of the following is the correct Lewis formula for chlorine fluoride (ClF)? F Cl ‚ f‚ ŠŠ F F 3. f N Explanation: The Lewis structure is B Cl f 8. ten 2. Ÿ Cl Ÿ  Cl 6. twenty ž 1. ž Cl ¡ | w 5. four ¡ yy 4. none x —x ~~ 3. one correct  — } } 2. thirteen › œ› € 1. two ›› €  –– 10. •  015 (part 1 of 1) 10 points How many lone pairs of electrons are on nitrogen in NF3 ? • 9. ““ 7. 8. ““ structure. ’’ 3− ’’ Explanation: O 0 −1 O P O 0 −1 −1 O ’’ 3. 1 correct correct ‘‘ 4. 4. 3  2.  1. 2 1. ŒŒ How many double bonds are present in the “best” resonance structure of the phosphate ion? 5 Kim, Eugene – Homework 5 – Due: Oct 2 2007, midnight – Inst: McCord ¢¢ ££ Cl 5. 5 ¤ ¤¤ ¤ Explanation: C → 4 valence e− O → 3 × 6 valence e− 2− → 2 e − Total = 24 valence e− One structure is · ·· C ·O ·· ¥¥ ¥ Cl ¥ 6. ¤ Cl ¤ 5. 4. 4 B Cl B Cl Cl ¦ ¦ —¦ ¦¦ Cl ¦ B Cl ¦¦ 7. ¦ ¦ ¦ ¦¦ §§ § f§ B Cl § correct §§ Cl § § f§ § § Cl § § § 8. § §§ B Cl ¨ Cl C ·· ·· O· · ·O· · ·· · ··O·· for a total of 3 resonance structures. 019 (part 1 of 1) 10 points Resonance is a concept that describes the bonding in molecules Cl ¨ ¨ 9. ·· · O· ·O· · ·· · which can also be drawn as ·· · or ··O · ·· O· C ·O ·· ·· ·· ¦ Cl 6 ¨ ¨¨ 1. by asserting that double bonds “flip” or resonate between two locations in the molecule. © —© Cl © Cl © 10. B Cl © © ©© Explanation: B contributes 3 valence e− and each Cl contributes 7 valence e− for a total of 24 e− . B is a known exception to the octet rule and can form stable with 6 valence e− : ªª ª ª —ª Cl ª B ªª Cl ª ª ª ª ª ªª 018 (part 1 of 1) 10 points The CO2− ion has how many resonance con3 figurations? 1. It does not exhibit resonance. 2. 2 3. 3 correct 3. where there is more than one choice of location for a double bond as deduced from Lewis dot structures. The true bonding is the average over all possible double-bond locations. correct ª ª —ª Cl 2. by asserting that electrons in a double bond can delocalize (spill over) onto adjacent single bonds to make a bond and a half. Explanation: In resonance, although the arrangement of all possible double bond locations may sometimes mean that a bond has a bond order of 1.5, other possibilities can occur such as 1.333 (1 and a third bonds). 020 (part 1 of 1) 10 points Which of the three Lewis structures is the most important? Kim, Eugene – Homework 5 – Due: Oct 2 2007, midnight – Inst: McCord 0 O B) −1 C +1 −1 N O C) −3 C +1 +1 N O « —« » o» N N O N O has formal charges out of the N O ÁÁ N À ¿ ↔ À ¾¾ à à ¬¬ ¼¼ ÆÆ ± f± ¯ ¯ Å fÅ N Ä Ä acceptable range. °° ¶¶ 022 (part 1 of 1) 10 points Calculate the formal charge on N in the molecule NH3 . ­­ ²² ´ f´ µµ ³ ³ 2. A and B only 3. None of these is important. 1. 1 4. A and C only 2. 2 5. C only 3. 3 6. B and C only 4. 4 7. A only 5. 0 correct 8. All of these are important. Explanation: H The Lewis structure is 0 Explanation: −1 is C N O probably the most important as it is the structure with the formal charges of the individual atoms closest to zero. +1 ¹ f¹ · −1 · ºº The Lewis structure 0 ¸¸ 021 (part 1 of 1) 10 points How many resonance structures can be drawn for N2 O? Disregard any structure with formal charges other than 0, +1, and −1. Ç fÇ ® o® 1. B only correct N 0 H H0 023 (part 1 of 1) 10 points Estimate the heat released when 1-butene (CH3 CH2 CH CH2 ) reacts with bromine to give CH3 CH2 CHBrCH2 Br. Bond enthalpies are C H : 412 kJ/mol; C C : 348 kJ/mol; C C : 612 kJ/mol; C Br : 276 kJ/mol; Br Br : 193 kJ/mol. 1. 288 kJ/mol 1. 2 correct 2. 181 kJ/mol 2. 3 3. 317 kJ/mol 3. 1 4. 507 kJ/mol 4. 0 5. 95 kJ/mol correct Explanation: ¿ N ½ o½ +1  — A) −2 C 7 Explanation: Kim, Eugene – Homework 5 – Due: Oct 2 2007, midnight – Inst: McCord H H C C C C H off energy). In a reaction, the bonds in the reactants are broken; the ones in the products are formed. The net energy flow determines if the overall reaction is exothermic or endothermic. H H H H + Br Br → H H ∆H = Ebreak − = (C C) + (Br − 2 (C H Br H C C C C H H 8 H H H 025 (part 1 of 1) 10 points Br Consider the reaction Emake CH4 (g) + I2 → CH3 I(g) + HI(g) . Br) Br) + (C C) = 612 kJ/mol + 193 kJ/mol − 2 (276 kJ/mol) + 348 kJ/mol = −95 kJ/mol , which means 95 kJ/mol of heat was released. 024 (part 1 of 1) 10 points The heat energy released or absorbed by a chemical reaction is generally determined by the difference between 1. the energy that is released upon breaking the bonds in the reactants and the energy that is released upon making the bonds in the products. 2. the energy that must be put in to break the bonds in the reactants and the energy that is released upon making the bonds in the products. correct 3. the energy that must be put in to break the bonds in the reactants and the energy that must be put in to make the bonds in the products. 4. the energy that is released upon breaking the bonds in the reactants and the energy that must be put in to make the bonds in the products. Explanation: Bond breaking is endothermic (takes in energy) and bond creation is exothermic (gives Bond energy tables give the following values: C H : 411 kJ/mol I I : 149 kJ/mol H I : 295 kJ/mol C I : 213 kJ/mol The change in enthalpy for this reaction is: 1. +52 kJ/mol correct 2. There is no way to tell. 3. −97 kJ/mol 4. +463 kJ/mol 5. −52 kJ/mol Explanation: H H C H +I −−→ −− I H I H C H +H H ∆H = BEreactants − BEproducts H) + (I I) = 4 (C − 3 (C H) − (C I) − (H I) = (C H) + (I I) I) − (H I) − (C = 411 kJ/mol + 149 kJ/mol − 213 kJ/mol − 295 kJ/mol = 52 kJ/mol I Kim, Eugene – Homework 5 – Due: Oct 2 2007, midnight – Inst: McCord 026 (part 1 of 1) 10 points Bond energies are approximate because 1. not all molecules burn in oxygen, making calorimetry difficult. 2. they are averages over a number of molecules. correct 3. they depend upon the physical state of the molecule, gas, liquid, or solid. 4. we cannot measure them accurately. 5. we cannot calculate them accurately. Explanation: Bond energies are averages over a number of molecules. The C H bonds in CH4 and CH3 OH are not the same and require different amounts of energy to break. 9 ...
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This note was uploaded on 11/01/2011 for the course CH 301 taught by Professor Fakhreddine/lyon during the Fall '07 term at University of Texas.

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