hw 5 answers - Iype, Shelby – Homework 5 – Due: Oct 2...

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Unformatted text preview: Iype, Shelby – Homework 5 – Due: Oct 2 2007, midnight – Inst: Vandenbout £ £ 1 −→ CaI − 3. None of these  Explanation: ¡¡ ¡ 1. Ca + I −→ Ca gives up two electrons to form the Ca2+ GG FF EE DD 003 (part 1 of 1) 10 points Use electron-dot notation to demonstrate the formation of ionic compounds involving the elements Ca and I. ¡¡ ¡ ¢¡ I + I + I Ca2 I3 FF FF EE EE DD DD Ca3+ + F ¢F E ¢E − − CC C BB B D ¢D AA A 10. Ca + I + I + I CC C ¤C BB B ¢B AA A ¤A @ @ ¢@ Explanation: E is aluminum; since it has three valence electrons it forms a +3 ion isoelectronic with Ne. −→ − 9 99 9 8 88 8 7 77 7 Ca +Ca +I +I +I 9 9 8 8 7 7 6. E(SO4 )3 3+ 3+ 9 ¢9 8 ¢8 2− 2− 2− Ca2 I3 6 6 6 5 5 5 4 4 4 9. Ca + Ca + 7 ¢7 I + I + I 6 5 4 3 2 5. E3 (SO4 ) 6 ¢6 1 5 ¤5 11 1 4. E(SO4 )2 2 ¢2 Ca + Ca + 4 ¤4 3 ¢3 I 1 1 + + 1 ¢1 0 3. E2 (SO4 ) 0 8. Ca + Ca + I 00 0 ¢0 2. E3 (SO4 )2 ) ( −→ 2− '' Ca + I '' '' 3+ ' ¢' 1. E2 (SO4 )3 correct & & 7. Ca + I −→ 3− $$ Ca & & ¤& % % ¢% + I $$ $$ 2+ $ ¢$ # 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. # 6. Ca + I −→ 2− −→ CaI −→ CaI −→ Ca2 I −→ −→ ! !! ! Ca + Ca + Ca + I + I ! ! ## # ¤# Explanation: 1 E ∝ , so the ion with the shortest radius r will have the greatest coulombic attraction. 2+ 2+ 2+ ! ¤! 3− 3−   3. Mg2+ , Se2−   5. Ca + Ca + Ca + ¢ I + I Ca3 I2   ¢   ¤ 2. Mg 2+ ,O 2− correct       Ca + Ca + Ca + I   + + +  ¢ 1. Mg2+ , S2−  4. Ca + Ca + Ca + I   ¢ ©©   ¨¨ ""  001 (part 1 of 1) 10 points Which of the following pairs of ions would have the greatest coulombic attraction in a solid compound? Ca + I + I ©© ©© ¨¨ ¨¨ 2+ § © ¤© §§ § ¦ ¦¦ ¨ ¢¨ ¦ 2. Ca + ¥ I + I −→ − −→ CaI2 correct ££ Ca + § § ¢§ ¦ ¦ ¤¦ ¥ I −→ 3− £ £ + £ ¤£ 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. − −→ Ca3 I −→ −→ −→ Iype, Shelby – Homework 5 – Due: Oct 2 2007, midnight – Inst: Vandenbout H H 2 − II Two iodine anions combine with one calcium cation to form an electrically neutral compound: RR R QQ Q Ca + I + 2+ I SS 004 (part 1 of 1) 10 points Which of the compounds below has bonds with the most covalent character? 1. MgCl2 2. BeCl2 correct 3. NaCl 4. CaCl2 5. LiCl 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 1. Ca(s) + Br2 ( ) → CaBr2 (s) 2. CaBr2 (s) → Ca2+ (g) + 2 Br− (g) correct 3. CaBr2 (s) → Ca(g) + 2 Br(g) 4. Ca(g) + 2 Br(g) → CaBr2 (g) 5. CaBr2 (s) → Ca(g) + Br2 (g) Explanation: 006 (part 1 of 1) 10 points How many valence electrons are in a I atom? 1. 15 2. 17; 6 3. 15; 5 4. 17; 8 5. 16; 6 6. 18; 8 7. 15; 7 8. 16; 7 9. 15; 6 10. 17; 7 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 TT SS Ca + I + I TT T ¤T TT SS S ¤S RR R ¤R QQ Q ¢Q II the I anion. HH cation, and I ¢I II I H H ¢H PP acquires an electron to form 2. 5 3. 3 4. 7 correct 5. 33 Explanation: Valence electrons are electrons that are found in the outermost energy levels of an atom. For I, Atomic number : 53 Electronic configuration : [Kr] 4d10 5s2 5p5 Shell Structure : 2 − 8 − 18 − 18 − 7 Therefore, I has 7 valence electrons. 007 (part 1 of 1) 10 points The P2− anion has how many total electrons and how many valence electrons? 1. 16; 5 − − − −→ CaI2 Iype, Shelby – Homework 5 – Due: Oct 2 2007, midnight – Inst: Vandenbout should appear in the dot formula for the chlorate ion ClO− ? 3 1. 32 2. 28 3. 26 correct 3. 4. 24 5. 30 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− 009 (part 1 of 1) 10 points Which of the following is the best representation of the compound calcium sulfide? 3− U U X X 3 Which of the following is the correct Lewis formula for oxygen (O2 )? yy O OO OO OO OO O ˆ ¤ˆ ‰‰ ‰‰ ‰‰ ‡‡ † ˆˆ ˆ ¤ˆ † ˆˆ‡ ‡ †† ‡‡ † † „ †† …… „ „„ 5. † O 1. 3 Ca2+ , 2 c c S YY ` ` 2− 2. 2 Ca , 3. Ca2+ , 2 + S S f bf dd g g − e e 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 4. None is appropriate because calcium sulfide is a covalent compound. 2− i 011 (part 1 of 1) 10 points Which of the following is the correct Lewis formula for acetylene (ethyne, C2 H2 )? ’ − s s •• • 010 (part 1 of 1) 10 points H •• • “ u“ “ “ 3. H C ” C ” ”” “ u“ 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. 2. H C C ’ H ’’ ’ ’ ’ ww v v 6. Ca , + S ’’ 1. H C C H correct ’ ’’ S rr t ut q i q 5. Ca 2+ , correct ‰‰ ‘‘ ‘ ‰‰ 8. ‡‡ 7. † 6. …… ƒƒ 4. ‚ OO ‚ ¤‚  € OO ‚ € ¤€ ‚ ¤‚ €   € € 2. € ¤€ yy xx 1. O yy  „„ xx xx correct   V WV hh a ba p bp Iype, Shelby – Homework 5 – Due: Oct 2 2007, midnight – Inst: Vandenbout –– – —– – u– 4 5. H ˜ ˜ C ˜ H C ™ ˜˜ H dd ™™ ™ 7. H C H e 9. H f f C f H ff Explanation: The Lewis formula for acetylene (ethyne, C2 H2 ) is 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. four; three 2. four; four 3. two; two ii i i H H i i 10. C C hh e H ee f —f e C f ee 8. H e C e C e ee d C 013 (part 1 of 1) 10 points Which of the following contains exactly one unshared pair of valence electrons? 1. H2 S 2. NaCl 3. PH3 correct 4. SiH4 5. C2 H4 Explanation: The PH3 molecule contains exactly 1 unshared pair of valence electrons: ·· P H H H 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. 014 (part 1 of 1) 10 points r ss uu pp O mm k O O O O O r ↔ t —t q uq v —v n on j uj l —l k ™™ d ™ e —e ™ gg ™ u™ 6. ™ H ™ C ™ ™™ – H –– – C 7. one; six ˜ –– – – – u– ™ u™ ™ 4. – H – C C 4. three; six – – 5. zero; eight 6. two; six correct 8. none; two 9. one; three Explanation: Iype, Shelby – Homework 5 – Due: Oct 2 2007, midnight – Inst: Vandenbout How many double bonds are present in the “best” resonance structure of the phosphate ion? 1. 0 2. 1 correct 3. 3 4. 2 Explanation: O 0 −1 O P O 0 −1 −1 O } }  — ~~ | |   zz € € { { x —x yy w w 5 F Cl F Cl  ŽŽ Œ ¤Œ  2. 3. ’’ F Cl F Cl ” ‘‘  ¤ ’’ 4. “ F Cl F Cl F Cl – •• • ¤• ” ¤” ““ “ 3− is the most plausible 6. 7. – F Cl F Cl F Cl ™™ ˜˜ ˜˜ –– —— – 1. ten 2. one correct 3. two 4. none 5. thirteen 6. twenty Explanation: The Lewis formula for chlorine fluoride (ClF) is F Cl 017 (part 1 of 1) 10 points Which of the following is the correct Lewis formula for boron trichloride (BCl3 )? Cl   › › › œ› › œ› šš šš š š 8. three N ‡ ž už Ÿ Cl ŸŸ Ÿ ŸŸ Ÿ Ÿ Ÿ uŸ 016 (part 1 of 1) 10 points Which of the following is the correct Lewis formula for chlorine fluoride (ClF)? 3. Ÿ Cl Ÿ B Cl Ÿ ŸŸ Ÿ uŸ F †† ˆ žž Ÿ ‰ ‹‹ ‡ ˆ „„ ƒ The Lewis structure is ž ž ‰ F F Cl žž 2. ŠŠ ž Cl ž ž B ž Cl ž ž Explanation: ‚ u‚ …… ƒ žž  Cl ž už   7. four 1. B  u ž už ™™ 10. —— 015 (part 1 of 1) 10 points How many lone pairs of electrons are on nitrogen in NF3 ? 9. – structure. 8. ” 5. ŒŒ  ŽŽ ‘‘  ¤ ” ¤” • ¤• ’’ ““ ––    ŒŒ 1. Œ ¤Œ correct Cl correct Iype, Shelby – Homework 5 – Due: Oct 2 2007, midnight – Inst: Vandenbout ¡¡ 6 7. Cl § § B § Cl ¨¨ §§ 8. Cl B Cl 019 (part 1 of 1) 10 points Resonance is a concept that describes the bonding in molecules Cl Cl © ©© © © 9. Cl Cl Cl B ©© © B ©© 10. 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− : ª Cl ª ª ª —ª ª B ª Cl ªª ª Cl ªª 018 (part 1 of 1) 10 points The CO2− ion has how many resonance con3 figurations? 1. 4 ª ªª © ¨ Cl ¨ ª —ª ¦ Cl ¦¦ ¦ Cl § ¦¦ 6. ¦ Cl ¦ B ¦ Cl ¦ ¥¥ ¦¦ ¥¥ Cl ¦ —¦ Cl ¥ ¥ 5. Cl B ¤¤ ¢¢ Cl ¢ ¢ ª ¢¢ ££ u 4. Cl B ¡ Cl 2. It does not exhibit resonance. ¡ u 3. 3 correct 4. 2 5. 5 Explanation: C → 4 valence e− O → 3 × 6 valence e− 2− → 2 e − Total = 24 valence e− One structure is ·· ·O C · ·· O· ·· · ·O· · ·· · which can also be drawn as ·· ·· ·· ·O O· C · ·· ·· · or ··O C O· · ·· ·· ·O· · ·· · ··O·· for a total of 3 resonance structures. 1. 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 2. by asserting that electrons in a double bond can delocalize (spill over) onto adjacent single bonds to make a bond and a half. 3. by asserting that double bonds “flip” or resonate between two locations in the molecule. 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 Iype, Shelby – Homework 5 – Due: Oct 2 2007, midnight – Inst: Vandenbout Which of the three Lewis structures is the most important? N +1 N +1 N O à à ÆÆ Ä Ä À ÁÁ ¿ 7 ↔ 1. C only 2. A and C only 3. None of these is important. 2. 3 4. A and B only 3. 2 5. All of these are important. 4. 1 6. B and C only 5. 4 7. A only 8. B only correct Explanation: −1 is O C N probably the most important as it is the structure with the formal charges of the individual atoms closest to zero. 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. 1. 0 2. 1 3. 2 correct 4. 3 Explanation: · · ¹ —¹ ¸¸ ºº The Lewis structure is 0 −1 +1 H N H0 The Lewis structure 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. 95 kJ/mol correct 2. 288 kJ/mol 3. 317 kJ/mol 4. 507 kJ/mol 5. 181 kJ/mol Explanation: Ç uÇ ¶¶ C) −3 C ´ u´ µµ ³ ³ +1 O 022 (part 1 of 1) 10 points Calculate the formal charge on N in the molecule NH3 . 1. 0 correct ¯ O °° ¯ ± u± B) −1 C ²² −1 N N O has formal charges out of the acceptable range. Explanation: 0 0 H À ¾¾ ¼¼ Å uÅ ¬¬ « —« A) −2 C ­­ ® o® +1 0 N N O N N O  — ¿ ½ o½ » o» Iype, Shelby – Homework 5 – Due: Oct 2 2007, midnight – Inst: Vandenbout H H C H H C H H C H C H H H C H ∆H = = (C Ebreak − C) + (Br H C H Br H C H C H Br 025 (part 1 of 1) 10 points Consider the reaction CH4 (g) + I2 → CH3 I(g) + HI(g) . + Br Br → 8 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. Emake Br) C) − 2 (C Br) + (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 must be put in to make 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 is released upon breaking the bonds in the reactants and the energy that is released upon making the bonds in the products. 4. 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. 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. −97 kJ/mol 2. +463 kJ/mol 3. −52 kJ/mol 4. +52 kJ/mol correct 5. There is no way to tell. Explanation: H H C H I H C H H +H I H +I I −−→ −− ∆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 Iype, Shelby – Homework 5 – Due: Oct 2 2007, midnight – Inst: Vandenbout 026 (part 1 of 1) 10 points Bond energies are approximate because 1. we cannot calculate them accurately. 2. not all molecules burn in oxygen, making calorimetry difficult. 3. they depend upon the physical state of the molecule, gas, liquid, or solid. 4. we cannot measure them accurately. 5. they are averages over a number of molecules. correct 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 03/30/2009 for the course CH 301 taught by Professor Fakhreddine/lyon during the Spring '07 term at University of Texas at Austin.

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