Download Exam 3 2013 KEY

CHE 211 EXAM 3 (Ch. 8-­‐10) KEY Answers in BOLD RED Each multiple-­‐choice question on this exam is worth three points (81 total) and each problem is worth 6 or 7 points, as indicated in the exam at the start of each chapter. Mark the letter of the BEST answer to multiple choice questions clearly on your answer sheet. CHAPTER 8 (Electron Configurations & Periodicity) 33 POINTS TOTAL (27 MC + 6 Problem points) 1. All of the elements in each Main Group period have electrons going into orbitals of the same a. principle quantum number b. azimuthal quantum number c. magnetic quantum number 2. How many electrons could have n = 3 and ml = +2? a. 0 b. 10 c. 6 d. 2 3. The effect of the mutual repulsion of electrons in the same orbital is to make the orbital a. smaller b. larger c. lower in energy d. nonspherical in shape 4. How many unpaired electrons should there be in lowest energy state gaseous atoms of the element P? a. 3 b. 2 c. 1 d. 0 5. How many core electrons are there in the element Sr (Z = 38)? a. 38 b. 36 c. 18 6. Of these four elements, all in the fourth period, which should have the SMALLEST atomic radius? a. K b. Ga c. Mn d. Br 7. A 2nd period Main Group element has successive ionization energies = 0.80, 2.43, 3.66, 25.02, & 32.82 MJ. It is: a. O b. B c. C d. N 8. Of the following isoelectronic ions, the largest one should be a. Al+3 b. Na+ c. O–2 9. Of the following elements, the most metallic based on standard trends in properties will be: a. Ba b. Zr c. Cu d. Ar PROBLEM I. d. 24 d. N–3 Answer fully as requested Draw spectroscopic notation (1s22s2…, etc.) electronic structures of the element V (Z = 23) and of its +2, +3, and +5 oxidation states (so you should draw four distinct sets of notations). CIRCLE any that would be para-­‐ magnetic; leave diamagnetic structures uncircled. (You may use the El[Noble gas]ns2np6 type notation.) V 1s22s22p63s23p64s23d3 [ or V [Ar]4s23d3and so on for the other three as well] V3+ 1s22s22p63s23p64s03d2 V2+ 1s22s22p63s23p64s03d3 [can simply omit the 4s also] V5+ 1s22s22p63s23p64s03d0 Note that only V+5 has no unpaired electrons, since all the d electrons are unpaired EXTRA CREDIT (5 points possible): Describe the bonding in the solid state of typical metals and use it to explain why metals change shape when struck or bent, but do not break apart like ionic or most covalent substances. The structure of a metal consists of metal cations in a sea of delocalized valence electrons (so Na is Na+ ions in such a sea, while Mg has Mg+2 ions in its sea). The bonding force is produced by mutual attraction between the metal cations and the delocalized electrons. The electrons can flow with and around the metal cations, so when a physical force is strong enough to move the atoms, the electrons flow with them. This allows the metal’s shape to be distorted without breaking up the bonding. No need for this part in your answer: Ionic substances fracture because any movement in the lattice puts like-­‐charge ions adjacent, so the structure will break apart as those charges repel one another. Covalent molecular substances that are solids have only weak forces between molecules, so they break apart when stressed. Covalent network substances like diamond do not bend or break unless a severe stress is placed upon them; it then breaks up covalent bonds, destroying the original substance (which was one huge networked molecule). CHE 211 Exam 3 CHAPTER 9 (BONDING BASICS) Page 2 34 POINTS TOTAL (27 MC + 7 problem points) 1. A substance is a solid with a 850°C melting point. It conducts electricity in solid or melted states; it’s probably a. a covalently bonded substance b. a metallic bonded substance c. an ionically bonded substance 2. The Main Group element S, when it is drawn in a Lewis structure, will have ____ electrons around the S symbol a. 2 b. 4 c. 6 d. 8 3. Which of these substances is likely to have the highest lattice energy per mole? Smallest, with highest charges a. NaI b. CaSe c. CaO d. NaCl 4. A bond with bond order = 3 should have a _________ exothermic heat of formation and _________ length than a bond with bond order = 1. a. more ; shorter b. more ; longer c. less ; shorter d. less ; longer 5. Element X forms an ionic compound with stoichiometry MgX and a covalent substance SX3. X is in group a. 17 b. 16 c. 15 d. 14 6. Benzene, C6H6, is a liquid at 20°C. It freezes at 5°C and boils at 80°C. When benzene boils, the vapor consists of a. ions and free atoms b. just ions c. just atoms d. just molecules 7. Using the table of electronegativities on the data sheet, which of these has polar covalent bonds? a. BH3 b. AlF3 c. Li3P d. SSe2 8. What is the value of ∆HRXN for the reaction CH4 + 4 Cl2 à CCl4 + 4 HCl, based upon the use of the bond energy values listed on the exam data sheet? (all answers in kJ for the stoichiometry shown in the equation) a. –110 b. –440 c. –5688 d. +5688 9. Which of these diatomic covalent molecules should have the strongest bond between its atoms? a. Cl2 b. F2 c. I2 d. H2 PROBLEM SHOW SET UP or WORK. NO CREDIT FOR ANSWER ONLY! The Born-­‐Haber Cycle can be used to calculate any of the thermodynamic quantities that combine to give us the ∆H°f for any ionic substance, provided that we know the others. Electron Affinity is sometimes hard to measure for highly reactive elements like Fluorine, but we can use the data for KF(s) to determine it. Here are the other data items: K(s) à K(g) ∆H = +90 kJ/mol K(g) à K+(g) + e– ∆H = +419 kJ/mol F2(g) à 2 F(g) ∆H = +159 kJ/mol F2 F(g) + e– à F–(g) ∆H = _____________kJ/mol (EA) K(s) + ½ F2(g) à KF(s) ∆H°f = –569 kJ K+(g) + F–(g) à KF(s) ∆H = –821 kJ/mol Using the values above and the concept of the Born-­‐Haber Cycle, calculate EA for fluorine. 1
ΔH of = ΔH VAP−K + ΔH ATOM −F2 + IEK + EAF + ΔH LATT 2
–569 kJ = 90 kJ + ½ (+159 kJ) + 419 kJ + EAF + (-­‐821 kJ) so EAF = –569 – 90 – ½(159) – 419 + 821 kJ = –336.5 = –336 kJ (no decimal sig figs) CHE 211 Exam 3 Page 3 CHAPTER 10 (VSEPR) 33 POINTS TOTAL (27 MC + 6 problem points) 1. How many lone pair electrons (NOT pairs; total electrons in lone pairs) does SbH3 have in its Lewis structure? a. 4 b. 2 c. 0 d. 6 2. Which of these molecules does not meet the octet rule? a. SO3 b. N2O (bonding = NNO) c. NO2 (odd electron) d. SF2 3. Which of the following species will require resonance to represent its structure well? a. CO2 b. OH– c. SO3 d. SiF4 4. The structure shown at the right is the Lewis structure of a small molecule found in photochemical smog. What is the formal charge on the N atom of this molecule? …of the O above the N? a. 0 ; 0 b. +1 ; -­‐2 c. +1 ; –1 d. –1 ; +1 5. The tetrafluoroiodate ion, IF4– has what electron geometry? a. tetrahedral b. octahedral c. trigonal bipyramidal 6. What will be the molecular geometry of the tetrafluoroiodate ion? (see problem 5 for formula) a. tetrahedral b. square planar c. seesaw d. trigonal bipyramidal 7. Based upon their Lewis structures, bond polarity, and VSEPR geometry, which of these has a dipole moment? a. CO2 b. SO2 c. BF3 d. PF5 8. Which of the following will have the largest bond angles between F atoms attached to its central atom? a. CF4 b. NF3 c. OF2 d. BF3 9. Which of the molecules shown here will have a V-­‐shaped (bent) molecular geometry? a. SF6 b. PCl3 c. BeF2 d. H2S H
O
H
C
C
H
O
N
O
O
O
d. square pyramidal PROBLEM DRAW the complete Lewis structures of TWO of the following (if you do all three only the first two will be graded). Include all LONE PAIR electrons. Then, state the: 1. Electron Geometry 2. Molecular Geometry 3. Bond angles between the atoms for each of the structures which you draw. ClO3– ion O
OCl2 O
Cl
Cl
O
O
F
Cl
Kr
KrF2 F
EG = tetrahedral EG = Tetrahedral EG = trigonal bipyramidal MG = Trigonal pyramidal MG = bent MG = linear <109.5° <109.5° c. 180° Note that especially chlorate ion has other ways of being drawn which have better formal charges. The chlorate structure shown better matches a textbook problem’s answer, but it has -­‐1 formal charges on each O and a +2 formal charge on Cl; Cl can expand beyond an octet also.