103AExam4Key - Chem 103A Examination IV July 7 2003 Page 1...

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Unformatted text preview: Chem 103A Examination IV July 7, 2003 Page 1 Exam 4 Grader Key 1. (5) This problem deals with the polyatomic ion shown below. The Lewis structure is correct. :(I): :.. — S — ° : o. I I. (a) What is the formal charge of sulfilr in this structure? +2 (b) What is the formal charge of oxygen in this structure? —1 (c) What value do you predict for the 080 angle? 109.5° (could be i 2 or 3 degrees) (d) What is the formal charge of fluorine in this structure? zero (e) What is the charge of the polyatomic ion? -1 2. (15) One of the three possible resonance Lewis structures for cyannte inn is Answer Structure 2: [til—CEO: ]“ Structure 35: [:N=C=O: ]" EXplanation (a) Draw the Lewis structures for l (a) The cyanate ion has 16 valence electrons: 5 from N, 4 from C, 6 from O, and 1 mac- count for the 1~ charge. Both Structures 2 and 5 have 16 valence electrons and dif- fer only in the placement of the multiple bonds, showing that they are correct resonance structures. (11) The formal charges on the atoms are shown in the table. (b) Use formal charges to determii Strum“: 2 Structure 3 , N c 0' N C O Valence electrons 5 4 6 5 4 6 The initial structure, 2 NEG—é: . is preferred. Lone pair electrons 6 0 2 4 0 4 if if 7 7 WW" 4 % shared electrons 1 4 3 2 4 2 Formal charge — 2 0 + 1 _ 1 0 0 The formal charges in Structures 2 and 3 add up to 1~—, as they must for the 1— Cyanate ion. Oxygen, the most electronegative atom, should bear the negative Charge. Structure 2 would be least preferred because it has an atom with a high for- mal charge (-2 on N) and a positive charge on oxygen. The initial structure [:NEC—Q: ]' is the one in which oxygen has the negative charge ( p. \3‘31). Thus, it is the preferred one. \— Chem 103A Examination IV July 7, 2003 Page 2 3. (20) (a) Polar molecule(s) CI H H I | I F CI Cl—C—CI H—C—H H—C—H | I | | I F—B=F F—B=F CI H Cl (b) Element(s) forming an ion with -3 charge Al I Ca s N Be K (c) Element with largest first ionization energy Na Mg Al Si P s Cl (d) Atom with largest atomic radius K Ca Ga Ge As Se Br (e) Element with largest electronegativity K Ca Ga Ge As Se Br (f) Atom with the largest atomic radius 0 S Se Te P 0 (g) Hypervalent molecule (oclet rule violated at the central atom) GaF3 ASF3 AlF3 PF; BrF 3 (h) Most polar bond O-F B-F N-F C-F F-F (i) Element(s) forming an ion with +3 charge SC I Ca s N Be K Chem 103A Examination IV July 7, 2003 Page 3 4. (10) Which of these molecules an have cis and trans isomers? For those that do, write the structural formulas for the two isomers and label them cis and trans. I Answer (b) and (C) H C CH Cl CH 3 \ — / 3 \ / 3 H3C\ /CH3 Br\ /CH3 (b) /C—C\ /C=C\ (C) /C=C\ C=C c1 c1 H3c Cl Br (:1 H3C/ \c1 cis trans cis trans Explanation . Because both of the groups on each carbon are the same, (a) cannot have cis and trans isomers. In (b), the two —CH3 groups and the two Cl atoms can both be on the same SldC of the C=C bond (the cis form) or on opposite sides (the trans form). The San-16 hOIdS ml“: for the “CH3 groups in (c). There are two CH3 groups on the same carbon 1n (d), so cis and trans isomers are not possible. 7" W w 7 7i , , , , 5. (10) Use the VSEPR model to predict the electron-pair geometry and the molecular geometry of Answer Electron-pair geometry Molecular geometry (a) tetrahedral tetrahedral (b) tetrahedral angular (c) triangular planar triangular planar (d) triangular planar triangular planar Explanation The Lewis structures are Iii + 3?: 1?: (a) H—P—H (b) .. ’0" .. (c) .. S .. (d) C | :c1/ \c1: :0/ \0: H/ \H H .. I. O. C. (a) The Lewis structure of PHI reveals that the central P atom has four electron pairs, 3“ bonding pairs to terminal hydrogen atoms. Consequently, the ion is an AX4EO'fYPC and has tetrahedral electron-pair and molecular geometries. (b) The central oxygen is surrounded by two bonding pairs and two lone pairs (AXZE2 type). These four electron pairs give a tetrahedral electron-pair geometry. The “1016‘ cular geometry is angular, not linear, because of lone pair-lone pair, lone pair—bonding pair, and bonding pair—bonding pair repulsions. The two lone Pm"S push the chlorine atoms closer together than the purely 109.5° tetrahedral angle ex- pected for four bonding pair- bonding pair repulsions. (C) The tWO bonding pairs in the 8:0 double bond are counted as one bond for deter- mining the geometry of the molecule, giving three bonding regions around 5 (AX3EO Chem 103A Examination IV July 7, 2003 Page 4 6. (10) Draw one correct Lewis structure for each molecule below. Use lines for bonds and dots (:) for unshared electron pairs. In no case can the octet rule be violated. N02‘ NCCN PO43' 3 points 4 points 3 points :"—N=" :NEC—CEN: :5: o. o. . l .. zq—¢—-= 7. (10) <ONO=60° 90° 109.50 120° 180° <NOH=60° 90° 109.50 120° 180° Chem 103A Examination IV July 7, 2003 Page 5 8. . (10) What is the major type of force that must be overcome in these changes? (a) The sublimation of solid iodine, 12 (b) The melting of propane (c) The decomposition of water into H2 and 02 (d) The evaporation of liquid PC13 Answer (a) and (b) London forces (c) Covalent bonds between 0 and H atoms (d) Dipole-dipole forces Explanation (a) Nonpolar 12 molecules in the solid are held to each other by London forces, which must be overcome for the 12 molecules in the solid to escape from each othcnnl become gaseous. (b) Solid propane melts when the thermal energy becomes great enough to overcome some of the noncovalent forces of attraction (London forces) among the nonpolar propane molecules (—187.7° C). Noncovalent forces still exist in the liquid propane. (c) To decompose water into its component elements requires breaking the 0—H cova— lent bonds in water molecules. Note that this change involves much greater energy (the bond energy) than that needed to melt or boil water, both of which involve overcoming noncovalent interactions. ((1) Dipole-dipole forces attract the polar PC13 molecules to each other in the lquid. These forces must be overcome so that PCI3 molecules can get away from their neighbors at the surface of the liquid and enter the gaseous state. Because of the rel- atively large number of electrons in each PC13 molecule, London forces are also im- portant in holding PC15 molecules close to each other and must be overcome. ...
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