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General Chemistry I

by: Maud Bins

General Chemistry I CH 221

Maud Bins
Central Oregon Community College
GPA 3.52

Zelda Ziegler

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Zelda Ziegler
Class Notes
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This 2 page Class Notes was uploaded by Maud Bins on Monday October 5, 2015. The Class Notes belongs to CH 221 at Central Oregon Community College taught by Zelda Ziegler in Fall. Since its upload, it has received 32 views. For similar materials see /class/218971/ch-221-central-oregon-community-college in Chemistry at Central Oregon Community College.

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Date Created: 10/05/15
CH 221WZIEgler Chapter 89 KEY CONCEPTS Class period 1 W Reading pp 275 278t 1 State the number ofvalence electrons for a maintroup element Problem 81 2 Draw the electroanot symbol for a maintroup element Problems 85 86 3 State the octet rule Problem 82 Class period 2 W Reading pp 280 W 291 4 Identify periodic trends in electronegativity Problems 833 5 Predict the polarity ofa bond Problem 835 836 6 Draw Lewis structures for molecules and polyatomic ions Problems 843 844 Class period 3 W Reading pp 292 297 315 W 323 328 W329 7 Use calculated Formal Charges to evaluate Lewis structures Problem 846 8 Write resonance forms Problems 847 848 a and b 9 Use the Lewis structure to predict molecular geometry and bond angles Note you will be required to do this only for molecules and ions in which the octet rule is satis ed for all atoms Problems 911 aWd 915 916 10 From the geometry ofa polyatomic ion or molecule predict whether it is polar or nonpolar Problems 922 925 927 Class period 4 W Reading pp 330 W 334b 336 W 339 11 State the hybridization in a covalently bonded species Problem 943 12 State the number ofpi and sigma bonds in a molecule or polyatomic ion Problems For the molecules in Problems 943 916 and 927 smte the number of sigma and pi bonds in each molecule Answers to Problems 81 See Text A8 82 The octet rule is the observation that atoms will gain lose or share electrons to achieve the stable con guration of 8 valence electrons The reason it is in quote marks is that it has so many exceptions Speci cally H He and pretty much everything in the lower periods 85 See Text A8 86 a 0M o 3 l g cgt 501307Scz b n 2 quot d Sez 833 See Text AW9 836 NOTE The answer book refers you 835 See Text AW9 to Figure 89 when they mean Fig 86 a OWF lt CWF lt BeWF b N7Br lt P7Br lt O7Br 843 See Text A 79 Q C7S lt N7O lt B7F As H O I O co IO 844 a H2CO H d A5053 H H O O 3 3 2 H b H202 H IOI IE f e H2S03 CCf 39 H CC H l c c2116 J i 0 sz 846 Lewis Structure Formal Charge calculation Oxidation s a S02 0 S O S O 6SO Ve7 6 6 6 4 72 there are 2 7 e on each atom 6 5 7 as per ch 4 resonance forms Formal Chg 0 1 71 bso3 o o o s o s o O Ve7 6 6 6 6 6 2 8 There are 3 7 e on ea atom 7 7 4 6 resonance form I 2 I Formal Chg 1 1 2 0 lt0 o o s o s o gol Ve7 6 6 6 6 4 2 c S052 7 e on eaatom 7 7 5 7 OI 239 Formal Chg 71 71 1 71 I l O O S O O S O o 2 Ve7 6 6 6 6 6 6 2 d SO44 S 7 e on ea atom 7 7 5 7 7 I El Formal Chg 1 1 2 1 1 O 848 a QNQ lt gt Q77NEO lt gt OEN77Q b Yes resonance structures are needed to describe the structure 916 1 10950 2 120 3 10950 4 120 5 10950 6 10950 7 180 8 10950 922 The 2 polar bonds in H20 would completely cancel each other out if the molecule were linear and then the net dipole moment would be zero Since water is polar the molecule can t be linear 11


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