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Lecture 16 - Topics for the day Administrative stuff Formal...

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Topics for the day Administrative stuff Formal charge one last time Valence shell electron pair repulsion (VSEPR) Lone pairs Bond Angles Resonance Multiple bonds Organic molecules Administrative stuff Sorry about office hours. (Snowday for my kids) Quest homework – 24 people have started it. By the end of this lecture you should be able to do up to Q17 of the 24 ALEKS homework (objective 2) - 24 finished/33 nearly done Extra sample problems to work aren’t up yet (see above) Lewis structures and VSEPR coming any day now. Reminder 70% on ALEKS prelim due 3/29... Where were we? When considering the bond strength and bond length of multiple bonds, triple bonds are stronger and shorter than double bonds, which are stronger and shorter than single bonds. When drawing Lewis structures for molecules without a central atom, the sequence of atoms in the molecular formula can be of help, as can knowing some of the standard functional groups; e.g., hydroxyl, carbonyl, amine, ether. Octet rule exceptions are for less than eight electrons (B, Be), more than eight electrons (n 3) and odd numbers of electrons (radicals). Formal charge can be used to determine the most likely Lewis structure – the one that minimizes the formal charge on each atom. The total formal charge is the same as the charge on the molecule (or ion). If two structures have the same formal charge arrangement, select the one with the negative formal charge on the more electronegative atom. The VSEPR model determines molecular shape based on minimizing electron-pair repulsions. 1 2 3
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Handy hint from Susan I gave the formal charge equation as this: Formal charge = # of valence electrons on free atom 2 e for each lone pair on atom 1 e for each bonding pair on atom From the CH301 class Susan TAed for last semester: Formal charge = # of valence electrons on free atom Total number of dots and dashes connected to atom (2 dots per loan pair, 1 dash per bonding pair) C O H H O FC(H) = 1 – 1 = 0 FC(H) = 1 – 1 = 0 FC(C) = 4 – 4 = 0 FC(O) = 6 – 7 = –1 FC(O) = 6 – 5 = +1 Total FC = 0 Question from last time Why did I choose to draw formic acid they way I drew it? C O H H O Couldn’t this be an option? C O H H O I drew it the first way, based on my knowledge of standard functional groups, as I had shown you on the previous slide in the last lecture: O H Hydroxy C O Carbonyl But formal charge considerations show us that the first configuration is the more likely. FC(H) = 1 – 1 = 0 FC(H) = 1 – 1 = 0 FC(C) = 4 – 4 = 0 FC(O) = 6 – 6 = 0 FC(O) = 6 – 6 = 0 Total FC = 0 FC(H) = 1 – 1 = 0 FC(H) = 1 – 1 = 0 FC(C) = 4 – 4 = 0 FC(O) = 6 – 7 = –1 FC(O) = 6 – 5 = +1 Total FC = 0 More use for Formal Charge Organic chemistry uses formal charge a lot, not so much for distinguishing between Lewis structures, but rather to identify the location of the formal charge of a polyatomic ion.
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