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[CHEM 2OA3] - Midterm Exam Guide - Everything you need to know! (74 pages long)
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by Kelsea Mann , Fall 2014
74 Pages
104 Views

Department
Chemistry
Course Code
CHEM 2OA3
Professor
Paul Harrison
Study Guide
Midterm

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Description
[CHEM 2OA3] Comprehensive Midterm Exam guide including any lecture notes, textbook notes and exam guides.find more resources at oneclass.com CHEM, chapter one 1-47 We can construct all compounds by combining atomic molecules, forming degenerate hybrid orbitals. All the bonds in this molecule has an antibonding MO Sigma bonds: low in energy, head-on overlap Sigma star: high in energy, complimentary Pi star: high in energy Strongest to weakest: Sigma, Pi, Pi star, Sigma star 1-48 Pi is weaker than sigma – if we start with an alkene, we destroy the pi bond during a reaction, but the sigma bond remains intact. Bond length: sigma bond between sp: s e closer to the nucleus. More s character: sp bond is shorter than an sp3 bond. 1-49 Bond length contracts as hybridization becomes more s in character Bond strength increases as the hybridization becomes more s in character Double bond is stronger than single bond 1-50 Valence shell electron pair repulsion (VSEPR) 1. Determine the steric number (number of substituents around the atom) Methane steric number: 4. Lone pairs count as steric numbers – it behaves as a substituent when finding steric number. When steric number is four, we need four orbitals (hybridize one s and 3 p in methane). When steric number is three, sp^2. When steric number is two, sp. 1-52 For any sp^3 hybridized atom, the 4 valence e pairs will form a tetrahedral e group geo Lone pairs repel bonded pairs more than bonded pairs repel bonded pairs. Number of lone pairs impacts the bond angles. 1-53 The geometry of ammonia: trigonal pyramidal. 1-54 BF3: six e species, strong lewis acid. Steric number of 3 – hybridization is sp2. Unlike ammonia, because there is no lone pair, there is a trigonal molecular geometry. 1-55 Boro▯ has a▯ u▯hy▯ridized p or▯ital, ▯ut it is e▯pty, so it does▯’t ▯ou▯t. It’s just a regio▯ of space where e could be if they wanted to. If we were to react BF3 with an e donor, the p orbital tells us where the e could go if we were to add them. 1-56 Lone pair, sigma bond to C on left, sigma to C on right, pi bond Sp^2 hybridized – one of the two sp^2 hybrids will accommodate the lone pair, the others will hang with C find more resources at oneclass.com
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