Lecture 18 - Topics for the day Administrative stuff...

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Topics for the day Administrative stuff Valence bond theory sp Hybridization sp 2 Hybridization sp 3 Hybridization More hybridization possibilities Lone pairs Multiple bonds Administrative stuff Quest HW2: By the end of *this* (not last) lecture you should be able to do up to Q20 of the 24 Due Saturday 5 pm Don’t guess. It takes off points. Practice problems: Lewis diagrams, VSEPR shapes, VB hybrids optional, no points answers available Fri 10 am (download problems before then) but can keep entering answers till exam time ALEKS homework (objective 2) over. Objective 3 due Monday after exam, but try to do most of it by exam time Review option. .. Administrative stuff Exam Two is next Wednesday May 10th, 7 - 9 pm, WEL 1.308. If you are expecting to take an earlier makeup, watch for an email from Susan con±rming you are on our list. If you haven’t heard from her by Wednesday, send us email. 1 2 3
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Help OfFce hours DrRuth: Tue 2 pm, Wed 3:30 pm, Thu 9:30 am Susan: Mon noon, Wed 9 am, ±ri 11 am Exam preparation Susan exam review Monday 4–5 pm. WEL 3.502 Sample exam posted soon. Mon/Wed next week overview of topics working problems from homeworks/sample exam Where were we? To determine the polarity of a molecule, determine the geometry, use Δ EN to Fnd each bond dipole size and direction, combine with lone pair positions to determine the overall polarity. All Fve standard molecular shapes (when there are no lone pairs) are symmetric and result in nonpolar molecules when all bonds are the same Δ EN. Valence bond theory explains bonding as electron pair sharing due to the overlap of two partially-Flled electron orbitals from two atoms. Hybridization combines at least two nonequivalent atomic orbitals to make new hybrid orbitals that lead to the electronic arrangements predicted by VSEPR. Each hybrid orbital holds a bonding pair or a lone pair of electrons. Each covalent bond forms from the overlap of two atomic orbitals, two hybrid orbitals, or one atomic and one hybrid orbital. ±C(O) = 6 – 7 = –1 ±C(O) = 6 – 5 = +1 ±C(O) = 6 – 6 = 0 Even though the triangle structure is favoured by ±ormal Charge arguments, it is not the molecule we observe experimentally. This can’t be explained
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Lecture 18 - Topics for the day Administrative stuff...

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