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# Ch 11 Example Problems Answers - HODSON CHE 1301 Petrucci...

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HODSON CHE 1301 Petrucci Ch11 Example Problems The following problems are taken from old exams ...... 1. Which of the following statements concerning the carbon dioxide molecule is correct? a) The molecule contains two σ bonds. b) The carbon is described by sp 2 hybridization. c)   Each oxygen is described by sp 3 hybridization. d) The molecule contains two lone pairs of valence electrons. e) The molecule contains four π bonds. Answer: A Explanation: We should be able to draw a Lewis structure for the carbon dioxide molecule. The formula is CO 2 , giving 4 + 2(6) valence electrons = 16. A skeletal diagram has the carbon atom in the center (less electronegative than O and written before O in the formula) and the two O atoms joined to it. We can arrange the remaining electrons (16 – 4 = 12) around the terminal atoms (the oxygens) to complete octets, and find we need to let the central carbon share two extra pairs on electrons in order to complete its octet. We can do this in three ways, giving three resonance forms, which we can assess using formal charges. The forms involving triple bonds are less favored, but still make a contribution in this molecule. The major form is that involving two double bonds. Remember that the ACTUAL molecule is a composite of the three forms (like the ‘labrasation’ cross breed). Anyway, what do we see ? According to VSEPR we have a central carbon atom with two electron groups around it, and so we have linear geometry (180° between the C-O bonds). (Remember that an electron group is a lone pair, single bonding pair or multiple bonding pairs of electrons, or even one electron in odd-electron species). We can eliminate d), as there are four lone pairs. Our best atomic model involves thinking of electrons in atoms as waveforms. We can describe the probability of finding electrons in given regions of an atom using pictorial representations called orbitals. The valence electrons on the central carbon atom in CO 2

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are in the 2s and 2p atomic orbitals (C 1s 2 2s 2 2p 2 ), but the geometries of these atomic orbitals do not provide the linear molecular geometry predicted by VSEPR (and seen in the real world) for CO 2 . Think about it – there are two p orbitals with one unpaired electron each, and one empty p orbital (draw the orbital diagram if you need to). If we pair each of the two unpaired electrons with the unpaired electron on an oxygen atom, we still have an empty p orbital (the carbon hasn’t gained a stable octet!) and the p atomic orbitals are at 90° to each other, not the 180° we need for CO 2 . What we can do is take the 2s and one of the 2p orbitals on carbon and combine them mathematically to form two hybrid orbitals. These are called sp hybrids because they were formed from an s and a p orbital (superscripts give the number of orbitals involved, and ‘1’ is assumed). The sp hybrid orbitals correspond more closely to the electron distribution around the carbon atom – they are 180° from each other, and so give linear geometry to our molecule! Two ‘electron groups’ becomes two sp ‘hybrid orbitals’. So,
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