Ch10 - CHAPTER 10 THE SHAPES OF MOLECULES 10.1 To be the...

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10-1 CHAPTER 10 THE SHAPES OF MOLECULES 10.1 To be the central atom in a compound, the atom must be able to simultaneously bond to at least two other atoms. He, F, and H cannot serve as central atoms in a Lewis structure. Helium is a noble gas, and as such, it does not need to bond to any other atoms. Hydrogen (1 s 1 ) and fluorine (1 s 2 2 s 2 2 p 5 ) only need one electron to complete their valence shells. Thus, they can only bond to one other atom, and it does not have d orbitals available to expand its valence shell. 10.2 Plan: Draw a Lewis structure and then see if additional structures may be drawn. Solution: Resonance must be present any time that a single Lewis structure is inadequate in explaining one or more aspects of a molecule or ion. The two N-O bonds in NO 2 are equivalent; no single Lewis structure shows this. The following Lewis structures may be drawn for NO 2 : ON O O O O The average of all of these structures gives equivalent N-O bonds with a bond length that is between N-O and N=O. 10.3 For an element to obey the octet rule it must be surrounded by 8 electrons. To determine the number of electrons present (1) count the individual electrons actually shown adjacent to a particular atom, and (2) add two times the number of bonds to that atom. Using this method the structures shown give: (a) 0 + 2(4) = 8; (b) 2 + 2(3) = 8; (c) 0 + 2(5) = 10; (d) 2 + 2(3) = 8; (e) 0 + 2(4) = 8; (f) 2 + 2(3) = 8; (g) 0 + 2(3) = 6; (h) 8 + 2(0) = 8. All the structures obey the octet rule except: c and g. 10.4 For an atom to expand its valence shell, it must have readily available d orbitals. The d orbitals do not become readily available until the third period or below on the periodic table. For the elements in the problem F, S, H, Al, Se, and Cl, the period numbers are 2, 3, 1, 3, 4, and 3, respectively. All of these elements, except those in the first two periods (H and F), can expand their valence shells. 10.5 Plan: Count the valence electrons and draw Lewis structures. Solution: Total valence electrons: SiF 4 has 32; SeCl 2 has 20; and COF 2 has 24. The Si, Se, and the C are the central atoms, because these are the elements in their respective compounds with the lower group number (in addition, we are told C is central). Place the other atoms around the central atoms and connect each to the central atom with a single bond. At this point the Si has an octet, and the remaining electrons are placed around the fluorines (3 pairs each). The 2 bonds plus 2 lone pairs complete the octet on Se and leave enough electrons to complete the Cl octet with 3 pairs each. The 3 bonds to the C leave it 2 electrons short of an octet. Forming a double bond to the O completes the C octet, and leaves sufficient electrons to form 2 lone pairs on the O and 3 lone pairs on each of the F’s. All atoms in each of the structures end with an octet of electrons.
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This note was uploaded on 06/08/2009 for the course BIOE 109 taught by Professor Pogson,g during the Spring '08 term at UCSC.

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Ch10 - CHAPTER 10 THE SHAPES OF MOLECULES 10.1 To be the...

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