159_fall_07_chapt_10_and_11_exampl

159_fall_07_chapt_10_and_11_exampl - Example 10.1 Use the...

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Example 10.1 Use the VSEPR method to predict the shape of the nitrate ion. Strategy The problem can be broken down into two parts. First, we must obtain an acceptable Lewis structure for the nitrate ion, using the general strategy. With this Lewis structure in hand, we focus on the central nitrogen atom to establish first the electron- group geometry and then the molecular geometry. Solution The formula for the nitrate ion is NO 3– . Step 1: Draw a plausible Lewis structure. The distribution of 24 valence electrons [that is, 5 + (3 x 6) + 1] gives each terminal O atom in the skeletal structure an octet, but the central N atom is lacking two electrons for a valence-shell octet. We must shift a lone pair of electrons from one of the O atoms into its bond with the nitrogen atom, forming a nitrogen-to-oxygen double bond. Step 2: Determine the number of electron groups. There are three electron groups around the N atom; all are bonding groups (one N-to-O double bond and two N-to-O single bonds). The VSEPR notation is therefore AX 3 . Step 3: Identify the electron-group geometry. The distribution of the three electron groups is trigonal planar.
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Assessment The Lewis structure obtained in step 1 is just one of three possible resonance structures, but it is clear that we reach the same conclusion about the molecular geometry no matter which of the three we use. Example 10.1 continued Solution continued Step 4: Identify the molecular geometry. For the structure AX 3 , the molecular geometry is the same as the electron- group geometry: trigonal planar. The N and O atoms are all in the same plane, and each O-N-O bond angle is 120°
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Example 10.2 Use the VSEPR method to predict the molecular geometry of XeF 2 . Strategy We can try the four-step method outlined and applied in Example 10.1 to see if it leads us to the molecular geometry. If it does not, we will need to consider the two additional factors cited above as well. Solution Step 1: Draw a plausible Lewis structure. There are 8 + (2 x 7) = 22 valence electrons in XeF 2 . We need only 20 electrons to attach two F atoms to a central Xe atom and to provide each of the three atoms with an octet. We must place the additional pair in an expanded valence shell on the Xe atom. The Lewis structure is therefore Step 2: Determine the number of electron groups. There are five electron groups around the Xe atom, two bonding pairs and three lone pairs. The VSEPR notation is AX 2 E 3 . Step 3: Identify the electron-group geometry. The five electron groups signify that the distribution of the electron groups is trigonal bipyramidal.
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Example 10.2 continued Solution continued Step 4: Identify the molecular geometry. We cannot decide on the molecular geometry at this point. There seem to be three possibilities, but only one can be correct: Structure I has one 120° LP–LP repulsion and two 90° LP–LP repulsions. Structure II has one 180° LP–LP repulsion and two 90° LP–LP
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159_fall_07_chapt_10_and_11_exampl - Example 10.1 Use the...

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