Resonance

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TITLE="Delocalized Orbitals"><U><U><IMG BORDER="0" SRC="Symb075c00b700c8ccffff00.gif" HEIGHT="13" WIDTH="24" ALIGN="bottom" HSPACE="0" VSPACE="0">Delocalized Orbitals</U></U></A></div> <bR> </td> <td valign="top" BGCOLOR="#000000" TEXT="#F7F7FF"> <div> <FONT SIZE="5" COLOR="#99CC00" FACE="Arial" ><B>University Blvd.</B></FONT>     <FONT SIZE="1" COLOR="#CCCC66" FACE="Arial" >|   <A HREF="index.htm" TARGET="_top" TITLE="University Blvd."><U><B>home</B></U></A></FONT></div> <div>                                                   </div> <div> <IMG SRC="03ff4010.gif" border=0 width="500" height="1" ALIGN="BOTTOM" HSPACE="0" VSPACE="0"></div> <div> Resonance: When Lewis structures Fail</div> <div>  There are some molecules for which we cannot write Lewis structures that agree with experimental observations. For example, the formate ion, CHO2-, if drawn using Lewis rules, would be:</div> <div>   <IMG SRC="0f64e500.gif" border=0 width="78" height="80" ALIGN="BOTTOM" HSPACE="0" VSPACE="0"></div> <div> But actually, experimental observation shows that the bond between the carbon and the two oxygen's are the same length. Since a CO double bond is a different length then the single bond, the Lewis structure must be wrong. Actually, observation shows that it is neither of the lengths of the single or double bond, but a resonance hybrid, and can be drawn like this:</div> <bR> <div> <IMG SRC="0f723500.gif" border=0 width="291" height="80" ALIGN="BOTTOM" HSPACE="0" VSPACE="0"></div> <div> The bond is neither a double nor single bond. Both the double and single bond "contribute" to the entire structure. The double-ended arrow shows that we are drawing resonance structures and implies that the true hybrid is a composite of the two resonance structures.</div> <div> A simple way of determining whether resonance should be applied is that when you must move electrons to create one more double bonds, the number of resonance structures is equal to the number of equivalent choices for the locations of the double bonds.</div> <div> For example, In a nitrate ion, NO3-, the double bond between N and the O can exist with any three of the oxygen's. The resonance structures would then be:</div> <bR> <div>  <IMG SRC="0f8e4580.gif" border=0 width="484" height="88" ALIGN="BOTTOM" HSPACE="0" VSPACE="0"></div> <div> The benefit of resonance hybrids in a molecule is that the total energy is lower than any one of its resonance structures, and therefore is more stable. A benzene molecule (see Organic Chemistry: Aromatic Hydrocarbons) is ring shaped and is drawn with alternating single or double bonds in between carbon atoms. It is a resonance hybrid though because the double and single bonds can alternate in a different fashion.</div> <bR> <div>  <IMG SRC="0f9963c0.gif" border=0 width="150" height="60" ALIGN="BOTTOM" HSPACE="0" VSPACE="0"></div> <div> The benzene ring is usually represented as a hexagon with a circle in the middle to show that the electron density of the three extra bonds is evenly distributed.</div> <bR> <div> <IMG SRC="0fa323c0.gif" border=0 width="50" height="60" ALIGN="BOTTOM" HSPACE="0" VSPACE="0"> </div> <div> The extra stability achieved is called the resonance energy. It does not behave like other organic compounds with double bonds because breaking those double bonds would mean breaking the resonance hybrid, giving it more energy, and destabilizing it. </div> <bR> </td> </tr></table></body>