9 Replies Latest reply on Feb 23, 2016 4:33 AM by karthiks94217840

    How to delete xml attributes by indesign file.

    karthiks94217840 Level 2

      Dear friends,

       

      I need to delete xml attributes only (Element "ul"), i have try to below coding and attached screenshot also, but i have doing  some mistake. i don't what i missing the below coding.

       

      Screen Shot 2016-02-22 at 12.57.49 PM.png

       

      My coding in the below.

       

      var elements = app.activeDocument.xmlElements[0].xmlAttributes.item["@new_class"] = "list-style-type:none";
          alert(elements.length);
          delete elements["@new_class"];
      

       

      please suggest friends,

       

      Advance Thanks,

        • 1. Re: How to delete xml attributes by indesign file.
          Ronald63 Level 4

          Hi,

          Try this ...

          var doc = app.activeDocument;
          
          removeAttributes(doc,'ul','style','list-style-type:none');
          
          function removeAttributes(source,element,attributename,attributevalue){
              for (var i = 0; i < source.xmlElements.length; i++){
                      try{
                          for(j=0; j<source.xmlElements[i].xmlAttributes.length; j++){
                              if(source.xmlElements[i].markupTag.name == element && source.xmlElements[i].xmlAttributes[j].name == attributename && source.xmlElements[i].xmlAttributes[j].value == attributevalue){
                                  source.xmlElements[i].xmlAttributes[j].remove();
                              }
                          }
                      }catch(e){}
                      removeAttributes(source.xmlElements[i],element,attributename,attributevalue);
              }
          }
          
          • 2. Re: How to delete xml attributes by indesign file.
            karthiks94217840 Level 2

            Dear Ronald,

             

            Thanks you so match Ronald your script working super.

             

            Again one more help Ronald,

             

            How to added xml attributes in the xml element "ul" list? I have attached screenshot for your reference.

             

            Step; 1

             

            I want added in the following attribues - ul list

             

            Screen Shot 2016-02-18 at 10.05.25 PM.png

            Advanced thanks Ronald friends,

             

            Thanks,

            • 3. Re: How to delete xml attributes by indesign file.
              Ronald63 Level 4

              Like this ...

               

              var doc = app.activeDocument;
              
              addAttributes(doc,'ul','class','ol_lower-alpha');
              
              function addAttributes(source,element,attributename,attributevalue){
                  for (var i = 0; i < source.xmlElements.length; i++){
                          try{
                              for(j=0; j<source.xmlElements[i].xmlAttributes.length; j++){
                                  if(source.xmlElements[i].markupTag.name == element ){
                                      source.xmlElements[i].xmlAttributes.add(attributename,attributevalue);
                                  }
                              }
                          }catch(e){}
                          addAttributes(source.xmlElements[i],element,attributename,attributevalue);
                  }
              }
              
              • 4. Re: How to delete xml attributes by indesign file.
                karthiks94217840 Level 2

                Dear Ronald,


                One more thanks friend,


                Your script is running good. But, this script not added in the attributes for empty tags (elements). I have attached screenshot four your reference.


                Screen Shot 2016-02-22 at 9.48.35 PM.png

                Please suggest me Ronald,


                Thanks

                • 5. Re: How to delete xml attributes by indesign file.
                  Ronald63 Level 4

                  Oups ...

                  var doc = app.activeDocument; 
                  
                  addAttributes(doc,'ul','class','ol_lower-alpha'); 
                  
                  function addAttributes(source,element,attributename,attributevalue){ 
                      for (var i = 0; i < source.xmlElements.length; i++){ 
                              try{ 
                                  if(source.xmlElements[i].markupTag.name == element ){ 
                                      source.xmlElements[i].xmlAttributes.add(attributename,attributevalue); 
                                  }  
                              }catch(e){} 
                              addAttributes(source.xmlElements[i],element,attributename,attributevalue); 
                      } 
                  }
                  
                  
                  • 6. Re: How to delete xml attributes by indesign file.
                    karthiks94217840 Level 2

                    Dear Ronald,


                    Very very fast and quick replay Ronald great work - super - super - super,


                    Again, I need your help Ronald,


                    How to Replace the "attributevalue" only? Same tags for ul list.


                    Example:

                    <ul class = "number"> change to <ul class = "title">

                     

                    Please one more help me friends,

                     

                    Thanks





                    • 7. Re: How to delete xml attributes by indesign file.
                      Ronald63 Level 4

                      Hi,

                       

                      Try this ...

                      var doc = app.activeDocument;
                      
                      updateAttributes(doc,'ul','class','number','title');
                      
                      function updateAttributes(source,element,attributename,attributevalue,newattributevalue){
                          for (var i = 0; i < source.xmlElements.length; i++){
                                  try{
                                      for(j=0; j<source.xmlElements[i].xmlAttributes.length; j++){
                                          if(source.xmlElements[i].markupTag.name == element && source.xmlElements[i].xmlAttributes[j].name == attributename && source.xmlElements[i].xmlAttributes[j].value == attributevalue){
                                              source.xmlElements[i].xmlAttributes[j].value= newattributevalue;
                                          }
                                      }
                                  }catch(e){}
                                  updateAttributes(source.xmlElements[i],element,attributename,attributevalue,newattributevalue);
                          }
                      }
                      
                      • 8. Re: How to delete xml attributes by indesign file.
                        karthiks94217840 Level 2

                        Dear Ronald,


                        Your script working super. one more thanks

                        Screen Shot 2016-02-23 at 5.57.20 PM.png

                         

                        Thanks

                        • 9. Re: How to delete xml attributes by indesign file.
                          karthiks94217840 Level 2

                          My Reference only,

                           

                           

                          <?xml version="1.0" encoding="utf-8"?>

                          <html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en"><?InDesign version="8.0"?><?InDesignPXID_Export version="2004"?><?docpage num="2" PAGENAME="2"?><?DOCUMENTINFO w="612.000" h="783.000" startpage="2" endpage="22" pagecount="21"?><head><title/><meta name="PXID_XML_Styler" content="v3019"/><meta charset="UTF-8"/><meta name="dcterms.conformsTo" content="PXE 1.39 ProductLevelReuse"/><meta name="generator" content="PXE Tools version 1.39.52"/></head><body><section class="chapter" id="ch01"><header><h1 class="title"><span class="pagebreak" title="2" id="page2"/><?REGION x="36.000" y="69.000" w="64.778" h="11.000" page="2" pagename="2"?><span class="label">Chapter </span><span class="number">1</span>

                          <?REGION x="36.000" y="82.000" w="324.683" h="11.000" page="2" pagename="2"?>Remembering General Chemistry: Electronic Structure and Bonding</h1></header><section class="frontmatter"/>

                          <section class="bodymatter" id="ch01bm"><section><header><h1 class="title"><?REGION x="36.000" y="108.000" w="80.838" h="11.000" page="2" pagename="2"?>Important Terms</h1></header>

                          <ul style="list-style-type:none"><li><p><b>antibonding molecular</b> a molecular orbital that results when two atomic orbitals with opposite phases orbital interact. Electrons in an antibonding orbital decrease bond strength.</p></li>

                          <li><p><b>atomic number</b> the number of protons (or electrons) a neutral atom has.</p></li>

                          <li><p><b>atomic orbital</b> an orbital associated with an atom; the three-dimensional area around its nucleus where electrons are most likely to be found.</p></li>

                          <li><p><b>atomic mass</b> the average mass of the atoms in the naturally occurring element.</p></li>

                          <li><p><b>aufbau principle</b> the principle that states that an electron will always go into the available orbital with the lowest energy.</p></li>

                          <li><p><b>bond dissociation energy</b> the amount of energy required to break a bond in a way that allows each of the atoms to retain one of the bonding electrons; or the amount of energy released when a bond is formed.</p></li>

                          <li><p><b>bonding molecular orbital</b> a molecular orbital that results when two atomic orbitals with the same phase interact. Electrons in a bonding orbital increase bond strength.</p></li>

                          <li><p><b>bond length</b> the internuclear distance between two atoms at minimum energy (maximum stability).</p></li>

                          <li><p><b>bond order</b> describes the number of covalent bonds shared by two atoms.</p></li>

                          <li><p><b>carbanion</b> a species containing a negatively charged carbon.</p></li>

                          <li><p><b>carbocation</b> a species containing a positively charged carbon.</p></li>

                          <li><p><b>condensed structure</b> a structure that does not show some (or all) of the covalent bonds.</p></li>

                          <li><p><b>core electrons</b> electrons in filled shells.</p></li>

                          <li><p><b>covalent bond</b> a bond created as a result of sharing electrons.</p></li>

                          <li><p><b>degenerate orbitals</b> orbitals that have the same energy.</p></li>

                          <li><p><b>dipole</b> a separation of positive and negative charges.</p></li>

                          <li><p><b>dipole moment</b> <span class="inlineequation">[&amp;~bf~(|bfmu|)~norm~&amp;]</span> a measure of the separation of charge in a bond or in a molecule.</p></li>

                          <li><p><b>double bond</b> a bond composed of a sigma bond and a pi bond.</p></li>

                          <?docpage num="3" PAGENAME="3"?><li><p><span class="pagebreak" title="3" id="page3"/><b>electronegative</b> describes an element that readily acquires an electron.</p></li>

                          <li><p><b>electronegativity</b> the tendency of an atom to pull electrons toward itself.</p></li>

                          <li><p><b>electrostatic attraction</b> an attractive force between opposite charges.</p></li>

                          <li><p><b>electrostatic potential map</b> a map shows that how electrons are distributed in a molecule.</p></li> <li><p>(potential map)</p></li>

                          <li><p><b>equilibrium constant</b> the ratio of products to reactants at equilibrium.</p></li>

                          <li><p><b>excited-state electronic</b> the electronic configuration that results when an electron in the ground state has configuration moved to a higher-energy orbital.</p></li>

                          <li><p><b>formal charge</b> the number of valence electrons <span class="inlineequation">[&amp;|minus|&amp;]</span> (the number of nonbonding electrons + the number of bonds).</p></li>

                          <li><p><b>free radical (radical)</b> a species with an unpaired electron.</p></li>

                          <li><p><b>ground-state electronic</b> a description of the orbitals the electrons of an atom occupy when they are all in configuration their lowest available energy orbitals.</p></li>

                          <li><p><b>Heisenberg uncertainty</b> a principle that states that both the precise location and the momentum of an principle atomic particle cannot be simultaneously determined.</p></li>

                          <li><p><b>Hund’s rule</b> a rule that states that when there are degenerate orbitals, an electron will occupy an empty orbital before it will pair up with another electron.</p></li>

                          <li><p><b>hybrid orbital</b> an orbital formed by hybridizing (mixing) atomic orbitals.</p></li>

                          <li><p><b>hydride ion</b> a negatively charged hydrogen (a hydrogen atom with an extra electron).</p></li>

                          <li><p><b>hydrogen ion (proton)</b> a positively charged hydrogen (a hydrogen atom without its electron).</p></li>

                          <li><p><b>ionic compound</b> composed of ions held together by electrostatic attraction.</p></li>

                          <li><p><b>ionic compound</b> a compound composed of a positive ion and a negative ion.</p></li>

                          <li><p><b>ionization energy</b> the energy required to remove an electron from an atom.</p></li>

                          <li><p><b>isotopes</b> atoms with the same number of protons but a different number of neutrons.</p></li>

                          <li><p><b>Kekulé structure</b> a model that represents the bonds between atoms as lines.</p></li>

                          <li><p><b>Lewis structure</b> a model that represents the bonds between atoms as lines or dots and the lone-pair electrons as dots.</p></li>

                          <li><p><b>lone-pair electrons</b> valence electrons not used in bonding.</p></li> <li><p>(nonbonding electrons)</p></li>

                          <?docpage num="4" PAGENAME="4"?><li><p><span class="pagebreak" title="4" id="page4"/><b>mass number</b> the number of protons plus the number of neutrons in an atom.</p></li>

                          <li><p><b>molecular mass</b> the sum of the atomic masses of all the atoms in the molecule.</p></li>

                          <li><p><b>molecular orbital</b> an orbital associated with a molecule that results from the combination of atomic orbitals.</p></li>

                          <li><p><b>molecular orbital (MO) </b> a theory that describes a model in which the electrons occupy orbitals as they do in theory atoms but the orbitals extend over the entire molecule.</p></li>

                          <li><p><b>node</b> a region within an orbital where there is zero probability of finding an electron.</p></li>

                          <li><p><b>nonbonding electrons</b> valence electrons not used in bonding.</p></li>

                          <li><p><b>nonpolar covalent bond</b> a bond formed between two atoms that share the bonding electrons equally.</p></li>

                          <li><p><b>octet rule</b> a rule that states that an atom will give up, accept, or share electrons in order to achieve a filled outer shell (or an outer shell that contains eight electrons) and no electrons of higher energy. Because a filled second shell contains eight electrons, this is known as the octet rule.</p></li>

                          <li><p><b>orbital</b> the volume of space around the nucleus where an electron is most likely to be found.</p></li>

                          <li><p><b>orbital hybridization</b> mixing of atomic orbitals.</p></li>

                          <li><p><b>organic compound</b> a compound that contains carbon.</p></li>

                          <li><p><b>Pauli exclusion principle</b> a principle that states that no more than two electrons can occupy an orbital and that the two electrons must have opposite spin.</p></li>

                          <li><p><b>pi</b> <span class="inlineequation">[&amp;~bf~(|bfpi|)~norm~&amp;]</span> <b>bond</b> a bond formed as a result of side-to-side overlap of <i>p</i> orbitals.</p></li>

                          <li><p><b>polar covalent bond</b> a bond formed between two atoms that do not share the bonding electrons equally.</p></li>

                          <li><p><b>potential map</b> a map that allows you to see how electrons are distributed in a molecule.</p></li> <li><p>(electrostatic potential map)</p></li>

                          <li><p><b>proton (hydrogen ion)</b> a positively charged hydrogen ion.</p></li>

                          <li><p><b>quantum mechanics</b> the use of mathematical equations to describe the behavior of electrons in atoms or molecules.</p></li>

                          <li><p><b>radical (free radical)</b> a species with an unpaired electron.</p></li>

                          <li><p><b>sigma</b> <span class="inlineequation">[&amp;~bf~(|bfsig|)~norm~&amp;]</span> <b>bond</b> a bond with a symmetrical distribution of electrons about the internuclear axis.</p></li>

                          <li><p><b>single bond</b> a pair of electrons shared between two atoms.</p></li>

                          <li><p><b>skeletal structure</b> shows the carbon–carbon bonds as lines, but does not show the carbons or the hydrogens that are bonded to the carbons.</p></li>

                          <?docpage num="5" PAGENAME="5"?><li><p><span class="pagebreak" title="5" id="page5"/><b>tetrahedral bond angle</b> the bond angle <span class="inlineequation">[&amp;~rom~(109.5|deg|)~norm~&amp;]</span> formed by an <span class="inlineequation">[&amp;sp^{3}&amp;]</span> hybridized atom that has no lone pairs.</p></li>

                          <li><p><b>tetrahedral carbon</b> a carbon that forms covalent bonds using four <span class="inlineequation">[&amp;sp^{3}&amp;]</span> hybrid orbitals.</p></li>

                          <li><p><b>trigonal planar carbon</b> an <span class="inlineequation">[&amp;sp^{2}&amp;]</span> hybridized carbon.</p></li>

                          <li><p><b>triple bond</b> composed of a sigma bond and two pi bonds.</p></li>

                          <li><p><b>valence electron</b> an electron in an outermost shell.</p></li>

                          <li><p><b>valence-shell electron-pair</b> a model for the prediction of molecular geometry based on the minimization of repulsion (VSEPR) model electron repulsion between bonding electrons and nonbonding electrons around an atom.</p></li>

                          <li><p><b>wave equation</b> an equation that describes the behavior of each electron in an atom or a molecule.</p></li>

                          <li><p><b>wave functions</b> a series of solutions to a wave equation.</p></li></ul></section>

                          <?docpage num="6" PAGENAME="6"?><section id="ch01probset1"><header><h1 class="title"><span class="pagebreak" title="6" id="page6"/><?REGION x="36.000" y="81.000" w="102.475" h="11.000" page="6" pagename="6"?><span class="label">Solutions to Problems</span></h1></header>

                          <ul style="list-style-type:none"><li class="general-problem">1. <div class="question" id="ch01qu01"><p>The atomic number <span class="inlineequation">[&amp;|eq|&amp;]</span> the number of protons.</p>

                          <p>The mass number <span class="inlineequation">[&amp;|eq|&amp;]</span> the number of protons + the number of neutrons.</p>

                          <p>All isotopes have the same atomic number; in the case of oxygen it is 8. Therefore:</p>

                          <ul style="list-style-type:none"><li><p>The isotope of oxygen with a mass number of 16 has 8 protons and 8 neutrons.</p></li>

                          <li><p>The isotope of oxygen with a mass number of 17 has 8 protons and 9 neutrons.</p></li>

                          <li><p>The isotope of oxygen with a mass number of 18 has 8 protons and 10 neutrons.</p></li></ul> </div></li>

                          <li class="general-problem">2. <div class="question" id="ch01qu02"><p>The number of protons an element has never changes. The number or electrons depends on the charge on the element.</p>

                          <ul style="list-style-type:none"><li><p> a. 11</p></li><li><p>   b. 18</p></li><li><p>   c. 17</p></li></ul>

                          <ul style="list-style-type:none"><li><p> a. 10</p></li><li><p>   b. 18</p></li><li><p>   c. 18</p></li></ul></div></li>

                          <li class="general-problem">3. <div class="question" id="ch01qu03"><p>(percentage of naturally occurring <span class="inlineequation">[&amp;^{35}~rom~Cl~norm~&amp;]</span> <span class="inlineequation">[&amp;|multins|&amp;]</span> atomic mass of <span class="inlineequation">[&amp;^{35}~rom~Cl~norm~&amp;]</span>) + (percentage of naturally occurring <span class="inlineequation">[&amp;^{37}~rom~Cl~norm~&amp;]</span> <span class="inlineequation">[&amp;|multins|&amp;]</span> atomic mass of <span class="inlineequation">[&amp;^{37}~rom~Cl~norm~&amp;]</span>)</p>

                          <p><span class="inlineequation"> [&amp;|pbo|.7577|multi|34.969|pbc||+||pbo|.2423|multi|36.966|pbc|&amp;]</span></p>

                          <p><span class="inlineequation"> [&amp;|pbo|26.496|+|8.957|pbc||=|35.453&amp;]</span></p></div></li>

                          <li class="general-problem">4. <div class="question" id="ch01qu04"><p>All four atoms have 2 core electrons in their filled first shell. (Notice that because the four atoms in the question are in the same row of the periodic table, they have the same number of core electrons.) The electrons that are not in a filled shell are valence electrons.</p>

                          <ul style="list-style-type:none"><li><p> a. 3   </p></li><li><p>b. 5</p></li><li><p>   c. 6</p></li><li><p>   d. 7</p></li></ul></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu05"><ul style="list-style-type:none"><li><p>5. a. Use the aufbau principle (electrons go into available orbitals with the lowest energy) and the Pauli exclusion principle (no more than two electrons are in each atomic orbital). The relative energies of the orbitals:</p>

                          <div class="informalequation">[&amp;1s|less|2s|less|2p|less|3s|less|3p|less|3d|less|4s|less|4p |less|4d|less|5s|less|5p&amp;]</div>

                          <p> Remember that each shell has one <i>s</i> atomic orbital and three degenerate <i>p</i> atomic orbitals. The third and fourth shells also have five degenerate <i>d</i> atomic orbitals.</p>

                          <div class="informalequation">[&amp;~rom~Cl|em||em|*AS**AP*~normal~1s^{2}|sp|2s^{2}|sp|2p^{6}| sp|3s^{2}|sp|3p^{5}~norm~&amp;]</div>

                          <div class="informalequation">[&amp;~rom~Br|em|*AS**AP*~normal~1s^{2}|sp|2s^{2}|sp|2p^{6}|sp|3 s^{2}|sp|3p^{6}|sp|3d^{10}|sp|4s^{2}|sp|4p^{5}~norm~&amp;]</div>

                          <div class="informalequation">[&amp;~rom~I|em||em|*AS**AP*~normal~1s^{2}|sp|2s^{2}|sp|2p^{6}|s p|3s^{2}|sp|3p^{6}|sp|3d^{10}|sp|4s^{2}|sp|4p^{6}|sp|4d^{10}|sp|5s^{2}|sp|5p^{5}~norm~&amp ;]</div></li>

                          <li><p> b. They each have 7 electrons in their outer shell; in each case, 2 are in an <i>s</i> orbital and 5 are in <i>p</i> orbitals. Notice that because the 3 elements all are in the same column of the periodic table, they have the same number of valence electrons.</p></li></ul></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu06"><p>6. The atomic numbers can be found in the periodic table on the last page of the text. Notice that elements in the same column of the periodic table have the same number of valence electrons, and their valence electrons are in similar orbitals.</p>

                          <ul style="list-style-type:none"><li><p> a. carbon (atomic number <span class="inlineequation">[&amp;|eq|&amp;]</span>  6;  2 core, 4 valence): <span class="inlineequation">[&amp;1s^{2}|sp|2s^{2}|sp|2p^{2}&amp;]</span>

                           

                           

                            silicon (atomic number <span class="inlineequation">[&amp;|eq|&amp;]</span> 14; 10 core, 4 valence): <span class="inlineequation">[&amp;1s^{2}|sp|2s^{2}|sp|2p^{6}|sp|3s^{2}|sp|3p^{2}&amp;]</span>< /p></li>

                          <?docpage num="7" PAGENAME="7"?><li><p><span class="pagebreak" title="7" id="page7"/> b. oxygen (atomic number <span class="inlineequation">[&amp;|eq|&amp;]</span>  8;  2 core, 6 valence): <span class="inlineequation">[&amp;1s^{2}|sp|2s^{2}|sp|2p^{4}&amp;]</span>

                           

                           

                            sulfur   (atomic number <span class="inlineequation">[&amp;|eq|&amp;]</span> 16; 10 core, 6 valence): <span class="inlineequation">[&amp;1s^{2}|sp|2s^{2}|sp|2p^{6}|sp|3s^{2}|sp|3p^{4}&amp;]</span>< /p></li>

                          <li><p> c. nitrogen  (atomic number <span class="inlineequation">[&amp;|eq|&amp;]</span>  7;  2 core, 5 valence): <span class="inlineequation">[&amp;1s^{2}|sp|2s^{2}|sp|2p^{3}&amp;]</span>

                           

                           

                            phosphorus (atomic number <span class="inlineequation">[&amp;|eq|&amp;]</span> 15; 10 core, 5 valence): <span class="inlineequation">[&amp;1s^{2}|sp|2s^{2}|sp|2p^{6}|sp|3s^{2}|sp|3p^{3}&amp;]</span>< /p></li>

                          <li><p> d. magnesium (atomic number <span class="inlineequation">[&amp;|eq|&amp;]</span> 12; 10 core, 2 valence): <span class="inlineequation">[&amp;1s^{2}|sp|2s^{2}|sp|2p^{6}|sp|3s^{2}&amp;]</span>

                           

                           

                            calcium    (atomic number <span class="inlineequation">[&amp;|eq|&amp;]</span> 20; 18 core, 2 valence): <span class="inlineequation">[&amp;1s^{2}|sp|2s^{2}|sp|2p^{6}|sp|3s^{2}|sp|3p^{6}|sp|4s^{2}&amp ;]</span></p></li></ul></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu07"><ul style="list-style-type:none"><li><p> 7. a. Potassium is in the first column of the periodic table; therefore, like lithium and sodium that are also in the first column, potassium has one valence electron.</p></li>

                          <li><p> b. It occupies a <span class="inlineequation">[&amp;4s&amp;]</span> orbital.</p></li></ul></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu08"><p> 8. The polarity of a bond can be determined by the difference in the electronegativities (given in Table 1.3 on page 10 of the text) of the atoms sharing the bonding electrons. The greater the difference in electronegativity, the more polar the bond.</p>

                          <ul style="list-style-type:none"><li><p> a. <span class="inlineequation">[&amp;~rom~Cl|bond|CH_{~normal~3}~norm~&amp;]</span></p></li><li>< p> b. <span class="inlineequation">[&amp;~rom~H|bond|OH~norm~&amp;]</span></p></li><li><p> c. <span class="inlineequation">[&amp;~rom~H|bond|F~norm~&amp;]</span></p></li><li><p><span class="inlineequation"> d. [&amp;~rom~Cl|bond|CH_{~normal~3}~norm~&amp;]</span></p></li></ul></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu09"><p> 9. The electronegativity differences in the four listed compounds are as follows:</p>

                          <div class="informalequation">[&amp;*AS*~rom~KCl|em|~normal~3.0|-|0.8*AP*|=|2.2~norm~&amp;]

                          [&amp;*AS*~rom~LiBr|em|~normal~2.8|-|1.0*AP*|=|1.8~norm~&amp;]

                          [&amp;*AS*~rom~NaI|em|~normal~2.5|-|0.9*AP*|=|1.6~norm~&amp;]

                          [&amp;*AS*~rom~Cl_{~normal~2}|em|3.0|-|3.0*AP*|=|0~norm~&amp;]</div>

                          <ul style="list-style-type:none"><li><p> a. KCl has the most polar bond because its two bonded atoms have the greatest differences in electronegativity.</p></li>

                          <li><p> b. <span class="inlineequation">[&amp;~rom~Cl_{~normal~2}~norm~&amp;]</span> has the least polar bond because the two chlorine atoms share the bonding electrons equally.</p></li></ul></div></li>

                          <li class="general-problem"><div class="question"><p>10. Solved in the text.</p></div></li>

                          <li class="general-problem"><div class="question"><p>11. To answer this question, compare the electronegativities of the two atoms sharing the bonding electrons using Table 1.3 on page 10 of the text.</p>

                          <ul style="list-style-type:none"><li><p> a. </p><figure class="informalfigure"><?docpage num="7" PAGENAME="7"?><?REGION x="127.250" y="510.658" w="50.346" h="20.178" xs="100" hs="100" page="7" pagename="7" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901001.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901001.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li><li><p> b. </p><figure class="informalfigure"><?docpage num="7" PAGENAME="7"?><?REGION x="235.426" y="509.716" w="37.459" h="21.12" xs="100" hs="100" page="7" pagename="7" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901002.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901002.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li><li><p> c. </p><figure class="informalfigure"><?docpage num="7" PAGENAME="7"?><?REGION x="327.634" y="509.229" w="50.08" h="22.807" xs="100" hs="100" page="7" pagename="7" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901003.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901003.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li><li><p> d. </p><figure class="informalfigure"><?docpage num="7" PAGENAME="7"?><?REGION x="433.866" y="509.084" w="50.83" h="23.052" xs="100" hs="100" page="7" pagename="7" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901004.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901004.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p> e. </p><figure class="informalfigure"><?docpage num="7" PAGENAME="7"?><?REGION x="126.634" y="548.287" w="51.552" h="20.3829999999999" xs="100" hs="100" page="7" pagename="7" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901005.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901005.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li><li><p> f.   </p><figure class="informalfigure"><?docpage num="7" PAGENAME="7"?><?REGION x="235.998" y="547.307" w="39.119" h="22.763" xs="100" hs="100" page="7" pagename="7" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901006.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901006.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li><li><p> g. </p><figure class="informalfigure"><?docpage num="7" PAGENAME="7"?><?REGION x="328.085" y="547.036" w="36.183" h="22.0340000000001" xs="100" hs="100" page="7" pagename="7" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901007.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901007.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li><li><p/> <figure class="informalfigure"><?docpage num="7" PAGENAME="7"?><?REGION x="433.866" y="548.373" w="56.047" h="21.497" xs="100" hs="100" page="7" pagename="7" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901008.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901008.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/>h. </figure></li></ul>

                          <p> (Notice that if the two atoms being compared are in the same row of the periodic table, the atom farther to the right is the more electronegative atom; if the atoms being compared are in the same column, the one closer to the top of the column is the more electronegative atom.)</p> </div></li>

                          <li class="general-problem"><div class="question" id="ch01qu12"><p>12. Solved in the text.</p></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu13"><p>13. The dipole moment is the magnitude of the charge times the distance between the charges. Because fluorine is more electronegative than Cl, the charge on H and F in HF is larger than the charge on H and Cl in HCl. The larger charge on F compared to the charge on Cl is more than enough to make up for the fact that <span class="inlineequation">[&amp;~rom~H|bond|F~norm~&amp;]</span> is a shorter bond than <span class="inlineequation">[&amp;~rom~H|bond|Cl~normal~.~norm~&amp;]</span></p></div></li>

                          <?docpage num="8" PAGENAME="8"?><li class="general-problem"><div class="question" id="ch01qu14"><ul style="list-style-type:none"><li><p><span class="pagebreak" title="8" id="page8"/>14. a. LiH and HF are polar (they have a red end and a blue end).</p></li>

                          <li><p> b. A potential map marks the edges of the molecule’s electron cloud. The electron cloud is largest around the hydrogen in LiH, because that hydrogen has more electrons around it than do the hydrogens in the other molecules.</p></li>

                          <li><p> c. Because the hydrogen of HF is blue, we know that this compound has the most positively charged hydrogen and, therefore, it will be most apt to attract a negatively charged species.</p></li></ul></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu15"><p>15. By answering this question, you will see that a formal charge is a book-keeping device. It does <i>not necessarily</i> tell you which atom has the greatest electron density or is the most electron deficient.</p>

                          <ul style="list-style-type:none"><li><p> a. oxygen</p></li><li><p> b. oxygen (it is more red)</p></li>

                          <li><p> c. oxygen</p></li><li><p> d. hydrogen (it is the deepest blue)</p></li></ul>

                          <p> Notice that in hydroxide ion, the atom with the formal negative charge <b>is</b> the atom with the greater electron density. In the hydronium ion, however, the atom with the formal positive charge <b>is not</b> the most electron-deficient atom.</p> </div></li>

                          <li class="general-problem"><div class="question" id="ch01qu16"><p>16. formal charge <span class="inlineequation">[&amp;|eq|&amp;]</span> number of valence electrons</p>

                          <p><span class="inlineequation">[&amp;|minus|&amp;]</span> (number of lone-pair electrons + the number of bonds)</p>

                          <p> In all four structures, every H is singly bonded and thus has a formal <span class="inlineequation">[&amp;~rom~charge|=|~normal~1|-||pbo|0|+|1|pbc||=|0.~norm~&amp;]</ span> Similarly, all <span class="inlineequation">[&amp;~rom~CH_{~normal~3}~norm~&amp;]</span> carbon atoms have four bonds and a formal <span class="inlineequation">[&amp;~rom~charge|=|~normal~4|-||pbo|0|+|4|pbc||=|0.~norm~&amp;]</ span> The formal charges on the remaining atoms:</p><ul style="list-style-type:none"><li><?docpage num="8" PAGENAME="8"?><p> a. </p><figure class="informalfigure"><?docpage num="8" PAGENAME="8"?><?REGION x="90.000" y="388.100" w="68.492" h="36.111" xs="100" hs="100" page="8" pagename="8" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901009.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901009.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure><p> formal charge on O</p> <p><span class="inlineequation">[&amp;6|-||pbo|2|+|3|pbc||=||plus|1&amp;]</span></p></li><li><?doc page num="8" PAGENAME="8"?><p> b. </p><figure class="informalfigure"><?docpage num="8" PAGENAME="8"?><?REGION x="208.600" y="388.500" w="46.315" h="36.155" xs="100" hs="100" page="8" pagename="8" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901010.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901010.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure><p> formal charge on C</p> <p><span class="inlineequation">[&amp;4|-||pbo|2|+|3|pbc||=||minus|1&amp;]</span></p></li><li><?do cpage num="8" PAGENAME="8"?><p> c. </p><figure class="informalfigure"><?docpage num="8" PAGENAME="8"?><?REGION x="322.232" y="372.800" w="69.398" h="53.041" xs="100" hs="100" page="8" pagename="8" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901011.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901011.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure><p> formal charge on N</p> <p><span class="inlineequation">[&amp;5|-||pbo|0|+|4|pbc||=||plus|1&amp;]</span></p></li><li><?doc page num="8" PAGENAME="8"?><p> d. </p><figure class="informalfigure"><?docpage num="8" PAGENAME="8"?><?REGION x="434.705" y="371.500" w="64.97" h="51.729" xs="100" hs="100" page="8" pagename="8" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901012.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901012.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure><p> formal charge on</p> <p>N: <span class="inlineequation">[&amp;5|-||pbo|0|+|4|pbc||=||plus|1&amp;]</span></p> <p>B: <span class="inlineequation">[&amp;3|-||pbo|0|+|4|pbc||=||minus|1&amp;]</span></p></li></ul> </div></li>

                          <li class="general-problem"><div class="question" id="ch01qu17"><p>17. The bond between two atoms can be shown by a pair of dots or by a line, so there are two ways each of the answers can be written. Remember that all lone pairs have to be shown.</p>

                          <ul style="list-style-type:none"><li><p> a. </p><figure class="informalfigure"><?docpage num="8" PAGENAME="8"?><?REGION x="89.500" y="494.100" w="63.853" h="98.5599999999999" xs="100" hs="100" page="8" pagename="8" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901013.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901013.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p> b. </p><figure class="informalfigure"><?docpage num="8" PAGENAME="8"?><?REGION x="89.500" y="620.097" w="60.616" h="83.7090000000001" xs="100" hs="100" page="8" pagename="8" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901014.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901014.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p> c. </p><figure class="informalfigure"><?docpage num="8" PAGENAME="8"?><?REGION x="201.000" y="494.100" w="58.051" h="108.586" xs="100" hs="100" page="8" pagename="8" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901015.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901015.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p> d. </p><figure class="informalfigure"><?docpage num="8" PAGENAME="8"?><?REGION x="201.000" y="615.123" w="64.97" h="108.863" xs="100" hs="100" page="8" pagename="8" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901016.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901016.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p> e. </p><figure class="informalfigure"><?docpage num="8" PAGENAME="8"?><?REGION x="330.000" y="494.100" w="64.97" h="108.896" xs="100" hs="100" page="8" pagename="8" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901017.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901017.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p> f. </p><figure class="informalfigure"><?docpage num="8" PAGENAME="8"?><?REGION x="330.000" y="620.449" w="58.023" h="84.047" xs="100" hs="100" page="8" pagename="8" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901018.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901018.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p> g. </p><figure class="informalfigure"><?docpage num="8" PAGENAME="8"?><?REGION x="459.000" y="494.100" w="70.47" h="87.538" xs="100" hs="100" page="8" pagename="8" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901019.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901019.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p/><figure class="informalfigure"><?docpage num="8" PAGENAME="8"?><?REGION x="459.000" y="623.810" w="50.217" h="93.3800000000001" xs="100" hs="100" page="8" pagename="8" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901020.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901020.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/> h. </figure></li></ul></div></li>

                          <?docpage num="9" PAGENAME="9"?><li class="general-problem"><div class="question" id="ch01qu18"><ul style="list-style-type:none"><li><p><span class="pagebreak" title="9" id="page9"/>18. a. </p><figure class="informalfigure"><?docpage num="9" PAGENAME="9"?><?REGION x="126.000" y="81.000" w="207.868" h="107.081" xs="100" hs="100" page="9" pagename="9" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901021.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901021.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p/><figure class="informalfigure"><?docpage num="9" PAGENAME="9"?><?REGION x="125.500" y="199.767" w="401.838" h="139.343" xs="100" hs="100" page="9" pagename="9" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901022.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901022.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/> b. </figure></li></ul></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu19"><p>19. Because the compounds are neutral, a halogen has 3 lone pairs, an oxygen has 2, a nitrogen has 1, and carbon or a hydrogen has no lone pairs.</p>

                          <ul style="list-style-type:none"><li><p> a. </p><figure class="informalfigure"><?docpage num="9" PAGENAME="9"?><?REGION x="127.250" y="370.403" w="54.049" h="13.69" xs="100" hs="100" page="9" pagename="9" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901023.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901023.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li> <li><p>b. </p><figure class="informalfigure"><?docpage num="9" PAGENAME="9"?><?REGION x="235.426" y="370.937" w="50.102" h="13.456" xs="100" hs="100" page="9" pagename="9" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901024.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901024.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li> <li><p>c. </p><figure class="informalfigure"><?docpage num="9" PAGENAME="9"?><?REGION x="345.384" y="370.585" w="50.056" h="13.608" xs="100" hs="100" page="9" pagename="9" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901025.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901025.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p> d. </p><figure class="informalfigure"><?docpage num="9" PAGENAME="9"?><?REGION x="127.866" y="392.003" w="43.022" h="13.49" xs="100" hs="100" page="9" pagename="9" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901026.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901026.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li> <li><p>e. </p><figure class="informalfigure"><?docpage num="9" PAGENAME="9"?><?REGION x="234.634" y="391.204" w="47.739" h="13.889" xs="100" hs="100" page="9" pagename="9" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901027.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901027.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li> <li><p/><figure class="informalfigure"><?docpage num="9" PAGENAME="9"?><?REGION x="343.998" y="391.829" w="32.715" h="13.664" xs="100" hs="100" page="9" pagename="9" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901028.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901028.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/>f. </figure></li></ul></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu20"><ul style="list-style-type:none"><li><p>20. a. </p><figure class="informalfigure"><?docpage num="9" PAGENAME="9"?><?REGION x="127.250" y="440.295" w="62.698" h="12.198" xs="100" hs="100" page="9" pagename="9" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901029.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901029.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li><li><p> b. </p><figure class="informalfigure"><?docpage num="9" PAGENAME="9"?><?REGION x="235.426" y="419.390" w="67.078" h="33.103" xs="100" hs="100" page="9" pagename="9" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901030.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901030.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li><li><p> c. </p><figure class="informalfigure"><?docpage num="9" PAGENAME="9"?><?REGION x="345.384" y="419.559" w="84.89" h="53.534" xs="100" hs="100" page="9" pagename="9" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901031.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901031.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li><li><p/><figure class="informalfigure"><?docpage num="9" PAGENAME="9"?><?REGION x="476.616" y="440.311" w="61.532" h="12.182" xs="100" hs="100" page="9" pagename="9" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901032.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901032.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/>      d. </figure></li></ul></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu21"><ul style="list-style-type:none"><li><p>21. a. the (green) chlorine atom</p></li>

                          <li><p> b. the (red) oxygen atoms</p></li>

                          <li><p>c.  the (blue) nitrogen atoms</p></li>

                          <li><p>d. the (black) carbon atoms and (gray) hydrogen atoms</p></li></ul></div></li>

                           

                           

                          <li class="general-problem"><div class="question" id="ch01qu22"><ul style="list-style-type:none"><li><p>22. a. </p><figure class="informalfigure"><?docpage num="9" PAGENAME="9"?><?REGION x="125.500" y="543.783" w="127.118" h="51.73" xs="100" hs="100" page="9" pagename="9" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901033.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901033.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p> b. </p><figure class="informalfigure"><?docpage num="9" PAGENAME="9"?><?REGION x="125.500" y="652.843" w="91.86" h="72.37" xs="100" hs="100" page="9" pagename="9" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901034.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901034.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p>c. </p><figure class="informalfigure"><?docpage num="9" PAGENAME="9"?><?REGION x="346.234" y="523.700" w="82.772" h="92.966" xs="100" hs="100" page="9" pagename="9" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901035.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901035.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p/><figure class="informalfigure"><?docpage num="9" PAGENAME="9"?><?REGION x="347.466" y="631.551" w="205.171" h="92.722" xs="100" hs="100" page="9" pagename="9" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901036.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901036.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/>d. </figure></li></ul></div></li>

                          <?docpage num="10" PAGENAME="10"?><li class="general-problem"><div class="question" id="ch01qu23"><ul style="list-style-type:none"><li><p><span class="pagebreak" title="10" id="page10"/>23. a. </p><figure class="informalfigure"><?docpage num="10" PAGENAME="10"?><?REGION x="91.250" y="100.591" w="44.363" h="15.602" xs="100" hs="100" page="10" pagename="10" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="6658501116.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/6658501116.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li><li><p> b. </p><figure class="informalfigure"><?docpage num="10" PAGENAME="10"?><?REGION x="163.426" y="82.190" w="55.65" h="37.003" xs="100" hs="100" page="10" pagename="10" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="6658501117.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/6658501117.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li><li><p> c. </p><figure class="informalfigure"><?docpage num="10" PAGENAME="10"?><?REGION x="270.634" y="82.000" w="63.85" h="45.093" xs="100" hs="100" page="10" pagename="10" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="6658501118.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/6658501118.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li><li><p/><figure class="informalfigure"><?docpage num="10" PAGENAME="10"?><?REGION x="379.866" y="103.417" w="36.28" h="14.976" xs="100" hs="100" page="10" pagename="10" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="6658501119.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/6658501119.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/> d. </figure></li></ul></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu24"><figure class="informalfigure"><?docpage num="10" PAGENAME="10"?><?REGION x="72.000" y="131.123" w="381.712" h="104.67" xs="100" hs="100" page="10" pagename="10" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901037.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901037.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/>24. </figure></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu25"><p><span class="inlineequation">25. [&amp;~rom~He_{~normal~2}^{|plus|}~norm~&amp;]</span> has three electrons. Using Figure 1.3 on page 23 of the text, two electrons will be in a bonding molecular orbital and one electron will be in an antibonding molecular orbital. Because there are more electrons in the bonding molecular orbital than in the antibonding molecular orbital, <span class="inlineequation">[&amp;~rom~He_{~normal~2}^{|plus|}~norm~&amp;]</span> exists.</p></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu26"><ul style="list-style-type:none"><li><p>26. a. <span class="inlineequation">[&amp;|pi||ast|&amp;]</span> This involves out-of-phase interaction of atomic orbitals (the interacting orbitals have different colors), leading to an antibonding molecular orbital. Because this example involves the side-to-side overlap of <i>p</i> orbitals, it is a <span class="inlineequation">[&amp;|pi||ast|&amp;]</span> antibonding molecular orbital.</p></li>

                          <li><p> b. <span class="inlineequation">[&amp;|pi|&amp;]</span> This involves in-phase overlap of atomic orbitals (the overlapping orbitals have the same color), leading to a bonding molecular orbital. Because this example involves the side-to-side overlap of <i>p</i> orbitals, it is a <span class="inlineequation">[&amp;|pi|&amp;]</span> bonding molecular orbital.</p></li>

                          <li><p> c. <span class="inlineequation">[&amp;|sig||ast|&amp;]</span> This involves out-of-phase interaction of atomic orbitals (the interacting orbitals have different colors), leading to an antibonding molecular orbital. Because this example involves the end-on overlap of atomic orbitals, it is a <span class="inlineequation">[&amp;|sig||ast|&amp;]</span> antibonding molecular orbital.</p></li>

                          <li> d. <p><span class="inlineequation">[&amp;|sig|&amp;] </span>This involves in-phase overlap of atomic orbitals (the overlapping orbitals have the same color), leading to a bonding molecular orbital. Because this example involves the end-on overlap of atomic orbitals, it is a <span class="inlineequation">[&amp;|sig|&amp;]</span> bonding molecular orbital.</p></li></ul></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu27"><ul style="list-style-type:none"><li><p>27. The 3 carbon-carbon bonds form as a result of <span class="inlineequation">[&amp;sp^{3}|emd|sp^{3}&amp;]</span> overlap.</p></li>

                          <li><p>The 7 carbon-hydrogen bonds form as a result of <span class="inlineequation">[&amp;sp^{3}|emd|s&amp;]</span> overlap.</p></li></ul></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu28"><p>28. The electron density of the large lobe of an <span class="inlineequation">[&amp;sp^{3}&amp;]</span> orbital (the lobe that overlaps the <i>s</i> orbital) is greater than the electron density of a lobe of a <i>p</i> orbital. Therefore, the overlap of an <i>s</i> orbital with an <span class="inlineequation">[&amp;sp^{3}&amp;]</span> orbital forms a stronger bond than the overlap of an <i>s</i> orbital with a <i>p</i> orbital.</p></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu29"><p>29. Solved in the text.</p></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu30"><ul style="list-style-type:none"><li><p>30. a. <b>One</b> <i>s</i> orbital and <b>three</b> <i>p</i> orbitals form <b>four</b> <span class="inlineequation">[&amp;sp^{3}&amp;]</span> orbitals.</p></li>

                          <li><p> b. <b>One</b> <i>s</i> orbital and <b>two</b> <i>p</i> orbitals form <b>three</b> <span class="inlineequation">[&amp;sp^{2}&amp;]</span> orbitals.</p></li>

                          <li><p><b> c. One</b> <i>s</i> orbital and <b>one</b> <i>p</i> orbital form <b>two</b> <i>sp</i> orbitals.</p></li></ul></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu31"><ul style="list-style-type:none"><li><p><b>31. a(1).</b> Solved in the text.</p></li>

                          <li><p><b> b(1).</b> Solved in the text.</p></li></ul></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu32"><ul style="list-style-type:none"><li><p><b>32. a(3).</b> The carbon forms four bonds, and each chlorine forms one bond.</p>

                          <figure class="informalfigure"><?docpage num="10" PAGENAME="10"?><?REGION x="260.463" y="670.349" w="55.074" h="53.8439999999999" xs="100" hs="100" page="10" pagename="10" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901038.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901038.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <?docpage num="11" PAGENAME="11"?><li><p><span class="pagebreak" title="11" id="page11"/><b>b(3).</b> The carbon uses <span class="inlineequation">[&amp;sp^{3}&amp;]</span> orbitals to form the bonds with the chlorine atoms, so the bond angles are all <span class="inlineequation">[&amp;109.5|deg|.&amp;]</span></p>

                          <figure class="informalfigure"><?docpage num="11" PAGENAME="11"?><?REGION x="291.873" y="105.711" w="64.254" h="50.973" xs="100" hs="100" page="11" pagename="11" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901039.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901039.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p><b>a(1).</b> The first attempt at drawing a Lewis structure results in a carbon that does not have a complete octet and does not form the needed number of bonds.</p>

                          <figure class="informalfigure"><?docpage num="11" PAGENAME="11"?><?REGION x="281.916" y="191.307" w="84.168" h="52.47" xs="100" hs="100" page="11" pagename="11" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901040.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901040.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure>

                          <p>Using one of oxygen’s lone pairs to put a double bond between the carbon and oxygen solves both problems.</p>

                          <figure class="informalfigure"><?docpage num="11" PAGENAME="11"?><?REGION x="281.930" y="270.341" w="84.14" h="52.647" xs="100" hs="100" page="11" pagename="11" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901041.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901041.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p><b>b(1).</b> The <span class="inlineequation">[&amp;sp^{3}&amp;]</span> hybridized <span class="inlineequation">[&amp;~rom~CH_{~normal~3}~norm~&amp;]</span> carbon uses <span class="inlineequation">[&amp;sp^{3}&amp;]</span> orbitals to form the four <i>s</i> bonds and has bond angles of 109.5°. The <span class="inlineequation">[&amp;sp^{2}&amp;]</span> hybridized <span class="inlineequation">[&amp;~rom~C|dbond|O~norm~&amp;]</span> carbon has <span class="inlineequation">[&amp;120|deg|&amp;]</span> bond angles, uses <span class="inlineequation">[&amp;sp^{2}&amp;]</span> orbitals to form the three s bonds and has bond angles of 120°.</p>

                          <figure class="informalfigure"><?docpage num="11" PAGENAME="11"?><?REGION x="277.636" y="367.218" w="92.728" h="61.466" xs="100" hs="100" page="11" pagename="11" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901042.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901042.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p><b>a(2).</b> In order to fill their octets and form the required number of bonds, carbon and nitrogen must form a triple bond.</p>

                          <figure class="informalfigure"><?docpage num="11" PAGENAME="11"?><?REGION x="299.875" y="455.878" w="48.25" h="10.806" xs="100" hs="100" page="11" pagename="11" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901043.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901043.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p><b>b(2).</b> Because the carbon is <i>sp</i> hybridized, the carbon used <i>sp</i> orbitals to form the two <i>s</i> bonds and <i>p</i> orbitals to form the two [&amp;|pi|&amp;] bonds. The bond angle is <span class="inlineequation">[&amp;180|deg|.&amp;]</span></p>

                          <figure class="informalfigure"><?docpage num="11" PAGENAME="11"?><?REGION x="300.111" y="503.690" w="47.778" h="26.994" xs="100" hs="100" page="11" pagename="11" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901044.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901044.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p><b>32. a(4).</b> The first attempt at drawing a Lewis structure (and remembering to avoid oxygen-oxygen single bonds) results in a carbon that does not have a complete octet and does not form the needed number of bonds.</p>

                          <figure class="informalfigure"><?docpage num="11" PAGENAME="11"?><?REGION x="281.640" y="564.332" w="84.72" h="32.352" xs="100" hs="100" page="11" pagename="11" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="6658501120.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/6658501120.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure>

                          <p>Using one of oxygen’s lone pairs to put a double bond between the carbon and the oxygen solves both problems.</p>

                          <figure class="informalfigure"><?docpage num="11" PAGENAME="11"?><?REGION x="281.640" y="616.332" w="84.72" h="32.352" xs="100" hs="100" page="11" pagename="11" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="6658501121.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/6658501121.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p><b>b(4).</b> The carbon uses <span class="inlineequation">[&amp;sp^{2}&amp;]</span> orbitals to form the three <i>[&amp;|sig|&amp;]</i> bonds and a <i>p</i> orbital to form the <i>[&amp;|pi|&amp;]</i> bond. The bond angles are 120°.</p>

                          <figure class="informalfigure"><?docpage num="11" PAGENAME="11"?><?REGION x="283.638" y="675.804" w="80.724" h="43.88" xs="100" hs="100" page="11" pagename="11" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="6658501122.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/6658501122.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li></ul></div></li>

                          <?docpage num="12" PAGENAME="12"?><li class="general-problem"><div class="question" id="ch01qu33"><ul style="list-style-type:none"><li><p><span class="pagebreak" title="12" id="page12"/>33. a. <span class="inlineequation">[&amp;120|deg|&amp;]</span></p></li><li><p>   b. <span class="inlineequation">[&amp;120|deg|&amp;]</span></p></li>

                          <li><p> c. Because the carbon is <span class="inlineequation">[&amp;sp^{3}&amp;]</span> hybridized and it has one lone pair, you can predict that the bond angle is similar to that in <span class="inlineequation">[&amp;~rom~NH_{3} |pbo|107.3|deg||pbc|.~norm~&amp;]</span></p></li></ul></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu34"><ul style="list-style-type:none"><li><p>34. The nitrogen atom has the greatest electron density.</p></li>

                          <li><p>The hydrogens are the bluest atoms. Therefore, they have the least electron density. In other words, they have the most positive (least negative) electrostatic potential.</p></li></ul></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu35"><ul style="list-style-type:none"><li><p>35. Water is the most polar—it has a deep red area and the most intense blue area.</p></li>

                          <li><p>Methane is the least polar—it is all nearly the same color (green) with no red or blue areas.</p></li></ul></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu36"><p>36. Solved in the text.</p></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu37"><p>37. Electrons in atomic orbitals farther from the nucleus form <b>longer</b> bonds; they also form <b>weaker</b> bonds due to less electron density in the region of orbital overlap. Therefore:</p>

                          <ul style="list-style-type:none"><li><p> a. <b>relative lengths</b> of the bonds in the halogens are: <span class="inlineequation">[&amp;~rom~Br_{~normal~2}|gtr|~rom~Cl_{~normal~2}~norm~&amp;]</spa n></p>

                          <p><b>relative strengths</b> of the bonds are: <span class="inlineequation">[&amp;~rom~Cl_{~normal~2}|gtr|~rom~Br_{~normal~2}~norm~&amp;]</spa n></p></li>

                          <li><p><b> b. relative lengths</b>: <span class="inlineequation">[&amp;~rom~CH_{~normal~3}|bond|~rom~Br|gtr|CH_{~normal~3}|bond|~ro m~Cl|gtr|CH_{~normal~3}|bond|~rom~F~norm~&amp;]</span></p>

                          <p><b>relative strengths</b>: <span class="inlineequation">[&amp;~rom~CH_{~normal~3}|bond|~rom~F|gtr|CH_{~normal~3}|bond|~rom ~Cl|gtr|CH_{~normal~3}|bond|~rom~Br~norm~&amp;]</span></p></li></ul></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu38"><ul style="list-style-type:none"><li><p>38. a. <b>longer: </b></p><ul style="list-style-type:none"><li><p><span class="inlineequation">1. [&amp;~rom~C|bond|I~norm~&amp;] </span></p></li><li><p><span class="inlineequation">2. [&amp;~rom~C|bond|Cl~norm~&amp;] </span></p></li><li><p><span class="inlineequation">3. [&amp;~rom~H|bond|Cl~norm~&amp;]</span></p></li></ul></li>

                          <li><p><b> b. stronger: </b></p><ul style="list-style-type:none"><li><p><span class="inlineequation">1. [&amp;~rom~C|bond|Cl~norm~&amp;] </span></p></li><li><p><span class="inlineequation">2. [&amp;~rom~C|bond|C~norm~&amp;] </span></p></li><li><p><span class="inlineequation">3. [&amp;~rom~H|bond|F~norm~&amp;]</span></p></li></ul></li></ul></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu39"><ul style="list-style-type:none"><li><p>39. a. CH<sub>3</sub>O<sup>−</sup></p>

                          <p> The carbon in C<sub>3</sub> is bonded to four atoms, so it uses four <span class="inlineequation">[&amp;sp^{3}&amp;]</span> orbitals.</p>

                          <p> Each carbon-hydrogen bond is formed by the overlap of an <span class="inlineequation">[&amp;sp^{3}&amp;]</span> orbital of carbon with the <i>s</i> orbital of hydrogen. The carbon-oxygen bond is formed by the overlap of an <span class="inlineequation">[&amp;sp^{3}&amp;]</span> orbital of carbon with an <span class="inlineequation">[&amp;~rom~s~normal~p^{3}~norm~&amp;]</span> orbital of oxygen. Because the four <span class="inlineequation">[&amp;sp^{3}&amp;]</span> orbitals of carbon orient themselves to get as far away from each other as possible, the bond angles are all <span class="inlineequation">[&amp;~bf~109.5|deg|~normal~.~norm~&amp;]</span></p>

                          <figure class="informalfigure"><?docpage num="12" PAGENAME="12"?><?REGION x="265.599" y="467.356" w="44.802" h="51.6799999999999" xs="100" hs="100" page="12" pagename="12" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901045.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901045.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure>

                          <div class="informalequation">[&amp;~rom~bond|sp|angles|=|~normal~109.5|deg|~norm~&amp;]</div> </li>

                          <li><p> b. <span class="inlineequation">[&amp;~bf~CO_{2}~norm~&amp;]</span></p>

                          <p> The carbon in <span class="inlineequation">[&amp;~rom~CO_{~normal~2}~norm~&amp;]</span> is bonded to two atoms, so it uses two <i>sp</i> orbitals. Each carbon-oxygen bond is a double bond. One of the bonds of each double bond is formed by the overlap of an <i>sp</i> orbital of carbon with an <span class="inlineequation">[&amp;sp^{2}&amp;]</span> orbital of oxygen. The second bond of the double bond is formed as a result of side-to-side overlap of a <i>p</i> orbital of carbon with a <i>p</i> orbital of oxygen. Because carbon’s two <i>sp</i> orbitals orient themselves to get as far away from each other as possible, the bond angle in <span class="inlineequation">[&amp;~rom~CO_{~normal~2}~norm~&amp;]</span> is <span class="inlineequation">[&amp;~bf~180|deg|~normal~.~norm~&amp;]</span></p>

                          <figure class="informalfigure"><?docpage num="12" PAGENAME="12"?><?REGION x="265.459" y="638.669" w="45.082" h="10.367" xs="100" hs="100" page="12" pagename="12" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901046.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901046.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/><div class="informalequation"> [&amp;~rom~bond|sp|angle|=|~normal~180|deg|~norm~&amp;]</div></figure></li>

                          <?docpage num="13" PAGENAME="13"?><li><p><span class="pagebreak" title="13" id="page13"/> c. <b>H<sub>2</sub>CO</b></p>

                          <p> The double-bonded carbon and the double-bonded oxygen in H<sub>2</sub>CO each uses <span class="inlineequation">[&amp;sp^{2}&amp;]</span> orbitals; thus, the bonds around the double-bonded carbon are all <span class="inlineequation">[&amp;120|deg|.&amp;]</span> Each carbon-hydrogen bond is formed by the overlap of an <span class="inlineequation">[&amp;sp^{2}&amp;]</span> orbital of carbon with the <i>s</i> orbital of hydrogen.</p>

                          <figure class="informalfigure"><?docpage num="13" PAGENAME="13"?><?REGION x="191.032" y="135.860" w="265.936" h="83.903" xs="100" hs="100" page="13" pagename="13" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901047.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901047.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p> d. <span class="inlineequation">[&amp;~bf~N_{2}~norm~&amp;]</span></p>

                          <p> The triple bond consists of one <span class="inlineequation">[&amp;|sig|&amp;]</span> bond and two <span class="inlineequation">[&amp;|pi|&amp;]</span> bonds. Each nitrogen has two <i>sp</i> orbitals; one is used to form the <span class="inlineequation">[&amp;|sig|&amp;]</span> bond, and the other contains the lone pair. Each nitrogen has two <i>p</i> orbitals that are used to form the two <span class="inlineequation">[&amp;|pi|&amp;]</span> bonds. A bond angle is the angle formed by three atoms. Therefore, there are no bond angles in this two-atom containing compound.</p>

                          <figure class="informalfigure"><?docpage num="13" PAGENAME="13"?><?REGION x="224.381" y="293.568" w="199.238" h="21.945" xs="100" hs="100" page="13" pagename="13" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901048.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901048.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure> </li>

                          <li><p><span class="inlineequation"> e. [&amp;~bf~BF_{3}~norm~&amp;]</span></p>

                          <p> Promotion gives boron three unpaired electrons, and hybridization gives it three <span class="inlineequation">[&amp;sp^{2}&amp;]</span> orbitals.</p>

                          <figure class="informalfigure"><?docpage num="13" PAGENAME="13"?><?REGION x="126.046" y="366.397" w="407.908" h="33.116" xs="100" hs="100" page="13" pagename="13" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901049.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901049.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure>

                          <p> Each <span class="inlineequation">[&amp;sp^{2}&amp;]</span> orbital of boron overlaps an <span class="inlineequation">[&amp;sp^{3}&amp;]</span> orbital of fluorine. The three <span class="inlineequation">[&amp;sp^{2}&amp;]</span> orbitals orient themselves to get as far away from each other as possible, resulting in bond angles of <span class="inlineequation">[&amp;~bf~120|deg|~normal~.~norm~&amp;]</span></p>

                          <figure class="informalfigure"><?docpage num="13" PAGENAME="13"?><?REGION x="285.226" y="439.182" w="77.548" h="64.331" xs="100" hs="100" page="13" pagename="13" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901050.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901050.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li></ul></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu40"><p>40. Solved in the text.</p></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu41"><p>41. We know that the <span class="inlineequation">[&amp;|sig|&amp;]</span> bond is stronger than the <span class="inlineequation">[&amp;|pi|&amp;]</span> bond, because the <span class="inlineequation">[&amp;|sig|&amp;]</span> bond in ethane has a bond dissociation energy of 90.2 kcal/mol, whereas the bond dissociation energy of the double bond <span class="inlineequation">[&amp;|pbo||sig||+||pi||pbc|&amp;]</span> in ethene is 174.5 kcal/mol, which is less than twice as strong.</p>

                          <p> Because the <span class="inlineequation">[&amp;|sig|&amp;]</span> bond is stronger, we know that it has more effective orbital-orbital overlap.</p> </div></li>

                          <li class="general-problem"><div class="question" id="ch01qu42"><p>42. Because electrons in an <i>s</i> orbital are closer on average to the nucleus than those in a <i>p</i> orbital, the greater the <i>s</i> character in the interacting orbitals, the stronger (and shorter) will be the bond. Therefore, the carbon-carbon <span class="inlineequation">[&amp;|sig|&amp;]</span> bond formed by <span class="inlineequation">[&amp;sp^{2}|emd|sp^{2}&amp;]</span> overlap is stronger (and shorter) than the carbon-carbon bond formed by <span class="inlineequation">[&amp;sp^{3}|emd|sp^{3}&amp;]</span> overlap, because an <span class="inlineequation">[&amp;sp^{2}&amp;]</span> orbital has 33.3% <i>s</i> character, whereas an <span class="inlineequation">[&amp;sp^{3}&amp;]</span> orbital has 25% <i>s</i> character.</p></div></li>

                          <?docpage num="14" PAGENAME="14"?><li class="general-problem"><div class="question" id="ch01qu43"><figure class="informalfigure"><?docpage num="14" PAGENAME="14"?><?REGION x="72.000" y="81.000" w="234.514" h="83.428" xs="100" hs="100" page="14" pagename="14" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901051.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901051.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/>43. </figure></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu44"><ul style="list-style-type:none"><li><p><span class="pagebreak" title="14" id="page14"/>44. a. </p><figure class="informalfigure"><?docpage num="14" PAGENAME="14"?><?REGION x="89.500" y="177.824" w="144" h="90.999" xs="100" hs="100" page="14" pagename="14" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901052.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901052.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p> b. </p><figure class="informalfigure"><?docpage num="14" PAGENAME="14"?><?REGION x="89.500" y="282.572" w="448.96" h="152.251" xs="100" hs="100" page="14" pagename="14" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901053.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901053.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p> c. the nitrogen and chlorine atoms</p></li></ul></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu45"><p>45. The bond angle depends on the central atom.</p>

                          <ul style="list-style-type:none"><li><p> a. <span class="inlineequation">[&amp;sp^{3}&amp;]</span> nitrogen with no lone pair: <span class="inlineequation">[&amp;109.5|deg|&amp;]</span></p></li>

                          <li><p> b. <span class="inlineequation">[&amp;sp^{3}&amp;]</span> nitrogen with one lone pair: <span class="inlineequation">[&amp;107.3|deg|&amp;]</span></p></li>

                          <li><p>c. <span class="inlineequation">[&amp;sp^{3}&amp;]</span> carbon with no lone pair: <span class="inlineequation">[&amp;109.5|deg|&amp;]</span></p></li>

                          <li><p><span class="inlineequation">d. [&amp;sp^{3}&amp;]</span> carbon with no lone pair: <span class="inlineequation">[&amp;109.5|deg|&amp;]</span></p></li></ul></div></li><li class="general-problem"><div class="question" id="ch01qu46"/></li>

                          <li class="general-problem"><div class="question" id="ch01qu48">46. <p><b>a</b>, <b>e</b>, <b>g,</b> and <b>h</b> have a dipole moment of zero because they are symmetrical molecules.</p>

                          <figure class="informalfigure"><?docpage num="14" PAGENAME="14"?><?REGION x="122.070" y="549.055" w="331.86" h="50.0920000000001" xs="100" hs="100" page="14" pagename="14" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901054.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901054.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu47"><p>47. The electrostatic potential map of ammonia is not symmetrical in the distribution of the charge—the nitrogen is more electron rich and, therefore, more red than the three hydrogens. Therefore, its shape, which indicates charge distribution, is not symmetrical.</p>

                          <p> The electrostatic potential map of the ammonium ion is symmetrical in the distribution of the charge, so its shape is symmetrical. Its symmetry results from the fact that nitrogen forms a bond with each of the four hydrogens and the four bonds point to the corners of a regular tetrahedron. The nitrogen in the ammonium ion has significantly lower electron density than the nitrogen in ammonia as a result of the lone pair having formed a bond to hydrogen.</p>

                          <?docpage num="15" PAGENAME="15"?><p><span class="pagebreak" title="15" id="page15"/>48. The atom with the greater electronegativity will decrease the electron flow toward the electronegative F atom, giving the compound a smaller dipole moment. Since <span class="inlineequation">[&amp;~rom~CH_{~normal~3}~rom~F~norm~&amp;]</span> has a smaller dipole moment than <span class="inlineequation">[&amp;~rom~CD_{~normal~3}~rom~F~normal~,~norm~&amp;]</span> we know that hydrogen is more electronegative than deuterium.</p></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu49"><ul style="list-style-type:none"><li><p>49. a. </p><figure class="informalfigure"><?docpage num="15" PAGENAME="15"?><?REGION x="127.250" y="128.703" w="137.958" h="32.668" xs="100" hs="100" page="15" pagename="15" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="6658501123.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/6658501123.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p> b. </p><figure class="informalfigure"><?docpage num="15" PAGENAME="15"?><?REGION x="127.426" y="173.313" w="126.422" h="33.4" xs="100" hs="100" page="15" pagename="15" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901056.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901056.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p>c. </p><figure class="informalfigure"><?docpage num="15" PAGENAME="15"?><?REGION x="345.634" y="131.280" w="141.124" h="12.233" xs="100" hs="100" page="15" pagename="15" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="6658501124.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/6658501124.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p>d. </p><figure class="informalfigure"><?docpage num="15" PAGENAME="15"?><?REGION x="346.866" y="158.633" w="131.255" h="12.88" xs="100" hs="100" page="15" pagename="15" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901058.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901058.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p/><figure class="informalfigure"><?docpage num="15" PAGENAME="15"?><?REGION x="345.634" y="182.379" w="126.35" h="12.846" xs="100" hs="100" page="15" pagename="15" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="6658501125.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/6658501125.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/>e. </figure></li></ul></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu50"><ul style="list-style-type:none"><li><p>50. a. <span class="inlineequation">[&amp;~rom~CH_{~normal~3}~rom~CH_{~normal~3}~norm~&amp;]</span></p ></li><li><p><span class="inlineequation">    b. [&amp;~rom~CH_{~normal~3}~rom~F~norm~&amp;]</span></p></li></ul></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu51"><p>51. If the central atom is <span class="inlineequation">[&amp;sp^{3}&amp;]</span> hybridized, the bond angle will depend on the number of lone pairs it has: no lone pairs[&amp;|=|&amp;]109.5°; one lone pair[&amp;|=|&amp;]107.3°; two lone pairs[&amp;|=|&amp;]104.5°</p>

                          <ul style="list-style-type:none"><li><p> a. <span class="inlineequation">[&amp;sp^{3},&amp;]</span> <span class="inlineequation">[&amp;107.3|deg|&amp;]</span></p></li>

                          <li><p> b. <span class="inlineequation">[&amp;sp^{2},&amp;]</span> <span class="inlineequation">[&amp;120|deg|&amp;]</span></p></li>

                          <li><p> c. <span class="inlineequation">[&amp;sp^{3},&amp;]</span> <span class="inlineequation">[&amp;107.3|deg|&amp;]</span></p></li>

                          <li><p> d. <span class="inlineequation">[&amp;sp^{2},&amp;]</span> <span class="inlineequation">[&amp;120|deg|&amp;]</span></p></li>

                          <li><p>e. <span class="inlineequation">[&amp;sp^{3},&amp;]</span> <span class="inlineequation">[&amp;109.5|deg|&amp;]</span></p></li>

                          <li><p>f. <span class="inlineequation">[&amp;sp^{2},&amp;]</span> <span class="inlineequation">[&amp;120|deg|&amp;]</span></p></li>

                          <li><p>g. <span class="inlineequation">[&amp;sp,|sp|180|deg|&amp;]</span></p></li>

                          <li><p>h. <span class="inlineequation">[&amp;sp^{3},&amp;]</span> <span class="inlineequation">[&amp;109.5|deg|&amp;]</span></p></li>

                          <li><p>i. <span class="inlineequation">[&amp;sp^{3},&amp;]</span> <span class="inlineequation">[&amp;107.3|deg|&amp;]</span></p></li>

                          <li><p>j. <span class="inlineequation">[&amp;sp^{2},&amp;]</span> <span class="inlineequation">[&amp;120|deg|&amp;]</span></p></li></ul></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu52"><ul style="list-style-type:none"><li><p>52. a. <span class="inlineequation">[&amp;~rom~CH_{~normal~3}~rom~CH_{~normal~2}~rom~CH_{~normal~3}~no rm~&amp;]</span></p></li><li><p> b. </p><figure class="informalfigure"><?docpage num="15" PAGENAME="15"?><?REGION x="228.426" y="336.784" w="62.531" h="12.529" xs="100" hs="100" page="15" pagename="15" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901059.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901059.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li><li><p/> <figure class="informalfigure"><?docpage num="15" PAGENAME="15"?><?REGION x="342.634" y="336.711" w="163.82" h="11.802" xs="100" hs="100" page="15" pagename="15" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901060.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901060.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/>c. </figure></li></ul></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu53"><p>53. The hybridization of the central atom determines the bond angle. If the hybridization is <span class="inlineequation">[&amp;sp^{3},&amp;]</span> the number of lone pairs on the central atom determines the bond angle.</p>

                          <ul style="list-style-type:none"><li><p> a. <span class="inlineequation">[&amp;109.5|deg|&amp;]</span></p></li><li><p> b. <span class="inlineequation">[&amp;104.5|deg|&amp;]</span></p></li><li><p> c. <span class="inlineequation">[&amp;107.3|deg|&amp;]</span></p></li><li><p><span class="inlineequation"> d. [&amp;107.3|deg|&amp;]</span></p></li></ul>

                          <p><span class="inlineequation"> [&amp;|ast|104.5|deg|&amp;]</span> is the correct prediction based on the bond angle in water.</p>

                          <p> However, the bond angle is actually somewhat larger <span class="inlineequation">[&amp;|pbo|108.2|deg||pbc|&amp;]</span> because the bond opens up to minimize the interaction between the electron cloud of the relatively bulky <span class="inlineequation">[&amp;~rom~CH_{~normal~3}~norm~&amp;]</span> group.</p> </div></li><li class="general-problem"><div class="question" id="ch01qu54"><p><span class="inlineequation"/></p>

                          <ul style="list-style-type:none"><li><p>54. </p><figure class="informalfigure"><?docpage num="15" PAGENAME="15"?><?REGION x="108.000" y="470.447" w="309.875" h="90.966" xs="100" hs="100" page="15" pagename="15" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901061.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901061.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li></ul></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu55"><ul style="list-style-type:none"><li><p>55. a. </p><figure class="informalfigure"><?docpage num="15" PAGENAME="15"?><?REGION x="125.500" y="576.544" w="127.22" h="51.631" xs="100" hs="100" page="15" pagename="15" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901062.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901062.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p> b. </p><figure class="informalfigure"><?docpage num="15" PAGENAME="15"?><?REGION x="125.500" y="638.305" w="64.815" h="13.2080000000001" xs="100" hs="100" page="15" pagename="15" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901063.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901063.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p> c. </p><figure class="informalfigure"><?docpage num="15" PAGENAME="15"?><?REGION x="380.500" y="591.710" w="125.66" h="36.5029999999999" xs="100" hs="100" page="15" pagename="15" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901064.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901064.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p/><figure class="informalfigure"><?docpage num="15" PAGENAME="15"?><?REGION x="380.500" y="634.204" w="136.602" h="36.009" xs="100" hs="100" page="15" pagename="15" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901065.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901065.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/> d. </figure></li></ul></div></li>

                          <?docpage num="16" PAGENAME="16"?><li class="general-problem"><div class="question" id="ch01qu56"><p><span class="pagebreak" title="16" id="page16"/></p><figure class="informalfigure"><?docpage num="16" PAGENAME="16"?><?REGION x="72.000" y="69.600" w="143.08" h="162.227" xs="100" hs="100" page="16" pagename="16" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901066.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901066.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/>56. </figure></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu57"><p>57. The greater the electronegativity difference between the two bonded atoms, the more polar the bond. (See Table <span class="number">1.3</span> on page 11.)</p>

                          <ul style="list-style-type:none"><li><p> a. <span class="inlineequation">[&amp;~rom~C|bond|F~norm~&amp;] [&amp;|gtr|&amp;] [&amp;~rom~C|bond|O~norm~&amp;] [&amp;|gtr|&amp;] [&amp;~rom~C|bond|N~norm~&amp;]</span></p></li>

                          <li><p> b. <span class="inlineequation">[&amp;~rom~C|bond|Cl~norm~&amp;] [&amp;|gtr|&amp;] [&amp;~rom~C|bond|Br~norm~&amp;] [&amp;|gtr|&amp;] [&amp;~rom~C|bond|I~norm~&amp;]</span></p></li>

                          <li><p> c. <span class="inlineequation">[&amp;~rom~H|bond|O~norm~&amp;] [&amp;|gtr|&amp;] [&amp;~rom~H|bond|N~norm~&amp;] [&amp;|gtr|&amp;] [&amp;~rom~H|bond|C~norm~&amp;]</span></p></li>

                          <li><p><span class="inlineequation"> d. [&amp;~rom~C|bond|N~norm~&amp;] [&amp;|gtr|&amp;] [&amp;~rom~C|bond|H~norm~&amp;] [&amp;|gtr|&amp;] [&amp;~rom~C|bond|C~norm~&amp;]</span></p></li></ul></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu58"><ul style="list-style-type:none"><li><p>58. a. </p><figure class="informalfigure"><?docpage num="16" PAGENAME="16"?><?REGION x="91.250" y="341.180" w="66.269" h="51.681" xs="100" hs="100" page="16" pagename="16" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901067.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901067.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li><li><p> b. </p><figure class="informalfigure"><?docpage num="16" PAGENAME="16"?><?REGION x="230.426" y="341.421" w="86.47" h="51.73" xs="100" hs="100" page="16" pagename="16" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901068.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901068.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li><li><p/><figure class="informalfigure"><?docpage num="16" PAGENAME="16"?><?REGION x="387.967" y="342.031" w="86.22" h="51.73" xs="100" hs="100" page="16" pagename="16" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901069.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901069.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/> c. </figure></li></ul></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu59"><figure class="informalfigure"><?docpage num="16" PAGENAME="16"?><?REGION x="72.000" y="410.689" w="216.138" h="38.372" xs="100" hs="100" page="16" pagename="16" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="6658501126.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/6658501126.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/>59. </figure></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu60"><ul style="list-style-type:none"><li><p>60. a. </p><figure class="informalfigure"><?docpage num="16" PAGENAME="16"?><?REGION x="91.250" y="464.582" w="61.992" h="45.579" xs="100" hs="100" page="16" pagename="16" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901070.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901070.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p> b. </p><figure class="informalfigure"><?docpage num="16" PAGENAME="16"?><?REGION x="91.426" y="535.419" w="56.484" h="36.1420000000001" xs="100" hs="100" page="16" pagename="16" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901071.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901071.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p>c. </p><figure class="informalfigure"><?docpage num="16" PAGENAME="16"?><?REGION x="224.634" y="464.482" w="49.721" h="45.579" xs="100" hs="100" page="16" pagename="16" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901072.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901072.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p>d. </p><figure class="informalfigure"><?docpage num="16" PAGENAME="16"?><?REGION x="225.866" y="527.357" w="56.61" h="43.904" xs="100" hs="100" page="16" pagename="16" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901073.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901073.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p>e. </p><figure class="informalfigure"><?docpage num="16" PAGENAME="16"?><?REGION x="394.634" y="464.974" w="68.609" h="45.087" xs="100" hs="100" page="16" pagename="16" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901074.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901074.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p/><figure class="informalfigure"><?docpage num="16" PAGENAME="16"?><?REGION x="393.248" y="525.787" w="60.141" h="45.774" xs="100" hs="100" page="16" pagename="16" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901075.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901075.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/>f. </figure></li></ul></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu61"><ul style="list-style-type:none"><li><p>61. a. <span class="inlineequation">[&amp;107.3|deg|&amp;]</span></p></li><li><p> b. <span class="inlineequation">[&amp;109.5|deg|&amp;]</span></p></li><li><p> c. <span class="inlineequation">[&amp;180|deg|&amp;]</span></p></li><li><p><span class="inlineequation"> d. [&amp;109.5|deg|&amp;]</span></p></li></ul></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu62"><ul style="list-style-type:none"><li><p>62. a. </p><figure class="informalfigure"><?docpage num="16" PAGENAME="16"?><?REGION x="91.250" y="610.518" w="41.6" h="19.443" xs="100" hs="100" page="16" pagename="16" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901076.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901076.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p> b. </p><figure class="informalfigure"><?docpage num="16" PAGENAME="16"?><?REGION x="91.426" y="635.822" w="39.184" h="20.139" xs="100" hs="100" page="16" pagename="16" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901077.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901077.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p>    c. </p><figure class="informalfigure"><?docpage num="16" PAGENAME="16"?><?REGION x="239.759" y="609.970" w="45.867" h="19.091" xs="100" hs="100" page="16" pagename="16" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901078.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901078.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p>    d. </p><figure class="informalfigure"><?docpage num="16" PAGENAME="16"?><?REGION x="240.991" y="635.674" w="26.04" h="17.787" xs="100" hs="100" page="16" pagename="16" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901079.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901079.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p>e. </p><figure class="informalfigure"><?docpage num="16" PAGENAME="16"?><?REGION x="394.634" y="610.229" w="45.52" h="19.1319999999999" xs="100" hs="100" page="16" pagename="16" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901080.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901080.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p/><figure class="informalfigure"><?docpage num="16" PAGENAME="16"?><?REGION x="393.248" y="635.764" w="58.741" h="19.797" xs="100" hs="100" page="16" pagename="16" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901081.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901081.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/>f. </figure></li></ul></div></li>

                          <?docpage num="17" PAGENAME="17"?><li class="general-problem"><div class="question" id="ch01qu63"><p><span class="pagebreak" title="17" id="page17"/>63. formal charge = number of valence electrons</p>

                          <p><span class="inlineequation">[&amp;|minus|&amp;]</span> (number of lone-pair electrons + the number of bonds)</p>

                          <p> In all four compounds, H has a single bond and is neutral and each C has four bonds and is neutral. Thus, the indicated formal charge is for O or N.</p>

                          <ul style="list-style-type:none"><li><p> a. </p><figure class="informalfigure"><?docpage num="17" PAGENAME="17"?><?REGION x="127.250" y="146.494" w="65.676" h="83.902" xs="100" hs="100" page="17" pagename="17" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901082.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901082.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li><li><p> b. </p><figure class="informalfigure"><?docpage num="17" PAGENAME="17"?><?REGION x="266.426" y="146.494" w="65.668" h="83.902" xs="100" hs="100" page="17" pagename="17" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901083.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901083.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li><li><p> c. </p><figure class="informalfigure"><?docpage num="17" PAGENAME="17"?><?REGION x="387.634" y="147.494" w="63.44" h="83.902" xs="100" hs="100" page="17" pagename="17" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901084.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901084.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li><li><p/><figure class="informalfigure"><?docpage num="17" PAGENAME="17"?><?REGION x="507.866" y="146.494" w="65.676" h="83.902" xs="100" hs="100" page="17" pagename="17" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901085.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901085.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/> d. </figure></li></ul> </div></li>

                          <li class="general-problem"><div class="question" id="ch01qu64"><p>64. The open arrow in the structures points to the shorter of the two indicated bonds in each compound.</p>

                          <ul style="list-style-type:none"><li><p> 1. </p><figure class="informalfigure"><?docpage num="17" PAGENAME="17"?><?REGION x="127.250" y="273.333" w="91.841" h="36.563" xs="100" hs="100" page="17" pagename="17" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901086.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901086.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p> 2. </p><figure class="informalfigure"><?docpage num="17" PAGENAME="17"?><?REGION x="127.250" y="334.251" w="58.971" h="59.145" xs="100" hs="100" page="17" pagename="17" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901087.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901087.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p> 3. </p><figure class="informalfigure"><?docpage num="17" PAGENAME="17"?><?REGION x="127.250" y="401.939" w="118.633" h="40.957" xs="100" hs="100" page="17" pagename="17" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901088.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901088.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p>4. </p><figure class="informalfigure"><?docpage num="17" PAGENAME="17"?><?REGION x="346.250" y="275.418" w="95.464" h="41.378" xs="100" hs="100" page="17" pagename="17" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901089.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901089.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p>5. </p><figure class="informalfigure"><?docpage num="17" PAGENAME="17"?><?REGION x="346.250" y="322.197" w="114.72" h="81.699" xs="100" hs="100" page="17" pagename="17" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901090.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901090.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p/><figure class="informalfigure"><?docpage num="17" PAGENAME="17"?><?REGION x="346.250" y="410.092" w="98.611" h="33.907" xs="100" hs="100" page="17" pagename="17" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901091.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901091.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/>6. </figure></li></ul>

                          <ul style="list-style-type:none"><li><p> For <b>1</b>, <b>2</b>, and <b>3</b>: A triple bond is shorter than a double bond, which is shorter than a single bond.</p></li>

                          <li><p> For <b>4</b> and <b>5</b>: The greater the <i>s</i> character in the hybrid orbital, the shorter is the bond formed using that orbital, because an <i>s</i> orbital is closer than a <i>p</i> orbital to the nucleus. Therefore, the bond formed by a hydrogen and an <i>sp</i> carbon is shorter than the bond formed by a hydrogen and an <span class="inlineequation">[&amp;sp^{2}&amp;]</span> carbon, which is shorter than the bond formed by a hydrogen and an <span class="inlineequation">[&amp;sp^{3}&amp;]</span> carbon. (See Table 1.7 on page 42 of the text.)</p></li>

                          <li><p> For <b>6</b>: Cl forms a bond using a <span class="inlineequation">[&amp;3sp^{3}&amp;]</span> orbital and Br forms a bond using a <span class="inlineequation">[&amp;4sp^{3}&amp;]</span> orbital. Therefore, the <span class="inlineequation">[&amp;~rom~C|bond|Cl~norm~&amp;]</span> bond is shorter.</p></li></ul> </div></li>

                          <?docpage num="18" PAGENAME="18"?><li class="general-problem"><div class="question" id="ch01qu65"><ul style="list-style-type:none"><li><p><span class="pagebreak" title="18" id="page18"/>65. a. </p><figure class="informalfigure"><?docpage num="18" PAGENAME="18"?><?REGION x="89.500" y="82.346" w="66.226" h="94.167" xs="100" hs="100" page="18" pagename="18" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901092.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901092.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p> b. </p><figure class="informalfigure"><?docpage num="18" PAGENAME="18"?><?REGION x="89.500" y="186.695" w="95.491" h="77.818" xs="100" hs="100" page="18" pagename="18" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901093.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901093.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p>c. </p><figure class="informalfigure"><?docpage num="18" PAGENAME="18"?><?REGION x="312.634" y="81.000" w="85.718" h="51.729" xs="100" hs="100" page="18" pagename="18" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901094.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901094.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure><p>

                          Each of the three carbons is <span class="inlineequation">[&amp;sp^{3}&amp;]</span> hybridized.</p>

                          <p>All the bond angles are <span class="inlineequation">[&amp;109.5|deg|.&amp;]</span></p></li>

                          <li><p/><figure class="informalfigure"><?docpage num="18" PAGENAME="18"?><?REGION x="313.866" y="169.267" w="79.162" h="65.462" xs="100" hs="100" page="18" pagename="18" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901095.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901095.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/>d. </figure><p>

                          Each of the four carbons is <span class="inlineequation">[&amp;sp^{2}&amp;]</span> hybridized.</p>

                          <p>All the bond angles are <span class="inlineequation">[&amp;120|deg|.&amp;]</span></p></li></ul></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu66"><ul style="list-style-type:none"><li><p>66. a. </p><figure class="informalfigure"><?docpage num="18" PAGENAME="18"?><?REGION x="91.250" y="280.200" w="115.705" h="92.964" xs="100" hs="100" page="18" pagename="18" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901096.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901096.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li><li><p> b. </p><figure class="informalfigure"><?docpage num="18" PAGENAME="18"?><?REGION x="233.000" y="300.850" w="108.086" h="71.932" xs="100" hs="100" page="18" pagename="18" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901097.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901097.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li><li><p/><figure class="informalfigure"><?docpage num="18" PAGENAME="18"?><?REGION x="360.000" y="280.388" w="194.677" h="92.308" xs="100" hs="100" page="18" pagename="18" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901098.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901098.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/> c. </figure></li></ul></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu67"><figure class="informalfigure"><?docpage num="18" PAGENAME="18"?><?REGION x="72.000" y="388.273" w="270.596" h="49.3049999999999" xs="100" hs="100" page="18" pagename="18" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="6658501127.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/6658501127.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/>67. </figure></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu68"><figure class="informalfigure"><?docpage num="18" PAGENAME="18"?><?REGION x="72.000" y="452.013" w="338.822" h="92.765" xs="100" hs="100" page="18" pagename="18" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901099.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901099.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/>68. </figure></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu69"><figure class="informalfigure"><?docpage num="18" PAGENAME="18"?><?REGION x="72.000" y="562.128" w="112.296" h="85.8499999999999" xs="100" hs="100" page="18" pagename="18" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901100.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901100.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/>69. </figure></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu70"><p/><figure class="informalfigure"><?docpage num="18" PAGENAME="18"?><?REGION x="72.000" y="659.448" w="357.993" h="51.73" xs="100" hs="100" page="18" pagename="18" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901101.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901101.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/>70. </figure></div></li>

                          <?docpage num="19" PAGENAME="19"?><li class="general-problem"><div class="question" id="ch01qu71"><p><span class="pagebreak" title="19" id="page19"/>71. In an alkene, six atoms are in the same plane: the two <span class="inlineequation">[&amp;sp^{2}&amp;]</span> carbons and the two atoms that are bonded to each of the two <span class="inlineequation">[&amp;sp^{2}&amp;]</span> carbons. The other atoms in the molecule will not necessarily be in the same plane with these six atoms.</p>

                          <figure class="informalfigure"><?docpage num="19" PAGENAME="19"?><?REGION x="208.261" y="123.372" w="153.247" h="102.312" xs="100" hs="100" page="19" pagename="19" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901102.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901102.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure><figure class="informalfigure"><?docpage num="19" PAGENAME="19"?><?REGION x="383.508" y="147.991" w="56.231" h="77.693" xs="100" hs="100" page="19" pagename="19" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901103.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901103.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/>  </figure>

                          <p> If you put stars next to the six atoms that lie in a plane in each molecule, you will be able to see more clearly whether the indicated atoms lie in the same plane.</p>

                          <figure class="informalfigure"><?docpage num="19" PAGENAME="19"?><?REGION x="196.426" y="276.307" w="165.017" h="70.377" xs="100" hs="100" page="19" pagename="19" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901104.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901104.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure><figure class="informalfigure"><?docpage num="19" PAGENAME="19"?><?REGION x="383.443" y="276.307" w="68.131" h="70.377" xs="100" hs="100" page="19" pagename="19" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901105.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901105.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/>  </figure> </div></li>

                          <li class="general-problem"><div class="question" id="ch01qu72"><ul style="list-style-type:none"><li><p>72. a. If the central atom is <span class="inlineequation">[&amp;sp^{3}&amp;]</span> hybridized and it does not have a lone pair, the molecule will have tetrahedral bond angles <span class="inlineequation">[&amp;|pbo|109.5|deg||pbc|.&amp;]</span> Therefore, only <span class="inlineequation">[&amp;^{|plus|}~rom~NH_{~normal~4}~norm~&amp;]</span> has tetrahedral bond angles. The following species are close to being perfectly tetrahedral: <span class="inlineequation">[&amp;~rom~H_{~normal~2}~rom~O~normal~,|sp|~rom~H_{~normal~3}~rom~ O^{|plus|}~normal~,|sp|~rom~NH_{~normal~3},|sp|^{|minus|}~rom~CH_{~normal~3}.~norm~&amp;]< /span> However, they all have bond angles slightly smaller than <span class="inlineequation">[&amp;109.5|deg|.&amp;]</span></p></li>

                          <li><p><span class="inlineequation"> b. [&amp;^{|plus|}~rom~CH_{~normal~3}~norm~&amp;]</span> and <span class="inlineequation">[&amp;~rom~BF_{~normal~3}~norm~&amp;]</span></p></li></ul></div></ li>

                          <li class="general-problem"><div class="question" id="ch01qu73"><p><span class="inlineequation">73. [&amp;~rom~CH_{~normal~3}~rom~CH_{~normal~2}~rom~Cl~norm~&amp;]</span> has the longer <span class="inlineequation">[&amp;~rom~C|bond|Cl~norm~&amp;]</span> bond because it is formed by the overlap of an <span class="inlineequation">[&amp;sp^{3}&amp;]</span> orbital of Cl with an <span class="inlineequation">[&amp;sp^{3}&amp;]</span> orbital of C, whereas the <span class="inlineequation">[&amp;~rom~C|bond|Cl~norm~&amp;]</span> bond in <span class="inlineequation">[&amp;~rom~CH_{~normal~2}|dbond|~rom~CHCl~norm~&amp;]</span> is formed by the overlap of an <span class="inlineequation">[&amp;sp^{3}&amp;]</span> orbital of Cl with an <span class="inlineequation">[&amp;sp^{2}&amp;]</span> orbital of C. (The more the <i>s</i> character, the shorter and stronger the bond.)</p></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu74"><p><span class="inlineequation">74. [&amp;~rom~CH_{~normal~2}~rom~Cl_{~normal~2}~norm~&amp;]</span> has the larger dipole moment because the two chlorines are withdrawing electrons in the same general direction, whereas in <span class="inlineequation">[&amp;~rom~CH_{~normal~3}~rom~Cl~norm~&amp;]</span> only one chlorine is withdrawing electrons.</p></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu75"><p>75. The bond angles at the triple-bonded carbons, when the bonding orbitals overlap maximally, are <span class="inlineequation">[&amp;180|deg|.&amp;]</span> A <span class="inlineequation">[&amp;180|deg|&amp;]</span> angle cannot fit into the ring structure. Therefore, the overlap between the <i>sp</i> orbital and the adjacent <span class="inlineequation">[&amp;sp^{3}&amp;]</span> orbital becomes distorted from the ideal end-on overlap. This poor overlap causes the compound to be unstable. (Compare the structure shown here with Figure 3.8 on page 122 of the text.)</p>

                          <figure class="informalfigure"><?docpage num="19" PAGENAME="19"?><?REGION x="303.217" y="599.305" w="41.566" h="72.379" xs="100" hs="100" page="19" pagename="19" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901108.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901108.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu76"><p>76. The dipole moment depends on the size of the charge and the distance between the bonded electrons. The longer <span class="inlineequation">[&amp;~rom~C|bond|Cl~norm~&amp;]</span> bond more than makes up for the greater charge on fluorine.</p></div></li>

                          <?docpage num="20" PAGENAME="20"?><li class="general-problem"><div class="question" id="ch01qu77"><ul style="list-style-type:none"><li><ul style="list-style-type:none"><li><p><span class="pagebreak" title="20" id="page20"/></p><figure class="informalfigure"><?docpage num="20" PAGENAME="20"?><?REGION x="108.000" y="81.000" w="67.215" h="50.612" xs="100" hs="100" page="20" pagename="20" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901109.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901109.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/>77. a. 1. </figure></li>

                          <li><p/><figure class="informalfigure"><?docpage num="20" PAGENAME="20"?><?REGION x="108.000" y="134.859" w="70.724" h="37.953" xs="100" hs="100" page="20" pagename="20" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901110.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901110.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/> 2. </figure></li>

                          <li><p/><figure class="informalfigure"><?docpage num="20" PAGENAME="20"?><?REGION x="278.000" y="92.000" w="52.858" h="18.365" xs="100" hs="100" page="20" pagename="20" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901111.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901111.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/> 3. </figure></li>

                          <li><p/><figure class="informalfigure"><?docpage num="20" PAGENAME="20"?><?REGION x="278.000" y="119.081" w="83.515" h="51.501" xs="100" hs="100" page="20" pagename="20" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901112.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901112.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/> 4. </figure></li>

                          <li><p/><figure class="informalfigure"><?docpage num="20" PAGENAME="20"?><?REGION x="448.000" y="93.000" w="51.557" h="17.586" xs="100" hs="100" page="20" pagename="20" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="6658501128.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/6658501128.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/> 5. </figure></li></ul></li><li><p>

                           

                           

                            b. 1. </p><ul style="list-style-type:none"><li><p/><figure class="informalfigure"><?docpage num="20" PAGENAME="20"?><?REGION x="108.000" y="187.800" w="44.852" h="21.646" xs="100" hs="100" page="20" pagename="20" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="6658501129.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/6658501129.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></li>

                          <li><p/><figure class="informalfigure"><?docpage num="20" PAGENAME="20"?><?REGION x="108.000" y="226.721" w="56.241" h="42.225" xs="100" hs="100" page="20" pagename="20" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="6658501130.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/6658501130.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/> 2. </figure></li>

                          <li><p/><figure class="informalfigure"><?docpage num="20" PAGENAME="20"?><?REGION x="278.000" y="189.800" w="50.412" h="17.053" xs="100" hs="100" page="20" pagename="20" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="6658501131.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/6658501131.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/> 3. </figure></li>

                          <li><p/><figure class="informalfigure"><?docpage num="20" PAGENAME="20"?><?REGION x="278.000" y="213.129" w="62.728" h="44.224" xs="100" hs="100" page="20" pagename="20" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="6658501132.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/6658501132.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/> 4. </figure></li>

                          <li><p/><figure class="informalfigure"><?docpage num="20" PAGENAME="20"?><?REGION x="448.000" y="187.800" w="48.08" h="29.714" xs="100" hs="100" page="20" pagename="20" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="6658501133.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/6658501133.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/> 5. </figure></li></ul></li>

                          <li><p/><figure class="informalfigure"><?docpage num="20" PAGENAME="20"?><?REGION x="89.500" y="282.808" w="13.869" h="15.038" xs="100" hs="100" page="20" pagename="20" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="6658501134.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/6658501134.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/> c. </figure></li></ul></div></li>

                          <li class="general-problem"><div class="question" id="ch01qu78"><p>78. Only the first structure has no dipole moment.</p>

                          <figure class="informalfigure"><?docpage num="20" PAGENAME="20"?><?REGION x="145.840" y="331.421" w="63.153" h="41.425" xs="100" hs="100" page="20" pagename="20" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901107.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901107.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/>     </figure><figure class="informalfigure"><?docpage num="20" PAGENAME="20"?><?REGION x="258.493" y="331.822" w="171.667" h="41.024" xs="100" hs="100" page="20" pagename="20" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901106.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901106.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure> </div></li></ul></section></section>

                          <?docpage num="21" PAGENAME="21"?><section class="backmatter"><section id="ch01probset2"><header><h1 class="title"><span class="pagebreak" title="21" id="page21"/><?REGION x="268.924" y="81.000" w="110.152" h="11.000" page="21" pagename="21"?><span class="label">Chapter </span><span class="number">1</span> Practice Test</h1></header>

                          <ul style="list-style-type:none"><li class="general-problem"><div class="question" id="ch01pt01"><p>1. Answer the following:</p>

                          <ul style="list-style-type:none"><li><p> a. Which bond has a greater dipole moment, a carbon-oxygen bond or a carbon-fluorine bond?</p></li>

                          <li><p> b. If <span class="inlineequation">[&amp;~rom~He_{~normal~2}^{|plus|}~norm~&amp;]</span> has three electrons in its molecular orbitals, how many electrons are in an antibonding molecular orbital?</p></li>

                          <li><p> c. Which is the longer bond, a carbon-hydrogen bond in ethene or a carbon-hydrogen bond in ethane?</p></li>

                          <li><p> d. Which is larger, the bond angle in water or the bond angle in ammonia?</p></li></ul></div></li>

                          <li style="list-style-type:none"><div class="question" id="ch01pt02"><p>2. What is the hybridization of the carbon atom in each of the following compounds?</p>

                          <div class="informalequation">[&amp;^{|plus|}~rom~CH_{~normal~3}|em||em|^{|minus|}~rom~CH_{~no rmal~3}|em||em||mdot|~rom~CH_{~normal~3}~norm~&amp;]</div></div></li>

                          <li class="general-problem"><div class="question" id="ch01pt03"><p>3. Draw the Lewis structure for <span class="inlineequation">[&amp;~rom~HCO_{~normal~3}^{|minus|}.~norm~&amp;]</span></p></div> </li>

                          <li class="general-problem"><div class="question" id="ch01pt04"><p>4. Circle the compounds below that have a dipole <span class="inlineequation">[&amp;~rom~moment|=|~normal~0.~norm~&amp;]</span></p>

                          <div class="informalequation">[&amp;~rom~CH_{~normal~2}~rom~Cl_{2}|em||em|CH_{~normal~3}~rom~C H_{~normal~3}|em||em|~rom~CH_{~normal~3}~rom~Cl|em||em|H_{~normal~2}~rom~C|dbond|O|em||em| CCl_{~normal~4}~norm~&amp;]</div></div></li>

                          <li class="general-problem"><div class="question" id="ch01pt05"><p>5. Which compound has greater bond angles, <span class="inlineequation">[&amp;~rom~H_{~normal~3}~rom~O^{|plus|}~norm~&amp;]</span> or <span class="inlineequation">[&amp;^{|plus|}~rom~NH_{4}~norm~&amp;]</span>?</p></div></li>

                          <li class="general-problem"><div class="question" id="ch01pt06"><p>6. Draw the structure for each of the following:</p>

                          <ul style="list-style-type:none"><li><p> a. methyl cation</p></li>

                          <li><p> b. a hydride ion</p></li>

                          <li><p> c. a bromine radical</p></li>

                          <li><p> d. an alkane with only primary carbons</p></li></ul></div></li>

                          <li class="general-problem"><div class="question" id="ch01pt07"><p>7. Draw the structure of a compound that contains five carbons, two of which are <span class="inlineequation">[&amp;sp^{2}&amp;]</span> hybridized and three of which are <span class="inlineequation">[&amp;sp^{3}&amp;]</span> hybridized.</p></div></li>

                          <li class="general-problem"><div class="question" id="ch01pt08"><p>8. What is the hybridization of each of the indicated atoms?</p>

                          <figure class="informalfigure"><?docpage num="21" PAGENAME="21"?><?REGION x="187.835" y="580.117" w="272.33" h="52.596" xs="100" hs="100" page="21" pagename="21" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901113.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901113.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></div></li>

                          <li class="general-problem"><div class="question" id="ch01pt09"><ul style="list-style-type:none"><li><p>9. a. What orbitals do carbon’s electrons occupy before promotion?</p></li>

                          <li><p> b. What orbitals do carbon’s electrons occupy after promotion and before hybridization?</p></li>

                          <li><p> c. What orbitals do carbon’s electrons occupy after hybridization?</p></li></ul></div></li>

                          <?docpage num="22" PAGENAME="22"?><li class="general-problem"><div class="question" id="ch01pt10"><p><span class="pagebreak" title="22" id="page22"/>10. Answer the following:</p>

                          <ul style="list-style-type:none"><li><p> a. What is the <span class="inlineequation">[&amp;~rom~H|bond|C|bond|O~norm~&amp;]</span> bond angle in <span class="inlineequation">[&amp;~rom~CH_{3}OH?~norm~&amp;]</span></p></li>

                          <li><p> b. What is the <span class="inlineequation">[&amp;~rom~H|bond|Be|bond|H~norm~&amp;]</span> bond angle in <span class="inlineequation">[&amp;~rom~BeH_{~normal~2}?~norm~&amp;]</span></p></li>

                          <li><p> c. What is the <span class="inlineequation">[&amp;~rom~H|bond|B|bond|H~norm~&amp;]</span> bond angle in <span class="inlineequation">[&amp;~rom~BH_{~normal~3}?~norm~&amp;]</span></p></li>

                          <li><p> d. What is the <span class="inlineequation">[&amp;~rom~C|bond|O|bond|H~norm~&amp;]</span> bond angle in <span class="inlineequation">[&amp;~rom~CH_{3}OH?~norm~&amp;]</span></p></li></ul></div></li>

                          <li class="general-problem"><div class="question" id="ch01pt11"><p>11. For each of the following compounds, indicate the hybridization of the atom to which the arrow is pointing.</p>

                          <figure class="informalfigure"><?docpage num="22" PAGENAME="22"?><?REGION x="185.777" y="196.012" w="204.446" h="44.051" xs="100" hs="100" page="22" pagename="22" filepath="/Applications/pxe_tools_1_39_69/content/9780134066585/printimages/printimages" filename="8265901114.eps"?><img alt="" width="100" height="100" data-profile-deliveryformat="print" src="../printimages/printimages/8265901114.eps" data-indesign-image-path-only="/Applications/pxe_tools_1_39_69/content/9780134066585/prin timages/printimages"/></figure></div></li>

                          <li class="general-problem"><div class="question" id="ch01pt12"><p>12. Indicate whether each of the following statements is true or false.</p>

                          <ul style="list-style-type:none"><li><p> a. A pi bond is stronger than a sigma bond. T F</p></li>

                          <li><p> b. A triple bond is shorter than a double bond. T F</p></li>

                          <li><p> c. The oxygen-hydrogen bonds in water are formed by the overlap of an <span class="inlineequation">[&amp;sp^{2}&amp;]</span> orbital of oxygen with an <i>s</i> orbital of hydrogen. T F</p></li>

                          <li><p> d. A double bond is stronger than a single bond. T F</p></li>

                          <li><p> e. A tetrahedral carbon has bond angles of <span class="inlineequation">[&amp;107.5|deg|.&amp;]</span> T F</p><p>ANSWERS TO ALL THE PRACTICE TESTS CAN BE FOUND AT THE END OF THE SOLUTIONS MANUAL.</p></li></ul></div></li></ul></section></section></section></body></html>