{"id":2037,"date":"2015-04-29T16:05:29","date_gmt":"2015-04-29T16:05:29","guid":{"rendered":"https:\/\/courses.candelalearning.com\/oschemtemp\/?post_type=chapter&#038;p=2037"},"modified":"2016-10-26T19:17:58","modified_gmt":"2016-10-26T19:17:58","slug":"multiple-bonds-2","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/chapter\/multiple-bonds-2\/","title":{"raw":"Multiple Bonds","rendered":"Multiple Bonds"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\nBy the end of this section, you will be able to:\r\n<ul>\r\n \t<li>Describe multiple covalent bonding in terms of atomic orbital overlap<\/li>\r\n \t<li>Relate the concept of resonance to \u03c0-bonding and electron delocalization<\/li>\r\n<\/ul>\r\n<\/div>\r\nThe hybrid orbital model appears to account well for the geometry of molecules involving single covalent bonds. Is it also capable of describing molecules containing double and triple bonds? We have already discussed that multiple bonds consist of \u03c3 and \u03c0 bonds. Next we can consider how we visualize these components and how they relate to hybrid orbitals. The Lewis structure of ethene, C<sub>2<\/sub>H<sub>4<\/sub>, shows us that each carbon atom is surrounded by one other carbon atom and two hydrogen atoms.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211902\/CNX_Chem_08_03_C4H4Lewis_img1.jpg\" alt=\"A Lewis structure is shown in which two carbon atoms are bonded together by a double bond. Each carbon atom is bonded to two hydrogen atoms by a single bond.\" data-media-type=\"image\/jpeg\" \/>\r\n\r\nThe three bonding regions form a trigonal planar electron-pair geometry. Thus we expect the \u03c3 bonds from each carbon atom are formed using a set of <em>sp<\/em><sup>2<\/sup> hybrid orbitals that result from hybridization of two of the 2<em>p<\/em> orbitals and the 2<em>s<\/em> orbital (Figure 1).\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"880\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211903\/CNX_Chem_08_03_sp3config1.jpg\" alt=\"A diagram is shown in two parts, connected by a right facing arrow labeled, \u201cHybridization.\u201d The left diagram shows an up-facing arrow labeled, \u201cE.\u201d To the lower right of the arrow is a short, horizontal line labeled, \u201c2 s,\u201d that has two vertical half-arrows facing up and down on it. To the upper right of the arrow are a series of three short, horizontal lines labeled, \u201c2 p.\u201d Above both sets of lines is the phrase, \u201cOrbitals in an isolated C atom.\u201d Two of the lines have vertical, up-facing arrows drawn on them. The right side of the diagram shows three short, horizontal lines placed halfway up the space and each labeled, \u201cs p superscript 2.\u201d An upward-facing half arrow is drawn vertically on each line. Above these lines is one other short, horizontal line, labeled, \u201cp.\u201d Above both sets of lines is the phrase, \u201cOrbitals in the s p superscript 2 hybridized C atom in C subscript 2 H subscript 4.\u201d\" width=\"880\" height=\"272\" data-media-type=\"image\/jpeg\" \/> Figure 1. In ethene, each carbon atom is <em>sp<\/em><sup>2<\/sup> hybridized, and the <em>sp<\/em><sup>2<\/sup> orbitals and the <em>p<\/em> orbital are singly occupied. The hybrid orbitals overlap to form \u03c3 bonds, while the <em>p<\/em> orbitals on each carbon atom overlap to form a \u03c0 bond.[\/caption]\r\n\r\nThese orbitals form the C\u2013H single bonds and the \u03c3 bond in the [latex]\\text{C}=\\text{C}[\/latex] double bond (Figure 2)<em data-effect=\"italics\">.<\/em> The \u03c0 bond in the [latex]\\text{C}=\\text{C}[\/latex] double bond results from the overlap of the third (remaining) 2<em data-effect=\"italics\">p<\/em> orbital on each carbon atom that is not involved in hybridization. This unhybridized <em data-effect=\"italics\">p<\/em> orbital (shown in red in Figure 2) is perpendicular to the plane of the <em data-effect=\"italics\">sp<\/em><sup>2<\/sup> hybrid orbitals. Thus the unhybridized 2<em data-effect=\"italics\">p<\/em> orbitals overlap in a side-by-side fashion, above and below the internuclear axis (Figure 2) and form a \u03c0 bond<em data-effect=\"italics\">.<\/em>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"881\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211905\/CNX_Chem_08_03_C2H4orbit1.jpg\" alt=\"Two diagrams are shown labeled, \u201ca\u201d and \u201cb.\u201d Diagram a shows two carbon atoms with three purple balloon-like orbitals arranged in a plane around them and two red balloon-like orbitals arranged vertically and perpendicularly to the plane. There is an overlap of two of the purple orbitals in between the two carbon atoms, and the other four purple orbitals that face the outside of the molecule are shown interacting with spherical blue orbitals from four hydrogen atoms. Diagram b depicts a similar image to diagram a, but the red, vertical orbitals are interacting above and below the plane of the molecule to form two areas labeled, \u201cOne pi bond.\u201d\" width=\"881\" height=\"282\" data-media-type=\"image\/jpeg\" \/> Figure 2. In the ethene molecule, C<sub>2<\/sub>H<sub>4<\/sub>, there are (a) five \u03c3 bonds shown in purple. One C\u2013C \u03c3 bond results from overlap of <em>sp<\/em><sup>2<\/sup> hybrid orbitals on the carbon atom with one <em>sp<\/em><sup>2<\/sup> hybrid orbital on the other carbon atom. Four C\u2013H bonds result from the overlap between the <em>sp<\/em><sup>2<\/sup> orbitals with <em>s<\/em> orbitals on the hydrogen atoms. (b) The \u03c0 bond is formed by the side-by-side overlap of the two unhybridized <em>p<\/em> orbitals in the two carbon atoms, which are shown in red. The two lobes of the \u03c0 bond are above and below the plane of the \u03c3 system.[\/caption]\r\n\r\nIn an ethene molecule, the four hydrogen atoms and the two carbon atoms are all in the same plane. If the two planes of <em>sp<\/em><sup>2<\/sup> hybrid orbitals tilted relative to each other, the <em>p<\/em> orbitals would not be oriented to overlap efficiently to create the \u03c0 bond. The planar configuration for the ethene molecule occurs because it is the most stable bonding arrangement. This is a significant difference between \u03c3 and \u03c0 bonds; rotation around single (\u03c3) bonds occurs easily because the end-to-end orbital overlap does not depend on the relative orientation of the orbitals on each atom in the bond. In other words, rotation around the internuclear axis does not change the extent to which the \u03c3 bonding orbitals overlap because the bonding electron density is symmetric about the axis. Rotation about the internuclear axis is much more difficult for multiple bonds; however, this would drastically alter the off-axis overlap of the \u03c0 bonding orbitals, essentially breaking the \u03c0 bond.\r\n\r\nIn molecules with <em>sp<\/em> hybrid orbitals, two unhybridized <em>p<\/em> orbitals remain on the atom (Figure 3). We find this situation in acetylene, [latex]\\text{H}-\\text{C}\\equiv \\text{C}-\\text{H},[\/latex] which is a linear molecule.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"650\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211906\/CNX_Chem_08_03_spC1.jpg\" alt=\"A diagram of a carbon atom with two balloon-like purple orbitals labeled, \u201csp\u201d arranged in a linear fashion around it is shown. Four red balloon-like orbitals are aligned in pairs in the y and z axes around the carbon and are labeled, \u201cunhybridized p orbital,\u201d and, \u201cSecond unhybridized p orbital.\u201d\" width=\"650\" height=\"312\" data-media-type=\"image\/jpeg\" \/> Figure 3. Diagram of the two linear <em>sp<\/em> hybrid orbitals of a carbon atom, which lie in a straight line, and the two unhybridized <em>p<\/em> orbitals at perpendicular angles.[\/caption]\r\n\r\nThe <em>sp<\/em> hybrid orbitals of the two carbon atoms overlap end to end to form a \u03c3 bond between the carbon atoms (Figure 4). The remaining <em>sp<\/em> orbitals form \u03c3 bonds with hydrogen atoms. The two unhybridized <em>p<\/em> orbitals per carbon are positioned such that they overlap side by side and, hence, form two \u03c0 bonds. The two carbon atoms of acetylene are thus bound together by one \u03c3 bond and two \u03c0 bonds, giving a triple bond.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"881\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211908\/CNX_Chem_08_03_C2H21.jpg\" alt=\"Two diagrams are shown and labeled, \u201ca\u201d and \u201cb.\u201d Diagram a shows two carbon atoms with two purple balloon-like orbitals arranged in a plane around each of them, and four red balloon-like orbitals arranged along the y and z axes perpendicular to the plane of the molecule. There is an overlap of two of the purple orbitals in between the two carbon atoms. The other two purple orbitals that face the outside of the molecule are shown interacting with spherical blue orbitals from two hydrogen atoms. Diagram b depicts a similar image to diagram a, but the red, vertical orbitals are interacting above and below and to the front and back of the plane of the molecule to form two areas labeled, \u201cOne pi bond,\u201d and, \u201cSecond pi bond,\u201d each respectively.\" width=\"881\" height=\"282\" data-media-type=\"image\/jpeg\" \/> Figure 4. (a) In the acetylene molecule, C<sub>2<\/sub>H<sub>2<\/sub>, there are two C\u2013H \u03c3 bonds and a C \u2261 C triple bond involving one C\u2013C \u03c3 bond and two C\u2013C \u03c0 bonds. The dashed lines, each connecting two lobes, indicate the side-by-side overlap of the four unhybridized p orbitals. (b) This shows the overall outline of the bonds in C<sub>2<\/sub>H<sub>2<\/sub>. The two lobes of each of the \u03c0 bonds are positioned across from each other around the line of the C\u2013C \u03c3 bond.[\/caption]\r\n\r\nHybridization involves only \u03c3 bonds, lone pairs of electrons, and single unpaired electrons (radicals). Structures that account for these features describe the correct hybridization of the atoms. However, many structures also include resonance forms. Remember that resonance forms occur when various arrangements of \u03c0 bonds are possible. Since the arrangement of \u03c0 bonds involves only the unhybridized orbitals, resonance does not influence the assignment of hybridization.\r\n\r\nFor example, molecule benzene has two resonance forms (Figure 5). We can use either of these forms to determine that each of the carbon atoms is bonded to three other atoms with no lone pairs, so the correct hybridization is <em>sp<\/em><sup>2<\/sup>. The electrons in the unhybridized <em>p<\/em> orbitals form \u03c0 bonds. Neither resonance structure completely describes the electrons in the \u03c0 bonds. They are not located in one position or the other, but in reality are delocalized throughout the ring. Valence bond theory does not easily address delocalization. Bonding in molecules with resonance forms is better described by molecular orbital theory.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"650\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211909\/CNX_Chem_08_03_C6H61.jpg\" alt=\"A diagram is shown that is made up of two Lewis structures connected by a double ended arrow. The left image shows six carbon atoms bonded together with alternating double and single bonds to form a six-sided ring. Each carbon is also bonded to a hydrogen atom by a single bond. The right image shows the same structure, but the double and single bonds in between the carbon atoms have changed positions.\" width=\"650\" height=\"256\" data-media-type=\"image\/jpeg\" \/> Figure 5. Each carbon atom in benzene, C<sub>6<\/sub>H<sub>6<\/sub>, is <em>sp<\/em><sup>2<\/sup> hybridized, independently of which resonance form is considered. The electrons in the \u03c0 bonds are not located in one set of p orbitals or the other, but rather delocalized throughout the molecule.[\/caption]\r\n\r\n<div class=\"textbox examples\">\r\n<h3>Example 1:\u00a0Assignment of Hybridization Involving Resonance<\/h3>\r\nSome acid rain results from the reaction of sulfur dioxide with atmospheric water vapor, followed by the formation of sulfuric acid. Sulfur dioxide, SO<sub>2<\/sub>, is a major component of volcanic gases as well as a product of the combustion of sulfur-containing coal. What is the hybridization of the S atom in SO<sub>2<\/sub>?\r\n\r\n[reveal-answer q=\"133420\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"133420\"]\r\n\r\nThe resonance structures of SO<sub>2<\/sub> are <img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211910\/CNX_Chem_08_03_SO2_img1.jpg\" alt=\"Two Lewis structures connected by a double-ended arrow are shown. The left structure shows a sulfur atom with one lone pair of electrons and a positive sign which is single bonded on one side to an oxygen atom with three lone pairs of electrons and a negative sign. The sulfur atom is double bonded on the other side to another oxygen atom with two lone pairs of electrons. The right-hand structure is the same as the left except that the position of the double bonded oxygen atom is switched. In both structures the attached oxygen atoms form an acute angle in terms of the sulfur atom.\" data-media-type=\"image\/jpeg\" \/>\r\n\r\nThe sulfur atom is surrounded by two bonds and one lone pair of electrons in either resonance structure. Therefore, the electron-pair geometry is trigonal planar, and the hybridization of the sulfur atom is <em>sp<\/em><sup>2<\/sup>.\r\n\r\n[\/hidden-answer]\r\n<h4>Check Your Learning<\/h4>\r\nAnother acid in acid rain is nitric acid, HNO<sub>3<\/sub>, which is produced by the reaction of nitrogen dioxide, NO<sub>2<\/sub>, with atmospheric water vapor. What is the hybridization of the nitrogen atom in NO<sub>2<\/sub>? (Note: the lone electron on nitrogen occupies a hybridized orbital just as a lone pair would.)\r\n\r\n[reveal-answer q=\"333632\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"333632\"]<em>sp<\/em><sup>2<\/sup>[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Key Concepts and Summary<\/h3>\r\nMultiple bonds consist of a \u03c3 bond located along the axis between two atoms and one or two \u03c0 bonds. The \u03c3 bonds are usually formed by the overlap of hybridized atomic orbitals, while the \u03c0 bonds are formed by the side-by-side overlap of unhybridized orbitals. Resonance occurs when there are multiple unhybridized orbitals with the appropriate alignment to overlap, so the placement of \u03c0 bonds can vary.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Exercises<\/h3>\r\n<ol>\r\n \t<li>The bond energy of a C\u2013C single bond averages 347 kJ mol<sup>-1<\/sup>; that of a [latex]\\text{C}\\equiv \\text{C}[\/latex] triple bond averages 839 kJ mol<sup>-1<\/sup>. Explain why the triple bond is not three times as strong as a single bond.<\/li>\r\n \t<li>For the carbonate ion, [latex]{\\text{CO}}_{3}^{2-},[\/latex] draw all of the resonance structures. Identify which orbitals overlap to create each bond.<\/li>\r\n \t<li>A useful solvent that will dissolve salts as well as organic compounds is the compound acetonitrile, H<sub>3<\/sub>CCN. It is present in paint strippers.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>Write the Lewis structure for acetonitrile, and indicate the direction of the dipole moment in the molecule.<\/li>\r\n \t<li>Identify the hybrid orbitals used by the carbon atoms in the molecule to form \u03c3 bonds.<\/li>\r\n \t<li>Describe the atomic orbitals that form the \u03c0 bonds in the molecule. Note that it is not necessary to hybridize the nitrogen atom.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>For the molecule allene, [latex]{\\text{H}}_{2}\\text{C}=\\text{C}={\\text{CH}}_{2},[\/latex] give the hybridization of each carbon atom. Will the hydrogen atoms be in the same plane or perpendicular planes?<\/li>\r\n \t<li>Identify the hybridization of the central atom in each of the following molecules and ions that contain multiple bonds:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>ClNO (N is the central atom)<\/li>\r\n \t<li>CS<sub>2\u00a0<\/sub><\/li>\r\n \t<li>Cl<sub>2<\/sub>CO (C is the central atom)<\/li>\r\n \t<li>Cl<sub>2<\/sub>SO (S is the central atom)<\/li>\r\n \t<li>SO<sub>2<\/sub>F<sub>2<\/sub> (S is the central atom)<\/li>\r\n \t<li>XeO<sub>2<\/sub>F<sub>2<\/sub> (Xe is the central atom)<\/li>\r\n \t<li>[latex]{\\text{ClOF}}_{2}^{+}[\/latex] (Cl is the central atom)<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Describe the molecular geometry and hybridization of the N, P, or S atoms in each of the following compounds. (a) H<sub>3<\/sub>PO<sub>4<\/sub>, phosphoric acid, used in cola soft drinks (b) NH<sub>4<\/sub>NO<sub>3<\/sub>, ammonium nitrate, a fertilizer and explosive (c) S<sub>2<\/sub>Cl<sub>2<\/sub>, disulfur dichloride, used in vulcanizing rubber (d) K<sub>4<\/sub>[O<sub>3<\/sub>POPO<sub>3<\/sub>], potassium pyrophosphate, an ingredient in some toothpastes<\/li>\r\n \t<li>For each of the following molecules, indicate the hybridization requested and whether or not the electrons will be delocalized:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>ozone (O<sub>3<\/sub>) central O hybridization<\/li>\r\n \t<li>carbon dioxide (CO<sub>2<\/sub>) central C hybridization<\/li>\r\n \t<li>nitrogen dioxide (NO<sub>2<\/sub>) central N hybridization<\/li>\r\n \t<li>phosphate ion [latex]\\left({\\text{PO}}_{4}^{3-}\\right)[\/latex] central P hybridization<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>For each of the following structures, determine the hybridization requested and whether the electrons will be delocalized:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>Hybridization of each carbon\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211922\/CNX_Chem_08_03_HybridCarb_img1.jpg\" alt=\"A Lewis structure is shown in which a carbon atom is single bonded to three hydrogen atoms and a second carbon atom. This second carbon atom is, in turn, double bonded to an oxygen atom with two lone pairs of electrons. The second carbon atom is also single bonded to another carbon atom that is single bonded to three hydrogen atoms.\" data-media-type=\"image\/jpeg\" \/><\/li>\r\n \t<li>Hybridization of sulfur\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211923\/CNX_Chem_08_03_HybridSulf_img1.jpg\" alt=\"A Lewis structure is shown in which a sulfur atom with two lone pairs of electrons and a positive sign is double bonded to an oxygen with two lone pairs of electrons. The sulfur atom is also single bonded to an oxygen with three lone pairs of electrons with a negative sign. It is drawn in an angular shape.\" data-media-type=\"image\/jpeg\" \/><\/li>\r\n \t<li>All atoms\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211925\/CNX_Chem_08_03_HybridAll_img1.jpg\" alt=\"A Lewis structure is shown in which a hexagonal ring structure is made up of five carbon atoms and one nitrogen atom with a lone pair of electrons. There are alternating double and single bonds in between each carbon atom. Each carbon atom is also single bonded to one hydrogen atom.\" data-media-type=\"image\/jpeg\" \/><\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Draw the orbital diagram for carbon in CO<sub>2<\/sub> showing how many carbon atom electrons are in each orbital.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"871433\"]Show Selected Answers[\/reveal-answer]\r\n[hidden-answer a=\"871433\"]\r\n\r\n1.\u00a0A triple bond consists of one \u03c3 bond and two \u03c0 bonds. A \u03c3 bond is stronger than a \u03c0 bond due to greater overlap.\r\n\r\n3. The answers are as follows:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211913\/CNX_Chem_08_03_Acetonitri_img1.jpg\" alt=\"A Lewis structure is shown in which a carbon atom is attached by single bonds to three hydrogen atoms. It is also attached by a single bond to a carbon atom that is triple bonded to a nitrogen atom with one lone electron pair. Below the structure is a right facing arrow with its head near the nitrogen and its tail, which looks like a plus sign, located near the carbon atoms. The arrow is labeled, \u201cdipole moment.\u201d\" data-media-type=\"image\/jpeg\" \/><\/li>\r\n \t<li>The terminal carbon atom uses <em>sp<\/em><sup>3<\/sup> hybrid orbitals, while the central carbon atom is <em>sp<\/em> hybridized. The terminal carbon atom forms four \u03c3 bonds (three to the hydrogen atoms and one to the carbon) while the central carbon forms two \u03c3 bond (one to carbon and one to nitrogen).<\/li>\r\n \t<li>Each of the two \u03c0 bonds is formed by overlap of a 2<em>p<\/em> orbital on carbon and a nitrogen 2<em>p<\/em> orbital.<\/li>\r\n<\/ol>\r\n5. The hybridization of each central atom is as follows:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li><em>sp<\/em><sup>2<\/sup><\/li>\r\n \t<li><em>sp<\/em><\/li>\r\n \t<li><em>sp<\/em><sup>2<\/sup><\/li>\r\n \t<li><em>sp<\/em><sup>3<\/sup><\/li>\r\n \t<li><em>sp<\/em><sup>3<\/sup><\/li>\r\n \t<li><em>sp<\/em><sup>3<\/sup><em>d<\/em><\/li>\r\n \t<li><em>sp<\/em><sup>3<\/sup><\/li>\r\n<\/ol>\r\n7. The answers are as follows:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li><em>sp<\/em><sup>2<\/sup>, delocalized\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211917\/CNX_Chem_08_03_ozone_img1.jpg\" alt=\"A pair of Lewis structures, connected by a double ended arrow, are shown. Each structure shows three oxygen atoms which are bonded together. In the first structure, the left oxygen, which has two lone electron pairs, is connected to the central oxygen which has one lone electron pair and a plus sign. The central oxygen is connected by a single bond to the right oxygen which has three lone electron pairs and a negative sign. The second structure is a mirror image of the first. Each structure has an angular shape.\" data-media-type=\"image\/jpeg\" \/><\/li>\r\n \t<li><em>sp<\/em>, localized\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211918\/CNX_Chem_08_01_CO2LewStr1_img3.jpg\" alt=\"A Lewis structure is shown in which a carbon atom is double bonded to an oxygen atom on each side. Each oxygen atom has two lone pairs of electrons.\" data-media-type=\"image\/jpeg\" \/><\/li>\r\n \t<li><em>sp<\/em><sup>2<\/sup>, delocalized\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211919\/CNX_Chem_08_03_NO2_img1.jpg\" alt=\"A pair of Lewis structures connected by a double ended arrow are shown. The first structure shows a nitrogen atom with a single lone electron pair and a positive sign that is singly bonded on the left to an oxygen. This oxygen atom has three lone electron pairs and a negative sign. The nitrogen atom is also double bonded to an oxygen atom on the right which has two lone electron pairs. The second image is a mirror image of the first and both are drawn in an angular shape.\" data-media-type=\"image\/jpeg\" \/><\/li>\r\n \t<li><em>sp<\/em><sup>3<\/sup>, delocalized\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211921\/CNX_Chem_08_03_PO43_img1.jpg\" alt=\"A Lewis structure is shown in which a phosphorus atom is single bonded to four oxygen atoms, each of which has three lone electron pairs and a negative sign.\" data-media-type=\"image\/jpeg\" \/><\/li>\r\n<\/ol>\r\n9.\u00a0Each of the four electrons is in a separate orbital and overlaps with an electron on an oxygen atom.\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211926\/CNX_Chem_08_02_CO2Diag1.jpg\" alt=\"A diagram is shown in two parts, connected by a right facing arrow labeled, \u201cHybridization.\u201d The left diagram shows an up-facing arrow labeled, \u201cE.\u201d To the lower right of the arrow is a short, horizontal line labeled, \u201c2 s,\u201d that has two vertical half-arrows facing up and down on it. To the upper right of the arrow are a series of three short, horizontal lines labeled, \u201c2 p.\u201d Above both sets of lines is the phrase, \u201cOrbitals in an isolated C atom.\u201d There are two upward facing arrows on two of these lines. The right side of the diagram shows two short, horizontal lines placed halfway up the space and each labeled, \u201cs p.\u201d An upward-facing half arrow is drawn vertically on each line. Above these lines are two other short, horizontal lines, each labeled, \u201c2 p,\u201d and which have two upward facing arrows on them. Above both sets of lines is the phrase, \u201cOrbitals in the s p hybridized C in C O subscript 2.\u201d\" width=\"880\" height=\"272\" data-media-type=\"image\/jpeg\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<p>By the end of this section, you will be able to:<\/p>\n<ul>\n<li>Describe multiple covalent bonding in terms of atomic orbital overlap<\/li>\n<li>Relate the concept of resonance to \u03c0-bonding and electron delocalization<\/li>\n<\/ul>\n<\/div>\n<p>The hybrid orbital model appears to account well for the geometry of molecules involving single covalent bonds. Is it also capable of describing molecules containing double and triple bonds? We have already discussed that multiple bonds consist of \u03c3 and \u03c0 bonds. Next we can consider how we visualize these components and how they relate to hybrid orbitals. The Lewis structure of ethene, C<sub>2<\/sub>H<sub>4<\/sub>, shows us that each carbon atom is surrounded by one other carbon atom and two hydrogen atoms.<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211902\/CNX_Chem_08_03_C4H4Lewis_img1.jpg\" alt=\"A Lewis structure is shown in which two carbon atoms are bonded together by a double bond. Each carbon atom is bonded to two hydrogen atoms by a single bond.\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p>The three bonding regions form a trigonal planar electron-pair geometry. Thus we expect the \u03c3 bonds from each carbon atom are formed using a set of <em>sp<\/em><sup>2<\/sup> hybrid orbitals that result from hybridization of two of the 2<em>p<\/em> orbitals and the 2<em>s<\/em> orbital (Figure 1).<\/p>\n<div style=\"width: 890px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211903\/CNX_Chem_08_03_sp3config1.jpg\" alt=\"A diagram is shown in two parts, connected by a right facing arrow labeled, \u201cHybridization.\u201d The left diagram shows an up-facing arrow labeled, \u201cE.\u201d To the lower right of the arrow is a short, horizontal line labeled, \u201c2 s,\u201d that has two vertical half-arrows facing up and down on it. To the upper right of the arrow are a series of three short, horizontal lines labeled, \u201c2 p.\u201d Above both sets of lines is the phrase, \u201cOrbitals in an isolated C atom.\u201d Two of the lines have vertical, up-facing arrows drawn on them. The right side of the diagram shows three short, horizontal lines placed halfway up the space and each labeled, \u201cs p superscript 2.\u201d An upward-facing half arrow is drawn vertically on each line. Above these lines is one other short, horizontal line, labeled, \u201cp.\u201d Above both sets of lines is the phrase, \u201cOrbitals in the s p superscript 2 hybridized C atom in C subscript 2 H subscript 4.\u201d\" width=\"880\" height=\"272\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. In ethene, each carbon atom is <em>sp<\/em><sup>2<\/sup> hybridized, and the <em>sp<\/em><sup>2<\/sup> orbitals and the <em>p<\/em> orbital are singly occupied. The hybrid orbitals overlap to form \u03c3 bonds, while the <em>p<\/em> orbitals on each carbon atom overlap to form a \u03c0 bond.<\/p>\n<\/div>\n<p>These orbitals form the C\u2013H single bonds and the \u03c3 bond in the [latex]\\text{C}=\\text{C}[\/latex] double bond (Figure 2)<em data-effect=\"italics\">.<\/em> The \u03c0 bond in the [latex]\\text{C}=\\text{C}[\/latex] double bond results from the overlap of the third (remaining) 2<em data-effect=\"italics\">p<\/em> orbital on each carbon atom that is not involved in hybridization. This unhybridized <em data-effect=\"italics\">p<\/em> orbital (shown in red in Figure 2) is perpendicular to the plane of the <em data-effect=\"italics\">sp<\/em><sup>2<\/sup> hybrid orbitals. Thus the unhybridized 2<em data-effect=\"italics\">p<\/em> orbitals overlap in a side-by-side fashion, above and below the internuclear axis (Figure 2) and form a \u03c0 bond<em data-effect=\"italics\">.<\/em><\/p>\n<div style=\"width: 891px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211905\/CNX_Chem_08_03_C2H4orbit1.jpg\" alt=\"Two diagrams are shown labeled, \u201ca\u201d and \u201cb.\u201d Diagram a shows two carbon atoms with three purple balloon-like orbitals arranged in a plane around them and two red balloon-like orbitals arranged vertically and perpendicularly to the plane. There is an overlap of two of the purple orbitals in between the two carbon atoms, and the other four purple orbitals that face the outside of the molecule are shown interacting with spherical blue orbitals from four hydrogen atoms. Diagram b depicts a similar image to diagram a, but the red, vertical orbitals are interacting above and below the plane of the molecule to form two areas labeled, \u201cOne pi bond.\u201d\" width=\"881\" height=\"282\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. In the ethene molecule, C<sub>2<\/sub>H<sub>4<\/sub>, there are (a) five \u03c3 bonds shown in purple. One C\u2013C \u03c3 bond results from overlap of <em>sp<\/em><sup>2<\/sup> hybrid orbitals on the carbon atom with one <em>sp<\/em><sup>2<\/sup> hybrid orbital on the other carbon atom. Four C\u2013H bonds result from the overlap between the <em>sp<\/em><sup>2<\/sup> orbitals with <em>s<\/em> orbitals on the hydrogen atoms. (b) The \u03c0 bond is formed by the side-by-side overlap of the two unhybridized <em>p<\/em> orbitals in the two carbon atoms, which are shown in red. The two lobes of the \u03c0 bond are above and below the plane of the \u03c3 system.<\/p>\n<\/div>\n<p>In an ethene molecule, the four hydrogen atoms and the two carbon atoms are all in the same plane. If the two planes of <em>sp<\/em><sup>2<\/sup> hybrid orbitals tilted relative to each other, the <em>p<\/em> orbitals would not be oriented to overlap efficiently to create the \u03c0 bond. The planar configuration for the ethene molecule occurs because it is the most stable bonding arrangement. This is a significant difference between \u03c3 and \u03c0 bonds; rotation around single (\u03c3) bonds occurs easily because the end-to-end orbital overlap does not depend on the relative orientation of the orbitals on each atom in the bond. In other words, rotation around the internuclear axis does not change the extent to which the \u03c3 bonding orbitals overlap because the bonding electron density is symmetric about the axis. Rotation about the internuclear axis is much more difficult for multiple bonds; however, this would drastically alter the off-axis overlap of the \u03c0 bonding orbitals, essentially breaking the \u03c0 bond.<\/p>\n<p>In molecules with <em>sp<\/em> hybrid orbitals, two unhybridized <em>p<\/em> orbitals remain on the atom (Figure 3). We find this situation in acetylene, [latex]\\text{H}-\\text{C}\\equiv \\text{C}-\\text{H},[\/latex] which is a linear molecule.<\/p>\n<div style=\"width: 660px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211906\/CNX_Chem_08_03_spC1.jpg\" alt=\"A diagram of a carbon atom with two balloon-like purple orbitals labeled, \u201csp\u201d arranged in a linear fashion around it is shown. Four red balloon-like orbitals are aligned in pairs in the y and z axes around the carbon and are labeled, \u201cunhybridized p orbital,\u201d and, \u201cSecond unhybridized p orbital.\u201d\" width=\"650\" height=\"312\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 3. Diagram of the two linear <em>sp<\/em> hybrid orbitals of a carbon atom, which lie in a straight line, and the two unhybridized <em>p<\/em> orbitals at perpendicular angles.<\/p>\n<\/div>\n<p>The <em>sp<\/em> hybrid orbitals of the two carbon atoms overlap end to end to form a \u03c3 bond between the carbon atoms (Figure 4). The remaining <em>sp<\/em> orbitals form \u03c3 bonds with hydrogen atoms. The two unhybridized <em>p<\/em> orbitals per carbon are positioned such that they overlap side by side and, hence, form two \u03c0 bonds. The two carbon atoms of acetylene are thus bound together by one \u03c3 bond and two \u03c0 bonds, giving a triple bond.<\/p>\n<div style=\"width: 891px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211908\/CNX_Chem_08_03_C2H21.jpg\" alt=\"Two diagrams are shown and labeled, \u201ca\u201d and \u201cb.\u201d Diagram a shows two carbon atoms with two purple balloon-like orbitals arranged in a plane around each of them, and four red balloon-like orbitals arranged along the y and z axes perpendicular to the plane of the molecule. There is an overlap of two of the purple orbitals in between the two carbon atoms. The other two purple orbitals that face the outside of the molecule are shown interacting with spherical blue orbitals from two hydrogen atoms. Diagram b depicts a similar image to diagram a, but the red, vertical orbitals are interacting above and below and to the front and back of the plane of the molecule to form two areas labeled, \u201cOne pi bond,\u201d and, \u201cSecond pi bond,\u201d each respectively.\" width=\"881\" height=\"282\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 4. (a) In the acetylene molecule, C<sub>2<\/sub>H<sub>2<\/sub>, there are two C\u2013H \u03c3 bonds and a C \u2261 C triple bond involving one C\u2013C \u03c3 bond and two C\u2013C \u03c0 bonds. The dashed lines, each connecting two lobes, indicate the side-by-side overlap of the four unhybridized p orbitals. (b) This shows the overall outline of the bonds in C<sub>2<\/sub>H<sub>2<\/sub>. The two lobes of each of the \u03c0 bonds are positioned across from each other around the line of the C\u2013C \u03c3 bond.<\/p>\n<\/div>\n<p>Hybridization involves only \u03c3 bonds, lone pairs of electrons, and single unpaired electrons (radicals). Structures that account for these features describe the correct hybridization of the atoms. However, many structures also include resonance forms. Remember that resonance forms occur when various arrangements of \u03c0 bonds are possible. Since the arrangement of \u03c0 bonds involves only the unhybridized orbitals, resonance does not influence the assignment of hybridization.<\/p>\n<p>For example, molecule benzene has two resonance forms (Figure 5). We can use either of these forms to determine that each of the carbon atoms is bonded to three other atoms with no lone pairs, so the correct hybridization is <em>sp<\/em><sup>2<\/sup>. The electrons in the unhybridized <em>p<\/em> orbitals form \u03c0 bonds. Neither resonance structure completely describes the electrons in the \u03c0 bonds. They are not located in one position or the other, but in reality are delocalized throughout the ring. Valence bond theory does not easily address delocalization. Bonding in molecules with resonance forms is better described by molecular orbital theory.<\/p>\n<div style=\"width: 660px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211909\/CNX_Chem_08_03_C6H61.jpg\" alt=\"A diagram is shown that is made up of two Lewis structures connected by a double ended arrow. The left image shows six carbon atoms bonded together with alternating double and single bonds to form a six-sided ring. Each carbon is also bonded to a hydrogen atom by a single bond. The right image shows the same structure, but the double and single bonds in between the carbon atoms have changed positions.\" width=\"650\" height=\"256\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 5. Each carbon atom in benzene, C<sub>6<\/sub>H<sub>6<\/sub>, is <em>sp<\/em><sup>2<\/sup> hybridized, independently of which resonance form is considered. The electrons in the \u03c0 bonds are not located in one set of p orbitals or the other, but rather delocalized throughout the molecule.<\/p>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 1:\u00a0Assignment of Hybridization Involving Resonance<\/h3>\n<p>Some acid rain results from the reaction of sulfur dioxide with atmospheric water vapor, followed by the formation of sulfuric acid. Sulfur dioxide, SO<sub>2<\/sub>, is a major component of volcanic gases as well as a product of the combustion of sulfur-containing coal. What is the hybridization of the S atom in SO<sub>2<\/sub>?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q133420\">Show Answer<\/span><\/p>\n<div id=\"q133420\" class=\"hidden-answer\" style=\"display: none\">\n<p>The resonance structures of SO<sub>2<\/sub> are <img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211910\/CNX_Chem_08_03_SO2_img1.jpg\" alt=\"Two Lewis structures connected by a double-ended arrow are shown. The left structure shows a sulfur atom with one lone pair of electrons and a positive sign which is single bonded on one side to an oxygen atom with three lone pairs of electrons and a negative sign. The sulfur atom is double bonded on the other side to another oxygen atom with two lone pairs of electrons. The right-hand structure is the same as the left except that the position of the double bonded oxygen atom is switched. In both structures the attached oxygen atoms form an acute angle in terms of the sulfur atom.\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p>The sulfur atom is surrounded by two bonds and one lone pair of electrons in either resonance structure. Therefore, the electron-pair geometry is trigonal planar, and the hybridization of the sulfur atom is <em>sp<\/em><sup>2<\/sup>.<\/p>\n<\/div>\n<\/div>\n<h4>Check Your Learning<\/h4>\n<p>Another acid in acid rain is nitric acid, HNO<sub>3<\/sub>, which is produced by the reaction of nitrogen dioxide, NO<sub>2<\/sub>, with atmospheric water vapor. What is the hybridization of the nitrogen atom in NO<sub>2<\/sub>? (Note: the lone electron on nitrogen occupies a hybridized orbital just as a lone pair would.)<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q333632\">Show Answer<\/span><\/p>\n<div id=\"q333632\" class=\"hidden-answer\" style=\"display: none\"><em>sp<\/em><sup>2<\/sup><\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Key Concepts and Summary<\/h3>\n<p>Multiple bonds consist of a \u03c3 bond located along the axis between two atoms and one or two \u03c0 bonds. The \u03c3 bonds are usually formed by the overlap of hybridized atomic orbitals, while the \u03c0 bonds are formed by the side-by-side overlap of unhybridized orbitals. Resonance occurs when there are multiple unhybridized orbitals with the appropriate alignment to overlap, so the placement of \u03c0 bonds can vary.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Exercises<\/h3>\n<ol>\n<li>The bond energy of a C\u2013C single bond averages 347 kJ mol<sup>-1<\/sup>; that of a [latex]\\text{C}\\equiv \\text{C}[\/latex] triple bond averages 839 kJ mol<sup>-1<\/sup>. Explain why the triple bond is not three times as strong as a single bond.<\/li>\n<li>For the carbonate ion, [latex]{\\text{CO}}_{3}^{2-},[\/latex] draw all of the resonance structures. Identify which orbitals overlap to create each bond.<\/li>\n<li>A useful solvent that will dissolve salts as well as organic compounds is the compound acetonitrile, H<sub>3<\/sub>CCN. It is present in paint strippers.\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Write the Lewis structure for acetonitrile, and indicate the direction of the dipole moment in the molecule.<\/li>\n<li>Identify the hybrid orbitals used by the carbon atoms in the molecule to form \u03c3 bonds.<\/li>\n<li>Describe the atomic orbitals that form the \u03c0 bonds in the molecule. Note that it is not necessary to hybridize the nitrogen atom.<\/li>\n<\/ol>\n<\/li>\n<li>For the molecule allene, [latex]{\\text{H}}_{2}\\text{C}=\\text{C}={\\text{CH}}_{2},[\/latex] give the hybridization of each carbon atom. Will the hydrogen atoms be in the same plane or perpendicular planes?<\/li>\n<li>Identify the hybridization of the central atom in each of the following molecules and ions that contain multiple bonds:\n<ol style=\"list-style-type: lower-alpha;\">\n<li>ClNO (N is the central atom)<\/li>\n<li>CS<sub>2\u00a0<\/sub><\/li>\n<li>Cl<sub>2<\/sub>CO (C is the central atom)<\/li>\n<li>Cl<sub>2<\/sub>SO (S is the central atom)<\/li>\n<li>SO<sub>2<\/sub>F<sub>2<\/sub> (S is the central atom)<\/li>\n<li>XeO<sub>2<\/sub>F<sub>2<\/sub> (Xe is the central atom)<\/li>\n<li>[latex]{\\text{ClOF}}_{2}^{+}[\/latex] (Cl is the central atom)<\/li>\n<\/ol>\n<\/li>\n<li>Describe the molecular geometry and hybridization of the N, P, or S atoms in each of the following compounds. (a) H<sub>3<\/sub>PO<sub>4<\/sub>, phosphoric acid, used in cola soft drinks (b) NH<sub>4<\/sub>NO<sub>3<\/sub>, ammonium nitrate, a fertilizer and explosive (c) S<sub>2<\/sub>Cl<sub>2<\/sub>, disulfur dichloride, used in vulcanizing rubber (d) K<sub>4<\/sub>[O<sub>3<\/sub>POPO<sub>3<\/sub>], potassium pyrophosphate, an ingredient in some toothpastes<\/li>\n<li>For each of the following molecules, indicate the hybridization requested and whether or not the electrons will be delocalized:\n<ol style=\"list-style-type: lower-alpha;\">\n<li>ozone (O<sub>3<\/sub>) central O hybridization<\/li>\n<li>carbon dioxide (CO<sub>2<\/sub>) central C hybridization<\/li>\n<li>nitrogen dioxide (NO<sub>2<\/sub>) central N hybridization<\/li>\n<li>phosphate ion [latex]\\left({\\text{PO}}_{4}^{3-}\\right)[\/latex] central P hybridization<\/li>\n<\/ol>\n<\/li>\n<li>For each of the following structures, determine the hybridization requested and whether the electrons will be delocalized:\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Hybridization of each carbon<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211922\/CNX_Chem_08_03_HybridCarb_img1.jpg\" alt=\"A Lewis structure is shown in which a carbon atom is single bonded to three hydrogen atoms and a second carbon atom. This second carbon atom is, in turn, double bonded to an oxygen atom with two lone pairs of electrons. The second carbon atom is also single bonded to another carbon atom that is single bonded to three hydrogen atoms.\" data-media-type=\"image\/jpeg\" \/><\/li>\n<li>Hybridization of sulfur<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211923\/CNX_Chem_08_03_HybridSulf_img1.jpg\" alt=\"A Lewis structure is shown in which a sulfur atom with two lone pairs of electrons and a positive sign is double bonded to an oxygen with two lone pairs of electrons. The sulfur atom is also single bonded to an oxygen with three lone pairs of electrons with a negative sign. It is drawn in an angular shape.\" data-media-type=\"image\/jpeg\" \/><\/li>\n<li>All atoms<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211925\/CNX_Chem_08_03_HybridAll_img1.jpg\" alt=\"A Lewis structure is shown in which a hexagonal ring structure is made up of five carbon atoms and one nitrogen atom with a lone pair of electrons. There are alternating double and single bonds in between each carbon atom. Each carbon atom is also single bonded to one hydrogen atom.\" data-media-type=\"image\/jpeg\" \/><\/li>\n<\/ol>\n<\/li>\n<li>Draw the orbital diagram for carbon in CO<sub>2<\/sub> showing how many carbon atom electrons are in each orbital.<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q871433\">Show Selected Answers<\/span><\/p>\n<div id=\"q871433\" class=\"hidden-answer\" style=\"display: none\">\n<p>1.\u00a0A triple bond consists of one \u03c3 bond and two \u03c0 bonds. A \u03c3 bond is stronger than a \u03c0 bond due to greater overlap.<\/p>\n<p>3. The answers are as follows:<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211913\/CNX_Chem_08_03_Acetonitri_img1.jpg\" alt=\"A Lewis structure is shown in which a carbon atom is attached by single bonds to three hydrogen atoms. It is also attached by a single bond to a carbon atom that is triple bonded to a nitrogen atom with one lone electron pair. Below the structure is a right facing arrow with its head near the nitrogen and its tail, which looks like a plus sign, located near the carbon atoms. The arrow is labeled, \u201cdipole moment.\u201d\" data-media-type=\"image\/jpeg\" \/><\/li>\n<li>The terminal carbon atom uses <em>sp<\/em><sup>3<\/sup> hybrid orbitals, while the central carbon atom is <em>sp<\/em> hybridized. The terminal carbon atom forms four \u03c3 bonds (three to the hydrogen atoms and one to the carbon) while the central carbon forms two \u03c3 bond (one to carbon and one to nitrogen).<\/li>\n<li>Each of the two \u03c0 bonds is formed by overlap of a 2<em>p<\/em> orbital on carbon and a nitrogen 2<em>p<\/em> orbital.<\/li>\n<\/ol>\n<p>5. The hybridization of each central atom is as follows:<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li><em>sp<\/em><sup>2<\/sup><\/li>\n<li><em>sp<\/em><\/li>\n<li><em>sp<\/em><sup>2<\/sup><\/li>\n<li><em>sp<\/em><sup>3<\/sup><\/li>\n<li><em>sp<\/em><sup>3<\/sup><\/li>\n<li><em>sp<\/em><sup>3<\/sup><em>d<\/em><\/li>\n<li><em>sp<\/em><sup>3<\/sup><\/li>\n<\/ol>\n<p>7. The answers are as follows:<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li><em>sp<\/em><sup>2<\/sup>, delocalized<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211917\/CNX_Chem_08_03_ozone_img1.jpg\" alt=\"A pair of Lewis structures, connected by a double ended arrow, are shown. Each structure shows three oxygen atoms which are bonded together. In the first structure, the left oxygen, which has two lone electron pairs, is connected to the central oxygen which has one lone electron pair and a plus sign. The central oxygen is connected by a single bond to the right oxygen which has three lone electron pairs and a negative sign. The second structure is a mirror image of the first. Each structure has an angular shape.\" data-media-type=\"image\/jpeg\" \/><\/li>\n<li><em>sp<\/em>, localized<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211918\/CNX_Chem_08_01_CO2LewStr1_img3.jpg\" alt=\"A Lewis structure is shown in which a carbon atom is double bonded to an oxygen atom on each side. Each oxygen atom has two lone pairs of electrons.\" data-media-type=\"image\/jpeg\" \/><\/li>\n<li><em>sp<\/em><sup>2<\/sup>, delocalized<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211919\/CNX_Chem_08_03_NO2_img1.jpg\" alt=\"A pair of Lewis structures connected by a double ended arrow are shown. The first structure shows a nitrogen atom with a single lone electron pair and a positive sign that is singly bonded on the left to an oxygen. This oxygen atom has three lone electron pairs and a negative sign. The nitrogen atom is also double bonded to an oxygen atom on the right which has two lone electron pairs. The second image is a mirror image of the first and both are drawn in an angular shape.\" data-media-type=\"image\/jpeg\" \/><\/li>\n<li><em>sp<\/em><sup>3<\/sup>, delocalized<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211921\/CNX_Chem_08_03_PO43_img1.jpg\" alt=\"A Lewis structure is shown in which a phosphorus atom is single bonded to four oxygen atoms, each of which has three lone electron pairs and a negative sign.\" data-media-type=\"image\/jpeg\" \/><\/li>\n<\/ol>\n<p>9.\u00a0Each of the four electrons is in a separate orbital and overlaps with an electron on an oxygen atom.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211926\/CNX_Chem_08_02_CO2Diag1.jpg\" alt=\"A diagram is shown in two parts, connected by a right facing arrow labeled, \u201cHybridization.\u201d The left diagram shows an up-facing arrow labeled, \u201cE.\u201d To the lower right of the arrow is a short, horizontal line labeled, \u201c2 s,\u201d that has two vertical half-arrows facing up and down on it. To the upper right of the arrow are a series of three short, horizontal lines labeled, \u201c2 p.\u201d Above both sets of lines is the phrase, \u201cOrbitals in an isolated C atom.\u201d There are two upward facing arrows on two of these lines. The right side of the diagram shows two short, horizontal lines placed halfway up the space and each labeled, \u201cs p.\u201d An upward-facing half arrow is drawn vertically on each line. Above these lines are two other short, horizontal lines, each labeled, \u201c2 p,\u201d and which have two upward facing arrows on them. Above both sets of lines is the phrase, \u201cOrbitals in the s p hybridized C in C O subscript 2.\u201d\" width=\"880\" height=\"272\" data-media-type=\"image\/jpeg\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-2037\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Chemistry. <strong>Provided by<\/strong>: OpenStax College. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/openstaxcollege.org\">http:\/\/openstaxcollege.org<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at https:\/\/openstaxcollege.org\/textbooks\/chemistry\/get<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17,"menu_order":4,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Chemistry\",\"author\":\"\",\"organization\":\"OpenStax College\",\"url\":\"http:\/\/openstaxcollege.org\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at https:\/\/openstaxcollege.org\/textbooks\/chemistry\/get\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-2037","chapter","type-chapter","status-publish","hentry"],"part":3005,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/pressbooks\/v2\/chapters\/2037","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/wp\/v2\/users\/17"}],"version-history":[{"count":10,"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/pressbooks\/v2\/chapters\/2037\/revisions"}],"predecessor-version":[{"id":6007,"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/pressbooks\/v2\/chapters\/2037\/revisions\/6007"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/pressbooks\/v2\/parts\/3005"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/pressbooks\/v2\/chapters\/2037\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/wp\/v2\/media?parent=2037"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/pressbooks\/v2\/chapter-type?post=2037"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/wp\/v2\/contributor?post=2037"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-chem-atoms-first\/wp-json\/wp\/v2\/license?post=2037"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}