{"id":3168,"date":"2018-06-22T20:10:48","date_gmt":"2018-06-22T20:10:48","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/?post_type=chapter&#038;p=3168"},"modified":"2020-06-23T02:26:25","modified_gmt":"2020-06-23T02:26:25","slug":"2-2-hybrid-orbitals","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/chapter\/2-2-hybrid-orbitals\/","title":{"raw":"2.2. Hybrid orbitals","rendered":"2.2. Hybrid orbitals"},"content":{"raw":"<section class=\"mt-content-container\">\r\n<div>\r\n<div class=\"textbox learning-objectives\">\r\n<h3 class=\"boxtitle\">Skills to Develop<\/h3>\r\n<ul>\r\n \t<li>Describe the hybrid orbitals used in the formation of bonding for each atom in some carbon containing compounds.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div id=\"section_2\">\r\n<h3 class=\"editable\"><strong><em>sp<sup>3<\/sup><\/em><\/strong> Hybrid orbitals and tetrahedral bonding<\/h3>\r\nNow let\u2019s look more carefully at bonding in organic molecules, starting with methane, CH<sub>4<\/sub>. Recall the valence electron configuration of a carbon atom:\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/04145949\/fig2-1-6.png\" alt=\"\" width=\"278\" height=\"111\" \/>\r\n\r\nThis picture is problematic when it comes to describing the <em>bonding<\/em> in methane. How does the carbon form four bonds if it has only two half-filled <em>p<\/em> orbitals available for bonding?\u00a0\u00a0 A hint comes from the experimental observation that the four C-H bonds in methane are arranged with <strong>tetrahedral<\/strong> geometry about the central carbon, and that each bond has the same length and strength.\u00a0 In order to explain this observation, valence bond theory relies on a concept called <strong>orbital hybridization<\/strong>.\u00a0 In this picture, the four valence orbitals of the carbon (one 2<em>s<\/em> and three 2<em>p<\/em> orbitals) combine mathematically (remember: orbitals are described by wave equations) to form four equivalent <strong>hybrid orbitals<\/strong>, which are called <strong><em>sp<sup>3<\/sup><\/em> orbitals <\/strong>because they are formed from mixing one <em>s<\/em> and three <em>p<\/em> orbitals. In the new electron configuration, each of the four valence electrons on the carbon occupies a single <em>sp<sup>3<\/sup><\/em> orbital.\r\n<table style=\"border-collapse: collapse;width: 100%\" border=\"1\">\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 33.3333%\"><img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/04145952\/fig2-1-7.png\" alt=\"\" width=\"252\" height=\"139\" \/><\/td>\r\n<td style=\"width: 33.3333%\">Click to see an <a class=\"external\" title=\"http:\/\/www.chemtube3d.com\/orbitalshybrid.htm\" href=\"http:\/\/www.chemtube3d.com\/orbitalshybrid.htm\" target=\"_blank\" rel=\"external nofollow noopener\">interactive 3D model<\/a>\r\n\r\n(<em>select 'load sp<sup>3<\/sup>' and 'load H 1s' to see orbitals<\/em>).<\/td>\r\n<td style=\"width: 33.3333%\"><img class=\"wp-image-4054 alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/26151242\/frame-6-150x150.png\" alt=\"\" width=\"150\" height=\"150\" \/><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThis geometric arrangement makes perfect sense if you consider that it is precisely this angle that allows the four orbitals (and the electrons in them) to be as far apart from each other as possible.\u00a0 This is simply a restatement of the Valence Shell Electron Pair Repulsion (VSEPR) theory that you learned in General Chemistry: electron pairs (in orbitals) will arrange themselves in such a way as to remain as far apart as possible, due to negative-negative electrostatic repulsion.\r\n\r\nEach C-H bond in methane, then, can be described as a sigma bond formed by overlap between a half-filled 1<em>s<\/em> orbital in a hydrogen atom and the larger lobe of one of the four half-filled <em>sp<sup>3<\/sup><\/em> hybrid orbitals in the central carbon. The length of the carbon-hydrogen bonds in methane is 109 pm.\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/04145955\/fig2-1-8.png\" alt=\"\" width=\"529\" height=\"179\" \/>\r\n\r\nWhile previously we drew a Lewis structure of methane in two dimensions using lines to denote each covalent bond, we can now draw a more accurate structure in three dimensions, showing the tetrahedral bonding geometry.\u00a0 To do this on a two-dimensional page, though, we need to introduce a new drawing convention: the solid \/ dashed wedge system.\u00a0 In this convention, a solid wedge simply represents a bond that is meant to be pictured emerging from the plane of the page.\u00a0 A dashed wedge represents a bond that is meant to be pictured pointing into, or behind, the plane of the page.\u00a0 Normal lines imply bonds that lie in the plane of the page.\u00a0 This system takes a little bit of getting used to, but with practice your eye will learn to immediately \u2018see\u2019 the third dimension being depicted.\r\n<div>\r\n<div id=\"exercise\">\r\n<div class=\"textbox examples\">\r\n<h3>Example<\/h3>\r\nImagine that you could distinguish between the four hydrogen atoms in a methane molecule, and labeled them H<sub>a<\/sub> through H<sub>d<\/sub>.\u00a0 In the images below, the <em>exact same<\/em> methane molecule is rotated and flipped in various positions.\u00a0 Draw the missing hydrogen atom labels. (It will be much easier to do this if you make a model.)\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/04145959\/figE2-1-1.png\" alt=\"\" width=\"393\" height=\"173\" \/>[reveal-answer q=\"129643\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"129643\"]\r\n\r\n<img class=\"internal default\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/107044\/E2-1S.png?revision=1&amp;size=bestfit&amp;width=488&amp;height=95\" alt=\"\" width=\"488\" height=\"95\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example<\/h3>\r\nWhat kind of orbitals overlap to form the C-Cl bonds in chloroform, CHCl<sub>3<\/sub>?\r\n[reveal-answer q=\"204888\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"204888\"]\r\n\r\nsp<sup>3<\/sup> orbital on carbon overlapping with an sp<sup>3<\/sup> orbital on chlorine.[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"mt-section\">\r\n\r\nDiamond is a crystal form of elemental carbon, and the structure is particularly interesting. In the crystal, every carbon atom is bonded to four other carbon atoms, and the bonds are arranged in a tetrahedral fashion. The bonding, no doubt, is due to the <em>sp<\/em><sup>3<\/sup> hybrid orbitals. The bond length of 154 pm is the same as the C-C bond length in ethane, propane and other alkanes.\r\n\r\nAn idealized single crystal of diamond is a gigantic molecule, because all the atoms are inter-bonded. The bonding has given diamond some very unusual properties. It is the hardest stone, much harder than anything else in the material world. It is a poor conductor, because all electrons are localized in the chemical bonds. However, diamond is an excellent heat conductor. A stone made of pure carbon is colorless, but the presence of impurities gives it various colors. The index of refraction is very high, and their glitter (sparkle or splendor) has made them the most precious stones.\r\n\r\n<\/div>\r\nHow does this bonding picture extend to compounds containing carbon-carbon bonds?\u00a0 In ethane (CH<sub>3<\/sub>CH<sub>3<\/sub>), both carbons are <em>sp<sup>3<\/sup><\/em>-hybridized, meaning that both have four bonds with tetrahedral geometry.\u00a0 The carbon-carbon bond, with a bond length of 154 pm, is formed by overlap of one <em>sp<\/em><sup>3<\/sup> orbital from each of the carbons, while the six carbon-hydrogen bonds are formed from overlaps between the remaining <em>sp<sup>3<\/sup><\/em> orbitals on the two carbons and the 1<em>s<\/em> orbitals of hydrogen atoms.\u00a0 All of these are sigma bonds.\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/04150006\/fig2-1-9.png\" alt=\"\" width=\"396\" height=\"171\" \/>\r\n\r\nBecause they are formed from the end-on-end overlap of two orbitals, <em>s<\/em><em>igma bonds are free to rotate<\/em>.\u00a0 This means, in the case of ethane molecule, that the two methyl (CH<sub>3<\/sub>) groups can be pictured as two wheels on an axle, each one able to rotate with respect to the other.\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/04150009\/fig2-1-10.png\" alt=\"\" width=\"331\" height=\"107\" \/>\r\n\r\nIn chapter 3 we will learn more about the implications of rotational freedom in sigma bonds, when we discuss the \u2018conformation\u2019 of organic molecules.\r\n\r\nThe <em>sp<sup>3<\/sup><\/em> bonding picture is also used to described the bonding in amines, including ammonia, the simplest amine.\u00a0 Just like the carbon atom in methane, the central nitrogen in ammonia is <em>sp<sup>3<\/sup>-<\/em>hybridized.\u00a0 With nitrogen, however, there are five rather than four valence electrons to account for, meaning that three of the four hybrid orbitals are half-filled and available for bonding, while the fourth is fully occupied by a nonbonding pair (lone pair) of electrons.\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/04150013\/fig2-1-11.png\" alt=\"\" width=\"530\" height=\"304\" \/>\r\n\r\nThe bonding arrangement here is also tetrahedral: the three N-H bonds of ammonia can be pictured as forming the base of a trigonal pyramid, with the fourth orbital, containing the lone pair, forming the top of the pyramid.\u00a0 Recall from your study of VSEPR theory in General Chemistry that the lone pair, with its slightly greater repulsive effect, \u2018pushes\u2019 the three N-H s bonds away from the top of the pyramid, meaning that the H-N-H bond angles are slightly less than tetrahedral, at 107.3\u02da rather than 109.5\u02da.\r\n\r\nVSEPR theory also predicts, accurately, that a water molecule is \u2018bent\u2019 at an angle of approximately 104.5\u02da. The bonding in water results from overlap of two of the four <em>sp<sup>3<\/sup><\/em> hybrid orbitals on oxygen with 1<em>s <\/em>orbitals on the two hydrogen atoms. The two nonbonding electron pairs on oxygen are located in the two remaining <em>sp<sup>3<\/sup><\/em>orbitals. <a class=\"thumb\" title=\"struc1fig66.png\" href=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/98074\/struc1fig66.png?revision=1\" rel=\"internal\"><img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/04150018\/struc1fig66.png\" alt=\"struc1fig66.png\" width=\"652\" height=\"351\" \/><\/a>\r\n<div>\r\n<div class=\"textbox examples\">\r\n<h3>Example<\/h3>\r\nDraw, in the same style as the figures above, orbital pictures for the bonding in a) methylamine (H<sub>3<\/sub>CNH<sub>2<\/sub>), and b) ethanol (H<sub>3<\/sub>C-CH<sub>2<\/sub>-OH.\r\n[reveal-answer q=\"686129\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"686129\"]\r\n\r\nBoth the carbon and the nitrogen atom in CH<sub>3<\/sub>NH<sub>2<\/sub> are sp<sup>3<\/sup>-hybridized. The C-N sigma bond is an overlap between two sp<sup>3<\/sup> orbitals.\r\n\r\nBoth the carbon and the nitrogen atom in CH<sub>3<\/sub>NH<sub>2<\/sub>\u00a0are\u00a0<em>s<\/em>p<sup>3<\/sup>-hybridized.\u00a0 The C-N\u00a0sigma\u00a0bond is an overlap between two\u00a0<em>s<\/em>p<sup>3\u00a0<\/sup>orbitals.\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/107045\/E2-3S.png?revision=1&amp;size=bestfit&amp;width=437&amp;height=191\" alt=\"\" width=\"437\" height=\"191\" \/>[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\nSee a <a class=\"link-https\" title=\"https:\/\/www.khanacademy.org\/science\/organic-chemistry\/gen-chem-review\/hybrid-orbitals-jay\/v\/sp3-hybridized-orbitals-and-sigma-bonds\" href=\"https:\/\/www.khanacademy.org\/science\/organic-chemistry\/gen-chem-review\/hybrid-orbitals-jay\/v\/sp3-hybridized-orbitals-and-sigma-bonds\" target=\"_blank\" rel=\"external nofollow noopener\">video tutorial on sp<sup>3<\/sup> orbitals and sigma bonds<\/a> (Note: This is the video linked to in the previous section)\r\n\r\n<\/div>\r\n<div id=\"section_3\">\r\n<h3 class=\"editable\"><strong><em>sp<sup>2<\/sup><\/em><\/strong> and <strong><em>sp<\/em><\/strong> hybrid orbitals and pi bonds<\/h3>\r\nThe valence bond theory, along with the hybrid orbital concept, does a very good job of describing double-bonded compounds such as ethene.\u00a0Three experimentally observable characteristics of the ethene molecule need to be accounted for by a bonding model:\r\n<ol>\r\n \t<li>Ethene is a planar (flat) molecule.<\/li>\r\n \t<li>Bond angles in ethene are approximately 120<sup>o<\/sup>, and the carbon-carbon bond length is 134 pm, significantly shorter than the 154 pm single carbon-carbon bond in ethane.<\/li>\r\n \t<li>There is a significant barrier to rotation about the carbon-carbon double bond.<\/li>\r\n<\/ol>\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/04150021\/fig2-1-13.png\" alt=\"fig2-1-13.png\" width=\"180\" height=\"166\" \/>\r\n\r\nClearly, these characteristics are not consistent with an <em>sp<sup>3<\/sup><\/em> hybrid bonding picture for the two carbon atoms.\u00a0Instead, the bonding in ethene is described by a model involving the participation of a different kind of hybrid orbital.\u00a0Three atomic orbitals on each carbon \u2013 the 2<em>s<\/em>, 2<em>p<\/em><sub>x<\/sub> and 2<em>p<\/em><sub>y<\/sub> orbitals \u2013 combine to form three <strong><em>sp<sup>2<\/sup><\/em> hybrids<\/strong>, leaving the 2<em>p<\/em><sub>z<\/sub> orbital unhybridized.\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/04150024\/fig2-1-14.png\" alt=\"fig2-1-14.png\" width=\"657\" height=\"192\" \/>\r\n\r\nThe three <em>sp<sup>2<\/sup><\/em> hybrids are arranged with trigonal planar geometry, pointing to the three corners of an equilateral triangle, with angles of 120\u00b0 between them.\u00a0The unhybridized 2<em>p<\/em><sub>z<\/sub> orbital is <em>perpendicular<\/em> to this plane (in the next several figures, <em>sp<sup>2<\/sup><\/em> orbitals and the sigma bonds to which they contribute are represented by lines and wedges; only the 2<em>p<\/em><sub>z<\/sub> orbitals are shown in the 'space-filling' mode).\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/04150031\/fig2-1-15.png\" alt=\"fig2-1-15.png\" width=\"248\" height=\"168\" \/>\r\n\r\nThe carbon-carbon double bond in ethene consists of one sigma bond, formed by the overlap of two <em>sp<sup>2<\/sup><\/em> orbitals, and a second bond, called a <strong>pi<\/strong><strong> bond<\/strong>, which is formed by the <em>side-by-side<\/em> overlap of the two unhybridized 2<em>p<\/em><sub>z<\/sub> orbitals from each carbon.\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/04150041\/fig2-1-16.png\" alt=\"fig2-1-16.png\" width=\"167\" height=\"132\" \/>\r\n<table style=\"border-collapse: collapse;width: 100%\" border=\"1\">\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 18.6743%;text-align: center\"><a class=\"link-https\" title=\"https:\/\/www.youtube.com\/watch?v=C2W-yDPcpl4\" href=\"https:\/\/www.youtube.com\/watch?v=C2W-yDPcpl4\" target=\"_blank\" rel=\"external nofollow noopener\">animation\u00a0<\/a><\/td>\r\n<td style=\"width: 25.4038%;text-align: center\"><a class=\"external\" title=\"http:\/\/andromeda.rutgers.edu\/~huskey\/images\/ethylene_bonding.jpg\" href=\"http:\/\/andromeda.rutgers.edu\/~huskey\/images\/ethylene_bonding.jpg\" target=\"_blank\" rel=\"external nofollow noopener\">spacefilling image<\/a><\/td>\r\n<td style=\"width: 24.0578%;text-align: center\"><a class=\"link-https\" title=\"https:\/\/www.khanacademy.org\/science\/organic-chemistry\/gen-chem-review\/hybrid-orbitals-jay\/v\/pi-bonds-and-sp2-hybridized-orbitals\" href=\"https:\/\/www.khanacademy.org\/science\/organic-chemistry\/gen-chem-review\/hybrid-orbitals-jay\/v\/pi-bonds-and-sp2-hybridized-orbitals\" target=\"_blank\" rel=\"external nofollow noopener\">\u00a0video tutoria<\/a>l<\/td>\r\n<td style=\"width: 31.8641%\">\r\n<p style=\"text-align: center\"><a href=\"http:\/\/www.chemtube3d.com\/orbitalsethene.htm\">interactive 3D model<\/a>\r\n<em>(select 'show resulting pi orbital')<\/em><\/p>\r\n<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 18.6743%\"><\/td>\r\n<td style=\"width: 25.4038%\"><\/td>\r\n<td style=\"width: 24.0578%\"><img class=\"wp-image-4055 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/26151546\/frame-7-150x150.png\" alt=\"\" width=\"150\" height=\"150\" \/><\/td>\r\n<td style=\"width: 31.8641%\"><img class=\"wp-image-4056 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/26151756\/frame-8-150x150.png\" alt=\"\" width=\"151\" height=\"151\" \/><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<em>\u00a0<\/em>This illustration (from University of Florida) shows the sigma and pi bonds in ethene.\r\n<p class=\"mt-align-center\"><img class=\"internal aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/22193029\/ethene.gif\" alt=\"ethene.gif\" \/><\/p>\r\nUnlike a sigma bond, a pi bond does <em>not<\/em> have cylindrical symmetry. If rotation about this bond were to occur, it would involve disrupting the side-by-side overlap between the two 2<em>p<\/em><sub>z<\/sub> orbitals that make up the pi bond.\u00a0 The presence of the pi bond thus \u2018locks\u2019 the six atoms of ethene into the same plane. This argument extends to larger alkene groups: in each case, six atoms lie in the same plane.\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/04150044\/fig2-1-17.png\" alt=\"fig2-1-17.png\" width=\"182\" height=\"148\" \/>\r\n<div class=\"mt-section\"><\/div>\r\n<div>\r\n<div class=\"textbox examples\">\r\n<h3>Example<\/h3>\r\nRedraw the structures below, indicating the six atoms that lie in the same plane due to the carbon-carbon double bond.\r\n[reveal-answer q=\"15124\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"15124\"]\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/04150048\/figE2-1-4.png\" alt=\"figE2-1-4.png\" width=\"299\" height=\"58\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example<\/h3>\r\nWhat is wrong with the way the following structure is drawn?\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/04150050\/figE2-1-5.png\" alt=\"figE2-1-5.png\" width=\"191\" height=\"103\" \/>\r\n\r\n[reveal-answer q=\"437064\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"437064\"]\r\n\r\nThe carbon atoms in an aromatic ring are sp2 hybridized, thus bonding geometry is trigonal planar: in other words, the bonds coming out of the ring are in the same plane as the ring, not pointing above the plane of the ring as the wedges in the incorrect drawing indicate. A correct drawing should use lines to indicate that the bonds are in the same plane as the ring:<img class=\"internal default aligncenter\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/107047\/E2-5S-b.png?revision=1&amp;size=bestfit&amp;width=624&amp;height=132\" alt=\"\" width=\"624\" height=\"132\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\nA similar picture can be drawn for the bonding in carbonyl groups, such as formaldehyde.\u00a0In this molecule, the carbon is <em>sp<sup>2<\/sup><\/em>-hybridized, and we will assume that the oxygen atom is also <em>sp<sup>2<\/sup><\/em>hybridized.\u00a0The carbon has three sigma bonds: two are formed by overlap between <em>sp<sup>2<\/sup><\/em> orbitals with 1<em>s <\/em>orbitals from hydrogen atoms, and the third sigma bond is formed by overlap between the remaining carbon <em>sp<sup>2<\/sup><\/em> orbital and an <em>sp<sup>2<\/sup><\/em> orbital on the oxygen.\u00a0The two lone pairs on oxygen occupy its other two <em>sp<sup>2<\/sup><\/em> orbitals.\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/04150054\/fig2-1-18.png\" alt=\"fig2-1-18.png\" width=\"489\" height=\"184\" \/>\r\n\r\n<a class=\"external\" title=\"http:\/\/www.chemtube3d.com\/orbitalsformaldehyde.htm\" href=\"http:\/\/www.chemtube3d.com\/orbitalsformaldehyde.htm\" target=\"_blank\" rel=\"external nofollow noopener\">interactive 3D model<\/a>\r\n\r\nThe pi bond is formed by side-by-side overlap of the unhybridized 2<em>p<\/em><sub>z<\/sub> orbitals on the carbon and the oxygen. Just like in alkenes, the 2<em>p<\/em><sub>z<\/sub> orbitals that form the pi bond are perpendicular to the plane formed by the sigma bonds.\r\n<div>\r\n<div class=\"textbox examples\">\r\n<h3>Example<\/h3>\r\na: Draw a diagram of hybrid orbitals in an <em>sp<sup>2<\/sup><\/em>-hybridized nitrogen.\r\n\r\nb: Draw a figure showing the bonding picture for the imine below.\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/04150059\/figE2-1-6.png\" alt=\"figE2-1-6.png\" width=\"84\" height=\"96\" \/>\r\n\r\nc: In your drawing for part b, what kind of orbital holds the nitrogen lone pair?\r\n[reveal-answer q=\"990885\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"990885\"]\r\n\r\na) The carbon and nitrogen atoms are both\u00a0<em>sp<\/em><sup>2<\/sup>\u00a0hybridized.\u00a0 The carbon-nitrogen double bond is composed of a\u00a0sigma\u00a0bond formed from two\u00a0<em>sp<\/em><sup>2<\/sup>\u00a0orbitals, and a\u00a0pi\u00a0bond formed from the side-by-side overlap of two unhybridized\u00a0<em>2p<\/em>\u00a0orbitals.\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/107048\/E2-6S.png?revision=1&amp;size=bestfit&amp;width=342&amp;height=192\" alt=\"\" width=\"342\" height=\"192\" \/>\r\n\r\nb) As shown in the figure above, the nitrogen lone pair electrons occupy one of the three\u00a0<em>s<\/em><em>p<sup>2<\/sup><\/em>\u00a0hybrid orbitals.\u00a0[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<h3 class=\"editable\">Bonding carbon atoms using <em>sp<\/em> hybrid orbitals<\/h3>\r\nConsider, for example, the structure of ethyne (common name acetylene), the simplest alkyne.\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/04150103\/fig2-1-20.png\" alt=\"fig2-1-20.png\" width=\"162\" height=\"73\" \/>\r\n\r\nBoth the VSEPR theory and experimental evidence tells us that the molecule is linear: all four atoms lie in a straight line.\u00a0The carbon-carbon triple bond is only 120 pm long, shorter than the double bond in ethene, and is very strong, about 837 kJ\/mol.\u00a0In the hybrid orbital picture of acetylene, both carbons are <strong><em>sp-<\/em>hybridized<\/strong>.\u00a0 In an <em>sp<\/em>-hybridized carbon,\u00a0 the 2<em>s<\/em> orbital combines with the 2<em>p<\/em><sub>x<\/sub> orbital to form two <em>sp<\/em> hybrid orbitals that are oriented at an angle of 180\u00b0 with respect to each other (eg. along the x axis).\u00a0The 2<em>p<\/em><sub>y<\/sub> and 2<em>p<\/em><sub>z<\/sub> orbitals remain unhybridized, and are oriented perpendicularly along the y and z axes, respectively.\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/04150108\/fig2-1-21.png\" alt=\"fig2-1-21.png\" width=\"594\" height=\"317\" \/>\r\n\r\nThe carbon-carbon sigma bond, then, is formed by the overlap of one <em>sp<\/em> orbital from each of the carbons, while the two carbon-hydrogen sigma bonds are formed by the overlap of the second <em>sp<\/em> orbital on each carbon with a 1<em>s<\/em> orbital on a hydrogen.\u00a0Each carbon atom still has two half-filled 2<em>p<\/em><sub>y<\/sub> and 2<em>p<\/em><sub>z<\/sub> orbitals, which are perpendicular both to each other and to the line formed by the sigma bonds.\u00a0These two perpendicular pairs of <em>p<\/em> orbitals form two pi bonds between the carbons, resulting in a triple bond overall (one sigma bond plus two pi bonds).\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/04150111\/fig2-1-22.png\" alt=\"fig2-1-22.png\" width=\"416\" height=\"111\" \/>\r\n\r\nWhen <em>sp<\/em> hybrid orbitals are used for the sigma bond, the two sigma bonds around the carbon are linear. Two other <em>p<\/em> orbitals are available for pi bonding, and a typical compound is the acetylene or ethyne HC\u2261CH. The three sigma and two pi bonds of this molecule can be seen in this diagram from <a class=\"external\" href=\"http:\/\/chem.ufl.edu\/%7Echm2040\/Notes\/Chapter_12\/multiple.html\" target=\"_blank\" rel=\"external nofollow noopener\">University of Florida: General chemistry<\/a> shown below.\r\n\r\n<img class=\"internal aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/22193027\/hcch.gif\" alt=\"\" width=\"432\" height=\"213\" \/>\r\n\r\nNote that molecules H-C\u2261C-H, H-C\u2261N, and \u00afC\u2261O<sup>+<\/sup> have the same number of electrons. Bonding in these molecules can be explained by the same theory, and thus their formation is no surprise.\r\n\r\n<a class=\"external\" title=\"http:\/\/www.chemtube3d.com\/orbitalsacetylene.htm\" href=\"http:\/\/www.chemtube3d.com\/orbitalsacetylene.htm\" target=\"_blank\" rel=\"external nofollow noopener\">interactive 3D model<\/a>\r\n<div class=\"mt-section\">\r\n<h3 class=\"editable\">Comparison of Structural Features<\/h3>\r\nSome typical bonding features of ethane, ethene, and ethyne are summarized in the table below:\r\n<table class=\"mt-responsive-table\">\r\n<thead>\r\n<tr>\r\n<th class=\"mt-align-center\" scope=\"col\">Systematic\r\nname<\/th>\r\n<th class=\"mt-align-center\" scope=\"col\">Ethane<\/th>\r\n<th class=\"mt-align-center\" scope=\"col\">Ethene<\/th>\r\n<th class=\"mt-align-center\" scope=\"col\">Ethyne<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>Hybrid orbitals\r\nof $$\\ce{C}$$<\/td>\r\n<th><em>sp<\/em><sup>3<\/sup><\/th>\r\n<th><em>sp<\/em><sup>2<\/sup><\/th>\r\n<th><em>sp<\/em><\/th>\r\n<\/tr>\r\n<tr>\r\n<td>Structural\r\nformula<\/td>\r\n<th>H H\r\n\\ \u00a0\/\r\nH--C---C--H\r\n\/ \u00a0\\\r\nH H<\/th>\r\n<th>H H\r\n\\ \u00a0\/\r\nC=C\r\n\/ \u00a0\\\r\nH H<\/th>\r\n<th>H-C<span class=\"st\">\u2261<\/span>C-H<\/th>\r\n<\/tr>\r\n<tr>\r\n<td>$$\\ce{C-C}$$\r\nBondlength pm<\/td>\r\n<th>154<\/th>\r\n<th>134<\/th>\r\n<th>120<\/th>\r\n<\/tr>\r\n<tr>\r\n<td>$$\\ce{C-H}$$\r\nBondlength pm<\/td>\r\n<th>112<\/th>\r\n<th>110<\/th>\r\n<th>106<\/th>\r\n<\/tr>\r\n<tr>\r\n<td>$$\\ce{H-C-C}$$\r\nbond angle \u00b0<\/td>\r\n<th>111<\/th>\r\n<th>121<\/th>\r\n<th>180<\/th>\r\n<\/tr>\r\n<tr>\r\n<td>$$\\ce{C-C}$$ bond\r\nenergy kJ\/mol<\/td>\r\n<th>368<\/th>\r\n<th>611<\/th>\r\n<th>820<\/th>\r\n<\/tr>\r\n<tr>\r\n<td>$$\\ce{C-H}$$ bond\r\nenergy kJ\/mol<\/td>\r\n<th>410<\/th>\r\n<th>451<\/th>\r\n<th>536<\/th>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nAs the bond order between carbon atoms increases from 1 to 3 for ethane, ethene, and ethyne, the bond lengths decrease, and the bond energy increases. Note that the bond energies given here are specific for these compounds, and the values may be different from the average values for this type of bonds.\r\n\r\n<\/div>\r\n<div class=\"mt-section\">\r\n<h2 class=\"editable\">Confidence building questions<\/h2>\r\n<ol>\r\n \t<li><strong>What hybrid orbitals are used by the carbon atoms in HC\u2261CH? <\/strong><strong>Discussion - <\/strong>\r\nLinear -C- bonds due to <em>sp<\/em> hybridized orbitals. This molecule is linear, and it consists of 3 sigma, \u03c3, bonds, and two pi, \u03c0, bonds. Compare the bonding of this with \u00afC\u2261O<sup>+<\/sup>, H-C\u2261N, and CH<sub>3<\/sub>-C\u2261N.<\/li>\r\n \t<li><strong>What hybrid orbitals are used by the carbon atoms in H<sub>2<\/sub>C=CH<sub>2<\/sub>? <\/strong><strong>Discussion - <\/strong>\r\nPlanar -C&lt; \u03c3 bonds due to <em>sp<\/em><sup>2<\/sup> hybridized orbitals. Another <em>p<\/em> orbital is used for the pi, \u03c0. How many sigma and pi bonds does this molecule have? Do all atoms in this molecule lie on the same plane?<\/li>\r\n \t<li><strong>What hybrid orbitals are used by the<\/strong><strong> carbon in CH<sub>3<\/sub>CH<sub>3<\/sub>? <\/strong><strong>Discussion - <\/strong>\r\nTetrahedral arrangement around C is due to <em>sp<\/em><sup>3<\/sup> hybridized orbitals.<\/li>\r\n \t<li><strong>What hybrid orbitals are used by the oxygen in CH<sub>3<\/sub>CH<sub>2<\/sub>OH? Discussion - <\/strong>\r\nTetrahedral arrangement around O is due to <em>sp<\/em><sup>3<\/sup> hybridized orbitals.<\/li>\r\n \t<li><strong>Which carbon to carbon bond length is the shortest in the following molecule: CH<sub>3<\/sub>-C\u2261C-CH<sub>2<\/sub>-CH=CH-CH<sub>2<\/sub>-OH? <\/strong><strong>Discussion - <\/strong>\r\nThe bond length decreases as the bond order increases, so the triple bond is the shortest.<\/li>\r\n \t<li><strong>How many carbon atoms make use of <em>sp<\/em><sup>2<\/sup> hybrid orbitals in this molecule: CH<sub>3<\/sub>-C\u2261C-CH<sub>2<\/sub>-CH=CH-CH<sub>2<\/sub>-COOH? <\/strong><strong>Discussion - <\/strong>\r\nRecognize that the type of bonding is important. For more guidance, see the next section in this book.<\/li>\r\n \t<li><strong>The molecular formula of caffeine is C<sub>8<\/sub>H<sub>10<\/sub>N<sub>4<\/sub>O<sub>2<\/sub>; draw a reasonable structure for it. <\/strong><strong>Discussion - <\/strong>\r\nThe structure is shown below. Can you sketch a bonding structure for caffeine? How many carbon atoms makes use of <em>sp<\/em><sup>2<\/sup> hybrid orbitals?<\/li>\r\n<\/ol>\r\n<img class=\"internal aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/22193033\/caffeine.gif\" alt=\"\" width=\"464\" height=\"464\" \/>\r\n\r\n<\/div>\r\n<div id=\"section_6\" class=\"mt-section\">\r\n<div>\r\n<div class=\"textbox examples\">\r\n<h3>Example<\/h3>\r\na) What kinds of orbitals are overlapping in bonds b-i indicated below?\u00a0 Be sure to distinguish between s and p bonds. An example is provided for bond 'a'.\r\n\r\nb) In what kind of orbital is the lone pair of electrons located on the nitrogen atom of bond a?\u00a0 Of bond e?\r\n\r\n<img class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/04150119\/fig2-1-24.png\" alt=\"fig2-1-24.png\" width=\"642\" height=\"343\" \/>\r\n\r\n[reveal-answer q=\"879188\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"879188\"]\r\n\r\na) bond b: Nsp2-Csp3 (this means an overlap of an sp2 orbital on N and an sp3 orbital on C)\r\n\r\nbond c: Csp2-Csp2 plus C2p-C2p (pi)\r\n\r\nbond d: Csp2-Csp3\r\n\r\nbond e: Csp3-Csp3\r\n\r\nbond f: Csp3-Csp3\r\n\r\nbond g: Csp2-Csp2 (s) plus C2p-C2p (pi)\r\n\r\nbond h: Csp2-H1s\r\n\r\nbond i: Csp2-Csp2\r\n\r\nb) bond a: lone pair on N occupies an sp2 orbital\r\n\r\nbond e: lone pair on N occupies an sp3 orbital[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n[embed]https:\/\/www.youtube.com\/watch?v=u1eGSL6J6Fo[\/embed]\r\n<h3><img class=\"size-thumbnail wp-image-4048 alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2017\/10\/26145837\/frame-4-150x150.png\" alt=\"\" width=\"150\" height=\"150\" \/><\/h3>\r\n<\/div>\r\n<\/section>","rendered":"<section class=\"mt-content-container\">\n<div>\n<div class=\"textbox learning-objectives\">\n<h3 class=\"boxtitle\">Skills to Develop<\/h3>\n<ul>\n<li>Describe the hybrid orbitals used in the formation of bonding for each atom in some carbon containing compounds.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div id=\"section_2\">\n<h3 class=\"editable\"><strong><em>sp<sup>3<\/sup><\/em><\/strong> Hybrid orbitals and tetrahedral bonding<\/h3>\n<p>Now let\u2019s look more carefully at bonding in organic molecules, starting with methane, CH<sub>4<\/sub>. Recall the valence electron configuration of a carbon atom:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/04145949\/fig2-1-6.png\" alt=\"\" width=\"278\" height=\"111\" \/><\/p>\n<p>This picture is problematic when it comes to describing the <em>bonding<\/em> in methane. How does the carbon form four bonds if it has only two half-filled <em>p<\/em> orbitals available for bonding?\u00a0\u00a0 A hint comes from the experimental observation that the four C-H bonds in methane are arranged with <strong>tetrahedral<\/strong> geometry about the central carbon, and that each bond has the same length and strength.\u00a0 In order to explain this observation, valence bond theory relies on a concept called <strong>orbital hybridization<\/strong>.\u00a0 In this picture, the four valence orbitals of the carbon (one 2<em>s<\/em> and three 2<em>p<\/em> orbitals) combine mathematically (remember: orbitals are described by wave equations) to form four equivalent <strong>hybrid orbitals<\/strong>, which are called <strong><em>sp<sup>3<\/sup><\/em> orbitals <\/strong>because they are formed from mixing one <em>s<\/em> and three <em>p<\/em> orbitals. In the new electron configuration, each of the four valence electrons on the carbon occupies a single <em>sp<sup>3<\/sup><\/em> orbital.<\/p>\n<table style=\"border-collapse: collapse;width: 100%\">\n<tbody>\n<tr>\n<td style=\"width: 33.3333%\"><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/04145952\/fig2-1-7.png\" alt=\"\" width=\"252\" height=\"139\" \/><\/td>\n<td style=\"width: 33.3333%\">Click to see an <a class=\"external\" title=\"http:\/\/www.chemtube3d.com\/orbitalshybrid.htm\" href=\"http:\/\/www.chemtube3d.com\/orbitalshybrid.htm\" target=\"_blank\" rel=\"external nofollow noopener\">interactive 3D model<\/a><\/p>\n<p>(<em>select &#8216;load sp<sup>3<\/sup>&#8216; and &#8216;load H 1s&#8217; to see orbitals<\/em>).<\/td>\n<td style=\"width: 33.3333%\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-4054 alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/26151242\/frame-6-150x150.png\" alt=\"\" width=\"150\" height=\"150\" \/><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>This geometric arrangement makes perfect sense if you consider that it is precisely this angle that allows the four orbitals (and the electrons in them) to be as far apart from each other as possible.\u00a0 This is simply a restatement of the Valence Shell Electron Pair Repulsion (VSEPR) theory that you learned in General Chemistry: electron pairs (in orbitals) will arrange themselves in such a way as to remain as far apart as possible, due to negative-negative electrostatic repulsion.<\/p>\n<p>Each C-H bond in methane, then, can be described as a sigma bond formed by overlap between a half-filled 1<em>s<\/em> orbital in a hydrogen atom and the larger lobe of one of the four half-filled <em>sp<sup>3<\/sup><\/em> hybrid orbitals in the central carbon. The length of the carbon-hydrogen bonds in methane is 109 pm.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/04145955\/fig2-1-8.png\" alt=\"\" width=\"529\" height=\"179\" \/><\/p>\n<p>While previously we drew a Lewis structure of methane in two dimensions using lines to denote each covalent bond, we can now draw a more accurate structure in three dimensions, showing the tetrahedral bonding geometry.\u00a0 To do this on a two-dimensional page, though, we need to introduce a new drawing convention: the solid \/ dashed wedge system.\u00a0 In this convention, a solid wedge simply represents a bond that is meant to be pictured emerging from the plane of the page.\u00a0 A dashed wedge represents a bond that is meant to be pictured pointing into, or behind, the plane of the page.\u00a0 Normal lines imply bonds that lie in the plane of the page.\u00a0 This system takes a little bit of getting used to, but with practice your eye will learn to immediately \u2018see\u2019 the third dimension being depicted.<\/p>\n<div>\n<div id=\"exercise\">\n<div class=\"textbox examples\">\n<h3>Example<\/h3>\n<p>Imagine that you could distinguish between the four hydrogen atoms in a methane molecule, and labeled them H<sub>a<\/sub> through H<sub>d<\/sub>.\u00a0 In the images below, the <em>exact same<\/em> methane molecule is rotated and flipped in various positions.\u00a0 Draw the missing hydrogen atom labels. (It will be much easier to do this if you make a model.)<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/04145959\/figE2-1-1.png\" alt=\"\" width=\"393\" height=\"173\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q129643\">Show Solution<\/span><\/p>\n<div id=\"q129643\" class=\"hidden-answer\" style=\"display: none\">\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/107044\/E2-1S.png?revision=1&amp;size=bestfit&amp;width=488&amp;height=95\" alt=\"\" width=\"488\" height=\"95\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example<\/h3>\n<p>What kind of orbitals overlap to form the C-Cl bonds in chloroform, CHCl<sub>3<\/sub>?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q204888\">Show Solution<\/span><\/p>\n<div id=\"q204888\" class=\"hidden-answer\" style=\"display: none\">\n<p>sp<sup>3<\/sup> orbital on carbon overlapping with an sp<sup>3<\/sup> orbital on chlorine.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"mt-section\">\n<p>Diamond is a crystal form of elemental carbon, and the structure is particularly interesting. In the crystal, every carbon atom is bonded to four other carbon atoms, and the bonds are arranged in a tetrahedral fashion. The bonding, no doubt, is due to the <em>sp<\/em><sup>3<\/sup> hybrid orbitals. The bond length of 154 pm is the same as the C-C bond length in ethane, propane and other alkanes.<\/p>\n<p>An idealized single crystal of diamond is a gigantic molecule, because all the atoms are inter-bonded. The bonding has given diamond some very unusual properties. It is the hardest stone, much harder than anything else in the material world. It is a poor conductor, because all electrons are localized in the chemical bonds. However, diamond is an excellent heat conductor. A stone made of pure carbon is colorless, but the presence of impurities gives it various colors. The index of refraction is very high, and their glitter (sparkle or splendor) has made them the most precious stones.<\/p>\n<\/div>\n<p>How does this bonding picture extend to compounds containing carbon-carbon bonds?\u00a0 In ethane (CH<sub>3<\/sub>CH<sub>3<\/sub>), both carbons are <em>sp<sup>3<\/sup><\/em>-hybridized, meaning that both have four bonds with tetrahedral geometry.\u00a0 The carbon-carbon bond, with a bond length of 154 pm, is formed by overlap of one <em>sp<\/em><sup>3<\/sup> orbital from each of the carbons, while the six carbon-hydrogen bonds are formed from overlaps between the remaining <em>sp<sup>3<\/sup><\/em> orbitals on the two carbons and the 1<em>s<\/em> orbitals of hydrogen atoms.\u00a0 All of these are sigma bonds.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/04150006\/fig2-1-9.png\" alt=\"\" width=\"396\" height=\"171\" \/><\/p>\n<p>Because they are formed from the end-on-end overlap of two orbitals, <em>s<\/em><em>igma bonds are free to rotate<\/em>.\u00a0 This means, in the case of ethane molecule, that the two methyl (CH<sub>3<\/sub>) groups can be pictured as two wheels on an axle, each one able to rotate with respect to the other.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/04150009\/fig2-1-10.png\" alt=\"\" width=\"331\" height=\"107\" \/><\/p>\n<p>In chapter 3 we will learn more about the implications of rotational freedom in sigma bonds, when we discuss the \u2018conformation\u2019 of organic molecules.<\/p>\n<p>The <em>sp<sup>3<\/sup><\/em> bonding picture is also used to described the bonding in amines, including ammonia, the simplest amine.\u00a0 Just like the carbon atom in methane, the central nitrogen in ammonia is <em>sp<sup>3<\/sup>&#8211;<\/em>hybridized.\u00a0 With nitrogen, however, there are five rather than four valence electrons to account for, meaning that three of the four hybrid orbitals are half-filled and available for bonding, while the fourth is fully occupied by a nonbonding pair (lone pair) of electrons.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/04150013\/fig2-1-11.png\" alt=\"\" width=\"530\" height=\"304\" \/><\/p>\n<p>The bonding arrangement here is also tetrahedral: the three N-H bonds of ammonia can be pictured as forming the base of a trigonal pyramid, with the fourth orbital, containing the lone pair, forming the top of the pyramid.\u00a0 Recall from your study of VSEPR theory in General Chemistry that the lone pair, with its slightly greater repulsive effect, \u2018pushes\u2019 the three N-H s bonds away from the top of the pyramid, meaning that the H-N-H bond angles are slightly less than tetrahedral, at 107.3\u02da rather than 109.5\u02da.<\/p>\n<p>VSEPR theory also predicts, accurately, that a water molecule is \u2018bent\u2019 at an angle of approximately 104.5\u02da. The bonding in water results from overlap of two of the four <em>sp<sup>3<\/sup><\/em> hybrid orbitals on oxygen with 1<em>s <\/em>orbitals on the two hydrogen atoms. The two nonbonding electron pairs on oxygen are located in the two remaining <em>sp<sup>3<\/sup><\/em>orbitals. <a class=\"thumb\" title=\"struc1fig66.png\" href=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/98074\/struc1fig66.png?revision=1\" rel=\"internal\"><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/04150018\/struc1fig66.png\" alt=\"struc1fig66.png\" width=\"652\" height=\"351\" \/><\/a><\/p>\n<div>\n<div class=\"textbox examples\">\n<h3>Example<\/h3>\n<p>Draw, in the same style as the figures above, orbital pictures for the bonding in a) methylamine (H<sub>3<\/sub>CNH<sub>2<\/sub>), and b) ethanol (H<sub>3<\/sub>C-CH<sub>2<\/sub>-OH.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q686129\">Show Solution<\/span><\/p>\n<div id=\"q686129\" class=\"hidden-answer\" style=\"display: none\">\n<p>Both the carbon and the nitrogen atom in CH<sub>3<\/sub>NH<sub>2<\/sub> are sp<sup>3<\/sup>-hybridized. The C-N sigma bond is an overlap between two sp<sup>3<\/sup> orbitals.<\/p>\n<p>Both the carbon and the nitrogen atom in CH<sub>3<\/sub>NH<sub>2<\/sub>\u00a0are\u00a0<em>s<\/em>p<sup>3<\/sup>-hybridized.\u00a0 The C-N\u00a0sigma\u00a0bond is an overlap between two\u00a0<em>s<\/em>p<sup>3\u00a0<\/sup>orbitals.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/107045\/E2-3S.png?revision=1&amp;size=bestfit&amp;width=437&amp;height=191\" alt=\"\" width=\"437\" height=\"191\" \/><\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>See a <a class=\"link-https\" title=\"https:\/\/www.khanacademy.org\/science\/organic-chemistry\/gen-chem-review\/hybrid-orbitals-jay\/v\/sp3-hybridized-orbitals-and-sigma-bonds\" href=\"https:\/\/www.khanacademy.org\/science\/organic-chemistry\/gen-chem-review\/hybrid-orbitals-jay\/v\/sp3-hybridized-orbitals-and-sigma-bonds\" target=\"_blank\" rel=\"external nofollow noopener\">video tutorial on sp<sup>3<\/sup> orbitals and sigma bonds<\/a> (Note: This is the video linked to in the previous section)<\/p>\n<\/div>\n<div id=\"section_3\">\n<h3 class=\"editable\"><strong><em>sp<sup>2<\/sup><\/em><\/strong> and <strong><em>sp<\/em><\/strong> hybrid orbitals and pi bonds<\/h3>\n<p>The valence bond theory, along with the hybrid orbital concept, does a very good job of describing double-bonded compounds such as ethene.\u00a0Three experimentally observable characteristics of the ethene molecule need to be accounted for by a bonding model:<\/p>\n<ol>\n<li>Ethene is a planar (flat) molecule.<\/li>\n<li>Bond angles in ethene are approximately 120<sup>o<\/sup>, and the carbon-carbon bond length is 134 pm, significantly shorter than the 154 pm single carbon-carbon bond in ethane.<\/li>\n<li>There is a significant barrier to rotation about the carbon-carbon double bond.<\/li>\n<\/ol>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/04150021\/fig2-1-13.png\" alt=\"fig2-1-13.png\" width=\"180\" height=\"166\" \/><\/p>\n<p>Clearly, these characteristics are not consistent with an <em>sp<sup>3<\/sup><\/em> hybrid bonding picture for the two carbon atoms.\u00a0Instead, the bonding in ethene is described by a model involving the participation of a different kind of hybrid orbital.\u00a0Three atomic orbitals on each carbon \u2013 the 2<em>s<\/em>, 2<em>p<\/em><sub>x<\/sub> and 2<em>p<\/em><sub>y<\/sub> orbitals \u2013 combine to form three <strong><em>sp<sup>2<\/sup><\/em> hybrids<\/strong>, leaving the 2<em>p<\/em><sub>z<\/sub> orbital unhybridized.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/04150024\/fig2-1-14.png\" alt=\"fig2-1-14.png\" width=\"657\" height=\"192\" \/><\/p>\n<p>The three <em>sp<sup>2<\/sup><\/em> hybrids are arranged with trigonal planar geometry, pointing to the three corners of an equilateral triangle, with angles of 120\u00b0 between them.\u00a0The unhybridized 2<em>p<\/em><sub>z<\/sub> orbital is <em>perpendicular<\/em> to this plane (in the next several figures, <em>sp<sup>2<\/sup><\/em> orbitals and the sigma bonds to which they contribute are represented by lines and wedges; only the 2<em>p<\/em><sub>z<\/sub> orbitals are shown in the &#8216;space-filling&#8217; mode).<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/04150031\/fig2-1-15.png\" alt=\"fig2-1-15.png\" width=\"248\" height=\"168\" \/><\/p>\n<p>The carbon-carbon double bond in ethene consists of one sigma bond, formed by the overlap of two <em>sp<sup>2<\/sup><\/em> orbitals, and a second bond, called a <strong>pi<\/strong><strong> bond<\/strong>, which is formed by the <em>side-by-side<\/em> overlap of the two unhybridized 2<em>p<\/em><sub>z<\/sub> orbitals from each carbon.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/04150041\/fig2-1-16.png\" alt=\"fig2-1-16.png\" width=\"167\" height=\"132\" \/><\/p>\n<table style=\"border-collapse: collapse;width: 100%\">\n<tbody>\n<tr>\n<td style=\"width: 18.6743%;text-align: center\"><a class=\"link-https\" title=\"https:\/\/www.youtube.com\/watch?v=C2W-yDPcpl4\" href=\"https:\/\/www.youtube.com\/watch?v=C2W-yDPcpl4\" target=\"_blank\" rel=\"external nofollow noopener\">animation\u00a0<\/a><\/td>\n<td style=\"width: 25.4038%;text-align: center\"><a class=\"external\" title=\"http:\/\/andromeda.rutgers.edu\/~huskey\/images\/ethylene_bonding.jpg\" href=\"http:\/\/andromeda.rutgers.edu\/~huskey\/images\/ethylene_bonding.jpg\" target=\"_blank\" rel=\"external nofollow noopener\">spacefilling image<\/a><\/td>\n<td style=\"width: 24.0578%;text-align: center\"><a class=\"link-https\" title=\"https:\/\/www.khanacademy.org\/science\/organic-chemistry\/gen-chem-review\/hybrid-orbitals-jay\/v\/pi-bonds-and-sp2-hybridized-orbitals\" href=\"https:\/\/www.khanacademy.org\/science\/organic-chemistry\/gen-chem-review\/hybrid-orbitals-jay\/v\/pi-bonds-and-sp2-hybridized-orbitals\" target=\"_blank\" rel=\"external nofollow noopener\">\u00a0video tutoria<\/a>l<\/td>\n<td style=\"width: 31.8641%\">\n<p style=\"text-align: center\"><a href=\"http:\/\/www.chemtube3d.com\/orbitalsethene.htm\">interactive 3D model<\/a><br \/>\n<em>(select &#8216;show resulting pi orbital&#8217;)<\/em><\/p>\n<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 18.6743%\"><\/td>\n<td style=\"width: 25.4038%\"><\/td>\n<td style=\"width: 24.0578%\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-4055 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/26151546\/frame-7-150x150.png\" alt=\"\" width=\"150\" height=\"150\" \/><\/td>\n<td style=\"width: 31.8641%\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-4056 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/26151756\/frame-8-150x150.png\" alt=\"\" width=\"151\" height=\"151\" \/><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><em>\u00a0<\/em>This illustration (from University of Florida) shows the sigma and pi bonds in ethene.<\/p>\n<p class=\"mt-align-center\"><img decoding=\"async\" class=\"internal aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/22193029\/ethene.gif\" alt=\"ethene.gif\" \/><\/p>\n<p>Unlike a sigma bond, a pi bond does <em>not<\/em> have cylindrical symmetry. If rotation about this bond were to occur, it would involve disrupting the side-by-side overlap between the two 2<em>p<\/em><sub>z<\/sub> orbitals that make up the pi bond.\u00a0 The presence of the pi bond thus \u2018locks\u2019 the six atoms of ethene into the same plane. This argument extends to larger alkene groups: in each case, six atoms lie in the same plane.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/04150044\/fig2-1-17.png\" alt=\"fig2-1-17.png\" width=\"182\" height=\"148\" \/><\/p>\n<div class=\"mt-section\"><\/div>\n<div>\n<div class=\"textbox examples\">\n<h3>Example<\/h3>\n<p>Redraw the structures below, indicating the six atoms that lie in the same plane due to the carbon-carbon double bond.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q15124\">Show Solution<\/span><\/p>\n<div id=\"q15124\" class=\"hidden-answer\" style=\"display: none\">\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/04150048\/figE2-1-4.png\" alt=\"figE2-1-4.png\" width=\"299\" height=\"58\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example<\/h3>\n<p>What is wrong with the way the following structure is drawn?<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/04150050\/figE2-1-5.png\" alt=\"figE2-1-5.png\" width=\"191\" height=\"103\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q437064\">Show Answer<\/span><\/p>\n<div id=\"q437064\" class=\"hidden-answer\" style=\"display: none\">\n<p>The carbon atoms in an aromatic ring are sp2 hybridized, thus bonding geometry is trigonal planar: in other words, the bonds coming out of the ring are in the same plane as the ring, not pointing above the plane of the ring as the wedges in the incorrect drawing indicate. A correct drawing should use lines to indicate that the bonds are in the same plane as the ring:<img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/107047\/E2-5S-b.png?revision=1&amp;size=bestfit&amp;width=624&amp;height=132\" alt=\"\" width=\"624\" height=\"132\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>A similar picture can be drawn for the bonding in carbonyl groups, such as formaldehyde.\u00a0In this molecule, the carbon is <em>sp<sup>2<\/sup><\/em>-hybridized, and we will assume that the oxygen atom is also <em>sp<sup>2<\/sup><\/em>hybridized.\u00a0The carbon has three sigma bonds: two are formed by overlap between <em>sp<sup>2<\/sup><\/em> orbitals with 1<em>s <\/em>orbitals from hydrogen atoms, and the third sigma bond is formed by overlap between the remaining carbon <em>sp<sup>2<\/sup><\/em> orbital and an <em>sp<sup>2<\/sup><\/em> orbital on the oxygen.\u00a0The two lone pairs on oxygen occupy its other two <em>sp<sup>2<\/sup><\/em> orbitals.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/04150054\/fig2-1-18.png\" alt=\"fig2-1-18.png\" width=\"489\" height=\"184\" \/><\/p>\n<p><a class=\"external\" title=\"http:\/\/www.chemtube3d.com\/orbitalsformaldehyde.htm\" href=\"http:\/\/www.chemtube3d.com\/orbitalsformaldehyde.htm\" target=\"_blank\" rel=\"external nofollow noopener\">interactive 3D model<\/a><\/p>\n<p>The pi bond is formed by side-by-side overlap of the unhybridized 2<em>p<\/em><sub>z<\/sub> orbitals on the carbon and the oxygen. Just like in alkenes, the 2<em>p<\/em><sub>z<\/sub> orbitals that form the pi bond are perpendicular to the plane formed by the sigma bonds.<\/p>\n<div>\n<div class=\"textbox examples\">\n<h3>Example<\/h3>\n<p>a: Draw a diagram of hybrid orbitals in an <em>sp<sup>2<\/sup><\/em>-hybridized nitrogen.<\/p>\n<p>b: Draw a figure showing the bonding picture for the imine below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/04150059\/figE2-1-6.png\" alt=\"figE2-1-6.png\" width=\"84\" height=\"96\" \/><\/p>\n<p>c: In your drawing for part b, what kind of orbital holds the nitrogen lone pair?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q990885\">Show Answer<\/span><\/p>\n<div id=\"q990885\" class=\"hidden-answer\" style=\"display: none\">\n<p>a) The carbon and nitrogen atoms are both\u00a0<em>sp<\/em><sup>2<\/sup>\u00a0hybridized.\u00a0 The carbon-nitrogen double bond is composed of a\u00a0sigma\u00a0bond formed from two\u00a0<em>sp<\/em><sup>2<\/sup>\u00a0orbitals, and a\u00a0pi\u00a0bond formed from the side-by-side overlap of two unhybridized\u00a0<em>2p<\/em>\u00a0orbitals.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/chem.libretexts.org\/@api\/deki\/files\/107048\/E2-6S.png?revision=1&amp;size=bestfit&amp;width=342&amp;height=192\" alt=\"\" width=\"342\" height=\"192\" \/><\/p>\n<p>b) As shown in the figure above, the nitrogen lone pair electrons occupy one of the three\u00a0<em>s<\/em><em>p<sup>2<\/sup><\/em>\u00a0hybrid orbitals.\u00a0<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<h3 class=\"editable\">Bonding carbon atoms using <em>sp<\/em> hybrid orbitals<\/h3>\n<p>Consider, for example, the structure of ethyne (common name acetylene), the simplest alkyne.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/04150103\/fig2-1-20.png\" alt=\"fig2-1-20.png\" width=\"162\" height=\"73\" \/><\/p>\n<p>Both the VSEPR theory and experimental evidence tells us that the molecule is linear: all four atoms lie in a straight line.\u00a0The carbon-carbon triple bond is only 120 pm long, shorter than the double bond in ethene, and is very strong, about 837 kJ\/mol.\u00a0In the hybrid orbital picture of acetylene, both carbons are <strong><em>sp-<\/em>hybridized<\/strong>.\u00a0 In an <em>sp<\/em>-hybridized carbon,\u00a0 the 2<em>s<\/em> orbital combines with the 2<em>p<\/em><sub>x<\/sub> orbital to form two <em>sp<\/em> hybrid orbitals that are oriented at an angle of 180\u00b0 with respect to each other (eg. along the x axis).\u00a0The 2<em>p<\/em><sub>y<\/sub> and 2<em>p<\/em><sub>z<\/sub> orbitals remain unhybridized, and are oriented perpendicularly along the y and z axes, respectively.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/04150108\/fig2-1-21.png\" alt=\"fig2-1-21.png\" width=\"594\" height=\"317\" \/><\/p>\n<p>The carbon-carbon sigma bond, then, is formed by the overlap of one <em>sp<\/em> orbital from each of the carbons, while the two carbon-hydrogen sigma bonds are formed by the overlap of the second <em>sp<\/em> orbital on each carbon with a 1<em>s<\/em> orbital on a hydrogen.\u00a0Each carbon atom still has two half-filled 2<em>p<\/em><sub>y<\/sub> and 2<em>p<\/em><sub>z<\/sub> orbitals, which are perpendicular both to each other and to the line formed by the sigma bonds.\u00a0These two perpendicular pairs of <em>p<\/em> orbitals form two pi bonds between the carbons, resulting in a triple bond overall (one sigma bond plus two pi bonds).<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/04150111\/fig2-1-22.png\" alt=\"fig2-1-22.png\" width=\"416\" height=\"111\" \/><\/p>\n<p>When <em>sp<\/em> hybrid orbitals are used for the sigma bond, the two sigma bonds around the carbon are linear. Two other <em>p<\/em> orbitals are available for pi bonding, and a typical compound is the acetylene or ethyne HC\u2261CH. The three sigma and two pi bonds of this molecule can be seen in this diagram from <a class=\"external\" href=\"http:\/\/chem.ufl.edu\/%7Echm2040\/Notes\/Chapter_12\/multiple.html\" target=\"_blank\" rel=\"external nofollow noopener\">University of Florida: General chemistry<\/a> shown below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/22193027\/hcch.gif\" alt=\"\" width=\"432\" height=\"213\" \/><\/p>\n<p>Note that molecules H-C\u2261C-H, H-C\u2261N, and \u00afC\u2261O<sup>+<\/sup> have the same number of electrons. Bonding in these molecules can be explained by the same theory, and thus their formation is no surprise.<\/p>\n<p><a class=\"external\" title=\"http:\/\/www.chemtube3d.com\/orbitalsacetylene.htm\" href=\"http:\/\/www.chemtube3d.com\/orbitalsacetylene.htm\" target=\"_blank\" rel=\"external nofollow noopener\">interactive 3D model<\/a><\/p>\n<div class=\"mt-section\">\n<h3 class=\"editable\">Comparison of Structural Features<\/h3>\n<p>Some typical bonding features of ethane, ethene, and ethyne are summarized in the table below:<\/p>\n<table class=\"mt-responsive-table\">\n<thead>\n<tr>\n<th class=\"mt-align-center\" scope=\"col\">Systematic<br \/>\nname<\/th>\n<th class=\"mt-align-center\" scope=\"col\">Ethane<\/th>\n<th class=\"mt-align-center\" scope=\"col\">Ethene<\/th>\n<th class=\"mt-align-center\" scope=\"col\">Ethyne<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>Hybrid orbitals<br \/>\nof $$\\ce{C}$$<\/td>\n<th><em>sp<\/em><sup>3<\/sup><\/th>\n<th><em>sp<\/em><sup>2<\/sup><\/th>\n<th><em>sp<\/em><\/th>\n<\/tr>\n<tr>\n<td>Structural<br \/>\nformula<\/td>\n<th>H H<br \/>\n\\ \u00a0\/<br \/>\nH&#8211;C&#8212;C&#8211;H<br \/>\n\/ \u00a0\\<br \/>\nH H<\/th>\n<th>H H<br \/>\n\\ \u00a0\/<br \/>\nC=C<br \/>\n\/ \u00a0\\<br \/>\nH H<\/th>\n<th>H-C<span class=\"st\">\u2261<\/span>C-H<\/th>\n<\/tr>\n<tr>\n<td>$$\\ce{C-C}$$<br \/>\nBondlength pm<\/td>\n<th>154<\/th>\n<th>134<\/th>\n<th>120<\/th>\n<\/tr>\n<tr>\n<td>$$\\ce{C-H}$$<br \/>\nBondlength pm<\/td>\n<th>112<\/th>\n<th>110<\/th>\n<th>106<\/th>\n<\/tr>\n<tr>\n<td>$$\\ce{H-C-C}$$<br \/>\nbond angle \u00b0<\/td>\n<th>111<\/th>\n<th>121<\/th>\n<th>180<\/th>\n<\/tr>\n<tr>\n<td>$$\\ce{C-C}$$ bond<br \/>\nenergy kJ\/mol<\/td>\n<th>368<\/th>\n<th>611<\/th>\n<th>820<\/th>\n<\/tr>\n<tr>\n<td>$$\\ce{C-H}$$ bond<br \/>\nenergy kJ\/mol<\/td>\n<th>410<\/th>\n<th>451<\/th>\n<th>536<\/th>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>As the bond order between carbon atoms increases from 1 to 3 for ethane, ethene, and ethyne, the bond lengths decrease, and the bond energy increases. Note that the bond energies given here are specific for these compounds, and the values may be different from the average values for this type of bonds.<\/p>\n<\/div>\n<div class=\"mt-section\">\n<h2 class=\"editable\">Confidence building questions<\/h2>\n<ol>\n<li><strong>What hybrid orbitals are used by the carbon atoms in HC\u2261CH? <\/strong><strong>Discussion &#8211; <\/strong><br \/>\nLinear -C- bonds due to <em>sp<\/em> hybridized orbitals. This molecule is linear, and it consists of 3 sigma, \u03c3, bonds, and two pi, \u03c0, bonds. Compare the bonding of this with \u00afC\u2261O<sup>+<\/sup>, H-C\u2261N, and CH<sub>3<\/sub>-C\u2261N.<\/li>\n<li><strong>What hybrid orbitals are used by the carbon atoms in H<sub>2<\/sub>C=CH<sub>2<\/sub>? <\/strong><strong>Discussion &#8211; <\/strong><br \/>\nPlanar -C&lt; \u03c3 bonds due to <em>sp<\/em><sup>2<\/sup> hybridized orbitals. Another <em>p<\/em> orbital is used for the pi, \u03c0. How many sigma and pi bonds does this molecule have? Do all atoms in this molecule lie on the same plane?<\/li>\n<li><strong>What hybrid orbitals are used by the<\/strong><strong> carbon in CH<sub>3<\/sub>CH<sub>3<\/sub>? <\/strong><strong>Discussion &#8211; <\/strong><br \/>\nTetrahedral arrangement around C is due to <em>sp<\/em><sup>3<\/sup> hybridized orbitals.<\/li>\n<li><strong>What hybrid orbitals are used by the oxygen in CH<sub>3<\/sub>CH<sub>2<\/sub>OH? Discussion &#8211; <\/strong><br \/>\nTetrahedral arrangement around O is due to <em>sp<\/em><sup>3<\/sup> hybridized orbitals.<\/li>\n<li><strong>Which carbon to carbon bond length is the shortest in the following molecule: CH<sub>3<\/sub>-C\u2261C-CH<sub>2<\/sub>-CH=CH-CH<sub>2<\/sub>-OH? <\/strong><strong>Discussion &#8211; <\/strong><br \/>\nThe bond length decreases as the bond order increases, so the triple bond is the shortest.<\/li>\n<li><strong>How many carbon atoms make use of <em>sp<\/em><sup>2<\/sup> hybrid orbitals in this molecule: CH<sub>3<\/sub>-C\u2261C-CH<sub>2<\/sub>-CH=CH-CH<sub>2<\/sub>-COOH? <\/strong><strong>Discussion &#8211; <\/strong><br \/>\nRecognize that the type of bonding is important. For more guidance, see the next section in this book.<\/li>\n<li><strong>The molecular formula of caffeine is C<sub>8<\/sub>H<sub>10<\/sub>N<sub>4<\/sub>O<sub>2<\/sub>; draw a reasonable structure for it. <\/strong><strong>Discussion &#8211; <\/strong><br \/>\nThe structure is shown below. Can you sketch a bonding structure for caffeine? How many carbon atoms makes use of <em>sp<\/em><sup>2<\/sup> hybrid orbitals?<\/li>\n<\/ol>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2018\/06\/22193033\/caffeine.gif\" alt=\"\" width=\"464\" height=\"464\" \/><\/p>\n<\/div>\n<div id=\"section_6\" class=\"mt-section\">\n<div>\n<div class=\"textbox examples\">\n<h3>Example<\/h3>\n<p>a) What kinds of orbitals are overlapping in bonds b-i indicated below?\u00a0 Be sure to distinguish between s and p bonds. An example is provided for bond &#8216;a&#8217;.<\/p>\n<p>b) In what kind of orbital is the lone pair of electrons located on the nitrogen atom of bond a?\u00a0 Of bond e?<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"internal default aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1518\/2017\/10\/04150119\/fig2-1-24.png\" alt=\"fig2-1-24.png\" width=\"642\" height=\"343\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q879188\">Show Solution<\/span><\/p>\n<div id=\"q879188\" class=\"hidden-answer\" style=\"display: none\">\n<p>a) bond b: Nsp2-Csp3 (this means an overlap of an sp2 orbital on N and an sp3 orbital on C)<\/p>\n<p>bond c: Csp2-Csp2 plus C2p-C2p (pi)<\/p>\n<p>bond d: Csp2-Csp3<\/p>\n<p>bond e: Csp3-Csp3<\/p>\n<p>bond f: Csp3-Csp3<\/p>\n<p>bond g: Csp2-Csp2 (s) plus C2p-C2p (pi)<\/p>\n<p>bond h: Csp2-H1s<\/p>\n<p>bond i: Csp2-Csp2<\/p>\n<p>b) bond a: lone pair on N occupies an sp2 orbital<\/p>\n<p>bond e: lone pair on N occupies an sp3 orbital<\/p><\/div>\n<\/div>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"sp3 hybridized orbitals and sigma bonds | Structure and bonding | Organic chemistry | Khan Academy\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/u1eGSL6J6Fo?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h3><img loading=\"lazy\" decoding=\"async\" class=\"size-thumbnail wp-image-4048 alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3369\/2017\/10\/26145837\/frame-4-150x150.png\" alt=\"\" width=\"150\" height=\"150\" \/><\/h3>\n<\/div>\n<\/section>\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-3168\">\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>Taken from 2.1: Valence Bond Theory. <strong>Authored by<\/strong>: Tim Soderberg. <strong>Provided by<\/strong>: University of Minnesota, Morris. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/chem.libretexts.org\/Textbook_Maps\/Organic_Chemistry\/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_(Soderberg)\/Chapter_02%3A_Introduction_to_organic_structure_and_bonding_II\/2.1%3A_Valence_Bond_Theory\">https:\/\/chem.libretexts.org\/Textbook_Maps\/Organic_Chemistry\/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_(Soderberg)\/Chapter_02%3A_Introduction_to_organic_structure_and_bonding_II\/2.1%3A_Valence_Bond_Theory<\/a>. <strong>Project<\/strong>: Organic Chemistry With a Biological Emphasis. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em><\/li><li>Taken from Hybrid Orbitals in Carbon Compounds. <strong>Authored by<\/strong>: Chung Chieh. <strong>Provided by<\/strong>: University of Waterloo, Ontario. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/www.science.uwaterloo.ca\/~cchieh\/cact\/\">http:\/\/www.science.uwaterloo.ca\/~cchieh\/cact\/<\/a>. <strong>Project<\/strong>: Libretexts. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em><\/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":311,"menu_order":2,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Taken from 2.1: Valence Bond Theory\",\"author\":\"Tim Soderberg\",\"organization\":\"University 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Ontario\",\"url\":\"http:\/\/www.science.uwaterloo.ca\/~cchieh\/cact\/\",\"project\":\"Libretexts\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-3168","chapter","type-chapter","status-publish","hentry"],"part":20,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/wp-json\/pressbooks\/v2\/chapters\/3168","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/wp-json\/wp\/v2\/users\/311"}],"version-history":[{"count":30,"href":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/wp-json\/pressbooks\/v2\/chapters\/3168\/revisions"}],"predecessor-version":[{"id":5092,"href":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/wp-json\/pressbooks\/v2\/chapters\/3168\/revisions\/5092"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/wp-json\/pressbooks\/v2\/parts\/20"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/wp-json\/pressbooks\/v2\/chapters\/3168\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/wp-json\/wp\/v2\/media?parent=3168"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/wp-json\/pressbooks\/v2\/chapter-type?post=3168"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/wp-json\/wp\/v2\/contributor?post=3168"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-potsdam-organicchemistry\/wp-json\/wp\/v2\/license?post=3168"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}