{"id":229,"date":"2017-10-04T15:03:30","date_gmt":"2017-10-04T15:03:30","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/?post_type=chapter&p=229"},"modified":"2017-10-18T18:46:51","modified_gmt":"2017-10-18T18:46:51","slug":"valence-bond-theory","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-mcc-organicchemistry\/chapter\/valence-bond-theory\/","title":{"raw":"Valence Bond Theory","rendered":"Valence Bond Theory"},"content":{"raw":"
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As we have been discussing how to use Lewis structures to depict the bonding in organic compounds, we have been very vague so far in our language about the actual nature of the chemical bonds themselves. We know that a covalent bond involves the \u2018sharing\u2019 of a pair of electrons between two atoms - but how does this happen, and how does it lead to the formation of a bond holding the two atoms together?<\/div>\r\n<\/div>\r\n
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\r\n\r\nValence bond theory<\/strong> is most often used to describe bonding in organic molecules.\u00a0 In this model, bonds are considered to form from the overlap of two atomic orbitals on different atoms, each orbital containing a single electron. In looking at simple inorganic molecules such as molecular hydrogen (H2<\/sub>) or hydrogen fluoride (HF), our present understanding of s <\/em>and p<\/em> atomic orbitals will suffice. In order to explain the bonding in organic molecules, however, we will need to introduce the concept of hybrid orbitals<\/strong>.<\/span><\/span>\r\n
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The sigma bond in the H2<\/sub> molecule<\/span><\/span><\/h4>\r\nThe simplest case to consider is the hydrogen molecule, H2<\/sub>.\u00a0When we say that the two hydrogen nuclei share their electrons to form a covalent bond, what we mean in valence bond theory terms is that the two spherical 1s<\/em> orbitals (the grey spheres in the figure below) overlap, and contain two electrons with opposite spin.<\/span>\r\n\r\n\"\"\r\n\r\nThese two electrons are now attracted to the positive charge of both<\/em> of the hydrogen nuclei, with the result that they serve as a sort of \u2018chemical glue\u2019 holding the two nuclei together.<\/span>\r\n\r\nHow far apart are the two nuclei? If they are too far apart, their respective 1s<\/em> orbitals cannot overlap, and thus no covalent bond can form - they are still just two separate hydrogen atoms.\u00a0 As they move closer and closer together, orbital overlap begins to occur, and a bond begins to form. This lowers the potential energy of the system, as new, attractive<\/em> positive-negative electrostatic interactions become possible between the nucleus of one atom and the electron of the second. <\/span>\r\n\r\nBut something else is happening at the same time: as the atoms get closer, the repulsive<\/em> positive-positive interaction between the two nuclei also begins to increase.\u00a0 <\/span>\r\n\r\n\"\"\r\n\r\nAt first this repulsion is more than offset by the attraction between nuclei and electrons, but at a certain point, as the nuclei get even closer, the repulsive forces begin to overcome the attractive forces, and the potential energy of the system rises quickly. When the two nuclei are \u2018too close\u2019, we have an unstable, high-energy situation. There is a defined optimal distance between the nuclei in which the potential energy is at a minimum, meaning that the combined attractive and repulsive forces add up to the greatest overall attractive force.\u00a0This optimal internuclear distance is the bond length<\/strong>. For the H2<\/sub> molecule, the distance is 74 pm (picometers, 10-12<\/sup> meters).\u00a0Likewise, the difference in potential energy between the lowest energy state (at the optimal internuclear distance) and the state where the two atoms are completely separated is called the bond dissociation energy, <\/strong>or, more simply, bond strength<\/strong>.\u00a0For the hydrogen molecule, the H-H bond strength is equal to about 435 kJ\/mol. <\/span>\r\n\r\nEvery covalent bond in a given molecule has a characteristic length and strength.\u00a0In general, the length of a typical carbon-carbon single bond in an organic molecule is about 150 pm, while carbon-carbon double bonds are about 130 pm, carbon-oxygen double bonds are about 120 pm, and carbon-hydrogen bonds are in the range of 100 to 110 pm.\u00a0The strength of covalent bonds in organic molecules ranges from about 234 kJ\/mol for a carbon-iodine bond (in thyroid hormone, for example), about 410 kJ\/mole for a typical carbon-hydrogen bond, and up to over 800 kJ\/mole for a carbon-carbon triple bond.\u00a0 <\/span>\r\n\r\nTable of bond lengths and bond energies<\/span><\/a>\r\n\r\nIt is not\u00a0 accurate, however, to picture covalent bonds as rigid sticks of unchanging length - rather, it is better to picture them as springs<\/em> which have a defined length when relaxed, but which can be compressed, extended, and bent.\u00a0\u00a0 This \u2018springy\u2019 picture of covalent bonds will become very important in chapter 4, when we study the analytical technique known as infrared (IR) spectroscopy<\/a>.\u00a0 <\/span>\r\n\r\nOne more characteristic of the covalent bond in H2<\/sub> is important to consider at this point.\u00a0The two overlapping 1s<\/em> orbitals can be visualized as two spherical balloons being pressed together. This means that the bond has cylindrical symmetry<\/strong>: if we were to take a cross-sectional plane of the bond at any point, it would form a circle. This type of bond is referred to as a\u00a0<\/span>\u03c3<\/strong> (sigma) bond<\/strong>.<\/span>\r\n\r\n\"\"\r\n\r\nA sigma bond can be formed by overlap of an s <\/em>atomic orbital with a p<\/em> atomic orbital.\u00a0 Hydrogen fluoride (HF) is an example:<\/span><\/span>\r\n\r\n\"\"\r\n\r\nA sigma bond can also be formed by the overlap of two p <\/em>orbitals.\u00a0 The covalent bond in molecular fluorine, F2<\/sub>, is a sigma bond formed by the overlap of two half-filled 2p<\/em> orbitals, one from each fluorine atom.\u00a0 <\/span>\r\n\r\n\"\"<\/span>\r\n\r\n<\/div>\r\n
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sp3<\/sup><\/em><\/strong> hybrid orbitals and tetrahedral bonding<\/span><\/h4>\r\nNow let\u2019s look more carefully at bonding in organic molecules, starting with methane, CH4<\/sub>. Recall the valence electron configuration of a carbon atom:<\/span>\r\n\r\n\"\"\r\n\r\nThis picture is problematic when it comes to describing the bonding<\/em> in methane. How does the carbon form four bonds if it has only two half-filled 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 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 orbital hybridization<\/strong>.\u00a0 In this picture, the four valence orbitals of the carbon (one 2s<\/em> and three 2p<\/em> orbitals) combine mathematically (remember: orbitals are described by wave equations) to form four equivalent hybrid orbitals<\/strong>, which are called sp3<\/sup><\/em> orbitals <\/strong>because they are formed from mixing one s<\/em> and three p<\/em> orbitals. In the new electron configuration, each of the four valence electrons on the carbon occupies a single sp3<\/sup><\/em> orbital.<\/span>\r\n\r\n\"\"\r\n\r\ninteractive 3D model<\/a>\r\n(select 'load sp3<\/sup>' and 'load H 1s' to see orbitals<\/em>)\r\n\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.\u00a0 <\/span>\r\n\r\nEach C-H bond in methane, then, can be described as a sigma bond formed by overlap between a half-filled 1s<\/em> orbital in a hydrogen atom and the larger lobe of one of the four half-filled sp3<\/sup><\/em> hybrid orbitals in the central carbon. The length of the carbon-hydrogen bonds in methane is 109 pm.<\/span>\r\n\r\n\"\"\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.<\/span>\r\n
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Example<\/h3>\r\nImagine that you could distinguish between the four hydrogen atoms in a methane molecule, and labeled them Ha<\/sub> through Hd<\/sub>.\u00a0 In the images below, the 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.)<\/span>\r\n\r\n\"\"[reveal-answer q=\"129643\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"129643\"]\r\n\r\n\"\"\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n
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Example<\/h3>\r\nWhat kind of orbitals overlap to form the C-Cl bonds in chloroform, CHCl3<\/sub>?<\/span>\r\n[reveal-answer q=\"204888\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"204888\"]sp3 orbital on carbon overlapping with 3p orbital on chlorine. [\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nHow does this bonding picture extend to compounds containing carbon-carbon bonds?\u00a0 In ethane (CH3<\/sub>CH3<\/sub>), both carbons are sp3<\/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 sp<\/em>3<\/sup> orbital from each of the carbons, while the six carbon-hydrogen bonds are formed from overlaps between the remaining sp3<\/sup><\/em> orbitals on the two carbons and the 1s<\/em> orbitals of hydrogen atoms.\u00a0 All of these are sigma bonds.<\/span>\r\n\r\n\"\"\r\n\r\nBecause they are formed from the end-on-end overlap of two orbitals, s<\/em>igma bonds are free to rotate<\/em>.\u00a0 This means, in the case of ethane molecule, that the two methyl (CH3<\/sub>) groups can be pictured as two wheels on an axle, each one able to rotate with respect to the other.\u00a0 <\/span>\r\n\r\n\"\"<\/span>\r\n\r\nIn chapter 3 we will learn more about the implications of rotational freedom in <\/span><\/span>sigma bonds, when we discuss the \u2018conformation\u2019 of organic molecules.\u00a0 <\/span>\r\n\r\nThe sp3<\/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 sp3<\/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.\u00a0 <\/span>\r\n\r\n\"\"\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 <\/span><\/span>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.\u00a0 <\/span>\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 sp3<\/sup><\/em> hybrid orbitals on oxygen with 1s <\/em>orbitals on the two hydrogen atoms. The two nonbonding electron pairs on oxygen are located in the two remaining sp3<\/sup><\/em>orbitals. <\/span>\"struc1fig66.png\"<\/a>\r\n
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Examples<\/h3>\r\nDraw, in the same style as the figures above, orbital pictures for the bonding in a) methylamine, and b) ethanol.<\/span><\/span>\r\n[reveal-answer q=\"686129\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"686129\"]Both the carbon and the nitrogen atom in CH3NH2 are sp3-hybridized. The C-N sigma bond is an overlap between two sp3 orbitals.\r\n\r\nBoth the carbon and the nitrogen atom in CH3<\/sub>NH2<\/sub>\u00a0are\u00a0s<\/em>p3<\/sup>-hybridized.\u00a0 The C-N\u00a0<\/span>sigma<\/span>\u00a0bond is an overlap between two\u00a0s<\/em>p3\u00a0<\/sup>orbitals.<\/span>\r\n

\"\"[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\nvideo tutorial on sp3<\/sup> orbitals and sigma bonds<\/a>\r\n\r\n<\/div>\r\n

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sp2<\/sup><\/em><\/strong> and sp<\/em><\/strong> hybrid orbitals and pi bonds<\/span><\/span><\/h4>\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:<\/span>\r\n
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  1. Ethene is a planar (flat) molecule.<\/span><\/li>\r\n \t
  2. Bond angles in ethene are approximately 120o<\/sup>, and the carbon-carbon bond length is 134 pm, significantly shorter than the 154 pm single carbon-carbon bond in ethane.<\/span><\/li>\r\n \t
  3. There is a significant barrier to rotation about the carbon-carbon double bond. <\/span><\/li>\r\n<\/ol>\r\n\"fig2-1-13.png\"\r\n\r\nClearly, these characteristics are not consistent with an sp3<\/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 2s<\/em>, 2p<\/em>x<\/sub> and 2p<\/em>y<\/sub> orbitals \u2013 combine to form three sp2<\/sup><\/em> hybrids<\/strong>, leaving the 2p<\/em>z<\/sub> orbital unhybridized. <\/span>\r\n

    \"fig2-1-14.png\"<\/p>\r\nThe three sp2<\/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 2p<\/em>z<\/sub> orbital is perpendicular<\/em> to this plane (in the next several figures, sp2<\/sup><\/em> orbitals and the sigma bonds to which they contribute are represented by lines and wedges; only the 2p<\/em>z<\/sub> orbitals are shown in the 'space-filling' mode).<\/span>\r\n\r\n\"fig2-1-15.png\"<\/span>\r\n\r\nThe carbon-carbon double bond in ethene consists of one <\/span><\/span>sigma bond, formed by the overlap of two sp2<\/sup><\/em> orbitals, and a second bond, called a<\/span> pi<\/strong><\/span> bond<\/strong>, which is formed by the side-by-side<\/em> overlap of the two unhybridized 2p<\/em>z<\/sub> orbitals from each carbon.\u00a0 <\/span>\r\n\r\n\"fig2-1-16.png\"\r\n

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    Example<\/h3>\r\nRedraw the structures below, indicating the six atoms that lie in the same plane due to the carbon-carbon double bond.<\/span><\/span>\r\n[reveal-answer q=\"15124\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"15124\"]\r\n\r\n\"figE2-1-4.png\"\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n
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    Example<\/h3>\r\nWhat is wrong with the way the following structure is drawn?<\/span><\/span> <\/span>\r\n\r\n\"figE2-1-5.png\"\r\n\r\n[reveal-answer q=\"437064\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"437064\"] 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:\"\"\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 sp2<\/sup><\/em>-hybridized, and we will assume that the oxygen atom is also sp2<\/sup><\/em>hybridized.\u00a0The carbon has three sigma bonds: two are formed by overlap between sp2<\/sup><\/em> orbitals with 1s <\/em>orbitals from hydrogen atoms, and the third sigma bond is formed by overlap between the remaining carbon sp2<\/sup><\/em> orbital and an sp2<\/sup><\/em> orbital on the oxygen.\u00a0The two lone pairs on oxygen occupy its other two sp2<\/sup><\/em> orbitals.<\/span>\r\n\r\n\"fig2-1-18.png\"\r\n

    interactive 3D model<\/a><\/p>\r\nThe pi bond is formed by side-by-side overlap of the unhybridized 2p<\/em>z<\/sub> orbitals on the carbon and the oxygen. Just like in alkenes, the 2p<\/em>z<\/sub> orbitals that form the pi bond are perpendicular to the plane formed by the sigma bonds.<\/span>\r\n

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    Example<\/h3>\r\na: Draw a diagram of hybrid orbitals in an sp2<\/sup><\/em>-hybridized nitrogen.<\/span><\/span>\r\n\r\nb: Draw a figure showing the bonding picture for the imine below. <\/span><\/span>\r\n\r\n\"figE2-1-6.png\"<\/span><\/span>\r\n\r\nc: In your drawing for part b, what kind of orbital holds the nitrogen lone pair?<\/span><\/span>\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\u00a0sp<\/em>2<\/sup>\u00a0hybridized.\u00a0 The carbon-nitrogen double bond is composed of a\u00a0sigma\u00a0<\/span>bond formed from two\u00a0sp<\/em>2<\/sup>\u00a0orbitals, and a\u00a0pi<\/span>\u00a0bond formed from the side-by-side overlap of two unhybridized\u00a02p<\/em>\u00a0orbitals.\r\n

    \"\"<\/p>\r\nb) As shown in the figure above, the nitrogen lone pair electrons occupy one of the three\u00a0sp2<\/sup><\/em>\u00a0hybrid orbitals.\u00a0[\/hidden-answer]\r\n\r\n<\/div>\r\n

    <\/p>\r\n\r\n<\/div>\r\nRecall that carbocations<\/a> are transient, high-energy species in which a carbon only has three bonds (rather than the usual four) and a positive formal charge.\u00a0 We will have much more to say about carbocations in this and later chapters. For now, though, the important thing to understand is that a carbocation can be described as an sp2<\/sup><\/em>-hybridized carbon with an empty<\/em><\/strong> p<\/em> orbital perpendicular to the plane of the sigma bonds.<\/span>\r\n\r\n\"fig2-1-19.png\"\r\n\r\nFinally, the hybrid orbital concept applies as well to triple-bonded groups, such as alkynes and nitriles.\u00a0 Consider, for example, the structure of ethyne (common\u00a0 name acetylene), the simplest alkyne.\u00a0 <\/span>\r\n\r\n\"fig2-1-20.png\"\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 sp-<\/em>hybridized<\/strong>.\u00a0 In an sp<\/em>-hybridized carbon,\u00a0 the 2s<\/em> orbital combines with the 2p<\/em>x<\/sub> orbital to form two sp<\/em> hybrid orbitals that are oriented at an angle of 180\u00b0 with respect to each other (eg. along the x axis).\u00a0The 2p<\/em>y<\/sub> and 2p<\/em>z<\/sub> orbitals remain unhybridized, and are oriented perpendicularly along the y and z axes, respectively.<\/span><\/span>\r\n\r\n\"fig2-1-21.png\"<\/span>\r\n\r\nThe carbon-carbon sigma bond, then, is formed by the overlap of one sp<\/em> orbital from each of the carbons, while the two carbon-hydrogen sigma bonds are formed by the overlap of the second sp<\/em> orbital on each carbon with a 1s<\/em> orbital on a hydrogen.\u00a0Each carbon atom still has two half-filled 2p<\/em>y<\/sub> and 2p<\/em>z<\/sub> orbitals, which are perpendicular both to each other and to the line formed by the sigma bonds.\u00a0These two perpendicular pairs of p<\/em> orbitals form two pi bonds between the carbons, resulting in a triple bond overall (one sigma bond plus two pi bonds).\u00a0 <\/span>\r\n\r\n\"fig2-1-22.png\"\r\n

    space-filling image<\/a><\/p>\r\n

    interactive 3D model<\/a><\/p>\r\n\r\n

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    Example<\/h3>\r\nLook at the structure of thiamine diphosphate in the 'structures of common coenzymes' table<\/a>.\u00a0 Identify the hybridization of all carbon atoms in the molecule.<\/span><\/span>\r\n[reveal-answer q=\"974002\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"974002\"]\r\n\"\"\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\nThe hybrid orbital concept nicely explains another experimental observation: single bonds adjacent to double and triple bonds are progressively shorter and stronger than single bonds adjacent to other single bonds.\u00a0Consider for example, the carbon-carbon single bonds in propane, propene, and propyne. <\/span>\r\n\r\n\"fig2-1-23.png\"\r\n\r\nAll three are single (sigma) bonds; the bond in propyne is shortest and strongest, while the bond in propane is longest and weakest. The explanation is relatively straightforward.\u00a0An sp<\/em> orbital is composed of one s<\/em> orbital and one p<\/em> orbital, and thus it has 50% s<\/em> character and 50% p<\/em> character.\u00a0 sp2<\/sup><\/em> orbitals, by comparison, have 33% s<\/em> character and 67% p<\/em> character, while sp3<\/sup><\/em> orbitals have 25% s<\/em> character and 75% p<\/em> character.\u00a0Because of their spherical shape, 2s<\/em> orbitals are smaller, and hold electrons closer and \u2018tighter\u2019 to the nucleus, compared to 2p<\/em> orbitals.\u00a0It follows that electrons in an sp<\/em> orbital, with its greater s<\/em> character, are closer to the nucleus than electrons in an sp2<\/sup><\/em> or sp3<\/sup><\/em> orbital.\u00a0Consequently, bonds involving sp<\/em>-sp<\/em>3<\/sup><\/em> overlap (as in propyne)\u00a0 are\u00a0 shorter and stronger than bonds involving sp2<\/sup><\/em>-sp3<\/sup><\/em> overlap (as in propene).\u00a0 Bonds involving sp3<\/sup>-sp3<\/sup><\/em> overlap (as in propane)\u00a0 are the longest and weakest of the three.<\/span><\/span>\r\n
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    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'.<\/span>\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? <\/span>\r\n\r\n\"fig2-1-24.png\"<\/span>\r\n\r\n[reveal-answer q=\"879188\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"879188\"]a) 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\nKahn Academy video tutorial on valence bond theory \/ hybrid orbitals<\/span><\/a>\r\n
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    Contributors<\/h3>\r\n

    animation <\/a>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0spacefilling image<\/a>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0video tutoria<\/a>l\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0interactive 3D model<\/a> (select 'show resulting pi orbital')<\/em><\/span><\/p>\r\nUnlike a sigma bond, a pi bond does 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 2p<\/em>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.<\/span>\r\n\r\n\"fig2-1-17.png\"\r\n