Orbitals and Molecular Structure

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9. Molecular Orbitals andMolecular Structure

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sp Hybrid Orbitals Most compounds of second- and third-row nonmetals involve tetrahedral ornear- tetrahedral geometry around a central atom. We already have discussed

methane, CH , which has ideal tetrahedral H-C-H angles of 109.5 , ammonia,

NH , with H-N-H angles of 107 ; and water with an H-O-H angle of 105 .Tetrahedral bonding geometry can be obtained by combining an s and three patomic orbitals of the central atom before bringing in other atoms, to produce a

set of four new orbitals called hybrid atomic orbitals, as at the right. Thesehybrid orbitals can be represented by t , t , t , and t , and can be writtenformally as

t = s + p + p + p

t = s + p -p -p

t = s -p + p -p

t = s -p -p + p

All four hybrid orbitals have an equal contribution from the spherical s orbital,but they point in different directions because they have different contributions

from p p and p The four hybrid orbitals extend out in the four directions of

the vertices of a tetrahedron, or to four nonneighboring comers of a cube. Thesigns of the p terms in the set of four equations above are, in effect, thecoordinates of each orbital The t orbital, for example, has its maximumelectron probability in the -x, +y, -z direction, as can be seen to the right.

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Page 1 of 1Foundations to Chemistry - adapted from "Chemistry, Matter and the Universe"


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sp Hybrid Orbitals Most compounds of second- and third-row nonmetals involve tetrahedral ornear- tetrahedral geometry around a central atom. We already have discussed

methane, CH , which has ideal tetrahedral H-C-H angles of 109.5 , ammonia,

NH , with H-N-H angles of 107 ; and water with an H-O-H angle of 105 .Tetrahedral bonding geometry can be obtained by combining an s and three patomic orbitals of the central atom before bringing in other atoms, to produce a

set of four new orbitals called hybrid atomic orbitals, as at the right. Thesehybrid orbitals can be represented by t , t , t , and t , and can be writtenformally as

t = s + p + p + p

t = s + p -p -p

t = s -p + p -p

t = s -p -p + p

All four hybrid orbitals have an equal contribution from the spherical s orbital,but they point in different directions because they have different contributions

from p p and p The four hybrid orbitals extend out in the four directions of

the vertices of a tetrahedron, or to four nonneighboring comers of a cube. Thesigns of the p terms in the set of four equations above are, in effect, thecoordinates of each orbital The t orbital, for example, has its maximumelectron probability in the -x, +y, -z direction, as can be seen to the right.

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sp Hybrid Orbitals These four tetrahedral hybrid atomic orbitals are less stable than the s andthree p orbitals from which they came, because a small amount of energy isrequired to bring the s-orbital energy up to the energy of p before they can behybridisied. This energy is regained several times over whn bonds are formedbetween these hybrid orbitals and orbitals from other atoms, so the mixing is

possible. The four tetrahedral orbitals are called sp hybrid atomic orbitals.Bonding in methane is illustrated at the right. Each of the four sp hybridorbitals can combine with a hydrogen 1s atomic orbital to form a localisedbonding and antibonding pair of MO's. The antibonding MO's are of noimportance for methane because thwey are never occupied. But when one ofthe bonding orbitals is filled with a pair of electrons, a bond is formed betweenC and H. The sp hybridisation leads to the observed molecular geometry withbond angles of 109.5. Formation of four such bonds uses all of the 1s orbitals

of the four hydrogen atoms, and the s, p , p and p orbitals from carbon.

Filling the bonding orbitals requires all four H electrons and all four second-shell electrons from C. The 1s orbital of carbon and its electron pair are notinvolved in the bonding process.

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P 1 f 1F d i Ch i d d f "Ch i M d h U i "


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P g 1 f 1F d ti t Ch i t d t d f "Ch i t M tt d th U i "


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Page 1 of 1Foundations to Chemistry adapted from "Chemistry Matter and the Universe"


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sp Hybrid Orbitals The Sp hybridization model also can be used for molecules involving N and O,with lone electron pairs filling some of the Sp orbitals. Thus an improvedpicture of water (below) employs two of the Sp orbitals in bonds with hydrogen,and the other two for the two lone electron pairs on 0. This Sp model predicts

a tetrahedral H-0-H bond angle of 109.51. The smaller observed angle of 105can be explained by the strong repulsion generated by the lone pairs, which arecloser to the oxygen atom and to each other than are the bonding electronpairs.

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Page 1 of 1Foundations to Chemistry - adapted from Chemistry, Matter and the Universe

Page 1 of 1Foundations to Chemistry - Chapter 11 - adapted from "Chemistry Matter and the Un


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sp Hybrid Orbitals The great simplification in this localized hybrid picture isthat each C-H bond involves only one hybrid orbitalfrom carbon, and what the other three Sp orbitals aredoing is irrelevant. Bonds can be considered one at atime, and it is not necessary to throw all five atoms intoone great mathematical pot. Consequently calculationsof electron density and energy are greatly simplified.The same Sp hybridization can be used for ethane, HC-CH and for a great many other carbon

compounds. In ethane, shown at the left, three of thefour Sp3 hybrid atomic orbitals on each carbon arecombined with atomic orbitals of hydrogen, as inmethane, and the fourth is combined with one sp3 fromthe other carbon atom. The sp3 hybrid orbitals extendout farther from the nucleus than the is orbitals ofhydrogen do, so a C-C bond is longer than a C-H bond:1.54 versus 1.09 . Bond angles throughout themolecule still have tetrahedral values of 109.5 . Thetwo ends of the molecule can rotate freely around theC-C bond, but the most stable arrangement ofhydrogen atoms by a small amount of energy is thatshown at the lower left. The hydrogen atoms are

"staggered" so that the hydrogen atoms on one carbonatom are as far as possible from the hydrogens on theother carbon atom.

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Page 1 of 1Foundations to Chemistry - Chapter 11 - adapted from Chemistry, Matter and the Un...

Page 1 of 1Foundations to Chemistry - Chapter 11 - adapted from "Chemistry, Matter and the Un...


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sp Hybrid Orbitals

The sp3 hybridization model also can be used formolecules involving N and 0, with lone electron pairsfilling some of the sp3 orbitals. Thus an improvedpicture of water (below) employs two of the sp3 orbitalsin bonds with hydrogen, and the other two for the twolone electron pairs on 0. This sp3 model predicts atetrahedral H-0-H bond angle of 109.51. The smaller

observed angle of 105 can be explained by the strongrepulsion generated by the lone pairs, which are closerto the oxygen atom and to each other than are thebonding electron pairs.

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Page 1 of 1Foundations to Chemistry Chapter 11 adapted from Chemistry, Matter and the Un...



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D o u b l e B o n d s a n d s p 2 H y b r i d i s at i o n Ethylene, H C-CH , is typical of carbon compounds that have double bondsbetween atoms. The carbon-carbon bond length in ethylene is 1.34,compared to 1.54 in ethane, and the bond energy, or energy required to pullthe carbon atoms apart, is 147 kcal mole , rather than 83 kcal mole forethane. Furthermore, the carbon-carbon double bond is rigid. No rotation ispossible around the bond, and the two carbon atoms and four hydrogen atomsall are constrained to lie in one plane. The H-C-H bond angle at either end of

the molecule is 117 . How can MO theory account for these characteristics ofethylene? Ethylene has 12 outer-shell atomic orbitals involved in bonding: ones and three p orbitals from each carbon, and a Is from each of the fourhydrogens. It also has 12 outer-shell electrons to place in MO's: four each fromthe carbons and one each from the hydrogens. The Is carbon orbitals are filledwith electron pairs, do not overlap appreciably, and play no part in bonding.One solution to the bonding problem would be to begin with sp3 hybrid orbitalsaround the carbons, and to assume that each carbon atom shares two suchtetrahedral orbitals with the other, as shown at the lower right. This is unlikely,because of the severely bent bonds that would result between carbons. It isalso wrong, because it predicts a H-C-H bond angle of 109.5 instead of theobserved 117 .

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D o u b l e B o n d s a n d s p 2 H y b r i d i s at i o n Ethylene, H C-CH , is typical of carbon compounds that have double bondsbetween atoms. The carbon-carbon bond length in ethylene is 1.34,compared to 1.54 in ethane, and the bond energy, or energy required to pullthe carbon atoms apart, is 147 kcal mole , rather than 83 kcal mole forethane. Furthermore, the carbon-carbon double bond is rigid. No rotation ispossible around the bond, and the two carbon atoms and four hydrogen atomsall are constrained to lie in one plane. The H-C-H bond angle at either end of

the molecule is 117 . How can MO theory account for these characteristics ofethylene? Ethylene has 12 outer-shell atomic orbitals involved in bonding: ones and three p orbitals from each carbon, and a Is from each of the fourhydrogens. It also has 12 outer-shell electrons to place in MO's: four each fromthe carbons and one each from the hydrogens. The Is carbon orbitals are filledwith electron pairs, do not overlap appreciably, and play no part in bonding.One solution to the bonding problem would be to begin with sp3 hybrid orbitalsaround the carbons, and to assume that each carbon atom shares two suchtetrahedral orbitals with the other, as shown at the lower right. This is unlikely,because of the severely bent bonds that would result between carbons. It isalso wrong, because it predicts a H-C-H bond angle of 109.5 instead of theobserved 117 .

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D o u b l e B o n d s a n d s p 2 H y b r i d i s at i o n Bond angles close to 120 suggest three equal orbitalsin a plane. This geometry can be obtained by combiningthe s. orbital and two of the three p orbitals of eachcarbon into a set of three Sp2 hybrid atomic orbitals.The third, unhybridized p orbital extends verticallyabove and below the plane of the page. In ethylene, twoof the three Sp2 orbitals on each carbon are combinedwith hydrogen Is orbitals, and the third is involved in thebond with the other carbon. These all are a- bondsbecause they are symmetrical about the individual bond

axes. This framework of the ethylene molecule usesten of the twelve available bonding electrons, and all ofthe outer-shell atomic orbitals except one unhybridized2p, orbital on each carbon.

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D o u b l e B o n d s a n d s p 2 H y b r i d i s at i o n The second half of the double bond in ethylene arises from a combination of

these last two p orbitals into a molecular orbital with z lobes of density of

opposite sign above and below the plane of the molecule. The double bond isshorter than a single bond because the p. orbitals must come closer together

before they overlap enough to bond. The orbital also forces the molecule to

be planar. Twisting about a bond axis is harmless to a symmetrical , bond,

but breaks a bond by pulling the p orbitals out of alignment. To twist one endof the ethylene molecule 90 relative to the other, one would have to supplyenergy equal to the difference between a C-C double bond and a single bond,

or 147 - 83 = 64 kcal mole . The ideal H-C-H bond angle of 120 at each end

of the ethylene molecule is decreased to 117 by electron-pair repulsionbetween the double bond and the two C-H single bonds. Double bonds are ofgreat importance in biological molecules, both because they help makeproteins and other molecules rigid and because of their unique ability to absorblight. We will come back to the structural rigidity aspects in the chapter onproteins, and to their light-absorbing properties in the postscript to this chapter.

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D o u b l e B o n d s a n d s p 2 H y b r i d i s at i o n The second half of the double bond in ethylene arises from a combination of

these last two p orbitals into a molecular orbital with z lobes of density of

opposite sign above and below the plane of the molecule. The double bond isshorter than a single bond because the p. orbitals must come closer together

before they overlap enough to bond. The orbital also forces the molecule to

be planar. Twisting about a bond axis is harmless to a symmetrical , bond,

but breaks a bond by pulling the p orbitals out of alignment. To twist one endof the ethylene molecule 90 relative to the other, one would have to supplyenergy equal to the difference between a C-C double bond and a single bond,

or 147 - 83 = 64 kcal mole . The ideal H-C-H bond angle of 120 at each end

of the ethylene molecule is decreased to 117 by electron-pair repulsionbetween the double bond and the two C-H single bonds. Double bonds are ofgreat importance in biological molecules, both because they help makeproteins and other molecules rigid and because of their unique ability to absorblight. We will come back to the structural rigidity aspects in the chapter onproteins, and to their light-absorbing properties in the postscript to this chapter.

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Tr i p l e b o n d s a n d s p h y b r i d i sa t i o n In a relatively small number of compounds, carbon is connected to anotheratom by a triple bond involving three electron pairs. This type of bond can bebuilt from sp hybrid orbitals involving one s and one p orbital on each carbonatom, as shown at the bottom of the page. Two sp hybrid atomic orbitals

extend out from an atom 180 apart, and the two remaining unhybridized porbitals are at right angles to these and to one another. In acetylene, H-C=C-H,each of the two carbon atoms uses one sp hybrid orbital in a C-H bond and theother in the bond between carbons. Three electron pairs are employed inholding this a-bonded framework together.The remaining two thirds of the triple bond involves the p orbitals. If the C-Cbond direction is chosen as the z axis, then the two p. orbitals on carbon

combine into one MO, and the two p , orbitals combine into another. This

means that the carbon atoms are held together by three electron pairs, one in

a bond and two in bonds. The and MO's taken together from a

symmetrical barrel of electron density around the carbon-carbon bond.

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Tr ip l e B o n d s an d s p H y b r id i s a t i o n

The acetylene molecule is linear, with all four atoms in a straight line. The C-Hbond lengths are little different from those of other molecules, but the threeelectron pairs in the triple bond pull the carbon atoms together until they areonly 1.21 apart, compared with 1.34 for a double bond and 1.54 for asingle bond. If a nitrogen atom replaces one carbon, a triple bond is stillpossible, but then the C-H with an electron pair bond is replaced by N:, with itslone electron pair, and the result is HCN (hydrogen cyanide), which is shown atthe right. If the other carbon atom also is replaced by N, the result is the triplybonded N~ molecule. This is the end of the road for bonds between mostatoms. Quadruple bonds involving s and p orbitals are geometricallyimpossible. An absolute requirement for bonding is that the AO's from the twoatoms overlap, and as the bond order increases from single to double to triple,the atoms have to be pushed closer together to achieve this overlap. No matterhow the s and three p orbitals are hybridized, the only way to make all fourorbitals overlap with the corresponding four from another atom is to push theatoms together until their nuclear centers coincide-an impossible thing to do.Hence, quadruply bonded C gas molecules, sharing all four bonding electronson one C with a single partner, are not found. This is one reason for theobserved dramatic difference in properties of the pure elements, between soliddiamond and N , O , and F gases. The C molecule can exist, but only withincomplete electron-sharing. This makesC very reactive and stable only at high temperatures.

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Tr ip l e B o n d s an d s p H y b r id i s a t i o n

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A r o m a t i c i t y : D e lo c al i sa t i o n a n d r e s o n an c e

What we have just gone through has been the outline of a very successfultheory of molecular structure and bonding. What has not been covered is thequantitative calculation of shapes and energies of molecular orbitals thatmakes MO theory even more attractive. We began by abandoning the conceptof a "bond" and looking at the entire molecule at once. It is perhaps fortunatefor our computational sanity that such a procedure usually is excessive withlarger molecules, and that the approximation of localized molecular orbitalsand two-atom bonds can be used instead. Adding the concepts of hybridizationand single, double, and triple bonds provides the framework of a theory thatcan explain most molecules. One important class of molecules that cannot beexplained in terms of two-atom bonds is the organic aromatic molecules,mentioned briefly in Chapter 4. Benzene, C H , is the most familiar example.It has six carbon atoms in a regular hexagonal ring, with all carbon-carbonbonds 1.39 long, which is intermediate between single and double bondlengths. Each carbon has one C-H bond of normal length. The skeleton ofbenzene is shown.

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As soon as we try to make a localized MO model of benzene, we run into

trouble. The planar hexagonal geometry of the molecule, with 120 bondangles, suggests sp2 hybridization around the carbons, with one spl orbitalfrom each C pointed toward an H, and the other two directed toward the

neighboring carbon atoms in the ring. This skeleton of bonds uses 24 of the30 bonding electrons (6 X 4 from carbons plus 6 x 1 from hydrogens), and allof the outer orbitals except the six p orbitals perpendicular to the plane of the

hexagon. This framework is shown. What should be done with the sixunused electrons and six remaining p orbitals? These are shown inperspective at the bottom left. Adjacent p orbitals could be combined in pairsaround the ring to make every other carbon-carbon bond a double bond. Thereare two ways of doing this, represented schematically below. These are knownas the Kekule structures after the man who first proposed them, but theycannot be correct because we know that all of the carbon-carbon bonds arethe same length. A somewhat less plausible way of pairing the p orbitals wouldbe to connect two across the ring, and then pair the remaining two at eitherside, as in the three Dewar structures shown below the Kekule rings.

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All five of these Kekule and Dewar structures, considered individually, arewrong. There is no way to write the correct benzene structure as a set of singleand double bonds, so the localized bond approximation, which has proven souseful for methane and other multi-atom molecules, breaks down. We mustretreat a little way from localized bonds toward the whole-molecule approachagain, at least where p electrons are concerned. We can combine the six porbitals around the benzene ring according to the rules of MO theory, and

produce the six full-ring MO's sketched above. In the lowest-energy orbital,

all six p AO's are combined with the same sign to produce two rings of

electron density above and below the plane of the benzene skeleton. There iszero electron probability in this plane because the original p orbitals had zeroprobability there. Like the two lobes of density in the p orbitals from which they

came, the two "doughnuts" of probability in the , MO have opposite wave-

function signs. The next most stable MO's, and , have the same energy

and the same shape, with one horizontal or vertical plane of zero probabilityperpendicular to the benzene ring. Two antibonding orbitals with the same

energy, and , each have two such zero-probability nodes at right

angles, and the least stable antibonding orbital, , has three such nodes.

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The energy-level diagram for these six MO's appears at the right. As we haveseen consistently in other AO's and MO's, the general principle holds that,other things being equal, the more nodes of zero probability in an orbital, thehigher its energy. Six p atomic orbitals are combined to form six molecularorbitals-three bonding and three antibonding. The six electrons not used in theo, framework of the benzene molecule are used to fill the three bonding MO's,and the antibonding orbitals are unused. The benzene ring therefore has three

more bonds in addition to its -bonded skeleton, but these three bonds arespread around the entire ring rather than being localized between pairs ofcarbon atoms, as the Kekule or Dewar models would predict. All of the carbon-carbon bonds in the ring are intermediate between "single" and "double," andtheir observed bond length verifies this.

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There is another way of looking at bonding in benzene that preserves thelanguage of single and double bonds. This is to say that an of the Keku16 andDewar structures are partially correct, but that no one of them alone is asufficiently good description of reality. The real benzene molecule in some wayis a combination of all of them, like a mule is a combination of a horse and adonkey. Unfortunately, the term "resonance" has become associated with thisviewpoint, and these partially correct structures are called resonance structures.

This term gives the quite erroneous idea of a flipping back and forth among theseveral structures. The benzene molecule contains features of all fiveresonance structures, but it no more flips back and forth between them, than amule "resonates" before your eyes from horse to donkey and back again.Nevertheless, the term resonance is so firmly embedded in the language ofchemistry that we shall use it too. The 40 kcal mole of extra stability of themolecules over that of a KekuIe structure is called the resonance energy of thebenzene molecules.

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Resonance structures provide a handy way of deciding how far delocalizationextends in a molecule. A set of resonance structures for a molecule must havethe atomic nuclei in the same places, and can differ only in the placement ofelectrons and hence of bonds and charges. When all possible resonancestructures have been written for a molecule that cannot be describedadequately by simple single and double bonds, then all atoms that areconnected by double bonds in at least one of the resonance structures are

involved in the delocalized electron system. For benzene these are the sixcarbon atoms, and the hydrogens play no part in delocalization. Delocalizationin carbon compounds almost always involves the combination of a set of p

orbitals perpendicular to the plane of a molecular skeleton connected by -bonds. The double bonds do not have to be alternating around a closed ring fordelocalization to occur. In the butadiene molecule shownon the next page, fourcarbon atoms are connected in a linear chain with two double bonds. Structure(b) is the one usually thought of for butadiene, but it cannot be completely rightbecause all ten atoms in the real molecule lie in a plane, and this would notnecessarily be true if the central carbon-carbon bond were a single bond.

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After making the skeleton of - bonds using sp2 hybridization around carbons,four electrons and four p orbitals in butadiene remain unused, and the porbitals are arranged as in the perspective drawing (a) at the upper left. As inbenzene, these p orbitals can be combined in several ways to produce variousresonance structures for the molecule. Four resonance structures are shown indrawings (b) - (e) at left. All four carbon atoms are linked in a delocalizedelectron system. Compounds with alternating single and double bonds, in

which p-electron delocalization can occur, are called conjugated molecules.Whether linear or in closed rings, conjugated molecules somewhat larger thanbutadiene or benzene have the useful property of absorbing visible light, as weshall see in the postscript. Linear conjugated molecules are used in thephotoreceptors of the eye, and both linear and aromatic conjugated moleculesare put to work in trapping light in photosynthesis.

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After making the skeleton of - bonds using sp2 hybridization around carbons,four electrons and four p orbitals in butadiene remain unused, and the porbitals are arranged as in the perspective drawing (a) at the upper left. As inbenzene, these p orbitals can be combined in several ways to produce variousresonance structures for the molecule. Four resonance structures are shown indrawings (b) - (e) at left. All four carbon atoms are linked in a delocalizedelectron system. Compounds with alternating single and double bonds, in

which p-electron delocalization can occur, are called conjugated molecules.Whether linear or in closed rings, conjugated molecules somewhat larger thanbutadiene or benzene have the useful property of absorbing visible light, as weshall see in the postscript. Linear conjugated molecules are used in thephotoreceptors of the eye, and both linear and aromatic conjugated moleculesare put to work in trapping light in photosynthesis.

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Delocalization has appeared several times in previouschapters, but only now can we give it an interpretationin terms of molecular orbitals. The benzene moleculewas introduced in Chapter 4, and the carbonate andnitrate ions in Chapter 5. The various Lewis dot models

for CO , which assigned single- or double-bond

character to different combinations of the three C-0bonds, were resonance structures for the carbonate ion,differing only in the positioning of electrons betweenatoms. The phosphate, sulfate, and perchlorate ionsdiscussed in Chapter 6 also were examples ofdelocalization. In all of these examples, delocalizationbrought extra stability to the ion, and it is a goodpractical rule of thumb that the more resonance

structures one can draw for a delocalized ion ormolecule, the more stability this delocalization creates.

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M o l e cu l a r o r b i t a l p i ct u r e o f b o n d s : a s u m m a r y

In MO theory a bond is formed when atomic orbitals of similar energy andsymmetry combine to form at least one molecular orbital of lower energy thanthat of the isolated AO's, and when that bonding MO then is filled by a pair ofelectrons. In principle, all bonds extend over the entire molecule, but in practiceit is usually possible to consider only two atoms at a time, and to think of thebond between them as being independent of all other bonds in the molecule.This localized-bond picture sometimes fails us, especially when p orbitals are

involved in delocalization along chains or rings of carbon atoms. When thisoccurs, the molecular skeleton can be treated as a set of a- bonds, and the porbitals can be treated separately. The filled inner shells in atoms can beignored in bonding, and only the outer orbitals and outer-shell electrons needbe considered. In localized, two-atom bonds, the s and three p atomic orbitalsusually are not the best starting points in bonding. All four orbitals can behybridized, before they are combined with orbitals from other atoms, to producea set of four identical sp3 hybrid orbitals pointing in tetrahedral directions.Alternatively, the s and two of the p orbitals can be hybridized into three Sp2

orbitals 120 apart in a plane; or the s and one p can be combined into two sporbitals pointing in opposite directions from the atom. The best hybridization touse depends on the actual geometry of the molecule, and on the presence ofdouble or triple bonds.

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M o l e cu l a r o r b i t a l p i ct u r e o f b o n d s : a s u m m a r y

Bonding orbitals in MO theory always concentrate an excess of electron densitybetween the atoms being bonded, and in this regard the MO theory is nodifferent from the old qualitative electron-pair ideas of a bond. But only MOtheory recognizes that electron pairs can tear a molecule apart as well, if theyare placed in antibonding orbitals. The essential factor in determining bonding isthe net excess of bonding electrons over antibonding electrons. If thesebonding and antibonding electrons counterbalance one another, then the

molecule will not form, as we saw for He , Be , and Ne . Nothing in MO theorycontradicts the simpler electron-pair theory, or we would have grave doubtsabout its validity. But even more, MO theory explains molecular propertiesabout which the simpler theory has nothing to say-for example, the magneticbehavior of the O molecule with its two unpaired electrons, and the planarity ofmolecules with double bonds. This chapter has been a pictorial and qualitativeintroduction to molecular orbital theory, but the theory also has a mathematicaland quantitative side that permits the calculation of energy levels and of

ionization energies, spectra, and reactivities of molecules.

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P o st scr i p t : D e l o ca l i s a t i o n a n d c o l o r

Aromatic compounds are carbon-framework moleculesin which the carbon skeleton is held together in flatrings by o- bonds, and the carbon p orbitalsperpendicular to the rings are involved in extensivedelocalized electron systems. Benzene is the simplestsuch molecule, but many rings can be fused together in

larger molecules. Naphthalene, C H , has two fused

rings; anthracene, C H has three, and many larger

molecules exist (see right). Replacing the peripheralhydrogens by other chemical groups gives rise to a richand varied branch of organic chemistry, which includesmany biologically important molecules, flavorings,dyestuffs, light receptors, and carcinogenic (cancer-producing) agents.

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A common property of aromatic molecules is their ability to absorbelectromagnetic radiation in the near ultraviolet or even in the visible range,thereby making the molecules brightly colored. The effect of combining porbitals in benzene into a set of delocalized MO's is to split the energy of theindividual p orbitals into four closely spaced energy levels, which werediagramed previously. In benzene, the three bonding orbitals are filled and theantibonding orbitals are empty. If the right frequency of radiation is supplied, a

benzene molecule can absorb it and promote one or more electrons frombonding to antibonding orbitals. The gap between levels is a measure of theenergy needed to make the transition from the ground state (lowest energy) toan excited electronic state. An excited molecule can emit this energy later as aphoton of radiation, and drop back to the ground state. Nonaromatic moleculesalso can be electronically excited, but larger amounts of energy are required,and this means that absorption and emission take place farther into the

ultraviolet. If enough energy is supplied, the - single bonds can be broken andthe molecules destroyed. The special property of aromatic molecules is thattheir orbital energy levels are closely spaced , which leads to absorption inthe lower-energy, longer-wavelength region.

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Delocalization leads to a lowering of energy levels and narrowing of thespacings between levels; and the larger the delocalized system, the greaterthis effect. This can be seen in the series of aromatic molecules compared atthe bottom of the opposite page. Benzene has six atoms in its delocalized -electron system, and the spacings between the -orbital energy levels aresuch that it absorbs energy at a set of wavelengths in the ultraviolet region,centered around 2550 . The visible wavelengths pass through the molecule

untouched, so benzene is colorless to our eyes. So are naphthalene andanthracene, which have 10 and 14 atoms in the delocalized system, althoughthe larger rings shift the absorption to longer wavelengths or lower energies:3150 and 3800 . In contrast, delocalization in naphthacene is so extensivethat the splitting between ir energy levels has narrowed to the point where bluelight around 4800 is absorbed. With the blue light absorbed, the remainingvisible wavelengths make naphthacene appear orange, the complement ofblue. In pentacene, which has five rings, absorption is shifted down to evenlower energies. Pentacene removes yellow light around 5800 and thereforeappears indigo. This "eyeball spectroscopy" is surprisingly informative inrevealing what aromatic molecules are doing. The visible spectrum is shown atthe right, with colors recorded as a function of wavelength from the ultravioletto the infrared. If any of these wavelengths is absorbed by a molecule, theremaining wavelengths give the molecule the complementary color. Removalof green wavelengths around 5300 makes a molecule appear purple. If themolecule absorbs red light at around 6800 , we will see it as blue-green. Bylooking at what is left of the visible spectrum after absorption, we can decideapproximately what visible wavelengths the compound is absorbing.

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Any attached side group on the ring that increases the number of atoms in thedelocalized system will shift the electronic absorption toward lower energy andlonger wavelengths. Early industrial dyestuff chemists knew by experience thatsome chemical groups such as -OH and -NH would shift the colors of aromaticmolecules down the series yellow-orange-red-purple-blue-green, thecomplement of the absorbed visible spectrum, violet-blue-green-yellow-orange-red, before they understood the relationship among color, frequency, and

energy. Adding an -OH group to benzene to produce phenol, C H OH, shiftsthe center of the absorption band slightly, from 2550 to 2700 , since theoxygen in the -OH group is rich in electrons. Phenol is an acid, and candissociate and lose the proton of the -OH to form the phenolate ion:

C H OH --> C H O + H+phenol phenolate ion

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In the phenolate ion the oxygen actually becomes part of the delocalizedsystem and enlarges it from six atoms to seven. You can see this by comparingthe various resonance structures for phenol and the phenolate ion at the left.Only benzenelike resonance structures are possible for phenol, since the -OHgroup really is not included in the delocalized system. The phenolate ion hasresonance structures in which the oxygen has three lone pairs and a negativecharge, and is attached to the ring by a single bond. But other resonance

structures can be drawn in which one of the three lone pairs is shifted into aC=0 double bond, and the negative charge shows up at different positions onthe ring. These resonance structures tell us that in the ion the oxygen is part ofthe general seven-atom delocalized system. With greater delocalization, thespacings between -orbital energy levels are decreased, and the electronicabsorption band shifts from 2700 to 2870 . We cannot see this changebecause our eyes are insensitive to the ultraviolet, but if all the wavelengthswere doubled, benzene would be red, phenol would be purple, and the

phenolate ion would be blue-violet. Phenol then would bean acid- baseindicator, revealing the acidity of its environment by its color. If added in smallamounts to a solution, it would appear purple in acid, where its own dissociationwas repressed by a plentiful H+ supply, and blue-violet in base, where most of ithad dissociated to the phenolate ion.

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Phenol is not a usable acid-base indicator because it absorbs in the wrong partof the spectrum, but the same principles explain why slightly larger aromaticmolecules make good indicators. Alizarin, shown at the top of the next page, isa derivative of anthracene. The two double-bonded oxygens on the center ringenlarge the delocalized system from 14 atoms to 16, and shift the center ofabsorption from 3800 in the ultraviolet to 4300 in the indigo part of thevisible spectrum. The unabsorbed wavelengths make alizarin yellow. The two -

OH groups on the outer ring do not participate in the delocalized system untilthey dissociate, as they do in basic solutions in which III ions are scarce. Whenthis happens, the two hydroxyl oxygens become just as much a part of thedelocalization as the original double-bonded oxygens. You can see this bydrawing other resonance structures that give the hydroxyl oxygens the doublebonds. This enlargement shifts the main absorption to around 4800 in theblue re Delocalization of phenol in acid (gray tint) and phenolate ion in base(color tint) gion, and makes the solution of ions orange. Alizarin is one of the

standard acid-base indicators, turning Yellow in acid and orange in base.Phenolphthalein, which is colorless in acid and deep red in base, and methylviolet, which is yellow in acid and violet in base, are other examples ofaromatic acid-base indicators.

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Linear conjugated molecules with alternating single and double bonds also willabsorb light if they are long enough. Carotenoids, which are "super butadienes"with 22 carbon atoms connected by alternating single and double bonds, areused as antennae by green plants to trap light and transfer the energy tochlorophyll for use in chemical syntheses. They are bright yellow-orangebecause they have evolved over the past 3.5 billion years to absorb light in the5000 wavelength region (blue-green), which is the most intense part of solar

radiation. Living organisms use aromatic and straight-chain conjugatedmolecules to trap light energy, to transfer this energy from one molecule toanother, and to recognize the presence of light so they can grow toward it, turntoward it, or use it for information gathering through vision. The radiation thatreaches the surface of our planet covers a relatively narrow range. Ozone, Oin the upper layers of the atmosphere absorbs almost all wavelengths shorterthan 2900 , and water vapor absorbs much of the infrared. Little infraredradiation longer than 13,000 remains by the time light reaches the surface of

the Earth, and only five meters below the surface of the ocean all radiationlonger than 8000 has been absorbed. The most intense radiation from thesun occurs in the blue-green region, around 5000 , the region for which thecarotenes have evolved to absorb light.

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Q u e s t i o n s

1. When atomic orbitals are combined into molecular orbitals, how does thenumber of MO's obtained compare with the original number of AO's?

2. For a multiatorn molecule, what approximation are we using when wecombine atomic orbitals from atoms two at a time in forming bonds betweenpairs of atoms? Why is this approximation useful? Give an example of amolecule for which this approximation is valid, and another for which it breaks

down.

3. What is the distinction between bonding and antibonding MO's? How is it thatmolecules such as N and F can have antibonding orbitals that are lower inenergy than some bonding orbitals?

4. How many electrons can each MO hold? What is the relationship betweenthe spins of electrons occupying the same MO? How does this compare with

the occupancy of AO's by electrons in isolated atoms?

5. In the hydrogenlike diatomic molecules , H , , and He , why do the

first two electrons increase the strength of the bond, and the third and fourthelectrons weaken the bond? What is the bond order in each of the abovemolecules or molecule-ions?

6. Compare the theoretical predictions in Question 5 with the observed bondlengths and bond energies. Does the bond length increase, or decrease, withincreasing bond order? Why?

7 . How do filled bonding MO's tend to hold the two bonded atomic nucleitogether? How do filled antibonding MO's tend to pull them apart? Explain interms of electron probability distributions.

8.What are the requirements in terms of location, energy, and symmetry of twoatomic orbitals if they are to be combined into molecular orbitals?

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Q u e s t i o n s

9. What is meant by and symmetry in molecular orbitals? How is thisnomenclature an obvious extension of the s and p notation for atomic orbitals?

Explain in terms of 180 rotations and the signs of the wave functions.

10. How is the bond energy of an electron-pair bond related relative energies ofthe bonding and antibonding MO's and the AO's from which they were derived?Explain with a diagram. If the energy difference between the AO's and the

bonding MO in H is x kcal mole , what is the bond energy of H ?

11. What is the relative order of increasing energies for the six MO's derivedfrom the six outer p AO's on tw o atoms in a diatomic molecule? How does thisenergy sequence account for the observed bond orders in the diatomicmolecules obtained from second-row elements?

12. For which of the second-row elements do diatomic molecules not exist?

Why, in terms of MO theory? For which of these elements do diatomicmolecules occur only at high temperatures? What is the their state at 298K, andwhy? For which of these elements is the diatomic molecule the stable form atroom temperature?

13. Which of the diatomic molecules of the second-row elements have unpairedelectrons? How many unpaired electrons do they have? Why are theseelectrons not paired in the same orbital? How does the Lewis electron-dotmodel account for these unpaired electrons?

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14. How do the observed bond lengths and bond energies of the second-rowdiatomic molecules correlate with the predicted bond order?

15. When AO's of different kinds of atoms are being combined to build an MO,what important energy criterion helps to decide which AO's on the two atomswill interact9 What symmetry considerations are important? Illustrate with theHF molecule.

16. Draw the Lewis electron-dot model for the HF molecule. The F atom shouldbe represented as having one bond to H, and three lone electron pairs. In MOtheory, what orbitals do these three electron pairs occupy? Which fluorine AO isinvolved in the H-F bond?

17. To which MO does the fluorine 2p AO make the greatest contribution, thebonding MO or the antibonding? What does this imply in terms of the location of

the bonding electron pair and the polarity of the bond? How is this related to therelative electronegativities of H and F?

18. How are the energy levels of the hydrogen is and fluorine 2p atomic orbitalsrelated to the first ionization energies of H and F? (The correlation isapproximate, but useful.)

19. In the limit of completely ionic bonding between two atoms of very high andvery low electronegativities, how would the bonding MO compare with the AO'sof the two atoms? What would this . I Illustrate imply about the location of thebonding electron pair. with NaCl.

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Q u e s t i o n s

27. Use the sp3 model and then the sp2 model to predict what would happen tothe double bond when the two ends of the ethylene molecule are twisted9 Howdoes the real molecule behave, and which model is most compatible withreality?

28. Sketch the o--bond skeleton of the ethylene molecule. Draw in the MO ofthe second bond. What is the electron probability density at each of the six

atoms in ethylene from the electrons in this MO? Draw the nodal surface forthis orbital.

29. Sketch the -bond skeleton of the acetylene molecule. Draw in the MO'sof the second and third bonds between carbon.

30. How does the carbon-carbon bond length change between ethane,ethylene, and acetylene? How would you expect the bond energies to change?

31 . Sketch the -bond skeleton of the benzene molecule. How many AO's andhow many bonding electrons are used in this - skeleton? What kind ofhybridization is used around the carbon atoms?

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32. How many bonding electrons, and how many and what kind of AO's, remainunused in benzene after the o-hond skeleton is built? How are these electronsand AO's used for further bonding in benzene? How does this resemble thesituation in ethylene and acety lene, and in what important way is it different?

33. How would the electrons and AO's not involved in the -bond skeleton beused in the Kekule bond model for benzene? In a perspective view of the

hexagonal C ring, sketch the bonding MO's.

34 . Repeat the preceding question for one of the Dewar models of benzene.

35. In a similar perspective view of the benzene ring, sketch the actual MO,

showing the rings of electron probability above and below the C benzene-ringplane.

36. For all of the MO's in benzene, what is the electron probability density atthe carbon and hydrogen atoms? How does this compare with the density at the

C and H atoms for the MO in ethylene? (See Question 28.)

37. In what sense does bonding in the benzene molecule represent a stepbackward from two-atom bonds to full-molecule orbitals? What is this called?What effect on the energy levels of a molecule accompanies such behavior?Give examples of inorganic oxygencontaining compounds that show the samebehavior.

38. In what sense is it possible to think of the actual bonding in benzene asbeing a "mixture" of Kekule and Dewar structures? What are these structurescalled? Is any actual alternation back and forth between structures implied?

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Q u e s t i o n s

39. In what ways can different resonance structures differ from one another,and in what ways must they be alike? How is the number of different resonancestructures that can be drawn for a given molecule correlated approximately withthe energy of the molecule?

40. Sketch the -bond skeleton of the butadiene molecule. How many electronsand AO's are used in this framework? What kind of hybridization is used around

the carbon atoms?

41. How many bonding electrons, and how many and what kind of AO's, remain

unused in butadiene after the -bond skeleton is built? How are these electronsand AO's used for further bonding in butadiene? Over which atoms in themolecule does delocalization extend?

42. Draw several resonance structures for butadiene. Show how each structure

assigns the four electrons not involved in - bonding among the four carbonatoms.

43. How do the energy levels in aromatic molecules change with an increase inthe extent of delocalization, and how is this reflected in the energy wavelengthsof absorption or radiation?

44. How is the color that we see in a chemical compound related to the energy

that it absorbs? If a molecule absorbs wavelengths in the blue region, why doesit appear orange to our eyes instead of blue?

45. Why does absorption shift to longer wavelengths when the phenol moleculedissociates to the phenolate ion? What prevents phenol from being a usefulacid-base indicator?

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46 . Why is alizarin a good acid-base indicator, whereas phenol is not?

47. Why is the ability to absorb visible light useful to living organisms? In whatway or ways is this ability especially useful to plants, and in what different wayis it useful to animals?

48. One objection to the supersonic transport (SST) is that waste oxides of

nitrogen emitted at high altitudes by the SST would combine with, and slowlydestroy, the high-altitude ozone layer around the planet. Recent calculationshave shown that a thermonuclear war would practically eliminate the ozonelayer. Why would the destruction of the ozone layer be dangerous to life (inaddition to the immediate hazards of the atomic war)?

49. What is the approximate wavelength range of the radiation that reaches thesurface of the Earth at the present time? What absorbs the longer and shorter

wavelengths and prevents them from reaching the surface?

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P r o b l e m s

6. The triple bond energies in acetylene, hydrogen cyanide, and nitrogen, are194, 213, and 226 kcal mole , respectively. Diagram these three molecules,showing the positions of all bonds and lone electron pairs. Why would oneexpect the observed progression of bond energies? In singly bonded -N-N-N-N-chains, it was stated previously that these were unstable in comparison with -C-C-C-C- chains because of repulsions between nitrogen lone electron pairs. Whyis this apparently not a dominant factor in the comparison of HC CH, HCN,and N ? Where are the lone pairs positioned in the N molecule?

7. The conventional numbering system for a three-ring carbon skeleton such asthat of alizarin is

Hence the undissociated alizarin molecule shown previously in this chapter canbe described as having carboryl (C=0) groups at Positions 9 and 10, and -OHgroups at Positions 1 and 2. The resonance model for the alizarin ion shown inthis chapter has -0 groups at Positions 1 and 2. Draw another resonancestructure with -0 groups at Positions 9 and 10 and carbonyl groups at 1 and 2.What would this imply about shifts of electrons to and from the various oxygenatoms?

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8. Continuing a search for resonance structures for the alizarin ion fromProblem 7, can you draw a structure in which carbonyl groups are present atPositions 9 and 2, and -0 at 1 and 10? What about carbonyl groups at 9 and 1,and -0 at 2 and 10? How many other resonance structures can you find thatdiffer in assignment of the negative charges to different pairs of oxygen?

9. How many different resonance structures can you draw for the alizarin ion inwhich the negative charges are located on carbon atoms rather than on oxygen

atoms?

10. As you try out various resonance models for the alizarin ion, you may comeacross the empirical observation that only those structures are possible forwhich the carbon atoms bearing negative charges (either directly or through anattached -0 ) are separated by an even number of other carbon atoms: 0, 2, 4, -- -. Can you explain this in terms of single and double bonds between carbonsin the three rings?

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Orbitals and Molecular Structure

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