The Localized Electron Model Chapter 9 Covalent Bonding .

The Localized Electron ModelChapter 9Covalent Bonding:Orbitals Drawthe Lewis structure(s) Determinethe arrangement of electron pairs(VSEPR model). SpecifyHybridizationValence Bond Theory Valence bond theory or hybrid orbitaltheory is an approximate theory to explain thecovalent bond from a quantum mechanicalview. According to this theory, a bond forms betweentwo atoms when the following conditions are met.(see Figures 10.21 and 10.22)1. Two atomic orbitals “overlap”2. The total number of electrons in both orbitalsis no more than two.the necessary hybrid orbitals. The mixing of atomic orbitals to formspecial orbitals for bonding. The atoms are responding as needed togive the minimum energy for the molecule.1

Figure 10.20: Formation of H2.Bond formed between two s orbitalsHybrid Orbitals One might expect the number of bonds formedby an atom would equal its unpaired electrons.Figure 10.21: Bonding in HCl.Bond formed between an s and p orbitalFigure 9.1: (a) The Lewis structureof the methane molecule. (b) Thetetrahedral molecular geometry ofthe methane molecule. Chlorine, for example, generally forms onebond and has one unpaired electron. Oxygen, with two unpaired electrons,usually forms two bonds. However, carbon, with only two unpairedelectrons, generally forms four bonds. Forexample, methane, CH4, is well known.2

Hybrid Orbitals Four unpaired electrons are formed as anelectron from the 2s orbital is promoted(excited) to the vacant 2p orbital. The following slide illustrates this excitation.(Recall that in the excited state for an element,a ground state electron is promoted to a higherorbital) More than enough energy is supplied for thispromotion from the formation of two additionalcovalent bonds.Energy The bonding in carbon might be explained asfollows:2p2p2s2s1s1sC atom (ground state)C atom (promoted)Hybrid OrbitalsHybrid Orbitals One bond on carbon would form using the 2sorbital while the other three bonds would usethe 2p orbitals. Hybrid orbitals are orbitals used to describebonding that are obtained by takingcombinations of atomic orbitals of an isolatedatom. This does not explain the fact that the fourbonds in CH4 appear to be identical. Valence bond theory assumes that the fouravailable atomic orbitals in carbon combine tomake four equivalent “hybrid” orbitals. In this case, a set of hybrids are constructedfrom one “s” orbital and three “p” orbitals, sothey are called sp3 hybrid orbitals. The four sp3 hybrid orbitals take the shape of atetrahedron (see Figure 10.23).3

Figure 9.2:The valenceorbitals on afree carbonatom: 2s,2px, 2py, and2pz.Figure 9.4:Crosssection ofan sp3orbital.Figure 9.3: The "native" 2s and three 2p atomic orbitals characteristic of a3free carbon atom are combined to form a new set of four sp orbitals. Thesmall lobes of the orbitals are usually omitted from diagrams for clarity.Figure 9.6: The tetrahedral set of four sp3 orbitals of thecarbon atom are used to share electron pairs with the four 1sorbitals of the hydrogen atoms to form the four equivalentC—H bonds. This accounts for the known tetrahedralstructure of the CH4 molecule.4

Figure 9.7: The nitrogen atom inammonia is sp3 hybridized.You can represent the hybridization of carbon inCH4 as follows.Energy2psp3sp3C-H bonds2s1sC atom(ground state)1sC atom(hybridized state)Figure 9.5: An energy-leveldiagram showing the formationof four sp3 orbitals.A Problem to Consider Describe the bonding in H2O according tovalence bond theory. Assume that themolecular geometry is the same as given bythe VSEPR model. From the Lewis formula for a molecule,determine its geometry about the central atomusing the VSEPR model.1sC atom(in CH4)5

A Problem to Consider Describe the bonding in H2O according tovalence bond theory. Assume that themolecular geometry is the same as given bythe VSEPR model. The Lewis formula for H2O isA Problem to Consider Describe the bonding in H2O according tovalence bond theory. Assume that themolecular geometry is the same as given bythe VSEPR model. Note that there are four pairs of electrons aboutthe oxygen atom. According to the VSEPR model, these are directedtetrahedrally, and from the previous table you seethat you should use sp3 hybrid orbitals.A Problem to Consider Describe the bonding in H2O according tovalence bond theory. Assume that themolecular geometry is the same as given bythe VSEPR model. From this geometry, determine the hybridorbitals on this atom, assigning its valenceelectrons to these orbitals one at a time.A Problem to Consider Describe the bonding in H2O according tovalence bond theory. Assume that themolecular geometry is the same as given bythe VSEPR model. Each O-H bond is formed by the overlap of a1s orbital of a hydrogen atom with one of thesingly occupied sp3 hybrid orbitals of theoxygen atom.6

You can represent the bonding to the oxygen atom inH2O as follows:Energy2psp3lonepairs2s1sO atom(ground state)sp31sO atom(hybridized state)O-HbondsFigure10.24:Bondingin H2O.1sO atom(in H2O)Figure 9.9: An orbital energy-leveldiagram for sp2 hybridization. Notethat one p orbital remainsunchanged.Figure 9.8: The hybridization of the s, px, and py atomic2orbitals results in the formation of three sp orbitals centeredin the xy plane. The large lobes of the orbitals lie in the planeat angles of 120 degrees and point toward the corners of atriangle.7

Figure 9.14: When one s orbital and onep orbital are hybridized, a set of two sporbitals oriented at 180 degrees results.Figure 9.21: A set of dsp3 hybrid orbitals on aphosphorus atom. Note that the set of five dsp3orbitals has a trigonal bipyramidal arrangement.(Each dsp3 orbital also has a small lobe that is notshown in this diagram.)Figure 9.22: (a) The structure of the PCI5 molecule. (b) Theorbitals used to form the bonds in PCl5. The phosphorus usesa set of five dsp3 orbitals to share electron pairs with sp3orbitals on the five chlorine atoms. The other sp3 orbitals oneach chlorine atom hold lone pairs.Figure 9.23: An octahedral set of d2sp3orbitals on a sulfur atom. The small lobeof each hybrid orbital has been omittedfor clarity.8

Hybrid Orbitals Note that there is a relationship between thetype of hybrid orbitals and the geometricarrangement of those orbitals. Thus, if you know the geometric arrangement,you know what hybrid orbitals to use in thebonding description. Figure 9.24 summarizes the types of hybridizationand their spatial arrangements.Hybrid rFigure 9.24:The relationshipof the numberof effectivepairs, theirspatialarrangement,and the hybridorbital setrequired.Hybrid OrbitalsNumber of ExampleOrbitals2Be in BeF2sp2Trigonal planar3B in BF3sp3Tetrahedral4C in CH4dsp3Trigonal bipyramidal5P in PCl5d2sp3Octahedral6S in SF6 To obtain the bonding description of any atomin a molecule, you proceed as follows:1. Write the Lewis electron-dot formula for themolecule.2. From the Lewis formula, use the VSEPR theoryto determine the arrangement of electron pairsaround the atom.9

Hybrid OrbitalsHybrid Orbitals To obtain the bonding description of any atomin a molecule, you proceed as follows: To obtain the bonding description of any atomin a molecule, you proceed as follows:3. From the geometric arrangement of the electronpairs, obtain the hybridization type.5. Form bonds to this atom by overlapping singlyoccupied orbitals of other atoms with the singlyoccupied hybrid orbitals of this atom.4. Assign valence electrons to the hybrid orbitals ofthis atom one at a time, pairing only whennecessary.A Problem to Consider Describe the bonding in XeF4 using hybridorbitals. From this geometry, determine the hybridorbitals on this atom, assigning its valenceelectrons to these orbitals one at a time.A Problem to Consider Describe the bonding in XeF4 using hybridorbitals. From the Lewis formula for a molecule,determine its geometry about the central atomusing the VSEPR model.10

A Problem to Consider Describe the bonding in XeF4 using hybridorbitals. The Lewis formula of XeF4 isA Problem to Consider Describe the bonding in XeF4 using hybridorbitals. Each Xe-F bond is formed by the overlap of axenon d2sp3 hybrid orbital with a singly occupiedfluorine 2p orbital. You can summarize this as follows:A Problem to Consider Describe the bonding in XeF4 using hybridorbitals. The xenon atom has four single bonds and twolone pairs. It will require six orbitals to describethe bonding. This suggests that you use d2sp3 hybrid orbitalson xenon.5d5p5sXe atom (ground state)11

5d5dd2sp3d2sp3lone pairsXe atom (hybridized state)Xe-F bondsXe atom (in XeF4)Multiple BondingMultiple Bonding According to valence bond theory, one hybridorbital is needed for each bond (whether asingle or multiple) and for each lone pair. To describe the multiple bonding in ethene,we must first distinguish between two kinds ofbonds. For example, consider the molecule ethene. A σ (sigma) bond is a “head-to-head” overlapof orbitals with a cylindrical shape about thebond axis. This occurs when two “s” orbitalsoverlap or “p” orbitals overlap along their axis. A π (pi) bond is a “side-to-side” overlap ofparallel “p” orbitals, creating an electrondistribution above and below the bond axis.12

(unhybridized)Multiple Bonding2p Each carbon atom is bonded to three otheratoms and no lone pairs, which indicates theneed for three hybrid orbitals.sp22sEnergy This implies sp2 hybridization. The third 2p orbital is left unhybridized andlies perpendicular to the plane of the trigonalsp2 hybrids. The following slide represents the sp2hybridization of the carbon atoms.2p1s1sC atom (ground state)C atom (hybridized)Figure 10.26Multiple Bonding Now imagine that the atoms of ethene moveinto position. Two of the sp2 hybrid orbitals of each carbonoverlap with the 1s orbitals of the hydrogens. The remaining sp2 hybrid orbital on eachcarbon overlap to form a σ bond.13

Multiple BondingA sigma (σ) bond centers along theinternuclear axis. The remaining “unhybridized” 2p orbitals oneach of the carbon atoms overlap side-to-sideforming a π bond.A pi (π) bond occupies the space aboveand below the internuclear axis.HHπC σCHH You therefore describe the carbon-carbondouble bond as one σ bond and one π bond.Figure 9.10: When an s and two p orbitals are mixed to form a2set of three sp orbitals, one p orbital remains unchanged and isperpendicular to the plane of the hybrid orbitals. Note that inthis figure and those that follow, the orbitals are drawn withnarrowed lobes to show their orientations more clearly.Figure 9.11: The s bonds inethylene. Note that for each bond theshared electron pair occupies theregion directly between the atoms.5514

Figure 9.12: A carbon-carbon double bond consists of a s bond and a pbond. In the s bond the shared electrons occupy the space directly betweenthe atoms. The p bond is formed from the unhybridized p orbitals on the twocarbon atoms. In a p bond the shared electron pair occupies the space aboveand below a line joining the atoms.Figure 9.15: The hybrid orbitalsin the CO2 molecule.Figure 9.13: (a) The orbitals used toform the bonds in ethylene. (b) TheLewis structure for ethylene.Figure 9.16: The orbital energylevel diagram for the formationof sp hybrid orbitals on carbon.15

Figure 9.17: The orbitals of ansp hybridized carbon atom.Figure 9.18: The orbitalarrangement for an sp2 hybridizedoxygen atom.Figure 9.19: (a) The orbitals used to form thebonds in carbon dioxide. Note that the carbonoxygen double bonds each consist of one sbond and one p bond. (b) The Lewis structurefor carbon dioxide.Figure 10.28: Bonding inacetylene.H-C/C-H16