CHAPTER NINE COVALENT BONDING: ORBITALS

1 212 CHAPTER NINECOVALENT bonding : energy is directly proportional to bond order. Bond length is inversely proportional to bondorder. Bond energy and bond length can be electrons in sigma bonding molecular ORBITALS are attracted to two nuclei, which is a lower, morestable energy arrangement for the electrons than in separate atoms. In sigma antibonding molecularorbitals, the electrons are mainly outside the space between the nuclei, which is a higher, less stableenergy arrangement than in the separated : Unpaired electrons are present. Measure the mass of a substance in the presence andabsence of a magnetic field. A substance with unpaired electrons will be attracted by the magneticfield, giving an apparent increase in mass in the presence of the field.

2 A greater number of unpairedelectrons will give a greater attraction and a greater observed mass that exhibit resonance have delocalized B bonding . In order to rationalize why the bondlengths are equal in molecules that exhibit resonance, we say that the B electrons are delocalized overthe entire surface of the Localized Electron Model and Hybrid has 2(1) + 6 = 8 valence has a tetrahedral arrangement of the electron pairs about the O atom that requires sp3hybridization. Two of the four sp hybrid ORBITALS are used to form bonds to the two hydrogen atoms3and the other two sp hybrid ORBITALS hold the two lone pairs of oxygen.

3 The two O S H bonds are3formed from overlap of the sp hybrid ORBITALS on oxygen with the 1s atomic ORBITALS on the 9 COVALENT bonding : has 4 + 4(7) = 32 valence has a tetrahedral arrangement of the electron pairs about the carbon atom which requires sp3hybridization. The four sp hybrid ORBITALS on carbon are used to form the four bonds to chlorine atoms also have a tetrahedral arrangement of electron pairs and we will assume thatthey are also sp hybridized. The C S Cl sigma bonds are all formed from overlap of sp hybrid33orbitals on carbon with sp hybrid ORBITALS on each chlorine has 2(1) + 4 + 6 = 12 valence central carbon atom has a trigonal planar arrangement of the electron pairs which requires sp2hybridization.

4 The two C S H sigma bonds are formed from overlap of the sp hybrid ORBITALS on2carbon with the hydrogen 1s atomic ORBITALS . The double bond between carbon and oxygen consistsof one F and one B bond. The oxygen atom, like the carbon atom, also has a trigonal planararrangement of the electrons which requires sp hybridization. The F bond in the double bond is2formed from overlap of a carbon sp hybrid orbital with an oxygen sp hybrid orbital. The B bond in22the double bond is formed from overlap of the unhybridized p atomic ORBITALS . Carbon and oxygeneach have one unhybridized p atomic orbital which are parallel to each other.

5 When two parallel patomic ORBITALS overlap, a B bond has 2(4) + 2(1) = 10 valence carbon atom in CH is sp hybridized since each carbon atom is surrounded by two effective pairsof electrons, , each carbon atom has a linear arrangement of electrons. Since each carbon atom issp hybridized, each carbon atom has two unhybridized p atomic ORBITALS . The two C S H sigma bondsare formed from overlap of carbon sp hybrid ORBITALS with hydrogen 1s atomic ORBITALS . The triple bondis composed of one F bond and two B bonds. The sigma bond between the carbon atoms is formed fromoverlap of sp hybrid ORBITALS on each carbon atom. The two B bonds of the triple bond are formed fromparallel overlap of the two unhybridized p atomic ORBITALS on each Exercises and for the Lewis structures.

6 To predict the hybridization, first determine theCHAPTER 9 COVALENT bonding : ORBITALS214arrangement of electron pairs about each central atom using the VSEPR model; then utilize theinformation in Figure of the text to deduce the hybridization required for that arrangement ofelectron ; C is sp ; P is sp ; C is sp ; N is sp + ; C is sp ; Se is sp ; C is sp ; Each O atom is sp ; Br is sp central N atom is sp hybridized in NO and NO. In NO, both central N2--atoms are sp OCN and SCN, the central carbon atoms in each ion are sp hybridized and in N, the---central N atom is also sp Exercises and for the Lewis the central atoms are sp the central atoms are sp the central atoms are sp O and in SO, the central atoms are sp hybridized and in SO, the central sulfur atom is2also sp Exercise for the Lewis :P is dsp :Be is sp :B is sp :Br is dsp :S is dsp :Xe is dsp :Cl is dsp.

7 S is dsp ClF, the central Cl atom is dsp hybridized and in BrF, the central Br atom is also dsp Exercise for the Lewis molecules in Exercise all have a trigonal planar arrangement of electron pairs about thecentral atom so all have central atoms with sp hybridization. The molecules in Exercise all 2have a tetrahedral arrangement of electron pairs about the central atom so all have central atoms withsp hybridization. See Exercises and for the Lewis molecules in Exercise all have central atoms with dsp hybridization since all are based on3 CHAPTER 9 COVALENT bonding : ORBITALS215the trigonal bipyramid arrangement of electron pairs.

8 The molecules in Exercise all have centralatoms with dsp hybridization since all are based on the octahedral arrangement of electron pairs. See23 Exercises and for the Lewis nonpolar< polar3 The angles in NF should be slightly less than because the lone pair requires more spacethan the bonding pairs. planarsp32 < polar120 nonpolara.. 120 , b.. 90 bipyramiddsplineardsp33a. 90 , b. 120 nonpolar180 nonpolari. 9 COVALENT bonding : ORBITALS216square planardspoctahedraldsp232390 nonpolar90 pyramiddspT-shapeddsp233. 90 polar. 90 sp2 Only one resonance form is shown.

9 Resonance does not change the position of the atoms. We canpredict the geometry and hybridization from any one of the resonance two other resonance structurestrigonal planar120 geometry about each S, ,sp hybrids; V-shaped arrangement about3peroxide O's, . , sp pyramid< CHAPTER 9 COVALENT bonding : V-shaped< see-saw. 90 , . 120 octahedral90 dspdsp 323 ). b). 90 c). 120 See-saw about S atom with one lone pair (dsp);3 bent about S atom with two lone pairs (sp) bipyramid90 and 120 , dsp3 CHAPTER 9 COVALENT bonding : the p- ORBITALS to properly line up to form the B bond, all six atoms are forced into the same the atoms were not in the same plane, the B bond could not form since the p- ORBITALS would no longerbe parallel to each , the CH planes are mutually perpendicular to each other.

10 The center C atom is sp hybridized andis involved in two B-bonds. The p- ORBITALS used to form each B bond must be perpendicular to each2other. This forces the two CH planes to be complete the Lewis structures, just add lone pairs of electrons to satisfy the octet rule for the atomswith fewer than eight (CHO) has 4(4) + 6(1) + 2(6) = 34 valence CCO angles are 120 . The six atoms are not in the same planebecause of free rotation about the carbon- carbon single (sigma) are 11 F and 2 B bonds in (CHO) has 4(4) + 8(1) + 2(6) = 36 valence carbon with the doubly-bonded O is sp other 3 C atoms are sp hybridized.