Molecular Structure I. Valence Bond Theory A. General 1. We will consider the covalent compounds formed by the interactions of nonmetals. 2. They interact by “sharing electrons” between them. 3. Two theories, the valence bond theory and the molecular orbital theory, have been developed to analyze these interactions and rationalize some important properties of covalent compounds. B. Valence Bond: Orbital overlap and the covalent bond. The H2 molecule 1. Consider two hydrogen atoms forming an H2 molecule. Originally each H has an electron in a spherically symmetric 1s orbital. Consider what happens as the two H atoms approach one another. Overlap region
HA 1sA1 Lewis diagram H•
H2 H:H or H–H
2. When the two atoms get close enough, their electron clouds can interpenetrate one another, that is, the orbitals can overlap so that there are common regions where both 1s wave functions are sizable. a. When an electron enters this region of overlap, it will lose its identify and could just as well move about nucleus A or B. In this way the two electrons are shared by both nuclei. b. This sharing results in a lowering of the energies of the electrons, because 1) the potential energies decrease because the electrons can be close to two nuclear centers. 2) the kinetic energies of the electrons decrease because they can occupy larger regions in space. c. The energy lowering gives rise to a directional force of attraction between the two atoms, called a covalent bond or an electron pair bond. 3. We can think of a covalent bond being formed by the overlap of two half-filled orbitals, one on each atom. It is also possible to form a covalent bond by the overlap of a vacant orbital on one atom with a filled orbital on another, such a covalent bond is called a
coordinate covalent bond. In either case, the covalent bond involves two orbitals and two electrons. 4. Since the two electrons in a bond can occupy the same orbital, their spins must be paired. C. Consider two F atoms forming an F2 molecule. 1. The electron configuration of F is 1s22s22p5. The valence electrons are distributed as ↑↓ ↑↓ ↑↓ ↑ 2s 2p 2. The two half-filled p orbitals can overlap to form a F–F bond.
– – or –F–F – 3. In the F2 molecule, each F atom will have four electron pairs associated with it, one F
shared pair and three other pair that are not shared. 4. The unshared pairs are called lone pairs. D. Summary. 1. Consider only the valence electrons. 2. Molecule is held together by adjacent atoms sharing pairs of electrons, giving a series of electron pair bonds. The shared electrons are said to be delocalized about the two atoms. 3. The sharing is accomplished by the overlap of an orbital on one atom and an orbital on an adjacent atom (two half-filled or one vacant and a completely filled orbital). The greater the overlap of the orbital, the stronger the bond. Since both electrons can occupy the same orbital, their spins must be paired. 4. Those electrons not involved in bonding are localized on the individual atom as lone pairs. 5. The bonding pattern is described by drawing Lewis diagrams (or electron-dot diagrams). II. Lewis Diagrams—Show the arrangement of the valence electrons, in terms of lone pairs and bonding pairs, about each atom in a molecule. A. Multiple bonding. 1. Types of bond possible. a. Single bond: atoms share one pair of electrons—utilizes one orbital from each atom. b. Double bond: atoms share two pairs of electrons-- utilizes two orbitals from each atom.
c. Triple bond: atoms share three pairs of electrons-- utilizes three orbitals from each atom. 2. For the same atoms, triple bonds are the strongest and shortest while single bonds are the weakest and give rise to the longest bond lengths. Example: Bond type single double triple
Bond C−C C=C C≡C
Bond Strength __(kJ/mol)____ 347 620 812
Bond distance (pm)____ 154 133 120
3. Valence electrons not involved in bonding are localized on the individual atoms as lone pairs. 4. In many compounds, the atoms utilize all of their valence s and p orbitals in bonding and thus obey the octet rule. a. Sum of lone pairs + bonding pairs = 8 (4 pairs). b. For hydrogen, two electrons (1 pair). 5. Coordinate covalent bond = bond formed by the overlap of a filled orbital on one atom and a vacant orbital on another atom.
H H H
coordinate covalent bond
6. There are a number of approaches to drawing Lewis diagrams. Two ways will be shown on the following pages. a. In one method one starts with the valence electrons placed about each atom and consider how they can be combined to form sufficient single double and triple bonds to give the final Lewis diagram of the molecule. b. In the other method, all valence electrons are formed into pairs and then the pairs are arranged as lone pair or bonding pairs about each atom in the molecule. This method makes no distinction between covalent and coordinate covalent bonds.
B. Some simple Lewis diagrams. 1. Neutral molecules—Nonmetal hydrides.
b. H2 O
e. C2 H6
H O H
H O H
H N H H
H H C H H
HH H C C H H H
H H C H
H C H H Single Bond
f. C2 H4
H C C H H H
H C H
C H H Double Bond
g. C2 H2
H C C
H H C
Triple Bond 2. Alternate approach a. Consider C2H6 2C = 8e 6H = 6e Total = 14e = 7 –
H H | | H–C–C–H | | H H
Consider C2H2 2C = 8e 2H = 2e Total = 10e = 5–
b. Complex ions. 2-
2|O| | |O–S–O| | |O|
S = 6e 4O = 24e 2 – = 2e 32e = 16 – N = 5e 2O = 12e + = -1e 16e = 8 –
[ O = N = O ]+
3. Nonoctet compounds a. Less than eight (electron deficient compounds) BeCl2 Be = 2e 2Cl = 14e | Cl – Be – Cl | 16e = 8 – BF3
B = 3e 3F = 21e 24e = 12 –
|F| | |F–B–F|
Even though there are enough electrons so that the central atoms could satisfy its octet through coordinate covalent bonding, these atoms are not electronegative enough to do this. b. Expanded octets ( hypervalent compounds) PF5
P = 5e 5F = 35e 40e = 20 –
|F| F| |F–P F| |F|
Br = 7e 3F = 21e 28e = 14 –
|F| Br – F | |F |
S = 6e 6F = 42e
48e = 24 –
|F |F| Note that in drawing the Lewis diagram the central atom will have the extra bonds or lone pairs, the atoms around the periphery obey the octet rule. The central atom will be in the third or higher period in the Periodic Table. c. Odd electron molecules. When there is an odd number of electrons, it is not possible for each atom to have its octet. An example of such a molecule is NO, which has 11 electrons. The best Lewis diagram for the molecule is
These compounds are better treated using Molecular Orbital theory. C. Formal Charge (FC) 1. The Formal Charge is a parameter that is assigned to each atom in a molecule that can help keep track of the types of bonds formed and aid in selecting the best of several possible Lewis diagrams. It is not a charge in the usual sense. 2. Formal Charge (FC) = number of valence electron – number of electrons in lone pairs – 1/2(number of electron in bonds) 3. Examples. a. FC’s in SO42– The Lewis diagram of SO42– is 2– O
FC(S) = 6 – 0 – 1/2(8) = +2
FC(O) = 6 – 6 –1/2(2) = –1
Note: 1) ∑FC’s = the charge on the formula. For SO42– ∑ = +2 + 4(–1) = –2 2) When an atom donates a pair of electrons in forming a coordinate covalent bond, its formal charge will go up by one. 3) In general, the FC’s should be as low as possible (ideally zero) and the more electronegative elements should not have positive FC’s.
4) Many times Lewis diagrams are drawn so as the minimize Formal Charges. For example, SO 2" 4 could be written as |O| |O| 2|| | |O—S—O| O=S=O ! || | |O| |O| With this diagram: FC(S) = 6 – ½(12) = 0; FC(O=) = 6 – 4- ½(4) = 0 FC(O-) = 6 – 6 – ½(2) = -1. Because of resonance, the O’s will have the same Formal Charge of -1/2. It should be emphasized that there is no experimental way to distinguish between the two Lewis structures, they both give rise to the same properties. b. Consider NO2+ _+ If the Lewis diagram is
FC(N) = 5 – 1/2(8) = 1,
FC(O) = 6 – 4 –1/2(4) = 0
If Lewis diagram is O=O=N , FC(N) = 5 – 4 –1/2(4) = –1
FC(O) = 6 – 0 –1/2(8) = +2 (middle)
FC(O) = 6 – 4 – 1/2(4) = 0 The – FC on N and the +2 FC on O is not reasonable. c. Consider NO If the Lewis diagram is •N=O|, FC(N)=5 – 3 – 1/2(4) = 0, FC(O)=6 – 4 – 1/2(4) = 0 If the Lewis diagram is |N=O•, FC(N)=5 – 4—1/2(4) = –1, FC(O)=6 –3 –1/2(4) = 1 The first Lewis diagram is more reasonable. C. Resonance 1. Consider SO3 (24e = 12 – ) O S O
a. This Lewis diagram shows one sulfur-oxygen double bond and two sulfur-oxygen single bonds. On this basis one would predict one short, strong bond and two longer, weaker bonds. b. Experimentally it is known that the three sulfur-oxygen bonds are identical. c. Note that equivalent Lewis diagrams could have been drawn with the double bond written between the sulfur and each oxygen. The only acceptable picture is that given by all three diagrams.
d. The molecule undergoes resonance. Each Lewis diagram is said to be a contributing form to the resonance hybrid. Note that SO3 has a single dynamic electronic structure; it requires three still pictures (Lewis diagrams) to describe it. e. SO3 is more stable than expected from one sulfur-oxygen double bond and two sulfur-oxygen single bonds. The molecule is said to be resonance stabilized. d. Whenever you can draw more than one equivalent Lewis diagram, by shifting double bonds, the substance exhibits resonance and all Lewis diagrams must be drawn and connected by double headed arrows. 2. Other examples. O a.
(18e = 9 – )
O| b. CO2 (16e = 8 – )
[ O=C=O ↔ |O–C≡O| ↔ |O≡C–O| ]
D. Lewis Diagram Summary 1. Determine the number of valence electron pairs that are to be used including any changes due to charges. You must use all of these pairs and only these pairs. 2. In substances with several different elements, the least electronegative element usually is the central atom. 3. First, try to draw the diagram using the octet rule, single bonds and lone pairs. If this is possible, then that is the Lewis diagram. a. If you are one pair short, the substance has a double bond. b. If you are two pairs short, the substance has a triple or two double bonds. 4. If there are electron pairs left over, they go as lone pairs around the central atom (the central atom will have an "expanded octet"). 5. Electron deficient compounds are found in some boron or metal containing compounds when an octet can be obtained only if the boron or metal accepts electron pairs in forming coordinate covalent bonds. Since these atoms are not electronegative enough to do this, the central boron or metal will not have its octet. 6. Calculate the FC’s and make sure they are reasonable. II. Shapes of molecules – Valence Shell Electron Pair Repulsion (VSEPR) Model. A. Approach. 1. Draw Lewis diagram. 8
2. Count the number of lone pairs (L) + atoms bonded(B) around each central atom. At this point, it does not matter whether the atoms are bonded by single, double, or triple bonds. 3. These groups will repel one another and arrange themselves about the central atom so as to get as far away from one another as possible. The order of repulsion is: lone pairs >> triple bonds > double bonds > single bonds.
4. If the sum of L+B about a central atom is equal to a. 2, the arrangement is linear.
b. 3, the arrangement is trigonal planar.
All three sites are equivalent c. 4, the arrangement is tetrahedral.
All four positions are equivalent.
d. 5, the arrangement is trigonal bipyramidal.
1) All five positions are not equivalent. 2) The three positions in the trigonal plane are the equatorial positions. 3) The two above and below the equatorial plane are the axial positions. e. 6, the arrangement is octahedral.
o 90 o
All six positions are equivalent. B. Examples. 1. B+L=2; Linear
2. L+B = 3 a. B = 3, L = 0
Example BF3 B
TRIGONAL PLANAR or PLANAR TRIANGULAR
b. B = 2, L = 1 Example NO2– (18 e- system)
V SHAPED or BENT
p 3 .6
NO2 (17 e- system)
p 9 .7
Note that one electron takes up less room than either a bonded pair or a lone pair.
3. L + B = 4; Tetrahedral. 109' 28"
CH4, SO 42-, NH 4+
a. B = 4, L = 0
b. B = 3, L = 1
NH3 SO32TRIGONAL PYRAMID
< 109' N
c. B = 2, L = 2 H2O V SHAPED or BENT
O H 104' 30"
4. L + B = 5; Trigonal Bipyramid. a. B = 5, L = 0
152.2 pm F
The F's are not equivalent. The two axial bonds are158 pm and the three equatorial bonds are152 pm
b. B = 4, L = 1 SF4
c. B = 3, L =2
181 pm F Br
d. B = 2, L = 3
5. L + B = 6; Octahedral a. B = 6, L = 0
90 ' F
b. B = 5, = 1 IF5
< 90 '
F 90 '
c. B = 4, L = 2
F 90 '
C. Other Considerations. 1. Electronegativity differences. a. Electron-electron repulsion increases the closer the electrons are to the central atom. The electron density in a bond formed between the central atom and a very electronegative atom will be polarized away from the central atom and will be less effective in repelling electrons than if the central atom was bonded to an element with lower electronegative.
b. Therefore, bond angles involving very electronegative will tend to be smaller than those involving less c. Example. O
H 118 '
2. When lone pairs are present, the bond angles decrease as the size of the central atom increases.
N H 107 '
H 94 '
As H 92 '
III. Hybridization A. General 1. Covalent bonds are formed by the overlap of orbitals on adjacent atoms. The greater the extent of overlap, the stronger will be the resulting bond. In comparing the shapes of molecules from VSEPR and the orientation of the atomic orbitals on a central atom that can be used in forming bonds (s, px, py, pz, dz2 , dx2-y2, dxy, dxz, dyz), it is apparent that the atomic orbitals alone will not give the most efficient overlap. a. Greater overlap, and hence stronger bonds, can be obtained by replacing the atomic orbitals with a set of orbitals whose lobes of maximum probability point directly along the bonding axes. b. These orbitals can be generated by taking suitable linear combinations of the atomic orbitals. Orbitals obtained in this way are called hybrid orbitals and the atom is said to be hybridized. 2. Since the lone pairs (L) and atoms bonded (B) to a central atom determine the geometry around the atom (from the VSEPR theory) the number of hybrid orbitals needed will be L+B. B. Types of hybrid orbitals. 1. L + B = 2. Linear arrangement. a. Will need two hybrid orbitals whose lobes of maximum probability point in opposite
directions. Assuming the bonding axis is the x axis, these hybrids can be obtained by adding and subtracting the s and the px orbitals as shown below.
! SP2 = "S + "Px
!SP1= "S - "Px
(The shaded areas represent a negative sign and the white a positive sign) b. Each hybrid orbital has 50% s and 50% p character. They are called sp hybrid orbitals and the atom is said to be sp hybridized. c. Consider BeCl2 it is a linear molecule. The Lewis diagram is | Cl−Be−Cl | The Be has two atoms bonded and no lone pairs (L+B=2). The Be is sp hybridized and the overlaps are as follows.
1) The bonding can be thought of as taking place through the steps:
__ __ __
2s ⇒ ↑
2p ↑ _ _
⇓ sp hybridize ↑ ↑ sp sp
_ _ py p z
The two half filled sp hybrid orbitals on the Be overlap with two half filled orbitals on the Cl's to give the Be−Cl bonds. 2) These bonds formed by the "head on" overlap of orbitals are called σ(sigma) bonds. d. Consider C2H2. The Lewis diagram is H−C≡C−H 1) Around each carbon L+B=2, therefore each C is sp hybridized as follows
2s : ↑↓
2p ↑ ↑ _
2s 2p ↑ ↑ ↑
⇓ sp hybridization ↑ ↑ sp sp
↑ ↑ py pz
2) The σ bonding is:
This accounts for the carbon-hydrogen single bonds and one of the carbon-carbon bonds. 3) The other two carbon-carbon bonds come about through the overlap of the unhybridized carbon p orbitals as depicted below.
4) A bond formed by this type of "sideways" overlap of two orbitals is called a π (pi) bond. The overlap is not nearly as efficient as that found in σ bonds. Therefore, π bonds are weaker than σ bonds. All single bonds are σ bonds. Triple bonds are composed of 1σ and 2π bonds.
Therefore, although a triple bond is stronger than a single bond, it is not three times as strong. d. Whenever L+B=2 for an atom it is sp hybridized. The atom may also form two π bonds. Examples: HCN, CO2 2. L + B = 3. Planar trigonal arrangement. a. Consider BF3. The Lewis diagram is |F| | B−F| | |F| 1) The boron has three atoms bonded and no lone pairs. The molecule is trigonal planar. The F−B−F bond angles are 120°. 2) Assuming that the molecular plane is the xy plane, The s, px and py orbitals can be combined to give three equivalent sp2 hybrid orbitals. Their lobes of maximum probability are in the same plane and are oriented at 120° from one another.
!sp2 (3) The hybridization sequence is 2s B(2s22p1): ↑↓
2p ↑ _ _
2s ⇒ ↑
2p ↑ ↑ _
⇓ sp2 hybridization _ pz
↑ ↑ ↑ sp2 sp2 sp2
The vacant pz orbital is oriented above and below the molecular plane. b. Consider C2H4. The Lewis diagram is
1) Each carbon is bonded to three other atoms and has no lone pairs. The carbons are sp2 hybridized in the same way as was the B in BF3. The difference is the each carbon has four electrons, three in the sp2 hybrid orbitals and one in the unhybridized pz orbital. C
↑ ↑ ↑ sp2 sp2 sp2
2) The σ bonding is H
3) The pz orbitals can overlap to form a π bond.
4) A double bond is composed of 1σ + 1π bond. c. Whenever L+B=3 the central atom is sp2 hybridized. Examples: SO3, NO2 , NO3 , BH3 3. L + B = 4. Tetrahedral arrangement. a. Consider CH4. The Lewis diagram is
H | H-C-H | H 1) The carbon has four atoms bonded and no lone pairs. The molecule is tetrahedral and the H−C−H bond angles are 109°28". The s, px, py, and pz orbitals can be combined
to give four equivalent sp3 hybrid orbitals oriented along the tetrahedral lines. 2) The hybridization sequence is 2s 2p 2 2 C2s 2p ↑↓ ↑ ↑ _ ⇒
2p ↑ ↑ ↑ ⇓ sp3 hybridization
↑ ↑ ↑ ↑ sp3 sp3 sp3 sp3 3) The carbon forms four σ bonds. b. Whenever L+B = 4 the central atom will be sp3 hybridized. Examples 2-
NH3, SO4 , NH4 , C2H6 (both C's sp3 hybridized), H2O, SO3 4. L + B = 5. Trigonal bipyramidal arrangement a. Consider PF5. The Lewis diagram is F F P
1) Phosphorus forms five single bonds and has no lone pairs. Therefore, it needs five hybrid orbitals. These can be obtained by hybridizing the 3s, 3p and one 3d atomic orbitals. 3s 2 3 0 P(3s 3p 3d ): ↑↓
3p ↑ ↑ ↑
3d _ _ _ _ _
⇓ ↑ ↑ ↑ ↑ _ _ _ _ 3 ⇓ dsp hybridization ↑ ↑ ↑ ↑ ↑ _ _ _ _ ↑
2) The five hybrid orbitals are not equivalent. The three in the equatorial plane are essentially sp2 hybrids and the two axials are dp hybrids. If the equatorial plane is the xy plane, then the s, px and py orbitals hybridize to form the sp2 equatorial hybrids, while the pz and dz2 orbitals combine to give the two axial dp hybrids. b. Whenever L+B=5 the central atom is dsp3 hybridized. Examples: BrCl3, XeF2, SCl4
5. L + B = 6. Octahedral arrangement. a. Consider SF6 . The Lewis diagram F F F
1) The sulfur forms four single bonds and has no lone pairs. Therefore, it needs six hybrid orbitals. These can be obtained by mixing the 3s, 3p and two of the 3d atomic orbitals to give six equivalent d2sp3 hybrid orbitals. 2) The hybridization sequence is: (assume that the F's are on the Cartesian axes) 3s 3p 3d 2 4 0 S 3s 3p 3d : ↑↓ ↑↓ ↑ ↑ _ _ _ _ _
⇓ ↑ ↑ _ _ _ ⇓ hybridization ↑ ↑ ↑ ↑ ↑ ↑ _ _ _ ↑
↑ ↑ ↑
d2sp3 hybrids d orbitals The two d orbitals used are the dz2 and the dx2-y2. 6. There are other ways to describe the bonding in the hypervalent main group compounds that might be better, but we will use the formalism of the electron pair bonds and dsp3 or d2sp3 hybridization. IV. Bond and molecular polarity. A. Bond polarity. 1. Whenever two unlike atoms share electrons they will not share them equally. a. The electron density in the bond is polarized towards the more electronegative element. b. The less electronegative atom will develop a partial positive charge (δ+) and the more electronegative atom will develop a partial negative charge (δ-). 2. Example: χF > χH (use
to show the direction of the polarization)
a. The bond is a polar bond and the covalent bond is said to possess
some ionic character. b. Can use the concept of resonance to explain this polarity by including an ionic form in the resonance hybrid. [ H − F ↔ H+ F- ] covalent ionic c. The two forms do not contribute equally, the greater the difference in electronegativity the more important will be the ionic term. 3. Ionic character in a bond tends to make the bond stronger. B. Polar molecules. 1. Bond polarity can give rise to an overall polarity in the molecule. a. One end of the molecule will be positive and the other end will be negative. b. The molecule is said to be a polar molecule. c. Polar molecules will tend to orient themselves in an electric field. This effect can be measured and the dipole moment, µ, of the molecule can be determined. For a system of two equal but opposite charges (δ) separated by a distance (r), the dipole moment is given by: +!
dipole moment = µ = !r
The dipole moment is a vector quantity pointing in the direction of the negative charge; this is the net direction of polarization. The unit of dipole moment is the debye (D). 1 D = 3.33x10-30 Cm d. Nonpolar molecules have µ = 0. 2. Diatomic molecules that have polar bonds will always be polar molecules. The polarity reflects the difference in electronegativities. For example, consider the hydrogen halides ( χH = 2.1) Compound
HF HCl HBr HI
4.0 3.0 2.8 2.5
1.92 1.08 0.78 0.38
3. For polyatomic molecules, the shape of the molecule, in addition to bond polarity, must be considered. In some cases the polarity of one bond will cancel the polarity of another. Therefore, it is possible for molecules that have polar bonds to be nonpolar.
a. Examples H
H polar µ = 1.46 D
polar µ = 1.87 D
polar µ = 1.60 D
b. All of the above molecules have polar bonds. However, in CO2 and SO3 the oxygen atoms are arranged symmetrically around the central atom and the polarities of the bonds cancel out. When lone pairs are present on the central atom, it usually means that the other atoms are not symmetrically distributed and the molecule will be polar. c. If more than three different elements are present, the molecule could be polar even though there are no lone pairs on the central atom. Example: H
H Polar µ=1.87 D
Valence Bond Theory Problems 1. Draw Lewis diagrams for the following covalent molecules and ions. 2. Some of the substances exhibit the phenomenon of resonance. Indicate which ones do and show the Lewis diagrams for the resonance forms. 3. State which hybrid orbitals each central atom uses in bonding. 4. Based on (3) above, describe or sketch the shape of each molecule or ion. Estimate the angles between bonds. SF3-
S2O32- (one S central)
5. Which of the above substances have polar bonds? Give the direction of polarization of these bonds. Which substances are polar.