Molecular Geometry and Bonding Theories

Multiple Bonds

The "internuclear axis" is the imaginary axis that passes through the two nuclei in a bond:

The covalent bonds we have been considering so far exhibit bonding orbitals which are symmetrical about the internuclear axis (either an s orbital - which is symmetric in all directions, or a p orbital that is pointing along the bond towards the other atom, or a hybrid orbital that is pointing along the axis towards the other atom)

In multiple bonds, the bonding orbitals arise from a different type arrangement:

• Multiple bonds involve the overlap between two p orbitals
• These p orbitals are oriented perpendicular to the internuclear (bond) axis

In p bonds:

• The overlapping regions of the bonding orbitals lie above and below the internuclear axis (there is no probability of finding the electron in that region)
• The size of the overlap is smaller than a s bond, and thus the bond strength is typically less than that of a s bond

Generally speaking:

• A single bond is composed of a s bond
• A double bond is composed of one s bond and one p bond
• A triple bond is composed of one s bond and two p bonds

C₂H₄ (ethylene; see structure above)

• The arrangement of bonds suggests that the geometry of the bonds around each carbon is trigonal planar
• Trigonal planar suggests sp² hybrid orbitals are being used (these would be s bonds)

What about the electron configuration?

Carbon: 1s² 2s² 2p²

• Thus, we have an extra unpaired electron in a p orbital available for bonding
• This extra p electron orbital is oriented perpendicular to the plane of the three sp² orbitals (to minimize repulsion):

• The unpaired electrons in the p orbitals can overlap one another above and below the internuclear axis to form a covalent bond

• This interaction above and below the internuclear axis represents the single p bond between the two p orbitals

Experimentally:

• we know that the 6 atoms of ethylene lie in the same plane.
• If there was a single s bond between the two carbons, there would be nothing stopping the atoms from rotating around the C-C bond.
• But, the atoms are held rigid in a planar orientation.
• This orientation allows the overlap of the two p orbitals, with formation of a p bond.
• In addition to this rigidity, the C-C bond length is shorter than that expected for a single bond.
• Thus, extra electrons (from the p bond) must be situated between the two C-C nuclei.

C₂H₂ (acetylene)

• The linear bond arrangement suggests that the carbon atoms are utilizing sp hybrid orbitals for bonding

• This leaves two unpaired electrons in p orbitals
• To minimize electron replusion, these p orbitals are at right angles to each other, and to the internuclear axis:

• These p orbitals can overlap two form two p bonds in addition to the single s bond (forming a triple bond)

localized electrons are electrons which are associated completely with the atoms forming the bond in question

In some molecules, particularly with resonance structures, we cannot associate bonding electrons with specific atoms

C₆H₆ (Benzene)

Benzene has two resonance forms

• The six carbon - carbon bonds are of equal length, intermediate between a single bond and double bond
• The molecule is planar
• The bond angle around each carbon is approximately 120°

The apparent hybridization orbital consistent with the geometry would be sp² (trigonal planar arrangement)

• This would leave a single p orbital associated with each carbon (perpendicular to the plane of the ring)

With six p electrons we could form three discrete p bonds

• However, this would result in three double bonds in the ring, and three single bonds
• This would cause the bond lengths to be different around the ring (which they are not)
• This would also result in one resonance structure being the only possible structure

The best model is one in which the p electrons are "smeared" around the ring, and not localized to a particular atom

• Because we cannot say that the electrons in the p bonds are localized to a particular atom they are described as being delocalized among the six carbon atoms

Benzene is typically drawn in two different ways:

• The circle indicates the delocalization of the p bonds

Structure of NO₃^-

The Lewis structure of NO₃^- ion suggests that three resonance structures describe the molecular structure

• For any individual Lewis resonance structure the electronic structure for the central N atom is predicted to be sp² hybrid orbitals participating in s bonds with each of the O atoms, and an electron in a p orbital participating in a p bond with one oxygen (forming a double bond)
• Two of the O atoms are predicted to have sp³ hybrid orbitals, with one orbital participating in a s bond with the central N atom and the other orbitals filled with non-bonding electron pairs. The other O atom is predicted to have sp² hybrid orbitals, with one orbital participating in a s bond with the central N and two orbitals filled with non-bonding pairs of electrons. Furthermore, this last O atom is participating in a double bond with the central N atom and therefore should have an electron in a p orbital to participate in a p bond with the central N

How will this arrangement look as far as the orbital diagrams?

• There are 24 valence electrons in the expected valence orbitals above
• Summing the valence electrons from the formula gives: (3 x 6) for O, plus 5 for N, plus 1 for ionic charge = 24

What might we expect for the electron configuration if we just started with the N atom?

• We would predict that the N can only make two s bonds, it would have one pair of non-bonding electrons, and a p electron left over to participate in a p bond with one of the s bonds
• This is different from what the Lewis structure shows, and from our prediction of hybrid orbitals from the expected geometry
• If we look at the sp³ O atoms above we see that they actually have 7 electrons (1 more than expected), while the sp² O atom has the expected 6. Furthermore, the N atom (in the correct sp² configuration) has 4 electrons (1 less than expected)
• The "extra" electron from the ionic charge is correctly accounted for in the summation of electrons

• begin by predicting the hybridization orbitals
• then determine lone pair arrangements and s and p bonding electrons for each atom
• confirm that all bonding electrons are correct and that the total of electrons is correct

1996 Michael Blaber