8.3 Orbital Overlaps & Types of Bonds

In this section we will see how these orbitals can be used to account for the bonds in different molecules.

In this lesson, we extend this idea to compounds containing double and triple bonds, and to those in which atomic d electrons are involved (and which do not follow the octet rule.)

1  Hybrid types and multiple bonds

We have already seen how sp hybridization in carbon leads to its combining power of four in the methane molecule. Two such tetrahedrally coordinated carbons can link up together to form the molecule ethane C₂H₆. In this molecule, each carbon is bonded in the same way as the other; each is linked to four other atoms, three hydrogens and one carbon. The ability of carbon-to-carbon linkages to extend themselves indefinitely and through all coordination positions accounts for the millions of organic molecules that are known.

Figure 8.48 Orbitals overlap in C₂H₆

Trigonal hybridization in carbon: the double bond

Carbon and hydrogen can also form a compound ethylene (ethene) in which each carbon atom is linked to only three other atoms. Here, we can regard carbon as being trivalent. We can explain this trivalence by supposing that the orbital hybridization in carbon is in this case not sp³, but is sp² instead; in other words, only two of the three p orbitals of carbon mix with the 2s orbital to form hybrids; the remaining p-orbital, which we will call the i orbital, remains unhybridized. Each carbon is bonded to three other atoms in the same kind of plane trigonal configuration that we saw in the case of boron trifluoride, where the same kind of hybridization occurs. Notice that the bond angles around each carbon are all 120°.

This alternative hybridization scheme explains how carbon can combine with four atoms in some of its compounds and with three other atoms in other compounds. You may be aware of the conventional way of depicting carbon as being tetravalent in all its compounds; it is often stated that carbon always forms four bonds, but that sometimes, as in the case of ethylene, one of these may be a double bond. This concept of the multiple bond preserves the idea of tetravalent carbon while admitting the existence of molecules in which carbon is clearly combined with fewer than four other atoms.

Figure 8.49 Sigma and Pi bonding in C₂H₆, C₂H₄ and C₂H₂

These three views of the ethylene molecule emphasize different aspects of the disposition of shared electron pairs in the various bonding orbitals of ethene (ethylene). (a) The “backbone” structure consisting of σ (sigma) bonds formed from the three sp²-hybridized orbitals on each carbon. (b) The π (pi) bonding system formed by overlap of the unhybridized p_z orbital on each carbon. The π orbital has two regions of electron density extending above and below the plane of the molecule. (c) A cutaway view of the combined σ and π system.

As shown above, ethylene can be imagined to form when two -CH₂ fragments link together through overlap of the half-filled sp² hybrid orbitals on each. Since sp² hybrid orbitals are always in the same plane, the entire ethylene molecule is planar. However, there remains on each carbon atom an electron in an unhybridized atomic p_z orbital that is perpendicular to the molecular plane. These two parallel p_z orbitals will interact with each other; the two orbitals merge, forming a sausage-like charge cloud (the π bond) that extends both above and belowthe plane of the molecule. It is the pair of electrons that occupy this new extended orbital that constitutes the “fourth” bond to each carbon, and thus the “other half” of the double bond in the molecule.

More about sigma and pi bonds

The σ (sigma) bond has its maximum electron density along the line-of-centers joining the two atoms (below left). Viewed end-on, the σ bond is cylindrically symmetrical about the line-of-centers. It is this symmetry, rather than its parentage, that defines the σ bond, which can be formed from the overlap of two s-orbitals, from two p-orbitals arranged end-to-end, or from an s– and a p-orbital. They can also form when sp hybrid orbitals on two atoms overlap end-to-end.

Pi orbitals, on the other hand, require the presence of two atomic p orbitals on adjacent atoms. Most important, the charge density in the π orbital is concentrated above and below the molecular plane; it is almost zero along the line-of-centers between the two atoms. It is this perpendicular orientation with respect to the molecular plane (and the consequent lack of cylindrical symmetry) that defines the π orbital. The combination of a σ bond and a π bond extending between the same pair of atoms constitutes the double bond in molecules such as ethylene.

Carbon-carbon triple bonds: sp hybridization in acetylene

We have not yet completed our overview of multiple bonding, however. Carbon and hydrogen can form yet another compound, acetylene (ethyne), in which each carbon is connected to only two other atoms: a carbon and a hydrogen. This can be regarded as an example of divalent carbon, but is usually rationalized by writing a triple bond between the two carbon atoms.

We assume here that since each carbon forms two geometrically equivalent bonds, this atom must be sp-hybridized in acetylene. On each carbon, one sp hybrid bonds to a hydrogen and the other bonds to the other carbon atom, forming the σ bond skeleton of the molecule.

In addition to the sp hybrids, each carbon atom has two half-occupied p orbitals oriented at right angles to each other and to the interatomic axis. These two sets of parallel and adjacent p orbitals can thus merge into two sets of π orbitals.

Figure 8.50 Sigma and Pi bonding

The triple bond in acetylene is seen to consist of one σ bond joining the line-of-centers between the two carbon atoms, and two π bonds whose lobes of electron density are in mutually-perpendicular planes. The acetylene molecule is of course linear, since the angle between the two sp hybrid orbitals that produce the s skeleton of the molecule is 180°.

Figure 8.51 Sigma and Pi bonding  Formation

Multiple bonds between unlike atoms

Multiple bonds can also occur between dissimilar atoms. For example, in carbon dioxide each carbon atom has two unhybridized atomic p orbitals, and each oxygen atom still has one p orbital available. When the two O-atoms are brought up to opposite sides of the carbon atom, one of the p orbitals on each oxygen forms a π bond with one of the carbon p-orbitals. In this case, sp-hybridization is seen to lead to two double bonds. Notice that the two C–O π bonds are mutually perpendicular.

Figure 8.52 Sigma and Pi bonding in HCN

Similarly, in hydrogen cyanide, HCN, we assume that the carbon is sp-hybridized, since it is joined to only two other atoms, and is hence in a divalent state. One of the sp-hybrid orbitals overlaps with the hydrogen 1s orbital, while the other overlaps end-to-end with one of the three unhybridized p orbitals of the nitrogen atom. This leaves us with two nitrogen p-orbitals which form two mutually perpendicular π bonds to the two atomic p orbitals on the carbon. Hydrogen cyanide thus contains one single and one triple bond. The latter consists of a σ bond from the overlap of a carbon sp hybrid orbital with a nitrogen p orbital, plus two mutually perpendicular π bonds deriving from parallel atomic p orbitals on the carbon and nitrogen atoms.

Figure 8.52 NO₃^– ion  structure

Pi bond delocalization furnishes a means of expressing the structures of other molecules that require more than one electron-dot or structural formula for their accurate representation. A good example is the nitrate ion, which contains 24 electrons:

Figure 8.53 Pi delocalization in NO₃^– ion

The electron-dot formula shown above is only one of three equivalent resonamce strutcures that are needed to describe trigonal symmetry of this ion.

Figure 8.54 Resonating forms in NO₃^– ion

Nitrogen has three half-occupied p orbitals available for bonding, all perpendicular to one another. Since the nitrate ion is known to be planar, we are forced to assume that the nitrogen outer electrons are sp²-hybridized. The addition of an extra electron fills all three hybrid orbitals completely. Each of these filled sp² orbitals forms a σ bond by overlap with an empty oxygen 2p_z orbital; this, you will recall, is an example of coordinate covalent bonding, in which one of the atoms contributes both of the bonding electrons. The empty oxygen 2p orbital is made available when the oxygen electrons themselves become sp hybridized; we get three filled sp hybrid orbitals, and an empty 2p atomic orbital, just as in the case of nitrogen.

The π bonding system arises from the interaction of one of the occupied oxygen sp orbitals with the unoccupied 2p_x orbital of the nitrogen. Notice that this, again, is a coordinate covalent sharing, except that in this instance it is the oxygen atom that donates both electrons.

Pi bonds can form in this way between the nitrogen atom and any of the three oxygens; there are thus three equivalent π bonds possible, but since nitrogen can only form one complete π bond at a time, the π bonding is divided up three ways, so that each N–O bond has a bond order of 4/3.

Figure 8.55 Pi bonding overlap in NO₃^– ion

e 8.62 Sigma and Pi bond in C₂H₄

Ref: commons.wikimedia.org/

Notice the overlap between the half-filled p orbitals on the carbon and oxygen atoms. When p orbitals overlap this way (side by side), the resulting bond is a pi bond (п ) bond and the electron density is above and below the internuclear axis.            When orbitals overlap end to end as in the case in all of the rest of the bonds in molecule the resulting bond is sigma bond.()

In case of ethane (C₂H₆) in the above figure, both C atoms of ethane are sp3 hybridized. The C-C bond involves the overlap of one sp3 orbital from each C, and each of the six C-H bonds involves overlap of a C sp3 orbital with an H 1s orbital. These bonds are formed as a result of end to end overlap. They are assigned as sigma  bond.

Figure 8.63(a)  Sigma and Pi bond in C₂H₄

Figure 8.63(b)  3D shape of C₂H₄

Ref: commons.wikimedia.org/

In case of ethylene, each C atom is sp2hybridized. Each C atom’s four valence electrons half-fill its three sp2 orbitals and its unhybridized 2p orbital, which is perpendicular

Figure 8.64 3D shape of C₂H₄

Ref: commons.wikimedia.org/

In case of acetylene, we see a triple bond C≡C consists of one sigma bond and two pi bond are formed. To maximize overlap in a linear shape, one s and one p orbital in each C atom form two sp hybrids, and two 2p orbitals remain unhybridized. Each C uses one of its sp orbitals to form a sigma bond with an H atom and uses the other form the C-C σ bond. Side to side overlap of one pair of 2p orbitals gives one п bond, and the other pair of 2p orbitals gives another п bond. The second pi bond is  perpendicular to the first pi bond. The electron density in the front and back of the sigma bond. The result is cylindrically symmetrical H-C≡C-H molecule.

In terms of strength, the double bond is about twice as strong as a single bond and a triple bond is about three times as strong.

Figure 8.65 3D shape of C₂H₂

Ref: commons.wikimedia.org/

The correspondence between valence bond theory and Lewis theory is striking here. In both models, the central carbon atom forms four bonds: two single bonds and one double bond. However, Valence bond theory gives us more insight into the bonds. According to Valence Bond theory, any double bond between two atoms consists of two different kinds of bonds-one sigma and one pi. In Lewis model two bonds within the double bond appear identical.

In general pi bonds are weaker than sigma bonds because side to side orbital overlap tends to be less efficient than end to end overlap. Consequently, the pi bond in a double bond is generally easier to break than sigma bond.

Valence bond theory allows us to see why the rotation about a double bond is severely restricted. Due to side wise overlap of the p-orbitals, the pi bond must essentially break for rotation to occur.

Although rotation about the double bond is highly restricted but rotation around the single bond is possible. For example, 1,2-dichloroenthane and 1,2-dichloroethene.

Figure 8.66 1,2-dichloroethane(left, rotation possible) 1,2-dichloroethene(right, rotation restricted)

Watch the following video for more explanation.

Sigma and Pi Bonds Explained, Basic Introduction, Chemistry

What you should be able to do

1. Make sure you thoroughly understand the following essential ideas which have been presented above.
2. Sketch out diagrams showing the hybridization and bonding in compounds containing single, double, and triple carbon-carbon bonds.
3. Define sigma and pi bonds.
4. Describe the hybridization and bonding in the benzene molecule.

Practice Questions:

1. Describe the types of bonds and orbitals in

a) acetone((CH₃)₂CO,

b) formaldehyde(H₂CO)

c) HCN

d) CO₂

Limitation:

The valence bond approach is especially useful in organic chemistry where so many molecules are built of tetrahedral C atoms, sp ³ hybridised. The concept of hybrids is intuitively very satisfying because they fit visually with our perceived picture of the shape of a molecule with its directed bonds between pairs of atoms. Unfortunately, the VB approach is not satisfactory for species like CO ₃ ⁼ , NO ₃ ⁻ , and benzene because the VB picture does not reflect the known chemical structure. A new concept of resonance hybrids must be introduced, and CO ₃ ⁼ must now be represented by a combination of three Lewis-octet structures. Worse still, the VB approach cannot easily give a satisfactory bonding picture for either of the important molecules O ₂ or CO.

In cases where the VB approach does not work well, the molecular orbital (MO) method is often more successful. The situation is best summarized by using the strengths of the VB approach where they are appropriate, as in CH ₄ , and using the MO approach where it is best suited, as in O ₂ and benzene. After all, each approach is an approximation, incomplete and imperfect.

Read more: http://www.chemistryexplained.com/Te-Va/Valence-Bond-Theory.html#ixzz6VbHDEIOG