Chapter 4 part 2

4.5. OTHER TYPES OF BONDING

4.5.1. Hydrogen bonds: Hydrogen bonds are well defined by a description of the hydrogen bonds in hydrogen fluoride (FIG. 4.5).

 

 

4.5.2. Van der Waals bonds are very weak forces of electrostatic attraction amongst instantaneous dipoles and induced dipoles.

At a given moment in time, the electron cloud surrounding even an inert gas atom may not be perfectly symmetrical. Thus one side will carry a relative positive charge and the other a relative negative charge. The separation of charge (dipole) can be almost negligible in small atoms such as helium. In addition to these instantaneous dipoles, there are dipoles induced by the original dipoles.

Since the dipoles are very small, the forces which exist between them are also very small. Even at low temperatures, the random motion of

group VIII atoms is vigorous enough to keep them gaseous. Moreover, on collision, repulsion between like charges is just as likely to occur as attraction between opposite charges.

However, as extremely low temperatures are reached, the random motion is not enough to break van der Waals attractions more often than they occur. Eventually, at low enough temperatures, the gas may liquefy, and the liquid may then even solidify. In energy terms, the kinetic energy is less than the bonding energy.

The larger an atom or molecule, the easier it is for dipoles to exist. In large atoms this is because the outer electrons are a long way from the control of the nucleus. Relatively simple large molecules (e.g. iodine) may solidify even at room temperature. In contrast, helium must be reduced to 4K, and a pressure of 103 atmospheres must be applied, in order for solidification to occur.

Van der Waals forces are particularly important in determining the shapes which macromolecules such as proteins adopt (section 25.2.5.). Here, the large number of bonds compensates for their extreme weakness. This may also be the case between very large molecules. Note that in DNA, large numbers of hydrogen bonds are the important bonds holding the two strands together (section 25.4.6.).

4.6. BONDING CHARACTERISTICS

4.6.1. The main type of bonding has effects on properties, as summarised in TABLE. 4.1.

 

 

The fact that ionic compounds have high melting and boiling points whereas covalent compounds have low melting and boiling points does not indicate that ionic bonds are stronger than covalent bonds.

The high melting and boiling points of ionic compounds do indicate that ionic bonds are strong i.e. the temperature must be high before ionic motion is vigorous enough to overcome the electrostatic forces of attraction.

However, covalent bonds are also generally very strong, but it is not these which break when small molecular compounds melt or boil. It is extremely weak van der Waals forces between one molecule and another which break.

4.7. INTERMEDIATE TYPES OF BONDING

4.7.1. The three main types of bonding, electrovalent, metallic, and covalent are extreme cases. As often as not, bonding is intermediate between two types.

4.7.2. Electrovalent vs. covalent: There is an initial problem here of deciding whether to treat the bonding as ionic with covalent character or covalent with ionic character. This must be judged from an overall assessment of properties.

For example, any compound which conducts electricity in the molten state is more likely to be judged as principally ionic than it is to be judged as principally covalent, and vice versa.

i) Ionic bonding with covalent character: When two types of ion come together to form a compound, there is likely to be a high degree of covalent character when:

First, the positive ion is: a) small...................................b) highly charged

Second, the negative ion is: a) large........................................b) highly charged

(These empirical rules are known as Fajan's rules.)

It is predicatable that a small, highly charged positive ion will have a high surface charge density, and will tend to attract electrons away from the negative ion into a covalent bond.

It is also predictable that it will be easiest to remove electrons from negative ions when those electrons are a long way from the negative ion's nucleus, and when there is a large number of "extra" electrons in the outer shell.

ii) Covalent bonding with ionic character: When a largely covalent bond forms between elements of differing electronegativity (most cases), the electron pair will not be shared equally.

The more electronegative element will have a greater share of the bonding pair, as in the case of HCl molecules:

If it seems like fantasy to consider this molecular gas as partially ionic, remember that it dissolves in water to produce hydrated H+ and Cl- ions.

Sometimes the polarity of a molecule will be enhanced by its shape and the existence of lone pairs:

The assymetric ammonia molecule is polar. Apart from discrete polar molecules, molecular solids may also exhibit ionic character. In fact, the mere existence of covalent molecules in a solid lattice is a step towards ionic character: the van der Waals bonding between molecules

is becoming less distinguishable from the covalent bonding within the molecules (e.g. section 14.3.1.).

4.7.3. Covalent vs. metallic bonding: This is most relelvant to the case of pure elements and is well illustrated by changes down group IV (section 17.2.1) and group VII (section 16.2.1.) and by changes across the period from sodium to chlorine. We shall discuss the changes across a period here.

Sodium (LHS) is electropositive and forms clear-cut metallic bonding. Chlorine (RHS) is electronegative and forms diatomic molecules with a precise covalent bond between the two atoms.

From sodium to aluminium, the increasing effective nuclear charge (as well as the increasing number of electrons contributed by each atom to the metallic bonding orbital) causes the metallic bonding to become stronger. (Note that this is not the same as saying the bonding becomes become more metallic.)

By silicon, the effective nuclear charge is high enough to hold the bonding electrons in fairly concentrated regions of electron density on the axes between the nuclei i.e. the bonds are more covalent than metallic, but silicon still has a lattice structure more comparable with a metal than with the discrete molecules normally associated with covalent bonding.

Moreover, silicon has a metallic lustre, suggesting that the bonding electrons are diffuse enough to reflect light. Some of silicon's semi-conductive properties may also be considered as intermediate between metallic and covalent properties.

The higher effective nuclear charges of phosphorus and sulphur result in even more coavlent character. Discrete molecules are formed, P4 and S8, but van der Waals forces between one molecule and another are strong enough for both elements to be solids at room temperature.

4.8. PREDICTING THE NUMBER OF BONDS

4.8.1. Having predicted the type of bonding, it is necessary to predict the number of bonds (charge in the case of ions).

4.8.2. The octet fable: In its least controversial form the octet "rule" observes that atoms often end up with eight outer electrons when they form bonds. In its most controversial form it suggests that atoms form bonds in order to achieve this configuration. The configuration acquires

the status of "stable octet" which is attributed with magical properties resulting from its similarity to the outer electron configuration of the "noble" gas atoms.

Perhaps it is the wishy-washy use of stability as an explanation which has allowed the continuation of the myth. In more cases than this, use of the word "stable" is an attempt to make "low energy" sound like a complete explanation.

The facts are even more disturbing. For one thing, in energy terms the octet is usually less stable than the electron configuration of the parent atom. This is indicated by the fact that even group I metal atoms (which are amongst the most able to lose electrons and form positive ions with an outer octet) require energy to remove the outer electron.

It is true that Cl-, with its outer octet, is more "stable" than the parent atom (electron affinities are often negative) but this is not because forming negative ions achieves the octet.

This is clearly illustrated by oxygen. An O- ion is more "stable" than an oxygen atom. The first electron affinity is negative i.e. the ion has less energy than the parent atom and energy is given out as heat when it forms. However, when O- goes on to achieve the octet by forming O2-, energy is required.

Elements which form negative ions are elements with high effective nuclear charges, and this is "why" first electron affinities can be negative. However, when a second electron is added to O-, it is repelled by the negative charge of the ion and must be added to a small shell which already contains seven, negatively charged, electrons.

Note that we have found it necessary to explain energy changes in terms of electrostatic forces.

4.8.3. Rubbing it in: In addition to all this, there are plenty of cases where bonding does not result in an outer octet:

i) More than the octet: a) Electrovalent: Fe3+ has 13 outer..........................................electrons in Fe2(SO4)3

......................................b) Covalent: Sulphur has 12 outer

..........................................electrons in SF6

ii) Fewer than the octet: a) Electrovalent: Pb2+ has 2 outer............................................electrons in PbSO4

........................................b) Covalent: Al has 6 outer electrons

............................................in AlCl3

The situation is more involved than the octet rule implies. There are far more events occurring than merely the formation, or otherwise, of octets. It is convenient to tackle the problem in terms of a more detailed look at energy changes, and to refer to the electrostatic forces involved in order to understand them.

4.8.4. Predicting the charge on a positive ion: Obviously this step comes after deciding that a compound is electrovalent and after deciding which element constitutes the positive ion.

i) The charge on a positive ion is equal to the number of electrons lost by the parent atom.

An atom is unlikely to lose more electrons than it has in the outer shell because the next shell is closer to ("and less screened from") the attraction of the nucleus.

An atom is unlikely to lose fewer electrons than it has present in the outer shell. This is because complete loss of the outer shell electrons not only makes the ion more highly charged, but also makes it much smaller. The resultant ion will therfore *form a much stronger lattice (in aqueous solution, it will be much more strongly hydrated).

With respect to calcium, try to picture the events and imagine the changing electrostatic forces when calcium atoms react with oxygen atoms. Then try to imagine the even more complex events when calcium metal reacts with oxygen gas.

(*In energy terms, the much higher lattice energy compensates for the extra ionisation energies.)

ii) Exceptions: Unlike the octet rule, this approach provides a basis for understanding exceptions. For example, elements at the bottom of group IV can form 2+ ions as well as the expected 4+ ions (section 17.3.2.iii).

The behaviour of d-block elements can also be understood. They can lose penultimate shell electrons as well as outer electrons because there is little difference in the strength with which their penultimate d-electrons and outer s-electrons are held by the nucleus (section 2.4.3.).

4.8.5. Predicting charge on a negative ion: When an electronegative atom gains electrons to form negative ions it fills any singly occupied orbitals in its outer shell.

It is unlikely to gain more than this number of electrons, since this would involve attracting electrons into a shell further ("and more screened from") the attraction of the nucleus.

It is unlikely to gain fewer electrons, because filling the singly occupied orbitals increases the charge but results in a relatively small increase in the overall size of the ion. The resultant ion will therefore *form a stronger lattice than an ion which still has a singly occupied orbital.

With respect to oxygen, try to picture the events and imagine the changing electrostatic forces when calcium atoms react with oxygen atoms. Then try to imagine the even more complex events when calcium metal react with oxygen gas.

(* In energy terms, the extra lattice energy compensates for the extra electron affinity.)

4.8.6. Predicting the number of covalent bonds:

i) Covalent bonds: Obviously, predicting the number of covalent bonds comes after deciding that the bonding is covalent. Covalent bonds are usually formed by the overlap of singly occupied atomic orbitals, and the number of singly occupied orbitals in atom gives a rough indication of the number of bonds it will form.

ii) Electron promotion: There are, as you will recall, complications. For example, carbon in its "ground state" has two singly occupied orbitals:

.

However, carbon almost exclusively forms four bonds. This is described in terms of a model in which one of the s electrons is promoted into the vacant 2p orbital, giving it four singly occupied orbitals:

The advantage of forming 4 bonds can be seen by considering methane. In each C-H bond the carbon and hydrogen nuclei are attracted to the high electron density of a bonding pair of electrons. This attraction is greater than the total attraction of a carbon nucleus for one of its outer electrons plus that of a hydrogen nucleus for its outer electron in an H2 molecule. In other words, the formation of two extra bonds outweighs the advantages of leaving carbon's s-electrons paired and closer to the nucleus than they would be in a p-orbital. Moreover, the electrons actually enter bonding orbitals rather than p-orbitals, as we shall see in section 4.8.

In energy terms, the two extra bond energies more than compensate for any promotion energy, and the H-H bond energy (c.f. section 17.3.1.ii.).

iii) d-orbitals: from the third period onwards, d-orbitals are available for electron promotion. Thus sulphur can form 2, 4, or 6 bonds:

In this electron promotion model, the number of bonds formed is dependent on the number of electrons promoted. This in turn depends on:

a) The sizes of the bonding atoms

b) The effective nuclear charges of the bonding atoms.

These can be deduced from the positions of the elements in the periodic table. Moreover, they affect two main aspects of covalent bonding which themselves have a bearing on the number of bonds formed:

a) First they affect the strengths of the bonds formed: then it can be predicted that the stronger the bonds, the larger the number likely to form.

b) Second, they affect the amount of space available for bond formation: then it can be predicted that the larger the bonding atoms

relative to the central atom, the smaller the number of bonds likely to form, owing to lack of space.

Thus, referring back to sulphur, this element forms SF6 but not SCl6. This is predicatble for two reasons: First, fluorine's nucleus is closer to ("and less screened from") the bonding electrons than is chlorine's nucleus. Fluorine therefore forms stronger bonds than chlorine with sulphur. Second, chlorine's larger size means that there is less room for 6 of its atoms to fit around the small central sulphur atom (FIG. 4.17.)

You are likely to develop a feel for this kind of reasoning only after considering a good number of examples. Note also:

c) Multiple bonding reduces the number of bonded atoms required to form the higher oxidation states.

iv) Dative covalent bonds: a final complication is that orbitals containing pairs of electrons may be involved in dative covalent bond formation by overlap with vacant atomic orbitals.

Moreover, singly occupied orbitals may become available for dative bond formation by spin-pairing:

It is possible to predict the liklihood of dative bond formation from its strength and the amount of room available, and thus from sizes and effective nuclear charges of the bonding atoms. Hence (ultimately) the prediction can be made from the positions of the bonding elements in the periodic table. Again you will develop a feel for this after considering a large number of cases.

In the N2O4 molecule cited above, it is found that in practice all the N-O bonds are equivalent. This emphasises the fact that we are using a only partially correct model in order to make predictions. The model is developed by considering one more piece of bonding theory, hybridisation.

4.9. HYBRIDISATION

4.9.1. The shapes of atomic orbitals come from the solution of complex wave equations. Such calculations also allow shapes to be assigned to bonding orbitals, and they rely heavily on energy considerations.

The wave equations are simplified by considering only the predicted shapes of orbitals which atoms could theoretically use to form bonds. Quantum mechanists can actually calculate the space around a nucleus in which an electron will most probably occur (FIG.2.3.). The shape is a direct function of the orbital's energy, and this can change if hybridisation is included in the quantum mechanical calculations.

Despite the useful outcome of these calculations, it is not wise to be generally bullied by mathemeticians. A cynic might say that they resort to mathematical models because they have as much difficulty understanding reality as we have understanding them. Mathematical models must, as they are here, be related to reality. The calculated shape of an orbital is very useful for developing the models we use to describe atoms and molecules.

4.9.2. Why we need the concept of hybrids: Even when carbon is described as having promoted its 2s electron into the vacant 2p orbital, it is not in an appropriate state to form bonds, according to our existing models. This is despite having four singly occupied orbitals.

According to our existing model, the four orbitals are not a convenient shape, nor are they conveniently arranged, to overlap with other atomic orbitals (FIG. 4.6.)

 

 

However, orbital shape is calculated as a function of energy. If the energy of an orbital were different from that considered as normal in a free atom, we would consequently predict a different shape. The energies of these orbitals would be different from normal if the distribution of energies amongst the orbitals were different from normal.

Moreover, the distribution of energy would be different from normal if the 2s orbital were equivalent in energy to one or more of the p-orbitals. As a predictive tool, it is suggested that this equalisation of energy does exist. It is known as hybridisation, and associated calculations describe some much more conveniently shaped (sp hybrid) orbitals (FIG. 4.7.).

 

 

There are three possibilities: sp1, sp2, and sp3 hybridisation.

4.9.3. sp3 hybridisation: In this case the s-orbital is considered as equivalent in energy to all three p-orbitals, giving four sp3 hybrid orbitals (FIG. 4.8.).

 

 

The four orbitals point to the four corners of a tetrahedron.

4.9.4. sp2 hybridisation: In this case the s-orbital is considered as equivalent in energy to two of the p-orbitlas, giving three sp2 hybrids, and leaving one p-orbital with a higher energy (FIG. 4.9.).

 

 

The three sp2 hybrids point to the three corners of an equilateral triangle.

4.9.5. sp1 hybridisation: In this case the s-orbital is considered as equivalent in energy to one of the p-orbitals, giving two sp1 hybrids, and leaving two p-orbitals with a higher energy, but with the same energy as each other (FIG. 4.10).

 

 

The sp1 orbitals are arranged linearly.

4.9.6. sp3 hybrids in methane (FIG. 4.11.):

 

 

In an s-bond, the electron density is concentrated along the axis between the centres of the two bonded atoms.

4.9.7. sp2 hybrids in ethane (FIG. 4.12):

 

 

4.9.8. sp2 hybrids in buta-1,3-diene (FIG. 4.13.):

 

 

Electrons are said to be delocalised when they are not confined to the positions indicated in a simple bonding diagram, but are spread out by, for example, p-orbital and p-bonding system overlap.

4.9.9. sp2 hybrids in benzene (FIG. 4.14.):

 

 

4.9.10. sp1 hybrids in ethyne (FIG. 4.15.):

 

4.10. SHAPES OF MOLECULES

4.10.1. Hybridisation theory allows the shapes of molecules to be predicted. Shape may also be predicted by remembering that electrons in one bond will repel electrons in another. The repulsive effect of multiple bonds will be greater than the repulsive effect of single bonds.

Also, lone pairs of electrons will repel electrons in bonds, and repel other lone pairs, as well. Lone pairs exert a greater repulsive force even than multiple bonds.

Thus the shape of a molecule will tend to achieve maximum separation of bonds and lone pairs, taking into consideration the relative strengths of the repulsive forces. These are summarised as decreasing in the order: lone pair > multiple bond > single bond. (This is a summary of, so-called, Sidgwick-Powell Theory). Some examples are shown in FIG 4.16 below.

(Lattice arrangements in ionic compounds, metals, and molecular crystals are summarised in chapter 5.)

4.11. QUESTIONS

1) In chapter 2 we implied that it was dangerous to talk about causes, and about answers to the question "why?". How do you equate this with sections 4.2.2. and 4.2.3.?

2) What makes helium unable to form He2 molecules?

3) A small highly charged positive ion is more likely than a big positive ion with a single charge to form a strong ionic lattice? Comment.

4) How can the forces which exist in an ionic lattice be responsible for the existence of ions within that lattice? Answer the question with reference to the formation of calcium oxide from i) calcium atoms and oxygen atoms ii) calcium metal and molecular oxygen gas.

5) Why is the phrase "less screened from" placed in parentheses in this chapter?

6) Predict the simplest formulae of all compounds formed between the following pairs of elements, and predict the type of bonding found in each case:

copper and zinc, lead and oxygen, potassium and bromine, chlorine and bromine, oxygen and sulphur, sulphur and chlorine, beryllium and chlorine, strontium and sulphur, hydrogen and sulphur, phosphorus and bromine, phosphorus and fluorine.

In three cases, explain your reasoning.

7) Predict the shapes of SO2, CO2, AlCl3, NCl3, the [Cu(H2O)6]2+ ion, and SCl4.

8) Draw simple bonding structures for NO, NO2, N2O4, N2O5, SO32-, SO4

2-, S2O3

2-, S4O62-, S2O8

2-, and S2O72-.