# Molecular Structure

Petrucci: Section 10-1 to 10-6, 16-9

## Lewis Dot structures

One of the early questions asked by scientists, once the concepts of atoms and molecules had been firmly established was "How are Atoms Bonded?"  We've developed many theories over the years in attempts to explain the bonding between atoms in various substances.  In some substances, our models explain the bonding that holds the atoms and molecules to each other quite well but as we explored more and more deeply, we have always been able to find examples of substances where the models don't work.  We will explore here, the development of models and the uses and limits to the models as we work through better and better models.

There are many different types of materials with properties that vary widely.  Yet even the simplest of models can be useful.

For example, Salt, NaCl,  liquid molecules, H2O, gaseous molecules, H2 and Cl2?

We say that there is a chemical bond between the atoms involved

Chemical Bonds: $\rightarrow$ Forces that hold togther atoms in a molecule.

Lewis Electron Dot Symbols Convenient for explaining chemical bonds. Very simple, theory but can be quite a useful visualization tool.

Recall the concept of the atomic core. We will assume that the inner (core) electrons in a molecule are not involved in chemical bonding since they are held tightly to the nucleus and are not involved in either electron sharing or electron transfer processes. As a result, Lewis Dot diagrams display only the outer or valence electrons.

OCTET RULE: In compound formation an atom tends to gain or loose electrons or to share electrons until there are eight electrons in its outer shell. (Often but not always the valence shell).

### Normal valence state

Valence is used for many purposes. The only usage I will give it is Valence = number of atoms that are bonded an atom. For example, Oxygen displays its valence of 2 in the compound H2O while nitrogen with a valence of 3 forms NH3. The valence of the elements can be equated to the number of unpaired electrons on the atom.  However, we must notice that the ground state of some of the elements (Be, B, C) is not the same as the valence state.  In other words, the element in free space in it's minimum energy configuration (ground state) is not necessarily the same as the element in a molecule.

To reconfigure an atom into its 'normal valence' state, we can split the non-bonding pair (lone pair) of electrons into two unpaired electrons.  In the case of Be, we split the two 2s electrons into two unpaired electrons (into orbitals not necessarily called 2s and 2p but we'll get to that later).

Similarly, we can expand the valence of B and C by splitting their lone pairs into single electrons. In each of these cases, we increase the "valence" by 2 for each lone pair we split up. We can do this because there are a maximum of 4 orbitals at the n=2 quantum level into which we can put electrons with minimal expense energy-wise.  If we tried to split the lone pair on Nitrogen into two unpaired electrons we would run into the limit of the number of orbitals.  N, has only 4 orbitals (places) to hold electrons.  The normal ground state of N already has at least 1 electron in each of the four orbitals.  If we tried splitting the lone pair we would have no place to put the extra electron without moving it to an n=3 orbital.  That would cost too much energy.

The energy we pay to move electrons around to form their normal valence state is returned to us when we form bonds.  We don't want to have to pay too much to move electrons around because the energy returns when we form a bond may not compensate.  That's another story we'll discuss later.

We see that this simple way of representing the valence electrons around the atom can actually help us understand the observed 'valence' of the atoms. NOTE that the valence indicated is an exact match with the number of unpaired electrons in the Lewis valence state structure.  We will use this concept of valence to build our Lewis Dot Structures.  Make sure you memorize (at least the last line of) this table!

Table of Valences
Group 1 2 3 4 5 6 7 8
#Valence e- 1 2 3 4 5 6 7 8
Valence 1 2 3 4 3 2 1 0
Lewis Dots
Groundstate
Li ·  Be : : B·
· C :
·
·
: N ·
·
·
: O :
·
··
: F :
·
··
: Ne :
··
Lewis  Dots
normal valence state

Li ·

· Be ·

·
· B ·

·
· C ·
·

·
: N ·
·

·
: O :
·

··
: F :
·

··
: Ne :
··

### Ionic Bonds

In an ionic bond, one atom looses all its outer electrons (leaving behind a filled inner shell) while another atom gains electron(s) to fill its valence shell. The two ions attract each other according to Coulombic interactions.

For example, consider sodium chloride.

The Lewis Structure for the Salt NaCl, shows two ions which have their (Now) outer shells of electrons filled with a complete octet. In the case of the sodium cation, the filled shell is the outermost of the 'core' electron shells. In the Chloride ion, the outer shell of valence electrons is complete with 8 electrons. The two ions are bonded with ionic bonds.

Consider the compound magnesium oxide: MgO.

Note that in both cases, the positive ions are metals and the negative ions are non-metals. This is generally true for ionic compounds.

In general Metals have lower ionization energies and form cations.

Non metals have high ionization energies (and high electron affinities and form anions.

 1 2 3 4 5 6 7 Li+ Ba2+ Na+ Mg2+ Al3+ K+ Ca2+ Rb+ Sr2+ Cs+ Ba2+

### Covalent Bonds

Covalent bonds involving the sharing of electrons to form 'bonding pairs' of electrons and satisfy their normal valences.  It turns out that when this is done, most commonly, there are eight electrons around each atom.  This is the so-called octet rule.

Take Cl2. Each Cl atom has one unpaired electron . It's Unlikely that an ionic type bond has formed since the two atoms are identical and such a transfer would leave one chlorine with only 6 electrons. More likely, they share a pair of electrons so that each can 'think' it has 8 electrons in its valence shell.  Note that Cl is below F in the periodic table and so has a normal valence of 1 with three extra lone pairs.

Mostly, this type of bonding occurs between non-metal atoms.

Consider PCl3:

Note: I hid the lone pairs on the chlorines to emphasize the electron configuration around the phosphorous.

The normal valence of P (like N) is 3 with one lone pair.  So, when combined with three Cl atoms, the structure as drawn above satisfies this valence requirement.  The P shares one of its unpaired electrons with each of the three Cl atoms to form three covalent bonds.  We see the normal valence of P = 3 and of Cl = 1 is satisfied with this structure.  All four atoms now have a completed octet.

It is not always necessary to use two dots to represent the pair of electrons. A dash can also represent the pair of electrons, especially in bonds. Consider the following examples

or using a  to represent a bonding pair

It is sometimes even possible to simplify even further by hiding the outer electrons on the outside atoms (but not the 'central' one), as long as this doesn't cause ambiguities.  One can also use a dash alongside the atom to represent lone pairs.

This method is quicker to draw, more clearly shows that electron pairs are bonding pairs and helps distinguish the bonding pairs from the non-bonding pairs more easily. It is also possible to use the dashes to represent both the bonding pairs and the non-bonding pairs of electrons. This is also an acceptable way of doing a Lewis dot structure for some purposes.

There are exceptions to the Octet rule. For example, H has only a 1s orbital occupied. The other orbitals are all higher in energy and not involved in the chemistry of hydrogen.

Consider water H2O.

There are only two electrons in the valence shell of hydrogen but that's all it takes to fill the first shell. Hence, the octet rule doesn't apply.

Other compounds where the Octet rule is not obeyed include some Boron compounds.

Boron has only three unpaired electrons to share.  It has a normal valence of 3 with no lone pairs according to the table of valences.

Here, Boron has only 6 valence electrons. It cannot complete its octet and remain a neutral compound. BF3 can combine with an F- ion to complete its octet as follows.

One of the lone pairs on the fluoride is shared with the boron and forms a bond.

The ammonium ion is another example of a complex ion that fills the octet rule. In this case, the nitrogen in ammonia has a completed octet but it combines with the hydrogen ion, which has an empty 1s orbital. In this case, the bonding pair of electrons comes completely from the nitrogen. The positive hydrogen ion is attracted to and then shares the (originally) non-bonding pair of electrons on the N to form a covalent bonding pair.

These polyatomic ions can form ionic compounds such as NH4+Cl- and Na+BF4-.

### Expanded (or hyper-) Valences

Elements in row 3,4,... have access to d orbitals (unlike row 2 elements that have only s and p orbitals) and can therefore have more than 8 electrons around them in certain compounds.  Recall that P has a normal valence of 3 with one lone pair.  If that lone pair is divided into two extra unpaired electrons to be used in creating bonds, we then have a hypervalence of 5.

By expanding the valence of P, we change it from Valence=3 with one lone pair to Valence=5 with no lone pairs.  PCl5 is one such compound that has P with a hypervalence of 5 and no lone pairs.

Here, we see that the valence of P is 5.  Since this is higher than it's normal valence, we call it the hypervalence. (hyper means higher than).

Other elements, like S can have more than one hypervalence.  S has a normal valence of 2 with two lone pairs.  If we split the lone pairs one at a time to create extra unpaired electrons for bonding, we can create hypervalences of 4 and 6.  We can see these at work in the following series of compounds. SF2, SF4, and SF6.  Try drawing these Lewis structures out, Note that the number of lone pairs on the S decreases by one as the valence goes up by 2.

At one time, it was believed that the octet rule was the overriding factor in determining bonding.  Using this logic, it was once thought that the Noble gases (He, Ne, ...) were totally unreactive since they already have a filled octet.  Using the concept of hypervalences, we see that we can expect that some noble-gas compounds can form.  We have in recent years produced compounds of Xe as follows:

XeF2, XeF4, and XeF6, XeO4.  All four of these  compounds are using a hypervalent state of the Xenon since it's normal valence is zero.

What possible valence states might you expect for iodine, bonded to one or more fluorines?  Is one of them a normal valence state?

The results of these expansions can be summarized in the table below.

Table of Normal and hypervalences
Elements Valence lone pairs Example
P, As, Sb... Normal 3 1
5 0
S, Se, Te, ... Normal 2 2
4 1
6 0
Cl, Br, I, At... Normal 1 3
3 2
5 1
7 0
Ar, Kr, Xe, Rn... Normal 0 4
2 3
4 2
6 1
8 0

A more compact view of this same table could be drawn like the following table.  There, are listed the valences and where applicable, the number of lone pairs.  for example, the fifth column normal valence has a 3,1 representing 3 bonds and 1 lone pair.  Notice that hypervalent states only occur where atoms have extra electrons in lone pairs that can be split up to make extra bonding sites.  This requires d orbitals and is therefore restricted to the n > 3 valence levels.

Table of Normal Valences and Hypervalences.
Group 1 2 3 4 5 6 7 8
Normal Valence states 1 2 3 4 3,1 2,2 1,3 0,4
Hyper-valent states         5,0 4,1 3,2 2,3
6,0 5,1 4,2
7,0 6,1
8,0

### Multiple Bonds

Some compounds have more than one pair of electrons shared by two atoms.

Since one pair of shared electrons is a bond, two pairs would be two bonds, etc. The number of bonds joining any two atoms is called the bond-order.

Take CO2, for example.  The normal valence of the C is 4 but the normal valence and the normal valence of O is 2. The only way we can connect these three atoms together is to connect them as follows.  Recall from the table of valences that C, in its normal valence state has 4 bonds and no lone pairs and O has 2 bonds and 2 lone pairs so that's the way we draw them in the Lewis structure.

The C=O bond order is 2

Now try C2H4.  Again, the normal valences of C (=4) and H (=1) can be satisfied with the following structure.

The C=C bond order is 2 whereas the C-H bond orders are all 1.

Now compare C2H2 and N2 (they're isoelectronic).  Here again, the normal valences of all the atoms is satisfied, only by creating multiple bonds.  Note that according to the table of valences that N has one lone pair.

### Formal Charge and Oxidation Number

There are several ways of accounting for electrons when we create these structures. We've seen that in the case of NaCl that the 'bonding' pair of electrons is completely on the chloride cation and hence, completely removed from the sodium cation. On the other hand, in Cl2, the bonding pair is exactly shared between the two identical chlorine atoms.

These represent two extremes of chemical bond formation. Most compounds form bonds which lie somewhere in between these two extremes, neither 100% covalent nor 100% ionic in nature.

However, for the purposes of bookkeeping, we need to make assumptions about the nature of the bonds. We will either assume 100% covalent or 100% ionic when we do the counting.

#### Oxidation Number

There is an alternate way to count electrons on molecules.  This results in a different individual number for each atom than we got using formal charges but like formal charges, the sum of these oxidation numbers must always add up to the overall charge on the species.  To count oxidation numbers, we assume 100% ionic bonding where electrons are completely transferred from the less electronegative atom to the more electronegative atom. For example, even though HCl is not ionic bonded, we make that assumption for the sake of accounting. (governments do it all the time when they make up new budgets)

In the case of HCl, Cl is more electronegative than H so it takes both electrons. Thus, to determine the oxidation number of the two atoms, we count the electrons differently.

(Higher electronegativity)

Oxidation # = Core Charge - # non-bonded electrons - # bonded electrons

(Lower electronegativity)

Oxidation # = Core Charge - # non-bonded electrons.

Thus, we can now determine the Ox#'s for HCl.

Oxidation # of Cl (more electronegative) = 7 - 6 - 2 = -1

Oxidation # of H (less electronegative) = 1 - 0 = +1

Notice that both the formal charge and the oxidation number of an atom is reported like a charge. Don't make the mistake of thinking that they really are charges. They are merely the result of the particular accounting method chosen.

Oxidation numbers will be revisited when we take up Oxidation/Reduction chemistry later.

#### Formal Charge

For the purposes of discussing covalent bonding and stability of molecules, we will do our accounting assuming that all bonds are 100% covalent, i.e., all bonding pairs of electrons are shared equally between the bonded atoms. Thus, in the molecule H-Cl, one electron from the bond is counted as belonging to the H and the other to the Cl.

 Half of the electrons in the bonding pairs go to one atom and the other half go to the other atom.

Recall that these structures ignore the core electrons. Hence, in our calculations of charge (either formal charge or oxidation number) we must do the same. So, to calculate the formal charge of the H and the Cl in HCl, we must first determine the core charge and then subtract from that the number of electrons in the outer shell (after the division).

More precisely,

Formal Charge = Core Charge - # non-bonded electrons - 1/2 # bonded electrons.

For hydrogen, the core charge is the atomic number Z = 1 since there are no inner electrons.   Recall that the Core charge is the column number of the periodic table (discounting the transition metals).

Hence,

Formal charge of H = 1 - 0 (non-bonded electrons) - 1/2 ×2 (bonded electrons) = 0

Similarly,

Core Charge of Cl = 17 - 10 = 7

Formal charge of Cl = 7 - 6(non-bonded electrons) - 1/2 ×2 (bonded electrons) = 0

#### The shortcut:

Most experienced chemists do not use the preceding equations.  They use their knowledge of the periodic table and experience to develop these numbers without calculating.  Here's how they do it.

1. The formal charge of any atom in it's normal valence state (or hypervalence state) is always zero.  So, if you were able to recognise that the Cl and the H were both in their normal valence states then you could have quickly stated that their formal charges were both zero.
2. If the element is not in its normal valence state (and not in a hypervalence state) then by comparison with neighbouring atom valence states, you can assign a formal charge.

Look at NH4+.  The nitrogen in this molecule (we've seen the Lewis structure above) has a valence state that seems to look like C.  It has a valence of 4 with no lone pairs.  Since C is one to the left of N in the periodic table, we assign a formal charge of +1 to the N since C has one less electron than N.  We verify that this structure is correct since the sum of all the formal charges must add up to the charge on the species.

When we draw the Lewis structure, we may indicate that the N has the formal charge of +1 by a plus sign in a circle next to the N.

Note that the number of bonds formed by an atom is related to its formal charge.

For example, the normal valence (number of bonds) of N is 3 but with a formal charge of +1, it can actually form 4 bonds.

You could think of it as

Now try OH-.  Again, we note that there is a charge of -1 on the species.  When we join these two atoms together, we can only have one bond between the O and the H since H has a valence of 1.   Now O looks like F which has a valence of 1 with 3 lone pairs.  F is one atom to the right so we assign a formal charge of -1 to the Oxygen since F has one more electron than O.

We know that we're correct since the sum of the formal charges equals the total charge.

### Using Formal charges in Lewis structures

For the our purposes, we will use formal charges to help us determine the best Lewis structure of covalent compounds (and covalently bonded polyatomic ions).  It is important to keep the formal charges as low as possible.  Real molecules are not very stable if they have high localized charges.  Our modeling should reflect this.  If you use a structure that gives you a formal charge of more than +1 then you have the wrong structure.

Let's look at SO42- as an example.  Lets try out several different models and eliminate all but the correct one using our rules as we know them so far.

In this first attempt we've met the normal valence of all the oxygens so they all have a formal charge of 0.  The S looks like one of the configurations of noble gas (valence of 8 with no lone pairs).  Since the noble gases are two columns to the right, that gives a formal charge of -2 on the S.  The sum of the formal charges adds up to the overall charge but there's a high charge on S which would not be stable.

In this second model, we can see that there is only 1 bond between the S and each O.  This makes the S look like a column 4 element (two columns to the left) and we assign a +2 formal charge and it makes the O atoms all look like F so we put 3 lone pairs and a formal charge of -1.  The sum of the formal charges adds up to the overall charge so we've put the correct number of electrons but there's a high charge on the S.

If we move to satisfy the normal valence of two of the oxygens and the S by making double bonds to two of the O atoms we get the structure shown on the right.  This time, two of the oxygens and the S are all in a normal (or hyper-) valence state so have a formal charge of zero.  Two of the oxygens still look like F and so have a formal charge of -1. The sum of the formal charges adds up to -2 which is the overall charge.  Thus, this structure works and it has the lowest formal charges of the three models we've tried here.

A good final check is to count the electrons in your structure and see if they add up to the sum of the valence electrons and the charge.

Count the electrons:

S:  6
4O: 24
Charge  2
total  32

Sulfate has 32 valence electrons that are paired up in 16 pairs.  Count the bars in our diagram representing pairs of electrons (either bonding or non-bonding) and we see 16 electron pairs are indeed drawn in on the diagram.  We have done a good Lewis-Dot structure.

The one problem left is that this model seems to make two of the oxygens different from the other two.  Experiment doesn't agree with this prediction.  We'll see later how we modify the model to accommodate this.

We'll see later an even greater reason for the stability of this molecule.

### Summary of the rules

So far, we've been exposed to the rules of creating Lewis structures one at a time.  Let's summarize here the rules we've learned so far.

1. Draw the structure of the species trying first to satisfy the normal or hypervalences of the elements.
2. If you must use a valence that is not a normal (or hyper-) valence for that particular atom then assign formal charges to the atoms based on comparison with the valence states of atoms from neighbouring columns in the periodic table.
3. If the formal charge on any one atom has a magnitude greater than 1 then try to rearrange pairs of electrons to lower the formal charges, keeping in mind the valence states as in step 2.
4. Sum up the formal charges.  They must equal the overall charge on the species.  If not, your structure is wrong, start again.
5. Assign lone pairs as a needed to make the atoms look like one of the images in the table of normal and hypervalences.  Thus, an atom with 3 bonds needs looks like N and so needs 1 lone pair (unless it's in group 3), etc.
6. If you can create more than one structure, the one with the smallest formal charges is the best one.

With practice, these rules become simply intuitive and you will see that eventually you will be able to quickly draw Lewis structures without reference to them any further.

Try using these rules to create Lewis structures for the following: CO32-, O2, AlCl3, IF3, XeF6. (Check here for the answers)

As one last example, let's look at Carbon monoxide and work its Lewis Structure out using the rules.

Since we have only two atoms, they must be bonded to each other and (with filled octets) they must be configured the same.

___________________________

If we try to make the atoms look like C we would come up with the incorrect structure shown here.

Here, the C looks OK, it has 4 bonds and no lone pairs, just like the table of valences describes so it has a formal charge of zero.  The Oxygen, of course, also has 4 bonds on it,  just like C which is two columns to the left so O has a formal charge of +2.  The overall +2 charge says that we have 2 too few electrons in this model of what should be neutral carbon monoxide.

___________________________

Next, we try to satisfy the normal valence state of O.  Again, an incorrect result:

This time, The O is neutral but the C has a -2 charge on it.  This model has 2 too many electrons on it, since carbon monoxide is a neutral molecule.  We can write off this picture too.

___________________________

So,  We were unable to come up with a model where the normal valence states of either atoms is satisfied.  Perhaps, if we "split the difference"...

This is the only Lewis structure that has a sum of formal charges that add up to zero, the charge on the molecule.  This one looks strange but is the only correct Lewis dot structure.

However, the Lewis-dot model does not properly predict atomic charges here. In this final model, we see that both the C and the O have a charge and that, in fact, the charges seem opposite of what we might have expected considering that O is very electronegative, compared to C.  In all likelihood, the O will pull the three bonding pairs closer to it, making an unevenly shared triple bond with most of the negative charge closer to the O.  That will bring the actual atomic charge closer to zero for each atom.

Remember, the Formal charge is not necessarily the real charge.  It's just the results of a very simple model for keeping track of electrons.

### Special cases

There are some few exceptions to the rules which may cause us trouble.  These are the radicals, In radicals, there is an odd number of electrons so no matter how many models we try where all the electrons are paired up, we'll never be able to make the model match the real molecule.  One such examples of this is NO2.  If we add up the electrons on this species, there are 5 from the N and 6 each from the two Os for a total of 17 electrons.  An odd number.  Now, pretend we didn't do that.  We don't know that we cannot use the normal rules to make a Lewis structure that will work.  Let's see what trouble we might get into.

 1 In this first model, we've satisfied the normal valence of one O and the N but one of the O atoms looks like F and we must give it a charge of -1.  Obviously wrong since the NO2 molecule is neutral. 2 Here, we see a second attempt.  This time, we find that the oxygens are both in their normal valence state and that there's a + charge on the N.  This doesn't work either. 3 Finally, we've given up trying to use pairs of electrons. We note that model 1 seems to have 1 electron too many.  By removing one electron from the N or from the O of model 1, we arrive at a structure whose formal charge adds up to the overall charge of zero. Now the question is: Which of these is the correct one.  You can't tell. This is the point where our rules for Lewis structures break down and we realize we need something better.  The latter of these has the lower formal charges so we might suspect that it is the favoured one.  However, EPR experimental data puts the unpaired electron almost exclusively on the N as shown in the former of the two.

#### Resonance

In some cases, there are more than one equivalent Lewis structures. In these cases, we must consider whether the multiple resonance structures are all different structures, representing chemically different molecules or whether there is some dynamic process going on that can allow the molecule to resonate back and forth amongst several electronic configurations. In our sulphate ion case, there are six identical structures and they an interchange one for the other as indicated below.

Since there are six structures which interchange on a rapid basis, we need to rethink our ideas of formal charges and of Bond order.

In this case, we can define average formal charge for any one atom to be:

Hence, for the top oxygen (pretend we have nailed down the ion so it can't rotate. That way we can keep track of the motion of the electron within the ion.) going clockwise, starting with the top-left structure we get:

The Avg. FC for the Sulphur, is obviously zero since it is so in all the structures.

We can also define an average bond order (BO) for any pair of atoms as:

Thus, for the pair SO (top) we can calculate as before

We can now draw a new type of structure using the average bond order and the average charges. This average model better represents the reality of the situation in the real sulphate ion.

The calculated average bond order of 1 1/2 implies that there is always 1 bond joining the SO pair and  half the time, there is a second bond joining that same pair (according to this resonance structures model).  Hence, the solid and the dotted lines joining the pair in the final "average" diagram.

#### Electron-deficient resonant structures.

In the previous example as in most resonance structures you will encounter, the average bond order is greater than one for the bonds where the resonance is actually occurring.  That is because, typically, one bond remains in place throughout the resonance process and a second (or third) bond moves from place to place in the different resonant structures.

In certain special situations, where there are insufficient electrons for normal bonding to occur, resonant structures can be formed between atoms of a species to form a bond order that is actually less than one.  In these cases, there must be at least one of the series of resonance structures where the bond is completely missing.  Obviously, these compounds would not be as stable as those we normally think of as having resonance but they do exist.

Take the molecule B2H6.  It does exist but if we try to draw a normal Lewis-dot structure, we cannot find a way to bond all the atoms together at one time.  Using resonance structures we show how the bonds can move in such a way as to create a bond for at least part of the time.

In this resonance model, we see that the B?H bond order to the central hydrogens is actually only 1/2, i.e., the bond exists only half the time in this picture.  An average picture that shows the entire structure as being connected might be

Here, we see that the central hydrogens are bridging between two borons in a curved, electron-deficient 3-centre bond sometimes called a banana bond.  This three-centre bond has only two electrons in it and hence the bond order of the individual B?H portions of this have a bond order of only 1/2.

Molecules with this kind of bonding are stable mostly in boron compounds although there are examples of other three-centre electron deficient bonds.  It is possible to create valid Lewis structure models of such bonding in any analogous compound made up of column 3 elements like Al, Ga, In, Tl.  We can draw the analogous structures but for other reasons, beyond the scope of this discussion, actual chemical analogues such as Al2H6 or Ga2H6 are not as stable and if created, some of these would quickly reform into other, more stable compounds.  We can find other examples of electron-deficient three-centre bonds that are stable and analogous to the B2H4molecule, for example, Al2Cl4 exists, is stable and used electron-deficient three-centre bonds that are quite analogous to those shown above.

#### Lewis Bases and Lewis Acids

We can now use the ideas developed here to describe a process where a bond is formed when a species with an extra electron pair forms a covalent bond with a species which is deficient in an electron pair.

This obvious acid-base reaction is exactly analogous to the reaction between ammonia and the hydrogen cation.

In this reaction, still would refer to the ammonia as a base and the hydrogen cation as the acid. but how about the following

In this case, the F- ion has the extra electron pair and uses it to bond with the BF3 in which the B is missing one electron pair to complete the octet. The Fluoride ion donates an electron pair to the Boron. The Fluoride acts just like the ammonia or the hydroxide ion did in the previous two examples where they clearly the base. Thus, the electron-pair donor is a Lewis Base. The electron pair Acceptor is the Lewis Acid (NOTE there is a typo in the book p. 240)

#### Hydrogen Bonding

One last type of bond can be explained now using the Lewis dot structures.  It is very similar in concept to the Lewis-acid/Lewis-base bond formation. In this case, however, a bond is not formed, merely a strong attraction.

This type of bond can only form between a hydrogen (which is bonded to a very small high-electronegative atom like N, O, or F) and some other electronegative atom (N, O, F) in which there are non-bonded electron pairs available.

In water, for example, the oxygen is bonded to two hydrogens but has two non-bonded electron pairs available to it. Hence, the H from one molecule may be able to join to the lone-pair of electrons from the O of a different molecule.

Hence, the molecules of water are attracted to each other strongly because of the phenomenon called Hydrogen bonding.

This explains the unusually high boiling point of water when compared to other molecules of similar molar mass such as CH4 which is a gas well below room temperature. It also explains the fact that ice floats. When crystals of ice form, the hydrogen bonding forces the molecules into a regular arrangement which is in fact spaced out further than when the molecules are in the liquid phase. Hence, ice is less dense than water and it floats.