Chapter 6:  Ionic Bonds

In this chapter we look at the dynamics of the formation of ionic bonds and look at some of the mechanics of same in detail.  The latter half of the chapter deals with some descriptive chemistry which, for now, is being culled from the course.  Heaven forbid that we actually teach any chemistry in this course (note sarcasm here...)!  The descriptive chemistry is pHun and some of the things we do with it will be discussed very briefly; but none of this really pHun stuff will be appearing on an exam.

So, we shall look at ions (cations and anions) and other things such as ionization energies, electron affinities, and lattice energies--along with other pHun stuff.

Ions and Their Electron Configurations:

Let us first look at the so-called representative elements, Groups 1A-8A.  How do they usually form ions?  The answer is realtively simple:
A-type elements are "happiest" when they can achieve an inert gas configuration.
For instance, Na "likes" to lose a single electron to achieve the same electron configuration as Neon.  That is,

On the other hand, F "likes" to gain an electron to achieve the same electron configuration as Neon.  In this case,

The main point here is that inert gas configurations are particularly stable!

In general, group 1A atoms lose one electron to get down to the previous inert gas.  That is,

Li loses one electron to
gain the [He] configuration.
Na loses one electron to
gain the [Ne] configuration.
K loses one electron to
gain the [Ar] configuration.
Rb loses one electron to
gain the [Kr] configuration.
Cs loses one electron to
gain the [Xe] configuration.
Fr loses one electron to
gain the [Rn] configuration.

We can say the same thing about group 2A metals.  That is,

Be loses two electrons to
gain the [He] configuration.
Mg loses two electrons to
gain the [Ne] configuration.
Ca loses two electrons to
gain the [Ar] configuration.
Sr loses two electrons to
gain the [Kr] configuration.
Ba loses two electrons to
gain the [Xe] configuration.
Ra loses two electrons to
gain the [Rn] configuration.

Similarly, with group 3A metals, we can start by noting that Al loses three electrons to gain the [Ne] configuration.  (Later down in the periodic table, things get a little more complicated, but this is a good start to our discussion!)

Much of what is said here is stated in what is sometimes called the octet rule.  That is, an inert gas configuration is of the form, ns2 nsp6. Such configurations are sometime written like this:

We shall discuss this later on when we get to Lewis structures.  For now, we note that type-A elements have this octet structure (except for H which obeys a "duet rule" which we write simply as H: for the hydride ion).

Things become more complicated (and, hence, more "pHun") when we look at atoms which have d-orbitals (either occupied or empty).  For instance, iron can lose either two or three electrons to form the follow two stable ions:

We might ask why these two.  Obviously, the two 4s electrons are in the outer shell and should go first.  But, then, why just one more d-electron?  The answer here is that we are left with a 3d5 configuration.  As stated in the previous chapter, a half-filled subshell is particularly stable!  That is, we have the following configuration redrawn to accentuate Hund's rule:


Having 5 unpaired electrons is a peculiarly stable situation!  Hence, this is what you see and get!

As we shall see shortly, ionization energies are positive (that is, it takes energy to remove the electron).  However, that is offset by the energy gained back from other things such as lattice energies.  We shall look at these ideas in much more detail shortly.

Ionic Radii:

We have seen how ions form.  Either electrons are lost to form cations or electrons are gained to form anions.  How big are these anions.  For a given atom, it is a good rule that cations are smaller than the neutral atom and anions are bigger.  With cations, the effective nuclear charge, Zeff , is made larger by having fewer electrons between an outer electron and the nucleus.  Thus, cations are most definitely smaller than the neutral atoms of the same element.  We show this in the following picture:

These are for +1 cations.  Look now at what happens with +2 cations:

Ionization Energy:
The ionization energy (aka "Ionization Potential") is the energy needed to remove an electron from an atom (which can be a neutral atom or ion in its own right).  For instance, consider the following process:

(Note that we are in the gas phase here--this is usual for the sort of "theoretical stuff" we are handling at the moment.)

We make note of one very important fact:

Even though a stable compound might form from the ionization (and this is certainly obvious for something like Na which is never found in its free state in nature) it still takes energy to remove the electron.  We note the following trends in any event: The energy needed to remove one or more electrons to make a stable compound is usually compenstated for by the lattice energy in a crystal or by the solvation energy of ions.  These topics will be discussed later.

Ionization energy follows periodic trends, as do most properties of elements.  For instance, here are plotted the ionization energies of the first 92 element (H - U):

Note here that it is quite easy to remove the first electron from elements such as Li, Na, K, Rb, Cs, and Fr (as conveniently noted on the diagram above).  It is also easy to get the first (and also a second electon) off of the elements Be, Mg, Ca, Sr, Ba, and Ra; these are all representative metals of groups 1A and 2A and this would be expected (getting the second electrons off of 2A metals is not shown in the above graph).

On the other hand, taking electrons away from noble gases is a NONO!  These are extremely stable configurations and removing electrons is not profitable from a chemical standpoint.  Much the same can also be said for attempts to remove electrons from halogens (group 7A) or group 6A elements (oxygen and the chalcogens, S, Se, and Te--no one in interested in what Po thinks about this...).

Ionization energies tend to increase as one goes from left to right in the periodic table and to decrease as one goes down.  The above graph gave some indication of this but the next one is even better:

There are occasional "bumps" in the observed ionization energies.  These usually occur because we are going from s- to p- or d-electrons in our configurations.  For instance, p- and d-electrons are more shielded than are s-electrons--so some little "burps" might be expected.  This is clearly shown for the first 20 elements (up through Ca); here just s- vs. p-effects are seen.  To see higher effects you need to go back to the graph of the first 92 elements (and pull out your magnifying glass)!  Anyway, here are the first 20 elements in all their "glory" as regards first ionization energies:

Higher ionization energies are well known for many elements.  In some cases, stable ionic compounds are formed; in other cases, these ionization energies are derived from spectroscopic methods and are of some interest from a theoretical viewpoint.  Luckily, we need not worry too much about them other than to give them in the following table.

Note the tremendous changes in energies when going from the blue to the red/orange part of the table.  This dividing line is very important for metals such as Mg and Al; even for higher elements, this large change is noted--however, elements such as Si, P, S, and Cl form covalent compounds primarily.  Ar couldn't care less about any of this!
Electron Affinity:
This is a complementary property to the ionization energy.  This is the energy given off when an atom gets another electron.  This is negative because the energy is released to the environment.  (We shall talk about these things later when we get into thermodynamics.)  Electrons affinities (EA's) are usually plotted upside down, however, so that--at least the plots--we get a "positive appearance."  Periodic trends for these parallel those of ionization energies, as is shown in the following graph.

Note that many of these approach zero since, in some cases, an atom simply does not "need" another electron.  However, as you can see, group 6A and 7A elements are very "happy" to take on electrons to form anions.

We note one special point:

The electron affinity of a given atom is just the ionization energy of its complementary species.  For instance, the electron affinity of chlorine is just the ionization energy of the chloride ion.  We'll let you think about this.
Ionic Bonds and the Formation of Ionic Solids--The Born-Haber Cycle:
Many metals react (sometimes violently) with nonmetals, the products usually being salts (except for strong bases such as oxides).  This process can be summed up into a series of steps which is sometimes called the Born-Haber cycle.  As we shall see later and note now, in passing, this is a corollary of the first law of thermodynamics.

As an example, we look at the formation of sodium chloride via the reaction

The energy of this reaction is easily measured directly and it gives off lot of heat (the word here is "exothermic"), which is often an indication that a stable compound is being formed.  Indeed NaCl is stable--as you should well know--whereas Na and Cl2 are not found in nature in the free state.  However, we know that it takes energy to remove an electron from the Na atom and to break the Cl-Cl bond in chlorine gas.  Where does the energy come from?

The various steps which can be delineated for this reaction are shown in the next figure.

We look at the various steps.  The first three consume energy when is then more than payed back in the last two steps.
Step 1:
The Na is converted from the solid to gaseous form.  107.3 kJ/mol are consumed in this process, called sublimation.
Step 2:
The Cl-Cl bond is broken to form gaseous Cl atoms.  The energy consumed here is 122 kJ/mol (per mole of Cl atoms produced).  Note that the energy for this process is harder to measure, as evidenced by having one less significant figure in this number!
Step 3:
The gaseous Na atoms are ionized at the cost of 495.8 kJ/mol (this ioization energy).
Step 4:
Finally, we start getting energy back.  The gaseous chlorine atoms become gaseous chloride ions and receive 348.6 kJ/mol of energy (with is the electron affinity).
Step 5:
The gaseous ions then condense to form the solid (actually, this process is called deposition, the opposite of sublimation).  This is rather large chunk of energy, namely 787 kJ/mol and is called the lattice energy.

Subtracting the energies of steps 4 and 5 from the sum of steps 1, 2, and 3 gives the total reaction energy of -411 kJ/mol of NaCl produced.

Please note that the energies of steps 4 and 5 are formally defined as negative in the process since, with each step, the system is losing energy.  As we shall discuss later in chapter 8, the energy change is positive for endothermic processes (the system gains heat energy) and negative for exothermic processes (the system loses hear energy).

Lattice energies may be caculated rather exactly by applying Coulomb's law to the crystal lattice (details on how this is done are left to more advanced courses).  Some typical lattice energies are tabulated below.

In the cases cited above, the calculated lattice energy is designated as -U (negatively defined so that we don't have to put minus signs in front of all the numbers in the table) and the formula is

where d is the distance of closest approach of a cation and anion, the z's are the ionic charges, and k is a constant for the type crystal lattice possessed by the salt.  (This is sometimes called a Madelung constant, which we now drop like a hot potato!)

Assigned Problems:

Problem 6.28 (p. 238):
Which of the following drawing is more likely to represent an ionic compound and which a covalent compound?

(a) is clearly a crystal lattice with ionic bonds.  The even packing of this crystal particle-by-particle should be a dead giveaway.  Clearly, then, (b) represents a covalent compound since each atom is clearly paired with just one of another type (and, thus, we can define individual molecules--something which cannot be done unambiguously with ionic compounds!).

Problem 6.31 (page 238):

Which of the following spheres is likely to represent a metal and which a nonmetal?  Which sphere in the product represents a cation and which an anion?

Here we reason backwards.  Clearly, the reactants are both the same size and it is impossible to tell which is a metal and which a nonmetal.  However, we do know that anions are larger than the atoms the come from and cations are smaller.  Hence, the metal atom is the blue one and the green atom is the nonmetal.

Problem 6.40 (p. 239):

Do ionization energies have a positive or negative sign?  Explain.

All ionization energies, as defined, must be positive.  In all such cases energy is required to remove the electron!  Hence, the system gains energy and the number must be positive.  This actually makes sense here but will be even clearer when we get to chapter 8!

Problem 6.42 (p. 239):
Which group of elements in the periodic table has the largest Ei1 and which has the smallest?  Explain.

We again look at the wondrous figure imported from the textbook:

Looking at this gives the answer immediately--although the figure should not be necessary if you understand the explanation.  We give the explanation now: So, the answer is simple and, in any event, the picture is certainly pretty!
Problem 6.44 (p. 239):
(a)  Which has the smaller second ionization energy, K or Ca?
Obviously, Ca.  K+ has the [Ar] inert gas configuration.  Ca would "like" like that, too, and is willing to lose another electron readily to get there.  However,  K+, has already been "fulfilled" and will hold on jealously to its remaining electrons.
(b)  Which has the larger third ionization energy, Ga or Ca?
Ca is also the "winner" here.  There is no way it would willingly surrender a third electron to lose the happiness of being "[Ar]."  However, Ga can attain "[Ar]-ness" by losing three electrons.  This, it is quite willing to do.  (And, it gets all that lattice energy, too!!!)

For both these problems, recall the following chart:

Luckily, we won't need to talk about this any more!  (Unless you enjoy stripping electrons off of poor, defenseless atoms!)
Problem 6.50 (p. 239):
What is the relationship between the electron affinity of a monocation such as Na+ and the ionization energy of the neutral atom?

These are actually the same number in magnitude.  That is, the EA is just the reverse of forming the cation.  In other words, the EA of the cation is simply the restoring of the ion to the neutral atom.  We can show this with the following simple diagram:

What we should note here is that IE (ionization energy) and EA (electron affinity) are simply complementary processes.  The language is set in terms of neutral atoms but, overall, the actual processes are, basically, just the changing of the number of electrons around the nucleus.

Problem 6.58 (p. 240):
Order the following compounds according to their expected lattice energies:  LiCl, KCl, KBr, MgCl2.

Here, we note that forces are greatest with the least distance between particles and the largest charges.  Also, lattice energies are negative but we give them in the order of their magnitudes.  This is the usual practice.

Given all that discussion, the ordering is just

MgCl2  >  LiCl  >  KCl  >  KBr.

Magnesium chloride is first because of the +2 charge on metal cation.  Then, we have anions which are the same or bigger than chloride.  So, we note that the smaller Li cation would have a greater lattice energy than KCl and that, finally, KCl > KBr because of the greater size of bromide.

Problem 6.62 (p. 240):
Find the lattice energy of LiBr in Table 6.3 and calculate the energy change (in kJ/mole) for the formation of solid LiBr from the elements.  (The sublimation energy of Li is +159.4 kJ/mol, the bond dissociation energy of Br2 is +224 kJ/mol, and the energy necessary to convert Br2(l) to Br2(g) is 30.9 kJ/mol.

This is much like the example given earlier except that there is an extra step:  the evaporation of liquid bromine.  We write the various steps and then sum to get the energy change for the net reaction:

There are also some other numbers you can get from the book.  First, the IE of Li is +520.2 (p. 216) and the EA for Br is -325 (p.226).  Both these numbers are in kJ/mol.  Now we add these value for the formation of solid LiBr.  The numbers below are added or subtracted according to what you would put in the Born-Haber cycle:

Energy (kJ/mol)
Sublimation of Li
Dissociation of Br2
(per Br atom produced)
Vaporization of Br2
Ionization energy, Li
Electron affinity, Br
Lattice energy, LiBr
Total Reaction Energy:

Note that the answer in the back of the book is wrong (I think that they meant for you to get the EA given the total reaction energy elsewhere--this is the way that many EA's are obtained!)  It is also possible that a couple of the numbers were obtained off of graphs and that they forgot that they were tabulated!  In any event, this is the right answer!