Chapter 6 Ionic Bonds and Compounds
OK. Now we have a handle on some of the properties of individual atoms. Let’s go the
next step up and look are the properties of molecules. We will start with the general
class of compounds called ionic compounds which are made from a metal combined
with a non-metal
6-1 The ionic bond and electrostatic forces
Ionic compounds are held together by ionic bonds
Ionic bonds have some very interesting, almost contradictory properties
Ionic solids are very strong. As seen by very high melting and boiling points
MP H2O = 0 oC
BP 100
MP NaCl = 800
BP 1413
Yet for all their strength, many ionic solids readily fall apart in water and break up
into aqueous ions: NaCl(s) 6Na+(aq) + Cl-(aq)
When this happens the water which is a non conductor, now starts to conduct
electricity
In contrast covalent compounds don’t so this
(DEMONSTRATION)
The key lies in the fact that ionic compounds are made of ions (DUH)
and it is the ions that carry the current in water. We will talk about the
interactions of ions in water in a later chapter. For now let’s focus on the ions
themselves
What is the electron configuration of Na+ and Cl-?
[Ne] and [Ar]
Clicker question 1; What are the electron configurations of Mg2+ and O2Clicker question 2: predict the ions that will come from elements achieving their noble
gas configurations
So ions are in stable noble gas configuration
Figure 6.3- Note you should memorize the charges of all the ions in this table!
But are also charged
It is the +/- charge interaction (electrostatic force) that hold molecule together
Key Terms:
The positively charged ions formed by metals are called cations
The negatively charged ions formed by non-metals are called anions
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Ionic Solids are so strong because aren’t really composed of molecules, but are
composed of alternating ions, and it is the sum of all these interactions that
makes the crystals so strong. (We will talk about this in more depth later)
Covalent solids, on the other hand are individual molecules held together by
much weaker forces (more later)
6-2 Ionic Charge and Chemical Formulas
Knowing the charges of the common ions we just examined, you should be able
to come up with the formula of any ionic compound made with these ions (Don’t
do names yet)
Key concept:
The net charge in any ionic compound is zero, so the sum of the charge of the
cations and the sum of the charge of the anions must cancel each other out
Examples:
Sodium Chloride
Na & Cl
Na+ and Cl1(+1) + 1(-1) =0
NaCl
Sodium oxide
Na & O
Na+ & O22(+1) + 1(-2) =0
Na2O
Scandium sulfide
Sc & S
Sc3+ & S22(+3) + 3(-2) = 0 Note that cross multiplying works nicely!
Sc2S3
Clicker questions: Rubidium chloride, Cesium nitride , Aluminum oxide
6-3 Ionic Charges in Transition Metals
Silver is a transition metal that is commonly found in a +1 ionic state, Ag+
Why?
While you might thing the electron configuration of Ag is [Kr]5s24d9
Aq is just like Cu above it in the periodic table and has the configuration
[Kr]5s14d10
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How would you get to an inert gas configuration? Lose 11 electrons or gain 7
electrons. Neither sounds very appealing
But if you lost 1 electron you would at least get to [Kr]4d10 and that configuration
would fill in the 4 shell (4s24p64d10) to achieve some stability
This explains the ions found in Figure 6.5
The block of atoms by Indium has another way to go.
The electron configuration of Indium is [Kr]5s24d105p1
By the logic we just looked at it can become a +3 ion by going to the
[Kr]4d10 configuration
But it can go to the +1 state by going to the [Kr]5s24d10 as well
So we have another block of common ions in this area
Figure 6.6
Note that In, Sc, Tl & Pb overlap with previous figure so there are multiple
ionic states for these element
This is part of a more general problem. Most of the transition metals can have 2
or more ionic states. How do you properly name and identify ionic compounds
when you have metal cations that can have different charged state?
6-4 Ionic Charges in Transition metals with >1 common charge
The answer to the above question is to use roman numerals to cearly identfy the
charge state you want.
Key Concept:
Metals with 1 common charge state so you DON”T use roman numerals
Group 1 (+1) Group 2 (+1) Al (+3) Zn(+2) Ag(+1)
All other metals need a roman numeral included in their name to identify the
charge on the ion
Examples
Give the name of CuCl2 (Copper(II) choride)
Give the formula of Cobalt(III) sulfide (Cu2S3)
Clicker questions: Names and formulas of several ionic compounds with and without
roman numerals
Note: ths book goes in to some older nomenclatures that I will skip!
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6-5 Filling order of Transition Metals
I think I will skip this section
6-6 Cations are smaller and Anions larger than parent atoms
Key concept:
Cations smaller than parent atom
Anions larger than parent atom
If you think about it this makes sense
What do you do to make a cation. You remove and electron from the outermost,
largest shell, so this naturally makes the cation smaller.
Not only that but you frequently go to a noble gas configuration, but the charge in
the nucleus is more positive than that of the noble gas you are mimicing, so this
pulls the electrons in even more tightly
To make an anion you add an electron to the outermost shell which will make the
ion larger than the original atom. Not only that, but all the electrons are repelled
from each other, to that makes the anion ‘swell’ up even larger.
This is shown graphically in figure 6.8
Practice problem
Sort the following atoms and ions from smallest to largest
Li+, N3-, Ne, Na+, S2-, Cl-, Ar
Clicker question Sorting atoms and ion by size
6-7 Coulomb’s Law and Ion Pair energy
Early in the chapter I said that ionic bonds were strong bonds, and that is why
ionic compounds have high melting points and boiling points
Let’s actually back that up with a calculation or two and get some real energies
to look at.
Consider the reaction:
Na(g) + Cl(g) 6Na+Cl- (g)
If this reaction releases energy, then the product NaCl (g) is stable because you
have to put that energy back into NaCl(g) to get it to change back into the
separate atoms of Na(g) + Cl(g)
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We can actually calculate the energy of this reaction by breaking it into individual
steps:
1.) Na(g) 6Na+ (g) + e2.) Cl(g) + e- 6Cl-(g)
3.) Na+(g) + Cl-(g) 6NaCl(g)
Reaction 1:
You should recognize the first reaction as the ionization energy. So we
can simply look this energy up. It is 0.824aJ
Reaction 2:
The second reaction is a new one. It is called the electron affinity. It is a
measure of how much energy is gained or lost when an atom grabs an
electron
Key Concept:
Electron affinity: Y(g) + e- 6Y-(g)
For Chlorine
Cl(g) + e- 6Cl-(g)
-0.58 0 aJ
The negative sign says that this process releases energy, thus Cl-(g) is
more stable than Cl(g) !
Just as there were 1st, 2nd, 3rd ... ionization energies, there are 1st ,2nd ,
3rd ...electron affinities.
Reaction 3:
Early in the chapter I said that ionic compounds were held together by
charge-charge - electrostatic interactions. Does anybody know the
equation for this kind of interaction?
It is called Coulomb’s law
Key Equation:
In this case Q1 and Q2 are the charges of the two ions, measured in units of the
charge of a proton, so Q1 for Na+ = +1, and Q2 for Cl- = -1
d is the distance between the centers of the ions in pm. In this case the radius of
Na+ is 102 pm and Cl- is 181 pm (See table 6.4 page 183) so d = 102+181 = 283
pm
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and k is a proportionality constant = 231 aJ@pm
So our energy is:
231 aJ@pm x (+1)(-1)/281 pm
= -.816 aJ
The - sign says that energy is released, so the ions of opposite charges are
attracted to each other (DUH!)
So let’s sum things up
RXN 1:
+0.824 aJ
RXN 2:
-0.580 aJ
RXN 3:
-0.816 aJ
Net:
-0.572 aJ
Negative sign, energy released, NaCl(g) is more stable than Na(g) + Cl(g)
Key calculation:
Given ionization energies, electron affinities and ionic radii be able to calculate
the energy X(g) + Y(g) 6XY(g). And any variation
At this point our story isn’t complete, because ionic compound are so stable, if is
very difficult to find them in the gas phase, they always occur as solids at room
temperature and pressure.
In an ionic solid we have the + and - ions in an alternating lattice, as shown in
figure 6.13
Look at the Na+ ion in the middle of this lattice. See how it is actually touching 6
nearest neighbor Cl- ions. That mean we have underestimated our energy in the
solid.
Key concept:
Lattice energy = E gained or lost in reaction
X(g) + Y(g) 6XY(s)
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The actual lattice energy for the NaCl solid is -1.28 aJ, more than twice our
simple estimate.
Calculating the true lattice energy that properly accounts for all neighbors around
each ion, and also includes the fact that the ions are not standing still but are
vibrating is beyond what you need in this course. It can be done, and the
calculation and the experimental number are in very good agreement.
(See table 6.6 in your text if you are interested)