Topic 4: Bonding
4.1 Ionic Bond
4.1.1. Describe the ionic bond as the result of the electron transfer leading to attraction
between oppositely charged ions.
An ion is an atom or group of atoms that has a net positive or negative charge. The best
known ionic compound is common table salt, or sodium chloride, which forms when
neutral chlorine and sodium react. With one electron stripped off, the sodium, with its 11
protons and only 10 electrons, not has a net 1+ charge – it has become a positive ion. A
positive ion is called a cation. The sodium ion is written as Na+ . If an electron is added
to chlorine, the 18 electrons produce a net 1- charge; the chlorine has become an ion with
a negative charge- an anion. The chloride ion is written as Cl-. Because anions and
cations have opposite charges, they attract each other. This force of attraction between
oppositely charged ions is called ionic bonding. Sodium metal and chlorine gas react to
form solid sodium chloride, which contains many Na+ and Cl- ions packed together. A
solid consisting of oppositely charged ions is called an ionic solid, or a salt. Just to
reemphasize because the objective wants you too, there is a transfer of an electron from
the sodium to the chloride.
4.1.2. Determine which ions will be formed when metals in groups 1, 2, and 3 loose
electrons.
In order to understand which ion would be formed, you have to understand the octet rule.
I talked about it earlier, but basically it states that every atom will normally try to gain 8
electrons in its highest energy level, and it will either gain or lose electrons to satisfy this
desire. So for the metals in group 1, they all have one valence electron. In order to reach
8 electrons in their outer energy level, the easiest way for them to do it would be to lose
that valence electron. If they do that, they have the electron arrangement of the noble gas
at the end of the previous period, which is where they want to be (noble gases all satisfy
the octet…under almost all circumstances.) So, all the elements in table 1 will have a 1+
charge in their ionic state. The alkaline earth metals (group 2) all have two valence
electrons. In their effort to satisfy the octet, the easiest thing for them to do is to lose two
electrons, and they’re at the same place as the noble gas again. So their ionic state is 2+.
And then the elements in the third group all have three electrons, and once again it’s a lot
easier to lose three electrons then to gain 5, so they all will lose their three electrons and
be back at the noble gas electron configuration, and will have a ionic charge of 3+. Note:
I am fighting myself to be sure to write the ionic charges right, because I naturally want
to write them +1, +2, +3. However, I strongly recommend you do not get into this bad
habit, because for right now you may not get in trouble, but later on 1+ and +1 can mean
very different things (oxidation states as opposed to ionic charges), so train yourself now
and save yourself the hassle!
4.1.3. Determine which ions will be formed when elements in groups 6 and 7 gain
electrons.
Once again, following the octet rule as stated in the previous objective, the elements in
group 6 have 6 valence electrons. In order to satisfy the octet, they would have to either
lose 6 electrons or gain 2, which would fill their outermost energy level. Which one is
more energy conserving? You guessed it! They will gain two electrons, making their
ionic charge 2-. And then for the elements in row 7, well all they have to do to satisfy an
octet is gain one electron, and that’s what they do, giving them all a charge of 1-.
4.1.4. State that transition metals can form more than one ion.
Restrict examples to simple ions like Fe2+ and Fe3+
This can get really complicated in understanding why certain elements take different
ionic charges while others don’t (I’ve never completely understood it), but at this point
all you have to know is that elements that are called transition metals (the ones in the D
block, that is, starting with Sc down to Ac and over to Zn and Hg, those are the D block
elements a.k.a. the transition metals) can take different ionic charges. An example of this
is Fe2+ and Fe3+. You’ll learn why later, but know that elements in the d block can take
ionic charges while elements outside of the d block will always have the same ionic
charge. When writing names for compounds with elements in the d block, you normally
write the ionic charge of the element after the element. For example, iron oxide is really
iron (II) oxide. If it were a compound with iron with an oxidation state of 3+, it would be
written iron (III) blah blah.
4.1.5. Predict whether a compound of two elements would be mainly ionic or mainly
covalent from the position of the elements in the periodic table, or from their
electronegativity values.
Basically, nonmetals and metals form ionic bonds, and nonmetals and nonmetals form
covalent bonds (usually). In ionic compounds, the nonmetal is normally the anion and
the metal is normally the cation. Remembering that nonmetals are the ones on the right
side of the table and metals are the ones of the left side of the table, you should be able to
tell how different things will bond. So you have potassium and chlorine, potassium is a
metal and chlorine is a nonmetal, so the two will form an ionic bond. Metals bonding to
metals is not really covered here, it REALLY does not happen often (but you’ll learn
about metallic bonds later).
4.1.6. Deduce the formula and state the name of an ionic compound formed from a
group 1, 2, or 3 metal and a group 5, 6, or 7 non-metal.
The rules to remember when deriving formula and stating the name of ionic compounds
are as follows. 1. Ionic compounds are always neutrally charged. So say you have a
cation with a charge of 3+ and an anion with a charge of 2-, the two charges have to
cancel each other out, so you will have to have 2 cations and 3 anions to have an equally
opposite charges and thus a neutral charge overall. A good rule of thumb when you have
odd and even charges pairing up or two odd charges pairing up is if you have the cation
Xz+ and Ab-, then your ionic compound will always be XbAz, where the subscripts denote
how many of those atoms you have (note: this rule is not necessary if you have two even
charges). Then the rules for naming binary ionic compounds are as follows…
1. The cation is always named first and the anion second.
2. A monatomic (meaning one-atom) cation takes its name from the name of the
element. For example, Na+ is called sodium in the names of compounds
containing this ion.
3. A monatomic anion is named by taking the root of the element name and adding –
ide. Thus the Cl- ion is called chloride.
So let’s name some ions. Say you have the ion of potassium and iodine. Well,
potassium, being a group 1 element, has a ionic charge of 1+, and iodine, being a group 7
element, has an ionic charge of 1-. That means that the two paired up will equal a neutral
charge just as they are, so the formula is KI and its name is potassium iodide. Now lets
try lithium and nitrogen. Lithium, once again being a group 1 element is going to have a
1+ charge, but nitrogen is a group 5 element, so it has a 3- charge. The two simply put
together will not add up to 0. So what we do is use that handy rule of thumb we learned
earlier, the charge on lithium is 1, the charge on nitrogen is 3, so the opposite atom’s
charge becomes the original atoms subscript, thus the formula is Li3N, or lithium nitride.
Notice that you do not use a prefix just because there are three lithium atoms. Since all
elements except the d block elements have only one ionic charge, by simply stating
lithium nitride, any chemist in the world will know that there are three lithium’s because
of the fact that the overall molecule has to be neutral.
4.2 Covalent Bond
4.2.1. Describe the covalent bond as the result of electron sharing.
The electron pair is attracted by both nuclei leading to a bond which is directional in
nature. Both single and multiple bonds should be considered . Dative covalent bonds
are not required.
When two atoms bond by sharing some of their outer electrons, the atoms create a
covalent bond, forming a molecule. To create a covalent bond, two atoms share a pair of
electrons; in most cases, each atom contributes one of the shared electrons. Each atom
becomes more stable, because the covalent bond has effectively provided each atom with
one more electron in its outer shell. This type of bond, in which one pair of electrons is
shared, is called a single bond. Sometimes, two atoms share two or three pairs of
electrons with each other. These bonds are called double or triple bonds, respectively.
4.2.2. Draw the electron distribution of single and multiple bonds in molecules.
Examples should include O2, N2, CO2, C2H4 and C2H2.
I don’t feel like drawing a Lewis dot, see your book.
4.2.3. State and explain the relationship between the number of bonds, bond length and
bond strength.
The comparison should include bond lengths and bond strengths of:
 Two carbon atoms joined by single, double and triple bonds.
 The carbon atom and the two oxygen atoms in the carboxyl group of a carboxylic
acid.
General rules of thumb are that the more shared electron pairs, the bond length shortens.
So single bonds are farther apart then double, which are farther apart then the triple
bonds. As for bond strengths, usually the more shared electrons the stronger the bond, so
double and triple bonds are stronger then single bonds. So, a carbon to carbon single
bond is weaker and longer then a carbon to carbon double bond, which is weaker and
longer then a carbon to carbon triple bond. In a carboxylic acid, the carbon has a single
bond to one oxygen and a double bond to another oxygen. The length of the single bond
is 143 pm, while the length of the double bond is 123 pm. The energy in the single bond
is 305 kJ/mol, while the energy in the double bond is 615 kJ/mol. So, the double bond is
stronger and shorter then the single bond.
4.2.4. Compare the relative electronegativity values of two or more elements based on
their positions in the periodic table.
Precise values of electronegativity are not required.
The trend for electronegativity is that is increases going to the right and up. So if you
have two elements on the periodic table and you want to compare their electronegativity,
you simply see which one is further up and to the right, and that one is more
electronegative. The farther apart the two are in the periodic table the larger the
difference will be in their electronegativity. See the topic on periodicity to understand
this trend for electronegativity.
4.2.5. Identify the relative polarity of bonds based on electronegativity values.
In a covalent bond, electron distribution may not be symmetrical and the electron pair
may not be equally shared.
There is a type of bond that is between the extreme of giving off electrons that is done by
ionic bonds and the complete sharing of electrons in a covalent bond that is called a polar
covalent bond. In a polar covalent bond, one part of the bond is able to attract the
electrons to its nucleus more so then the other parts of the bond, and so the electrons are
still being shared, but are being hogged you could say by the more electronegative atom.
An example of this type of bond occurs in hydrogen fluoride (HF) molecule. When a
sample of hydrogen fluoride gas is placed in an electric field, the molecules tend to orient
themselves with the fluoride end closest to the positive pole and the hydrogen end closest
to the negative pole. This result implies that the HF molecule has the following charge
distribution:
H—F
δ+ δwhere δ (lowercase delta) is used to indicate a fractional charge. The most logical
explanation for ht development of the partial positive and negative charges on the atoms
(bond polarity) in such molecules as HF and H2O is that the electrons in the bonds are no
shared equally. For example, we can account for the polarity of the HF molecule by
assuming the fluorine atom has a stronger attraction for the shared electrons than the
hydrogen atom does. Likewise, in the H2O molecule the oxygen atom appears to attract
the shared electrons more strongly then the hydrogen atoms do. Because bond polarity
has important chemical implications, we find it useful to quantify the ability of an atom to
attract shared electrons, i.e. electronegativity.
Generally speaking, atoms with similar electronegativity values do not form polar bonds,
because the electrons are shared almost equally. However, atoms that do not have similar
electronegativity values do have polar bonds because their electronegativity is so
different that the more electronegative atom is going to be able to almost steal the
electrons from the other atom. One way to do this is the rule that generally speaking, if
the difference in electronegativity is greater then .5, the bond will be polar. Another way
is if the molecule is not symmetrical, the bond will be polar because one part of the
molecule will be able to draw electrons to it much more strongly then the other. So, say
we are taking hydrogen and chlorine. The electronegativity values are greater then .5 and
they are from far enough apart on the periodic table to cause HCl to be a polar molecule.
On the other hand, if the electronegativity values are not so strongly different, it may
form a polar bond but it may not be as polar as some others.
4.2.6. Draw and deduce Lewis (electron dot) structures of molecules and ions for up to
four electron pairs on each atom.
A pair of electrons can be represented by dots, crosses, a combination of dots and
crosses or by a line. Note: Cl-Cl is not a Lewis Structure.
The Lewis structure of a molecule shows how the valence electrons are ranged among the
atoms in the molecule. The rules for writing Lewis structures are based on observations
of thousands of molecules. From experiment, chemists have learned that the most
important requirement for the formation of a stable compound is that the atoms achieve
noble gas electron configurations (the octet rule). (Note: The exception to this rule is
hydrogen, which wants two electrons, thus it follows the duet rule). The rules for writing
Lewis structures are as follows:
1. Sum the valence electrons from all the atoms. Do not worry about keeping track
of which electrons come from which atoms. It is the total number of electrons
that is important.
2. Use a pair of electrons to form a bond between each pair of bound atoms.
3. Arrange the remaining electrons to satisfy the duet rule for hydrogen and the octet
rule for all other elements.
We will practice this first with water.
Step 1: First we sum the valence electrons from H2O as shown:
1(H) + 1(H) + 6 (O) = 8 valence electrons.
Step 2: Using a pair of electrons per bond, we draw in the two O—H single bonds:
H—O—H
Note that a line instead of a pair of dtos is used to indicate each pair of bonding electrons.
This is the standard notation.
Step 3: We distribute the remaining electrons to achieve a noble gas electron
configuration for each atom. Since four electrons have been used in forming the two
bonds, four electrons remain to be distributed. Hydrogen is satisfied with two electrons
(duet rule), but oxygen needs eight electrons to have a noble gas configuration. Thus the
remaining four electrons are added to oxygen as two lone pairs. Dots are used to
represent the lone pairs:
Then we check it. Does each hydrogen have two electrons? Yes. Does oxygen satisfy
the octet rule? Yes. Are there 8 electrons? Yes. Thus this is the correct Lewis Dot
structure for water.
Now let’s try the Lewis structure of the CN- (cyanide) ion. Summing the valence
electrons, we have
4(C)+ 5(N) + 1(the -) = 10 electrons
Note that the negative charge means an extra electron must be added. After drawing a
single bond (C—N), we distribute the remaining electrons to achieve a noble gas
configuration for each atom. Eight electrons remain to be distributed. We can try
various possibilities, for example:
Let’s test this now. Yes there is ten electrons, but each atom only has six electrons
instead of eight. The correct structure is:
(Satisfy yourself that both carbon and nitrogen have eight electrons.)
4.2.7. Predict the shape and bond angles for molecules with four charge centers on the
central atom.
Use the valence shell electron pair repulsion (VSEPR) theory to predict the shapes and
bond angles of molecules and ions having four pairs of electrons (charge centers) around
the central atom. Suitable examples are NH3, H2O and alkanes (eg CH4).
The structures of molecules play a very important role in determining their chemical
properties. As we will see later, this is particularly important for biological molecules; a
slight change in the structure of a large biomolecule can completely destroy its usefulness
to a cell or may even change the cell from a normal one to a cancerous one. Many
accurate methods now exist for determining molecular structure, the three-dimensional
arrangement of the atoms in a molecule. These methods must be used if precise
information is required. However, it is often useful to be able to predict the approximate
molecular structure of a molecule. In this section we consider a simple model that allows
us to do this. This model, called the valence shell electron pair repulsion (VSEPR)
model, is useful in predicting the geometries of molecules formed from nonmetals. The
main postulate of this model is that the structure around a given atom is determined
principally by minimizing electron-pair repulsions. The idea here is that the bonding and
nonbonding pairs around a given atom will be positioned as far apart as possible. The
steps to finding the VSEPR model are as follows
1. Draw the Lewis structure for the molecule.
2. Count the electron pairs and arrange them in the way that minimizes repulsion
(that is, put the pairs as far apart as possible.)
3. Determine the positions of the atoms from the way the electron pairs are shared.
4. Determine the name of the molecular structure from the positions of the atoms.
The other important concept to keep in mind is that lone electron pairs, that is, electron
pairs that are not being shared, repel more then shared electron pairs. At this point, you
only need to know molecules with four charge centers on the central atom, ie, there are
four electron pairs around the central atom, but it will not always be that easy.
This idea is really very graphic dependent, so instead of attempting to create wonderful
and beautiful graphics here I am simply going to recommend you check out the webpage
http://dbhs.wvusd.k12.ca.us/VSEPR/VSEPR.html . First read the introduction to
VSEPR, then, to understand what this specific objective is asking, take a look at the
webpage under four electron domains (charge centers). (I am pretty sure that with two
electron domains/charge centers, the shape will always be linear and 180 degrees…I
believe.)
4.2.8. Identify the shape and bond angles for species with two and three negative
charge centers.
Examples should include species with non-bonding as well as bonding electron pairs, eg
CO2, SO2, C2H2, C2H4, CO32- and NO2.
Look under the webpages for two and three electron domains on the previous webpage.
(I am pretty sure that with two electron domains/charge centers, the shape will always be
linear and 180 degrees…I believe.)
4.2.9. Predict molecular polarity based on bond polarity and molecular shape.
The polarity of a molecule depends on its shape and on the electronegativities of its
atoms, eg CO2, H2O.
Polarity of a molecule depends on both bond polarity and its molecular shape. If there
are no polar bonds, i.e. no one atom is much more electronegative then another, then the
molecule is not polar. If there are polar bonds, but they are symmetrical then the
molecule is not polar (think about it like 3D vector addition...if they add to zero, then it's
not polar). If there are polar bonds and the molecule is not symmetrical, then the
molecule is polar.
4.3 Intermolecular Forces
4.3.1. Describe the types of intermolecular force (hydrogen bond, dipole-dipole
attraction and van der Waals’ forces) and explain how they arise from the
structural features of molecules.
All these intermolecular forces are weaker than covalent bonds. For substances of
similar molar mass, hydrogen bonds are stronger then dipole-dipole attractions which
are stronger than van der Waals’ forces. Van der Waals’ forces arise from the
electrostatic attraction between temporary induced dipoles in both polar and non-polar
molecules.
As we saw in 4.2.5., molecules with polar bonds often behave in an electric field as if
they had a center of positive charge and a center of negative charge. That is, the exhibit a
dipole movement. Molecules with dipole movements can attract each other
electrostatically by lining up so that the positive and negative ends are close to each
other. This is called a dipole-dipole attraction. In a condensed state such a liquid, where
many molecules are in close proximity, the dipoles find the best compromise between
attraction and repulsion. That is, the molecules orient themselves to maximize the (+)--(-) interactions and to minimize the (-)---(-) and (+)---(+) interactions. Dipole-dipole
forces are typically only about 1% as strong as covalent or ionic bonds, and they rapidly
become weaker as the distance between the dipoles increases. At low pressures in the gas
phase, where the molecules are far apart, these forces are relatively unimportant. These
arise from the structural features of molecules in that polar molecules have the dipoles,
and remember how polarity has a relationship with structure, thus structure has a
relationship with the creation of dipole-dipole attractions.
Particularly strong dipole-dipole forces, however, are seen among molecules in which
hydrogen is bound to a highly electronegative atom, such as nitrogen, oxygen, or
fluorine. Two factors account for the strengths of these interactions: the great polarity of
the bond and the close approach of the dipoles, allowed by the very small size of the
hydrogen atom. Because dipole-dipole attractions of this type are so unusually, strong,
they are given a special name – hydrogen bonding. This arises from the structure of
molecules because it is dependent on the need for the strong polar bond with a hydrogen
atom.
Even molecules without dipole movements must exert forces on each other. We know
this because all substances-even the noble gases-exist in the liquid and solid states under
certain conditions. The relatively weak forces that exist among noble gas atoms and nonpolar molecules are called Van der Waals’ forces, or London Dispersion Forces. To
understand the origin of these forces, let’s consider a pair of noble gas atoms. Although
we usually assume that the electrons of an atom are uniformly distributed about the
nucleus, this is apparently not true at every instant. As the electrons move about the
nucleus, a momentary nonsymmetrical electron distribution can develop that produces a
temporary dipolar arrangement of charge. The formation of this temporary dipole can, in
turn, affect the electron distribution of a neighboring atom. That is, this instantaneous
dipole that occurs accidentally in a given atom can then induce a similar dipole in a
neighboring atom. This phenomenon leads to an interatomic attraction that is relatively
weak and short lived, but which can be very significant for large atoms. For these
interactions to become strong enough to produce a solid, the motions of the atoms must
be greatly slowed down. This explains, for instance, why the noble gas elements have
such low freezing points. Something to observe is that the freezing point rises going
down the group. The principal cause for this trend is that as the atomic number increases,
the number of electrons increases, and there is an increased chance of the occurrence of
momentary dipole interactions. We describe this phenomenon using the term
polarizability, which indicates the ease with which the electron cloud of an atom can be
distorted to give a dipolar charge distribution. Thus we say that large atoms with many
electrons exhibit a higher polarizability than small atoms. This means that the
importance of van der Waals’ forces increases greatly as the size of the atom increases.
4.3.2. Describe and explain how intermolecular forces affect the boiling points of
substances.
The hydrogen bond can be illustrated by comparing the physical properties of
 H2O and H2S
 NH3 and PH3
 C3H8, CH3CHO and C2H5OH
Intermolecular forces have a very strong affect on the boiling points of substances.
Why? Well, boiling point is the point where enough energy is added a substance to
break the forces of attraction between the different molecules, i.e. the intermolecular
forces. If a substance has a particularly strong intermolecular force holding it
together, then it will have a very high boiling point because it will take a lot of energy
to separate those molecules due to their high attraction. However, if a substance has
particularly weak intermolecular forces holding it together, then it will have a low
boiling point because it will not take much energy to separate the different molecules.
This is how intermolecular forces affect the boiling points of substances.
Examples of this can be easily shown with hydrogen bonds. Remember that oxygen
and sulfur are in the same group and thus have very similar chemical properties.
However, for some reason H2O has a boiling point of 100 degrees Celsius while H2S
has a boiling point of less then 0 degrees Celsius (it’s actually lower then -50
degrees). What’s the explanation for this? Hydrogen bonding! H2O exemplifies
hydrogen bonding because oxygen is electronegative enough to cause the H-O bonds
to be so strongly polar that it is a hydrogen bond. Hydrogen bonds are much stronger
the plain dipole-dipole attractions which holds H2S together, so that explains why
H2O has such a higher boiling point. See if you can explain how hydrogen bonding
affects the other substances mentioned in the objective. Note: I would highly
recommend memorizing which elements can demonstrate hydrogen bonding, they
include nitrogen, oxygen, and fluorine.
4.4 Metallic Bond
4.4.1. Describe metallic bond formation and explain the physical properties of
metals.
Metallic bonding is explained in terms of a lattice of positive ions surrounded by
delocalized valence electrons. The delocalized electrons should be related to the high
electrical conductivity, malleability and ductility of metals.
Any successful bonding model for metals must account for the typical physical
properties of metals: malleability, ductility, and the efficient and uniform conduction
of heat and electricity in all directions. Although the shapes of most pure metals can
be changed relatively easy, most metals are durable and have high melting points.
These facts indicate that the bonding in most metals is both strong and nondirectional.
That is, although it is difficult to separate metal atoms, it is relatively easy to move
them, provided the atoms stay in contact with each other. The simplest picture that
explains these observations is the electron sea model, which envisions a regular array
of metal cations in a “sea” of valence electrons. The mobile electrons can conduct
heat and electricity, and the metal ions can be easily moved around as the metal is
hammered into a sheet or pulled into a wire. Note: This is all that IB really required
for this objective, but if you would like to know more there is much more complicated
models that are somewhat more interesting and not so simplistic, but it is not
necessary. If you do want to know, look up the “band model” or “Molecular Orbital
(MO) model” for metals.
4.5 Physical Properties
4.5.1. Compare and explain the following properties of substances resulting from
different types of bonding: melting and boiling points, volatility, conductivity and
solubility.
Consider melting points, boiling points and volatility of similar substances, such as
F2, Cl2, Br2, and I2, and substances with different types of bonding and different
intermolecular forces. Students should be aware of the effect of impurities on the
melting point of a substance.
The solubility’s of compounds in non-polar and polar solvents should be compared
and explained. Consider also the solubility’s of alcohols in water as the length of the
carbon chain increases.
Melting points and boiling points largely depend on the strength of the intermolecular
bonds. Stronger intermolecular bonds such as hydrogen bonds have very high
melting points and boiling points, while weaker intermolecular bonds have much
lower melting points and boiling points.
4.5.2. Predict the relative values of melting and boiling points, volatility,
conductivity and solubility based on the different types of bonding in substances.
This is basically a culmination of everything in this chapter. If the intramolecular
forces are strong, then it will have a high melting and boiling point and low volatility.
If there are ions or electrons free to be charged and decharged, then it will be very
conductive and if not, it won’t be so conductive. If it is polar it will be soluble, if not
it won’t be.