Bonding
Topics 4 and 14
IB Chemistry Year 1 HL
The Brooklyn Latin School
1
REVIEW: Ionic
Bonding
Wednesday, December 3, 2014
2
4.1 Ionic Bonding &
Structure
Understandings
• Positive ions (cations) form by metals losing valence
electrons.
• Negative ions (anions) form by non-metals gaining
electrons.
• The number of electrons lost or gained is determined by
the electron configuration of the atom.
• The ionic bond is due to electrostatic attraction between
oppositely charged ions.
• Under normal conditions, ionic compounds are usually
solids with lattice structures.
3
4.1 Ionic Bonding &
Structure
Applications
• Deduction of the formula and name of an ionic
compound from its component ions, including polyatomic
ions.
Guidance
• Students should be familiar with the names of these
polyatomic ions: NH4+, OH–, NO3–, HCO3–, CO32–, SO42–,
and PO43–
• Explanation of the physical properties of ionic compounds
(volatility, electrical conductivity, and solubility) in terms of
their structure.
4
Valence Electrons
• Valence electrons are the electrons in the outermost
principal energy level
• All atoms want to be like the Noble Gases, in Group 18,
and have a filled s and p sublevel in the valence shell
5
Ions
• Atoms can gain or lose valence electrons to get a stable
octet.
• Elements that have a small number of electrons in their
outer shells (Groups 1, 2, and 13) will lose those electrons
and form positive ions called cations. These elements are
the metals.
• Elements that have higher numbers of electrons in their
outer shells (Groups 15, 16, and 17) will gain electrons and
form negative ions called anions. These elements are the
non-metals.
• Group 14 elements do not tend to form ions.
6
Sneak Peak: Redox
•
•
•
•
Formation of ionic compounds are always redox reactions
Oxidation is losing electrons
Reduction is gaining electrons
LEO the Lion goes GER
7
Families
• The group or family on the Periodic Table helps you
predict what kind of ions will be formed (unlike last year,
you will not be given this info!)
8
Let’s Practice
• What kind of ion is formed by the following elements?
1. Lithium
2. Sulfur
3. Argon
4. Oxygen
5. Nitrogen
NOTE: When writing an ion, always put the charge number
first as the superscript and then the + or -. Example: Na3+ NOT
Na+3. This is important for IB!!
9
Transition Metals
• Remember from Unit 2 Periodic Trends that the transition
metals can form more than one type of ion (lose from the
s and d sublevels)
• Refer back to what we learned in Unit 2!
10
Unusual Ions
• Lead, Pb, despite being in Group 14, forms a stable ion
Pb2+
• Tin, Sn, also in Group 14, can form Sn4+ and Sn2+
• Silver, Ag, forms the ion Ag+
• Hydrogen, H, can form H– (hydride) as well as the more
common H+.
11
Naming Ions
• Cations, or positive ions, keep their name. If an element
can form more than one positive ion, a Roman numeral is
written to indicate the charge
o E.g Iron +3 is Iron (III) where Iron +2 is Iron (II)
• Anions, or negative ions, get their ending changed to –ide
o E.g. Oxygen -2 is oxide and Nitrogen -3 is nitride
12
Polyatomic Ions
• Polyatomic ions are groups of atoms that are covalently
bound together (i.e. sharing electrons) with an overall
change.
• When naming them, you simply say their name.
• You do have to memorize the names and charges of
several polyatomic ions!!!
13
Memorize!
14
Ionic Compounds
• Ionic compounds are formed when there is a transfer of
electrons from metals to nonmetals (or to a group of
atoms in the case of polyatomic ions)
• The resulting positive and negative ions are attracted to
each other via electrostatic attraction
• The ions now have the electron configuration of Noble
gases
15
Writing Ionic Compound
Formulas
• Since electrons are transferred from one set of atoms to
another, the overall ionic compound must remain
electrically neutral
• When we are determining the formula for ionic
compounds, we simply have to balance to positive and
negative charges
• Last year, we called this the kriss kross method!
16
Writing Ionic Formulas
1. Check the periodic table for the ions that each
element will form.
Al is in group 3, Oxygen is in group 6
2. Write the number of the charge above the ion.
3+
Al
2O
3. “Criss cross” the numbers to balance the charges.
3
Al
2
O
4. Write the final formula using subscripts to show the
number of each ion.
Al2O3
Let’s Practice
• Write the formula for the compound that forms between
magnesium and nitrogen.
• Step 1: Determine the charge each ion forms
Mg +2 N -3
• Step 2: Write the cation first and the anionl second
• Step 3: Using subscripts, put the number of ions of each
element that allows the charges to balance out; overall
charge should equal ZERO!
Mg +2 N -3
Mg3N2
18
Reminder
• Ionic formulas are ALWAYS empirical because ionic
compounds form crystal structures of repeating ions; this
simplest ratio of ions is called the formula unit which simply
states the ratio of ions in the crystal
• When using the kriss kross method, only bring down the
absolute value; there are no signs in the subscripts
• If polyatomic ions are in the compound, put them in
parenthesis and put the subscript outside of that; do not
need if the subscript is 1 (one)
19
Let’s Practice
• Write the formula for ammonium phosphate.
20
Lies, oversimplifications and
more…
• In Regents Chemistry, we sometimes oversimplified
complex concepts to help students understand what was
going on; now you need to know the WHY.
• When we are going over bonding now, I am going to try
to rid you of the technically incorrect, oversimplified
models you may have from last year.
21
22
The Truth is Out There
• Ionic bonding and covalent bonding are not “two
different types of bonds.”
• Sodium atoms do not “want to lose an electron.” In fact,
ionization is always endothermic, meaning it actually
takes energy to remove the electron from sodium. The
fact is, chlorine wants that electron more and takes it from
sodium resulting in two stable electron configurations.
Energy is release when chlorine takes that electron AND
energy is released when the resulting positive ions form
the crystal lattice structure. Reactions are all about how
much energy you put in and how much you get out!
23
More Lies
Most textbooks will do the following which ignores
the truth:
• Showing how sodium atoms give an electron to a
chlorine atom when the reaction is actually
between sodium metal and gaseous chlorine
molecules, Cl2 (see the image on the right
showing sodium metal burning in chlorine gas).
• Showing atoms of sodium and chlorine to both be
the same size, i.e. with identical atomic radii.
• Showing a chloride ion the same size as a
chlorine atom, and in some cases the same size
as a sodium ion.
• Showing all the energy levels in both the sodium
and chlorine atoms to be equally spaced.
• Showing the electrons separately in the first
energy level and in pairs in the subsequent
energy levels.
• Ignoring the changes of state that take place
during the reaction.
Source: ThinkIB.net
24
Typical Way Ionic Bonding is
Shown
What is wrong here?
25
Let’s Practice
1 Write the formula for each of the compounds in the table
on page 3 of your notes with polyatomic ions.
2 Write the formula for each of the following compounds:
(a) potassium bromide (b) zinc oxide (c) sodium sulfate
(d) copper(II) bromide (e) chromium(III) sulfate (f)
aluminium hydride
3 Name the following compounds:
(a) Sn3(PO4)2 (b) Ti(SO4)2 (c) Mn(HCO3)2
(d) BaSO4 (e) Hg2S
4 What are the charges on the positive ions in each of the
compounds in Q3 above?
5 What is the formula of the compound that forms from
element A in Group 2 and element B in Group
15?
6 Explain what happens to the electron configurations of
Mg and Br when they react to form the compound
magnesium bromide.
26
27
28
Lesson 2:
Advanced Ionic
Bonding
Thursday, December 4th, 2014
29
Ionic Bonding
Ionic bonds result from the attractions
between positive and negative
ions.
Ionic bonding involves 3 aspects:
1. Loss of an electron(s) by one element.
2. Gain of electron(s) by a second element.
3. Attraction between positive and negative ions.
30
Stable Octet Rule
• Atoms tend to either gain or lose
electrons in their highest energy level to
form ions
• Atoms prefer having 8 electrons in their
highest energy level
Examples
Na atom
Cl atom
octet
Na+ Ion
Cl- Ion
1s2 2s2 2p6 3s1
1s2 2s2 2p6 3s2 3p5
1s2 2s2 2p6
1s2 2s2 2p6 3s2 3p6
One electron extra
One electron short of a stable
Stable octet
Stable octet
Positive ions attract negative ions forming ionic
bonds.
31
Ionic Bonding
Ionic substances are made of repeating arrays of
positive and negative ions.
An ionic crystal lattice
32
Ionic Bonding
The array is repeated over and over to form the
crystal lattice.
Model of a
Sodium
chloride
crystal
Each Na+ ion is surrounded by 6 other Cl- ions. Each Clion is surroundedby 6 other Na+ ions
33
Ionic Compound Structure
• In Chemistry, opposites attract, meaning the ions want to
surround themselves with other ions of opposite charge
• The ions take on a predictable three-dimensional
crystalline structure known as the ionic lattice (see below)
• The coordination number of the lattice tells you how many
ions each ion in the crystal is surrounded by
For example, in the sodium
chloride lattice, the coordination
number is six because each Na+ ion
is surrounded by six Cl– ions and
each Cl– ion is surrounded by six
Na+ ions.
34
Lattice Energy
• Lattice energy is a measure of the strength of attraction
between ions in the lattice of an ionic compound
• Lattice energy is higher for ions that are small and highly
charged and weaker for ions that are larger and have a
lower charge
NaCl
35
Ionic Bonding
• The shape and form of the crystal lattice depend on
several factors:
1. The size of the ions
2. The charges of the ions
3. The relative numbers of
positive and negative
ions
36
Characteristics of Ionic Bonds
1. Crystalline at room
temperatures
2. Higher melting points
and boiling points than
covalent compounds
3. Conduct electrical
current in molten or
solution state but not
in the solid state
4. More soluble in polar
solvents such as water
5. Brittle
Water solutions of ionic
compounds are
usually electrolytes.
37
Physical Properties –
A Closer Look
Melting and Boiling Point
• The electrostatic force of attraction that holds together all
the positive and negative ions in the lattice are very
strong; it takes a lot of energy to separate the ions
• Example: The melting point of NaCl is over 800oC.
• MP and BP is generally higher when the charge on ions
are greater
• Ionic compounds have low volatility (tendency to
vaporize)
38
Coulomb’s Law
• In Physics, we learned that the force between two
charged particles is equal to:
• Where q1 and q2 are the charges, r is the distance
between the two charged particles
39
Solubility
• Solubility is determined by the degree to which the two
substances mixing are attracted to each other (“like
dissolves like”)
• Ionic compounds can dissolve in polar substances
because the positively and negatively charged ions are
attracted to the partially positive and negative regions in
the polar covalent compound
• Example: When NaCl dissolves in water, the partially
negative oxygen atoms in the molecule can dislodge the
positive ions from the crystal structure
40
41
Solvation of Ionic
Compounds
• When an ionic compound dissolves in a polar liquid, the
ions get dislodged from the crystal lattice structure.
• In Water: As these ions separate from the lattice, they
become surrounded by water molecules and are said to
be hydrated and the state symbol (aq) is used.
• In Other Polar Solvent: If a liquid other than water is able
to dissolve the solid, the ions are said to be solvated and
an appropriate state symbol to denote the solvent is used.
• In the case of solvents like oil or hexane, C6H14, which are
non-polar and so have no charge separation, there is no
attraction between the liquid and the ions. So here the
ions remain tightly bound to each other in the lattice, and
the solid is insoluble.
42
Electrical Conductivity
• Substances can conduct electricity when they are able to
move a charge
• Ionic compounds in the solid phase are locked into place
and therefore cannot conduct electricity
• Ionic compounds can only conduct charge in the liquid
(molten) phase or when dissolved in a polar solvent
43
Brittle
• Ionic compounds are brittle, meaning they easily shatter
when a force is applied.
• Take a look at the picture below. Why do you think this is?
44
Ionic Character
• Ionic compounds typically form between metals, which
have a tendency to lose electrons, and nonmetals, which
have a tendency to gain electrons
• In order for two elements to form a binary ionic
compound, they must have two very difference
electronegativity values
• In general, an electronegativity difference of greater than
1.8 is considered to be ionic
• The larger the electronegativity, the more ionic the bond.
45
Bond Continuum
• We are going to jump into covalent bonding next.
• We will see that the distinction between ionic and
covalent bonds is not black and white, but is best
described as a bonding continuum with all intermediate
types possible.
• In general, the larger the electronegativity difference, the
more ionic the bond. The smaller the en difference, the
more covalent the bond.
46
47
Let’s Practice
• Explain which of the following pairs will be most likely to
form an ionic bond.
A Be and F
B Si and O
C N and Cl
D K and S
Consider the difference in electronegativity of each pair:
A 1.6 and 4.0 B 1.9 and 3.4 C 3.0 and 3.2 D 0.8 and 2.6
D has the greatest difference, so the compound K2S will be
the most ionic.
48
More Practice
7 Which fluoride is the most ionic?
A NaF B CsF C MgF2 D BaF2
8 Which pair of elements reacts most readily?
A Li + Br2 B Li + Cl2 C K + Br2 D K + Cl2
9 You are given two white solids and told that only one of
them is an ionic compound. Describe three tests you could
carry out to determine which it is.
49
Answers
• 9. Test the melting point: ionic solids have high melting
points. Test the solubility: ionic compounds usually dissolve
in water but not in hexane. Test the conductivity: ionic
compounds in aqueous solution are good conductors, as
are ionic compounds when they are molten
50
Lesson 3:
Review
Covalent
Compounds
Friday, December 5, 2014
51
Covalent Bonding
Understandings:
• A covalent bond is formed by the electrostatic attraction
between a shared pair of electrons and the positively
charged nuclei.
• Single, double, and triple covalent bonds involve one,
two, and three shared pairs of electrons respectively.
• Bond length decreases and bond strength increases as
the number of shared electrons increases.
• Bond polarity results from the difference in
electronegativities of the bonded atoms.
Guidance
• Bond polarity can be shown either with partial charges,
dipoles, or vectors.
52
Covalent Bonding
Applications and skills:
• Deduction of the polar nature of a covalent bond from
electronegativity values.
Guidance
• Electronegativity values are given in section 8 of the data
booklet.
53
REVIEW: Covalent Bonding
• In a covalent bond, both elements are trying to gain
electrons in order to achieve the stable noble gas
configuration
• Instead of a transfer of electrons, an electron pair is
shared between the two nuclei.
• A group of atoms that are held together by covalent
bonds are called a molecule
54
Energetics of Bond
Formation
• The formation of a covalent bond stabilizes the two atoms
entering into the bond so energy is released.
• Energy is always required to break a covalent bond.
• The two nuclei are both attracted to the electron pair
while also repelling each other; the bond forms at an
equilibrium point that balances the attraction and
repulsion
55
56
Lewis Dot Structures
• Lewis dot structures are helpful to visualize molecules and
covalent bonds
• Atoms enter into covalent bond to try to get a total of
eight valence electrons (octet rule) so we only show
valence electrons in the Lewis Dot structure
• Bonds are shown as a line between the two bonded
atoms and lone pairs of electrons that do not enter into
the bond are shown as dots
57
Multiple Bonds
• A single bond is a sharing of two electrons
• A double bond is a sharing of two pairs of, or four,
electrons
• A triple bond is a sharing of three pairs of, or six, electrons
• You can never have a quadruple bond!
58
Bond Properties
• Bond length: a measure of the distance between the two
bonded nuclei.
• Bond strength: usually described in terms of bond
enthalpy, is a measure of the energy required to break
the bond.
• As we go down a group, molecules form longer bond
lengths
• As bond length increases, bond enthalpy decreases
59
Multiple Bond
• Double bonds are shorter and stronger than single bonds
and triple bonds are shorter and stronger than double
bonds
60
Even in the same molecules, double
bonds are shorter and stronger!
61
Polar Bonds
• A bond is polar when the two elements sharing electrons
share them unevenly
• Elements with different electronegativity values will share
electrons unevenly
• The term dipole is often used to indicate the fact that this
type of bond has two separated opposite electric
charges.
• The more electronegative atom with the greater share of
the electrons, has become partially negative or 𝛅–, and
the less electronegative atom has become partially
positive or 𝛅+.
62
63
Electronegativity Difference
• As the difference in electronegativity between the two
element increase, the bond becomes more polar
• NOTE: This does not mean the overall molecule is polar!
More on this later!
64
Let’s Practice
• Use the electronegativity values to put the following
bonds in order of decreasing polarity.
a) N–O in NO2
b) N–F in NF3
c) H–O in H2O
d) N–H in NH3
65
Challenge Question
• Oxygen is a very electronegative element with a value of
3.4 on the Pauling scale. Can you determine the formula
of a compound in which oxygen would have a partial
positive charge?
66
Polar or Nonpolar?
• In Regents Chemistry, we stated that any bonds with an
electronegativity difference of < 0.4 were considered
nonpolar.
• In IB Chem, the only bonds that are considered truly
nonpolar are between the same element (i.e. N2, O2 etc.).
These are called pure covalent molecules.
• However, there are some bonds, notables C-H that have
an EN difference of < 0.4 that behave basically nonpolar
(even though carbon is slightly more electronegative).
Remember – it is all a continuum!
• Polar bonds have more ionic character than nonpolar
bonds!
67
Bond Continuum
68
Let’s Practice
10 Which substance contains only ionic bonds?
A NaNO3 B H3PO4 C NH4Cl D CaCl2
11 Which of the following molecules contains the shortest
bond between carbon and oxygen?
A CO2 B H3COCH3 C CO D CH3COOH
12 For each of these molecules, identify any polar bonds
and label them using δ+ and δ– appropriately:
(a) HBr (b) CO2 (c) ClF (d) O2 (e) NH3
13 Use the electronegativity values in Section 8 of the IB
data booklet to predict which bond in each of
the following pairs is more polar.
(a) C–H or C–Cl (b) Si–Li or Si–Cl (c) N–Cl or N–Mg
69
Solutions
10. D
11. C
70
Lesson 4:
Covalent
Bonding
Structures
Tuesday, December 9, 2014
71
4.3 Covalent Structures
Understandings:
• Lewis (electron dot) structures show all the valence electrons in a
covalently bonded species.
Guidance:
• The term ‘electron domain’ should be used in place of ‘negative
charge centre’.
• Electron pairs in a Lewis (electron dot) structure can be shown as
dots, crosses, a dash, or any combination.
• Coordinate covalent bonds should be covered.
• The ‘octet rule’ refers to the tendency of atoms to gain a valence
shell with a total of 8 electrons.
• Some atoms, like Be and B, might form stable compounds with
incomplete octets of electrons.
• Resonance structures occur when there is more than one possible
position for a double bond in a molecule.
• Shapes of species are determined by the repulsion of electron pairs
according to VSEPR theory.
• Carbon and silicon form giant covalent/network
covalent/macromolecular structures.
72
4.3 Covalent Structures
Guidance:
• Allotropes of carbon (diamond, graphite, graphene, C60
buckminsterfullerene) and SiO2 should be covered.
Applications and skills:
• Deduction of Lewis (electron dot) structure of molecules and
ions showing all valence electrons for up to four electron pairs
on each atom.
• The use of VSEPR theory to predict the electron domain
geometry and the molecular geometry for species with two,
three, and four electron domains.
• Prediction of bond angles from molecular geometry and
presence of non-bonding pairs of electrons.
• Prediction of molecular polarity from bond polarity and
molecular geometry.
• Deduction of resonance structures, including C6H6, CO32–,
and O3.
• Explanation of the properties of giant covalent compounds in
terms of their structures.
73
Lewis Dot Diagrams
• Lewis Dot diagrams are a way to show how atoms are
bonded in a molecule
• Also called a “dot and cross” diagram
74
Steps For Drawing Lewis
Dot Structures
1. Calculate total number of valence electrons in the
molecule. Consider these the money you have at the
bank. You cannot spend more than you have and you
have to spend it all!
2. Draw the skeletal structure to show how the atoms are
linked together.
3. Use a pair of crossed, a pair of dots or a straight line
between all of the atoms bonded together
4. Add more electron pairs around the atoms until everyone
has 8 valence electrons (except for H which only gets 2
electrons!)
5. Check to make sure everyone has an octet! If they do
not, make double and/or triple bonds!
6. Make sure you are showing the # of electrons from Step75 1
Skeleton Help
• Often, the most difficult part is deciding how to arrange
the atoms. When we jump into the HL level material, we
will learn about formal charge and how that can help us
out with more complicated diagrams
• For now, try to put the least electronegative element in
the middle
1. Halogens are rarely in the center
2. Hydrogen is never in the center
3. If you have carbon, put that in the center
4. Try to give Group 17 elements one bond, Group 16
elements two bonds, 15 three bonds and 14 four bonds.
5. Try to make the molecule somewhat symmetrical if
possible
76
Easy Example
Draw the Lewis structure for the molecule CCl4.
Step 1: Determine # of valence e’s. 12+(7*4) = 32
Step 2: Set up skeleton structure
Step 3: Use a straight line to represent each bond. You may
also use dots or x’s
Step 4: Add the remaining amount of electrons.
Step 5: Check to make sure everyone has an octet! If not,
start to make double and triple bonds!
Step 6: Check that you are showing exactly 32 e’s
77
Let’s Practice
Draw the following Lewis Dot diagrams
1. CH4
2. NH3
3. H2O
4. CO2
5. HCN
6. OH- (HINT: What does the -1 charge mean?)
7. SO42-
78
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Coordinate Covalent Bonds
• In a coordinate covalent bond, one atoms donates BOTH
of the electrons in the covalent bond
• These are also called dative bonds
• Sometimes, you might show this in a Lewis Dot structure
with an arrow pointing from the element donating both
electrons
80
Octet Rule Exceptions
Element
Exception
Example
Be
Can be happy with 2 bonds
BeCl2
B
Can be happy with 3 bonds
BF3
S
Can expand octet; form 6
bonds
SF6
P
Expands octet, forms 5 bonds PCl5
Xe
Can bond, forms 6 bonds
XeF4
81
Less Than An Octet
• Atoms that have less than a stable octet are said to be
electron deficient and serve as excellent Lewis Acids
(electron pair accepter)
82
Let’s Practice!
14 Draw the Lewis structures of:
(a) HF (b) CF3Cl (c) C2H6 (d) PCl3 (e) C2H4 (f) C2H2
15 How many valence electrons are in the
following molecules?
(a) BeCl2 (b) BCl3 (c) CCl4 (d) PH3 (e) SCl2 (f) NCl3
16 Use Lewis structures to show the formation of a
coordinate bond between H2O and H+.
17 Draw the Lewis structures of:
(a) NO3– (b) NO+ (c) NO2– (d) O3 (e) N2H4
83
Lesson 5: VSEPR
Wednesday, December 10, 2014
84
Molecule Shapes
• Lewis Dot structures are helpful because they let us know
how the atoms are arranged and where there are lone
pairs of electrons but tell us very little about the shape of
the molecules
• Why is shape important? Structure determines
functionality!
• Enzymes work because they “fit” another molecule in.
• We touched on molecule shapes last year but will now
dive deeper!
• What determines molecule shapes? Electron repulsion!
Like charges repel!
85
Valence Shell Electron Pair
Repulsion (VSEPR) Theory
• This theory is based on the simple notion that because
electron pairs in the same valence shell carry the same
charge, they repel each other and so spread themselves
as far apart as possible.
• When we said “pair” of electrons that is an
oversimplification; in fact when you have multiple bonds
they often act together. It is more appropriate to say
electron domain so we will be using this term in class
• What matters in determining shape is the total number of
electron domains, and this can be determined from the
Lewis structure.
86
Central Atoms
• We often look at the central atom for simple compounds
to determine the shape of the molecule
• How many electron domains exist in the central atom of
the following molecules whose Lewis structures are
shown?
87
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VSEPR
• The repulsion applies to electron domains, which can be
single, double, or triple bonding electron pairs, or nonbonding pairs of electrons.
• The total number of electron domains around the central
atom determines the geometrical arrangement of the
electron domains.
• The shape of the molecule is determined by the angles
between the bonded atoms.
• Non-bonding pairs (lone pairs) have a higher
concentration of charge than a bonding pair because
they are not shared between two atoms, and so cause
slightly more repulsion than bonding pairs. The repulsion
decreases in the following order:
lone pair–lone pair > lone pair–bonding pair > bonding pair–
89
bonding pair
Lone Pairs vs. Bonded Pairs
• Since lone pairs have a higher concentration of charge,
molecules with lone pairs will have slight distortions in the
expected bonding angles because they will push away
the other electron domains more!
90
Two Electron Domains
• The furthers away two electron domains can be from
each other is 180 degrees, giving those molecules a linear
shape
91
Three Electron Domains
• Molecules with three electron domains will position them
at 120° to each other, giving a triangular planar shape to
the electron domain geometry.
• If only two of the three domains are bonding, the shape
of the molecule will be bent or V-shaped and the bond
angle is 117 degrees.
92
Four Electron Domains
• Molecules with four electron domains will position them at
109.5° to each other, giving a tetrahedral shape to the
electron domains
• If all four sides are bonding electrons, the molecule will
also have a tetrahedral shape (e.g. CH4) with bond angles
of 109.5°
• If three sides are bonding, the molecule will have a
trigonal pyramidal shape (e.g. NH3) with bond angles of
107°
• If two sides are bonding, the molecule will have a bent or
V-Shape (e.g. H2O) with bond angles of 105°
93
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Summary
# of
# of
Electron Bonded
Domains Domains
# of Long
Pairs
Shape
Bond Angles
2
3
3
4
4
4
5
6
0
0
1
0
1
2
0
0
Linear
Trigonal Planar
Bent or T-Shaped
Tetrahedral
Trigonal Pyramidal
Bent/T-Shaped
Trigonal Bipyramidal
Octahedral
180
120
117
109.5
107
105
90 & 120
90
2
3
2
4
3
2
5
6
95
96
Expanded Octets
• More on the shapes where there are 5 and 6 electron
domains later in the unit when we go into the HL level
material
• In order to get those shapes, the atoms need expanded
octets
97
Steps To Determine Shape
1 Draw the Lewis structure
2 Count the total number of electron domains on the
central atom.
3 Determine the electron domain geometry as follows:
2 electron domains → linear
3 electron domains → triangular planar
4 electron domains → tetrahedral
4 Determine the molecular geometry from the number of
bonding electron domains.
5 Consider the extra repulsion caused by the lone pairs and
adjust the bond angles accordingly.
98
Let’s Practice!
18 Predict the shape and bond angles of the following
molecules:
(a) H2S (b) CF4 (c) HCN (d) NF3 (e) BCl3 (f) NH2Cl (g) OF2
19 Predict the shape and bond angles of the following ions:
(a) CO32– (b) NO3– (c) NO2+ (e) ClF2+ (d) NO2– (f) SnCl3–
20 How many electron domains are there around the
central atom in molecules that have the following shapes?
(a) tetrahedral (b) bent (c) linear (d) trigonal pyramidal (e)
triangular planar
99
100
Polarity
• Bond polarity is whether the pair of electrons are shared
evenly among the two nonmetals covalently bonded
together
o A bond is considered polar if the electronegativity difference between the two
elements is less than 0.4
• Molecular polarity refers to whether there is an overall
uneven charge distribution in the molecule called a
dipole moment
o A molecule with all nonpolar bonds will always be nonpolar overall
o A molecule with polar bonds can be nonpolar overall IF the molecule is
symmetrical, meaning the forces pulling on the electron pairs all balance out
leading to no net force pulling on the bonding pairs
101
Molecular Polarity
Molecular polarity depends on:
1.
2.
The polarity of the bonds
The shape of the molecule
• When molecules are symmetrical, even though the bonds
are polar, the charge distributions effectively cancel each
other out; see below
102
What shapes can be
nonpolar?
• In order for a molecule to have polar bonds but be
nonpolar overall, it has to be symmetrical, meaning that
the same atom is attached to the central atom
• In addition, there cannot be lone pairs attached to the
central atom as those electron domains are different from
the bonded domain
• The shapes we have learned so far that can even be
nonpolar when the bonds are polar are linear, trigonal
pyramidal and tetrahedral
• Trigonal pyramidal and bent molecules can not be
nonpolar with polar bonds.
103
Net Dipole
• If either the molecule contains bonds of different polarity,
or its bonds are not symmetrically arranged, then the
dipoles will not cancel out, and the molecule will be polar.
• Another way of describing this is to say that it has a net
dipole moment, which refers to its turning force in an
electric field.
104
It helps to think about dipole
in terms of vectors
(FLASHBACK to Freshman
year Physics!)
If the Fnet of all the resultant
force vectors is equal to zero,
there is no overall dipole
moment and the molecule is
nonpolar.
Another way to conceptualize
this is by thinking about a
“tug of war” (see left)
105
Let’s Practice
• 21 Predict whether the following will be polar or non-polar
molecules:
a) Polar
(a) PH3
b) Nonpolar
(b) CF4
c) Polar
(c) HCN
d) Nonpolar
(d) BeCl2
e) Nonpolar
(e) C2H4
f) Polar
(f) ClF
g) Nonpolar
(g) F2
h) Nonpolar
(h) BF3
106
Trickier Question
22 The molecule C2H2Cl2 can exist as two forms known as
cis–trans isomers, which are shown below. (Sneak Preview for
Orgo: The double bond locks this molecule in place
preventing the atoms from rotating around each other.
Can’t wait for Orgo!)
Nonpolar
Polar
Determine whether either of these has a net dipole moment.
107
Lesson 6: Lewis
Dot Structures
and Resonance
Wednesday, December 17
108
Warm-Up
Try to draw the following Lewis Dot structures:
1. CO322. O3
What do you notice?
109
Delocalized Electrons
• In these Lewis Dot structures, we find that we can draw
more than one type of structure.
• Which one is correct?
• Turns out…neither of them is exactly correct. These
structures suggest that the molecule should contain one
oxygen–oxygen double bond and one oxygen–oxygen
single bond, which we would expect to be different
in bond length and bond strength. However, experimental
data reveal that ozone actually contains two equal
oxygen–oxygen bonds, intermediate in length and
110
strength between single and double bonds.
111
Delocalized Electrons
• What is going on here?
• In some molecules, bonding electrons are less restricted
and are shared between more than one bonding
position.
• These electrons are said to be delocalized. Free from the
constraints of a single bonding position, delocalized
electrons spread themselves out, giving greater stability to
the molecule or ion.
112
Resonance
• Resonance occurs when more than one valid Lewis
structure can be drawn for a particular molecule. The true
structure is an average of these, known as a resonance
hybrid.
• Let’s try to draw the resonance structures for the
carbonate ion CO32–.
113
Benzene
• Benzene is an important organic molecule that has
delocalized electrons
• The resonance structure gives this molecule additional
stability that would otherwise not be predicted
114
Benzene
• When we learn about sigma and pi bonds later in the unit
when we move onto the HL level, we will see that the
electrons in the double bonds of all the carbons are all
delocalized and spread out among the carbon atoms
• Bond order is a measure of the number of electrons
involved in bonds between two atoms. Values for bond
order are: single bonds = 1, double bonds = 2, triple bonds
= 3. Resonance hybrids have fractional values of bond
order. The carbon-carbon bonds in benzene have a bond
order of 1.5.
115
Let’s Practice
• Compare the structures of CH3COOH and CH3COO– with
reference to their possible resonance structures.
• Put the following species in order of increasing carbon–
oxygen bond length: CO CO2 CO32– CH3OH
• By reference to their resonance structures, compare the
nitrogen–oxygen bond lengths in nitrate(V) (NO3–) and
nitric(V) acid (HNO3).
116
Network Solids
• Network solids are a special type of covalent compound
where there are no molecules or discrete particles but
rather, all the atoms are covalently bonded together into
one giant crystal
• It is referred to as a giant molecular or network covalent
structure or macromolecular structure.
• These network solids have very different characteristics
than regular covalent compounds (i.e. higher boiling
points, higher melting points, very hard, etc.)
117
Two Network Solids To
Know!!!!
• You need to know the following two network solids for the
IB exam:
1. Diamond – in diamond, each carbon atom is covalently
bonded to four other carbon atoms in a continuous
tetrahedral shape
2. Silicon Dioxide – SiO2 (commonly referred to as silica or
quartz) where each Si atom is covalently bonded to four
O atoms, and each O to two Si atoms
118
Quartz/Silica
• SiO2 refers to the ratio of atoms within the giant molecule –
it is an empirical formula and the actual number of atoms
present will be a very large multiple of this. As the atoms
are strongly held in tetrahedral positions that involve all
four silicon valence electrons, the structure has the
following properties:
• strong;
• insoluble in water;
• high melting point;
• non-conductor of electricity.
• These are all properties we associate with glass and sand
– different forms of silica.
119
Allotropes
• Allotropes are different forms of the same elements that
have different structures in the same phase
• The different structures mean the different allotropes have
different physical and chemical properties
• For example, O2 gas and O3 gas are bonded differently
and have very different characteristics
• You need to know the four allotropes of carbon for the IB
exam – diamond, graphite, fullerene, and graphene
120
Characteristic
Graphite
Diamond
Fullerene
Graphene
Structure
Each C is sp2
hybridized,
bonded to 3
other Cs and
found in
layers that can
slide of each
other; bond
angle 120
Each C is sp3
hybridized
and bonding
to 4 other C
atoms; bond
angle 109.5
Each C atom is
sp2 hybridized
and bonded in
a sphere of 60
carbon atoms,
consisting of 12
pentagons and
20 hexagons.
Each C atom is
covalently
bonded to 3
others, as
in graphite,
forming
hexagons with
bond angles of
120°. Single
layer.
Electrical
Conductivity
Good
None
Semiconductor
Very good
Thermal
Conductivity
Poor
Very good
Very low
Very good;
best known
Special
Properties
Soft
Hardest
known
Light and
strong
Thinnest
material ever
Uses
Pencils!
Cutting class
Jewels
Nanotech
Lots of new
stuff!
121
122
123
124
Graphene
• Since graphene is relatively new (discovered in 2004), the
IB might want you to know more about it. It is at the
current edge of technology
125
Let’s Practice
25 Describe the similarities and differences you would
expect in the properties of silicon and diamond.
26 Explain why graphite and graphene are good
conductors of electricity whereas diamond is not (HINT:
which ones have resonance?).
126
Answers
25 Similarities: strong, high melting points, insoluble in water,
non-conductors of electricity, good thermal conductors.
Differences: diamond is stronger and more lustrous; silicon
can be doped to be an electrical conductor.
26 Graphite and graphene have delocalized electrons that
are mobile and so they conduct electrical charge. In
diamond all electrons are held in covalent bonds and so are
not mobile.
127
Lesson 7:
Intermolecular
Forces
Thursday, December 18
128
Warm-Up
• What were the five intermolecular forces we learned
about in Regents Chemistry last year?
129
Topic 4.4 – Intermolecular
Forces
Understandings:
• Intermolecular forces include London (dispersion) forces,
dipole–dipole forces, and hydrogen bonding.
Guidance
• The term ‘London (dispersion) forces’ refers to
instantaneous dipole–induced dipole forces that exist
between any atoms or groups of atoms and should be
used for non-polar entities. The term ‘van der Waals’ is an
inclusive term, which includes dipole–dipole, dipole–
induced dipole, and London (dispersion) forces.
• The relative strengths of these interactions are London
(dispersion) forces < dipole–dipole forces < hydrogen
130
bonds.
Topic 4.4 – Intermolecular
Forces
Applications and skills:
• Deduction of the types of intermolecular force present in
substances, based on their structure and chemical
formula.
• Explanation of the physical properties of covalent
compounds (volatility, electrical conductivity, and
solubility) in terms of their structure and intermolecular
forces.
131
Intermolecular Forces
• Covalent bonds hold together atoms within a molecule
(intramolecular)
• However, in between molecules, intermolecular forces
hold them together and allow them to enter into the liquid
and solid phase
• The type of intermolecular force depends on the polarity
of the molecule; the strength of the intermolecular force
depends on the type and the size of the molecules
• The strength of intermolecular forces determines the
physical properties of a substance. Volatility, solubility,
and conductivity can all be predicted and explained
from knowledge of the nature of the forces between
molecules.
132
London Dispersion Forces
• London dispersion forces form between nonpolar
molecules and are the result of temporary or induced
dipole moments
• Nonpolar molecules do not have a dipole moment,
meaning there is no partial positive or negative side;
HOWEVER, since electrons are always moving, at any
given moment there may be an uneven distribution of
charge giving one side a partial negative and the other
side a partial positive.
• This in turn can affect the electron clouds of the nearby
atoms causing an induced dipole
133
London Dispersion Forces
134
London Dispersion
• The intermolecular force is the weak attraction between
these temporary dipoles
• These forces increase as molecules get larger; larger
electron clouds increase the probability of a charge
imbalance
135
136
Properties
Since London Dispersion Forces are so weak, the nonpolar
molecules that are held together by them have:
1. Low melting points
2. Low boiling points
3. High vapor pressure
• Polar molecules also have London dispersion forces but
these are far weaker than some of the other forces at
work!
137
Dipole-Dipole Attractions
• Form between polar molecules
• In polar molecules, one end of the molecule is electron
deficient with a partial positive charge (δ+), while the
other end is electron rich with a partial negative charge
(δ–). This is known as a permanent dipole.
• It results in opposite charges on neighboring molecules
attracting each other, generating a force known
as a dipole–dipole attraction.
138
Strength of Dipole-Dipole
• Dipole-dipole attractions are stronger than London
dispersion forces meaning these substances have a higher
bp and mp
• The strength of the dipole-dipole attractions is due to the
degree of polarity, size of the molecule, orientation and
more
139
Van der Waals’ Forces
• You may also hear the term van der Waals’ forces used
when talking about intermolecular attraction
• This is an umbrella term used to describe any
intermolecular force not due to full ion-ion attraction or
covalent bonding
• Sometimes, in the case of very large molecules, these
forces will occur intramolecularly (i.e. large proteins)
140
Hydrogen Bonding
• Hydrogen bonds are a special, exceptionally strong type
of intermolecular force that only occurs when you have
hydrogen covalently bonding to an extremely
electronegative element, nitrogen, fluorine or oxygen.
• Hydrogen bonding is stronger than any van der Waals’
force and accounts for the exceptionally high boiling
points of water, ammonia (NH3), alcohols and HF.
• This type of intermolecular force is extremely important
when looking at Biochemistry as it holds together our DNA
and is important in the folding of proteins!
141
Why so strong?
• The large electronegativity of the O, N or F make the
hydrogen atom bonded to them extremely electron
deficient.
• The hydrogen atom has no other electrons other than the
one in that bonding pair. It now exerts a very strong
attractive force on another bonding domain from a
neighboring molecule.
142
Hydrogen bonds explain
boiling point trends
143
H2O
• Without hydrogen bonding, water would be a gas at
room temperature meaning life as we know it on this
planet could not exist
• Also, hydrogen bonds account for the fact that solid ice is
one of the only substances in the universe LESS DENSE than
liquid water. Without them, ice would not float and our
oceans would freeze from the bottom up!
144
Biochemistry
• The hydrogen bond is important in the Chemistry of life.
• TIP FOR NEXT YEAR: This will be covered in Biology but is
also an option for next year – Option B: Biochemistry. Keep
in mind that since you are covering this in Bio, it means
that if we cover a different option in Chem, you have two
different choices for Paper 3.
• Hydrogen bonding cause our proteins to fold and holds
together our very DNA!
• The strands of our DNA are held together by hydrogen
bonds. These bonds are strong enough to hold the strands
together but not so strong that they cannot be separated
when we need to replicate a gene!
145
Advance Molecular Orbital
Theories on H-Bonds
• The extreme electronegativity different between F, O and
N does not fully explain the strength and behavior of
hydrogen bonds.
• In reality, hydrogen bonds go beyond just being a
stronger version of dipole-dipole forces; they have some
properties of covalent bonds.
• This is an area Chemistry still being investigated!!
146
All of these forces are weaker than bonds!!
147
Let’s Practice
1. Methoxymexane (CH3–O–CH3) boils at a much lower
temperature than ethanol (CH3CH2–O–H). Use your
knowledge of intermolecular forces to explain why.
2. Put the following molecules in order of increasing boiling
point and explain your choice: CH3CHO, CH3CH2OH and
CH3CH2CH3
148
Let’s Practice
What is the strongest intermolecular force holding together
the following substances?
1. NH3
2. CCl4
3. C4H9OH
4. N2
5. PH3
6. CO2
7. CH3F
8. HF
9. Na2O
149
10. SiO2
Lesson 8: Physical
Properties of
Covalent
Compounds
Friday, December 19
150
Properties of Covalent
Compounds - Solubility
• Solubility – polar covalent compounds are generally
soluble in other polar covalent compounds and can mix
with ionic compounds; nonpolar covalent compounds
are soluble in other nonpolar covalent compounds. “Like
dissolves like”
• The larger the polar covalent compounds, the less soluble
they become in other polar covalent compounds
because the dipole becomes a less important
component of the molecule; the nonpolar parts of the
molecule overwhelm the dipole-dipole attractions
• Network solids are not soluble!
151
Electrical Conductivity
• Covalent compounds do not contain ions so they are not
able to conduct electricity
• HOWEVER…some covalent compounds can ionized under
certain conditions (HINT: remember our acids!) and can
conduct electricity in solution. Examples include HCl,
H2SO4, etc.
• Some network solids such as graphite and graphene are
electrical conductors due to their mobile electrons.
152
Summary
153
Let’s Practice
154
Answers
155
QUIZ!!
20 minutes!
156
Lesson 9:
Metallic
Bonding
Tuesday, December 23
157
Topic 4.5 Metallic Bonding
Understandings:
• A metallic bond is the electrostatic attraction between a
lattice of positive ions and delocalized electrons.
• The strength of a metallic bond depends on the charge of
the ions and the radius of the metal ion.
• Alloys usually contain more than one metal and have
enhanced properties.
Guidance
• Examples of various alloys should be covered.
158
Topic 4.5 Metallic Bonding
Applications and skills:
• Explanation of electrical conductivity and malleability in
metals.
• Explanation of trends in melting points of metals.
Guidance
• Trends should be limited to s- and p-block elements.
• Explanation of the properties of alloys in terms of nondirectional bonding.
159
Metallic Bonding - Intro
• Metals tend to have low ionization energies and tend to
lose electrons in order to gain Noble gas configurations.
• In the elemental state, when there is no other element
present to accept the electrons and form an ionic
compound, the outer electrons are held only loosely by
the metal atom’s nucleus and so tend to ‘wander off’ or,
more correctly, become delocalized.
• The metal atoms become positively charged ions and
form a regular lattice structure through which these
electrons can move freely.
• There is a force of electrostatic attraction between the
lattice of cations and the delocalized electrons, and this is
known as metallic bonding.
160
Referred to as “sea of electrons”
161
Metallic Bond Strength
• Metallic bond strength is determined by:
1. The number of delocalized electrons
2. The charge on the cation
3. The radius of the cation
Example 1: If we compare the melting point of Na which
gets 1 delocalized electron per atom and Mg which gets 2
delocalized electrons per atom, Na melts at 98 degree C
and Mg melts at 650 degrees C.
Example 2: Melting points decrease down a group as the
attraction between the nucleus and the delocalized
electrons decrease.
162
Metallic Properties
• These delocalized electrons give metallic
substances very unique properties including:
o Good electrical conductivity
o Good thermal conductivity
o Malleable
o Ductile
o High melting points
o Shiny, lustruous
163
164
Alloys
• When metals form homogeneous solutions, they are
called alloys
• Alloys are made by mixing the metals together while they
are in the molten or liquid state
• Even if the positive cations have different sizes, since they
are not held in a strict lattice
• Alloys often have different characteristics than the original
metals including being more stable and stronger; it is
more difficult for the atoms to move over each other when
they are different sizes
165
Some Common Alloys
Alloy
Brass
Steel
Dental
Amalgam
Composition
Copper and
zinc
Iron, Carbon
and other
metals
Mercury, silver
and tin
Uses
Door handles,
screws
Bridges,
buildings
Used to be
used in teeth
fillings
166
167
Alloys
Small amounts of a another
element added to a metal
can change its overall
properties.
For example, adding a small
amount of carbon to iron,
will significantly increase its
hardness and strength
forming steel.
16
Semimetals
The electrons in semimetals are much less mobile than
in metals, hence they are semiconductors
16
Comparison of Types of Bonding
Ionic
Covalent
Metallic
Formation
Anion & cation
Transferred
electrons
Shared electrons
Cations in a sea of
mobile valence
electrons
Source
Metal + nonmetal
Two nonmetals
Metals only
Melting point
Relatively high
Relatively low
Generally high
Solubility
Dissolve best in
water and polar
solutions
Dissolve best in nonpolar solvents
Generally do not
dissolve
Conductivity
Water solutions
conduct
electricity
Solutions conduct
electricity poorly or
not at all
Conduct
electricity well
Other
properties
Strong crystal
lattice
Weak crystal
structure
Metallic
properties; luster,
malleability etc.
17
Bonding Types Are Continuous
• There are no clear
boundaries between
the three types of
bonding.
• Chemical bonding
may be thought of
as a triangle.
• Each vertex
represents one of
the three types of
chemical bonds.
• There are all
degrees of bonding
types between these
extremes.
17
Let’s Practice
31 Which is the best definition of metallic bonding?
A
B
C
D
the attraction between cations and anions
the attraction between cations and delocalized electrons
the attraction between nuclei and electron pairs
the attraction between nuclei and anions
32 Aluminium is a widely used metal. What properties make
it suitable for the following applications?
(a) baking foil (b) aircraft bodywork (c) cooking pans
(d) tent frames
33 Suggest two ways in which some of the properties of
aluminium can be enhanced.
172
173
Lesson 10: HL
Expanded
Octets and
Shapes
Tuesday, January 6
174
14.1 Further Aspects of
Covalent Bonding - HL
Understandings:
• Covalent bonds result from the overlap of atomic orbitals. A sigma bond
(σ) is formed by the direct head-on/end-to-end overlap of atomic
orbitals, resulting in electron density concentrated between the nuclei of
the bonding atoms. A pi bond (π) is formed by the sideways overlap of
atomic orbitals, resulting in electron density above and below the plane
of the nuclei of the bonding atoms.
Guidance:
• The linear combination of atomic orbitals to form molecular orbitals
should be covered in the context of the formation of sigma (σ) and pi (π)
bonds.
• Formal charge (FC) can be used to decide which Lewis (electron dot)
structure is preferred from several. The FC is the charge an atom would
have if all atoms in the molecule had the same electronegativity. FC =
(number of valence electrons) – 1⁄2(number of bonding electrons) –
(number of non-bonding electrons). The Lewis (electron dot) structure
with the atoms having FC values closest to zero is preferred.
• Exceptions to the octet rule include some species having incomplete
octets and expanded octets.
175
14.1 Further Aspects of
Covalent Bonding - HL
Guidance
• Molecular polarities of geometries corresponding to five
and six electron domains should also be covered.
• Delocalization involves electrons that are shared
by/between all atoms in a molecule or ion as opposed to
being localized between a pair of atoms.
• Resonance involves using two or more Lewis (electron
dot) structures to represent a particular molecule or ion. A
resonance structure is one of two or more alternative
Lewis (electron dot) structures for a molecule or ion that
cannot be described fully with one Lewis (electron dot)
structure alone.
176
14.1 Further Aspects of
Covalent Bonding - HL
Applications and skills:
• Prediction whether sigma (σ) or pi (π) bonds are formed from
the linear combination of atomic orbitals.
• Deduction of the Lewis (electron dot) structures of molecules
and ions showing all valence electrons for up to six electron
pairs on each atom.
• Application of FC to ascertain which Lewis (electron dot)
structure is preferred from different Lewis (electron dot)
structures.
• Deduction using VSEPR theory of the electron domain
geometry and molecular geometry with five and six electron
domains and associated bond angles.
• Explanation of the wavelength of light required to dissociate
oxygen and ozone.
• Description of the mechanism of the catalysis of ozone
depletion when catalysed by CFCs and NOx.
177
Expanded Octets
• While the octet is the most common arrangement for
atoms when entering into covalent bonds, there are some
exceptions to the octet rule that include expanded octets
utilizing the d-sublevel
• Since the atom needs to use the d sublevel to expand its
octet, only elements in Row 3 and below can have
expanded octets
• Elements such a sulfur and phosphorus can create
compounds with 5 or 6 electrons.
178
5 Electron Domains
• The electron domain in a 5 electron domain compound,
such as PCl5, will position themselves in a trigonal
bipyramidal shape with bond angles of 90°, 120°, and
180°
• As with the other shapes, the molecular shape will be
slightly different depending on how many domains on
bonded vs. lone pairs
179
Additional Shapes –
5 Electron Domains
• 5 bonded domains – electron configuration triangular
bipyramidal and molecular configuration triangular
bipyramidal
• 4 bonded domains - electron configuration triangular
bipyramidal and molecular configuration unsymmetrical
tetrahedron or see-saw
• 3 bonded domains - electron configuration triangular
bipyramidal and molecular configuration T-shaped
structure
• 2 bonded domains - electron configuration triangular
bipyramidal and molecular configuration linear
180
Shapes
See-saw
T-Shaped
Linear
181
6 Electron Domains
• Molecules with six electron domains will position them in
an octahedral shape with angles of 90°.
182
6 Electron Domains –
Molecular Shapes
• 6 bonded domains – electron configuration
octahedral and molecular configuration
octahedral
• 5 bonded domains – electron configuration
octahedral and molecular configuration
square pyramidal
• 4 bonded domains – electron configuration
octahedral and molecular configuration
square planar
183
Shapes
Square Pyramidal
Square Planar
184
185
186
187
Molecular Polarity
• If there are no lone pairs and all the atoms attached to
the central atom are the same, the molecules are nonpolar as there is no net dipole for 5 and 6 electron domain
shapes.
o For example, PCl5 and SF6 are both non- polar.
• However, here the molecule may be non-polar even with
different atoms bonded to the central atom depending on
HOW they are bonded!
o For example, SBrF5 has a net dipole and is polar; PCl3F2 has no net dipole due to
cancellation, so it is non-polar.
188
Molecular Polarity (cont.)
• Whereas before, the presence of lone pairs always meant
the molecule was polar, if the lone pairs are on opposite
sides of the molecule and the same element is bound to
the central atom, the molecule can be nonpolar!
189
Let’s Practice
190
Answers
191
Lesson 11: HL
Formal Charge
Wednesday, January 7
192
Warm-Up
Pair Share
When we can draw more than one Lewis Dot structure for a
compound, how do we know which one is correct?
193
Sulfur Dioxide
• Let’s take a look at SO2, sulfur dioxide. Knowing that sulfur
can sometimes expand its octet, we can draw the Lewis
Dot structure two ways:
Way 1
Way 2
Which one is correct?
194
Formal Charge
• Formal charge is used to help us determine which
structure is most stable
• Formal charge treats each covalent bond as an equal
electron distribution with each atom “owning” 1 of the
electrons in the bond.
• Each atom also “owns” every electron that is in a lone
pair around it
• When calculating formal charge we count how many
electrons each atom “owns” and compare that to the
original number of valence electrons it started with; it is
generally more favorable if those numbers are the same!
195
Calculating Formal Charge
• The number of valence electrons (V) is determined from
the element’s group in the Periodic Table.
• The number of electrons assigned to an atom in the Lewis
(electron dot) structure is calculated by assuming that:
o (a) each atom has an equal share of a bonding electron pair (one electron per
atom), even if it is a coordinate bond (1⁄2B);
o (b) an atom owns its lone pairs completely (L).
• This means that the number of electrons assigned = 1⁄2
number of electrons in bonded pairs (1⁄2B) + number of
electrons in lone pairs (L)
• So overall:
FC = V – (1⁄2B + L)
196
197
Looking Back at SO2
198
SO2
• We can conclude that structure (ii) where all atoms have
a formal charge of zero is the most stable structure for
SO2.
199
Let’s Practice
• Use the concept of formal charge to determine which of
the following Lewis (electron dot) structures for XeO3 is
preferred?
200
Other Considerations
• In addition to formal charge, when looking at compounds
where multiple Lewis Dot structures can be drawn, we
should also look at electronegativity
• So, a useful guideline to follow is that the most stable of
several Lewis (electron dot) structures is the structure that
has:
• the lowest formal charges and
• negative values for formal charge on the more
electronegative atoms.
• Which is the correct structure here?
201
Let’s Practice
39 Use the concept of formal charge to explain why BF3 is
an exception to the octet rule.
40 Draw two different Lewis (electron dot) structures for
SO42–, one of which obeys the octet rule for all its atoms, the
other which has an octet for S expanded to 12 electrons.
Use formal charges to determine which is the preferred
structure.
202
Answers
203
Lesson 12:
Sigma and Pi
Bonds
Thursday, January 8
204
Molecular Orbitals
• When atoms covalently bond together, the s and p
electron orbitals actually change and hybridize in order to
form molecular or bonding orbitals
• The molecular orbital is formed from one electron from
each atom overlapping and combining in an orbital with
lower energy than the two atomic orbitals where they
started out
205
Sigma Bonds
• When two atomic orbitals overlap along the bond axis –
an imaginary line between the two nuclei – the bond is
described as a sigma bond, denoted using the Greek
letter σ.
• This type of bond forms by the overlap of s orbitals, p
orbitals, and hybrid orbitals (to be described in the next
section) in different combinations.
• It is always the bond that forms in a single covalent
bond!!!
206
Sigma Bond Picture
207
Sigma Bonds
• s orbital
overlap
• p orbital
overlap
• s and p
overlap
Sigma Bonds (σ) bond
• All single bonds are sigma
• In order to be a sigma bond, the bond must overlap
along the bond axis.
Pi Bonds
• When two p orbitals overlap sideways, the electron
density of the molecular orbital is concentrated in two
regions, above and below the plane of the bond axis.
• This is known as a pi bond, denoted using the Greek letter
π.
• This type of bond only forms by the overlap of p orbitals
alongside the formation of a sigma bond.
• In other words, pi bonds only form within a double bond or
a triple bond.
210
Pi Bond Picture
211
pi (π) bond
• Two p orbitals overlap sideways
• Electron density becomes concentrated in two regions.
Where?
pi (π) bond
• Pi bonds can only form alongside formation of a sigma
bond.
• Pi bonds therefore only form within a double or triple
bond.
Pi Bonds vs. Sigma Bonds
• Pi bonds are weaker than sigma bonds, as their electron
density is further from the nucleus.
Lesson 13:
Hybridization
Friday, January 9
215
14.2 Hybridization
Understandings:
• A hybrid orbital results from the mixing of different types of
atomic orbitals on the same atom.
Applications and skills:
• Explanation of the formation of sp3, sp2, and sp hybrid
orbitals in methane, ethene, and ethyne.
• Identification and explanation of the relationships
between Lewis (electron dot) structures, electron
domains, molecular geometries, and types of
hybridization.
Guidance
• Students need only consider species with sp3, sp2, and sp
hybridization.
216
Hybridization
• We know that both s and p atomic orbitals take part in
covalent bonding
• However, they are (a) of different energy levels with s
being lower in energy and (b) have very different shapes.
This would lead to uneven bonds for an atom that can
make four bonds! So what happens?
217
Closer Look: Carbon
• Carbon starts with 2 electrons in the 2s sublevel and 2
electrons in the 2p sublevel. But we know that they need
to form 4 covalent bonds to get a stable octet so
something has to happen to these orbitals to get each
electron into an orbital alone!
• Step 1: A process known as excitation occurs in which an
electron is promoted within the atom from the 2s orbital to
the vacant 2p orbital. Now each atomic orbital has 1
electron and can bond with another atom.
• Step 2: Atomic orbitals hybridize to form four bonding
orbitals each with the same amount of energy called sp3
orbitals
218
219
Hybridization
• The details of hybridization are complex and depend on
an understanding of quantum mechanics, but in essence
unequal atomic orbitals within an atom mix to form new
hybrid atomic orbitals which are the same as each other,
but different from the original orbitals.
• This mixing of orbitals is known as hybridization.
• Hybrid orbitals have different energies, shapes, and
orientation in space from their parent orbitals and are
able to form stronger bonds by allowing for greater
overlap.
• You only need to know sp, sp2 and sp3 hybrid orbitals for
IB but know that there are also sp3d and sp3d2
220
Advanced Work
• We are only going to learn the valence bond theory and
not the more robust molecular orbital theory
• In reality, the hybridization is much more complex and
when orbitals overlap they interfere constructively forming
bonding orbitals and destructively forming anti-bonding
orbitals
• All this is way beyond even the scope of IB but may be
required in advance Chemistry courses in college!
221
S and P Hybrid Orbitals
• Hybridization of one s orbital with three p orbitals
produces four so-called sp3 hybrid orbitals that are equal
to each other. Their shape and energy have properties of
both s and p, but are more like p than s.
• Hybridization of one s orbital with two p orbitals will
produce three equal sp2 hybrid orbitals. In this case, one p
orbital is left unhybridized and can participate in pi
bonding. You will see this when there are double bonds.
• Finally, hybridization of one s orbital with one p orbital will
produce two equal sp hybrid orbitals. In this case, two p
orbitals are left unhybridized and can participate in pi
bonding. You will see this when there are triple bonds.
222
sp3 hybridization
• When carbon forms four single bonds, it undergoes sp3
hybridization, producing four equal orbitals.
• These orbitals orientate themselves at 109.5°, forming a
tetrahedron. Each hybrid orbital overlaps with the atomic
orbital of another atom forming four sigma bonds.
• For example, methane, CH4.
223
224
sp2 hybridization
• When carbon forms a double bond, it undergoes sp2
hybridization, producing three equal orbitals.
• These orbitals orientate themselves at 120°, forming a
triangular planar shape. Each hybrid orbital overlaps with
a neighboring atomic orbital, forming three sigma bonds.
• For example, ethene, C2H4.
225
226
sp hybridization
• When carbon forms a triple bond, it undergoes sp
hybridization, producing two equal orbitals.
• These orbitals orientate themselves at 180°, giving a linear
shape. Overlap of the two hybrid orbitals with other
atomic orbitals forms two sigma bonds.
• For example, ethyne, C2H2.
227
228
Triple Bonds
• In a triple bond, each carbon has two unhybridized p
orbitals that are orientated at 90° to each other. As these
overlap each other sideways, two pi bonds form
representing four lobes of electron density around the
atoms.
• These coalesce into a cylinder of negative charge around
the atom, making the molecule susceptible to attack by
electrophilic reagents (those that are attracted to
electron-dense regions).
229
Expanded Octets
• If an atom has 5 electron domains, one d orbital gets
involved and the hybridization is sp3d which produces five
equivalent orbitals orientated to the corners of a
triangular bipyramid.
• If an atom has 6 electron domains, two d orbitals get
involved and the hybridization is sp3d2 which produces six
equivalent orbitals orientated to the corners of a
octahedral.
230
Lone Pairs
• Lone pairs DO get counted in the electron hybridization
• For example, in ammonia, NH3, the non-bonding pair on
the N atom resides in a sp3 hybrid orbital.
231
Molecule Shapes
• Although we have focused mostly here on examples from
organic chemistry (those involving carbon), this concept
of hybridization can be used to explain the shape of any
molecule.
• And conversely the shape of a molecule can be used to
determine the type of hybridization that has occurred. The
relationship is as follows:
• tetrahedral arrangement ↔ sp3 hybridized
• triangular planar arrangement ↔ sp2 hybridized
• linear arrangement ↔ sp hybridized
232
233
Let’s Practice
• Urea (see below) is present in solution in animal urine.
What is the hybridization of C and N in the molecule, and
what are the approximate bond angles?
234
235
More Practice
236
Even More Fun!
237
238
Lesson 15:
Hybridization
and Stuctures
Tuesday, January 13, 2015
239
Looking Back – Delocalized
Electrons
Now that we have learned about hybridization, we are
going to take a look back at some of the more advance
chemical topics.
Resonance
• Now we can see that resonance generally occurs when
there is sp2 or sp3 hybridization
• The sigma bonds form in the same plane whereas the
unhybridized p-orbitals
240
Nitrate Ion - Example
• NO3+
241
242
Benzene
• In Benzene, all of the Carbon atoms have sp2
hybridization and the unhybridized p orbitals form pi
bonds above and below the bonding plane
• These electrons become delocalized and are shared
among all the carbon atoms!
243
Lesson 16:
Ozone
Wednesday, January 14, 2015
244
Lesson 18:
Review
Thursday, January 15, 2014
245
Lesson 18:
Review
Friday, January 16, 2015
246