Unit: Atomic and Molecular Structure
Unit Review Part B Key
Part E: Valence Bond Theory and Orbital Hybridization
1. Consider a central atom with hybrid sp3d orbitals.
a) What type of orbitals and how many of each type combine to make sp 3d hybridized orbitals?
b)
c)
One s and three p orbitals combine to make four equivalent sp3 hybridized orbitals to make 4 equivalent single,
sigma bonds.
How many sp3d orbitals exist on the atom? Four (see the answer above.)
What VSEPR formula and shape would be characteristic of a molecule with a central atom exhibiting sp 3d
hybridization if the central atom had
i) no lone pairs? Tetrahedral, AX4
ii) 1 lone pair? Trigonal pyramidal, AX3E
Lone pair
iii) 2 lone pairs? Bent, AX2E2
d)
e)
iv) 3 lone pairs? Linear, AXE3
Underneath each of your answers to part c) above, draw a 3-D diagram to represent each of the molecular
shapes. (See table above.)
Why do all of the central atoms in all of the molecules in part c) have the same orbital hybridization? All of
the central atoms need to be able to accommodate four sets of electrons around them so they will all need four
equivalent orbitals to house them. This means all of the central atoms from #3 need sp 3 hybridization.
2. There are five major VSEPR shapes from which the rest are derived when one or more lone pairs are added to
the central atom. These five major shapes are: linear, trigonal planar, tetrahedral, trigonal bipyramidal and
octahedral. Each of these five major VSEPR shapes has its own characteristic type of hybridization on the
central atom and characteristic bond angles. Match each of the five major shapes below with the name of the
shape of the hybrid orbitals on the central atom.
The Five Major VSEPR Shapes
Orbital Hybridization
A.
B.
__D__ sp3d
__E__ sp
__B__ sp3d2
Trigonal planar
C.
octahedral
__A__ sp2
D.
__C__ sp3
Trigonal bipyramidal
tetrahedral
E.
linear
3.
In a general sense, what effect do lone pairs on a central atom have on the bond angles of trigonal planar, tetrahedral, trigonal bipyramidal
and octahedral VSEPR shapes? Explain why the effect occurs. Lone pairs decrease bond angles because they repel one another
and bonding pairs more than bonding pairs repel one another.
4. Label each of the bonds in the following molecule as a sigma bond, σ,or as a pi bond,π. Label each carbon atom with the hybridization its
orbitals exhibit.
One of these bonds is σ and the
The carbon on the far left needs to make four equivalent σ bonds so it is
2
other one is π.
sp3 hybridized. Its neighbor has only 3 atoms attached to its three σ
2
5. Describe
twoC atom
differences
between
bonds are made
because the
is sp hybridized
and sigma
the left and
over ppi bonds.
One of these bonds is σ and the
3
orbital makes the π bond. The carbon up top has the same hybridization
other two are π.
as the second carbon from the left. The last two carbons on the right are
both sp hybridized. They need to be able to make two σ bonds with
these hybrid orbitals and have two p orbitals left to make the two π
bonds.
CH
HC
C
C
CH
5. Describe two differences between sigma and pi bonds. (See page 236.)
Sigma bonds make single bonds. Pi bonds are involved in multiple bonds.
Sigma bonds are end-to-end overlaps of orbitals. Pi bonds are side-to-side overlaps.


6. Define the terms “hybridization of orbitals” and “hybrid orbitals”. (See page 234.) When and why does hybridization
occur? (Read page 233, “Hybrid Orbitals”.)
Part F: Lewis Structures, VSEPR Theory, Molecular Polarity (There is another page to the answer key. The answers for #1
and #2. are there.
3. Textbook questions: #36, 37, 38 page 287 (Answers on page 816.)
Part G: Intermolecular Forces
1.
2.
3.
Explain the difference between intermolecular forces and intramolecular forces. Intramolecular forces are covalent
bonds within a molecule. They are the bonds within the molecule that hold the atoms of the molecule together.
Sometimes ionic bonds are considered intramolecular bonds, too, however ionic compounds are generally not molecules.
Intermolecular forces are the forces of attraction between neighboring molecules.
List the 3 different types of intermolecular forces and distinguish between them. The three types of intermolecular
forces are London forces, dipole-dipole forces and a special type of dipole-dipole force called hydrogen bonding. All are
covered in your grade 11 notes, in the grade 11 review notes given to you in this unit, on your second quiz and on pages
257-264 of the textbook.
How are intermolecular forces linked to the properties of molecules? Again – this stuff can be found in the same places
as the answers for #2. Summary: The stronger the intermolecular forces between neighboring molecules of the same
type, the more energy required to separate the molecules from one another when changing state. So, generally, boiling
points increase with the strength of intermolecular forces. Generally, increased intermolecular forces mean the
molecules involved are more polar. The more polar the molecule, the higher its solubility tends to be in a polar solvent
like water and the less soluble it tends to be in an non-polar solvent like a hydrocarbon (e.g. hexane(l)).
4.
In Part F, question #1 you drew a 3-D diagram for SF6. Add bond dipoles and partial charges to a copy of the 3 – D
diagram. State whether this molecule would be polar or non-polar and explain your reasoning.
The bonds in this molecule are all polar covalent with the central atom, S, slightly
positive compared to each F atom (all F atoms are slightly negative. All bond dipoles
will move from the central S atom toward each F atom.) Because the molecule is
symmetrical (all bonds involving S are to F atoms) and the central atom has no lone
pairs, the bond dipoles will cancel. This molecule will be non-polar.
5.
In Part F, question #1 you drew a 3-D diagram for BrF5. Add bond dipoles and partial charges to a copy of the 3 – D
diagram. State whether this molecule would be polar or non-polar and explain your reasoning.
The bonds in this molecule are all polar covalent with the central atom, Br, slightly
positive compared to each F atom (all F atoms are slightly negative. All bond
dipoles will move from the central Br atom toward each F atom.) In this case, the
central atom has one lone pair so the molecule is no longer symmetrical and the
bond dipoles do not cancel out. So – there is a net bond dipole and the molecule is
considered a polar molecule.
Lone pair
6a)
Write in the bond dipoles and partial charges as necessary on the bonds in the molecules drawn below. State whether
each molecule would be polar or non-polar. Identify all of the intermolecular forces that would exist between neighboring
molecules of the same type in each case.
Molar mass = 46 g/mol
Molar mass = 46 g/mol
Non-
δ-
symmetrical
molecule with
a net dipole
so is a polar
molecule.
O
H
Polar molecule –
same reasoning as
for the first one.
δ-
C
H
Intermolecular forces between neighboring molecules of
this type: London forces and dipole-dipole forces (because
of the polar C=O bond)
H
C
H
N
δ-
H
Also a polar molecule – same
reasoning as above.
O
Intermolecular forces between neighboring molecules of
this type: London forces, dipole-dipole forces and
hydrogen bonding (because of the O-H bond)
δ-
O
δ-
Molar Mass = 45 g/mol
Intermolecular forces
between neighboring
molecules of this type:
6b) Rank the three compounds from part a) from
highest to lowest boiling point. Explain
your answer. These molecules are all about the
London forces, dipole-
neighbors in samples of each molecule are about the
diple and hydrogen
same. (Review the factors affecting the strength of
bonding (because of the
London forces if this does not make sense to you.)
N-H bond within the
So, London force is not likely the deciding factor here
molecule)
in terms of predicting relative boiling points. Likely
same size suggesting that the London forces between
HCOOH will have the highest boiling point because its
O-H bond is more electronegative than the N-H bond
in the amide. They’ll likely be pretty close, though.
The ketone will have the lowest of the three boiling
points because although its does show dipole-dipole
attraction between neighbors, there is no hydrogen
bonding.
7. Textbook questions: page 266 #1-4; page 281 #18,
7. Textbook questions: page 266 #1-4 (See answers below.) ; page 281 #18, 20 (Answers in back of text.) ; page 289 #27 (an
all-inclusive question) (See answers below.)
Page 289 #27