Chemistry 103
Chapters 6&7 (Chemical Bonding and Molecular Structure)
Class Notes
Suggested Reading: All of Chapters 6 and 7 (We will be moving back-and-forth between
these two chapters. Use this notes packet as a guide to the sequence.)
I. Quick Review of Ionic Bonding
-
Bonds are formed by electrostatic attraction between ions of opposite charge.
-
Ionic compounds result from metal/nonmetal combinations and are electrically
neutral. (Charges must balance.)
-
Anion = negative ion;
-
Sizes of ions:
-
Formula Grid:
Cation = positive ion
Anion = larger than neutral atom
Cation = smaller than neutral atom
M+
M2+
M3+
X‾
MX
MX2
MX3
X2‾
M2X
MX
M2X3
X3‾
M3X
M3X2
MX
II. Lattice Energy – Heats of Formation for Ionic Compounds
-
When an ionic compound is formed from the elements of which it is composed, the
reaction is always exothermic!
-
Reminders:
-
If a formation reaction is exothermic, where does the released heat come from? To
illustrate how this works, we will use the formation of NaCl from its elements.
1)
endothermic = heat is absorbed
exothermic = heat is released
enthalpy = heat energy
Elemental substances (in stable states at 25ºC) are converted to gases:
Na (s) +
½ Cl2 (g) →
Na (g) + Cl (g)
∆H = + 230 kJ
(Note that this is an endothermic step.)
6/7-2
2)
Electron transfer occurs between gaseous atoms to form ions:
Na (g) +
3)
Cl (g) → Na+(g)
+
Cl‾(g)
∆H = + 147 kJ
The gaseous ions combine to form the ionic solid:
Na+(g)
+ Cl‾(g) → NaCl (s)
∆H = – 788 kJ
(Note that this is a highly exothermic step.)
-
Now we can use Hess’ Law to find the equation and ∆H for the overall process:
(1)
(2)
Na (s) + ½ Cl2 (g) → Na (g) + Cl (g)
Na (g) + Cl (g) → Na+(g)
+ Cl‾(g)
∆H = + 230 kJ
∆H = + 147 kJ
(3)
Na+(g) + Cl‾(g) → NaCl (s)
∆H = – 788 kJ
----------------------------------------------------------------------------------------------Na (s) + ½ Cl2 (g) → NaCl (s)
∆H = – 411 kJ
-
A similar process could be repeated for any ionic compound that is formed from its
elements, and it will always be negative!
-
The 3rd step in the example above illustrates a chemical concept called ―lattice
energy‖:
-
Lattice energy = the energy (enthalpy) change when a crystal lattice is formed
from gaseous ions.
-
Lattice energy is always a negative quantity. (Energy is given off during
formation, indicating that the crystal is more stable than the individual atoms/ions.
If it weren’t, the crystal lattice wouldn’t exist.)
-
Lattice energy is a measure of the strength of the ionic bond.
-
In general, the strength of ionic bonds increases with:
a) smaller ions
b) larger charge
6/7-3
III. Nature of the Covalent Bond
Covalent bond = a chemical bond produced by shared electrons in the region
between two atoms. The covalent bond forms because of mutual attraction by the two
nuclei for the shared electrons.
-
London-Heitler Model for H2: (Sketch Diagram)
_____ Attractive Forces
_____ Repulsive Forces
-
Note that this creates an ―overlap orbital‖
A. Representing Covalent Bonds
H–H
Lewis structure (In this case, it is also a structural formula)
H:H
Dot diagram
-
Remember that we should think of the shared electrons as being in positions of high
probability, and not having fixed positions.
-
Sharing electrons gives greater stability! A bond forms (and continues to exist)
because it is easier in terms of energy! “It is easier to live together than it is apart.”
-
By sharing we can increase electron totals to stabilize orbitals. With main-group
elements, this would give them a Noble gas configuration.
H ↑↓ H
← By sharing , each hydrogen has a helium configuration.
B. Bond Length (See Handout)
-
The bond length is determined by ―least energy‖. (The place where if is easiest to exist.)
6/7-4
C. Lewis Structures
-
Lewis structures show the distribution of outer (valence) electrons in an atom,
molecule, or polyatomic ion. Unshared valence electrons are shown as dots, bonds
(the shared electron pair) are shown as straight lines.
Comparison Table (Add appropriate “dots”)
Product
Reactants
Lewis Structure
“Dot Diagram”
H
2H
-
+
+
F
→
O →
H–F
H F
O–H
O H
H
H
HOW TO DRAW LEWIS STRUCTURES
1) Count the valence electrons available. (The number of valence electrons
contributed by a non-metal atom is equal to its ―Roman Numeral Group‖ number
on the Periodic Table.) Important: Add or subtract electrons to account for charge.
OCl‾ ion:
CH3OH molecule:
SO32‾ ion:
2) Draw a skeleton structure using single bonds. Note: Carbon almost always
forms four bonds. Also, oxygen atoms are usually bonded to a central atom rather
than to each other.
Space for skeleton structures:
6/7-5
3) Deduct two electrons for each single bond in the skeleton structure:
OCl‾ ion:
CH3OH molecule:
SO32‾ ion:
4) Distribute the remaining electrons to give appropriate atoms an octet (this
gives the atoms in Groups IV, V, VI, and VII a noble gas configuration).
Final Lewis structures:
Extra Hints:
-
Atoms with higher electronegativities are usually ―terminal‖. (The closer an atom
is to the middle of the Periodic Table, the closer it usually is to the middle of the
molecule.)
-
Hydrogen is a terminal atom.
-
Use multiple bonds (double or triple) to get the noble gas structure (octet), if
needed.
Ex. #1: Give the Lewis Structure for the nitrate ion, NO3‾
6/7-6
Ex. #2: Give the Lewis Structure for each of the following molecules and ions
a) N2
b) NOCl
c) ClO4‾
d) PH3
e) NO2‾
f) NH4+
g) H2NNF2
h) CN‾
i) HOCO2‾
j) N2O
k) N2O2
l) HOC+
(an “oddity)
6/7-7
IV. Resonance Forms
-
Back to the NO3‾ example: (Bottom of page 5)
-
Actual experimental evidence shows that the bond lengths are all equal in the
nitrate ion! This wouldn’t be true if it were a conventional double bond.
-
Three possible structures could be drawn: (Add appropriate “dots”)
O–N–O
–
O=N–O
–
O–N=O
║
│
│
O
O
O
–
-
It looks as if all three structures are the same. However, each oxygen atom could
be ―tagged‖ as a different isotope of oxygen. Then, (for the sake of argument) we
could say there are three different structures!
-
It used to be thought that the double bond ―resonated‖ within the structure. In the
nitrate example, we could imagine the double bond rotating from one N – O bond
to the next. The name resonance structure has stuck, even though science no longer
believes that this is what actually happens.
-
It is more likely that the electron pair that forms the
―double bond‖ is distributed throughout the molecule
or polyatomic ion: (We will talk more about this later
in the chapter.)
O–N–O
–
│
O
More Notes on Resonance Structures:
-
Resonance can be expected when it is possible to draw more that one structure that
follows the octet rule.
-
Resonance forms are obtained by moving electrons, not atoms! (If you move
atoms, you have made isomers.)
-
A double bond can sometimes be written as a single-and-triple combination.
-
Resonance structures are still useful as convenient models, which is why we still
learn how to do them.
6/7-8
Ex. #3: Give all possible resonance forms for each of the following:
a)
(Add appropriate “dots”)
SO32‾
O=S–O
2-
│
O
↔
O–S=O
│
O
2-
O–S–O
↔
║
O
b) SeO2
c)
CO2
d) C2O42‾
(Hint: four possible structures)
V. Exceptions to the Octet Rule
A. Odd-Electron Species (Free Radicals)
Examples: (These are actual molecules, but they have unpaired electrons!)
NO (11 valence electrons)
NO2 (17 valence electrons)
O
N=O
N
O
Note: Experiments confirm that these molecules are paramagnetic! Confirm this by
adding dots for the remaining valence electrons.
2-
6/7-9
B. Central Atom Surrounded by Less than 4 Pairs of Electrons
In Groups 2 (II) and 13 (III), there just aren’t enough electrons to follow the octet
rule. This does not make them unstable!
Two Classic Examples: (Once again, add dots for the remaining valence electrons)
F – Be – F
F
F
B
│
F
C. Central Atom Surrounded by More than 4 Pairs of Electrons
Examples:
PCl5
Note: We will study ―expanded octets‖ a little later in this unit.
6/7-10
SUMMARY OF “HINTS” FOR LEWIS STRUCTURES
1. Atoms usually have an ―octet‖ of electrons around them in a correct Lewis structure.
Exceptions:
a) Atoms in Groups 1 (I), 2 (II), and 13 (III)
b) Free Radicals
c) Expanded Octets
2. Use multiple bonds when necessary.
3. Multiple bonds usually form with the least electronegative atoms. For example, a
multiple bond is more likely between C and N than between C and O. However, keep
in mind that only a few elements form multiple bonds. (See Hint #4)
4. Only a few elements commonly form multiple bonds: C, O, N, S, and P.
(These also happen to be the elements associated with ―life‖ as we know it.)
On rare occasions, the following can also form multiple bonds: Si, Se, Te, Cl,
and As.
5. The atom closest to the middle of the Periodic Table (lowest electronegativity) is
usually closest to the middle of the molecule or polyatomic ion.
6. Atoms with higher electronegativities are usually terminal.
7. After drawing the skeleton structure, begin placing needed electrons on the most
electronegative atoms.
8. Always consider nature’s tendency to form symmetrical molecules and polyatomic
ions.
9. Carbon almost always forms four bonds.
10. Oxygen atoms are normally bonded to a central atom rather than to each other.
11. Hydrogen is always terminal
12. Resonance forms are obtained by moving electrons, not atoms. (If you move atoms,
you have made isomers.)
6/7-11
VI. Formal Charge
-
―Formal Charge‖ can be used to help you decide between two or more possible
Lewis structures.
-
Formal Charge = the number of valence electrons in a given atom minus the
number of electrons assigned to that atom in the Lewis structure.
FC = EV – ½ EB – EU
EV = valence electrons
EB = shared electrons
EU = unshared electrons
-
Ex. #4: Which of these is the most likely structure?
H – C ≡ N:
or
H – N ≡ C:
6/7-12
-
“Rules” for Formal Charges
(1) Usually, the most plausible Lewis structure is the one with no formal
charges (that is, with formal charges of zero on all atoms).
(2) Where formal charges are required, these should be a small as possible.
(3) Negative formal charges should appear on the most electronegative atoms.
(4) Formal charges on the atoms in a Lewis structure must total to zero for a
neutral molecule, or to the net charge on a polyatomic ion.
-
Follow-Up
.. .. ..
: Cl – O – S :
·· │ ··
: Cl :
··
Or?
.. .. ..
: Cl – S – O :
·· │ ··
: Cl :
··
-
WARNING: There are always exceptions to every ―rule‖ when we talk about
―mother nature‖ (and therefore chemistry). ―Formal charge‖ is not a foolproof
method for deciding on Lewis structures.
-
MORAL: It may be best to rely on your experience and your understanding of the
―hints‖ on page 6/7-10. Use formal charge in ―trouble spots‖, but keep in mind that
there are some exceptions!
6/7-13
-
Practice with Formal Charge:
Ex. #5: Which of the following is the correct structure for NOCl?
a)
.. .. ..
N = O – Cl :
··
··
FC N =
FC O =
FC Cl =
b)
.. .. ..
O = N – Cl :
··
··
FC N =
FC O =
FC Cl =
c)
.. .. ..
: O – Cl = N
··
··
FC N =
FC O =
FC Cl =
Best Structure: _________
VII. Covalent Bond Properties
A. Bond Polarity (Remember that ―bond polarity‖ means between two atoms.)
-
Two Types of Polarity
(1) Bond Polarity = between two atoms
(Now)
Note: This is based on electronegativity difference.
(2) Molecular Polarity = whole molecule (Later)
Note: This is based on overall symmetry.
6/7-14
-
Some Definitions:
electronegativity = measure of an atom’s ability to attract an electron pair to itself
in a bonding situation. (Compare to ―electron affinity‖.)
See “Electronegativity Table” or your SW Periodic Table
ionic character = extent (percent) to which a bond resembles an ionic bond
electric dipole = formation of positive and negative poles, but has an overall net
charge of zero.
nonpolar bond = no electric dipole is produced. (The electron density around the
two atoms is symmetrical.)
WARNING: Just because the word ―symmetrical‖ is used here, do NOT confuse with molecular polarity,
which is based on the overall symmetry of the whole molecule.
polar bond = an electric dipole is produced. (The electron density is
asymmetrical.)
-
Reminder: What do we know about bonding?
Ionic Bonding = electron(s) transferred to create positive and negative ions
and the attraction of oppositely charged ions forms the bond.
Covalent Bonding = bond is formed by sharing electrons
Nonpolar Covalent = equal sharing (a dipole is NOT formed)
Polar Covalent = unequal sharing (a dipole is formed)
-
Comparison Sketches:
Nonpolar Covalent
Polar Covalent
δ+
δ–
covalent bonding
polar bonds
Ionic
+
–
6/7-15
-
Where are the boundaries? How do we make distinctions between the three types
of bonds?
We use ―electronegativity difference‖. The difference in
electronegativity compares the ―pulling power‖ of two atoms. The greater the
difference, the more polar the bond.
ANSWER:
-
Ideal Situations:
Perfect Nonpolar Covalent Bond = This would happen between two
identical atoms, and the electronegativity difference would be zero.
Such as: H – H, Cl – Cl, or O = O
Perfect Ionic Bond = electron is completely transferred from one atom to
another. (In other words, there is absolutely no sharing.)
-
Making Distinctions Between Categories
Type
Nonpolar Covalent
Polar Covalent
Ionic
Electronegativity
Difference
< 0.5
0.5 – 1.7
> 1.7
Percent Ionic
Character
< 5%
5% – 50%
> 50%
Note: A value of 1.7 represents only 50% ionic character. However, this is the ―cutoff‖ usually used because combinations of Group 1 or 2 metals with oxygen or fluorine
usually yield a difference greater than 1.7, an these combinations are considered to be
―ionic‖. (They act as if they are ionic bonds.)
Suggestion: Label the “cut-off” points on your SW Periodic Table.
Exceptions: Although the table above is convenient, there are some exceptions.
a) C – H = 2.5 – 2.1 = 0.4 = nonpolar (most organic chemists would consider
this to be ―slightly polar‖).
b) H – F = 4.0 – 2.1 = 1.9 = ionic (because of hydrogen, which is not
considered to be a metal, this is usually described as ―strongly polar‖.)
6/7-16
Ex. #6: Complete the following Table:
Bond
Electronegativity
Difference
Percent Ionic
Character
Bond Polarity
(―nonpolar covalent‖, ―polar covalent‖, or ―ionic‖)
C–O
Li – O
Rh – O
C–S
Reminder: Bond polarity is determined by ―electron greed‖.
B. Estimating Bond Lengths
-
Reminder: Bond lengths are measured from the centers of atoms!
-
Distances for Multiple Bonds:
a)
Multiple bonds have shorter bond lengths than their corresponding single
bonds. (Why?)
b) Compare the bond lengths of C – C, C = C, and C ≡ C on your reference table.
-
To get quick estimates of bond lengths:
a)
Draw the Lewis structure of the molecule (if necessary) to identify single vs.
multiple bonds.
b) Take the sum of the appropriate covalent radii, as obtained from your Periodic
table or an appropriate reference table.
Ex. #7: Determine the bond length between Zn and S in ZnS.
(Use your Periodic Table.)
6/7-17
Ex. #8: Estimate all the bond lengths in a molecule of formic acid.
Bond Length
Periodic Table
Bond Table
Bond
(Use Table values.)
Structural Formula
C–H
O
H –– C
C=O
O –– H
C–O
O–H
Note Difference (Why?)
C. Bond Energies
-
Bond energy (bond enthalpy) = the energy (ΔH) needed to break a bond (in the
gas phase) Note: This is usually a “per mole” unit.
-
The bond energy is a direct measure of bond strength! (The higher the bond
energy, the stronger the bond.)
-
Examples:
H2 (g) → 2 H (g)
ΔH = +436 kJ
Cl2 (g) → 2 Cl (g)
ΔH = +243 kJ
-
The examples above show that the bond between hydrogen atoms in H2 is stronger
than the bond between chlorine atoms in Cl2.
-
Two Factors Can Affect Bond Energy
1) Number of Electron Pairs
-
The more electron pairs shared, the stronger the bond. So, triple bonds
are stronger than double bonds, and double bonds are stronger than
single bonds. (See “Bond Energy Tables” on Blue Sheet)
-
A rough estimate is that the bond energy for a double bond is about
twice the bond energy of a single bond, and the bond energy of a triple
bond is about three times the bond energy of a single bond. (See “Bond
Energy Tables” on Blue Sheet)
6/7-18
2) Polarity
-
The greater the difference in electronegativity (the more polar the bond),
the greater the ―extra‖ bond energy.
-
In other words, electrostatic attractions add to the ―normal‖ amount of
energy that would be needed to break the bond.
-
Example: The B.E.H – F is greater than the average of B.E.H – H and
B.E.F – F .
D. Using Bond Energy to Estimate ∆H for a Reaction
-
IMPORTANT:
-
Steps:
This only works if all reactants and products are in gaseous state!
1) Break necessary bonds of reactants.
∆H1 = ∑ B.E.Reactants
2) Form necessary bonds for products.
∆H2 = – ∑ B.E.Products
3) Use Hess’ Law to get estimate of ∆H
∆H = ∆H1 + ∆H2
Ex. #9: Estimate ∆H for: H2O (g) +
Cl2 (g) → HOCl (g) + HCl (g)
First determine what actually happens:
H–O–H
+
Cl – Cl
→
H – O – Cl
+
H – Cl
6/7-19
Ex. #10: Estimate ∆H for: CH3OH (g) + HBr (g) → CH3Br (g) + H2O (g)
-
Why are the ∆H values we have found only ―estimates’?
The bond energies on you table are an average for a variety of molecules
where the particular bond occurs. The amount of energy required to break a bond
also depends on the composition of the rest of the molecule.
ANSWER:
For example: Breaking an O – H bond would require slightly different amounts of
energy for:
Cl
│
Cl – C – O – H
and
H–O–H
WHY?
│
Cl
-
Answer: Since the bond energies we used for calculating ∆H are only averages, our
determination can only be an estimate.
VIII. Molecular Geometry
-
Understanding Lewis Structures can help us with:
a)
Shapes of Molecules
 bond angles
 orientation around the central atom
 polarity
b) Valence Electrons
 bonding versus nonbonding electrons
 which orbitals are occupied
6/7–20
-
VSEPR Theory (Valence Shell Electron Pair Repulsion)
= electron pairs surrounding an atom, repel each other and will tend to be as far
from each other as possible. (This is the underlying concept we used for predicting
shapes in the first-year chemistry course.)
-
Quick Review of ―Architecture‖ from your First-Year Course:
Example
Lewis Structure
Structural Formula
Bond
Type
Shape
BeF2
..
..
: F – Be – F :
F – Be – F
sp
linear
sp3
bent
(or angular)
∙∙
H2O
∙∙
..
H–O–H
∙∙
BF3
..
..
:F–B–F:
∙∙
O
H
H
F
F
B
│
F
∙∙
:F:
∙∙
NH3
..
H–N–H
│
H
H
sp3
pyramidal
3
(p )
H
H
│
│
C
H
(planar
triangular)
H
H–C–H
│
sp2
trigonal
planar
N
H
CH4
(p2)
H
sp3
H
H
tetrahedral
6/7-21
-
Summary Table from First-Year Course:
1 (I)
18 (VIII)
H
2 (II)
13 (III)
14 (IV)
15 (V)
(filler)
Be
B
C
N
O
F
bent
(filler)
sp3
(p2)
sp3
(p)
~~~~~~
(s)
linear
trigonal
sp
sp2
tetrahedral pyramidal
sp3
sp3
(p3)
He
16 (VI) 17 (VII)
(No
Bonds)
A. Hybridization and Geometry
-
We will now take a closer look at the ―classic‖ examples.
- Some Terms:
electron promotion = shift of an electron within an energy level to maximize
bonding capacity.
hybridization = re-shaping of atomic orbitals to make orbitals of equal energy
BeF2
4Be
=
↑↓
↑↓
1s
2s
↑↓
↑
1s
__ __ __
(ground state)
2p
↑
2s
__ __
(after electron promotion)
2p
The orbital notation after electron promotion suggests a shape such as:
This would suggest that the Be–F bond lengths are different
in the BeF2 molecule. However, actual experimental
evidence tells us the bond lengths are the same!
What actually happens:
Result:
↑↓
↑ ↑
__ __
1s
sp
2p
(after hybridization)
Linear Shape = sp bonding = 180º bond angle
6/7-22
BF3
5B
=
↑↓
↑↓
↑ __ __
1s
2s
2p
↑↓
↑
1s
(ground state)
↑ ↑ __
2s
(after electron promotion)
2p
↑↓
↑ ↑ ↑
__
1s
sp2
2p
(after hybridization)
What actually happens:
Trigonal Planar Shape = sp2 bonding = 120º bond angle
CH4
6C
=
↑↓
↑↓
↑ ↑ __
1s
2s
2p
↑↓
↑
↑ ↑ ↑
1s
2s
(ground state)
(after electron promotion)
2p
↑↓
↑ ↑ ↑ ↑
1s
sp3
(after hybridization)
What actually happens:
Tetrahedral Shape = sp3 bonding = 109.5º bond angle
H2O
8O
=
↑↓
↑↓
↑↓ ↑ ↑
1s
2s
2p
(ground state)
Your first-year may have called this ―p2 bonding‖, which would imply 90º bond angles
since p orbitals are aligned along the axes. Actual experimental data indicates that the
actual bond angle is approximately 109.5º, which suggests sp3 bonding!
8O
=
↑↓
↑↓ ↑↓ ↑ ↑
1s
sp 3
(after hybridization)
6/7-23
Note: Even though this is a tetrahedral ―base structure‖,
it still results in a ―bent‖ shape. The unshared electron pairs
act like ―bond sites‖.
(Sketch of H2O)
NH3
7N
=
↑↓
1s
↑↓
↑ ↑ ↑
2s
(ground state)
2p
Your first-year may have called this ―p3 bonding‖, which would again imply 90º bond
angles. However, actual experimental data indicates that the bond angle is again close
to 109.5º, which suggests sp3 bonding!
7N
=
↑↓
↑↓ ↑ ↑ ↑
1s
sp 3
(after hybridization)
Note: Even though this is another tetrahedral ―base structure‖,
it results in a ―pyramidal‖ shape. The unshared electron pairs
are acting like ―bond sites‖.
(Sketch of NH3)
-
SEE SUMMARY TABLE ON NEXT PAGE
-
Table Notes:
X = central atom
Y = attached atom
E = unshared electron pair
Revised Summary Table
I (1)
6/7-24
II (2)
III (13)
IV (14)
V (15)
VI (16)
VII (17)
VIII (18)
sp3
sp3
―sp ‖
―sp ‖
180º
N/A
―Linear‖
N/A
Dot
Structure
Orbital
Picture
(Transition
Metals)
Bond
Type
―s‖
sp
sp2
sp3
Bond
Angle
180º
180º
120º
109.5º
<109.5º <109.5º
―Linear‖
Linear
Trigonal
Planar
Tetrahedral
Pyramid
Bent
3
3
Example
Shape
(―Two Points‖)
(―Two Points‖)
Species
Type
XY
XY2
XY3
XY4
XY3E
XY2E2
XYE3
XE4
Unshared
Electron
Pairs
0
0
0
0
1
2
3
4
6/7-25
B. Actual Bond Angles
-
Why are the bond angles for H2O (XY2E2) and NH3 (XY3E) actually a little less
than 109.5º?
-
Orbitals where electron pairs are shared get ―stretched‖, since the pair is attracted to
both nuclei.
-
Unshared pairs therefore get pulled closer to the nucleus; this ―flattens‖ out the
orbital.
2 e– in each orbital
- Repulsive forces between the electrons in the flattened
orbitals (E’s) and the electrons in the bonded atoms (Y’s)
push the bonded atoms a little closer than 109.5º.
IX. Hybridization in Molecules Containing Multiple Bonds
A. What about C2H4?
H
C=C
H
Our first assumption might be that this is sp 3
bonding, since that is what carbon atoms are
typically known to do.
H
?
H
However, experimental data indicates that the bond angle is 120º!
(This implies sp2 bonding!)
If there are 120º bond angles, (due to sp3 bonding), then each carbon atom must be:
6C
= ↑↓
1s
↑ ↑ ↑
sp2
Each carbon atom in C2H4 would have the
hybridized sp2 orbitals in the same plane,
―unhybridized‖ p orbital would be perpendicular
to the plane.
↑
2p
(after hybridization)
6/7-26
Examine the orbital notation:
1H
Hydrogens have to be bonded to sp2
orbitals because of the 120º angles.
↑
↑
1H
1s
6C
6C
↑↓
↑
↑
↑
1s
sp2
sp2
sp2
↑↓
1s
1H
↑
1s
.
1s
↑
↑
↑
↑
sp2
sp2
sp2
1H
carbon-carbon bond
.
2p
↑
What about this?
.
2p
↑
.
1s
Drawing it out:
Former p orbitals
form a ―π bond‖
Combined sharing
above and below the
plane constitute one
π bond
-
The p orbitals above and below the plane of the molecule can share (see arrows
above). Clearly, this is a different type of bonding situation, known as: π bonding.
Two Kinds of Bonds:
Sigma Bond (σ) = high electron density along an inter-nuclear axis.
Pi Bond (π) = high electron density perpendicular to an inter-nuclear axis.
6/7-27
Summary:
-
There is only one π bond in C2H4.
-
The extra electron pair(s) in any multiple bond are not hybridized. (In the C2H4
example, we needed 3 sigma bonds for each carbon: 2 for the hydrogen atoms, and
one to bond to the other carbon. Each carbon atom had one electron in the
perpendicular p orbital to share and create the pi bond.)
-
The perpendicular (extra) electron pair(s) do not affect the geometry (shape) of the
molecule! (Nature does not need the sp3 as in CH4, so an extra p electron is
available from each carbon atom in C2H4 to form the pi bond.) In other words,
pi bonds do not change the shape of the molecule!
B. What about C2H2?
H–C ≡ C–H
← Since the bond angle is 180º, each carbon must be
using sp bonding!
Examine the orbital notation:
1H
↑
.
1s
6C
6C
1H
↑↓
↑
↑
1s
sp
sp
↑↓
↑
↑
1s
sp
sp
↑
↑
.
↑
.
2p
↑
2p
↑
1s
(See model or text for illustration of resulting structure.)
C. Summary (important)
Single Bond = one sigma bond
Double Bond = one sigma bond, one pi bond
Triple Bond = one sigma bond, two pi bonds
.
6/7-28
Ex. #10: Give the number of sigma and pi bonds in each of the following:
.. .. ..
..
..
..
+
a) O = N – Cl :
b) O = C = O
c) H – O – H
∙∙
∙∙
∙∙
∙∙
H
σ = ______
σ = ______
σ = ______
π = ______
π = ______
π = ______
H
d)
N ≡ N–O
e)
H
C
C
σ = ______
H
σ = ______
C
π = ______
║
C
π = ______
H
C
C
N–O
║
H
O
Note: The sum of the sigma and pi bonds must equal the total number of bonds!
Predicting Geometry, Bond Angles, and Bond Hybridization
Hybrid.
Angle
Basic Shape
sp
180º
sp2
120º
trigonal planar
3
sp3
109.5º
tetrahedral
4
linear
*Bond Sites
2
*Note: a) Unshared electron pairs DO count as bond sites
b) π bonds DO NOT count as bond sites
6/7-29
Ex. #11: Complete the following table.
Bond
Example
Lewis Structure
Angles
Hybrid.
e – pair
Geometry
Molecular
Geometry
a) AlCl3
b) K2S
c) PH3
d) CaBr2
e) KBr
f) SiF4
------------------------------------------------------------------------------------------------------------g) GeF2
Note: Even though Ge is in Group IV (14), it did not hybridize the way the carbon
family usually does. This is a XY2E species we have not discussed yet.
6/7-30
Ex. #12: Give the indicated bond angle and hybridization for each of the following:
Structure
..
..
a) : Cl – Be – Cl :
∙∙
Hybridization
________
________
________
________
________
________
________
________
________
________
________
________
________
________
________
________
________
________
∙∙
..
..
b) : O – S = O
∙∙
Bond Angle
∙∙
∙∙
..
c) : N ≡ N – O :
∙∙
d)
..
..
:O–N–O:
∙∙
∙∙
:O:
..
e)
:O:
.. | ..
: O – Si – O :
∙∙
|
∙∙
:O:
∙∙
.. .. ..
f) : O – O – O :
∙∙
g)
∙∙
∙∙
H–C ≡ C–H
.. .. ..
h) : O – N = O
∙∙
∙∙
..
:O:
i)
..
H–C=O
│
∙∙
..
j) : Cl
..
: Cl
..
C=O:
________
________
∙∙
..
:O:
k)
.. │ ..
: O – Br – O :
∙∙
∙∙
∙∙
________
________
6/7-31
Ex. #13: For each of the following, give the angle of each of the indicated bonds.
.. ..
a)
:Cl: :Cl:
(1) ________
│
│
(1)
H–C=C–H
(2) ________
.
(2)
b)
H
(1) ________
(3)
│
H–C–C≡ C–H
(1)
(2) ________
│
H
(3) ________
(2)
(6)
c)
H H
│
│ (2)
∙∙
(4)
∙∙
:O:
║
(1) ________
∙∙
H–C–C–C–O–O–N–O:
(1)
│
│
║
∙∙
H H :O:
∙∙
(2) ________
∙∙
(5)
(3) ________
(4) ________
(3)
(5) ________
(6) ________
X. Molecular Polarity (Whole Molecule)
- Molecular Polarity (whether or not the molecule is a dipole) is based on symmetry.
symmetrical = nonpolar
asymmetrical = polar (dipole)
-
Reminder: “bond polarity” is between two atoms, and is determined by electronegativity difference.
A. Automatically Polar (because they can never be symmetrical)
- bent
- pyramid
B. Could be nonpolar (if the attached atoms are all the same)
-
linear
trigonal planar
tetrahedral
6/7-32
-
Examples:
Shape
Nonpolar (Symmetrical)
Polar (Asymmetrical)
Cl2
HCl
BeF2
BeClF
BF3
BeClF2
CH4
CH3Cl
Linear
Linear
Trigonal Planar
Tetrahedral
Ex. #14: Identify each of the following as either polar (dipole) or nonpolar.
a)
H – C ≡ N:
____________
..
+
H–O–H
b)
____________
H
H
..
..
c) : O – S – O :
∙∙
║
_____________________
:O:
H
..
:S = O:
e)
____________
H–N–H
d)
∙∙
+
____________
│
..
..
f) O = C = O
∙∙
____________
∙∙
:O:
∙∙
g)
..
..
: F – Al – F :
∙∙
____________
h)
.. .. ..
: F – Ge – F :
∙∙
∙∙
____________
∙∙
:F:
∙∙
..
..
O=N=O
i)
∙∙
k)
____________
j)
∙∙
H
____________
∙∙
H
C=C
H
..
H–S–H
l)
..
Cl :
H
∙∙
____________
C=C
H
: Cl :
∙∙
H
____________
6/7-33
-
Reminders:
a) Symmetry can make a molecule nonpolar, regardless of what bond polarities are
within the molecule.
b) Experimentally, molecular polarity can be determined by checking a molecule’s
orientation in an electric field, If it is polar, it will align with the field.
XI. Isomers
Isomer = substances that have the same chemical formula, but a different
arrangement of atoms. (The ―Octet Rule‖ can be satisfied in more than one way.)
Compare to:
Resonance = same formula, but different bonding arrangement. (The ―Octet Rule‖
can be satisfied by more than one arrangement of multiple bond(s).)
-
Example: C2H4Br2
Br H
‫׀ ׀‬
H–C–C–H
‫׀ ׀‬
Br H
Two Br on one C
Or?
H H
‫׀ ׀‬
Br – C – C – Br
‫׀ ׀‬
H H
same


H H
‫׀ ׀‬
H–C–C–H
‫׀ ׀‬
Br Br
One Br on each C
└──────┬──────┘
These two structures are not isomers of each other;
only a rotation of the central axis – SEE MODELS
-
Isomers do have different properties. (melting point, density, reactions, etc.)
-
When drawing isomers, it is often acceptable to omit ―terminal‖ hydrogens. This
makes it easier to compare molecules because we are focusing on the skeleton
structures.
6/7-34
Ex. # 15: Consider the following ways of representing the hexane molecule, C6H14.
How many different isomers are shown below?
C–C–C–C–C–C
C
‫׀‬
C–C–C–C–C
C–C–C–C–C
‫׀‬
C
C–C–C
‫׀‬
C–C–C
C–C–C–C–C
‫׀‬
C
C
‫׀‬
C–C–C
‫׀‬
‫׀‬
C
C
C C–C
‫׀ ׀ ׀‬
C–C C
ANSWER:
-
______
Remember: Look for the longest continuous chain. Compare ―branches‖ by
counting in from the ends of the chain.
A. Types of Isomers
1. Structural Isomers = structural formula is different
a) Rearrangement of Atoms (See Above and page 6/7-33)
Example: C4H10 (Both of the structures below have the same chemical
formula, but differ in structure.
H H H H
│
│
│
│
H–C–C–C–C–H
│
│
│
H H H
│
│
│
H–C–C–C–H
│
│
│
H H H H
H
H
H–C–H
│
H
6/7 - 35
b) Movement of a Functional Group
Note: functional group = a group of atoms that determines the ―chemistry‖
of the molecule.
H H H
‫׀ ׀ ׀‬
H – C – C – C – OH
‫׀ ׀ ׀‬
H H H
H H
‫׀‬
‫׀‬
H–C – C –
‫׀‬
‫׀‬
H OH
n–propyl alcohol
c)
H
‫׀‬
C–H
‫׀‬
H
isopropyl alcohol
Re-positioning of a Multiple Bond
(Note that this is different from resonance.)
Example: C4H8
H H H H
│
│
│
H H
│
│
H–C=C–C–C–H
│
│
H H
│
│
H–C–C=C–C–H
│
H H
│
│
H
H
2. Stereoisomers = different spatial arrangement of groups
a) Geometric Isomers (cis- and trans- isomers)
= restricted rotation due to double or triple bond makes molecules
different.
Cl
H
\
/
C == C
/
\
Cl
H
Cl
Cl
\
/
C == C
/
\
H
H
Cl
H
\
/
C == C
/
\
H
Cl
Cis-
Trans-
└──────┬──────┘
Geometric Difference
└──────┬──────┘
Structural Difference
6/7- 36
b) Optical Isomers = have non-superimposable mirror images
(can exist in left-handed and right-handed forms)
(See Models)
Some terms:
Enantiomers = a pair of non-superimposable molecules
Chiral = describes molecules (and other objects) that have nonsuperimposable mirror images
B. Practice
Ex. #16: Draw the three structural isomers of C5H12.
Ex. #17: Draw the isomers of C2H2F2, and label any geometric isomers.
Ex. #18: Draw all of the isomers of C4H6.
6/7-37
Ex. #19: Lactic acid isomers are non-superimposable. In the lactic acid molecule
there are four different groups (–H, – OH, – CH3, and – COOH) attached to the corners
of a tetrahedral carbon. Add these groups to the sp3 tetrahedron of the central carbon
to show the enantiomers of lactic acid
C
C
XII. Expanded Octets
A. Overview
-
Consider the IF3 molecule
-
Counting valence electrons: IF3 = 7 + 21 = 28 e‾
(Add appropriate “dots” to structure.)
F–I–F
│
F
-
We haven’t done anything wrong! This is a situation where nature forms an
―expanded octet‖, meaning that there are more than 4 pairs of electrons around the
central atom.
-
Previous Examples (page 6/7-9)
PCl5
Progress Check
Example
SO2
ClO4‾
AsO3‾
NI3
C2HCl
C2H2Cl2
Lewis
Structure
Hybridization
Bond
Angles
6/7–38
e Pair Geometry
Molecular Geometry
–
Sigma
Pi
Molecular
Bonds Bonds Dipole? Resonance? Isomers?
6/7-39
B. Hybrid Orbitals
-
For PCl5:
P = [10Ne] ↑↓
↑ ↑ ↑
3s
(ground state)
3p
= [10Ne] ↑
↑ ↑ ↑
3s
↑
(after e‾ promotion)
3p
3d
Note: A ―d‖ orbital was used instead of the 4s on the next energy level!
= [10Ne] ↑ ↑ ↑ ↑ ↑
(after hybridization)
sp3d
3d
↑
sp d hybrids
3
-
↑
(only four d orbitals remaining)
For SF6:
S = [10Ne] ↑↓
↑↓ ↑ ↑
3s
(ground state)
3p
= [10Ne] ↑
↑ ↑ ↑
3s
3p
↑
↑
(after e‾ promotion)
3d
= [10Ne] ↑ ↑ ↑ ↑ ↑ ↑
(after hybridization)
sp3d2
3d
↑
sp d hybrids
↑
3 2
(only three d orbitals remaining)
C. Geometry of Expanded Octets (See Handout)
-
Base Structures
5 pairs:
6 pairs:
│
│
X—
X
│
│
trigonal bipyramid
octahedral
6/7-40
-
Important: The shape of the molecule depends on how many atoms are
attached to the central atom! (See Handout)
-
―Walk-Down‖ the shapes of each of the following:
6 e‾ pairs
- octahedral
5 e‾ pairs
- trigonal bypyramid
4 e‾ pairs
- tetrahedral
- square pyramid
- distorted tetrahedron (see-saw)
- pyramid
- square planar
- T-shaped
- bent
- _________________
- __________________
- linear
- _________________
- __________________
- _________________
3 e‾ pairs
2 e‾ pairs
- trigonal planar
- linear
- _________________
- __________________
- _________________
-
Consider the two possible structures of XeF4 :
2e‾
F
2e‾ │
Xe
2e‾ │
F
F
F
│
F
Xe
F
F
│
F
2e‾
Why this one?
Answer: ___________________________________________________________
___________________________________________________________________
6/7-41
D. “Short-Cut” Method of Drawing Lewis Structures for Expanded Octets
Short-Cut Method:
Add one electron for each attached atom to the Group number
(Roman Numeral) of the central atom. Correct for charge if
necessary.
-
Back to the IF3 Example:
a) Using short-cut: IF3 = 7 + 3 = 10 e‾ = 5 pairs
(trigonal pyramid base structure)
│
b) Draw skeleton structure:
I
│
c) Fill in the attached atoms in their proper positions:
Result: T-Shaped
..
:F:
2e‾
│
I
F:
2e‾
│
∙∙
: F:
∙∙
∙∙
6/7-42
Ex. #20: Complete the table below for each of the given species.
Species
BrF5
AsF5
BrF4–
TeF6
TeF4
e – Pairs
Around
Central
Atom
Lewis Structure
e – Pair
Geometry
Molecular
Geometry
Central Atom
Hybridization
Molecular
Polarity
6/7-43
Ex. #20: (Continued)
Species
XeF6
I3–
SbCl5
SnCl62–
XeF2
e – Pairs
Around
Central
Atom
Lewis Structure
e – Pair
Geometry
Molecular
Geometry
Central Atom
Hybridization
Molecular
Polarity