David Tsang-Per.1-Ch.8 Summary

Electronegativity
-Higher the difference in electronegativity, the more polar the compound becomes.
-Electronegativity increases across a period and decreases going down a group.
0 Intermediate Large Difference in electronegativity
Bond Type Covalent Polar Covalent Ionic

Bond Polarity and Dipoles
- Atoms with a higher electronegativity tend to draw electrons closer to itself, causing dipoles.
-Dipoles signified by +

O = C = O

+ +

-Atoms pulling in opposite directions cancel out dipoles.

Ions: Electron configuration and sizes
-Elements in Group I, II, XIII, XV, XVI, XVII tend to have electron configurations of the closest noble gases (in terms of atomic number) when they are ions. For example Na + and F - have the same electron configuration as Ne.
-Size of Ions
Atom > Cation (# of protons ># of electrons decreasing electron repulsion)
Anion > Atom (# of electrons ># of protons, increasing electron repulsion)

Formation of Binary Ionic Compounds
-Lattice Energy- the change in energy that takes place when separated gaseous ions are packed together to form an ionic solid.
M + (g) + X - (g)

MX(s)
Theoretical Process
M(s) + X(g)

MX(s)
1.
M(s)

M(g) ( enthalpy of sublimation for M)
2.
M(g)

M + (g) + e -
3.
X(g)+ e -
4.
M +

X (g)
(ionization energy for M)
(electron affinity of X)
(g) + X - (g)

MX(s) (lattice energy reaction)
5.
Combine all subreactions and cancel out any of the same substance on both sides. This should leave you with the main reaction.
6.
Add up the energy change from each subreaction and this should give you the ∆H° f
for MX(s)

Ionic Character
-Trend: As the difference between the electronegativity of two atom increases, ionic character (having the properties of an ionic compound) increases.

Covalent Bond Energies and Chemical Reactions
-Bond Energy of X-Y= (Bond Energy of X-X +Bond Energy of Y-Y) / 2
-Triple bonds are stronger than double bonds and require more energy to break a double bond than to break a double bond. Double bonds are stronger than single bonds and require more energy to break a double bond than to break a single bond.
-The stronger the bond, the shorter the bond length.
-For a given reaction
∆H= ∑D (bonds broken) - ∑D(bonds formed)
D represents bond energy per mole of bond.
-See pg. 373 for table of average bond energies


Lewis Structure
-The Lewis Structure of a molecule shows how the valence electrons are arranged among the atoms in a molecule.
-Typically atoms are stable when it has achieved noble gas configurations.
Examples
Na F Ne NaF F
2
CH
4
CN -
-Hydrogen is stable when it shares 2 electrons (the duet rule). This is because H atom can only hold 2 atoms in it 1s orbital.
-Helium do not form bonds because it already filled up its 1s orbital (duet rule) and thus is stable.
- Typically nonmetals share electrons in order to achieve an octet (8 valence electrons around the atom) when they are bonded (See F
2
).
Pairs of electrons not involved in bonding are called long electron pairs.
-Noble gases do not react with other atoms because they achieved the octet (See Ne)
-For molecules with a charge, draw brackets around the compound and draw the charge (see CN ). Add electrons to the structure according the charge number for molecules with a negative charge. Subtract the electrons if the molecule has a positive charge (For CN , add 1 electron to the structure. For NH
4
+ , subtract 1 electron from the structure).
Rules for writing lewis structures
-Sum the valence electrons from all the atoms. Do not worry about keeping track of which electrons come from which atoms. It is the total number of electrons that is important
-Use a pair of electrons to form a bond between each pair of bound atoms.
-Arrange the remaining electrons to satisfy the duet rule for hydrogen and the octet rule for the second-row elements
Example
H
2
O
1.
1+1+6=8 valence electrons.
2.
Draw two O-H bonds.
3.
Draw in the lone electron pairs
H-O-H
H-O-H
CO
2
1.
4+6+6= 16 valence electrons.
2.
Draw two C-O bonds. O-C-O
3.
Draw in the lone electrons
4.
We used up 16 valence electrons.
But carbon doesn’t fulfill the octet rule.
O-C-O
5.
Use double bonds instead of a single bond O=C=O

Exceptions to the Octet Rule
-The second-row elements C,N, O, F should always be assumed to obey the octet rule.
-The second-row elements B and Be often have fewer than 8 electrons around them in their compounds.
-The second-row elements never exceed the octet rule since they only have 2s and 2p orbital can only hold 8 valence electrons
-Third-row and heavier element often satisfy the octet rule but can exceed the octet rule by using their empty valance d orbitals.
-When writing the Lewis structure for a molecule, satisfy the octet rule for the atoms first. If electrons remain after the octet rule has been satisfied, then place them on the elements having available d orbitals.
SF
6
ICl
3
- RnCl
2
BeCl
2
ICl
4
-

Resonance
-Resonance occurs when some molecules made contain more than one valid Lewis structures. These molecules usually contain more than one type of bond.
NO
3
-

-Formal Charge
When there are more than one different Lewis structure for a molecule, use formal charge to determine which one is correct.
-To calculate formal charge on an atom.
1.Take the sum of the lone pair electrons and ½ the shared electrons.
2. Subtract the number of assigned electrons from the number of valence electrons on the free, neutral atom to obtain the formal charge.
-The sum of the formal charges of all atoms in a given molecule or ion must equal the overall charge on that species.
-If the sum of formal charges do not equal the charge of the species, those with formal charges closest to zero and with any negative formal charges on the most electronegative atoms are considered to best describe the bonding in the molecule or ion.
2
3
3

VSEPR Model (Valence Shell Electron Pair Repulsion Model)
No. of
Electron
Pairs
Molecular
Geometry
# of
Lone e -
Pairs
Theoretical
Formula
Bond
Angle
Example
4
Linear
Bent
0
1
Trigonal
Planar
0
Tetrahedral 0
AB
2
AB
2
E
AB
3
AB
4
180
120
120
CO
2
NO
2
-
CO
3
2-
4 1 AB
3
E
109.5 CH
4
107 NH
3
4
Trigonal
Pyramidal
Bent 2 AB
2
E
2
5
5
5
5
6
6
Trigonal
Bipyramidal
See-saw
T-Shape
Linear
0
1
2
3
Octahedral 0
1
AB
5
AB
4
E
AB
3
E
2
AB
2
E
3
AB
6
AB
5
E
104.5 H
2
O
90,
120
PCl
5
SF
4
90,
120
90,
180
BrF
3
ICl
2
180
90
90
SF
6
BrF
5
6
Square
Pyramidal
Square
Planar
2 AB
4
E
2
90 ICl
4
-
Go to http://www.molecules.org/VSEPR_table.html
to see a really good 3-D representation for each type of geometry in the VSERP Theory

Multiple Choice
1.
Type of Bond
O-H
C-C
O=O
C=O
C-H
Bond Energy (kJ/mol)
467
347
495
799
413
C-O 358
Using the table of bond energies, calculate ∆H the following reaction.
C
2
H
5
OH + 3O
2

2CO
2
+ 3H
2
O
A) 1083 B) -996
C) 1276 D) -1083
E) -1276
2. For the following bonds, place them in order of increasing polarity.
Cs-F, H-Cl, I-I, S-O, Li-Br
A) H-Cl < Li-Br < Cs-F < I-I < S-O
C) Li-Br < S-O < I-I < Cs-F < H- Cl
E) I-I< H-Cl < S-O < Li-Br < Cs-F
B) I-I < S-O < H-Cl < Li-Br < Cs-F
D) Cs-F < Li-Br < S-O < H-Cl < I-I
3. What is the geometry of SiO
3
2.
A) Trigonal Planar B) Linear
C) Bent D) Trigonal Pyramidal
E) Tetrahedral
4. Which one of these does not exhibit resonance?
A) SO
2
C) CH
3
Br
E) NO
3
2-
B) SO
3
D) CO
3
2-
5. Use the following data to estimate ∆H°
Li (s) + ½ F
2
(g)

LiF (s) f
for sodium chloride
Lattice energy -1047 kJ/mol
Ionization energy for Li
Electron affinity of F
520 kJ/mol
-328 kJ/mol
Bond energy for F
2
Enthalpy of sublimation for Li
A) -394 kJ/mol B)-1002 kJ/mol
C)-729 kJ/mol D)-617 kJ/mol
E)-502 kJ/mol
Free response
77 kJ/mol
161 kJ/mol
Answer the following questions in terms of atomic/molecular structure and chemical bonding. You must include references to both substances. a) Draw the complete Lewis electron dot structure for the molecules H
2
O and H
3
O + and indicate its geometry and bond angles. b) Explain why the carbon-carbon bond energy in C
2
H
4
is greater that of C
2
H
6 c) Consider the molecules SF
4
and CF
4
i) Draw the complete Lewis electron dot structure for both molecules
ii) In terms of molecular geometry, account for the fact that the molecule is nonpolar CF
4 whereas the SF
4 molecule is polar

Multiple Choice Answers
1.
E
∆H=1mol X 347 kJ/mol C-C + 5 mol X 413 kJ/mol C-H + 1mol X 358 kJ/mol C-O + 1mol X 467 kJ/mol
O-H +3mol X 495 kJ/mol O=O – 4 mol X 799 kJ/mol C=O – 6 mol X 467 kJ/mol O-H
∆H=-1276 kJ/mol
2.
B
3.
A
4.
C
5. D
This is the only way to draw CH
3
Br (Shifting the Br around the C does not count as reasonance)
Process
Li(s)

Li(g)
Li(g)

Li + (g) + e -
1/2 F
2
(g)

F (g)
F(g) + e
Li +
- 
F - (g)
(g) + F - (g)

LiF(s)
Energy charge (kJ)
161
520
77
-328
-1047
--------------------------------------------------------------
Li (s) + ½ F
2
(g)

LiF (s) -617 kJ (per mole of LiF)
Free Response Answer a) b)
C
2
H
4
C
2
H
6
The carbon-to-carbon bond in C
2
H
6 is a single bond while the carbon-to-carbon bond in C
2
H
4 is a double bond. More energy is needed to break the double bond in C
2
H
4 than to break the single bond in C
2
H
6
; therefore the double bond in C
2
H
4 is stronger and has greater bond energy than the single bond in C
2
H
6.

c)
i)
CF
4
SF
4
ii)CF
4 has a tetrahedral geometry, so the bond dipoles cancel out each other. This leads to nonpolar molecule.
SF
4 has 5 pairs of electrons around the central atom (S) and exhibits a trigonal bipyramidal geometry, with the lone pair of electrons. In this configuration, the bond dipoles do not cancel and the molecule is polar.