Energy, Bonds
& Chemical Structure
Chapter 7, 8 & 9
Determination of Atomic Radius:
• Half of the distance between nuclei in
covalently bonded diatomic molecule
"covalent atomic radii"
Determination of Atomic Radius:
• Periodic Trends in Atomic Radius
– Radius decreases from left to right
across a period
• Increased effective nuclear charge
due to decreased shielding
– Radius increases down a group
• Addition of principal quantum
levels
Table of Atomic Radii
Ionic Radii
• Cations
– Positively charged ions
– Lose electrons
– Smaller than the corresponding atom
• Anions
– Negatively charged ions
– Gains electrons
– Larger than the corresponding atom
Table of Ion Sizes
Ionization Energy
• The energy required to remove one mole
electrons from a mole of gaseous atoms to
produce one mole of gaseous ions
• M(g)  M+(g) + e• Second ionization energy – energy change
accompanying
M+(g)  M2+(g) + e• Measured in kJ/mol
• Positive energy values – requires energy to
remove electrons (endothermic)
Ionization Energy
• Magnitude is determined by the attraction
of the positive nucleus for the negative
electrons that are being removed
• Attraction is dependant on
– Nuclear energy
– Shielding effect of inner electrons
Ionization of Magnesium
Mg + 738 kJ  Mg+ + eMg+ + 1451 kJ  Mg2+ + eMg2+ + 7733 kJ  Mg3+ + e-
Electron Affinity
• The energy change when one mole of
gaseous atoms gains one mole of
electrons to form one mole of gaseous
ions
X(g) + e-  X• Measured in kJ/mol
Electron Affinity
• Affinity tends to increase across a period
• Affinity tends to decrease as you go down
in a period
– Electrons farther from the nucleus
experience less nuclear attraction
– Some irregularities due to repulsive
forces in the relatively small p orbitals
Lattice Energy
•
•
•
•
•
Atomization of Lithium
Atomization of fluorine
Ionization of lithium
Electron affinity of fluorine
Lattice Energy for LiF
+155.2 kJ/mol
+150.6 kJ/mol
+520 kJ/mol
-328 kJ/mol
-1016 kJ/mol
Electronegativity
• A measure of the ability of an atom in a
chemical compound to attract electrons
• Electronegativities tend to increase
across a period
• Electronegativities tend to decrease
down a group or remain the same
Periodic Tables of
Electronegativities
Example
• Which bond is more polar? H—F Na—Cl
C—H
Ionic bonds
• Electrons are transferred
• Electronegativity differences are
generally greater than 2.0
• The formation of ionic bonds is always
exothermic!
General Rule
• In most chemical compounds, atoms are
reacting so that they have the same electronic
structure as a noble gas (they are “isoelectronic”
with a noble gas).
• General Rule For any set of isoelectronic
atoms, the highest nuclear charge (Z) will be
smallest, because that nucleus can pull on the
electron clouds the hardest.
• Ex Put these species in order of increasing size:
Se2-, Br-, Rb+ and Sr2+
Covalent Bonds
Covalent bonds – sharing of electrons to form
discrete molecules
Polar-Covalent bonds
• Electrons are unequally shared
• Electronegativity difference between 0 and
2.0
Nonpolar-Covalent bonds
• Electrons are equally shared
• Electronegativity difference of 0
Calculating Bond Energy
• Values come from a table (Pg 372 in your
book)
• Energy (∆H) is the sum of the energy
required to break old bonds plus the sum
of the energy required to make new bonds
– Old bonds have a positive sign (require
energy to break)
– New bonds have a negative sign (take in
energy to make)
Bond Energy
• Bond energy is given in kJ/mol
• More bonds (double or triple) contain more
energy and are harder to break
• If the overall value of ∆H is positive,
energy must be added into the reaction
(endothermic)
• If the overall value of ∆H is negative,
energy is released into the environment
(exothermic)
Example
•
1.
2.
3.
4.
Methane is burned in oxygen gas
creating water and carbon dioxide.
Calculate the ∆H for the reaction.
Write the reaction.
Figure out what bonds are in each
molecule.
Figure out how many of each bond there
are and what the energy for each is (Use
Pg. 372)
∆H = Σ(reactants) – Σ(products)
Lewis Structures
• Shows how valence
electrons are
arranged among
atoms in a
molecule.
• Reflects central
idea that stability of
a compound relates
to noble gas
electron
configuration.
The Octet Rule
• Combinations of elements tend to form
so that each atom, by gaining, losing,
or sharing electrons, has an octet of
electrons in its highest occupied
energy level.
Bonding for fluorine
Drawing Lewis Structures
1. Calculate the total number of valence electrons
for the compound
- don’t forget to take charge into account
2. Decide which atom is the central atom
- if this is not obvious, use the least
electronegative atom available
3. Use a line to indicate covalent bonds to the
outer atoms
4. Arrange the rest of the electrons around the
atoms to form octets
5. Form multiple bonds (double or triple
bonds/lines) if necessary to complete the
octets
Completing a Lewis Structure
• Draw Lewis structures for the following
– Fluorine gas
– Hydrochloric acid
– Carbon tetrabromide
– PCl6– NH4+
Comments on the Octet Rule
• 2nd row elements C, N, O, F observe the octet
rule (HONC rule as well).
• 2nd row elements B and Be often have fewer
than 8 electrons around themselves - they are
very reactive.
• 3rd row and heavier elements CAN exceed the
octet rule using empty valence d orbitals.
• When writing Lewis structures, satisfy octets
first, then place electrons around elements
having available d orbitals.
Multiple Covalent Bonds:
Double Bonds
• Two pairs of shared electrons
Multiple Covalent Bonds:
Triple Bonds
• Three pairs of shared electrons
Resonance
• Resonance is invoked when more than one
valid Lewis structure can be written for a
particular molecule.
• The actual structure is an average of the
resonance structures.
• Benzene, C6H6
• The bond lengths in the ring are identical,
and between those of single and double
bonds.
Resonance, Bond Length and
Bond Energy
• Resonance bonds are shorter and
stronger than single bonds.
• Resonance bonds are longer and
weaker than double bonds.
Resonance in Ozone, O3
• Neither structure is correct.
• Oxygen bond lengths are identical, and
intermediate to single and double bonds
Resonance in Polyatomic Ions
• Resonance in a carbonate ion:
• Resonance in an acetate ion:
Formal Charge
• Formal Charges a method of determining
– a. which resonance structure is most likely
– b. where the electrons are most concentrated
in a structure
• Formal charge of an atom is
(Family # ) – ( # of lone pair e-) - (# of
bonds on that atom)
Formal Charge
• Structures that have formal charges
closest to zero and any negative formal
charges on the most electronegative
atoms are most likely to be stable
Example
• Ex Which is the most likely Lewis
structure for CO2 ?
• Ex For the SCN- ion?
Localized Electron Model
• Lewis structures are an application of
the “Localized Electron Model”
– L.E.M. says: Electron pairs can be
thought of as “belonging” to pairs of
atoms when bonding
– Resonance points out a weakness in
the Localized Electron Model.
VSEPR – Valence Electron Pair
Repulsion
X+E
Overall Structure
Forms
2
3
4
5
6
Linear
Trigonal Planar
Tetrahedral
Trigonal bipyramidal
Octahedral
AX2
AX3 , AX2E
AX4 , AX3E , AX2E2
AX5 , AX4E , AX3E2 , AX2E3
AX6 , AX5E , AX4E2
A = central atom
X = atoms bonded to A
E = nonbonding electron pairs on A
VSEPR - Linear
AX2
Example:
CO2
VSEPR – Trigonal Planar
AX3
BF3
AX2E
SnCl2
VSEPR - Tetrahedral
AX4
CCl4
AX3E
PCl3
AX2E2
Cl2O
VSEPR – Trigonal Bipyramidal
AX5
PCl5
AX4E
SF4
AX3E2
ClF3
AX2E3
I3-
VSEPR - Octahedral
AX6
SF6
AX5E
BrF5
AX4E2
ICl4-
Sigma and Pi Bonds
• Sigma () bonds exist in the region directly
between two bonded atoms.
• Pi () bonds exist in the region above and
below a line drawn between two bonded
atoms.
Single Bond
1 sigma bond
Double Bond
1 sigma, 1 pi bond
Triple Bond
1 sigma, 2 pi bonds
Sigma and Pi Bonds
Single Bonds
1  bond
Sigma and Pi Bonds
Double Bonds
1  bond
H
H
C
H
C
H
1  bond
Sigma and Pi Bonds
Triple Bonds
1  bond
1  bond
1  bond
The De-Localized Electron Model
• Pi bonds () contribute to the delocalized
model of electrons in bonding, and help
explain resonance
• Electron density from  bonds can be
distributed symmetrically all around the ring,
above and below the plane.
Hybridization
• The blending of orbitals
poodle
+ cocker spaniel
cockapoo
=
What proof exists for hybridization?
• We have studied electron configuration
notation and the sharing of electrons in the
formation of covalent bonds.
• Lets look at a molecule
of methane, CH4.
• Methane is a simple natural gas. Its molecule
has a carbon atom at the center with four
hydrogen atoms covalently bonded around
it.
Carbon Ground State Configuration
• What is the expected orbital notation of
carbon in its ground state?
Can you see a problem with this?
• (Hint: How many unpaired electrons does
this carbon atom have available for
bonding?)
Carbon’s Bonding Problem
• You should conclude that carbon only
has TWO electrons available for
bonding. That is not enough!
• How does carbon overcome this
problem so that it may form four
bonds?
Carbon’s Empty Orbital
• The first thought that chemists had was
that carbon promotes one of its 2s
electrons…
A Problem Arises…
• However, they quickly recognized a
problem with such an arrangement…
1s
1s
2s
1s
1s
1s
2p
2p
2p
• Three of the carbon-hydrogen bonds
would involve an electron pair in which
the carbon electron was a 2p, matched
with the lone 1s electron from a
hydrogen atom.
Unequal Bond Energy
• This would mean that three of the
bonds in a methane molecule would be
identical, because they would involve
electron pairs of equal energy.
• But what about the fourth bond…?
Unequal Bond Energy
• The fourth bond is between a 2s
electron from the carbon and the lone
1s hydrogen electron.
1s
1s
2s
1s
1s
1s
2p
2p
2p
• Such a bond would have slightly less
energy than the other bonds in a
methane molecule.
Unequal Bond Energy
• This bond would be slightly different in
character than the other three bonds in
methane.
• This difference would be measurable to
a chemist by determining the bond
length and bond energy.
• But is this what they observe?
Enter Hybridization
• The simple answer is, “No”.
• Measurements show that all four
bonds in methane are equal. Thus, we
need a new explanation for the bonding
in methane.
• Chemists have proposed an
explanation – they call it Hybridization.
• Hybridization is the combining of two
or more orbitals of nearly equal energy
within the same atom into orbitals of
equal energy.
• In the case of methane, they call the
hybridization sp3, meaning that an s
orbital is combined with three p orbitals
to create four equal hybrid orbitals.
1s
2sp3
• These new orbitals have slightly MORE
energy than the 2s orbital…
… and slightly LESS energy than the 2p
orbitals.
sp3 Hybrid Orbitals
sp3 hybrid orbitals
• Here is another way to look at the sp3
hybridization and energy profile…
sp hybrid orbitals
• While sp3 is the hybridization observed in
methane, there are other types of
hybridization that atoms undergo.
• These include sp hybridization, in which one
s orbital combines with a single p orbital.
• This produces two hybrid orbitals, while
leaving two normal p orbitals
sp2 hybrid orbitals
• Another hybrid is the sp2, which
combines two orbitals from a p
sublevel with one orbital from an s
sublevel.
• One p orbital remains unchanged
Hybridization Involving “d” Orbitals
• Beginning with elements in the third
row, “d” orbitals may also hybridize
• dsp3 = five hybrid orbitals of equal
energy
• d 2sp3 = six hybrid orbitals of equal
energy
Hybridization and Molecular
Geometry
Forms
Overall Structure
AX2
Linear
AX3, AX2E
AX4, AX3E, AX2E2
AX5, AX4E, AX3E2,
AX2E3
AX6, AX5E, AX4E2
Trigonal planar
Tetrahedral
Trigonal
bipyramidal
Octahedral
Hybridization of
“A”
sp
A = central atom
X = atoms bonded to A
E = nonbonding electron pairs on A
sp2
sp3
dsp3
d2sp3