CH 8 CHEMICAL BONDING
AND MOLECULAR STRUCTURE
AP CHEMISTRY 2013-2014
BONDS ARE ATTRACTIVE FORCES THAT
HOLD GROUPS OF ATOMS TOGETHER AND
MAKE THEM FUNCTION AS A UNIT.
• Being bound requires less energy than
existing in the elemental form. Energy
is released when a bond is formed;
energy is required to break a bond.
The energy required to break a bond is
called the bond energy.
• Ionic bonds
• Covalent bonds
• Coulomb’s Law is used to calculate the
energy of an ionic bond (see equation on
the right; k = 2.31 x 10-19 Jnm, Q = charge
on each ion)
• The energy of an ionic bond will be negative; it
indicates an attractive force so that the ion
pair has lower energy than the separated ions.
• Chemical bond formation
• When two hydrogen atoms approach
each other, two repulsions & one
attraction occur
• Electron/electron repulsion
• Proton/proton repulsion
• Proton/electron attraction
• When the attractive forces offset the
repulsive forces, the energy of the two
atoms decreases and a bond is formed.
Bond length is the distance between two
nuclei where the energy is at a minimum.
• The electrons and nucleus of one atom
strongly perturb or change the spatial
distribution of the other atom’s valence
electrons. A new orbital (wave function) is
needed to describe the distribution of the
bonding electrons bond orbital. The
energy of the electrons in a bond orbital is
lower that their energy in valence electron
orbitals when they are in isolated atoms.
• In an ionic bond, the bonding orbital is
strongly displaced toward one nucleus.
• In a covalent bond, the bond orbital is
more or less evenly distributed and the
electrons are shared by two nuclei.
• Most chemical bonds are somewhere
between purely ionic and purely
covalent.
• Electronegativity is the ability of an atom in a
molecule to attract shared electrons to itself.
• Fluorine is the most electronegative (4.0) due to
highest effective nuclear charge (Zeff) and smallest
radius—so that the nucleus is closest to the “action”
• Francium is the least (0.7) due to lowest Zeff and
largest radius so that the nucleus is farthest from the
“action”
• This atomic trend is only used when
atoms form molecules.
• Ionic: ΔEN >1.67
• Covalent: ΔEN < 1.67
• Nonpolar covalent: ΔEN < 0.4
COVALENT BONDING
• Most compounds are covalently bonded,
especially carbon compounds.
• Localized electron (LE) bonding model: assumes
that a molecule is composed of atoms that are
bound together by sharing pairs of electrons using
the atomic orbitals of the bound atoms. Electron
pairs are assumed to be localized on a particular
atom [lone pairs] or in the space between two
atoms [bonding pairs].
• Lewis structures describe the valence electron arrangement
• Geometry of the molecule is predicted with VSEPR
• Number of bond pairs/Octet rule
• Single and multiple bonds
• Single bond: one pair of electrons shared
(sigma σ bond)
• Double bond: two pairs of electrons shared
(one sigma bond, one pi bond)
• Triple bond: three pairs of electrons shared
(one sigma bond, two pi bonds)
DOUBLE BOND (ETHENE)
DOUBLE BOND (CARBON DIOXIDE)
TRIPLE BOND (ETHYNE)
• Pi bonds are weaker than sigma
bonds but never exist alone.
Triple bonds are stronger than
double bonds are stronger than
single bonds.
• Multiple bonds are most often
formed by C, N, O, and P.
• Multiple bonds increase the electron
density between two nuclei and therefore
decrease the nuclear repulsions while
enhancing the attraction between
nucleus/electrons; either way, the nuclei
move closer together and the bond
length is shorter for a double bond than a
single bond.
• Exceptions to the Octet rule
• Fewer than eight: H, Be, B
• Expanded valence: can only happen if the central atom has dorbitals and can thus be surround by more than four valence
pairs in certain compounds.
• Odd-electron compounds: ex. NO, NO2, ClO2
EXERCISE 1 RELATIVE BOND POLARITIES
Order the following bonds according to
polarity: H-H, O-H, Cl-H, S-H, and F-H
• Bond polarity and electronegativity:
electronegativity determines polarity since it
measures a nucleus’s attraction or “pull” on the
bonded electron pair.
• When two nuclei are the same, the sharing is
equal and the bond is described as nonpolar.
• When two nuclei are different, the electrons are
not shared equally, setting up slight +/- poles, and
the bond is described as polar.
• When the electrons are shared very unequally, the
bond is described as ionic.
IONIC BONDING
• The final result of ionic bonding is a solid, regular
array of cations and anions called a crystal lattice.
• Enthalpy of dissociation: energy required to
decompose an ion pair (from a lattice) into ions; a
measure of the strength of the ionic bond (related
to Coulomb’s law)
• The energy of attraction depends directly on the
magnitude of the charges and inversely on the distance
between them (related to the size of the ion).
EXERCISE 2 COMPARING LATTICE ENERGY
Which compound in each pair will have
the higher lattice energy?
NaF or RbF
MgO or LiCl
DRAWING LEWIS STRUCTURES
• H is always a terminal atom
• Atom with lowest EN goes in center
• Find the total number of valence
electrons by adding together the
valence electrons of every atom in the
compound
• For ions, add for negative charges and
subtract for positive charges
• Place one pair of electrons, a sigma bond,
between each pair of bonded atoms.
• Complete the octets of all atoms with lone
pairs. Leftover pairs are assigned to the
central atom if it can accommodate them.
Double/triple bonds may need to be used
(pi bonds).
EXERCISE 3
WRITING LEWIS STRUCTURES
Give the Lewis structure for each of the following.
a) HF
EXERCISE 3
WRITING LEWIS STRUCTURES
Give the Lewis structure for each of the following.
b) N2
EXERCISE 3
WRITING LEWIS STRUCTURES
Give the Lewis structure for each of the following.
c) NH3
EXERCISE 3
WRITING LEWIS STRUCTURES
Give the Lewis structure for each of the following.
d) CH4
EXERCISE 3
WRITING LEWIS STRUCTURES
Give the Lewis structure for each of the following.
e) CF4
EXERCISE 3
WRITING LEWIS STRUCTURES
Give the Lewis structure for each of the following.
f) NO+
EXERCISE 4
LEWIS STRUCTURES FOR MOLECULES THAT
VIOLATE THE OCTET RULE
Give the Lewis structure for each of the following.
a) PCl5
EXERCISE 4
LEWIS STRUCTURES FOR MOLECULES THAT
VIOLATE THE OCTET RULE
Give the Lewis structure for each of the following.
b) ClF3
EXERCISE 4
LEWIS STRUCTURES FOR MOLECULES THAT
VIOLATE THE OCTET RULE
Give the Lewis structure for each of the following.
c) XeO3
EXERCISE 4
LEWIS STRUCTURES FOR MOLECULES THAT
VIOLATE THE OCTET RULE
Give the Lewis structure for each of the following.
d) RnCl2
EXERCISE 4
LEWIS STRUCTURES FOR MOLECULES THAT
VIOLATE THE OCTET RULE
Give the Lewis structure for each of the following.
e) BeCl2
EXERCISE 4
LEWIS STRUCTURES FOR MOLECULES THAT
VIOLATE THE OCTET RULE
Give the Lewis structure for each of the following.
f) ICl4-
RESONANCE STRUCTURES
• Ex. ozone has equal bond lengths and equal bond
strengths, implying that there are an equal number
of bond pairs on each side of the central oxygen
atom. The Lewis structure does not agree with this;
instead, we have to use a composite to describe
the reality. This composite depicts the blending of
resonance structures for ozone. Instead of truly
having a single bond and a double bond, both of
its C-O bonds O-O bonds could be thought of as “a
bond and a half”.
• Resonance structures differ only in the
assignment of electron pair positions,
never atom positions. They differ in the
number of bond pairs between a
given pair of atoms.
• Note that the resonance structures
and composite are drawn with
brackets (required for full credit on AP
exam).
EXERCISE 5
RESONANCE STRUCTURES
Draw every resonance structure for the carbonate
ion. Also draw the composite structure.
BOND PROPERTIES
• Bond order: simply the number of bonding electron
pairs shared by two atoms in a molecule.
•
•
•
•
1 = one shared pair; sigma bond between two atoms
2 = two shared pairs; sigma bond and pi bond
3 = three shared pairs; sigma bond and two pi bonds
Fractional for resonance structures (3/2 for ozone, 2/3 for
carbonate)
Bond order = number of shared pairs linking X and Y
number of X-Y links
• Bond length: the distance between the nuclei
of two bonded atoms
• Higher bond order = shorter length
• Bond energy: the greater the number of
electron pairs between a pair of atoms, the
shorter the bond. This implies that atoms are
held together more tightly when there are
multiple bonds, so there is a relationship
between bond order and the energy required
to break a bond.
• Bond dissociation energy (D): the energy
supplied to break a chemical bond
• Endothermic; D is positive
• Bonds in reactants are broken while bonds in products
are formed. Energy released is greater than energy
absorbed in exothermic reactions. The converse is also
true.
ΔH°rxn = ΣmD(bonds broken) - ΣnD(bonds made)
ΔH°rxn = reactants(E cost) - products(E payoff)
• Note that this is “backwards” from thermodynamics. First we
must break the bonds of the reactants (costs energy) then
subtract the energy gained by forming new bonds in the
products.
EXERCISE 6
ΔH FROM BOND ENERGIES
Using the bond energies from the table on the right,
calculate ΔH° for the reaction of methane with
chlorine and fluorine to give Freon-12 (CF2Cl2).
CH4 + 2Cl2 + 2 F2  CF2Cl2 + 2 HF + 2 HCl
CH4 + 2Cl2 + 2 F2  CF2Cl2 + 2 HF + 2 HCl
FORMAL CHARGE
• Formal charge is the difference between the
number of valence electrons on a free element,
and the number of electrons assigned to the atom
once it is in a molecule.
• Formal charge = group number – [# of lone
electrons – 2(# of bonding electrons)]
• The idea of formal charge allows us to determine the
most favored structure out of a set of nonequivalent
Lewis structures.
• Oxidation states of more than +/- are questionable,
while formal charges are more realistic.
• The sum of the formal charges on an ion must equal the
ion’s overall charge.
• Use formal charges along with the following to
determine resonance structure
• Atoms in molecules (or ions) should have formal charges that are
as small (close to zero) as possible
• A molecule (or ion)is most stable when any negative formal
charge resides on the most electronegative atom.
• Ex. There are three possible structures for the
sulfate ion shown below (note that these are
not resonance structures). The third is the most
valid of the three; it results in the fewest (and
smallest) formal charges.
EXERCISE 7
FORMAL CHARGES
Give possible Lewis structures for XeO3, an explosive
compound of Xenon. Which Lewis structure or
structures are most appropriate according to the
formal charges?
VALENCE SHELL ELECTRON PAIR
REPULSION THEORY (VSEPR)
• Molecular shape changes with the number of
sigma bonds plus lone pairs about the central atom
• Molecular geometry is the arrangement in space of
the atoms bonded to a central atom
• lone pairs take up more space around an atom than bonds
• Each lone pair or bond pair repels all other electron pairs;
they try to avoid each other making as wide an angle as
possible.
• Ex. Water: the two lone pairs on oxygen “warp” the normal 109.5
angle through repulsion, resulting in a bond angle of 104.5.
TO DETERMINE MOLECULAR
GEOMETRY
• Sketch the Lewis dot structure
• Describe the structural pair or electronic geometry (the
shape of the molecule considering both its bonds and
lone pairs)
• Focus on the bond locations (ignore lone pairs) and
assign a molecular geometry based on their locations
• Molecular geometry and electronic geometry are only
the same in the absence of lone pairs on the central
atom.
• Works well for elements of the s and p-blocks; does not
apply to transition element compounds (exceptions)
• Molecular shapes for central atoms
with normal valence
• No more than 4 structural pairs if the
atom obeys the Octet rule
• The combination of s and p orbitals
provides four bonding sites
• Molecular shapes for central atoms with
expanded valence
• Only elements with a principal energy level of 3 or
higher can expand their valence and violate the
octet rule on the high side. This is because d orbitals
are needed for expansion to a 5th or 6th bonding
location.
• Lone pairs “want” to be
as far apart as possible.
For example, look at the
two possible structures
for XeF4 on the right. The
equatorial configuration
is favored (lower
energy) because it
allows the lone pairs to
be as far from each
other as possible.