CH 8 BONDING
AP Chemistry
2014-2015
BONDS
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 (endothermic).
Ionic bond: electrostatic attraction between
oppositely-charged ions; results when a metal
reacts with a nonmetal
Covalent bond: type of bond in which bonding
electrons are shared
 Unequal sharing results in polar covalent bonds;
equal sharing results in nonpolar (or “pure” covalent
bonds. We can categorize a bond qualitatively by
looking at the periodic table, or quantitatively using
electronegativity (more on this later). Most bonds are
somewhere between purely ionic and purely covalent.
EXERCISE 1 BOND T YPES
 The following electrostatic potential maps
correspond to H 2 , LiH, and HF. Determine which map
belongs to each, and explain why.
The map on the left belongs to LiH. LiH is ionic, so the sharing of electron
density will be very unequal.
The map in the middle belongs to H2, because it is nonpolar, so the electron
density is shared equally.
The map on the right belongs to HF. Its electron density is also shared
unequally, but because it is polar covalent, the sharing is more equal than
for LiH.
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 (joules x nanometers), Q
= charge on each ion, r =
distance between ion centers
in nanometers)
 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.
IONIC BONDING
 The final result of ionic bonding is a solid, regular
array of cations and anions called a crystal lattice.
 Lattice energy or lattice enthalpy: energy required to
decompose an ion pair (from a lattice) into ions; a
measure of th e 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
Coulomb’s Law can also be used to
calculate the repulsive energy when two
like-charged ions are brought together.
The energy of this type of interaction
would be positive.
In the top scenario, the energy of
the bond is negative (an
attraction) because of the
opposite charges.
In the bottom scenario, the
energy of the “bond” is positive
(a repulsion) because both
charges are negative.
 Continuing from the idea of positive repulsions and
negative attractions… bond length is the distance
where the sum of the attractive energy forces
(negative sign) and repulsive energy forces (positive
sign) is at a minimum. The attractive forces are due
to the attraction of each nucleus for the electrons in
the other atom in the bond. The repulsive forces are
due to electron-electron repulsion and proton-proton
repulsion.
ELECTRONEGATIVIT Y
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 (Z eff ) and smallest
radius.
 Francium is the least (0.7) due to lowest Z eff and
largest radius.
 This atomic trend is only used when atoms form
molecules.
 Ionic: ΔEN >1.67
 Covalent: ΔEN < 1.67
 Nonpolar covalent: ΔEN < 0.4
BOND POLARIT Y AND
ELECTRONEGATIVIT Y
 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.
EXERCISE 3 RELATIVE BOND POLARITIES
Order the following bonds according to polarity:
H-H, O-H, Cl-H, S-H, and F-H
H-H
δ-
δ+
S-H
δ-
δ+
Cl-H
δ- δ+
O-H
δ- δ+
F-H
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 (valence shell electron pair repulsion)
OCTET RULE
Atoms of main-group elements tend to
combine in such a way that each atom has
eight electrons in its valence shell. This
configuration can be achieved by sharing
or transferring electrons.
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)
 Pi bonds are weaker than sigma bonds but never
exist alone. Triple bonds are stronger than double
bonds are stronger than single bonds.
EXERCISE 4 T YPES OF BONDS
Compare the types of bonds found in ethane (C 2 H 6 ),
ethene (C 2 H 4 ), and ethyne (C 2 H 2 ). Also determine
which molecule would have the strongest carbon carbon bond.
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 d-orbitals and can thus be surround by
more than four valence pairs in certain compounds.
 Odd-electron compounds: ex. NO, NO 2 , ClO 2
DRAWING LEWIS STRUCTURES
For atoms
 Write the symbol
 Determine the number of valence
electrons in the element
 Place the valence electrons (dots)
around the symbol—in order
 Exceptions—carbon and silicon form
hybrid orbitals and do not follow the
usual order
EXERCISE 5 WRITING LEWIS
STRUCTURES (SINGLE ATOMS)
Give the Lewis structure for each of the following.
Sulfur
Chlorine
Silicon
Hydrogen
Nitrogen
Boron
Xenon
Fluorine
FOR COMPOUNDS
 H is always a terminal atom
 Atom with lowest EN goes in center (or, the atom capable of
forming the largest number of bonds)
 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).
 Watch out for charges —you might have to give the Lewis
structure for an ion
EXERCISE 6 WRITING LEWIS
STRUCTURES
Give the Lewis structure for each of the
following.
a) HF
b) N 2
c) NH 3
d) CH 4
e) CF 4
f) NO +
EXERCISE 7 LEWIS STRUCTURES FOR
MOLECULES THAT VIOLATE THE OCTET
RULE
Give the Lewis structure for each of the
following.
a) PCl 5
b) ClF 3
c) XeO 3
d) RnCl 2
e) BeCl 2
f) ICl 4 -
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 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 8 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, 4/3 for
carbonate)
 Bond order = number of shared pairs linking X and Y
number of X-Y links
BOND PROPERTIES
 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.
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 ideal 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 9 FORMAL CHARGES
 Give possible Lewis structures for XeO 3 , 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 .
 VSEPR 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 5 th or 6 th
bonding location.
Lone pairs “want” to
be as far apart as
possible. For example,
look at the two
possible structures for
XeF 4 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.
In general, when
you need to
determine the
most stable
shape for a
molecule, you
should think
about repulsions.
Everything wants
to be as far from
everything else
as possible
(“everything”
being electrons,
of course).
Another example: SF 4 has one lone pair and
four S-F bonds. Configuration (a) is favored
because it minimizes total repulsions (two
90° and two 129° lone-bonding repulsions vs.
three 90° and one 180° lone-bonding
repulsions)—see saw shape.
EXERCISE 10 PREDICTION OF
MOLECULAR STRUCTURE I
Which of the three following arrangements do you
predict to be the most stable for I 3 - ?
EXERCISE 11 PREDICTION OF
MOLECULAR STRUCTURE II
Draw and describe the molecular structure of
the water molecule.
EXERCISE 12 PREDICTION OF
MOLECULAR STRUCTURE III
When phosphorous reacts with excess chlorine gas, the
compound phosphorous pentachloride is formed. In the
gaseous and liquid states, this substance consists of PCl 5
molecules, but in the solid state it consists of a 1:1 mixture of
PCl 4 + and PCl 6 - ions. Predict the geometric structures of PCl 5 ,
PCl 4 + , and PCl 6 - .
MOLECULAR POLARIT Y
 Bonds can be polar while the entire molecule isn’t, and vice
versa.
 Dipole moment: separation of the charge in a molecule;
product of the size of the charge and the distance of
separation.
 Molecules align themselves with an electric field
 Molecules align with each other in the absence of an electric field
 The direction of the “arrow” indicating the dipole moment always
points to the negative pole with the cross hatch on the arrow at the
positive pole.
In water, two lone
pairs establish a
strong negative pole.
Similarly, a negative
pole is established
by the one lone pair
in ammonia (NH 3).
This negative pole is
flipped if we
substitute the
hydrogen atoms for
fluorine to make
nitrogen trifluoride.
If the Octet rule is obeyed and all the surrounding
bonds are the same, then the molecule is
nonpolar since all the dipole moments cancel
each other out.
 Ex. carbon dioxide is nonpolar since the dipole moments
cancel
Ex. methane, CH4, and carbon tetrachloride, CCl4,
are both nonpolar since the dipole moments in
each molecule cancel. However, the molecules
“between” these two (shown below) all have net
dipoles and are therefore polar.
EXERCISE 13 BOND POLARIT Y AND
DIPOLE MOMENT
For each of the following molecules, show the
direction of the bond polarities and indicate which
ones have a dipole moment: HCl, Cl 2 , SO 3 , and H 2 S.
EXERCISE 14 PREDICTION OF
MOLECULAR STRUCTURE IV
Predict the molecular structure of the sulfur
dioxide molecule. Is this molecule expected
to have a dipole moment?
1) VSEPR predicts an SbF 5 molecule will
be which of the following shapes?
A.tetrahedral
B.trigonal bipyramidal
C. square pyramid
D.trigonal planar
E. square planar
2) The shortest bond would be present in
which of the following substances?
A. I2
B. CO
C. CCl4
D. O22–
E. SCl 2