AP Chemistry
Basics of Chemical Bonding
Properties of substances are largely dependent
on the bonds holding the material together.
Basics of Bonding
A chemical bond occurs when atoms or ions are
strongly attached to each other.
Ionic bonds involve the transfer of e– and the
subsequent electrostatic attractions.
-- “metal/nonmetal” (“cation/anion”)
Covalent bonds involve the sharing of e–
between two atoms.
-- “nonmetal/nonmetal”
metallic bonds: each metal atom is
bonded to several neighboring atoms
-- bonding e– (i.e., valence e–) are free to move
throughout the material
Lewis symbols show ONLY the
valence e– (i.e., the ones
involved in bonding).
C
N
S
octet rule: atoms “want” 8 v.e–
-- several exceptions (a topic for later) Gilbert Lewis
(1875–1946)
Ionic Bonding
Na(s) + ½ Cl2(g)
Na + Cl
NaCl(s)
Na+ + [ Cl ]
lattice energy: the energy required
to separate 1 mole of solid ionic
compound into gaseous ions
-- a measure of stability
NaCl(s)
Na+(g) + Cl–(g)
DHlatt = +788 kJ/mol
“Salts” are brittle
– solids with high
melting points.
DHfo = –410.9 kJ/mol
enthalpy
(i.e., heat)
of formation
In general, ionically bonded substances have...
…big, (+) lattice energies.
Because lattice energies are electrostatic in nature,
two variables are involved in how big they are:
1. the magnitude of the charges
(“wears the pants”)
2. the separation between the ions
Charge can easily be twice as large
(i.e., 2+ and 2– vs. 1+ and 1–), but the
separation never varies by that much.
Put the following in order of
increasing lattice energy:
LiBr, FeN, CdO.
LiBr < CdO < FeN
Now these:
MgS, MgCl2, MgO.
MgCl2 < MgS < MgO
With transition-metal ions, the e– lost first come
from the subshell with the largest value of n.
e.g.,
Ni = [Ar] 4s2 3d8
Ni2+ = [Ar] 3d8
Ni3+ = [Ar] 3d7
Recall that polyatomic ions are groups of atoms that
stay together and have a net charge.
-- e.g., NO3–
CH3COO–
-- their atoms are held to each other by… covalent
bonds
Cations are smaller than the neutral atoms from
which they are derived.
e.g.,
Li
1s2 2s1
Li+
1s2
-- orbitals might be completely vacated
-- e–/e– repulsion always decreases
Fe
Fe2+
Fe3+
Anions are larger than the neutral atoms from
which they are derived.
-- more e–/e– repulsion
Cl
p+,
17
17 e–
Cl–
17 p+,
18 e–
An isoelectronic series is a list of species having an
identical electron configuration.
e.g., N3–, O2–, F–, Ne, Na+, Mg2+, Al3+
(big radius)
(small radius)
EX. Which has the
largest radius?
Rb+
Sr2+
(smallest Zeff)
Y3+
(largest Zeff)
Covalent Bonding
-- atoms share e–
-- covalent (molecular)
compounds tend to be
solids with low melting
points, or liquids or gases
+
+
(–) charge density
-- one shared pair
= a single covalent bond
of e– (i.e., 2 e–)
-- two shared pairs = a double covalent bond
of e– (i.e., 4 e–)
-- three shared pairs = a triple covalent bond
of e– (i.e., 6 e–)
bond polarity: describes the sharing of e– between
atoms
nonpolar covalent bond: e– shared equally
polar covalent bond: e– NOT shared equally
electronegativity (EN): the ability of an atom
in a molecule to
attract e– to itself
-- A bonded atom w/a large EN
has a great ability to attract e–. max. EN = 4.0 (F)
-- A bonded atom w/a small EN
min. EN = 0.7 (Cs)
–
does not attract e very well.
-- EN values have been tabulated. (see p. 308)
The DEN between bonded atoms approximates
the type of bond between them.
DEN < 0.5
0.5 < DEN < 2.0
DEN > 2.0
nonpolar covalent
polar covalent
ionic bond
As DEN increases,
bond polarity... increases.
DEN for a C–H bond = 0.4, so
all C–H bonds (such as those
in candle wax) are nonpolar.
Dipole Moments
Polar covalent molecules have a partial (–) and
a partial (+) charge and are said to have a
dipole moment.
d+ d–
H–F
partial
charge
H–F
d+
H
d+
H
O
d–
big polarity = _____
big dipole moment
big DEN = _____
Polar molecules tend to align themselves with each
other and with ions.
H–F
NO3–
NH4+
H–F
–
NH
NO3
4
+
** Nomenclature tip: For binary compounds, the less
electronegative element comes first.
-- Compounds of metals w/high ox. #’s (e.g., 4+ or
higher) tend to be molecular rather than ionic.
e.g., TiO2, ZrCl4, Mn2O7
Lewis Structures (or “electron-dot structures”)
1. Sum the valence e– for all atoms. If the species is an
ion, add one e– for every (–); subtract one e– for every (+).
2. Write the element symbols and connect the
symbols with single bonds.
3. Complete octets for the atoms on the exterior of
the structure, but NOT for H.
4. Count up the valence e– on your L.S. and compare
that to the # from Step 1.
-- If your LS doesn’t have enough e–, place as many e–
as needed on central atom.
-- If LS has too many e– OR if central atom doesn’t have
an octet, use multiple bonds.
Draw Lewis structures for the following species.
10 e–
..
HCN
..
..
..
26 e–
..
PCl3
.. .. ..
Cl–
.. P–Cl
..
Cl
..
..
H–C–N
..
H–C=N
..
H–C–N
..
..
..
[
]
..
..
..
..
PO43– 32 e–
O
..
..
O– P –O
..
..
O
..
3–
H2
CH3CH2OH
2 e–
20 e–
H–H
H H ..
H–C–C–O–H
..
H H
This CO bond is longer (and
weaker) than this CO bond.
.. ..
O–C–O
.. ..
..
16 e–
..
CO2
O=C=O
As the # of bonds between two atoms increases,
the distance between the atoms... decreases.
formal charge: the charge a bonded atom would have
if all the atoms had the same
electronegativity; to find it, you must
first draw the Lewis structure
# of v.e– in
# of e– assigned
formal = the isolated – to the atom in the
charge
atom
Lewis structure
When several Lewis structures are possible,
the most stable is the one in which:
–
all
unshared
e
(1) the atoms have the smallest
plus half of the
formal charges, and
bonding e–
(2) the (–) charges reside on the
most electronegative atoms
Find the formal charge on each
atom in the following species.
v.e–
assigned e–
16 e–
]
..
N–C–S
..
[
..
..
NCS–
5 4 6
7 4 5
2– 0 1+
[
N=C=S
5 4 6
6 4 6
1– 0 0
O
6
7
+1 –1
(N is more EN than S.)
–
]
[
–
N–C–S ]
5 4 6
5 4 7
0 0 1–
..
S: +1
O: –1
–
S
6
5
2–
..
]
..
..
e–
..
..
26
[
.. .. ..
O– S –O
..
..
O
..
..
..
SO3
2–
resonance structures: the two or more Lewis
structures that are equally correct for a species
..
..
Resonance structures are
a blending of two or more
Lewis structures.
–
] [
..
..
O– N–O
..
..
]
..
] [
..
O= N–O
..
O
..
..
–
..
..
[
..
O– N=O
..
O
..
24 e–
..
..
e.g., For NO3–...
O
In drawing resonance
structures, e–s move.
Atoms don’t move.
–
Aromatic compounds are based on
resonance structures of the benzene molecule.
H
H
C
H
C
C
H
H
C
C
C
H
H
H
C
C
C
C
H
In the interest of efficiency,
benzene is often
represented like this…
C
H
C
H
H
Exceptions to the Octet Rule
There are a few cases (other than for H) in
which the octet rule is violated. These are:
1. particles with an odd number of valence e–
-- e.g., ClO2, NO, NO2
2. atoms with less than an octet
-- e.g., B, Be
..
..
..
..
F= B –F
..
..F
-- would require resonance
-- F is WAY more EN than B
and won’t share extra e–
-- inconsistent w/chemical behavior
B
..F
..
..
(24 e–) BF3
3. atoms with more than an octet
-- this occurs when an atom gains an expanded
valence shell
(40 e–)
-- other e.g., AsF6–, ICl4–, SF4
P
Cl
..
..
..
PCl5
Expanded valence shells occur only for atoms
in periods > 3.
-- unfilled d orbitals are involved
-- large central atom = more space to accept
e– pairs
-- small exterior atoms = fit more easily around
central atom
The strengths of a molecule’s covalent bonds are
related to the molecule’s stability and the amount of
energy required to break the bonds.
bond enthalpy: the DH req’d
to break 1 mol of a particular
bond in a gaseous substance
-- big DH = strong bond
(g)
..
..
-- e.g.,
.. ..
..I–I..
At room temp., I2 is
a silver-purple solid.
2
..
..I
(g)
..
.
-- always +
DH = +149 kJ/mol
.
..
.
..
. .
..
atomization: the process of breaking a molecule into
its individual atoms
H ..
..
H–C–Cl
C (g) + 3 H (g) + Cl
.. (g)
.. (g)
H
methylchloride
(chloromethane)
-- DH for a given bond (e.g., the
C–H bond) varies little between
compounds.
e.g., C–H bonds in CH4 vs. those
in CH3CH2CH3 have about
the same DH
CH3Cl was once used
as a refrigerant, but it is
toxic and so is now used
primarily in the production
of silicone polymers. It is
also used as an
organic solvent.
-- Typical values of bond enthalpies for specific
bonds have been tabulated. (p. 326)
-- To find bond enthalpy for atomization, add
up bond enthalpies for each bond broken.
Calculate the bond enthalpy for the
atomization of dichloromethane.
Cl
H–C–Cl
H
(i.e., break 2 C–H
bonds and
2 C–Cl bonds)
DHatm = 2 (413 kJ/mol)
+ 2 (328 kJ/mol)
=
1482 kJ/mol
You can approximate reaction enthalpy using
Hess’s law and tabulated bond enthalpies.
DHrxn  S(DHbroken bonds) – S(DHformed bonds)
Approximate the reaction enthalpy for the combustion
of propane. C3H8 + 5 O2
4 H2O + 3 CO2
Formed:
O=O
2 C–C: 2 (348)
8 C–H: 8 (413)
5 O=O: 5 (495)
8 O–H: 8 (463)
6 C=O: 6 (799)
O=C=O
..
Broken:
H
H–O
..
H H H
H–C–C–C–H
H H H
6475 kJ
–2023 kJ
(Actual = –2220 kJ)
8498 kJ
bond length: the center-to-center distance
between two bonded atoms
-- fairly constant for a given bond (e.g.,
the C–H bond), no matter the compound
e.g., C–H bonds in CH4 are about the same length
as those in CH3CH2CH3
-- Average bond lengths have been
tabulated for many bonds. (p. 329)
-- As the number of bonds between two atoms
increases, bond length…
and bond enthalpy…
e.g., C–C
1.54 A
348 kJ/mol
C=C
1.34 A
614 kJ/mol
C–C
1.20 A
839 kJ/mol