2. Chemical Bonds
• Three basic types of
bonds:
– Ionic
• Electrostatic attraction
between ions
– Covalent
• Sharing of electrons
– Metallic
• Metal atoms bonded to
several other atoms
4. Energetics of Ionic Bonding
As we saw in the last
chapter, it takes 495
kJ/mol to remove
electrons from
sodium.
5. Energetics of Ionic Bonding
We get 349 kJ/mol
back by giving
electrons to chlorine.
6. Energetics of Ionic Bonding
• But these numbers
don’t explain why the
reaction of sodium
metal and chlorine gas
to form sodium
chloride is so
exothermic!
7. Energetics of Ionic Bonding
• There must be a third
piece to the puzzle.
• What is as yet
unaccounted for is the
electrostatic attraction
between the newly
formed sodium cation
and chloride anion.
8. Lattice Energy
• This third piece of the puzzle is the lattice energy:
The energy required to completely separate a mole of a
solid ionic compound into its gaseous ions.
• The energy associated with electrostatic
interactions is governed by Coulomb’s law:
Eel = κ
Q1Q2
d
9. Lattice Energy
• Lattice energy, then, increases with the charge on the
ions.
• It also increases
with decreasing size
of ions.
10. Energetics of Ionic Bonding
By accounting for all
three energies
(ionization energy,
electron affinity, and
lattice energy), we can
get a good idea of the
energetics involved in
such a process.
11. Energetics of Ionic Bonding
• These phenomena also
helps explain the “octet
rule.”
• Metals, for instance, tend to stop losing electrons
once they attain a noble gas configuration because
energy would be expended that cannot be overcome
by lattice energies.
12. Covalent Bonding
• In these bonds atoms share
electrons.
• There are several electrostatic
interactions in these bonds:
– Attractions between electrons
and nuclei
– Repulsions between electrons
– Repulsions between nuclei
13. Polar Covalent Bonds
• Although atoms often
form compounds by
sharing electrons, the
electrons are not always
shared equally.
• Fluorine pulls harder on the electrons it shares
with hydrogen than hydrogen does.
• Therefore, the fluorine end of the molecule has
more electron density than the hydrogen end.
14. Electronegativity is the ability of an atom to attract
toward itself the electrons in a chemical bond.
Electron Affinity - measurable, Cl is highest
Electronegativity - relative, F is highest
X (g) + e-
X-
(g)
9.5
15. Electronegativity:
• The ability of atoms in a
molecule to attract
electrons to itself.
• On the periodic chart,
electronegativity
increases as you go…
– …from left to right across
a row.
– …from the bottom to the
top of a column.
16. Polar Covalent Bonds
• When two atoms share
electrons unequally, a bond
dipole results.
• The dipole moment, µ,
produced by two equal but
opposite charges separated by a
distance, r, is calculated:
µ = Qr
• It is measured in debyes (D).
17. Polar Covalent Bonds
The greater the
difference in
electronegativity,
the more polar is
the bond.
19. Covalent
share e-
Polar Covalent
partial transfer of e-
Ionic
transfer e-
Increasing difference in electronegativity
Classification of bonds by difference in electronegativity
Difference Bond Type
0 Covalent
≥ 2 Ionic
0 < and <2 Polar Covalent
9.5
20. Classify the following bonds as ionic, polar covalent,
or covalent: The bond in CsCl; the bond in H2S; and
the NN bond in H2NNH2.
Cs – 0.7 Cl – 3.0 3.0 – 0.7 = 2.3 Ionic
H – 2.1 S – 2.5 2.5 – 2.1 = 0.4 Polar Covalent
N – 3.0 N – 3.0 3.0 – 3.0 = 0 Covalent
9.5
22. Writing Lewis Structures
1. Find the sum of valence
electrons of all atoms in
the polyatomic ion or
molecule.
– If it is an anion, add one
electron for each
negative charge.
– If it is a cation, subtract
one electron for each
positive charge.
PCl3
5 + 3(7) = 26
23. Writing Lewis Structures
2. The central atom is the
least electronegative
element that isn’t
hydrogen. Connect the
outer atoms to it by
single bonds.
Keep track of the electrons:
26 − 6 = 20
24. Writing Lewis Structures
3. Fill the octets of the
outer atoms.
Keep track of the electrons:
26 − 6 = 20 − 18 = 2
25. Writing Lewis Structures
4. Fill the octet of the
central atom.
Keep track of the electrons:
26 − 6 = 20 − 18 = 2 − 2 = 0
26. Writing Lewis Structures
5. If you run out of electrons
before the central atom
has an octet…
…form multiple bonds until
it does.
27. Writing Lewis Structures
• Then assign formal charges.
– For each atom, count the electrons in lone pairs and half
the electrons it shares with other atoms.
– Subtract that from the number of valence electrons for
that atom: The difference is its formal charge.
28. Writing Lewis Structures
• The best Lewis structure…
– …is the one with the fewest charges.
– …puts a negative charge on the most
electronegative atom.
29. A resonance structure is one of two or more Lewis structures
for a single molecule that cannot be represented accurately by
only one Lewis structure.
O O O
+ -
OOO
+-
O C O
O
- -
O C O
O
-
-
OCO
O
-
- 9.8
What are the resonance structures of the
carbonate (CO3
2
-) ion?
30. Exceptions to the Octet Rule
The Incomplete Octet
H HBe
Be – 2e-
2H – 2x1e-
4e-
BeH2
BF3
B – 3e-
3F – 3x7e-
24e-
F B F
F
3 single bonds (3x2) = 6
9 lone pairs (9x2) = 18
Total = 24
9.9
31. Exceptions to the Octet Rule
Odd-Electron Molecules
N – 5e-
O – 6e-
11e-
NO N O
The Expanded Octet (central atom with principal quantum number n > 2)
SF6
S – 6e-
6F – 42e-
48e-
S
F
F
F
F
F
F
6 single bonds (6x2) = 12
18 lone pairs (18x2) = 36
Total = 48
9.9
32. The enthalpy change required to break a particular bond in one
mole of gaseous molecules is the bond energy.
H2 (g) H (g) + H (g) ∆H0
= 436.4 kJ
Cl2 (g) Cl (g) + Cl (g) ∆H0
= 242.7 kJ
HCl (g) H (g) + Cl (g) ∆H0
= 431.9 kJ
O2 (g) O (g) + O (g) ∆H0
= 498.7 kJ O O
N2 (g) N (g) + N (g) ∆H0
= 941.4 kJ N N
Bond Energy
Bond Energies
Single bond < Double bond < Triple bond
9.10
33. Average bond energy in polyatomic molecules
H2O (g) H (g) + OH (g) ∆H0
= 502 kJ
OH (g) H (g) + O (g) ∆H0
= 427 kJ
Average OH bond energy =
502 + 427
2
= 464 kJ
9.10
34. Bond Energies (BE) and Enthalpy changes in reactions
∆H0
= total energy input – total energy released
= ΣBE(reactants) – ΣBE(products)
Imagine reaction proceeding by breaking all bonds in the
reactants and then using the gaseous atoms to form all the
bonds in the products.
9.10
36. Use bond energies to calculate the enthalpy change for:
H2 (g) + F2 (g) 2HF (g)
∆H0
= ΣBE(reactants) – ΣBE(products)
Type of
bonds broken
Number of
bonds broken
Bond energy
(kJ/mol)
Energy
change (kJ)
H H 1 436.4 436.4
F F 1 156.9 156.9
Type of
bonds formed
Number of
bonds formed
Bond energy
(kJ/mol)
Energy
change (kJ)
H F 2 568.2 1136.4
∆H0
= 436.4 + 156.9 – 2 x 568.2 = -543.1 kJ
9.10
