1. Chapter 8: Ionic and Covalent Bonding
RVCC Fall 2009
CHEM 103 – General Chemistry I
Chapter 8: Ionic and Covalent Bonding
Chemistry: The Molecular Science, 3rd Ed.
by Moore, Stanitski, and Jurs

2. Bonding – What holds atoms together?
Octet rule: To form bonds, atoms gain, lose, or share e- to achieve a valence shell of 8 (or isoelectronic with a noble gas).
Ionic bond –
an electrostatic attraction between a cation and an anion that forms when electrons transfer from one atom to another.
Covalent (Molecular) bond –
the net attractive force that results from the sharing of electrons between atoms.

3. : . . Ionic Bonds Cl Na Na+ Cl-
An ionic bond is formed by the transfer of electrons from one atom (metal with low EA) to another (nonmetal with high EA). The resultant ions are held together by electrostatic attraction. : Cl . Na .
[Ne]3s1 [Ne]3s23p5
Na+ Cl-
[Ne] [Ne]3s23p6 = [Ar]
Each atom has satisfied the octet rule. 2

4. Ionic Bonds
MgF2 : F . : F . Mg . : - : - F Mg 2+ F 2

5. Ionic Compounds - Properties
crystalline
high melting point
high boiling point
soluble in water
electrolytes
Crystal Lattice
hard
brittle

6. Covalent Bonding - G.N. Lewis (1916)
Some atoms share e- to form bonds. When two nonmetals bond, they often share electrons since they have similar attractions (EA) for them. This sharing of valence electrons is called the covalent bond.
Attraction Stable bond Repulsion
Number of bonds = Number shared e- pairs.

7. Covalent Bonds H2 . . : + H H H H 2

8. Covalent Bonds 2

9. Covalent Bonds - Epotential “well”

10. Single Covalent Bonds H H H − H Lewis structures:
show ALL valence electrons
dot = 1 e- line = 1 pair of e- H H
H − H
single bond- one shared pair of e-

11. Single Covalent Bonds 4A 4 4 C in CH4 5A 5 3 N in NF3 6A 6 2 O in H2O
# of e- shared
Group # of to form an octet Example
valence e- (8 - A group#)
4A 4 4 C in CH4
5A 5 3 N in NF3
6A 6 2 O in H2O
7A 7 1 F in HF
# of e- shared
to form an octet
(8-A group#) H
H – C – H
F – N – F F ..
H – O – H ..
H – F ..
H – F .. ..

12. : : : : : Lewis Structures H Cl H Cl
Lewis electron-dot formulas or Lewis structures.
bonding pair : H Cl
lone pairs : :
bonding pair : H Cl :
lone pair
An electron pair is either a bonding pair (shared between two atoms) or a lone pair (an electron pair that is not shared). 2

13. Multiple Bonds
In the molecules described so far, each of the bonds has been a single bond, that is, a covalent bond in which a single pair of electrons is shared.
It is possible to share more than one pair. A double bond involves the sharing of two pairs between atoms.
C has octet.
H OK with 2. C : H or 2

14. Elements that form multiple bonds: C, O , N, S
Triple bonds are covalent bonds in which three pairs of electrons are shared between atoms. C or H :
:::
Elements that form multiple bonds: C, O , N, S 2

15. The Procedure
Using the molecular formula, count the total number of valence electrons available (bonding + lone pairs).
Valence electrons for each atom corresponds to group #
Adjust for charge (add electron for each minus, delete electron for each plus)
2(1) + 6
3(1)

16. The Procedure
Make a skeleton by connecting the atoms with single bonds only. When connecting atoms, remember…
Put the least electronegative atom in the center. (Usually the first listed in the chemical formula.)
Hydrogen is ALWAYS a terminal atom. More electronegative atoms are terminal (F, O…)
Make the structure symmetric.
4 pairs pairs
2 pairs left pair left

17. The Procedure
Put the left over electrons as lone pairs, preferably on the more electronegative atoms Is the octet rule satisfied?
If YES, then you’re done…

18. The Procedure
If you are electron-deficient (not enough electrons to complete an octet), then some atoms must share more than two electrons. “If you have a lone pair, make those two atoms share.”
Ex. C2H4

19. The Procedure
If you have excess electrons, at least one atom must have an expanded valence
Must be element from third period or lower
Usually the central atom
e.g. SF4

20. General Rule
The total number of valence electrons on an atom (from bonds & lone pairs) cannot exceed that atom’s maximum valence.
First period: 2 electrons (s)
Second period: 8 electrons (s,p)
Third period & below: prefer to have 8, but can expand when necessary (s,p,d)

21. Writing Lewis Dot Formulas
20 e- total or 10 pairs
SCl2
8 left
0 left : : : : : Cl S Cl : : : 2

22. Writing Lewis Dot Formulas
24 e- total pairs
COCl2 :
9 left Cl C O
0 left : :
Note that the carbon has only 6 electrons. 2

23. Writing Lewis Dot Formulas
COCl2
12 pairs
9 e- left : Cl C O
0 e- left : :
To fulfill the octet rule…
“If you have a lone pair, make those two atoms share!” 2

24. Writing Lewis Dot Formulas
COCl2
24 e- total
18 e- left : O :
0 e- left C : : Cl Cl
Note that the octet rule is now obeyed. 2

25. Writing Lewis Dot Formulas Practice
N2 SF4
O2 ClO3-
HCN ClO2-1
PO4-3 NO3-1

26. Question no no no no no First evaluate the total valence electrons:
26 e-, wrong
24 e-, looks OK
24 e-, one F has too many
24 e-, N not enough
24 e-, but least electronegative has to be in the center
24 e-, no bond between two N.

27. Exceptions to the Octet Rule
Although many molecules obey the octet rule, there are exceptions where the central atom has less or more than eight electrons.
Incomplete octet – B, H
Boron has 3 valence electrons
BF3
:F – B – F:
:F: .. 2

28. Exceptions to the Octet Rule
If a nonmetal is in the third period or greater it can accommodate as many as twelve electrons as the central atom.
PF5
: F : :
F : :
: F : P
F : : :
: F : 2

29. Exceptions to the Octet Rule
In sulfur tetrafluoride, SF4, the sulfur atom must accommodate two extra lone pairs for a total of 5 electron pair (10 electrons)
: F :
F : : : S
: F :
F : : 2

30. Formal Charge and Lewis Structures
In certain instances, more than one feasible Lewis structure can be illustrated for a molecule. For example, : : H C N H N C or
The concept of “formal charge” can help us decide which structure is correct. 2

31. Formal Charge and Lewis Structures
formal charge = valence e- before bonding– valence e- after bonding
= valence e- - [1/2 bonding e- + lone pair e-] : : H C N H N C or
H: 1-½(2) = 0
H: 1-½(2) = 0
C: 4 - ½(8) = 0
C: 4 – (½(6)+2) = -1
N: 5 – (½(8) + 2) = 0
N: 5 – (½(8)) = +1 2

32. Formal Charge and Lewis Structures
Smaller formal charges are more favorable
More electronegative (or higher EA) atom should have negative formal charges
Like charges should not be on adjacent atoms
Net formal charge should be the overall charge on the molecule/ion. or H C N :
formal charges +1 -1 2

33. Practice
Determine the most stable structure for dinitrogen oxide. (All structures have 16 valence electrons.)
N=N=O N-N≡O N ≡ N - O
formal charge= valence e- - [1/2 bonding e- + lone pair e-]

34. Practice - Formal Charge
Which structure is correct? : O N Cl O N Cl or -1 +1

35. Delocalized Bonding: Resonance
The structure of ozone, O3, can be represented by two different Lewis electron-dot formulas. O : O :
The bond lengths for the above structures are:
O – O 132 pm O = O 112 pm
However, experiments show that both bonds are identical. 2

36. Delocalized Bonding: Resonance
According to theory, one pair of bonding electrons is spread (delocalized) over the region of all three atoms. O
In fact, the actual bond length is pm (in between 132 and 112pm).
The actual molecule is a hybrid or composite structure and not different structures that change back and forth… although, we often represent it that way. 2

37. Delocalized Bonding: Resonance
Lewis resonance structures, have the same atoms in the same positions. Only an electron pair position is different. O
H O N O
H O N O
H O N 2

38. Resonance Structures O=S O O S=O S=S O S O=S
Which pair does NOT represent resonance structures?
O=S O O S=O
S=S O S O=S

39. Resonance Structures O O O C O C O O C
Draw resonance structure(s) for the following: -2 O
O C -2 O
O C O
O C -2

40. “All covalent bonds are created equal but some are more equal than others.”
(We assumed equal sharing when we calculated formal charge.)

41. Electronegativity…
…is a measure of the ability of an atom in a molecule to draw bonding electrons to itself when bonded.
Periodic Trend - Electronegativity
increases
across a period
decreases
down a group

42. Electronegativity Notice, there are NO values for
EN for the noble gases.

43. Types of Bonds Ionic: Na+ Cl- Covalent: C - H Metallic: Zn Zn
electron transfer
Covalent:
ΔEN <1.8
electron sharing
Metallic:
electron-sea model or band theory
Na+ Cl-
C - H
Zn Zn

44. Covalent Bonds (EN<1.8)
Non-polar covalent - ΔEN = 0 – 0.5
Examples: H-H, Cl-Cl, C-H bonds
Polar covalent - ΔEN = 0.5 – 1.8
Examples: H-O, C-Cl, C-O bonds 2

47. : : H :Cl: Bond Polarity d+ d-
In HCl we have a partial negative charge on the chlorine (denoted d-) and a partial positive charge on the hydrogen (denoted d+)
The bond is polar covalent. d+ d- :
H :Cl: :

48. Bond Polarity I Br Cl F non polar most polar
Arrange the following bonds from the most to the least polar:
HH, HCl, HF, HI, HBr
Compare the electronegativity of Cl, F, I and Br:
Least Most
Electronegative Electronegative
I Br Cl F
Determine polarity:
HH HI HBr HCl HF
non polar most polar

49. Practice Which of the following bonds in each pair are more polar?
C-S or C-O
Cl-Cl or O=O
N-H or C-H

50. Bond Length
Bond length (or bond distance) is the distance between the nuclei of two bonded atoms (the sum of atomic radii). 2

51. Bond Length The size of atoms that form the bond determine the length.
C - N 147 pm
C - C 154 pm
C - P 187 pm (P, period 3)

52. Bond Length – Multiple Bonds
As the electron density between atoms increases the bond
lengths decrease; the atoms are pulled together more strongly.
decrease

53. Bond length Bond enthalpy
Bond Enthalpy – the enthalpy change that occurs when the bond between two bonded atoms in the gas phase is broken and the atoms are separated completely at constant pressure.
As the electron density between two atoms increases, the bond
gets shorter and stronger.
Bond length Bond enthalpy C
154 pm
134 pm
120 pm
835 kJ/mol
602 kJ/mol
346 kJ/mol
NaCl
lattice energy
-786 kJ/mol
MgO
-3791 kJ/mol

54. Bond Enthalpy
Bond Enthalpies can be used to calculate the standard enthalpies of reaction (gas phase, STP)
DHº = ∑[(moles of bonds) × D(bonds broken)] -
∑[(moles of bonds) × D(bonds formed)]
DHº - standard enthalpy of reaction

55. Bond Enthalpy Estimate the DHº for the following reaction:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) : : : : :
+ 2 :O = O:  : O = C = O: + 2H – O - H :
4 C – H bonds O = O bond 2 C = O bond H – O bond
per molecule per molecule per molecule per molecule

56. + 2 :O = O:  : O = C = O: + 2H – O - H :
4 C – H bonds O = O bond 2 C = O bonds H – O bonds
per molecule per molecule per molecule per molecule
4 C – H bonds O = O bonds C = O bonds H – O bonds
DHº = ∑[(moles of bonds) × D(bonds broken)] -
∑[(moles of bonds) × D(bonds formed)]
= [4 × D(C-H) + 2 × D (O=O)] – [2 × D (C=O) + 4 × D(H-O)] =
[4 × × 498] – [2 × × 467] = -814 kJ