Lecture 9: Covalent Bonding: Part 1 Lewis
Structures (Ch. 7)
Recap
 Ionic compounds are formed by the transfer of electrons from a metal to a
nonmetal
 The result is a cation-anion pair. Each ion has reached its nearest noble gas
configuration
Na ([Ne] 3s1) + Cl ([Ne] 3s2 3p5) ---> Na+ Cl[Ne]
• The Coulombic attraction between the oppositely
charged ions is what holds the molecule together.
This is why ionic compounds are solids at room
temperature.
[Ar]
Na+ Cl-
Covalent Bonding
 Covalent bonding is electron sharing (not transfer) between nonmetals
 Consider Cl2(g)
 Each Cl has 7 valence electrons
[Ne]3s23p5
Cl
Cl
[Ne]3s23p5
 Chlorine is 1 electron short of a full octet ([Ar] configuration)
 Since both atoms are the same, they have the same electron affinities and
ionization energies, so 1 chlorine will not donate an electron to the other.
Covalent Bonding
 For both atoms to achieve an [Ar] configuration (full octet), they share a
pair of electrons between them. Elements in a covalent bond will react so
that at least eight electrons occupy the valence shell. This is the OCTET
RULE.
[Ar]
Cl Cl
[Ar]
Covalent bond
 To indicate the covalent bond, we use a solid line. This is a single bond
(2 electrons). Bonding represents an overlap of the valence orbitals.
Cl
Cl
 The electrons not involved in bonding are called lone pairs.
Notes About the Octet Rule
 Nonmetals with valence shells having a principle quantum number
(n) of 2 must obey the octet rule.
 They will bond to obtain exactly 8 electrons in their valence shell.
NO MORE, NO LESS. (Boron is a metalloid and is exempt).
 n=2, L = 0, 1 (ex. C, N, O, F)
 These elements possess both valence s and valence p orbitals
 s orbitals hold a max of 2 electrons, p orbitals can hold 6 (8 total)
 Elements beyond the 2nd row may violate the octet rule if need be.
This will be discussed later.
Covalent Bonding
 Ex. OF2
 Oxygen has 6 valence electrons (needs 2)
 Fluorine has 7 (each F needs 1)
F
[He]2s22p5
F
O
[He]2s22p4
[He]2s22p5
 To achieve octets, each F will share its electron with O
F O F
F
O
F
[Ne]
[Ne]
[Ne]
Electronegativity
 Before we learn about how to draw a Lewis structure, it is very
important to consider what makes ionic and covalent bonds so
different: electronegativity
 Electronegativity is the ability of an atom to attract electrons to itself.
A more electronegative atom is in greater need of electrons, and will
attract them more strongly than a less electronegative atom (think
“tug-of-war”)
 Thus, when there is a difference in electronegativities between atoms
in a molecule, the electrons are NOT equally shared.
Table of Electronegativities
Electronegativity of
atoms increases up
and to the right.
Electronegativity
 Although we have discussed ionic and covalent bonds, most chemical
bonds are neither purely ionic or purely covalent
 Most compounds are an intermediate between the two.
Bond Type
Covalent
Polar Covalent
(Partially polar)
Ionic (Totally Polar)
Difference In
Electronegativity
< 0.4
0.4 – 2.0
> 2.0
Electronegativity
 Let’s consider NaCl. The difference in electronegativity between Na
and Cl is:
3.16 – 0.93 = 2.23
ionic
 Because the difference in electronegativity is so big, the Na electron is
completely pulled away by the Cl atom. So, the molecule is totally
polar (ionic)
Na+
Cl
 This is why there is no actual bond in ionic compounds, only coulombic
attraction.
Electronegativity
• Let’s look at another molecule, like HCl:
3.16 – 2.1 = 1.06
polar covalent
• The HCl molecule is not purely ionic or covalent, but BOTH.
• The electron density is unevenly distributed, such that more of the
electron cloud is on the Cl than on the H.
• Therefore, the H and Cl have partial charges. The arrow depicts the
direction of electron “pull”, or dipole. Any molecule with a net dipole is
polar. We will discuss dipoles later.
+
δ
Partial positive character
H
Cl
δ
Partial negative character
General Rules of Drawing Covalent Lewis Structures
1.
Arrange atoms together. Put the least electronegative atom in the
center. Hydrogen is an exception to this rule. Hydrogen atoms are
always terminal (at the ends of molecules), as hydrogen only
possesses a 1s orbital and can therefore accommodate a maximum
of two electrons (one bond)
2.
Compute the total number of valence electrons. Account for
charges.
3.
Represent bonds with solid lines. Ensure octets around all atoms,
except hydrogen. Hydrogens can only accommodate 2 electrons.
4.
Add in remaining lone pairs
5.
If there are not enough electrons to complete an octet, consider
multiple bonds or formal charges.
Other Rules
1.
Fluorine is always a terminal atom.
2.
Terminal halogens will only make single bonds.
3.
Carbon atoms will usually have four bonds around them (no lone
pair). A few exceptions exist, like CO, CN-, and carbon ions (you will
learn about these in organic chemistry)
4.
Never draw a Lewis structure possessing an unpaired electron,
unless told explicitly that the molecule is a radical!
Example
 Draw the Lewis structure of CCl4
 Carbon is the least electronegative atom (also, there is only one C), so C
is the central atom
 Compute the valence electrons
 C: [He] 2s2 2p2
Cl: [Ne] 3s2 3p5
• Each Cl needs 1, the C needs 4. Therefore, the C
will share each of its 4 valence electrons with Cl
Cl
Cl
C
Cl
Cl
Cl
Cl C Cl
Cl
Cl
Cl
C
Cl
Cl
Hydrogen
 Hydrogen CAN ONLY ACCOMMODATE 2 ELECTRONS
 Ex. H2(g)
H H
H
H
 HX where X = F, Cl, Br, I (halogens)
H X
H X
 In chemical structures, hydrogens are always terminal atoms, meaning
that they are at the ends of a molecule
Organic Molecules (Carbon-based)
 Ex. Methanol, CH3OH
H
H H
H
O
C
Hydrogens are terminal, which means there is a C—O bond in the
center
C O
Now, C needs 3 more electrons around it to achieve an octet. O needs 1.
H
H C O H
H
H
H
C
H
O
H
Double and Triple Bonds
 There are instances where single bonds (2 e- bonds) are not enough to
satisfy the octet rule.
 Ex. C2H4 (ethene)
C
C H
H H
H C C H
H H
H
• As you can see, each C only possesses 7 total electrons, including a lone
electron. Never draw a Lewis structure with unpaired electrons.
• Lone electrons migrate into the C-C bond to form a 4e- double bond.
H C C H
H H
H C
H
C H
H
H
H
C
C
H
H
Double and Triple Bonds
 6e- triple bonds are also possible, as in the case of N2 (g)
N
N
N N
Two unpaired electrons each
N
N
N
N
6 electron Triple Bond
• The N atoms are sharing 6 electrons, so each N now has a full octet.
Group Examples
 Draw the following Lewis Structures. Include all lone pairs.
 CH3Cl
 NH3
 H2O
 CO2
 C2H2
 H2CO
 NOF
Formal Charges
 So far, all of the atoms we’ve seen in covalent compounds have had a zero
charge.
 However, nonmetals can assume positive or negative charges in order to
facilitate the formation of a covalent bond
 Thus, to satisfy the octet rule in certain cases, formal charges have to be
applied. This is most commonly observed in polyatomic ions.
Formal Charges
 Consider ammonium, NH4+
-Lets start by looking at the valence electrons of the neutral atoms
H H H H
N
- As is, N can only make 3 bonds.
H N H
H
X
H
- N loses an electron to accommodate the 4th H, attains a charge of 1+.
N
+
H+
H N H
H
H
H
+
N H
H
Formal Charge Allows Us to Rule Out Structures
 Formal charges can help us determine if we’ve arranged a molecule
incorrectly
The preferred arrangement of atoms in a molecule is always the
arrangement that requires the least amount of formal charge
 If formal charges must exist, the more electronegative elements prefer
negative formal charges, and less electronegative elements prefer
positive ones.
Ambiguity in Atomic Arrangement
 Ex. Hydrogen cyanide, HCN
 We know that C is the central atom in HCN. But how can we rule out HNC?
Lets consider both possibilities
2
total
H
8
total
C
8
total
N
• As we determined earlier, we have a CN
triple bond and an HC bond. All octets
are satisfied.
• There are no formal charges required
Ambiguity in Atomic Arrangement
2
total
H
2
valence
H
8
total
7
total
N
C
8
valence
8
total
N
C
+
+
-
 If we rearrange N and C, we have the following
structure. As shown, C is 1 electron short of an
octet.
 N can not make anymore bonds because it
can not exceed an octet. N has a 1+ formal
charge because it must lose a valence
electron.
 C must take on a 1- formal charge to reach
an octet by accepting the N electron. Now,
all the atoms are satisfied, and the molecule
is neutral.
Ambiguity in Atomic Arrangement
 Now, our two possibilities are:
2
total
H
8
total
C
8
total
N
2
total
H
8
total
8
total
N
C
+
-
 Both structures satisfy the octet rule, and both are charge neutral, but
the HNC structure has formal charges on both N and C, while the HCN
structure has zero formal charges. Thus, the HCN structure is
preferred.
Some Hints For Drawing Polyatomic Ions
 For anionic molecules containing oxygens, the negative charge will most
likely be on the oxygens. A negatively charged O will only require a single
bond.
 Positive charges are most likely found on the central atom, which is the least
electronegative (not including hydrogen)
 Remember, terminal halogens only require one electron. They will not make
multiple bonds or take on formal charges.
Group Examples
 Ex. Nitrite (NO2-)
6 valence
8 total
0 charge
5 valence
8 total
0 charge
7 valence
8 total
charge = 1-
Single bonded oxygen requires an
additional electron (blue) to reach
octet. Now has 7 valence electrons
instead of 6, giving it a 1- charge.
 Draw the Lewis structures of carbonate
 The correct structure of the NOCl is a central N with a double bond to
O, a single bond to Cl, and a lone pair on the N.
 Use formal charges to show why O is not the central atom.
Resonance
 When a covalent molecule has an overall charge, the charge is said to be
delocalized, meaning that the charge is spread over the whole molecule, as
opposed to being localized on a single atom (wave-particle duality)
 Ex. Nitrite NO2
• You can’t distinguish between one O and the other. So, you could write
either of the following:
-
N
O
O
-
O
N
O
Resonance
• These structure are called resonance structures. We can account for both
structures by writing the resonance form, as shown below. All resonance
structures possess a multiple bond.
O
N
O
• Since the double bond exists simultaneously at both locations, we can
imagine that there are 1 ½ bonds at each site. Therefore, we would
say that the BOND ORDER (average number of bonds per site) is 1 ½.
Expansion into the d-orbital
 Nonmetals with available d-orbitals (n> 3) can use these orbitals to hold
more than 8 electrons. Unless Carbon is present, assume that these atoms
are the central atoms.
 When drawing Lewis structures of molecules with expanded octets, put
octets around the outer atoms first, then account for any remaining
electrons on the central atom.
F
 Ex. SF4
• After completing octets
on the F atoms, there is a
lone pair remaining on S
F
S
F
F
10 electrons
around Sulfur
Group Examples
 Draw the Lewis structures for the following “expanded octet”
molecules
 PCl5
 SF6
 IF5
Always be mindful when
dealing with atoms of n>3 of
the possibility of surpassing
8 valence electrons.
• Draw the following polyatomic ions and report the bond orders:
• sulfite
• phosphate
• perchlorate
Revisiting Dipoles: Polar vs. Nonpolar Molecules
 Dipole moments describe the coulombic attraction that is formed between
the partial positive and negative charges on nonmetals when electrons are
unevenly shared.
 Dipole moments are vectors, quantities with both magnitude and
direction. Imagine pulling on a rope. The force is in the direction of the
pull.
 If two people pull on opposing ends of a rope with equal force, the net
force is zero.
 Polar molecules have an overall dipole moment, nonpolar molecules
don’t.
Table of Electronegativities
Electronegativity of
atoms increases up
and to the right.
Here, electronegativity is labeled as β
δ
-
O
δ
+
C
O
δ
-
Equal force of attraction in opposite
directions: Non polar
+
Δβ = 0.89
Δβ = 0.89
δ
-
O
Δβ = 0.89
=0
δ
+
C
S
Δβ = 0.03
δ
-
Unequal dipole moments. Thus, there is
an overal dipole: Polar
+
=