EXAMPLE 5.1
Lewis Symbols
Without referring to Table 5.1, give Lewis symbols for magnesium, oxygen, and phosphorus. You may use the
periodic table.
Solution
Magnesium is in group 2A, oxygen is in group 6A, and phosphorus is in group 5A. The Lewis symbols
therefore have two, six, and five dots, respectively. They are
•Mg•
•
:O• :
•
:P• •
Exercise 5.1
Without referring to Table 5.1, give Lewis symbols for each of the following elements. You may use the
periodic table.
a. Ar
b. Ca
c. F
d. N
e. K
f. S
EXAMPLE 5.2
Electron Transfer to Form Ions
Use Lewis symbols to show the transfer of electrons from sodium atoms to bromine atoms to form ions with
noble gas configurations.
Solution
Sodium has one valence electron, and bromine has seven. Transfer of the single electron from sodium to
bromine leaves each with a noble gas configuration.
: :
: :
Na• + •Br:  Na+ + :Br:–
Exercise 5.2A
Use Lewis symbols to show the transfer of electrons from lithium atoms to fluorine atoms to form ions with
noble gas configurations.
Exercise 5.2B
Use Lewis symbols to show the transfer of electrons from rubidium atoms to iodine atoms to form ions with
noble gas configurations.
EXAMPLE 5.3
Electron Transfer to Form Ions
Use Lewis symbols to show the transfer of electrons from magnesium atoms to nitrogen atoms to form ions
with noble gas configurations.
Solution
Mg2+
Mg
Mg +
Mg
N
N
 Mg2+ +
Mg2+
N
3–
N
3–
Each of three magnesium atoms gives up two electrons (a total of six), and each of the two nitrogen atoms
acquires three (a total of six). Notice that the total positive and negative charges on the products are equal
(6+ and 6–). Magnesium reacts with nitrogen to yield magnesium nitride (Mg 3N2).
Exercise 5.3
Use Lewis symbols to show the transfer of electrons from aluminum atoms to oxygen atoms to form ions with
noble gas configurations.
EXAMPLE 5.4
Determining Formulas by Electron Transfer
What is the formula of the compound formed by the reaction of sodium and sulfur?
Solution
Sodium is in group 1A; the sodium atom has one valence electron. Sulfur is in group 6A; the sulfur atom has
six valence electrons.
Na
S
Sulfur needs two electrons to gain an argon configuration, but sodium has only one to give. The sulfur atom
therefore must react with two sodium atoms.
Na
Na
+
S

Na+
+
S
2–
Na+
The formula of the compound, called sodium sulfide, is Na 2S.
Exercise 5.4A
What are the formulas of the compounds formed by the reaction of (a) calcium with fluorine, and (b) lithium
with oxygen?
Exercise 5.4B
Use data in Figure 5.4 to predict the formulas of the two compounds that can be formed from iron (Fe) and
chlorine.
EXAMPLE 5.5
Determining Formulas from Ionic Charges
Give the formulas for (a) calcium chloride and (b) aluminum oxide.
Solution
a.
First, write the symbols for the ions. (We write the charge on chloride ion explicitly as “1–” to
illustrate the crossover method. You may omit the “1” when you are comfortable with the process.)
Ca2+ Cl1–
Then cross over the numbers as subscripts.
Then rewrite the formula, dropping the charges. The formula for calcium chloride is
Ca1Cl2 or (dropping the “1”) simply CaCl2
b.
Write the symbols for the ions.
Al3+ O2–
EXAMPLE 5.5
Determining Formulas from Ionic Charges continued
Cross over the numbers as subscripts.
Then rewrite the formula, dropping the charges. The formula for aluminum oxide is
Al2O3
Exercise 5.5
Give the formulas for (a) potassium oxide, (b) calcium nitride, and (c) calcium sulfide.
EXAMPLE 5.6
Naming Ionic Compounds
What are the names of (a) MgS and (b) FeCl3?
Solution
a.
From Table 5.2 we can determine that MgS is made up of Mg2+ (magnesium ion) and S2–
(sulfide ion). The name is simply magnesium sulfide.
b.
From Table 5.2 we can determine that the ions in FeCl3 are
Fe3+
Cl–
How do we know the iron ion in FeCl3 is Fe3+ and not Fe2+? Because there are three Cl– ions, each 1–,
the one Fe ion must be 3+ because the compound FeCl3 is neutral. The names of these ions are
iron(III) ion (or ferric ion) and chloride ion. Therefore, the compound is iron(III) chloride (or, by the
older system, ferric chloride).
Exercise 5.6
What are the names of (a) CaF2 and (b) CuBr2?
EXAMPLE 5.7
Naming Covalent Compounds
What are the names of (a) SCl2 and (b) SF6?
Solution
a.
With one sulfur atom and two clorine atoms, SCl2 is sulfur dichloride.
b.
With one sulfur atom and six fluorine atoms, SF6 is sulfur hexafluoride.
Exercise 5.7A
What are the names of (a) BrF3 and (b) BrF5?
Exercise 5.7B
What are the names of (a) N2O and (b) N2O5?
EXAMPLE 5.8
Formulas of Covalent Compounds
Give the formula for tetraphosphorus hexoxide.
Solution
The tetra- indicates four phosphorus atoms, and the hex- specifies six oxygen atoms. The formula is P4O6.
Exercise 5.8A
Give the formulas for (a) phosphorus trichloride and (b) dichlorine heptoxide.
Exercise 5.8B
Give the formulas for (a) nitrogen triiodide and (b) disulfur dichloride.
EXAMPLE 5.9
Covalent Bonds from Lewis Structures
Use Lewis structures to show the formation of a covalent bond (a) between two fluorine atoms and (b)
between a fluorine atom and a hydrogen atom.
Solution
a.
F + F
F F
bonding pair
b.
H + F H F
bonding pair
Exercise 5.9
Use Lewis structures to show the formation of a covalent bond between (a) two bromine atoms, (b) between a
hydrogen atom and a bromine atom, and (c) between an iodine atom and a chlorine atom.
EXAMPLE 5.10
Covalent Bonds from Lewis Structures
Use data from Figure 5.5 to classify bonds between each of the following pairs of atoms as nonpolar covalent,
polar covalent, or ionic:
a.
H, H
b.
O, H
c.
C, H
Solution
a.
Two H atoms have exactly the same electronegativity; the electronegativity difference is 0; the bond
is nonpolar covalent.
b.
The electronegativity difference is 3.5 – 2.1 = 1.4; the bond is polar covalent.
c.
The electronegativity difference is 2.5 – 2.1 = 0.4; the bond is nonpolar covalent.
Exercise 5.10A
Use data from Figure 5.5 to classify bonds between each of the following pairs of atoms as nonpolar covalent,
polar covalent, or ionic:
a.
H, Br
b.
Na, O
c.
C, C
Exercise 5.10B
Use a periodic table to classify the following bonds as nonpolar covalent or polar covalent:
a.
C—N
b.
C—O
c.
C—
—C
EXAMPLE 5.11
Formulas Using Polyatomic Ions
What is the formula for ammonium sulfide?
Solution
Ammonium ion is found in Table 5.4; sulfide ion is a sulfur atom (group 6A) with two additional electrons.
The ions are
NH4+ S2–
Crossing over,
we get
(NH4+)2S12–
The parentheses with a subscript 2 indicate that the entire ammonium unit is taken twice; there are two
nitrogen atoms and eight (4 x 2 = 8) hydrogen atoms.
EXAMPLE 5.11
Formulas Using Polyatomic Ions continued
Exercise 5.11A
What are the formulas for (a) calcium acetate, (b) ammonium nitrate, and (c) potassium permanganate?
Exercise 5.11B
How many (a) nitrogen atoms are in the formula for ammonium nitrate; and (b) carbon atoms are in the
formula for calcium acetate?
EXAMPLE 5.12
Naming Compounds with Polyatomic Ions
What is the name of the compound NaCN?
Solution
The ions are Na+, sodium ion, and CN–, cyanide ion (found in Table 5.4). The compound is sodium cyanide.
Exercise 5.12A
What are the names of (a) CaCO3 and (b) Mg3(PO4)2?
Exercise 5.12B
What are the names of (a) K2CrO4 and (b) (NH4)2Cr2O7?
EXAMPLE 5.13
Lewis Formulas
Give Lewis formulas for (a) methanol, CH3OH, (b) the BF4– ion, and (c) carbon dioxide, CO2.
Solution
a. We start by following the preceding rules:
1.
The total number of valence electrons is 4 + (4 x 1) + 6 = 14.
2.
The skeletal structure must have all the H atoms on the outside. That means the C and H atoms
must be bonded to each other. A reasonable skeletal structure is
H
H
C
O
H
H
3.
Now, we count five bonds with two electrons each, making a total of ten electron. Thus four of
the 14 valence electrons are left to be assigned. They are placed (as two lone pairs) on the
oxygen atom.
H
H
C
O
H
H
(The remaining steps are not necessary; both carbon and oxygen have octets of electrons.)
EXAMPLE 5.13
Lewis Formulas continued
b. Again, we start by applying the preceding rules:
1.
There are 3 + (4 x 7) + 1 = 32 electrons.
2.
The skeletal structure is
F
F
B
F
F
3.
Place three lone pairs on each fluorine atom.
–
F
F
B
F
F
4.
We have assigned 32 electrons. None remain to be assigned. A negative sign is added to show
that this is the structure for an anion, not a molecule.
EXAMPLE 5.13
Lewis Formulas continued
c. Again, we start by applying the rules:
1.
There are 4 + (2 x 6) = 16 valence electrons.
2.
The skeletal structure is O—C—O.
3.
Place three lone pairs on each oxygen atom.
O—C—O
4.
5.
We have assigned 16 electrons. None remain to be placed.
The central carbon atom has only four electrons. It needs to form two double bonds in order to
have an octet. Move a lone pair from each oxygen atom to the space between the atoms to form a
double bond on each side of the carbon atom.
O
C
O
Exercise 5.13A
Give Lewis formulas for (a) oxygen difluoride, OF2, and (b) methyl chloride, CH3Cl.
Exercise 5.13B
Give Lewis formulas for (a) azide ion, N3–, and (b) nitryl fluoride, NO2F (O—N—O—F skeleton).
EXAMPLE 5.14
Shapes of Molecules
What are the shapes of (a) the H2CO molecule and (b) SCl2 molecule?
Solution
a.
We follow the preceding rules, starting with the Lewis structure.
1. The Lewis structure is
H
C
O
H
2. There are three sets to consider: two C—H single bonds and one C O double bond.
3. The three sets get as far apart as possible, giving a triangular arrangement of the sets.
H
120°
C
O
H
4. All the sets are bonding pairs; the molecular shape is triangular, the same as the arrangement of the
electrons.
EXAMPLE 5.14
Shapes of Molecules continued
b. Again, we follow the rules, starting with the Lewis structure.
1. The Lewis structure is
S
Cl
Cl
2. There are four sets on the sulfur atom to consider.
3. The four sets get as far apart as possible, giving a tetrahedral arrangement of the sets about the
central atom.
4. Two of the sets are bonding pairs and two are LP. Ignore the LP; the molecular shape is bent, with a
bond angle of about 109.5°.
Exercise 5.14A
What are the shapes of (a) the PH3 molecule and (b) the nitrate ion (NO3–)?
Exercise 5.14B
What are the shapes of (a) the carbonate ion (CO32–) and (b) the hydrogen peroxide (H2O2) molecule?