UNIT (2) ATOMS AND ELEMENTS
2.1 Elements
An element is a fundamental substance that cannot be broken down by chemical means
into simpler substances.
Each element is represented by an abbreviation called the symbol of the element. The
first letter in the symbol of the element is always capitalized, however the second letter
(if present), is never capitalized.
The following table lists the names and symbols of some common elements.
You are expected to learn the name and the symbol of these elements:
Element
Aluminum
Argon
Barium
Boron
Bromine
Calcium
Carbon
Chlorine
Chromium
Cobalt
Copper
Fluorine
Symbol
Element
Symbol
Element
Symbol
Al
Ar
Ba
B
Br
Ca
C
Cl
Cr
Co
Cu
F
Gold
Helium
Hydrogen
Iodine
Iron
Lead
Lithium
Magnesium
Manganese
Mercury
Neon
Nickel
Au
He
H
I
Fe
Pb
Li
Mg
Mn
Hg
Ne
Ni
Nitrogen
Oxygen
Phosphorous
Potassium
Silicon
Silver
Sodium
Strontium
Sulfur
Tin
Xenon
Zinc
N
O
P
K
Si
Ag
Na
Sr
S
Sn
Xe
Zn
2.2 The Periodic Law and the Periodic Table
The periodic law states that the properties of elements exhibit a repeating pattern when
arranged according to increasing atomic number.
Periodic table: A chart of the elements arranged in order of increasing atomic number.
The elements are arranged in rows (periods) that create vertical columns of elements
(group) that exhibit similar chemical properties.
A box differentiates each element and contains its atomic number, atomic symbol, and
atomic mass. For example, locate nitrogen on the periodic table. You will find it in the
second horizontal row and fifth vertical column; it is therefore a second period, group VA
element. The symbol for nitrogen is (N), the atomic number is 7 and the atomic mass is
14.01 amu. (We will see more on atomic number and atomic mass shortly).
2-1
The Periodic Table
Group
A-group
Period
B-group
Main-group
elements
A-group
Main-group elements
Transition elements
Lanthanides and Actinides
Periods and Groups
A period is a horizontal row of elements and a group is a vertical column of elements.
The eight A-groups (two on the left and six on the right) contain the main group
elements. The ten B-groups, located between the two A-groups, contain the transition
elements.
Certain groups have common names that you should learn:
Group IA are the alkali metals (except for hydrogen).
Group IIA are the alkaline earth metals.
Group VIIA are the halogens.
Group VIIIA are known as noble gases (or inert gases).
Metals, Nonmetals, and Metalloids
Metals occupy the left side of the table and they have similar properties. They are good
conductors of heat and electricity and most of the elements are metals.
Nonmetals occupy the upper right side of the table and have more varied properties.
Metalloids are sometimes called semiconductors because of their intermediate electrical
conductivity, which can be controlled and changed.
Metalloids lie along the stepwise diagonal line beginning at boron (B) and extending
downward through polonium/astatine (Po/At). This line separates the metals to the left
and nonmetals to the right.
The following seven elements exist as diatomic molecules:
H2
N2
O2
F2
Cl2
Br2
I2
2-2
Practice 2-1
Write the name and the symbol of the element that fits each of the following
descriptions:
a) the alkaline earth metal in the sixth period.
b) the metalloid in the third period.
c) the nonmetal in group IVA.
d) the halogen that is liquid at room temperature.
e) the group-VIIIB transition metal with properties similar to Ru.
f) the third-period element that exists as diatomic molecule.
Answer
a) Barium, Ba
b) Silicon, Si
c) Carbon, C
d) Bromine, Br
e) Iron, Fe
f) Chlorine, Cl
2.3 Composition of the Atom
Atom: The smallest unit of an element that retains the properties of that element.
Atoms are composed of three subatomic particles: protons, neutrons, and electrons.
Protons and neutrons are located in the center of the atom and form a compact core
termed the nucleus. The electrons are located in the considerably larger space outside the
nucleus.
Table 2.1 Properties and Location of Subatomic Particles
Subatomic
Particle
Proton
Neutron
Electron
Charge
+1
0
-1
Mass
(g)
1.67 x 10-24
1.67 x 10-24
9.11 x 10-28
Mass
(amu)*
1.01
1.01
0.000548
Location
nucleus
nucleus
outside nucleus
* Atomic mass unit (abbreviated amu) is convenient for representing the mass of very
small particles.
1 amu = 1.67 x 10-24 g
2-3
Each nucleus is characterized by 2 quantities: the atomic number and the mass number.
The atomic number is the number of protons contained in the nucleus of an atom.
The mass number is the total number of protons and neutrons in the nucleus of an atom.
Atoms are electrically neutral; therefore the number of electrons (negative charges) is
equal to the number of protons (positive charges). Given this fact, the atomic number also
gives the number of electrons for neutral atoms.
Consider the element sodium with atomic number of 11 and the mass number of 23.
The mass number is written in the upper-left corner (as a superscript) of the symbol for
the element, and the atomic number in the lower-left corner (as a subscript).
mass number 23
Na
atomic number
11
Worked Example 2-1
State the number of protons, neutrons, and electrons in an atom of each of the
following:
Solution
The subscript value refers to the atomic number (p+), and the superscript value
refers to the mass number (p+ and n0).
Thus,
has 3 p+ and 4 n0 (7 – 3 = 4)
In a neutral atom, the number of protons in the nucleus is exactly equal to the
number of electrons outside the nucleus.
Thus,
has 3 electrons.
# protons
3
7
12
12
12
# neutrons
4
7
12
13
14
# electrons
3
7
12
12
12
2-4
Worked Example 2-2
The nucleus of an atom contains 15 protons and 16 neutrons. Write the symbol for
the atom.
Solution
The atomic number equals the number of protons (15) and the mass number
equals the sum of the protons and neutrons (15 + 16 = 31).
The element with an atomic number of 15 is phosphorous (see the periodic table).
Therefore the symbol of the atom is
mass number 31
P
atomic number
15
Practice 2-2
The nucleus of an atom contains 27 protons and 33 neutrons. Write the symbol for
the atom.
Answer
The atomic number equals the number of protons (27) and the mass number
equals the sum of the protons and neutrons (27 + 33 = 60).
The element with an atomic number of 27 is cobalt (see the periodic table).
Therefore the symbol of the atom is
mass number 60
Co
27
atomic number
2-5
2.4 Isotopes and Atomic Masses
A magnesium atom will always have 12 protons; however, the number of neutrons can
vary. Some magnesium atoms found in nature have 12 neutrons, some have 13, and
some have 14.
Atoms with the same number of protons but different numbers of neutrons are called
isotopes. Isotopes are the same element with different atomic masses.
Most elements found in nature exist in isotopic forms. We often refer to an isotope by
stating the elements name followed by its mass number, for example, magnesium-24.
All naturally occurring magnesium atoms contain 78.99% magnesium-24, 10.00%
magnesium-25, and 11.01% magnesium-26. The atomic mass of an element reported on
the periodic table is the weighted average mass of all naturally occurring isotopes of the
elements.
Natural abundance and atomic masses of several isotopes are listed in the following table:
Isotope
Atomic mass
(amu)
Percent abundance
in nature
1.0078
99.985
2.0140
0.015
23.9850
78.99
24.9858
10.00
25.9826
11.01
57.9353
68.27
58.9302
26.10
60.9310
1.13
61.9283
3.59
63.9280
0.91
2-6
The following worked example illustrates the calculation of atomic mass from isotopic
mass and abundance data.
Worked Example 2-3
Calculate the atomic mass of magnesium using the three naturally occurring
isotopes below:
(atomic mass 23.9850 amu, abundance 78.99%)
(atomic mass 24.9858 amu, abundance 10.00%)
(atomic mass 25.9826 amu, abundance 11.01%)
Solution:
The atomic mass of magnesium is called a weighted average of the atomic masses
of these three isotopes. To calculate the atomic mass, simply multiply the atomic
mass by the percent abundance in decimal form (the percent divided by 100), then
add the results.
Fraction Mg-24 =
78.99
100
Fraction Mg-25 =
10.00
= 0.1000
100
Fraction Mg-26 =
11.01
= 0.1101
100
= 0.7899
(23.9850 amu x 0.7899) + (24.9858 amu x 0.1000) + (25.9826 amu x 0.1101) = 24.31 amu
The “average” mass of a magnesium atom is 24.31 amu, although you should note
that no magnesium atom actually exists with a mass of 24.31 amu. In nature,
magnesium atoms weigh 23.9850 amu or 24.9858 amu or 25.9826 amu. A mass of
24.31 amu represents the mass of a hypothetical average magnesium atom.
Mathematics tells that an average of data points is a more reliable representation of
the data than any individual value; we remain consistent with that principle here.
The mass reported on the periodic table for magnesium is 24.31 amu.
See the periodic table.
2-7
Practice 2-3
Calculate the atomic mass of chlorine given the two naturally occurring isotopes
below.
(atomic mass 34.969 amu, abundance 75.77%)
(atomic mass 36.966 amu, abundance 24.23%)
Answer
(34.969 amu x 0.7577) + (36.966 amu x 0.2423) = 35.45 amu
2.5 Electron Arrangement
Energy Levels
Atoms have electrons in several different energy levels. These energy levels are
symbolized by n; n can equal 1, 2, 3, 4, … Level 1 corresponds to n = 1, level 2
corresponds to n = 2, and so on. The energy of the level increases as the value of n
increases.
The maximum number of electrons per energy level = 2n2.
For example, the third energy level (n = 3) holds a maximum of 18 electrons (2 x 32 =
18).
Sublevels (Subshells)
Each energy level is divided into sublevels, symbolized as s, p, d, and f.
Energy level 1 consists of one sublevel, labeled 1s.
Energy level 2 consists of two sublevels, labeled 2s and 2p.
Energy level 3 consists of three sublevels, labeled 3s, 3p, and 3d.
Energy level 4 consists of four sublevels, labeled 4s, 4p, 4d, and 4f.
2-8
Orbitals
An atomic orbital is a specific region of a sublevel containing a maximum of two
electrons.
The orbital on sublevel s is called s orbital. There is only one possible s orbital.
The orbitals on sublevel p are called p orbitals. There are three possible p orbitals.
The orbitals on sublevel d are called d orbitals. There are five possible d orbitals.
The orbitals on sublevel f are called f orbitals. There are seven possible f orbitals.
Orbital
types
s
p
d
f
number of
orbitals
1
3
5
7
max number
of electrons
2
6
10
14
Shape of orbitals:
2-9
Practice 2-4
How many orbitals exist in the third energy level? What are they?
Answer
Total of 9 orbitals.
3s (ONE orbital), 3p (THREE orbitals), and 3d (FIVE orbitals).
Electron Configuration
The distribution of electrons in atomic orbitals is called the atom electron configuration.
Orbitals fill in the order of increasing energy from lowest to highest. See Figure 2.1.
Orbital energies: s < s < p < s < p< s< d < p < s < d < p < s……
____ ____ ____
5p
5p
5p
____ ____ ____ ____ ____
4d
4d
4d
4d 4d
____
5s
____ ____ ____
4p
4p
4p
____ ____ ____ ____ ____
3d
3d
3d
3d 3d
e
n
e
r
g
y
____
4s
____ ____ ____
3p
3p
3p
____
3s
____ ____ ____
2p
2p
2p
____
2s
____
1s
Figure 2.1
Approximate relative energies of atomic orbitals. Electrons fill orbitals in the
order of increasing energy.
2-10
Writing Electron Configurations (Orbital Notations)
We describe an electron configuration by listing the symbols for the occupied sublevels
one after another, adding a superscript to each symbol to show the number of electrons in
the sublevel. For example, nitrogen in its lowest energy state has two electrons in its 1s
sublevel, two electrons in its 2s sublevel, and three electrons in its 2p sublevel.
The notation for this configuration is: 1s22s22p3.
Worked Example 2-4
Write the electron configuration for an atom of bromine (Br).
Solution
The correct filling order of atomic orbitals from Fig 2.1 is
1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p ………….
An s orbital can hold 2 electrons, p orbitals 6, d orbitals 10, and f orbitals 14
electrons.
Bromine is element 35 and therefore contains 35 electrons. The electron
configuration is
1s2 2s2 2p6 3s2 3p6 4s23d104p5
Practice 2-5
Give the electron configuration for each of the following: Mg, P, Cl, and Ca.
Answer
Mg: 1s2 2s2 2p6 3s2
P: 1s2 2s2 2p6 3s2 3p3
Cl: 1s2 2s2 2p6 3s2 3p5
Ca: 1s2 2s2 2p6 3s2 3p6 4s2
2-11
Orbital Diagrams
We can represent the electrons in orbitals by means of orbital diagrams.
We use a box to show an orbital, an arrow, pointing upward, to represent a single
electron, and a pair of arrows pointing in opposite directions to represent two electrons.
See worked example 2-5.
When drawing orbital diagrams, use the following rules:
Orbitals fill in the order of increasing energy. See Figure 2.1.
Each orbital can hold a maximum of two electrons.
When there is a set of orbitals of equal energy, each orbital is occupied by a single
electron before a second electron enters. For example, all three p orbitals must be
half-filled (contain one electron) before the second electron enters.
Worked Example 2-5
Write the electron configuration and draw orbital diagrams for the first eight
elements in the periodic table.
Solution
atom
H
electron
configuration
1s1
He 1s2
orbital diagram
1s
1s
1s22s1
1s
2s
Be 1s22s2
1s
2s
Li
B
1s22s22p1
1s
2s
2p
C
1s22s22p2
1s
2s
2p
N
1s22s22p3
1s
2s
2p
O
1s22s22p4
1s
2s
2p
2-12
Practice 2-6
For the element sulfur:
a) write the electron configuration,
b) draw the orbital diagrams,
c) how many unpaired electrons are in the sulfur atom?
Answer
a) S: 1s2 2s2 2p6 3s2 3p4
b)
1s2
2s2
2p6
3s2
3p4
c) Two unpaired electrons in 3p orbitals.
Shorthand Electron Configurations
An electron configuration can also be written in an abbreviated (shorthand) configuration.
In this method the electron configuration of the preceding noble gas is replaced by
writing its symbol inside brackets.
B: [He] 2s22p1
Si: [Ne] 3s23p2
Ca: [Ar] 4s2
Cs: [Xe] 6s1
2.6 Valence Electrons and Lewis Structures
The electrons in the highest energy level of an atom are called valence electrons. For
example, phosphorous, which has the electron configuration: 1s22s22p63s23p3, has
electrons in the first, second, and the third energy level. The third energy level, the
highest energy level, has a total of five electrons (3s23p3), so phosphorous has five
valence electrons.
For all group A elements, the number of valence electrons is equal to the group number.
For example, the elements in Group IIA such as Be, Mg, Ca, Sr, and Ba, all have two
electrons in the highest energy level.
Be: 1s22s2
Mg: 1s22s22p63s2
Ca: 1s22s22p63s23p64s2
Sr: 1s22s22p63s23p64s23d104p65s2
Ba: 1s22s22p63s23p64s23d104p65s24d105p66s2
2-13
Periodic
table group
Group IA
Group IIA
Group IIIA
Group IVA
Group VA
Group VIA
Group VIIA
Group VIIIA
Example
Na
Mg
Al
Si
P
S
Cl
Ar
Electron
configuration
1s22s22p63s1
1s22s22p63s2
1s22s22p63s23p1
1s22s22p63s23p2
1s22s22p63s23p3
1s22s22p63s23p4
1s22s22p63s23p5
1s22s22p63s23p6
Valence Group
electrons number
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
8
Worked Example 2-6
Give the number of valence electrons for each the following: K, Ba, O, Br, and P.
Solution
For group A elements, you can predict the number of valence electrons by noting
the group number of each element from the periodic table.
K s n roup IA …...
Ba is in group IIA ….
O s n roup VIA ….
Br s n roup VIIA…
P s n roup VA …...
K has 1 valence electron.
Ba has 2 valence electrons.
O has 6 valence electrons.
Br has 7 valence electrons.
P has 5 valence electrons.
Lewis Structures (Electron-dot Formulas)
To keep track of valence electrons we use a notation called electron-dot formula. In this
notation, the outer electrons are shown as dots on the sides of the symbol for the element.
Each dot corresponds to an outer-shell electron. A thou h dot p acement sn’t cr t ca , one
method is to begin at the top of the symbol and move clockwise around it placing dots on
each side of the symbol one at a time. You must not pair any electrons until all four sides
have one dot.
2-14
Electron Configuration and the Periodic Table
The periodic table can be divided into four blocks (s block, p block, d block, and f block).
s block: the first two columns on the left side
p block: the six columns on the right side
d block: the transition elements
f block: the lanthanides and actinides
Table 2.2 The blocks of elements in the periodic table.
We can use the general pattern of the periodic table to obtain the electron configuration
for an atom.
Beginning at the top left and going across successive rows of the periodic table provides
a simple way of remembering the order of orbital filling:
1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p
This is identical to that shown in Fig 2.1.
2-15
Worked Example 2-7
Use the periodic table to write the electron configuration for arsenic, As.
Solution
Arsenic is element 33 in Period 4, Group VA, three places after the transition
elements. It must contain three electrons in the 4p orbitals.
Its electron configuration is:
As: 1s22s22p63s23p64s23d104p3
Practice 2-7
Use the periodic table to write both the electron configuration and the shorthand
notation for iodine.
Answer
I: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p5
I: [Kr] 5s2 4d10 5p5
2-16
Homework Problems
2.1 For elements (Br, Ca, Fe, Na, S, Si, and Xe) which of the following term(s) apply?
a. metal
b. nonmetal
c. metalloid
d. noble gas
e. halogen
f. alkali metal
g. alkaline earth metal
h. transition metal
i. main group element
j. gas at room temperature
2.2 Identify the following atoms using their symbols:
a. contains 20 protons.
b. contains 12 electrons.
c. has atomic number = 30.
d. the fifth-period transition metal with properties similar to copper.
e. the group VA metalloid in the fifth period.
f. the lightest metallic element in group IVA.
g. the fourth-period element with properties similar to oxygen.
h. the metalloid in the second period.
i. the element with the fewest number of electrons in the sixth period.
j. the third-period element with the largest number of electrons.
2.3 In nature, gallium (Ga) is found as two isotopes. Calculate the atomic mass of this
element.
69
31Ga
(atomic mass 68.926 amu, abundance 60.11%)
71
31Ga
(atomic mass 70.925 amu, abundance 39.89%)
2.4 Write the complete electron configuration for the following:
a. sulfur
b. potassium
c. nickel
d. xenon
2-17
2.5 Use orbital diagrams to show the distribution of electrons in the orbitals of:
a. the 5p sublevel of iodine
b. the 3d sublevel of iron
c. the 4s sublevel of potassium
d. the 3s and 3p sublevel of silicon
2.6 Write the shorthand electron configurations for each of the following:
a. iodine
b. zinc
c. phosphorous
d. barium
2-18
UNIT (3) COMPOUNDS
Substances are either elements or compounds. In unit 2 we studied elements, and in this
unit we will study compounds.
A compound is a substance that consists of two or more different elements.
The elements in a compound are not just mixed together. Their atoms are bonded
together in a specific way.
The forces that hold atoms together in a compound are called chemical bonds.
We will study ionic and covalent bonds.
An ionic bond involves the transfer of electrons from a metal to a nonmetal.
A covalent bond consists of a pair of electrons shared between two nonmetals.
Metals lose their valence electrons and nonmetals gain electrons to satisfy the octet rule.
Electron sharing satisfies the octet rule.
3.1 The Octet Rule (Rule of 8)
In the formation of either ionic bond or covalent bond, atoms lose, gain, or share
electrons to achieve an electron configuration identical to the noble gas nearest them in
the periodic table. These noble gas configurations have eight electrons in their valence
shells (except for helium, which has two electrons).
The octet rule states that atoms tend to combine in such a way that each has eight
electrons in their valence shells identical to the noble gas nearest them in the periodic
table.
3.2 Ions and the Octet Rule
As you studied in unit 2, atoms are neutral because they have equal numbers of electrons
and protons. By losing or gaining one or more electrons, an atom can be converted into a
charged particle called an ion.
The loss of electron(s) from a neutral atom gives a positively charged ion called cation
(pronounced cat-ion).
The gain of electron(s) by a neutral atom gives a negatively charged ion called anion
(pronounced an-ion).
For most of s block and p block elements, the charge on an ion can be predicted from the
position of the element on the periodic table.
The metals (on the left-hand side of the table) lose electrons to form cations.
The Group IA (lose ONE electron), Group IIA (lose TWO electrons), and Group IIIA
(lose THREE electrons).
3-1
The nonmetals (on the right-hand side of the table) gain electrons to form anions.
The Group VIIA (gain ONE electron), Group VIA (gain TWO electrons), and Group VA
(gain THREE electrons).
Some transition metals and metals in Group IVA have variable charges (more than one
positive ion). See Figure 3.1.
Figure 3.1
Some common ions and their locations on the periodic table are given. The table lists
only the elements that you need to memorize.
IA
IIA
IIIA
IVA VA VIA VIIA
+
H
Li+
Be2+
Na+
Mg2+
K+
Ca2+
Rb+
Sr2+
Cs+
Ba2+
Al3+
Fe2+
Fe3+
Co2+
Co3+
Ni2+
Cu+
Cu2+
Zn2+
Ag+
Hg22+
Hg2+
N3-
O2-
F-
P3-
S2-
ClBr-
Sn2+
Sn4+
Pb2+
Pb4+
I-
(Note: charges are written numbers first then sign).
3.3 Ionic Bond Formation
An ionic bond forms by transfer of electron(s) from the metals to the nonmetals.
The result is a formation of an ionic compound.
Lewis structures (electron-dot symbols) are helpful in visualizing the formation of ionic
compounds.
Using Lewis symbols, the formation of the ionic compound NaCl from the elements
sodium and chlorine can be shown as follows:
Sodium needed to lose one electron for octet formation (the neon electron configuration),
chlorine needed to gain one electron for octet formation (the argon electron
configuration). The electron transfer required 1:1 ratio of reacting atoms 1 Na to 1 Cl.
3-2
Worked Example 3-1
Use the electron-dot symbols to write the equation for the formation of the ionic
compound formed between barium and iodine.
Solution
Barium has to lose two electrons for octet formation (the xenon electron
configuration).
Iodine has to gain one electron for octet formation (the xenon electron
configuration).
The transfer of two electrons from barium requires the acceptance of those two
electrons by two iodine atoms.
The electron transfer requires 1:2 ratio of reacting atoms 1 Ba to 2 I.
Practice 3-1
Use the electron-dot symbols to write the equation for the formation of the ionic
compound formed between aluminum and fluorine.
Answer
F
Al
F
F
F
Al3+
F
F
Formula (AlF3)
-
3-3
3.4 Writing Formulas for Ionic Compounds
Ionic compounds are electrically neutral. Therefore, when writing formulas, the cations
(positive) and anions (negative) must combine to produce a net charge of zero. Formulas
for ionic compounds are called formula units.
The correct combining ratio when Na+ ions and Cl- ions combine is: NaCl (one to one).
The correct combining ratio when Na+ ions and O2- ions combine is: Na2O (two to one).
The correct combining ratio when Na+ ions and P3- ions combine is: Na3P (three to one).
Worked Example 3-2
Write the formula for the ionic compound that is formed when each of the
following pairs of ions interact:
a) K+ and S2b) Mg2+ and O2c) Ca2+ and Id) Li+ and N3e) Al3+ and S2Solution
a) The cation has a charge of 1+ and anion has a charge of 2-. Thus two positive
ions are required for each negative ion in a neutral formula unit.
The formula is K2S.
b) The cation has a charge of 2+ and anion has a charge of 2-. The ratio is 1:1.
The formula is MgO.
c) The cation has a charge of 2+ and anion has a charge of 1-. Two negative ions
are required for each positive ion. The formula is CaI2.
d) The cation has a charge of 1+ and anion has a charge of 3-. Three positive ions
are required for each negative ion. The formula is Li3N.
e) The cation has a charge of 3+ and anion has a charge of 2-. Two positive ions
are required for three negative ions. The formula is Al2S3.
K+
and S22+
MgO
-
CaI2
Mg and O
2+
K2S
2-
Ca
and I
Li+
and N3- Li3N
Al3+ and S2-
Al2S3
3-4
3.5 Naming Ions
Names of cations and anions are formed by a system developed by the International
Union of Pure and Applied Chemistry (IUPAC).
A) Names of Cations From Metals That Form Only One Type of Positive Ion:
Elements in Groups IA, IIA, and IIIA and some transition elements form only one type of
cations. For these ions the name of the cation is the name of the metal followed by the
word “ion”:
Na+
Al3+
sodium ion
aluminum ion
K+ potassium ion
Ag+ silver ion
Mg2+
Zn2+
magnesium ion
zinc ion
B) Names of Cations From Metals That Form Two different Positive ions.
Metals in Group IVA and most transition metals form more than one type of cation, so
the name of the cation must show its charge. For these ions the charge on the ion is given
as a Roman numeral in parentheses right after (with no space) the metal name.
Sn2+
Pb2+
Cu+
Fe2+
Co2+
Hg22+
tin(II)
lead(II)
copper(I)
iron(II)
cobalt(II)
mercury(I)
Sn4+
Pb4+
Cu2+
Fe3+
Co3+
Hg2+
tin(IV)
lead(IV)
copper(II)
iron(III)
cobalt(III)
mercury(II)
C) Names of Anions:
Anions are named by replacing the ending of the element name with –ide
followed by the word “ion”:
F- fluoride ion
O2- oxide ion
ClS2-
chloride ion
sulfide ion
Br- bromide ion
N3- nitride ion
IP3-
iodide ion
phosphide ion
D) Names of Polyatomic Ions:
A polyatomic ion is an ion that contains two or more elements. You must memorize the
names and the formulas of the following polyatomic ions:
NH4+
ammonium
SO32sulfite
CNcyanide
SO42sulfate
OH
hydroxide
HSO3 hydrogen sulfite
C2H3O2- acetate
HSO4- hydrogen sulfate
2CrO4
chromate
PO33phosphite
Cr2O72- dichromate
PO43phosphate
2MnO4
permanganate
HPO4 hydrogen phosphate
NO2nitrite
ClOhypochlorite
NO3
nitrate
ClO2chlorite
2CO3
carbonate
ClO3
chlorate
HCO3hydrogen carbonate
ClO4perchlorate
3-5
The common name for HCO3-, HSO3-, and HSO4- are bicarbonate, bisulfite, and
bisulfate respectively.
3.6 Naming Ionic Compounds
I: Binary ionic compounds from metals that form only one type of positive ion:
These compounds contain only two elements a metal ion and a nonmetal ion.
The chemical name is composed of the name of the metal followed by the name of the
nonmetal, which has been modified with the suffix –ide.
Worked Example 3-3
Name the following binary ionic compounds:
NaCl
MgBr2
AlP
K2S SrF2 ZnI2
Solution
NaCl
MgBr2
AlP
sodium chloride
magnesium bromide
aluminum phosphide
K2S
SrF2
ZnI2
potassium sulfide
strontium fluoride
zinc iodide
Practice 3-2
Name the following binary ionic compounds:
BaO
Ca3P2
Sr3N2
Ag2S LiBr NiCl2
Answer
BaO
Ca3P2
Sr3N2
barium oxide
calcium phosphide
strontium nitride
Ag2S
LiBr
NiCl2
silver sulfide
lithium bromide
nickel chloride
3-6
II: Binary ionic compounds from metals that form two different positive ions:
To name these compounds we must include the charge on the cation as a Roman numeral
in parentheses right after (with no space) the metal name, followed by the name of the
anion.
Worked Example 3-4
Name the following binary ionic compounds:
FeBr3 CoF2
SnO
PbI4 HgS Cu3P
Solution
FeBr3
CoF2
SnO
iron(III) bromide
cobalt(II) fluoride
tin(II) oxide
PbI4
HgS
Cu3P
lead(IV) iodide
mercury(II) sulfide
copper(I) phosphide
Practice 3-3
Name the following binary ionic compounds:
SnS2
PbI2
Hg2O
CuCl2
FeN
Co2O3
Answer
SnS2
PbI2
Hg2O
tin(IV) sulfide
lead(II) iodide
mercury(I) oxide
CuCl2
FeN
Co2O3
copper(II) chloride
iron(III) nitride
cobalt(III) oxide
III: ionic compounds that include polyatomic ions
Naming these compounds is similar to naming binary compounds. The cation is named
first, followed by the name for the negative polyatomic ion.
3-7
Worked Example 3-5
Name the following polyatomic ionic compounds:
Ca(NO3)2
ZnSO4
NH4CN
Li3PO4
Na2CO3
Mg(HCO3)2
Solution
Ca(NO3)2
ZnSO4
NH4CN
Li3PO4
Na2CO3
Mg(HCO3)2
calcium nitrate
zinc sulfate
ammonium cyanide
lithium phosphate
sodium carbonate
magnesium hydrogen carbonate
Practice 3-4
Name the following polyatomic ionic compounds:
Sr(ClO4)2 Na2CrO4 KH2PO4
Ag2SO4 Co(OH)3
Cu(C2H3O2)2
Answer
Sr(ClO4)2
Na2CrO4
KH2PO4
Ag2SO4
Co(OH)3
Cu(C2H3O2)2
strontium perchlorate
sodium chromate
potassium hydrogen phosphate
silver sulfate
cobalt(III) hydroxide
copper(II) acetate
3-8
3.7 Covalent Bond Formation and Lewis Structure
The Covalent Bond Model (Molecules)
● Molecular compounds are compounds formed between two or more nonmetals.
● Molecular compounds are made up of discrete units called molecules.
● Within the molecule nonmetal atoms are held together by covalent bonds.
A covalent bond forms when electron pairs are shared between two nonmetals.
In general, the number of covalent bonds that a nonmetal atom forms is the same as the
number of electrons it needs to have an octet.
Nonmetals may share all or some of their valence electrons. The shared electrons are
called bonding electrons and the unshared electrons are called lone pairs. The bonding
electrons are shown as dashes and lone pairs as dots.
According to the octet rule the total number of (bonds + lone pairs) should equal four,
which is a total of 8 electrons.
The following table summarizes the typical number of bonds and lone pairs for the atoms
of interest:
Atoms
Number of
bonds
Number of
lone pairs
F, Cl, Br, I
1
3
O, S
2
2
N, P
3
1
C, Si
4
0
H
1
0
Bonds and
lone pairs
A molecular representation that shows both the connections among atoms and the
locations of lone pairs are called Lewis structure.
3-9
Worked Example 3-6
Draw a Lewis structure for each of the following: H2, HF, O2, and N2.
Solution
Worked Example 3-7
Draw Lewis structure for each of the following: NH3, CF4, H2O, and CO2.
Solution
Notice that the central atom in Lewis structures is the atom that appears only once in the
formula.
3-10
3.8 Naming Molecular Compounds
We will only consider binary (two element) molecular compounds.
The names of binary compounds (molecules) are written as two words.
First word: Full name of the first nonmetal in the formula; a Greek numerical prefix is
used to show the number of atoms.
Second word: The stem of the name of the second nonmetal in the formula with the
suffix-ide; a Greek prefix is used to show the number of atoms.
Greek prefixes
1 (mono-)
2 (di-)
6 (hexa-)
7 (hepta-)
3 (tri-)
8 (octa-)
4 (tetra-)
9 (nona-)
5 (penta-)
10 (deca-)
Worked Example 3-8
Name the following binary molecular compounds:
N2O5 CO2 P4S3 XeF6 ICl
NH3 I4O9 CO
H2O
H2O2
Solution
N2O5
dinitrogen pentoxide
NH3
nitrogen trihydride (ammonia)
CO2
P4S3
XeF6
ICl
*carbon dioxide
tetraphosphorous trisulfide
*xenon hexafluoride
iodine monochloride
I4O9
CO
H2O
H2O2
tetraiodine nonoxide
*carbon monoxide
dihydrogen monoxide (water)
dihydrogen dioxide
*When only one atom of the first nonmetal is present, omit the initial prefix
mono-. That is, the prefix mono- is never used with the cation.
Oxides of carbon are named by using “mono-” and “di-” to distinguish between
the two oxides. Notice that we say monoxide rather than monooxide.
When the prefix ends in a or o and the element name begins with a or o, for the
ease of pronunciation, the vowel of the prefix is dropped.
3-11
Practice 3-5
Name each of the following
S2F10
SiI4
P2O5
B4Cl4
P4S7
NBr3
I2Cl6
SI5
Answer
S2F10
SiI4
P2O5
B4Cl4
P4S7
NBr3
I2Cl6
SI5
disulfur decafluoride
silicon tetraioidide
diphosphorous pentoxide
tetraboron tetrachloride
tetraphosphorous heptasulfide
nitrogen tribromide
diiodine hexachloride
sulfur pentaiodide
3.9 The Shape of Molecules: Molecular Geometry
We can predict the molecular geometry using the Valence Shell Electron Pair Repulsion
(VSEPR) model:
According to VSEPR, “electron groups”, as bonding pairs or lone pairs, stay as far apart
as possible so that electron-electron repulsions are at minimized.
We will examine only five cases:
I) 2 electron groups (2 bonding pairs with no lone pairs)
II) 3 electron groups (3 bonding pairs with no lone pairs)
III) 4 electron groups (4 bonding pairs with no lone pairs)
IV) 4 electron groups (3 bonding pairs and 1 lone pair)
V) 4 electron groups (2 bonding pairs and 2 lone pairs)
3-12
See the table:
molecules bonded
atoms
BeH2
2
lone
pairs
0
lone
pairs +
bonded
2
molecular shape
Linear
BF3
3
0
3
Trigonal planar
CH4
4
0
4
NH3
3
1
4
Tetrahedral (Td)
Trigonal pyramidal
H2 O
2
2
4
Angular (bent)
3-13
3.10 Electronegativity
Electronegativity is the measure of the ability of an atom to attract bonding electrons.
Electronegativity displays a periodic trend. The element fluorine (F) is the most
electronegative atom and is assigned 4.0, and all other elements are assigned values in
relation to fluorine. Electronegativity generally increases from left to right across a row
(period) of the periodic table and increases from bottom to top within a column (group).
In general, the closer an element is to fluorine, the greater its electronegativity.
element
H
electronegativity 2.1
element
electronegativity
C
2.5
Si
1.8
N
3.0
P
2.2
O
3.5
S
2.5
F
4.0
Cl
3.0
3.11 Bond Polarity
The bonding electrons are shared equally between two nonmetal atoms of identical
electronegativity. When electrons in a covalent bond are shared by atoms with different
electronegativities, the atoms “pull” on the electrons with different strengths very similar
to a tug-of-war. The slight winner will be the more electronegative atom (one closer to
fluorine). This unequal sharing of electrons gives the bond a partially positive end (at the
less electronegative atom) and a partially negative end (at the more electronegative
atom).
Nonpolar covalent bond: bonding electrons are shared equally.
Examples of nonpolar bonds: H2, N2, O2, F2, Br2, CH4.
Polar covalent bond: bonding electrons are not shared equally due to differences in
electronegativity. The unequal sharing means that the bonding electrons spend more time
near the more electronegative atom and less time near the less electronegative atom. As a
result the more electronegative atom carries a partial negative charge, represented by δ(delta minus), and the less electronegative atom carries a partial positive charge,
represented by δ+.
Examples of polar bonds for HF and H2O molecules:
The distinctions between nonpolar covalent, polar covalent, and ionic bonds are not
always clear. In general, we find guidance by examining the difference in
electronegativity between the bonded atoms.
Nonpolar covalent: Electronegativity difference between 0 to 0.4
3-14
Polar covalent: Electronegativity difference between 0.5 to 1.9
Ionic bond: Electronegativity difference larger than 1.9
Worked Example 3-9
Determine whether each of the following covalent bonds is polar or nonpolar.
a) N – H bond
b) S – O bond
c) C – S bond
Solution
a) N – H polar bond
b) S – O polar bond
c) C – S nonpolar bond
An electronegativity difference is (3.0 - 2.1 = 0.9).
An electronegativity difference is (3.5 - 2.5 = 1.0).
An electronegativity difference is (2.5 - 2.5 = 0).
3.12 Molecular Polarity
Molecules can exhibit polarity similar to bond polarity. A molecule may be nonpolar
despite the presence polar bonds.
We must consider both the geometry of the molecule and the polarity of bonds to
determine whether or not a molecule is polar.
If a molecule has no polar bonds, then the molecule is nonpolar (nonpolar bond /
nonpolar molecule).
If a molecule has polar bond(s), then the molecule is polar only if the centers of positive
and negative charges don’t coincide.
In a polar molecule, one side has a partial positive charge and the other has a partial
negative charge.
Worked Example 3-10
Determine whether H2, HF, and CO2 are polar molecules.
Solution
H2 is a diatomic element; therefore the H – H bond is nonpolar and the H2
molecule is nonpolar.
The linear HF molecule contains a polar bond (electronegativity difference of
1.9). The hydrogen side of molecule is positive and the fluorine side of the
molecule is negative. Thus, HF is a polar molecule.
The CO2 molecule contains polar bonds (electronegativity difference of 1.0).
However, the linear shape of molecule causes the two polar bonds to oppose and
cancel one another and the molecule is nonpolar.
3-15
Homework Problems
3.1 Use the electron-dot symbols to write the equation for the formation of the ionic
compounds from each of the following pairs:
a. K and F
b. Na and O
c. Ca and P
d. Al and S
3.2 Write the formula for the ionic compound that is formed from each of the
following pairs of ions:
a. Na+
b. Ca2+
c. Pb4+
d. Zn2+
e. NH4+
f. Al3+
g. Mg2+
h. Pb2+
i. Cu2+
j. Fe3+
and
and
and
and
and
and
and
and
and
and
FOHO2C2H3O2S2HPO42CO32IN3SO32-
3.3 Write chemical formula for the following compounds:
a. barium nitrate
b. strontium chlorate
c. ammonium phosphate
d. cobalt(II) sulfite
e. mercury(II) iodide
f. copper(I) cyanide
g. magnesium phosphide
h. potassium sulfide
i. zinc hydroxide
j. silver chromate
k. iron(III) oxide
l. lead(II) permanganate
3-16
3.4 Name each of the following compounds:
a. KClO4
b. Co3N2
c. NiF2
d. NH4OH
e. NaNO2
f. Sn(C2H3O2)2
g. Ca(MnO4)2
h. FeCr2O7
i. CuHPO4
j. Al(HCO3)3
k. MgH2
l. PbS
3.5 Name the following molecular compounds:
a. CI4
b. P2I4
c. Br3O8
d. N2O3
e. BCl3
f. N2O5
g. P4O6
h. O2F2
i. IF7
j. SiBr4
k. H2S
l. P4S10
3.6 For the SCl2 molecule:
a. draw the Lewis structure
b. use VSEPR to predict the shape of the molecule
c. are there any polar bonds? Explain.
d. is the molecule polar? Explain.
3-17
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