Valence number tutorial
Today we started our lesson on valence numbers and how we can use them
to write chemical formulas for compounds.
1. Some vocabulary to remember:
Valence-the ability of an atom to become stable through bonding
Valence number-the number of electrons lost, gained, or shared to become
stable.
2. We have already learned the difference between metals and nonmetals
with respect to atomic structure and the tendency to bond.
a. Metals with less than 4 valence electrons will lose electrons, form ionic
bonds, and become positive ions.
b. Metals are found in Groups I, II, III (groups give us the number of
valence electrons).
c. Nonmetals have more than 4 valence electrons, will gain electrons, can
form ionic bonds with metals, but covalent bonds with metalloids,
Hydrogen, and nonmetals.
d. Nonmetals are found in Groups V, VI,VII.
e. Noble gases are inert or unreactive under most conditions. Therefore,
they do not have a valence or an ability to bond in order to become
stable (they are already stable). Their valence number is 0.
Let’s take a look at specific atoms.
a.
b.
c.
d.
e.
f.
g.
Lithium is found in Group I.
It has an atomic number of 3.
Its electron configuration is 2-1
Because it has 1 valence electron it will be found in Group I
It is a metal
It will lose one electron on its L shell
This means it has one too many electrons-it wants to get rid of this
electron.
h. According to the definition for valence number- it is the number of
electrons lost in order to become stable-lithium will lose 1 electrons to
make it stable.
i. Lithium’s valence number is +1.
j. When we write a valence number it is to the right and above the symbol—
Li+1
a.
b.
c.
d.
e.
f.
g.
Aluminum is found in Group III
It has an atomic number of 13
Its electron configuration is 2-8-3
Because it has 3 valence electrons it will be found in Group III
It is a metal
It will lost 3 electrons on its M shell
This means it has 3 too many electrons-it will want to get rid of those 3
electrons.
h. Aluminum will lose 3 electrons to make it stable.
i. Aluminum’s valence number is +3
j. Al+3
Similarly the same can be done for nonmetals.
a.
b.
c.
d.
e.
f.
g.
Nitrogen is found in Group V
It has an atomic number of 7
Its electron configuration is 2-5
Because it has 5 valence electrons it will be found in Group V
It is a nonmetal
It will need to gain 3 electrons in order to become stable
This means it is missing 3 electrons to have the complete octet on the L
shell
h. According to the definition for valence number-it is the number of
electrons gained in order to become stable-nitrogen will gain 3 electrons
to make it stable.
i. Nitrogen’s valence number if -3
j. N-3
a. Fluorine is found in Group VII
b.
c.
d.
e.
f.
g.
It has an atomic number of 9
Its electron configuration is 2-7
Because it has 7 valence electrons it will be found in Group VII
It is a nonmetal
It will need to gain 1 electron in order to become stable
This means it is missing 1 electron to have the complete octet on the L
shell
h. Fluorine’s valence number if -1
i. F-1
Now how can we use these valence numbers? Well, before to show how an atom
became stable, we would draw the Bohr model of the atoms involved, show the
transfer or sharing of electrons and then write a formula.
So, to show the bond between sodium and oxygen the following would be done:
Sodium Na: atomic number 11
Electron configuration 2-8-1
Oxygen O: atomic number 8
electron config 2-6
+
Na will need to
Lose 1 electron
Oxygen will need
to gain 2 electrons
2-
+
Now oxygen and sodium are stable. The formula is Na2O
That is a lot of work. Now that we know valence numbers, we can write a formula
with a lot less work. Let’s see how that works.
Remember the valence number tells us how many electrons will be lost or gained
in order for the atom to become stable.
Looking at sodium Na, Na is a Group I atom, it has 1 valence electron-1 electron
too many to be stable. Na wants to get rid of that electron. (we saw this when we
drew the Na atom). Na has a valence number of +1.
Oxygen, O is a Group VI atom, it has 6 valence electrons-missing 2 electrons to be
stable. O wants to gain 2 electrons. (we saw this when we drew the O atom) O
has a valence number of -2.
If we take the valence numbers and criss cross them, we can write a correct
formula for this compound. The valence numbers become the subscripts.
Na+1 O-2
Na2O
Let’s look at another example:
Al a group III atom has a valence number of+3
O a group VI atom has a valence number of -2
Al+3
O-2
Al2O3
This process works with both ionic bonds and covalent
bonds.
The previous examples were both ionic bonds. Let’s look at a
covalent bond.
H+1 O-2
C+-4 Cl-1
H2O
CCl4
Now let’s take a look at a special situation. Remember
according to John Dalton, atoms bond in specific ratio.
Ratios must be in a reduced state-so if subscripts can be
reduced, they must be reduced.
Mg+2 O-2
Mg2O2
Divide each
Subscript by 2
MgO
Now that we can write a formula using the valence numbers for each atom, we
can take a look at specific reactions. There are four types of reactions:
synthesis, decomposition, single replacement and double replacement
1. Synthesis reaction: atoms combine to form one new compounds
General formula:
A +
B
A B
Metal nonmetal
compound
Mg
+
Mg+2Cl-1 =MgCl2
Cl2
**Remember the valence numbers come from the valence electrons found
on the valence shell of each atom. Mg has 2 valence electrons, which it will
want to lose in order to become stable. Its valence number will be +2 to
show that it is a metal that will lose two electrons. Having 2 electrons on
the valence shell will also place Mg in Group II. The same can be said about
Cl. It is a nonmetal that has 7 electrons on its valence shell. This means it
will gain 1 electron. Its valence number would be -1 to show that it is
missing 1 electron to be stable; it will gain one electron. It would be found
in Group VII.
Now that we have the correct formula for the reaction, we need to take a
look at the reaction and make sure it obeys the Law of Conservation of
Matter which states that matter cannot be created nor destroyed.
Mg
+
Cl2
MgCl2
If we look at the reaction, we can see that the number of magnesium and
chlorine are the same on both sides of the arrow. 1 Mg on the left and the
right and 2 Cl on the left and the right. The reaction obeys the Law of
Conservation of Matter.
Now let’s take a look at one that will not obey the Law.
K
+
F2
KF
(valence number for K is +1, F is -1)
Now if we look at the nonmetal first, we can see that there are 2 F on the left
side of the arrow but only 1 F on the right side. According to the Law of
Conservation of Matter, this cannot be. We need to use coefficients in order to
balance the reaction. Balancing a reaction means that the number of atoms on
the left will equal the number of atoms on the right.
A couple of rules to remember:
1. Use valence numbers to write the correct formula
2. Write a work line. This line will list the number of atoms on both sides of the
arrow.
3. Place brackets around the compound.
4. Use coefficients where necessary to balance the reaction.
Work line:
K
1K
+
F2
2F
[KF]
1K 1F
Start with the nonmetal. There are 2 F on the left and only 1 on the right.
Rule 3 Place brackets around the compound.
Rule 4 Use coefficients to balance the reaction. If there are 2 fluorine on the left
there must be 2 fluorine on the right. Putting a ‘2’ in front of KF will give us 2
Fluorine on the right.
Work line:
K
1K
+
F2
2F
2 [KF]
1K 1F
2 2
Now we have 2 F on both sides of the arrow. However, because coefficients will be
distributed over everything inside the brackets, we now have 2 K on the right but only
1 on the left.
2 K
1K
2
Work line:
+
F2
2F
2 [KF]
1K 1F
2 2
Rule 4, still using coefficients, we need to show that we have 2K on the left.
Now we have 2 K on both sides and 2F on both sides and the reaction is balanced.
Synthesis reactions are the first reactions we looked at. There are three more.
2. Decomposition reactions are the opposite of synthesis reactions.
Decomposition reactions break down compounds into individual atoms of
different atoms.
The general formula for a decomposition reaction is:
AB
A
Compound
+
metal
B
nonmetal
A couple of things to remember: 1. You do NOT use valence numbers in a
decomposition. The formula for the compound, AB, will always be given to you in its
correct form. 2. The nonmetal will almost always be a diatom, so you MUST put a
subscript of 2 on the symbol.
Let’s look at an example:
Li2O
This is what you will be given. Li2O is a correctly written formula. I have already
crisscrossed valence numbers.
Following the general formula of AB
the compound between the atoms.
Li2O
Li
+
A + B, you finish the reaction by splitting
O
Li is the metal, O in the nonmetal
Remember in a decomposition, the nonmetal will almost always be a
diatom, which means you must write oxygen as a diatom.
Li2O
Li
+
O2
From this point on, you balance the reaction the same as a synthesis.
Work line:
Li2O
2 Li 1 O
Li +
1 Li
O2
2O
Starting with the nonmetal, we see that we started with 1 oxygen and ended with
2 oxygen. If we ended with 2 we must have started with 2 oxygen. Remember to
bracket the compound.
To show that we started with 2 oxygen a coefficient of 2 must be placed in front
of Li2O.
2 Li2O
Update the work
Line:
2 Li 1 O
Li
+
O2
1 Li
2O
4 Li 2 O
Checking both sides of the arrow, we now have 2 oxygens on both sides. Good! If
we look at Li we see that we started with 4 and ended with 1. This cannot be. If
we started with 4 Li we have to end with 4 Li. A coefficient of 4 in front of Li on
the right side of the arrow will give us the 4 Li we need to balance the equation.
2 Li2O
:
4 Li
2 Li 1 O
1 Li
4 Li 2 O
4 Li
+
O2
2O
Looking at both sides of the arrow, we now have a balanced equation. 4 Li on
both sides of the arrow and 2 O on both sides of the arrow.