A beginner’s formulation guide
With the exception of Group 8, all elements in the periodic table are assigned oxidation
numbers. These numbers are a way of showing the number of electrons that are lost or gained
when that atom forms a chemical bond (ionic or metallic). In most cases, this number will
reflect how many electrons are lost or gained to fill the outer shell of the atom.
Beryllium, in Group 2, has only 2 electrons in the outer shell. When it
forms a chemical bond it will try to lose these 2 electrons so that it then
has a full outer shell (in this case the first shell would become the outer
shell). The neutral Be atom would then become an ion with 2 positive
charges (2+):
Be  Be2+
For this reason we give Be the oxidation number +2. (Notice how the oxidation number has
the + or – sign and then a number).
Flourine, in Group 7, has 7 electrons in its outer shell. To form a chemical
bond it will try to gain 1 electron so that it fills the second shell. In this case
the F atom would become an ion with 1 negative charge (1-):
F  FFor this reason we give F the oxidation number -1. (Notice how the oxidation number has the
+ or – sign and then a number).
Writing the formula of a compound…
When formulating compounds, the aim is to make sure that overall the oxidation numbers add
up to equal 0. If we were asked to formulate beryllium fluoride:
Ox. number Be = +2
So I need 1 x Be and 2 x F  BeF2
Ox. number F = -1
+2
-1
-1
=0
Easy oxidation numbers to remember:
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Group 1 elements = +1
Group 2 elements = +2
Group 3 elements = +3
Oxygen = -2
Group 7 = -1
(Hydrogen = +1 or -1)
Writing the name of a compound…
The simplest compounds are made of a metal and a non-metal such as NaCl. To name these
compounds we start with the name of the metal i.e sodium and then add –ide to the end of
the non-metal i.e chloride.
NaCl  sodium chloride
LiBr  lithium bromide
MgO  magnesium oxide
BeF2  magnesium fluoride
If the metals have more than 1 oxidation number then we must indicate which number it is in
the name. For example, with FeO the iron must have an oxidation number of +2 so that when
we add it to the -2 of the oxygen we overall get 0.
FeO
 +2 -2 = 0
 iron(II) oxide
We use roman numerals to show the oxidation number – I, II, III, IV, V. If we are asked to
name Fe2O3 then the iron must now have an oxidation number of +3 so that the total of the
oxidation numbers equals 0.
Fe2O3  +3 +3 -2 -2 -2 = 0  iron(III) oxide
Using polyatomic ions…
We must also learn the names of different polyatomic ions. These are ions that have already
been formed by 2 or more atoms e.g. the sulfate ion  SO42-. When formulating with these,
we can treat the charge as the total oxidation number for that ion and follow the same rule as
before:
Example 1:
Lithium sulfate – Lithium is in group 1 so will have an oxidation state of +1. The sulfate ion (SO42-) has
a charge of 2-. So, to formulate this compound I will need 2 lithium atoms and 1 sulfate ion…
+1
+1
-2
=
0
Li2SO4
Example 2:
Gold(III) silicate – In this case gold has an oxidation state of +3 (remember it can also have +1). The
silicate ion (SiO44-) has a charge of 4-. So, to formulate this compound I will need 4 gold atoms and 3
silicate ions…
+3
+3
+3
+3
-4
-4
-4
=
0
Au4(SiO4)3
These are the key oxidation numbers for commonly used elements:
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Group 1 elements (Li, Na, K)
Group 2 elements (Be, Mg, Ca)
Group 3 elements (B, Al)
Group 7 elements (F, Cl, Br)
Hydrogen
 +1 or -1
 +1
 +2
 +3
 -1
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Nitrogen
Phosphorus
Oxygen
Sulfur
Chromium
Manganese
Iron
Cobalt
Nickel
Copper
Zinc
 -3
 -3
 -2
 -2
 +2 or +3 or +6
 +2 or +3 or +7
 +2 or +3
 +2 or +3
 +2 or +3
 +1 or +2
 +2
Key ions:
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Carbonate
Sulfate
Nitrate
Phosphate
Hydroxide
 CO32 SO42 NO3 PO43 OH-
These common molecules have been given special names because we talk about
them so often. You just have to remember them as they follow no rules:
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NH3
CH4
BH3
B2H6
H2O
ammonia
methane
borane
diborane
water
(because it is formed by 2 boranes)
The full list of ox. numbers and ions you need to know for Selectividad is on the science
weebly.
For more advanced practice try the “How to become a formulation hero” document.