4. FORMATION OF COMPOUNDS
The noble (inert) gases – 2He, 10Ne, 18Ar, 36Kr, 54Xe – are so called because they are chemically
unreactive. In science, this is indicative of stability. The noble gases are stable so they do not need to
react chemically to gain stability. We know that the reason these gases are stable is because of their
electron count. You can think of the electron count of each of these elements as magic numbers: 2, 10,
18, 36, 54, which confers stability on the atoms.
In nature, all systems tend to move to a state having the greatest stability. Compounds are formed
because atoms “want” to attain a maximum stability – namely attaining the electron count of the nearest
noble gas.
Atoms can do this in two ways – by forming ionic bonds or by covalent bonds.
4.1 Ionic Bonds
Recall that an atom consists, in part, of protons and neutrons densely packed into a nucleus. The
electrons are moving rapidly outside of, and relatively far away from the nucleus. Thus for example, if the
nucleus of a hydrogen atom (located in downtown Montreal) is blown up to the size of a tennis ball, the
electron will be, on the average, in Beaconsfield. Because of these distances, the electron can “forget”
atom to which it belongs and attach itself to other atoms. The electrons of the atom can become “lost”.
Thus, atoms are capable of losing and gaining electrons.
If an atom gains an electron (which, of course, carries a negative charge), it then acquires an overall
negative charge. Similarly, if an atom loses an electron, it acquires a positive charge. Substances may
lose or gain more than one electron. Atoms gain charges by gaining or losing electrons. They do not
change their charge by gaining or losing protons.
e.g.
+
Na + e
Na
–
Cl + e
Zn
Fe
Fe
+ 2e
6+
+ 3e
+ 2e
+3e
–
O+2e
e.g.
3+
2+
Fe
Cr
Cl
2+
Zn
–
–
–
–
Fe
–
–
Cr
O
3+
2–
What are the charges of the substance when the specified numbers of electrons are added or
removed?
1. One electron is added to a bromine atom.
2. Three electrons are added to a nitrogen atom.
3. One electron is removed from a hydrogen atom.
Chemical substances which have charges are called ions.
3–
2–
–
Anions are chemical substances which have negative charges. e.g., N , O , I .
+
+
Cations are chemical substances which have positive charges. e.g., H , Na , Ca
2+
Typically:
Metals will lose electrons thereby forming cations.
Nonmetals tend to gain electrons thereby forming anions.
Again note that ions are NEVER formed by the addition or removal of protons.
page 30
The Periodic Table is useful for determining the type of ion that will form.
He
Li+
2+
Be
Be2
Be2
+
+
+
+
Be2
+
3+
2–
O 2
Be
+
+
Ne
Al
S
Cl
–
Ar
K+
Ca2+
Ga3+
Se2–
Br–
Kr
Rb+
Sr2+
In3+
Te2–
I–
Xe
Cs+
Ba2+
Transition metals will form cations having various charges
2–
–
F2
Be
Mg
Na
2+
Be2
Rn
The reason that metals tend to lose electrons is that they are to the left of the Periodic Table – it is simpler
for them to lose electrons to acquire the electron count to the noble gas of the preceding row. Thus, for
–
example, it is easier for 3Li to lose one electron to attain the electron count of He (2) than to gain 7 e to
–
acquire the electron count of Ne (10). Likewise a nonmetal such as 9F will prefer to gain one e to attain
–
the Ne electron count (10) than to lose 7 e to get to He (2).
To be Memorised
Thus:
•
Group IA elements: Li, Na, K, Rb, Cs will always form cations having (+1)
charge.
Group IIA elements: Be, Mg, Ca, Sr, Ba will always form cations having (+2)
charge.
Group IIIA elements: Al, Ga, In will always form cations having (+3) charge.
•
•
Generally (but not always):
•
•
Group VIA elements: O, S, Se, Te form anions having (–2) charge.
Group VIIA elements: F, Cl, Br, I form anions having (–1) charge.
In general, substances will not give up electrons unless there are other atoms ready to receive those
electrons. Likewise, a substance cannot accept electrons without other substances ready to supply those
electrons.
Thus, for example,
+
Na
–
Na + e
Cl + e
+
–
–
Cl
–
Na ------Cl
The attraction of positive to negative results in the formation of an ionic bond. The resulting neutrally
charged compound is known as an ionic compound. Ionic compounds are crystalline.
Ionic compounds can be recognised because only they have metal atoms.
page 31
4.2 Covalent Bonds
–
We have already said that nonmetals tend to gain e to attain the electron count of the nearest noble gas
–
(to the right). This is fine when there is a nearby atom willing to supply e ‘s. If, however, no such supplier
–
exists in the container, another method of acquiring e ‘s must be found.
–
Consider the case of a sample consisting only of 9F atoms. 9F requires one e to acquire the electron
count of the nearest noble gas, 10Ne. One can think of F as having an electron hole that needs to be filled.
.
°.F
F
°
Each fluorine atom seeks to seize one electron from the other fluorine atom while retaining all of its own 9
electrons. The result is that the two atoms are drawn together with the two electrons, one from each
fluorine atom, located between the two nuclei.
F
x
•
F
This type of bond is known as a covalent bond. In covalent bonds, the electrons are shared between two
atoms.
Covalent bonds are formed between nonmetal atoms.
page 32
4.3 Nomenclature
Nomenclature is a system of naming compounds so that:
•
•
given the name of the compound, you can write the formula of the compound uniquely; or
given the formula of the compound, you can write the name uniquely.
Before giving the system for naming compounds, it is necessary to know something about the nature of
the compound.
4.3.1
BINARY IONIC COMPOUNDS
The term binary compound means that the compound consists only of two different types of atoms.
Exercise:
State whether the following compounds are binary.
4. CO
5. CO2
6. C2H2Cl2
7. CH3CH3
8. HCN
Binary ionic compounds are formed from single atom (metal) cation and single atom (non-metal) anions.
Binary ionic compound = metal + non-metal
Whenever one of the atoms is a metal the compound is an ionic compound.
e.g.
Which of the following compounds are binary ionic compounds?
CO2
Review:
You must be able to recognize
a metal from its location in the
Periodic Table
CaO
CaCO3
FeCl3
NaNO2
CCl4
4.3.1.1
e.g.
Simple Binary Ionic Compounds
Name each of the following compounds.
metal + non-metal + ide
The suffix –ide is very important. It is
the signature of binary compounds.
Thus there are two things to look out for:
1. Is it binary?
2. Is it ionic?
look for the suffix –ide.
look to see if there is a metal.
page 33
Exercise:
NaBr
CaCl2
BaO
Li3N
NaNO3
4.3.1.2
Determine whether each of the following is a binary ionic or binary covalent compound.
Chemical Formulas from Names
In the previous section, we have been given the formula and from the formula we provided the name. A
functional system for naming compounds requires that the reverse also be true. That is, given the name of
the compound, we must be able to unambiguously write the chemical formula.
Name
Rule:
Formula
When we put together a formula from the name, the final formula must have
an overall charge of zero (0).
To put together the formula of a binary ionic compound you must:
1. Identify the charge of the cation and the anion.
2. Multiply each cation and each anion by small integers so that the total charge supplied by all the
cations equals the total charge supplied by all the anions.
3. Write the formula of the compound with the
cation + number of cations (subscripted) + anion+ number of anions (subscripted)
e.g.
Write the formula for each of the following compounds.
1. Potassium iodide
a. Identify the type of compound
This is ionic – potassium is a metal. It is binary – it has the suffix –ide.
b. Write the charges of each ion:
K belongs to Group IA so it has a charge of +1:
I belongs to Group VIIA so it has a charge of –1:
c.
K
–
I
+
Write an equation showing the gain or loss of electrons by each type of ion:
K
K + + e–
I + e–
I–
d. Multiply each half equation by small integers so that the number of electrons produced in the 1st
equation equals the number of electrons absorbed in the 2nd equation
+
This one is easy. One electron is produced to form K and one electron is absorbed to
–
form I .
e. Put together the formula
KI
2. Calcium chloride
a. Identify the type of compound
This is ionic – calcium is a metal. It is binary – it has a the suffix –ide.
b. Write the charges of each ion:
Ca belongs to Group IIA so it has a charge of +2:
Cl belongs to Group VIIA so it has a charge of –1:
c.
2+
Ca
–
Cl
Write an equation showing the gain or loss of electrons by each type of ion:
page 34
Ca
Ca
Cl + e
–
2+
–
+ 2e
Cl
–
d. Multiply each half equation by small integers so that the number of electrons produced in the 1st
equation equals the number of electrons absorbed in the 2nd equation
Each Cl absorbs only one e–. Thus we need 2 Cl’s to absorb both electrons produced by the Ca.
Ca
Ca
2+
2 x [Cl + e
–
+ 2e
–
–
Cl ]
e. Put together the formula
CaCl2
Exercise:
Write the formula for each of the following compounds.
magnesium chloride
barium sulphide
aluminium selenide
rubidium oxide
strontium fluoride
calcium bromide
indium oxide
page 35
4.3.2
BINARY IONIC COMPOUNDS: CATIONS WITH VARIOUS CHARGES
Most metal atoms can show several charges in different compounds. Thus, for example, iron is capable of
forming two types of compounds when combined with oxygen having completely different physical and
chemical properties.
FeO
grey colour
Fe2O3 reddish brown (rust)
According to the rules learned in the previous section, the name of both of these compounds would be:
iron oxide (the compound is binary). This is not adequate however, because the name would not give us a
unique chemical formula.
The nomenclature rule for these types of compounds is as follows:
metal (charge of the metal in roman numerals) anion-ide
Thus in the examples cited above:
FeO would be named iron (II) oxide.
Fe2O3 would be named iron (III) oxide.
The difficulty with this system is that you have to either know or calculate the charges.
The charges carried by atoms in a compound are known as oxidation numbers (or oxidation states).
4.3.2.1
Rules for Assigning Oxidation States to Atoms
These are to be memorised in the sequence provided.
1. The sum of the oxidation numbers of all atoms in a chemical substance must add up to the overall
charge of the substance.
2. Group IA elements: Li, Na, K, Rb, Cs will always form cations having (+1) charge.
3. Group IIA elements: Be, Mg, Ca, Sr, Ba will always form cations having (+2) charge.
4. Group IIIA elements: Al, Ga, In will always form cations having (+3) charge.
5. Ag always has (+1)
3+
Zn always has (+2)
Al
Cd always has (+2)
2+
3+
Zn
Ga
Ag
+
2+
Cd
3+
In
6. H will usually have (+1) charge. The exception is when H is bonded to a metal; in this case it will
have an oxidation number of (–1).
7. Usually, but not always the Group A nonmetals will have a charge associated with the group
number. Thus:
•
•
•
Group VA elements (Nitrogen group) will have (–3).
Group VIA elements (Oxygen group) will have (–2).
Group VIIA elements (Halogens) will have (–1).
Care should be utilised here. There are many exceptions.
8. The oxidation number of all other atoms must be calculated.
************
Once the oxidation states of all atoms in the compound have been calculated, you can use them to name
the compound.
page 36
e.g., Write the name of each of the following compounds.
1. Cr2O3
a. Determine the oxidation state of each element in the compound
•
Use Rule 1 to set up an equation
2(Cr) + 3 (O) = 0
•
This is an equation with two unknowns. You must choose a value of oxidation number for
one of the atoms. The priority is given to Group A atoms. Choose oxygen according to its
group number.
(O) = –2
•
Substitute this into the equation above.
2(Cr) + 3 (–2) = 0
2(Cr) = +6
(Cr) = +3
b. Write the name of the compound.
It is a binary, ionic compound.
chromium (III) oxide
Cr has a charge of +3
2. CoBr6
a. Determine the oxidation state of each element in the compound
•
Use Rule 1 to set up an equation
2(Co) + 6 (Br) = 0
•
This is an equation with two unknowns. You must choose a value of oxidation number for
one of the atoms. The priority is given to Group A atoms. Choose bromine according to its
group number.
(Br) = –1
•
Substitute this into the equation above.
(Co) + 6 (–1) = 0
(Co) = +6
b. Write the name of the compound.
cobalt (VI) bromide
It is a binary, ionic compound.
Co has a charge of +6
3. SnS2
a. Determine the oxidation state of each element in the compound
•
Use Rule 1 to set up an equation
(Sn) + 2 (S) = 0
•
This is an equation with two unknowns. You must choose a value of oxidation number for
one of the atoms. The priority is given to Group A atoms. Choose sulphur according to its
group number.
(S) = –2
•
Substitute this into the equation above.
(Sn) + 2 (–2) = 0
(Sn) = +4
b. Write the name of the compound.
tin (IV) sulphide
It is a binary, ionic compound.
Sn has a charge of +4
page 37
e.g.,
Write the formula for each of the following compounds
1. Iron (III) chloride
a. Identify the charges of the ions
(Fe) = +3
(Cl) = –1
b. Combine the two types of atoms so there is a net charge of zero
Fe
Fe
3+
3 x [ Cl + e
c.
Write the formula:
–
+ 3e
–
–
Cl ]
FeCl3
2. Manganese (V) oxide
a. Identify the charges of the ions
(Mn) = +5
(O) = –2
b. Combine the two types of atoms so there is a net charge of zero
Mn5+ + 5 e– ]
2 x [ Mn
5 x [ O + 2 e–
c.
Write the formula:
O2– ]
Mn2O5
3. Mercury (II) nitride
a. Identify the charges of the ions
(Hg) = +2
(N) = –3
b. Combine the two types of atoms so there is a net charge of zero
3 x [ Hg
Hg
2+
–
2 x [ N + 3e
c.
Write the formula:
–
+ 2e ]
3–
N
]
Hg3N2
***********************************
General Rule for the Nomenclature of all Binary Ionic Compounds
Metal
F Charge of I anion
GH metal JK
ide
Do not put this in if the metal has only one
oxidation state. See Rules 2, 3, 4 and 5.
e.g.,
Write the names of each of the following compounds
K2S
W 2O5
CdO
ZnCl2
AgCl
page 38
4.3.3
BINARY COVALENT COMPOUNDS
These can be recognised because they contain no metal atoms. For our purposes, only nonmetals form
covalent bonds.
nonmetal + nonmetal
Binary covalent compounds:
Binary covalent compounds are named quite differently from binary ionic compounds. In the case of
binary ionic compounds, once the oxidation numbers of all species are known, there is only one way to
put the compound together. In the case of binary covalent compounds, the situation is completely
different. The same pair of atoms may form several covalent compounds. The case of the reactions of
nitrogen (N2) with oxygen (O2) demonstrates this. Possible combinations include:
NO
N2O
NO2
N2O3
N2O4
N2O5
Since the products are not predictable when nonmetals react, we must use a different system of
nomenclature. We use numerical prefixes to state explicitly the number of each type of atom present in
the molecule.
Memorise the following list of numerical prefixes
1
mono-
2
di-
3
tri-
4
tetra-
5
penta-
6
hexa-
7
hepta-
8
octa-
9
nona-
10 decaNomenclature:
(prefix) nonmetal (prefix) nonmetal -ide
Number of
first nonmetal
atom (omit if it
is mono)
First nonmetal
is named. It is
the one
closest to the
metal side
Signature of
binary
compounds
page 39
1. NO
a. Is it ionic or covalent? Is it binary?
b. Number of N-atoms = 1
Number of O-atoms = 1
c. Name the compound
(it is covalent and binary)
prefix = mono
prefix = mono
mononitrogen monoxide
nitrogen monoxide
2. N2O
a. Is it ionic or covalent? Is it binary?
b. Number of N-atoms = 2
Number of O-atoms = 1
c.
Name the compound
prefix = di
prefix = mono
dinitrogen monoxide
3. NO2
a. Is it ionic or covalent? Is it binary?
b. Number of N-atoms = 1
Number of O-atoms = 2
c. Name the compound
(it is covalent and binary)
(it is covalent and binary)
prefix = mono
prefix = di
mononitrogen dioxide
nitrogen dioxide
4. N2O3
a. Is it ionic or covalent? Is it binary?
b. Number of N-atoms = 2
Number of O-atoms = 3
(it is covalent and binary)
prefix = di
prefix = tri
c. Name the compound
dinitrogen trioxide
5. N2O4
a. Is it ionic or covalent? Is it binary?
(it is covalent and binary)
b. Number of N-atoms = 2
Number of O-atoms = 4
prefix = di
prefix = tetra
c. Name the compound
dinitrogen tetroxide
6. N2O5
a. Is it ionic or covalent? Is it binary?
(it is covalent and binary)
b. Number of N-atoms = 2
Number of O-atoms = 5
prefix = di
prefix = penta
c. Name the compound
dinitrogen pentoxide
7. P4O6
a. Is it ionic or covalent? Is it binary? (it is covalent and binary)
b. Number of P-atoms = 4
Number of O-atoms = 6
c.
Name the compound
prefix = tetra
prefix = hexa
tetraphosphorous hexoxide
page 40
POLYATOMIC IONS
NH 4+
Ammonium
C2 H3O2
Acetate
Hg22+
Mercury (I)
C2 O42
Oxalate
MnO4
Permanganate
OH
Hydroxide
Cr2 O72
Dichromate
CN
Cyanide
CrO42
Chromate
SCN
Thiocyanate
SO42
Sulfate
ClO4
Perchlorate
SO32
Sulfite
ClO3
Chlorate
PO43
Phosphate
ClO2
Chlorite
PO33
Phosphite
ClO
Hypochlorite
NO3
Nitrate
CO32
Carbonate
NO2
Nitrite
4.3.4
Naming Acids
+
Acids (Arhennius definition) – substances which, when dissolved in water, produce H ions.
+
By implication then, acids are ionic compounds in which the cations are all protons, H .
Nomenclature of Acids is based on the nomenclature of the anion as there is no need to reference the
cation which is always H+.
Anion
ide
Corresponding Acid
hydro
ic acid
Example
Anion
Corresponding Acid
Cl–, chloride
HCl, hydrochloric acid
ate
ic acid
ClO3 , chlorate
HClO3 , chloric acid
ite
ous acid
ClO , hypochlorite
HClO, hypochlorous acid
Examples:
HI
hydroiodic acid
HCN
hydrocyanic acid
HIO4
periodic acid
HBrO2 bromous acid
page 41
4.3.5
Acid anions
For anions having multiple charges, there is no reason for the cations to be identical. It is perfectly
legitimate to have an ionic compound with a formula that combines several different cations. For example,
the combination of lithium and ammonium cations with phosphate can have:
(NH4)Li2PO4
Ammonium dilithium phosphate
(NH4)2LiPO4
diammonium lithium phosphate
Note that the sequence in which the cations are named is alphabetical.
+
As you would expect, ionic compounds having one or more H cations, but not all, are special cases. Thus
for example, compounds such as:
Na2HPO4
NaH2PO4
and
Ba(HCO2)2
are possible. The nomenclature is based on the anion root.
Cation (Cation Charge) + (prefix for the number of hydrogen – omit mono) hydrogen + root anion.
Thus
Na2HPO4
sodium hydrohen phosphate
NaH2PO4
sodium dihydrogen phosphate
Ba(HCO2)2 barium hydrogen carbonite
Note that although the name is based on the anion root, the anion in these hydrogenated anions includes
the hydrogens. Thus the following substances decompose as follow:
Na2 HPO4
2 Na + + HPO42
Anion = HPO42 not PO43
NaH 2 PO4
Na + H 2 PO4
Anion = H 2 PO4 not PO43
+
page 42
Exercises on Nomenclature
For each substance whose name is given, write the formula, if the formula is given, provide the name.
Use the Periodic Table as your main reference. If you encounter ions with which you are unfamiliar, you
may look these up in your textbook.
This assignment will count as part of the quiz grade. It is due on Thursday, March 18, 2004.
1. carbon dioxide
2.
BaCl2
3. calcium hypochlorite
4.
KMnO4
5. barium hydroxide
6.
FePO4
7. cobalt (II) acetate
8.
SnS
9. calcium fluoride
10. CBr4
11. mercury (II) chlorate
12. Al2(Cr2O7)3
13. Ammonia
14. Al2(SO4)3
15. manganese (II) oxalate
16. CdCl2
17. sodium carbonate
18. KH2PO4
19. mercury (I) chloride
20. Al(ClO2)3
21. aluminium oxide
22. Zn(NO3)2
23. gallium selenate
24. CaCl2 6H2O
25. iron (III) sulphide
26. NH4NO2
27. iron (II) sulphate heptahydrate
28. H3PO3 (aq)
29. barium bromate
30. H2O2
31. lead (II) iodide
32. Ca(OH)2
33. carbonic acid
34. K2SO3
35. sodium sulphite
36. Hg(BrO3)2
37. periodic acid
38. HCN (aq)
39. magnesium hydrogen sulphate
40. Cr2O3
.
page 43