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Structure
& bonding
Francine
Taylor-Campbell, Contributor
YOU
SHOULD BE ABLE TO:
- Explain
the formation of ionic and covalent
bonds
- Predict
the formation of ionic and covalent
bonds based on atomic structure
- Write
formulae to represent ions and molecules
MAIN
POINTS
- Atoms
participate in bonding in order
to obtain a filled outer electron
shell similar to that of noble gases
in Group 8/O.
- To
bond, atoms may either lose, gain
or share electrons. Losing or gaining
electrons results in the formation
of an ionic or electrovalent bond
and occurs between metals and non-metals.
Sharing of electrons results in
the formation of a covalent bond
and this occurs between non-metals.
- Metals
with one, two or three electrons
in their outer shell will readily
give up or lose their electrons
(and form positively charged ions
called cations) in order to be stable
or to have a filled outer shell.
- Non-metals
with five, six or seven electrons
in their outer shell will readily
accept or gain electrons (and form
negatively charged ions called anions)
to acquire a noble gas configuration.
- Two
or more non-metals with four - seven
electrons in their outer shell may
share their electrons in order to
have a stable arrangement.
Let
us examine the periodic table again.
If you take out your table you will
see that the elements are divided
into groups and periods. For the first
20 elements (H to Ca), there are eight
groups (1 to 8). The elements in group
1 have one outer electron (write out
the electronic structure) and the
elements in group 7 have seven electrons
in the outer shell. This can be said
for all the elements; the number of
outer shell electrons determines which
group they are in.
Take
a look at magnesium: E.C. = 2:8:2
(atomic # = 12)
This
means that magnesium has 12 electrons
with 2 in the outer shell. Magnesium
is, therefore, in group 2 but since
the outer shell can hold a maximum
of eight electrons for it to be filled,
then magnesium is not stable and either
needs to gain six more electrons (which
is harder to do) or lose the two outer
electrons to be stable (this is more
favourable). When magnesium loses
the two electrons, its electronic
configuration will be: E.C. = 2:8
(# electrons =10 and atomic number/#
protons = 12)
Remember,
only electrons are involved in bonding.
Protons remain unchanged in the nucleus.
Since
magnesium has 10 electrons (with negative
charge -10) and 12 protons (with positive
charge +12), then the overall charge
on the magnesium ion is +2 (+12
10 = +2). In other words, you are
saying that Mg has 2 more protons
than electrons; Mg2+.
Now
positive ions generally exist with
negative ions around it so Mg would
have to give up those 2 electrons
to another element that needs it to
be stable. For example, oxygen has
an electronic configuration E.C. =
2:6 (atomic
# =8). This means that oxygen is in
group 6 and needs 2 more electrons
in order to have 8 in the outer shell
and become stable.
When
O gains or accepts 2 electrons, it
will now have 10 electrons (-10) and
8 protons (+8). The overall charge
on O is now 2 (+8 10
= -2). O2- has an E. C. = 2:8.
An
ionic or electrovalent bond is formed
between Mg and O
Mg**
→ Mg2+ + 2e- (lost
2 electrons)
12p,
12e → 12p, 10e
xx
x O x + 2e → O2- (gained
2 electrons)
xx
8p,
8e → 8p, 10e
So,
the formula of the compound formed
between magnesium and oxygen is Mg2+
O2- = MgO Magnesium oxide
(+2-2 = 0 so the charges cancel out)
Try
to form the bond between magnesium
and chlorine.
Consider
the bond between carbon and oxygen.
It is difficult and energetically
unfavourable for C 2:4 or O 2:6 to
give up any of their electrons. It
is easier for them to share these
outer electrons. O needs 2 and C needs
4, hence
2 oxygen atoms can share with one
carbon atom.
C
- 2:4 + O - 2:6 =
CO2.
The
molecule is carbon dioxide.
This
sharing of electrons between two non-metals
results in the formation of a covalent
bond.
Try
to show the bond formed between Phosphorous
and Chlorine and give the formula
of the compound formed.
Here
is a list of common cations and anions.
By knowing the formulae of these ions
and their charges you can combine
them to form compounds. The whole
aim of this exercise is to balance
the charges so that they will cancel
out to zero. Note that groups of atoms
with a charge are called radicals
eg NH4+ the ammonium ion.
| Names/Cations |
Formulae |
Names/Anions |
Formula |
| Hydrogen |
H+ |
Fluoride |
F- |
| Lithium |
Li+ |
Chloride |
Cl- |
| Sodium |
Na+ |
Bromide |
Br- |
| Potassium |
K+ |
Iodide |
I- |
| Copper(I) |
Cu+ |
Hydride |
H- |
| Silver |
Ag+ |
Hydroxide |
OH- |
| Ammonium |
NH4+ |
Nitrite |
NO2- |
| Magnesium |
Mg2+ |
Nitrate |
NO3- |
| Calcium |
Ca2+ |
Manganate
(VII) |
MnO4- |
| Barium |
Ba2+ |
Hydrogen
(bi) carbonate |
HCO3- |
| Iron(II) |
Fe2+ |
Hydrogen
sulphate |
HSO4- |
| Copper(II) |
Cu2+ |
Ethanoate |
CH3COO- |
| Zinc |
Zn2+ |
Oxide |
O2- |
| Tin(II) |
Sn2+ |
Sulphide |
S2- |
| Lead(II) |
Pb2+ |
Carbonate |
CO32- |
| Iron(III) |
Fe2+ |
Sulphite |
SO32- |
| Aluminum |
Al3+ |
Sulphite |
SO42- |
|
|
Dichromate |
Cr2O72- |
|
|
Phosphate |
PO42- |
To
form the compound sodium nitrate:
Ions
Na+ and NO3-
The charges cancel out (+1-1) = 0
Formula = NaNO3
For
potassium sulphate: K+ and SO42-
-ve charges = -2, +ve charges = +1
For these to cancel out we need +1
more
+ve charge. Hence we need 2K+
and SO42-. Formula
= K2SO4
Francine
Taylor-Campbell teaches at Jamaica
College. Send questions and comments
to kerry-ann.hepburn@gleanerjm.com
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