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Structure
and bonding
Francine
Taylor Campbell, Contributor
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| Boys
from Kingston and St. Andrew and
sections of St.Thomas which falls
within the boundaries of Region
1 in the Ministry of Education
and who are in need of help with
maths and English attend a three-day
camp at the St. Josephs Teachers'
College. The boys were tutored
in mathematics and English by
teachers from the region. - Ian
Allen Photo |
POINTS
TO NOTE
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Atoms participate in bonding in order
to obtain a filled outer electron
shell similar to that of noble gases
in Group 8/0.
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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.
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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 shells.
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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.
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Two or more non-metals with four to
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 (one to eight). The elements
in group one have one outer electron
(write out the electronic structure)
and the elements in group seven have
seven electrons in the outer shell.
This can be said for all the elements
the number of outer shell electrons
determine 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 two. However,
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
(# of electrons =10 and atomic number/#
of protons = 12)
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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 two more protons
than electrons, hence Mg2+.
Now,
positive ions generally exist with
negative ions around it. So, Mg would
have to give up those two electrons
to another element that needed it
to be stable. For example: Oxygen
has an electronic configuration E.C.
= 2:6 (atomic # = 8). This means that
oxygen is in group six and needs two
more electrons in order to have eight
in the outer shell and become stable.
When
O gains or accepts two 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
Mgxx
Mg2+
+ 2e- (lost two electrons)
12p,
12e
12p, 10e
xx
x
O x + 2e O2-
(gained two e-)
xx
8p,
8e
8p, 10e
So,
the formula of the compound formed
between magnesium and oxygen is Mg2+
O2- = MgO (+2-2 = 0 the charges cancel
out) magnesium oxide.
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 two and C
needs four, hence two oxygen atoms
can share with one carbon atom. See
Diagram 1.
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.
See
Chart 1 for 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, e.g., NH4+ the
ammonium ion.
Chart 1
| 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- |
| Iron(II) |
Ba2+ |
Hydrogen
(bi) carbonate |
HCO3-
|
| Barium |
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) |
Fe3+ |
Sulphite |
SO32- |
| Aluminium |
Al3+ |
Sulphate |
SO42- |
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|
Dichromate |
Cr2O72- |
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|
Phosphate |
PO43- |
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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
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*
Francine Taylor-Campbell is
an independent contributor.
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