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CSEC>> Chemistry

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Structure and bonding
Francine Taylor Campbell, Contributor

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

* Atoms participate in bonding in order to obtain a filled outer electron shell similar to that of noble gases in Group 8/0.

* 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 shells.

* 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 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)

* 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-
    Dichromate Cr2O72-
    Phosphate PO43-

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 is an independent contributor.

 
 
 
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