Structure and bonding
Francine Taylor-Campbell, Contributor

Radha Kannegatuti (right), a student from St. Hilda's High School, makes a point during the governor-general's Youth Award for Excellence Youth Consultation forum at Breezes Runnaway Bay, recently. - Ian Allen/Staff Photographer

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.

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/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 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. 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 (Mg): E.C. = 2:8:2 (atomic # = 12)

This means that magnesium has 12 electrons with two in the outer shell. Mg is therefore in group two, but since the outer shell can hold a maximum of eight electrons for it to be filled, then Mg 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 Mg loses the two electrons, its electronic configuration will be: E.C. = 2:8 (# electrons =10 and atomic number/# protons = 12)

Fig 2


Remember, only electrons are involved in bonding, protons remain unchanged in the nucleus.

Since Mg has 10 electrons (with negative charge -10) and 12 protons (with positive charge +12) then the overall charge on the Mg ion is +2 (+12 - 10 = +2). In other words, you are saying that Mg has two more protons than electrons: 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 (O) 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 eight 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 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 (+2-2 = 0 the charges cancel out) magnesium oxide

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 figure two)

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 aim of this exercise is to balance the charges so they will cancel out to zero. Note that groups of atoms with a charge are called radicals. An example of this is NH4+ the ammonium ion. (See figure 1)

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.