IB chemistry IB syllabus > bonding (sl) > 4.1 

These notes were written for the old IB syllabus (2009). The new IB syllabus for first examinations 2016 can be accessed by clicking the link below.

IB syllabus for first examinations 2016

4.1 - The ionic bond


4.1.1: Describe the ionic bond as electronic attraction between oppositely charged ions.


Electrostatic attraction

This is the force experience by oppositely charged particles. Its magnitude is proportional to the magnitude of the charges and inversely proportional to the square of the distance between them, i.e. it rapidly gets weaker with distance.

Ions are charged particles and attract one another to form giant ionic lattices in which the postive ions are surrounded by the negative ions and vice versa.

This is known as ionic bonding.


4.1.2: Describe how ions can be formed as a result of electron transfer.


Metal atoms

The metal atoms lose the outer shell of electrons and transfer them to the non-metal atoms. This results in the formation of ions - positive ions from the metal atoms and negative ions from the non-metal atoms. Remember that the total charge on an atoms is the number of protons (positive) balanced by the number of electrons (negative) in a neutral atom the number of protons exactly balances the number of electrons.

Example:

Sodium atoms contain 11 protons in the nucleus and have 11 electrons in the shells around the nucleus. These 11 positive charges are cancelled out by the 11 negative charges an the atom is neutral electronically. Sodium ions contain 11 protons in the nucleus and have 10 electrons in the shells around the nucleus. These 11 positive charges are not wholly cancelled out by the 10 negative charges and the one positive charge remains overall.
Sodium atom 11(+) + 11(-) = zero Sodium ion 11(+) + 10(-) = 1+

When these sodium ions react with non-metal atoms the outer (valence electrons) are transferred to the non metal atoms. This leaves the sodium particles with one electron less than the number of protons. Overall it will have an electrical charge of 11(+) + 10(-) = 1+. This particle is now called a sodium ion.


The non metal atoms

These receive the electrons transferred from the metals atoms...

Chlorine atoms contain 17 protons in the nucleus and have 17 electrons in the shells around the nucleus. These 17 positive charges are cancelled out by the 17 negative charges and the atom is electronically neutral Chloride ions contain 17 protons in the nucleus and have 18 electrons in the shells around the nucleus. These 17 positive charges only cancel out 17 of the negative charges leaving one negative in excess. The ion has a negative charge overall
Chlorine atom 17(+) + 17(-) = zero Chloride ion 17(+) + 18(-) = 1-

The non-metal atoms (in this case chlorine) receive electrons (from the metal atoms) into their outer shells making the outer shell full, a negatively charged ion is formed. Notice that the ending of the particle is now changed to -ide.

The positive ions formed by the metal atoms are now attracted to the negative ions formed by the non-metal atoms and the ions pack tightly together into a giant ionic structure.


Ionic bonding summary

Overall


4.1.3: Deduce which ions will be formed when elements in groups 1, 2 and 3 lose electrons.


Formation of ions from metals in groups 1, 2, and 3

The formation of ions is a process that makes the atoms electronic structure more stable. Metal atoms form groups 1, 2 and 3 lose all of their outer shell electrons when forming ions. As each electron corresponds to a negative charge then the loss of one electron produces a positive ion with one positive charge. The loss of two electrons produces a positive ion with two positive charges.. etc

Note that these rules do not apply to the transition metals.


4.1.4: Deduce which ions will be formed when elements in groups 6 and 7 gain electrons.


Negative ions from groups 6 and 7

Atoms in groups 6 and 7 gain electrons to obtain a full outer shell. This means that atoms in group 6 form ions with a double negative charge and atoms in group 7 form ions with a single negative charge.

Oxygen (configuration 2, 6) forms oxide ions (configuration 2,8), O2-

Sulfur (configuration 2, 8, 6) forms sulfide ions (configuration 2, 8, 8), S2-

Fluorine (configuration 2, 7) forms fluoride ions (configuration 2,8), F-

Chlorine (configuration 2, 8, 7) forms chloride ions (configuration 2, 8, 8), Cl-

Non-metals gain just enough electrons to fill their outer shells. It may be seen from the electronic configuration of group 6 elements that their outer shells with 6 electrons need another two electrons to be filled. Each electron gained corresponds to a negative charge and so group 6 non-metals form double negative ions.

Sulphur (2,8,6) --> [sulphide ion (2,8,8)]2-

Similarly group 7 elements need only one electron for a full outer shell and so form single negative ions.

Chlorine (2,8,7) --> [chloride ion (2,8,8)]-

Notice again the slight change of name for the negative ions to - ide (the ionic form)


4.1.5: State that transition metals can form more than one ion. Include examples such as Fe2+ and Fe3+. .


Variable ion formation

As stated above the normal 'rules' do not apply to the transition metals as they can form more than one type of ion. The reason for this lies in the electronic configuration which involves electrons from the 3rd level called 'd' electrons.

Examples: Fe2+ and Fe3+

In these two ions the Iron atoms have lost either two electrons (in the case of Fe2+) or three electrons (in the case of Fe3+). This type of behaviour is possible for all of the transition metals. Iron II compounds (containing Fe2+ ions) are usually pale green in colour whereas Iron III compounds (containing Fe3+ ions) are usually yellow, orange or red.

Iron II sulphate solution Iron III sulphate solution

Cu2+ and Cu+

These two copper ions have lost two electrons and one electron respectively. Copper II compounds (containing Cu2+ ions) are blue or green and soluble whereas copper I compounds (containing Cu+ ions) are white and insoluble.

Copper II sulphate solution Copper I sulphate (insoluble)

4.1.6: Predict whether a compound of two elements would be ionic from the position of the elements in the periodic table, or from their electronegativity values.


Prediction of bonding type

Compounds in which the bonded atoms have a large electronegativity difference for ionic compounds. Where the difference is slight, they are covalent. There is no hard and fast value at which the change occurs. Rather there is a greater and greater degree of covalency as the values become closer together.

Perhaps the closest to a 'cut off' is the compound formed between aluminium and chlorine. In the solid state at 0ºC there is considerable evidence that it is ionic, but at room temperature it seems to be covalent. At higher temperatures it sublimes as a dimer with the formula Al2Cl6.

Aluminium has an electronegativity of 1.5 and chlorine 3.0. That makes the difference in electronegativity = 1.5 units on the Pauling scale. This is a good value to use as a 'rule of thumb'.

Greater than 1.5 units = ionic

Less than 1.5 units = covalent.

Ionic or covalent bond formation

It was stated above that metals reacting with non-metals form ionic bonds. This is the GENERAL rule. It is really the ease with which a metal can lose electrons coupled with the attraction that a non-metal has for more electrons.

When an element loses electrons easily and another element has a high attraction for electrons then this encourages ionic bond formation. As metals lose electrons easily they are said to be electropositive, or to put it another way, to have low electronegativity values.

Non-metals tend to attract electrons and the non-metals closest to group 7 have higher electronegativity values. Electronegativity increases from bottom left to top right in the periodic table.

When the DIFFERENCE between the electronegativity values is large, ionic bonds are formed.

It is usually the case that the difference in electronegativity values between metals and non-metals is large enough to cause ionic bond formation (transfer of electrons etc etc), however this is not the case for all metal:non-metal combinations.


4.1.7: State the formula of common polyatomic ions formed by the non-metals in periods 2 and 3. Examples include NO3-, OH-, SO42-, CO32-, PO43-, NH4+, HCO3-.


Formulae of polyatomic ions

Type of ion Formula
Nitrate NO3-
Hydroxide OH-
Sulfate SO42-
Carbonate CO32-
Phosphate PO43-
Ammonium NH4+
Hydrogen carbonate HCO3-

4.1.8: Describe the lattice structure of ionic compounds. Students should be able to describe the structure of sodium chloride and an example of an ionic lattice .


The bulk giant ionic structure

Sodium chloride crystal structure Ionic properties

This sodium chloride structure is typically ionic with billions upon billions of oppositely charged ions arranged into a lattice (network) in which each positive ion is surrounded by negative ions and each negative ion is surrounded by positive ions.

Ionic structures are usually hard, brittle, high melting point and conduct electricity when melted (molten) or when dissolved in water.

The lattice energies of ionic compounds are relatively large. The lattice energy of NaCl, for example, is 787.3 kJ/mol, which is only slightly less than the energy given off when natural gas burns.

The bond between ions of opposite charge is strongest when the ions are small.

The lattice energies for the alkali metal halides is therefore largest for LiF and smallest for CsI, as shown in the table below.

Lattice Energies of Alkali Metals Halides (kJ/mol)

F- Cl- Br- I-
Li+ 1036 853 807 757
Na+ 923 787 747 704
K+ 821 715 682 649
Rb+ 785 689 660 630
Cs+ 740 659 631 604

The ionic bond should also become stronger as the charge on the ions becomes larger. The data in the table below show that the lattice energies for salts of the OH- and O2- ions increase rapidly as the charge on the ion becomes larger.

Lattice Energies of Salts of the OH- and O2- Ions (kJ/mol)

OH- O2-
Na+ 900 2481
Mg2+ 3006 3791
Al3+ 5627 15916

Constructing ionic formulae

Deducing formula units requires a knowledge of the charges on the ions (see above). All chemical compounds are electronically neutral i.e. they have no overall positive or negative charges. This means that the charges carried by the ions must be cancelled out by the charges carried on the oppositely charged ions.

Example: Sodium Chloride

Sodium ions have a single positive charge

Chloride ions have a single negative charge

Hence the charge on one sodium ion Na+ is canceled out by the charge on one chloride ion Cl-. The compound formed, sodium chloride, has one sodium ion for every one chloride ion and is given the formula NaCl. Remember that the ionic charges are never written in the compound formula.

When an ion has more than one charge, it associates with enough of the opposite charged ion to cancel out the charge overall.

Example: Calcium Fluoride

As calcium is in group 2 it has a double positive ion Ca2+

Fluorine is in group 7 and forms a single negative charges ion F-

In order to cancel out the two positives of calcium we need to use two fluoride ions and so the formula unit is CaF2

Remember that these formula units do not represent molecules, ionic compounds do not form molecules, they represent the simplest ratio of the positive ions to the negative ions in the giant ionic lattice.

Sodium chloride is made up of billions of sodium ions and billions of chloride ions all arranged in a giant crystal. If they were to be counted however, it would be found that for every one sodium ion there is one chloride ion and so the formula is NaCl.


solid NaCl liquid NaCl gaseous NaCl

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