Working out the shapes of ions using VSEPR theory

This page follows on from an earlier page on using VSEPR to work out the shapes of simple molecules. Before trying to work out the shapes of ions you should already know the 5 basic outlines for the shapes of all the molecules you are likely to meet. These are shown below for reference. To work out the shape of ions using the valence shell electron pair repulsion theory (VSEPR) it is a simple matter of identifying the central atom in a molecule and working out the number of bonding and lone pairs (non-bonding pairs) of electrons that are present in the molecule and then taking into account the size of the charge present on the ion.

Using VSEPR rules to work out the shapes of ions

The best way I think to understand how to use the VSEPR model to work out the shapes of ions is by simply trying a few examples, so let's get started!

Carry out the following steps in order to find the shapes of the ions below:

1. Identify the central atom and the number of valence electrons it has. This is easily done using the periodic table to find which group the central atom is in, and this will give the number of valence electrons.

group in periodic table where element is found	1	2	3	4	5	6	7	8
number of valence electrons	1	2	3	4	5	6	7	8

2. Count the number of atoms bonded to the central atom; each atom bonded to the central atom in the molecule or molecular ion will contribute 1 electron to form a covalent bond.
3. Add up the total number of electrons and divide by 2 to get the number of electron pairs. Some of these electron pairs could be bonding pairs of electrons and some could be lone pairs or non-bonding pairs of electrons.
4. To work out the shapes of ions simply add or subtract electrons from your running total of electrons calculated so far, depending on whether the ion is a cation or an anion. Remember that a cation, that is a positively charged ion, is formed when a species loses an electron and anions, that is negatively charged ions, are formed by the addition of electrons to a molecule.

Shapes of ions

Simply follow the rules above and those we used previously to work out the shape of ions.

Example 1: What is the shape of an ammonium ion (NH₄⁺)?

As before using VSEPR rules:

1. Nitrogen is the central atom and it is in group 5 of the periodic table so it has 5 valence electrons.
2. Four hydrogen atoms are covalently bonded to the central nitrogen atom and each contributes 1 electron to the outer valence shell. So we have 4 electrons in total.
3. The molecular ion has a charge of +1; this means it has lost an electron.
4. The total number of electrons in the valence shell is therefore 8 electrons; dividing by 2 gives 4 electron pairs; so the shape will be based on a tetrahedral structure. There are 4 bonding pairs of electrons from the 4 N-H bonds and this means there are NO lone pairs of electrons in this ion.

Example 2 - What is the shape of the bromine difluoride ion (BrF₂⁺) ?

As before use the VSEPR rules to work out the shape of this ion:

1. Bromine is the central atom and it is a halogen found in group 7 of the periodic table; it has 7 valence electrons.
2. Two fluorine atoms are covalently bonded to the central bromine atom; with each fluorine atom contributing 1 electron. So we have 2 electrons in total from the two fluorine atoms.
3. The molecular ion has a charge of +1; this means it has lost an electron.
4. The total number of electrons in the valence shell is 8 electrons; dividing by 2 gives 4 electron pairs; so the shape of this molecular ion will be based on a tetrahedral structure. There are 2 bonding pairs of electrons from the 2 Br-F covalent bonds so this means there are two lone pairs of electrons.

So we can say that the shape of the BrF₂⁺ ion will be based on a tetrahedral arrangement around the central bromine atom, with 2 lone pairs and 2 bonding pairs of electrons. However the overall shape of the ion will be based only on the two bonding pairs of electrons. As we have seen before the presence of the 2 lone pairs will reduce the bond angle between the bromine and the fluorine atoms down from the expected 109.5° in a tetrahedral arrangement to 104.5° due to the extra space requirements lone pairs have.

Example 3: What is the shape of an amide ion (NH₂^-)?

As before using VSEPR rules:

1. Nitrogen is the central atom in this molecular ion and it is found in group 5 of the periodic table; it will therefore have 5 valence electrons.
2. Two hydrogen atoms are covalently bonded to the central nitrogen atom with each one contributing 1 electron to the covalent bond; so we have 2 electrons in total from these two hydrogen atoms in the outer valence shell.
3. The molecular ion has a charge of -1; this means it has gained an electron.
4. The total number of electrons in the valence shell is 8 electrons; dividing by 2 gives 4 electron pairs, so the shape will be based on a tetrahedral structure, there are 2 bonding pairs of electrons from the 2 N-H bonds, this means there are 2 lone pairs of electrons in this ion.

So we can say the shape of the amide ion will be based on a tetrahedral arrangement around the central nitrogen atom, with 2 lone pairs and 2 bonding pairs of electrons. The overall shape of the ion will be based only on the bonding pairs. However the presence of the 2 lone pairs will reduce the bond angle between the nitrogen and the hydrogen atoms down from the normal 109.5° in a tetrahedral arrangement to 104.5°, this is outlined in the image below:

Example 4: What is the shape of thallium(III) bromide ion (TlBr₃^2-)?

As before using VSEPR rules:

1. Thallium is the central atom in this molecular ion and it is in group 3 of the periodic table, so it has 3 valence electrons.
2. Three bromine atoms are covalently bonded to the central atom and each contributes 1 electron to the outer valence shell of electrons. So we have 3 electrons in total.
3. The molecular ion has a charge of -2; this means it has gained 2 electrons.
4. The total number of electrons in the valence shell is 8 electrons; dividing by 2 gives 4 electron pairs; so the shape will be based on a tetrahedral structure. There are 3 bonding pairs of electrons from the 3 Tl-Br bonds so this means there is one lone pair of electrons in this ion.

So we can say that the shape of the thallium(III) bromide molecular ion will be based on a tetrahedral arrangement around the central thallium atom, with 1 lone pair and 3 bonding pairs of electrons. The overall shape of the ion will be based only on the three bonding pairs. However the presence of the 1 lone pair will reduce the bond angle between the thallium and the bromine atoms down from the normal 109.5° in a tetrahedral arrangement to 107.5°.

Example 5: What shape is the tetrafluorobromate(V) ion (BrF₄^-)?

As before using VSEPR rules:

1. Bromine is the central atom in this molecular ion and it is a halogen found in group 7 of the periodic table; so it has 7 valence electrons.
2. Four fluorine atoms are covalently bonded to the central atom and each contributes 1 electron to the outer valence shell. So we have 4 electrons in total.
3. The molecular ion has a charge of -1; this means it has gained an electron.
4. The total number of electrons in the valence shell is therefore 12 electrons; dividing by 2 gives six electron pairs; so the shape will be based on an octahedral structure. There are 4 bonding pairs of electrons from the 4 Br-F bonds so this means there are 2 lone pairs of electrons in this ion.

So we can say that the shape of the BrF₄^- ion will be based on an octahedral arrangement around the central bromine atom, with 2 lone pairs and 4 bonding pairs of electrons. The overall shape of the ion will be based only on the four bonding pairs; this is outlined in the image below:

Shape finder

Use the activity below to work out the shapes of various ions. Simply select the group from the periodic table in which the central atom in the molecular ion is found, then select the number of atoms bonded to it and finally the charge on the ion. Work out the shape of the ion yourself before you press the calculate shape button. Remember the ions cannot have an odd number ofelectrons since there are 2 electrons in each bonding and lone pair.

Key Points

• To find the shape of a molecule simply identify the central atom; this is usually obvious from the formula of the substance.
• Each atom bonded to the central atom will contribute one electron to the covalent bond formed. Simply add the number of bonded atoms to the number of valence electrons in the central atom; this will give you the number of electrons in the outer or valence shell of the molecule or molecular ion.
• Simply divide your total number of electrons by 2 and this will give you the number of electron pairs around the central atom in the molecule or molecular ion.
• If the molecular ion has a positive charge, that is it is a cation, then subtract the size of the charge from the number of valence shell electrons. If the ion has a negative charge, that is it is an anion, add the size of this charge to the number of valence electrons.
• If the ion has lone pairs or non-bonding pairs of electrons these are NOT used in determining the overall shape of the ion. However they will influence bond angles present in the ion since lone pairs of electrons take up more space than bonding pairs of electrons.