shapes of ions

Working out the shapes of ions using VSEPR theory

This page follows on from an earlier page on using VSEPR to work out the shape of simple molecule. 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 in account the size of the charge present on the ion.

basic outlines of molecular shapes

Using VSEPR rules to work out the shapes of ions

The best way I think to understand how to use the VSEPR model to workout the shapes of ions is by simply doing examples, so lets get started!


Carry out the following steps in order to find the shape of the ion:

  1. Identify the central atom and the number of valency electrons it has. This is easily done, just use the periodic table to find what group the central atom is in and this will give the number of valency electrons.
  2. group in periodic table where element is found 1 2 3 4 5 6 7 8
    number of valency electrons 1 2 3 4 5 6 7 8
  3. Count the number of atoms bonded to the central atom; each atom bonded to the central atom will contribute 1 electron each to form a covalent bond.
  4. 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.
  5. To work out the shapes of ions. Remember that a cation, that is a positively charged ion is formed when a species loses an electron and anions 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 workout the shape of ions.

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

model showing the shape of an ammonium ion As before using VSEPR rules:

  1. Nitrogen is the central atom and it is in group 5. It has 5 valency electrons.
  2. Four hydrogen are bonded to the central atom, each contributes 1 electron. So we have 4 electrons in total.
  3. Charge of +1, this means it has lost an electron.
  4. The total number of electrons in the valency shells is 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, this means there are NO lone pairs in this molecule.

Example 2 - What is the shape of the BrF2+ ion?

As before use the VSEPR rules to work out the shape of this ion:
  1. Bromine is the central atom and it is in group 7. It has 7 valency electrons
  2. Two fluorine are bonded to the central atom, each contributes 1 electron. So we have 2 electrons in total.
  3. Charge of +1, this means it has lost an electron.
  4. The total number of electrons in the valency shells 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 from the 2 Br-F covalent bonds, this means there are two lone pairs of electrons.

shape of the BrF2 ion

Example 3: What is the shape of an amide ion, NH2-?

As before using VSEPR rules:

  1. Nitrogen is the central atom and it is in group 5. It has 5 valency electrons.
  2. Two hydrogen are bonded to the central atom, each contributes 1 electron. So we have 2 electrons in total.
  3. Charge of -1, this means it has gained an electron.
  4. The total number of electrons in the valency shells 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 in this molecule.

Example 4: What is the shape of thallium (III) bromide, TlBr32-?

As before using VSEPR rules:

  1. Thallium is the central atom and it is in group 3. It has 3 valency electrons.
  2. Three bromine atoms are covalently bonded to the central atom, each contributes 1 electron. So we have 3 electrons in total.
  3. Charge of -2, this means it has gained 2 electrons.
  4. The total number of electrons in the valency shells 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, this means there is one lone pair of electrons in this ion.

Example 5: What shape is the tetrafluoro bromate(V) ion (BrF4-)?

As before using VSEPR rules:

  1. Bromine is the central atom and it is in group 7. It has 7 valency electrons.
  2. Four fluorine atoms are covalently bonded to the central atom, each contributes 1 electron. So we have 4 electrons in total.
  3. Charge of -1, this means it has gained an electron.
  4. The total number of electrons in the valency shells is 12 electrons, dividing by 2 gives 6 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, this means there are 2 lone pairs in this molecule.
shape of the tetrafluoro bromate(V) ion

Key Points

Practice questions

Check your understanding - Questions on shape of molecules and ions

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