Common Exam Questions on Dot and Cross Diagrams
Practise how to draw dot and cross diagrams for molecules that are commonly featured in schools exams and at the O Level.
Q1. Draw a dot and cross diagram to show the bonding in water, H2O.

- Place hydrogen at the ends and oxygen at the centre, as hydrogen can only form 1 bond and hence cannot be the central connector
- Each hydrogen forms a single bond with oxygen, which is represented by a pair of dot and cross
- After bonding, hydrogen has no other electron, while oxygen has 4 non-bonding electrons
Q2. Draw a dot and cross diagram to show the bonding in methane, CH4.

- Like in water, place hydrogen at the ends and carbon at the centre
- Each hydrogen forms a single bond with carbon, which is represented by a pair of dot and cross
- After bonding, hydrogen and carbon alike have no other electron
Q3. Draw a dot and cross diagram to show the bonding in hydrogen chloride gas, HCl.

- Each hydrogen forms a single bond with chlorine, which is represented by a pair of dot and cross
- After bonding, hydrogen has no other electron, while chlorine has 6 non-bonding electrons
Q4. Draw a dot and cross diagram to show the bonding in carbon dioxide, CO2.

- Place carbon at the centre and oxygen at the ends
- Each oxygen forms double bond with carbon, which is represented by 2 pairs of dot and cross
- After bonding, carbon has no other electron, while oxygen has 4 non-bonding electrons
Q5: Challenging Question from Prelim Paper
A non-metallic element Z in the third period forms a bromide with the formula ZBr3. Using the formula, infer the bonding in ZBr3 by drawing a dot-and-cross diagram showing the valence electron only.

Before picking up your pencil to draw, consider if the bond is covalent or ionic. Because both Z and bromine are non-metallic elements, they form a covalent bond together.
- Place Z at the centre and bromine at the ends
- Since bromine has a valency of 1, each bromine atom can only form a single bond with Z, which is represented by a pair of dot and cross
- After bonding, Z only has 6 electrons. Assuming that Z would have achieved the electronic configuration of a noble gas, we infer that Z must have 2 more non-bonding valence electrons.
Therefore, Z is from Group VI, as it has 5 valence electrons. 3 are used in covalent bonding, while 2 remain as non-bonding electrons in ZBr3.
Marking Points
- Correct number of shared electrons and non-bonding electrons
- Correct relative size of valence shell (like how the valence shell of chlorine is bigger than that of hydrogen)
- Legend included