Lewis electron dot structures are crucial because they visually represent the valence electrons of atoms, helping to predict molecular bonding, structure, and reactivity in chemistry. This resource contains all you need to know to get started with these structures.
Lewis structures, also known as electron dot structures, are diagrams that represent the valence electrons of atoms within a molecule. These structures are used to show how the electrons are arranged around individual atoms in a molecule. The main purpose of Lewis structures is to predict the number and type of bonds that may form around an atom.
Example 1 – drawing Lewis structures of molecular compounds
Draw the Lewis structure of sulfur dioxide, \(\ce{SO}_{2}\).
Step 1: Count the total number of valence shell electrons of all the atoms involved in the molecule.
Sulfur dioxide has two sulfur atoms and one oxygen atom.
Atom
Atomic number
Electron configuration
Valence shell electrons
\(\ce{S}\)
\(16\)
\(1s^{2}\,2s^{2}\,2p^{6}\,3s^{2}\,3p^{4}\)
\(6\)
\(\ce{O}\)
\(8\)
\(1s^{2}\,2s^{2}\,2p^{4}\)
\(6\)
The total number of valence shell electrons is:
\[6+\left(6\times2\right)=18\]
Step 2: Predict the central atom. This is the atom that can form most bonds. If two different atoms can form the same number of bonds, then the least electronegative atom should be the central atom.
Among sulfur and oxygen, sulfur is the least electronegative atom. Therefore, it is the central atom.
Step 3: Draw single bonds connected to the central atom.
Step 4: Add the lone pairs of electrons to the atoms connected to the central atom. Calculate the remaining valence electrons after drawing bonds.
Four valence electrons are used for two bonds. The remaining valence electrons are:
\[=18-4=14\]
These valence electrons can be distributed among atoms connected to the central atom as lone pairs to complete the octet.
Step 5: Assign lone pairs of electrons to the central atom if any valence electrons remain.
To assign lone pairs to the connected atoms, we have used 12 electrons. Therefore, we can calculate the electrons remaining:
\[14-12=2\]
These electrons can be allocated to the central atom as a lone pair.
Step 6: If the central atom still lacks the octet, create multiple bonds with connected atoms using lone pairs of electrons.
Since sulfur has not achieved an octet, a double bond must be created with one of the oxygen atoms using a lone pair. With the double bond, all three atoms have acquired an octet.
Step 7: Calculate the formal charges for each atom. Formal charges should be closer to zero for each atom. The formula to calculate the formal charges is:
For the single-bonded \(\ce{O}\), formal charge \(=6-6-1=-1\).
For the double-bonded \(\ce{O}\), formal charge \(=6-4-2=0\).
Example 2 – drawing Lewis structures of polyatomic ions
Draw the Lewis structure of the carbonate ion, \(\ce{CO}_{3}^{2-}\).
Step 1: Count the total number of valence shell electrons of all the atoms involved in the molecule, including the charge of the ion.
The carbonate ion is composed of three oxygen atoms and one carbon atom. It also has a \(2-\) charge, so two extra electrons.
Atom
Atomic number
Electron configuration
Valence shell electrons
\(\ce{C}\)
\(6\)
\(1s^{2}\,2s^{2}\,2p^{2}\)
\(4\)
\(\ce{O}\)
\(8\)
\(1s^{2}\,2s^{2}\,2p^{4}\)
\(6\)
The total number of valence shell electrons is:
\[=4+\left(6\times3\right)+2=24\]
Step 2: Predict the central atom.
Among carbon and oxygen, carbon is the least electronegative atom. Therefore, it is the central atom.
Step 3: Draw single bonds connected to the central atom.
Step 4: Add the lone pairs of electrons to the atoms connected to the central atom. Calculate the remaining valence electrons after drawing bonds.
Six valence electrons are used for three bonds. The remaining valence electrons are:
\[=24-6=18\]
These valence electrons can be distributed among atoms connected to the central atom as lone pairs to complete the octet.
Step 5: Assign lone pairs of electrons to the central atom if any valence electrons remain.
To assign lone pairs to the connected atoms, we have used 18 electrons. We have used all the valence electrons.
Step 6: If the central atom still lacks the octet, create multiple bonds with connected atoms using lone pairs of electrons.
Since carbon has not achieved an octet, a double bond must be created with one of the oxygen atoms using a lone pair. With the double bond, all four atoms have acquired an octet.
Step 7: Calculate the formal charges for each atom. Allocate the two negative charges.
For \(\ce{C}\), formal charge \(=4-0-4=0\).
For the double-bonded \(\ce{O}\), formal charge \(=6-4-2=0\).
For the single-bonded \(\ce{O}\)s, formal charge \(=6-6-1=-1\).
Draw the Lewis structure of the ammonium ion, \(\ce{NH}_{4}^{+}\).
Step 1: Count the total number of valence shell electrons of all the atoms involved in the molecule, including the charge of the ion.
The ammonium ion is composed of one nitrogen atom and four hydrogen atoms. It also has a \(1+\) charge, so one fewer electron.
Atom
Atomic number
Electron configuration
Valence shell electrons
\(\ce{N}\)
7
\(1s^{2}\,2s^{2}\,2p^{3}\)
\(5\)
\(\ce{H}\)
\(1\)
\(1s^{1}\)
\(1\)
The total number of valence shell electrons is:
\[=5+\left(1\times4\right)-1=8\]
Step 2: Predict the central atom.
Since hydrogen can not handle many bonds, nitrogen is the central atom.
Step 3: Draw single bonds connected to the central atom.
Step 4: Add the lone pairs of electrons to the atoms connected to the central atom. Calculate the remaining valence electrons after drawing bonds.
Eight valence electrons are used for four bonds. The remaining valence electrons are:
\[=8-8=0\]
We have used all the valence electrons.
Step 5: Assign lone pairs of electrons to the central atom if any valence electrons remain.
We have used all the valence electrons.
Step 6: If the central atom still lacks the octet, create multiple bonds with connected atoms using lone pairs of electrons.
The central atom has already achieved an octet.
Step 7: Calculate the formal charges for each atom. Allocate the one positive charge.
For \(\ce{N}\), formal charge \(=5-0-4=+1\).
For \(\ce{H}\), formal charge \(=1-0-1=0\).
Example 4 – using Lewis structures to determine molecular shape
Predict the molecular shape of \(\ce{NH3}\).
Step 1: Draw the Lewis structure of the molecule and identify the central atom.
The central atom is \(\ce{N}\) as it forms most bonds.
Step 2: Sum up the number of bonds and lone pairs connected to the central atom to get the total number of electron clouds around the central atom.
The central atom \(\ce{N}\) forms three bonds—one with each \(\ce{H}\) atom—and one lone pair. Therefore there are four electron clouds (three bonds and one lone pair) around the central atom.
Step 3: Recall the VSEPR theory to determine how the electron clouds will be arranged.
According to VSEPR theory, the four electron clouds around the \(\ce{N}\) will arrange in such a way that each cloud stays as far as possible from the others to minimise the repulsion forces between them. This will result in a tetrahedral shape. Molecules with a tetrahedral shape have a bond angle of \(109^{\circ}\). However, due to the presence of a lone pair on the central atom, the overall shape of ammonia becomes pyramidal rather than tetrahedral with a bond angle of \(107^{\circ}\).
Molecular geometry diagram
Tetrahedron
A diagram of a tetrahedral shape with a central atom labeled "C" (in red) and surrounding atoms labeled "A" (in blue). The legend indicates that "C" is the central atom and "A" represents other atoms.
Ammonia – Pyramid Shape
The chemical structure of ammonia (NH₃), showing a nitrogen atom bonded to three hydrogen atoms in a pyramidal shape. The nitrogen has a lone pair represented by a grey lobe.
A 3D model of the ammonia molecule with a blue nitrogen atom and white hydrogen atoms arranged in a pyramidal shape.
