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Learning Objectives
- Draw Lewis structures for covalent compounds.
The following procedure can be used to draw Lewis structures for molecules and polyatomic ions.
How-to: Constructing Lewis electron structures
1. Determine the total number of valence electrons in the molecule or ion.
- Add together the valence electrons from each atom. (Recall that the number of valence electrons is indicated by the position of the element in the periodic table.)
- If the species is a polyatomic ion, remember to add or subtract the number of electrons necessary to give the total charge on the ion.
For CO32−, for example, we add two electrons to the total because of the −2 charge.
2. Arrange the atoms to show specific connections.
- When there is a central atom, it is usually the least electronegative element in the compound. Chemists usually list this central atom first in the chemical formula (as in CCl4 and CO32−, which both have C as the central atom), which is another clue to the compound’s structure.
- Hydrogen and the halogens are almost always connected to only one other atom, so they are usually terminal rather than central.
3. Place a bonding pair of electrons between each pair of adjacent atoms to give a single bond.
- In H2O, for example, there is a bonding pair of electrons between oxygen and each hydrogen.
4. Beginning with the terminal atoms, add enough electrons to each atom to give each atom an octet (two for hydrogen).
- These electrons will usually be lone pairs.
5. If any electrons are left over, place them on the central atom.
- We will explain later that some atoms are able to accommodate more than eight electrons.
6. If the central atom has fewer electrons than an octet, use lone pairs from terminal atoms to form multiple (double or triple) bonds to the central atom to achieve an octet.
- This will not change the number of electrons on the terminal atoms.
7. Final check
- Always make sure all valence electrons are accounted for and that each atom has an octet of electrons, except for hydrogen (with two electrons).
- The central atom is usually the least electronegative element in the molecule or ion; hydrogen and the halogens are usually terminal.
Now let’s apply this procedure to some particular compounds, beginning with one we have already discussed.
Example \(\PageIndex{1}\): Water
Write the Lewis Structure for H2O.
Solution
| Steps for Writing Lewis Structures | Example \(\PageIndex{1}\) |
|---|---|
| 1. Determine the total number of valence electrons in the molecule or ion. | Each H atom (group 1) has 1 valence electron, and the O atom (group 16) has 6 valence electrons, for a total of 8 valence electrons. |
| 2. Arrange the atoms to show specific connections. | H O HBecause H atoms are almost always terminal, the arrangement within the molecule must be HOH. |
| 3. Place a bonding pair of electrons between each pair of adjacent atoms to give a single bond. 4. Beginning with the terminal atoms, add enough electrons to each atom to give each atom an octet (two for hydrogen). | Placing one bonding pair of electrons between the O atom and each H atom gives H -O- Hwith 4 electrons left over. Each H atom has a full valence shell of 2 electrons. |
| 5. If any electrons are left over, place them on the central atom. | Adding the remaining 4 electrons to the oxygen (as two lone pairs) gives the following structure:
|
| 6. If the central atom has fewer electrons than an octet, use lone pairs from terminal atoms to form multiple (double or triple) bonds to the central atom to achieve an octet. | Not necessary. |
| 7. Final check. | The Lewis structure gives oxygen an octet and each hydrogen 2 electrons. |
Example \(\PageIndex{2}\)
Write the Lewis structure for the \(CH_2O\) molecule
Solution
| Steps for Writing Lewis Structures | Example \(\PageIndex{2}\) |
|---|---|
| 1. Determine the total number of valence electrons in the molecule or ion. | Each hydrogen atom (group 1) has 1 valence electron, carbon (group 14) has 4 valence electrons, and oxygen (group 16) has 6 valence electrons, for a total of [(2)(1) + 4 + 6] = 12 valence electrons. |
| 2. Arrange the atoms to show specific connections. |
Because carbon is less electronegative than oxygen and hydrogen is normally terminal, C must be the central atom. |
| 3. Place a bonding pair of electrons between each pair of adjacent atoms to give a single bond. | Placing a bonding pair of electrons between each pair of bonded atoms gives the following:
6 electrons are used, and 6 are left over. |
| 4. Beginning with the terminal atoms, add enough electrons to each atom to give each atom an octet (two for hydrogen). | Adding all 6 remaining electrons to oxygen (as three lone pairs) gives the following:
Although oxygen now has an octet and each hydrogen has 2 electrons, carbon has only 6 electrons. |
| 5. If any electrons are left over, place them on the central atom. | Not necessary. There are no electrons left to place on the central atom. |
| 6. If the central atom has fewer electrons than an octet, use lone pairs from terminal atoms to form multiple (double or triple) bonds to the central atom to achieve an octet. | To give carbon an octet of electrons, we use one of the lone pairs of electrons on oxygen to form a carbon–oxygen double bond:
|
| 7. Final check | Both the oxygen and the carbon now have an octet of electrons, so this is an acceptable Lewis electron structure. The O has two bonding pairs and two lone pairs, and C has four bonding pairs. This is the structure of formaldehyde, which is used in embalming fluid. |
Exercise \(\PageIndex{1}\)
Write Lewis electron structures for CO2 and SCl2.
- Answer CO2
-
.
- Answer SCl2
-
.
The United States Supreme Court has the unenviable task of deciding what the law is. This responsibility can be a major challenge when there is no clear principle involved or where there is a new situation not encountered before. Chemistry faces the same challenge in extending basic concepts to fit a new situation. Drawing of Lewis structures for polyatomic ions uses the same approach, but tweaks the process a little to fit a somewhat different set of circ*mstances.
Writing Lewis Structures for Polyatomic Ions(CK-12)
Recall that a polyatomic ion is a group of atoms that are covalently bonded together and which carry an overall electrical charge. The ammonium ion, \(\ce{NH_4^+}\), is formed when a hydrogen ion \(\left( \ce{H^+} \right)\) attaches to the lone pair of an ammonia \(\left( \ce{NH_3} \right)\) molecule in a coordinate covalent bond.
When drawing the Lewis structure of a polyatomic ion, the charge of the ion is reflected in the number of total valence electrons in the structure. In the case of the ammonium ion:
\(1 \: \ce{N}\) atom \(= 5\) valence electrons
\(4 \: \ce{H}\) atoms \(= 4 \times 1 = 4\) valence electrons
subtract 1 electron for the \(1+\) charge of the ion
total of 8 valence electrons in the ion
It is customary to put the Lewis structure of a polyatomic ion into a large set of brackets, with the charge of the ion as a superscript outside of the brackets.
Exercise \(\PageIndex{2}\)
Draw the Lewis electron dot structure for the sulfate ion, SO42−.
- Answer(CK12 License)
Summary
Lewis structures are a model of how the valence electrons in a molecule (or polyatomic ion) are shared. Lewis structures show electrons bonding pairs, which are shared by two atoms, and lone pairs, which are not shared between atoms.
Contributions and Attributions
- edited by Spencer Berger
