
# 10.5: Writing Lewis Structures for Covalent Compounds


Learning Objectives

• Draw Lewis structures for covalent compounds.

The following procedure can be used to construct Lewis electron structures for more complex molecules and 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 H

Because 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- H

with 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, a vile-smelling, unstable red liquid that is used in the manufacture of rubber.

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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 circumstances.

### Writing Lewis Structures for Polyatomic Ions

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.

Figure $$\PageIndex{3}$$: The ammonium ion.

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.

### Exceptions to the Octet Rule

As important and useful as the octet rule is in chemical bonding, there are some well-known violations. This does not mean that the octet rule is useless—quite the contrary. As with many rules, there are exceptions, or violations.

There are three violations to the octet rule. Odd-electron molecules represent the first violation to the octet rule. Although they are few, some stable compounds have an odd number of electrons in their valence shells. With an odd number of electrons, at least one atom in the molecule will have to violate the octet rule. Examples of stable odd-electron molecules are NO, NO2, and ClO2. The Lewis electron dot diagram for NO is as follows:

Although the O atom has an octet of electrons, the N atom has only seven electrons in its valence shell. Although NO is a stable compound, it is very chemically reactive, as are most other odd-electron compounds.

Electron-deficient molecules represent the second violation to the octet rule. These stable compounds have less than eight electrons around an atom in the molecule. The most common examples are the covalent compounds of beryllium and boron. For example, beryllium can form two covalent bonds, resulting in only four electrons in its valence shell:

Boron commonly makes only three covalent bonds, resulting in only six valence electrons around the B atom. A well-known example is BF3:

The third violation to the octet rule is found in those compounds with more than eight electrons assigned to their valence shell. These are called expanded valence shell molecules. Such compounds are formed only by central atoms in the third row of the periodic table or beyond that have empty d orbitals in their valence shells that can participate in covalent bonding. One such compound is PF5. The only reasonable Lewis electron dot diagram for this compound has the P atom making five covalent bonds:

Formally, the P atom has 10 electrons in its valence shell.

Example $$\PageIndex{3}$$: Octet Violations

Identify each violation to the octet rule by drawing a Lewis electron dot diagram.

1. ClO
2. SF6

Solution

a. With one Cl atom and one O atom, this molecule has 6 + 7 = 13 valence electrons, so it is an odd-electron molecule. A Lewis electron dot diagram for this molecule is as follows:

b. In SF6, the central S atom makes six covalent bonds to the six surrounding F atoms, so it is an expanded valence shell molecule. Its Lewis electron dot diagram is as follows:

Exercise $$\PageIndex{3}$$: Xenon Difluoride

Identify the violation to the octet rule in XeF2 by drawing a Lewis electron dot diagram.