How to draw Lewis structures with exceptions to the octet rule

Lewis structures are representations of chemical compounds based on the distribution of valence electrons of the different atoms that make them up. These structures serve to both predict and explain the structures of different compounds, as well as their molecular geometry, leading to important predictions about polarity, solubility, melting and boiling points, and other important properties.

We already covered in a previous article the detailed procedure for drawing the Lewis structures of compounds whose atoms satisfy the octet rule. This paper seeks to show how to draw Lewis structures in compounds that do not obey this rule for one of three different reasons:

• They have an odd number of electrons.
• They have an incomplete octet.
• They have an expanded octet.

Review of the procedure for drawing Lewis structures

As we saw in our first article on Lewis structures, the procedure for drawing them consists of six steps. A short summary of these steps follows, and most apply, with some modifications, in cases where the compound does not follow the octet rule.

• Step 1: Count the total number of valence electrons. This step involves multiplying the number of atoms of each type by the number of valence electrons in its group on the periodic table, and then subtracting the total charge of the chemical species (in the case of an ion).
• Step 2: Write the fundamental structure of the molecule. This means partitioning the atoms to assign connectivity between them. The common thing is that the least electronegative atom is always located in the center (unless it is hydrogen) while the most electronegative ones are located on the periphery.
• Step 3: Draw single covalent bonds between all the atoms that are linked together. If it is a covalent compound, all atoms must have at least one single covalent bond with a neighboring atom.
• Step 4: Fill in the octets with the remaining valence electrons, starting with the most electronegative. This step seeks to satisfy the octet rule first for the atoms with the greatest tendency to retain electrons which are those with the highest electronegativity.
• Step 5: Complete the octet of the central atom by forming pi bonds if necessary. Only once the octet rule has been satisfied for electronegative atoms is it considered complete for less electronegative atoms. If there are no more electrons to share, then this is achieved by sharing a pair of electrons from a neighboring atom with the central atom.
• Step 6: Calculate formal charges. One of the important stability criteria of a Lewis structure is the distribution of formal charges. For this reason, it is always advisable to determine and draw on the structure the formal charge of each atom. In addition, the sum of all formal charges must equal the net charge of the molecule or ion in question, so it is a handy way to verify that the structure has the correct number of valence electrons. The formula to calculate formal charge is CF=valence electrons – unshared electrons -1/2 shared electrons.

Exceptions to the octet rule

As can be seen in the previous section, when drawing a Lewis structure, the main criteria to take into account when distributing the valence electrons are electronegativity and the octet rule, which is verified in steps 4 and 5. However, there are situations in which this is not possible, such as when the total number of electrons is odd, which makes it impossible for all atoms to be surrounded by 8 electrons.

Another similar situation occurs when the number of valence electrons is simply not enough to complete the octet of all atoms. On the other hand, there are situations where there are too many valence electrons and a coherent structure cannot be drawn without violating the octet rule.

Below are three examples of Lewis structures in which the octet rule is not satisfied, and how to proceed in such cases.

odd number of electrons

The simplest situation in which it is recognized that the octet rule cannot be fulfilled occurs when there are an odd number of electrons. An example of these compounds are nitric oxide (NO) and nitrogen dioxide (NO ₂ ). Let’s see how the Lewis structure of the second would be drawn following the steps described above:

Step 1:

Nitrogen has 5 valence electrons and oxygen has 6, so the total number of valence electrons is 1 x ( 5 ) + 2 x ( 6 ) = 17 eV

As can be seen, the number of electrons is odd, so it is impossible to complete the octet with the three atoms of the molecule.

Step 2:

Nitrogen is less electronegative than oxygen, so a structure can be considered in which nitrogen is in the center surrounded by the two oxygen atoms:

Step 3:

We now place single bonds between each oxygen and nitrogen.

Step 4:

So far we have drawn only 4 valence electrons which are found in the two sigma bonds. This means that we still have 13 electrons to share among the three atoms. First we complete the octet of the two oxygens, which carries 12 electrons, so the last one is placed on nitrogen.

Step 5:

Nitrogen has only 5 electrons around it, so it has a very incomplete octet. The next step is for one of the two oxygens to give up a pair of electrons to form a pi bond , thus contributing two more electrons. This brings nitrogen to 7 electrons, while both oxygens have full octets.

There are two additional structures in which single-bonded oxygen gives up one of its electrons to form, together with the unpaired nitrogen electron, a second pi bond between these two atoms. However, these structures have the unpaired electron and incomplete octet on oxygen atoms instead of nitrogen, which is unfavorable.

Step 6:

The calculation of the formal charge is carried out for each atom that has a different electronic environment, in this case, for all three atoms:

CF _{Single bond oxygen} = 6 – 6 – ½ x 2 = -1

CF _{Oxygen double bond} = 6 – 4 – ½ x 4 = 0

CF _Nitrogen = 5 – 1 – ½ x 6 = +1

The following figure shows the final two Lewis structures of nitrogen dioxide.

incomplete octets

Many compounds have an atom that does not complete the octet either because there are not enough electrons or because completing it is unfavorable since it would provide a positive charge on a very electronegative atom. A typical example of the first case is borane (BH ₃ ) and of the second is boron trifluoride (BF ₃ ).

Let’s see how the Lewis structure of the second is built to illustrate structures that have an incomplete octet despite having enough electrons to complete them.

Step 1:

Fluorine has 7 valence electrons and boron has 3, so the total number of valence electrons is 3 x ( 7 ) + 1 x ( 3 ) = 24 eV

Step 2:

Boron is less electronegative than fluorine, so a structure is proposed in which boron is in the center surrounded by the three fluorine atoms:

Step 3:

We now place single bonds between each fluorine and boron.

Step 4:

We still have 18 valence electrons left over to share (since 6 of them are in single bonds). We use these to complete the octet to the three fluorine atoms that are the most electronegative.

Step 5:

As can be seen, fluorine atoms all have their full octet but boron does not. In this step, we should take an unshared pair of electrons from any of the three fluorine atoms to form a pi bond. This would result in three resonance structures that would be:

In all three resonance structures the octet is satisfied for all atoms present, which is desirable and is the purpose of step 5. However, in the next step a considerable problem arises that we have not yet addressed.

Step 6:

There are three different types of atoms with different electronic environments, two of them fluorine and the third the boron atom:

CF _{Single bond fluorine} = 7 – 6 – ½ x 1 = 0

CF _{Fluorine double bond} = 7 – 4 – ½ x 4 = +1

CF _Boron = 3 – 0 – ½ x 8 = -1

The following figure shows the three resonance structures with the formal charges.

The problem with these structures is that they all have a fluorine atom with a partial positive charge while boron has a negative charge. Considering that fluorine is the most electronegative element in the periodic table, it is very difficult for boron to be able to remove enough electron density to leave fluorine with a positive charge.

For this reason, none of these three resonance structures has any chance of adequately representing BF ₃ . Consequently, it is much more likely that the correct structure is the one we drew in step 3, which has a boron with the incomplete octet.

expanded octets

Just as there are cases in which differences in electronegativities and formal charges make structures with incomplete octets preferable to those that comply with this rule, the same can happen in the opposite direction. It sometimes happens that, in a compound, all the atoms follow the octet rule after step 3, but when calculating the formal charges we see a large charge separation that can be lightened by forming additional pi bonds, thus surrounding the compound. central atom with more than 8 electrons.

This type of violation of the octet rule can only occur in elements from the third period onwards, since the only way to expand its octet is if the atom still has unoccupied atomic orbitals in which it can accommodate the extra electrons. This only happens for atoms that have vacated d orbitals in their valence shell, and, by the rules of quantum numbers , this is only possible for elements whose valence shell is at the third energy level or higher.

A typical example of this situation is the sulfate ion (SO ₄ ^2- ). In this case, both oxygen and sulfur each have 6 valence electrons, so the total number of electrons is 5 x ( 6 ) – (–2) = 32 eV , where the charge of the ion is subtracted, which is – 2.

If we followed the 6 steps to the letter to build the structure of this ion, we would get the following:

Despite the fact that in this structure all the atoms follow the octet rule, the most important problem is that there is too great a separation of formal charges. In fact, not only do all atoms have non-zero formal charges, but also the central sulfur atom has a +2 charge. All this makes this structure considerably unstable.

However, this problem can be easily solved by considering that sulfur, as it belongs to the third period, has the possibility of expanding its octet by means of its empty 3d orbitals. Today it is accepted that the actual structure of the sulfate ion is the resonance hybrid between all the different Lewis structures that can be posited in which sulfur forms two double and two single bonds with oxygen atoms, as shown in the following structures:

References

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Chang, R., Manzo, Á. R., Lopez, PS, & Herranz, ZR (2020). Chemistry (10th ed.). New York City, NY: MCGRAW-HILL.

Exceptions to the Octet Rule. (2021, June 16). Retrieved from https://chem.libretexts.org/@go/page/25290

Lever, ABP (1972). Lewis structures and the octet rule. An automatic procedure for writing canonical forms. Journal of Chemical Education , 49 (12), 819. Retrieved from https://sci-hub.do/https://pubs.acs.org/doi/abs/10.1021/ed049p819

lumen. (nd). Exceptions to the Octet Rule | Chemistry for Non-Majors. Retrieved from https://courses.lumenlearning.com/cheminter/chapter/exceptions-to-the-octet-rule/

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