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Lewis structure: what it is, how to do it, examples

But what is a covalent bond? It is the sharing of a pair of electrons (or points) between any two atoms of the periodic table. With these diagrams many skeletons can be drawn for a given compound. Which of them is correct will depend on the formal charges and the chemical nature of the atoms themselves.

Compound 2-bromopropane. By Ben Mills [Public domain], from Wikimedia Commons.

In the image above you have an example of what a Lewis structure is. In this case the represented compound is 2-bromopropane. You can see the black dots corresponding to the electrons, both those that participate in the bonds and those that are not shared (the only pair just above Br).

If the pairs of dots “:” were replaced by a long dash “-“, then the carbon skeleton of 2-bromopropane would be represented as: C – C – C. Why instead of the drawn “molecular framework” it couldn’t be C – H – H – C? The answer lies in the electronic characteristics of each atom.

For this reason, the Lewis structures where C and H intervene must be coherent and respect what is governed by their electronic configurations. In this way, if carbon has more than four bonds, or hydrogen more than one, then the sketch can be discarded and a new one more in line with reality can be started.

It is here that one of the main motifs or endorsements of these structures appear, introduced by Gilbert Newton Lewis in his search for molecular representations faithful to experimental data: molecular structure and formal charges.

All existing compounds can be represented by Lewis structures, giving a first approximation to how the molecule or the ions could be.

It is a representative structure of the valence electrons and the covalent bonds in a molecule or ion that serves to get an idea of ​​its molecular structure.

However, this structure fails to predict some important details such as molecular geometry regarding an atom and its environment (if it is square, trigonal plane, bipyramidal, etc.).

Likewise, it does not say anything about what is the chemical hybridization of its atoms, but it does say where the double or triple bonds are located and if there is resonance in the structure.

With this information, one can argue about the reactivity of a compound, its stability, how and what mechanism the molecule will follow when it reacts.

For this reason, Lewis structures never cease to be considered and are very useful, since new chemical learning can be condensed in them.

How do you do it?

To draw or sketch a structure, formula or Lewis diagram, the chemical formula of the compound is essential. Without it, you cannot even know which are the atoms that make it up. Once with it, the periodic table is used to locate which groups they belong to.

For example, if you have the compound C 14 O 2 N 3 then you would have to look for the groups where carbon, oxygen and nitrogen are. Once this is done, no matter what the compound is, the number of valence electrons remains the same, so sooner or later they are memorized.

Thus, carbon belongs to group IVA, oxygen to group VIA and nitrogen to VA. The group number is equal to the number of valence electrons (points). They all have in common the tendency to fill the valence layer octet.

What is the octet rule?

This says that there is a tendency for atoms to complete their energy level with eight electrons to achieve stability. This applies to all non-metallic elements or those found in the sop blocks of the periodic table.

However, not all elements obey the octet rule. Particular cases are the transition metals, whose structures are based more on formal charges and their group number.

Number of electrons in the valence shell of non-metallic elements, those in which the Lewis structure can be operated.

Applying the mathematical formula

Knowing which group the elements belong to, and therefore the number of valence electrons available to form bonds, we proceed with the following formula, which is useful for drawing Lewis structures:

C = N – D

Where C means shared electrons , that is, those that participate in covalent bonds. Since each bond is made up of two electrons, then C / 2 is equal to the number of bonds (or dashes) that must be drawn.

N are the necessary electrons , those that the atom must have in its valence shell to be isoelectronic to the noble gas that follows it in the same period. For all elements other than H (since it requires two electrons to compare to He) they need eight electrons.

D are the available electrons , which are determined by the group or numbers of valence electrons. Thus, since Cl belongs to group VIIA, it must surround itself with seven black dots or electrons, and keep in mind that a pair is needed to form a bond.

Having the atoms, their points and the number of C / 2 bonds, a Lewis structure can then be improvised. But additionally, it is necessary to have a notion of other “rules”.

Where to place the least electronegative atoms

The least electronegative atoms in the vast majority of structures occupy the centers. For this reason, if you have a compound with P, O and F atoms, the P must therefore be located in the center of the hypothetical structure.

Also, it is important to note that hydrogens normally bind to highly electronegative atoms. If you have Zn, H and O in a compound, H will go together with O and not with Zn (Zn – O – H and not H – Zn – O). There are exceptions to this rule, but it generally occurs with non-metallic atoms.

Symmetry and formal charges

Nature has a high preference for creating molecular structures that are as symmetrical as possible. This helps to avoid creating messy structures, with the atoms arranged in such a way that they do not obey any apparent pattern.

For example, for the compound C 2 A 3 , where A is a fictitious atom, the most likely structure would be A – C – A – C – A. Note the symmetry of its sides, both reflections of the other.

Formal charges also play an important role when drawing Lewis structures, especially for ions. Thus, bonds can be added or removed so that the formal charge of an atom corresponds to the total charge exhibited. This criterion is very helpful for transition metal compounds.

Limitations on the octet rule

Representation of aluminum trifluoride, a compound that is unstable. Both elements are made up of six electrons, which generates three covalent bonds, when they should be eight to achieve stability. Source: Gabriel Bolívar

Not all the rules are followed, which does not necessarily mean that the structure is wrong. Typical examples of this are observed in many compounds where group IIIA elements (B, Al, Ga, In, Tl) are involved. Aluminum trifluoride (AlF 3 ) is specifically considered here .

Applying then the formula described above, we have:

D = 1 × 3 (one aluminum atom) + 7 × 3 (three fluorine atoms) = 24 electrons

Here 3 and 7 are the respective groups or numbers of valence electrons available for aluminum and fluorine. Then, considering the necessary electrons N:

N = 8 × 1 (one aluminum atom) + 8 × 3 (three fluorine atoms) = 32 electrons

And therefore the shared electrons are:

C = N – D

C = 32 – 24 = 8 electrons

C / 2 = 4 links

Since aluminum is the least electronegative atom, it must be placed in the center, and fluorine only forms one bond. Considering this, we have the Lewis structure of AlF 3 (upper image). Shared electrons are highlighted with green dots to distinguish them from unshared ones.

Although calculations predict that 4 bonds must be formed, aluminum lacks enough electrons and there is also no fourth fluorine atom. As a result, aluminum does not comply with the octet rule and this fact is not reflected in the calculations.

Examples of Lewis structures

Iodine

Nonmetals of iodine have seven electrons each, so by sharing one of these electrons each, they generate a covalent bond that provides stability. Source: Gabriel Bolívar

Iodine is a halogen and therefore belongs to group VIIA. It then has seven valence electrons, and this simple diatomic molecule can be represented improvising or applying the formula:

D = 2 × 7 (two iodine atoms) = 14 electrons

N = 2 × 8 = 16 electrons

C = 16 – 14 = 2 electrons

C / 2 = 1 link

As of 14 electrons 2 participate in the covalent bond (green dots and dash), 12 remain as non-shared; and since they are two iodine atoms, 6 must be divided for one of them (its valence electrons). In this molecule only this structure is possible, whose geometry is linear.

Ammonia

Nitrogen has 5 electrons, while hydrogen only 1. Enough to achieve stability by establishing three covalent bonds, composed of one electron from N and another from H Source: Gabriel Bolívar

What is the Lewis structure for the ammonia molecule? Since nitrogen belongs to the group VA, it has five valence electrons, and then:

D = 1 × 5 (one nitrogen atom) + 1 × 3 (three hydrogen atoms) = 8 electrons

N = 8 × 1 + 2 × 3 = 14 electrons

C = 14 – 8 = 6 electrons

C / 2 = 3 links

This time the formula is correct with the number of links (three green links). As 6 of the 8 available electrons participate in the bonds, there remains an unshared pair that is located above the nitrogen atom.

This structure says everything that needs to be known about the ammonia base. Applying the knowledge of TEV and TRPEV, it is deduced that the geometry is tetrahedral distorted by the free pair of nitrogen and that the hybridization of this is therefore sp 3 .

2 H 6 O

Source: Gabriel Bolívar

The formula corresponds to an organic compound. Before applying the formula, it must be remembered that hydrogens form a single bond, oxygen two, carbon four, and that the structure must be as symmetrical as possible. Proceeding as in the previous examples, we have:

D = 6 × 1 (six hydrogen atoms) + 6 × 1 (one oxygen atom) + 4 × 2 (two carbon atoms) = 20 electrons

N = 6 × 2 (six hydrogen atoms) + 8 × 1 (one oxygen atom) + 8 × 2 (two carbon atoms) = 36 electrons

C = 36 – 20 = 16 electrons

C / 2 = 8 links

The number of green dashes correspond to the 8 calculated links. The proposed Lewis structure is that of ethanol CH 3 CH 2 OH. However, it would also have been correct to propose the structure of dimethyl ether CH 3 OCH 3 , which is even more symmetric.

Which of the two is “more” correct? Both are equally so, since the structures arose as structural isomers of the same molecular formula C 2 H 6 O.

Permanganate ion

Source: Gabriel Bolívar

The situation is complicated when it is desired to make Lewis structures for transition metal compounds. Manganese belongs to group VIIB, likewise, the electron of the negative charge must be added among the available electrons. Applying the formula we have:

D = 7 × 1 (one manganese atom) + 6 × 4 (four oxygen atoms) + 1 electron times charge = 32 electrons

N = 8 × 1 + 8 × 4 = 40 electrons

C = 40 – 32 = 8 shared electrons

C / 2 = 4 links

However, transition metals can have more than eight valence electrons. Furthermore, for the MnO  ion to exhibit the negative charge, it is necessary to decrease the formal charges of the oxygen atoms. How? Through the double bonds.

If all the bonds of MnO  were simple, the formal charges of the oxygens would be equal to -1. Since there are four of them, the resulting charge would be -4 for the anion, which is obviously not true. When the double bonds are formed, it is guaranteed that a single oxygen has a negative formal charge, reflected in the ion.

In the permanganate ion it can be seen that there is resonance. This implies that the single Mn – O single bond is delocalized between the four O atoms.

Dichromate ion

Source: Gabriel Bolívar

Finally, a similar case occurs with the dichromate ion (Cr 2 O 7 ). Chromium belongs to group VIB, so it has six valence electrons. Applying the formula again:

D = 6 × 2 (two chromium atoms) + 6 × 7 (seven oxygen atoms) + 2 electrons times the divalent charge = 56 electrons

N = 8 × 2 + 8 × 7 = 72 electrons

C = 72 – 56 = 16 shared electrons

C / 2 = 8 links

But there are not 8 bonds, but 12. For the same reasons found, in the permanganate ion two oxygens with negative formal charges must be left that add up to -2, the charge of the dichromate ion.

Thus, as many double bonds as necessary are added. In this way we arrive at the Lewis structure of the image for Cr 2 O 2– .

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