Introduction

In chemical bonding, there are some trends that apply to the majority of chemical compounds. One of these is known as the Octet Rule. This rule states that an atom will always attempt to have a full set of eight valence electrons by bonding with other atoms either covalently or ionically. The way that the Octet Rule works varies for different types of compounds, so in this article we will explore the different possibilities in an organized fashion. Overall, the Octet Rule is one of the most important fundamentals of chemical bonding.

 

General Trends 

The amount of electrons an element will tend to gain or lose depends on the number of electrons in the atom's valence shell. An atom will always try to ionize to its stable state with the transfer of the fewest electrons possible.

Example 1: A fluorine atom can get a full valence shell by either gaining one more electron, or by losing seven electrons. The former requires the transfer of less electrons, so the fluorine atom will try to gain one electron first. Therefore, \(F^{-}\) ions are more common than \(F^{7+}\) ions.

Example 2: A magnesium atom can get a full valence shell bt either losing two electrons, or by gaining six. The former requires the transfer of fewer electrons, so \(Mg^{2+}\) ions are more common than \(Mg^{6-}\) ions. 

Example 3: Note that a carbon atom's valence shell has \(4\) electrons. It can either lose four electrons or gain four electrons to get a full valence shell. Note that either transition requires the transfer of the same number of electrons, so which ion results depends on the other chemicals involved. Note that in an ionic compound comprised of a metal and carbon, carbon will form the \(-4\) carbide ion and form compounds such as magnesium carbide \(Mg_2C\) and sodium carbide \(Na_4C\).

Here is a table of the group that a chemical is in and its most common transaction of gaining or losing electrons to gain a full valence shell.

Group          Electrons gained/lost

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1                   Lose 1 electron

2                   Lose 2 electrons

3                   Lose 3 electrons

4                   Lose 4 electrons/gain 4 electrons, depends on situation

5                  Gain 3 electrons

6                  Gain 2 electrons

7                  Gain 1 electrons

8                  These chemicals, known as the noble gases, already have a full valence shell and thus tend to be unreactive (inert).

 

Binary Ionic Compounds

In an ionic compound, electrons are not shared between ions. Each valence electron is designated to belong to a specific ion in the bond. In a binary ionic compound (one with only two different types of ions), each valence electron belongs to either the cation or anion. When either the cation or anion is polyatomic, it becomes more complicated. We can say that the electron belongs to either the cation or anion, but then we need to determine how it is shared between atoms that comprise the polyatomic ion. We will worry about this special case later, in the final section of the article.

Example 4: Consider Sodium Chloride, \(NaCl\). Sodium atoms have one valence electron and chlorine atoms have seven. If a single electron is transferred from the sodium atom to the chlorine atom, then we get \(Na^{+}\) and \(Cl^{-}\), two ions in stable states. Thus we only need one of each atom to form this stable ionic compound, and both ions now satisfy the Octet Rule.

Example 5: Consider Aluminum Bromide, \(AlBr_3\). Aluminium is in Group III, and it tends to lose three electrons to form an ion with a full octet (meaning it obeys the Octet Rule). Bromine atoms tend to gain just one electron to get to a full octet, as Bromine is in Group VII. A chemical consisting of an aluminum ion and a bromide ion in their stable states would be \(AlBr^{2+}\), but it is not an ionic compound because it has a charge. Therefore, we need more negatively charged bromide ions to get an ionic compound. Each such ion has a charge of \(-1\), so we need two more to get a neutral compound. Thus aluminum bromide is \(AlBr_3\), with each ion being in its stable octet form.

Example 6: Consider Magnesium Nitride, \(Mg_3N_2\). Magnesium is in Group II and has two electrons in its valence shell. Thus it tends to lose two electrons. In this case, the next set of electrons closer to the nucleus is the new valence shell, and it is full. Nitrogen, on the other hand, is in Group V and has five valence electrons, so it needs to gain three electrons to get a full valence shell. Thus the most stable state for nitrogen ions is the \(N^{3-}\) ion. 

 

Covalently Bonded Compounds

In covalently bonded compounds, electrons are shared between two atoms through the bonds between them. Every such [single] bond consists of two electrons, which count towards both atoms' full set of eight valence electrons. Thus in a compound where two atoms have a single bond to each other, if both atoms are stable, then each will have six additional valence electrons.

Example 7: Consider \(F_2\), fluorine's elemental form. Use the Octet Rule to explain why fluorine is a diatomic molecule (exists naturally in groups of two atoms bonded together).

Solution: Fluorine is in Group VII, and a single fluorine atom has seven valence electrons. However, by the Octet Rule it would like to gain one electron to get a full octet of valence electrons. Two fluorine atoms can each "sacrifice" one of their valence electrons to form a single bond between the atoms. Now each fluorine atom has six valence electrons to itself. However, the bond consists of two additional valence electrons for each fluorine atom (as the electrons are shared), so both fluorine atoms now have a full octet. Thus fluorine is more stable as a diatomic molecule.

Below is a Lewis Structure for elemental fluorine for reference. Note that only valence shell electrons are included in the picture. This is a general convention for drawing Lewis Structures.

Now, recall that double bonds and triple bonds exist. These have slightly different advantages compared to single bonds. In particular, they allow even more electrons to be shared between two atoms. If a double bond exists between two atoms, they share four valence electrons; if there is a triple bond, the two atoms share six valence electrons. The Octet Rule still applies regardless of which type of covalent bond is involved. 

Example 8: Consider \(O_2\), oxygen's elemental form. Lone oxygen atoms have six valence electrons. However, oxygen is even more stable in groups of two atoms, because each oxygen atom can "sacrifice" two valence electrons, giving a total of four electrons used to bond the two atoms. Since there are four valence electrons to share, this is a double bond. Each oxygen atom has four valence electrons to itself and four that are shared through the double bond. Thus each oxygen atom now has a full octet. Below is the Lewis Structure.

Example 9: Consider \(N_2\), nitrogen's elemental form. Unlike oxygen, nitrogen atoms have five valence electrons, so they need three more to get a full octet. This is where a triple bond comes into play. Nitrogen atoms share three of their valence electrons, and in return get three more that initially belonged to another nitrogen atom. Thus there are two valence electrons that belong solely to one of the nitrogen atoms, and there are six that are shared, so each nitrogen atom has eight valence electrons. Below is the Lewis Structure.

Ionic Compounds with a Polyatomic Ion

When a molecule has a nonzero charge, it becomes a polyatomic ion. One or more atoms in the molecule will have violated the Octet Rule, making them unstable unless they are in the presence of a different ion of the opposite charge. Most polyatomic ions are anions. In this case, one or more atoms will have one of the following: more lone pairs than they should, or an expanded octet, a set of more than eight valence electrons. This section is quite complicated and its treatment is often limited in high school chemistry classes. 

Example 10: Consider magnesium sulfate, \(MgSO_4\). The sulfate ion's structure is shown below:

Two of the oxygen atoms (the left and bottom ones) have an extra lone pair, but still have a full octet and no extra valence electrons. These oxygen atoms have a negative formal charge, meaning they hold some electrons longer than they would in the perfectly stable case (two lone pairs, two bonding pairs). In lieu of a second bonding pair, these oxygen atoms have a third lone pair. These electrons are not shared with other atoms, so they further contribute to negative charge around the oxygen atom.

When assigning formal charges, we assume that the two atoms in a bond share bonding pair electrons equally, so each bonding pair electron essentially contributes \(\frac{1}{2}\) the charge of an electron to each atom involved in the bond. Therefore, when two electrons go from being shared to being held only by the oxygen atom, their "charge contribution" to the oxygen atom increases by \(\frac{1}{2}\) an electron's charge each, increasing the charge around the oxygen atom by \(-1\). Thus the oxygen atom's formal charge is \(+1\).

Next, note that the top and right oxygen atoms have two lone pairs and two bonding pairs (as we discussed earlier as typical) and obey the Octet Rule. 

Now, let's talk about the Sulfur atom. It has an expanded octet, as it has twelve apparent valence shell electrons (six single bonds' worth; count the bonds). However, in an expanded octet, some electrons from the shell of electrons inside the valence shell participate in bonding. That is why it APPEARS sulfur has twelve valence electrons in this case, but this is just an illusion. 

This is a confusing subject and you likely have unanswered questions at this point, but do not worry! It is suggested from here that you do further reading (either through OpenCurriculum or another resource) on Octet Rule exceptions and formal charges. The discussion gets even more technical and will be omitted here.

Now let's talk about where the Octet Rule does apply: the ionic bond. The magnesium atom used to form this compound still wants a full octet, and as it is in Group II, it achieves this most easily by losing two valence electrons, giving this magnesium ion a charge of \(+2\). The sum of the formal charges of the atoms of the sulfate ion is \(-2\), so the charge of the ion is \(-2\) (this will always be true for a polyatomic ion). Thus the magnesium cation and sulfate ion match up in a 1:1 ratio, giving \(MgSO_4\). 

 

Thanks to http://www.wolframalpha.com/widgets/view.jsp?id=689aa5a01c216d8b16ed0250cebdc702 for the Lewis Structure diagram pictures.