Ions and the Octet Rule - Online Tutor, Practice Problems & Exam Prep

Main group elements strive for stability by achieving a filled outer shell of 8 valence electrons. Metals, like sodium, lose electrons to resemble the noble gas preceding them, while nonmetals, such as sulfur, gain electrons to mimic the following noble gas. This electron transfer leads to the formation of cations and anions, enhancing stability and reducing reactivity. For cations, electrons are removed from the highest shell number first, following the order of subshells. Understanding these principles is crucial for grasping chemical reactivity and the formation of ionic compounds.

Ions and the Octet Rule Concept 1

There's a tendency by main group elements in achieving 8 valence electrons or a filled outer shell, but undergoing chemical reactions. Main group metals lose electrons to be like the noble gas that is before them in the periodic table. For example, let's say that we're dealing with the sodium atom. Sodium has an atomic number of 11. The noble gas that came before it is neon, which has an atomic number of 10. So sodium wants to lose 1 electron so that it has the same number of electrons as neon, 10. Main group non-metals, however, tend to gain electrons to be like the noble gas that is after them in the periodic table. For example, let's say we have sulfur. Sulfur has an atomic number of 16. It wants to be like the noble gas that comes after it, which is argon. So sulfur would want to gain 2 electrons, and in that way have 18 electrons just like argon. Now why are they doing this? Well, they do this in order to create totally filled s and p subshells because this will lead to greater stability and help to lower further chemical reactivity.

So if we take a look here, we have lithium and fluorine. Lithium here is s 2 2 1 . Here, this electron is in its outer shell, so that's why it's being shown there. This is its outer shell electron. Fluorine is s 2 2 2 p 5 . It has this as its 2nd shell electrons because they're both 2, so in total, that's 7 electrons. Lithium wants to lose 1 electron so that its s subshell is completely filled because remember, s can hold a maximum of 2 electrons. Doing this gives it a similar electron configuration as helium. Fluorine gains that one electron that came from lithium, and when it gained that electron it becomes negative. But more importantly, when it gained that electron, it went to this p subshell. So now it's p 6 . Remember, p subshells can hold a maximum of 6 electrons, so now it's completely filled too. Doing this gives it a similar electron configuration to neon. So now both of my original elements have been turned into ions that have similar electron configurations to their noble gases near them. This is the whole point. This is what they want to do. Noble gases are very stable because of their filled outer shells, and doing this allows lithium and fluorine to mimic those noble gases.

Ions and the Octet Rule Example 1

How many electrons must the magnesium atom, which has an atomic number of 12, lose in order to obtain a filled outer shell? So with this element, you could take the approach of doing the electron configuration. That would work. You would see that you have s12, s22, p26, s32. And if you were to eliminate these electrons here, that'd give you a similar electron configuration to a noble gas of neon. You could also say that magnesium has an atomic number of 12, and remember, we said that the metals want to lose electrons to be like the noble gases that came before them. The noble gas before it is neon. Neon has an atomic number of 10. To have 10 electrons just like neon, magnesium would need to lose 2 electrons. So you could also take this approach. Either way, you still get the same answer. Option C, magnesium wants to lose 2 electrons so that it can have a filled outer shell like a noble gas. So here, this is a giveaway that I want you to make it into a noble gas.

Ions and the Octet Rule Concept 2

With a metal cation, we remove electrons from the highest shell number. Remember, a high shell number just means n value. The n value provides the shell number or energy level of the electron. So, for example, we have s12, s22, p26, s32, p36. The number that is before my subshell letters of s and p, those numbers are connected to my n value. So in s12, n is 1, meaning these electrons are in the first shell. In s22 and p26, because both have 2 in them, that means all these electrons are in the second shell where n equals 2. And then in s32 and p36, all those electrons are in the third shell where n equals 3. So if we started taking electrons away, they'd have to come from the highest shell number, so they'd start coming from here, the third shell. And we would remove them in the order that we have them here. So we'd remove p36 before we ever touched s32.

Ions and the Octet Rule Example 2

Here we're told to write the condensed electron configuration for the sodium ion. Here, so step 1, we have to provide the electron configuration for the neutral form of the element. So neutral sodium has an atomic number of 11. That means its electron configuration is s12, s22, p26s31. Step 2, we begin by removing electrons from the highest numbered shell to obtain the desired charge. Now recall, each electron removed causes the ion charge to increase by plus 1. Alright. So plus 1 here means we have lost 1 electron. So we remove that 1 electron from the highest shell number. Here, since this is a 1, n equals 1. Both of these are 2, so n equals 2. And since this is a 3, n equals 3. The highest shell number we have here is n equals 3, which would mean I'm taking the electron from this s3 orbital here or subshell here. That would mean that my sodium ion would have an electron configuration, a condensed one now, remember it has to be condensed, it would be here neon because if you look, what's left behind is s12, s22, p26. That's the same electron configuration as neon. Now if you wanted, you could also, which is allowed, you could also say helium, which represents the s12 portion here, s22p26. This way is less common. This would be the better way of writing it because this is the most condensed it can be. Our sodium+1 ion has the same electron configuration now as the neon element. Right? So here, brackets with neon would be the most acceptable answer.

Ions and the Octet Rule Example 3

Now with a nonmetal anion, we add an electron or electrons to the orbitals with available space. So here we have to write the full electron configuration for the nitride ion. So first we're going to start off by providing the electron configuration for the neutral form of the element. Nitrogen has an atomic number of 7. Its electron configuration is s122, s222, p232. Step 2, we add electrons to the orbitals that can accommodate more electrons. Alright. So minus 3 here means that we have gained 3 electrons, So we need to add 3 more electrons. So here for a nitride ion, s2 is already filled, it can only hold up to 2, but remember your p2 can hold up to 6. So the extra 3 that we need to add, we can add them there. So the nitride ion would be s122, s222, p262. We need to write the full electron configuration and this would be it. This would be a similar electron configuration to neon, but again they're not asking for the condensed, they're asking for the full electron configuration.

Determine the electron configuration for the Cl^– ion.

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Here’s what students ask on this topic:

What is the octet rule and why is it important in chemistry?

The octet rule states that atoms tend to gain, lose, or share electrons to achieve a full set of eight valence electrons, resembling the electron configuration of noble gases. This rule is crucial because it explains the stability of noble gases and the reactivity of other elements. By achieving an octet, atoms lower their energy and become more stable. This principle helps predict the formation of ions and the types of bonds atoms will form, which is fundamental in understanding chemical reactions and the properties of compounds.

How do metals and nonmetals achieve a stable electron configuration?

Metals achieve a stable electron configuration by losing electrons to form cations, thereby resembling the noble gas that precedes them in the periodic table. For example, sodium (Na) loses one electron to have the same electron configuration as neon (Ne). Nonmetals, on the other hand, gain electrons to form anions, mimicking the noble gas that follows them. For instance, sulfur (S) gains two electrons to achieve the electron configuration of argon (Ar). This electron transfer results in filled outer shells, enhancing stability and reducing reactivity.

Why do noble gases have low chemical reactivity?

Noble gases have low chemical reactivity because they possess a complete set of valence electrons, typically eight, which makes their outer electron shell fully filled. This stable electron configuration means they have little tendency to gain, lose, or share electrons, resulting in minimal chemical reactivity. Their stability is a key reason why other elements strive to achieve a similar electron configuration through the gain or loss of electrons, as described by the octet rule.

How are electrons removed from metal cations?

When forming metal cations, electrons are removed from the highest shell number first, which corresponds to the highest principal quantum number (n). For example, in the electron configuration 1s² 2s² 2p⁶ 3s² 3p⁶, electrons would be removed from the 3p subshell before the 3s subshell. This order ensures that electrons are taken from the outermost energy level, which is the least stable and most easily lost, leading to the formation of a positively charged ion.

What is the significance of achieving a noble gas electron configuration?

Achieving a noble gas electron configuration is significant because it leads to greater stability for the atom or ion. Noble gases are inherently stable due to their filled valence shells, which means they have minimal chemical reactivity. When other elements gain or lose electrons to achieve a similar configuration, they also attain a lower energy state and increased stability. This concept is fundamental in understanding why atoms form ions and how ionic compounds are created, as it explains the driving force behind chemical reactions and bonding.