Ions and the Octet Rule - Video Tutorials & Practice Problems

Ions and the Octet Rule Concept 1

there's a tendency by main group elements in achieving eight valence electrons or a filled outer shell. But undergoing chemical reactions. Now, main group metals lose electrons to be like the noble gas that is before them, that is before them in the periodic table. So for example, let's say that we're dealing with the sodium adam. So 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 one electron so that it has the same number of electrons as neon. 10. Main group main group non metals though they tend to gain electrons to be like an old gas that is after them In the periodic table. So for example, let's say we have here, Sulfur, Sulfur has an atomic number of 16. It wants to be like the noble gas that comes after it which is are gone. So sulfur would want to gain two electrons and in that way have 18 electrons just like are gone now. Why are they doing this? Well they do this in order to create totally filled S. And P sub shells 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 we have flooring, lithium here is one s. 2 to s. one here. This is electron is in its outer shell. So that's why it's being shown there. This is its outer shell electron. Flooring is one S 22 S 22 P five. It has this as its second shell electrons because they're both too. So in total about seven electrons. lithium wants to lose one electron so that it's S sub shell is completely filled because remember s can hold a maximum of two electrons doing this gives it a similar electron configuration as helium Florin gained that one electron that came from lithium and when again that electronic becomes negative. But more importantly, when it gained that electron it went to this p sub shell. So now it's two P six. Remember P sub shots can hold a maximum of six electrons. So now it's completely filled into doing this gives it a similar electron configuration to neon. Yeah. 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 flooring 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 will see that you have one S 22 S 22 P S two. And if you were eliminated these electrons here that 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 medals 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 two electrons. So you could also take this approach. Either way, you still get the same answer option. C magnesium wants to lose two electrons so that it can have a filled outer shell, like a noble gas. So here this is a giveaway that they want you to make it into a noble gas.

Ions and the Octet Rule Concept 2

with a metal cat eye on. We remove electrons from the highest shell number. Remember high shell number just means end value. The end value provides the shell number or energy level of the electron. So for example, we have one S 22 S 22 P 63 S 23 P six. The number that is before my sub shell shell letters of S and P. Those numbers are connected to my end value. So one has to and is one meaning these electrons are in the first show to S two and two P. Six because both have two of them, That means all these electrons are in the second shell where end equals two and then three years, two and three P six. All those electrons are in the third show where and equals three. So if we started taking electrons away, they'd have to come from the highest shell number. So they start coming from here. The shell of the third show. And we would remove them in the order that we have them here. So we remove three P. Before we ever touched three S two

Ions and the Octet Rule Example 2

here. We're told to write the condensed electron configuration for the sodium. I'll here. So, step one, 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 one s. 2 to S to two P six three S one. Step two. We begin by removing electrons from the highest numbered shell to obtain a desire charge. Now recall each electron remove, causes the ion charge to increase by plus one. All right, So plus one here means we have lost one electron. So we removed that one electron from the highest shell number Here. Since this is a one and equals one. Both of these are two, so and equals two. And since this is a three N equals three, the highest shell number we have here is N equals three. Which would mean I'm taking the electron from this three S. Orbital here or sub shell here. That would mean that my sodium ion would have an electron configuration. A condensed one now remember has to be condensed. It would be here Neon because if you look what's left behind is one s. 2 To us 2 2 p six. 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 one has to portion here to us 22 P six. 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 plus one 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 the non metal and ion we add an electron or electron to the orbital's with available space. So here we have to write the full electron configuration for the nitride ion. So first we're gonna start off by providing the electron configuration for the neutral form of the element Nitrogen has an atomic number of seven. Its electron configuration is one s. 2 To us to to P three Step two. We add electricity orbital's that can accommodate more electrons. Alright, so -3 here means that we have gained three electrons. So we need to add three more electrons. So here for nitrite ion, so as is already filled, it can only hold up to two. But remember your P can hold up to six. So the extra three that we need to add, we can add them there. So the nitride ion would be one S two, two, S 22 P six. 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 condenser, asking for the full electron configuration.

Determine the electron configuration for the Cl^– ion.

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