Electron Configuration of Elements - Online Tutor, Practice Problems & Exam Prep

The Electron Configuration

In this video we're going to take a look at the concept of the electron configuration. Now, electron configurations represent the arrangement of electrons within shells and orbitals. In order to connect the periodic table to the electron configuration of elements and ions, we need to have a redesign of the periodic table itself. We will imagine the periodic table being composed of different blocks. The first two columns, which are these blue columns here, these blue boxes, which also include this one here, represent our s block. Our purple boxes here represent our d block. Our yellow ones here represent our p block. And finally, our 2 rows here represent our f block. We can further label this periodic table, stating that these first two boxes represent our 1s orbitals. In the 2nd row, this represents 2s, and in the 3rd row, this is 3s, then 4s, 5s, 6s, and 7s. Moving to the d block, after 4s, we go down to 3d, 4d, 5d, and then 6d. In the p block, being in the 2nd row, this represents 2p, 3p, 4p all the way down to 7p. In our f block, the first row is 4f and the bottom row is 5f.

Each letter s, d, p, and f can hold a maximum number of electrons. S can hold up to 2 electrons, which explains why our s block is composed of 2 slots per row. D can hold up to 10 electrons, and P can hold up to 6 electrons. Meanwhile, f can hold up to 14 electrons per row. Understanding how the periodic table itself is arranged within s, d, p, and f blocks, and how each of those blocks has different orbitals like 1s, 2p, 4f, etc., is the key to understanding how to determine the electron configuration of any element or ion.

All you have to do is remember, just reimagine the periodic table in this order and then apply it. Now that we've seen this periodic table, we can use it to answer this example question. We're asked to find and write the full ground state electron configuration for the atomic number of silicon. Remember, your atomic number is unique to an element; it gives us the number of protons for that element. Silicon has already been labeled on the periodic table. So here goes silicon: To count to silicon, we go 1s and there are 2 elements that are 1s, so s12, then s22, p26, next is s32. And then we have to land on silicon, which is in the 3p row and it's in slot p32. So, this would be the full ground state electron configuration for silicon: s12, s22, p26, s32, p32. As long as you can remember the order of the periodic table, just apply it to find the electron configuration of any element or ion that's given to you.

Now that you've seen how to do a simple basic ground state electron configuration of an element, click on to the next video where we talk about the condensed electron configuration of elements.

The Electron Configuration

So let's take a look at condensed electron configurations. With condensed electron configurations, we start at the last noble gas, before the desired element. In this question it says, write the condensed electron configuration for the following element. Here we have to write the condensed electron configuration of calcium. Here we're given an atomic number of 20 which would mean in terms of our periodic table, calcium would fall here. The last noble gas that I pass before I get to calcium is argon. Remember, our noble gases are in group 8. So here our condensed electron configuration would start out with argon. Then we just have to count to calcium. Calcium is in the 4s row and then it's in slot 2. So that'd be argon 4s². Condensed electron configurations are huge time savers when it comes to electron configuration in general.

That's because if I wanted to write the full ground state electron configuration of calcium, I'd have to start out with 1s and count all the way to calcium. What would that look like? Well, that would mean that I'd have 1s²2s²2p⁶, 3s², 3p⁶ and then 4s². All of this here can be greatly shortened by just substituting in argon. It means the same exact thing. Again, when it comes to the full ground state electron configuration, only give that if they're asking for it specifically. Most of the time when it comes to electron configurations, we're writing them in the condensed form because again, it saves a lot of time.

Now that you've seen these 2 different types of electron configurations, click on to the 3rd video to take a look at our charged electron configurations and the rules that are applied when it comes to the electron configurations of these types of ions.

The Electron Configuration

With charged electron configurations. Now, with a cation, remember a cation is a positive ion, we first remove electrons. Our n value represents the principal quantum number and deals with the atomic orbital's size and energy. Furthermore, we're going to say that the principal quantum number provides the shell number or energy level of the electron. For example, if we're looking at electrons within our s1 orbital, because the number here is 1, that means we're dealing with electrons in the first shell. So n=1. Electrons in s2 or p2, we're dealing with electrons in the second shell because the number is 2 so n=2. And then based on the pattern, we can see that here because these numbers are 3, we're dealing with electrons in the 3rd shell, so n=3.

Now, it says, write the condensed electron configurations for the element and its ion. So here we're looking at vanadium. If we glance back up to our periodic table, vanadium would fall right here. So we want the condensed electron configuration of vanadium. The last noble gas we pass before we get to vanadium is argon. So we'd start out with argon, then here we're at the s4 row. So s42 and then we're here in the d3 row and it's 3d^3. So, it would be Ar s42 d33. Coming back down here we'd say that the electron configuration of vanadium is Ar s42 d33.

Now, here we need to figure out the electron configuration vanadium 3 ion. It's plus 3. Plus 3 means that we've lost 3 electrons. Remember, when you're losing negative electrons, you become more positive. If we have to lose 3 electrons, they're going to have to come from the highest shell number, which is what we set up above. So if we're looking at the electron configuration of neutral vanadium, the electrons lost are coming from either s4 or d3. Because the number here is 4, so that means n=4. And because the first electrons would have to come from the s4 orbital. The first electrons would have to come from the s4 orbital because its n value is largest. So we'd have argon. s4 is now gone because we had to lose both those electrons. But remember, we're not losing just 2 electrons. We're losing 3 electrons. One more electron has to be lost and it'll be coming from the d3 orbital here. We'd have left d32. Again, remember when it comes to charged electron configurations, when it comes to cations, we're losing electrons from the highest shell number first. Then if you need to lose any additional ones, you take them away from the next orbitals next to it. So in this case, d3. So we lost 2 from s4 and then 1 from d33 leaving us with, at the end, argon d32 as the electron configuration of vanadium 3 ion.

Again, guys, just apply the concepts that we've seen here to give either the full electron configuration, the condensed electron configuration, or our charged electron configuration of any element or ion. All of it starts with you remembering how to write the periodic table properly with its different s blocks and p blocks as well as orbitals, such as p2 and f4.

Electron Configurations of Transition Metals

In this video, we're going to take a look at the electron configurations of specifically transition metals. Now remember, we can reimagine the periodic table in terms of s, p, d, and f blocks. Doing this reimagining, we can then make a connection between electron configurations and the periodic table.

Now when it comes to our transition metals, they're located in 2 places. We have our transition metals that are within our d block, and then we have our transition metals which are in our f block. For the most part, we're going to ignore the transition metals in the f block. That's because they have atomic numbers that are really large and as a result they tend to have weird characteristics such as some of them being radioactive, they have odd electron configuration characteristics, so we're going to tend to ignore those. Also, a lot of the reactions that we're going to see in organic chemistry that deal with transition metals are really reserved to the transition metals found within our d block. So, we're going to be concerned with only the transition metals in our d block. As a result, we're going to say here that the transition metal elements or transition metals are the elements in the d block. We're going to ignore the f block.

Our main group elements, which are elements from groups 1 to 8, we coined them as being our group a elements. Transition metals, specifically the ones in the d block, we refer to them as group b elements. That's just some convention that was came up with long ago. It doesn't really affect anything that we're going to see in terms of reactions. Just realize that main group elements, we label them as group a elements and transition metals as group b elements.

Now with the transition metals we have are group 3b to 4b all the way up to 8b. 8b is actually these 3 columns together. And then we go to 1b and 2b here. So just realize that our transition metals are a little bit different. Now within a given period, remember a period is a row, a transition metal fills its s orbital in the n quantum level followed by its d orbital. So for example, if we're looking at vanadium like we did previously, remember vanadium's electron configuration, we had said was Ar 4s23d3. So with our typical d block transition metals, what did we do? We wrote out our 4s orbitals first, so we filled in our s orbital first before we moved on to our d orbital electrons.

Also, recall we said that when it comes to our different sublevels s, p, d, and f, they can hold a certain number of electrons. That's because each one has a certain number of electron orbitals. So our s sublevel can hold up to 2 with opposite spins, our p can hold up to 6 electrons because it has 3 electron orbitals, d can hold up to 10 because it had 5 electron orbitals, and f can hold up to 14 because it has 7 electron orbitals. These electron orbitals are important because later on, we'll not only need to give the electron configuration of an element or ion, but we may be asked to draw its electron orbital diagram.

So just remember, when it comes to the transition metals, we're going to pay close attention to the transition metals within our d block because they more greatly relate to organic chemistry reactions than the transition metals we would see in the f block. Also, remember, typical of transition metals, we fill up the s orbital then we move over to the d orbital. Now that we've laid the groundwork for some basic ideas of electron configurations, we're asked to not only give the full electron configuration of titanium but we're also asked to do the electron orbital diagram for the element.

Electron Configurations of Transition Metals Exercise 1

Alright. So here it says, write the full ground state electron configuration and electron orbital diagram for the following element. So we're looking at titanium. Here we're given its atomic number as being 22. So if we come up here, I've already marked off where titanium is. It's right here. So all we have to do is count to it starting with 1s. So we'd say s12, s22, p26, which gives us all 6 of these slots. s32, p36, s42, d32. So that would be my full ground state electron configuration of titanium.

Now, we have to give the electron orbital diagram for the following element. Now typically for electron orbital diagrams, we'd write the condensed electron configuration and then give the electron orbital diagram. So if we're talking about condensed electron configuration, we'd say that the last noble gas that I pass before I get to titanium is argon. So for the condensed, we'd start out with argon. And then we'd say s42d32.

Now, we're gonna say for our electron orbital diagram, our noble gas would fit within these brackets. Then remember, an s sublevel has one electron orbital. And then a d sublevel has 5 electron orbitals. So remember, that's coming from this section here. So that's why we have 5 boxes here because they're part of 3d. Alright. So, we'd totally fill in 4s. It has 2 electrons in it, one spins up, one spins down. And then within 3d we have 2 electrons. Following Hund's rule, we first half fill. So, we have 2 electrons: up, up. And, we stop because we don't have any more electrons. So, this would represent the electron orbital diagram for titanium.

So, it's just what we've seen previously in terms of electron configurations, but now we're paying close attention to our transition metal.

Now that you've seen this example, attempt to do the practice one on your own. Here, you're just asked to write the condensed electron configuration for the cobalt (I) ion. So remember the rules we talked about in terms of charged electron configurations and apply them to this example. I'll give you a huge hint. It's best to write the electron configuration of the neutral form of cobalt first, then that gives you an indication of what to do to write the electron configuration of its ionic form. Once you've done that, come back and see if your answer matches up with mine. Good luck, guys.

Write the condensed configuration for the following element: 

Co⁺ (Z = 27)

Alright. So a huge hint that I gave to you was, first, write the electron configuration of the neutral form of cobalt. Cobalt, I've highlighted as being right here. Now, we want to write its condensed electron configuration. So, the last noble gas I passed before I get to cobalt was argon. And then we just have to count to cobalt. So 4s2 3d7. This would be the Electron Configuration of neutral cobalt. So, we're going to come down here. So neutral cobalt has this configuration, but we want cobalt plus 1. Plus 1 means we've lost 1 electron. Based on the rules that we know for charged electron configurations, we know that the electron has to come from the highest n value which represents our highest shell number. So here, 4 means that the electrons here are in the 4th shell so n equals 4. 3 here means the electrons are in our 3rd shell where n equals 3. So the electron has to come from the 4th shell. So for cobalt, its condensed electron configuration would become [Ar]4s13d7. Hopefully, you guys were able to get the correct answer. Just remember, we're applying the concepts that we learned earlier about charged electron configurations and now we're just applying that concept to transition metals. Transition metals can get tricky. Later on, we'll look at some exceptions to their electron configurations. But for now, just realize, utilize what we know in terms of charged electron configurations, rethink the reimagined periodic table and apply them to find the electron configurations of either a neutral element or a charged element.

Electron Configuration Exceptions

In this video, we're going to take a look at electron configuration exceptions. Starting from the element chromium, as the atomic number z increases, exceptions to electron configurations can be observed. There are two major types of exceptions that occur when we have electrons from our s orbital being promoted up to our d orbitals.

So, when does this happen? Well, for the first one, s orbitals can be promoted to create half-filled orbitals with \( d^4 \) elements. Essentially, this exception happens with transition metals that are neutral and end with \( d^4 \) in their electron configurations. For example, if we look at chromium on our periodic table, we initially see it as \( \text{[Ar]} \, 4s^2 \, 3d^4 \). If we fill out its electron orbital diagram, remember, your s sublevel has one electron orbital and d has five. \( 4s^2 \) means we have two electrons, one up, one down. \( 3d^4 \) so we have four electrons and we half-fill first, following Hund's rule. Here, the \( 3d \) orbitals would be even more stable if they were half-filled. Unfortunately, we're one electron short. To deal with this and to become more stable, what happens is that one of our electrons from the \( 4s \) orbital gets promoted up to this d orbital. So, the correct configuration of chromium is actually \( \text{[Ar]} \, 4s^1 \, 3d^5 \).

Our second type of exception arises when s orbitals can be promoted to create totally filled orbitals with \( d^9 \) elements. In this case, considering the transition metal copper, we initially see \( \text{[Ar]} \, 4s^2 \, 3d^9 \). We have our \( 4s \) orbital here and our five \( 3d \) orbitals here. \( 4s^2, 3d^9 \). Following Hund's rule, we half-fill and then come back around. We fill in nine electrons. The d orbitals are most favorable and stable when they're either half-filled or totally filled. In this case, we're one electron short of having this orbital completely filled. The actuality is \( \text{[Ar]} \, 4s^1 \, 3d^{10} \). So there goes the one electron that I still have in \( 4s \), and my \( 3d \) orbitals. Now, I have ten electrons. So there goes my original nine, and then we have our tenth electron there.

If we look at our periodic table, focusing on our d-block transition metals, when it comes to our \( d^4 \) elements, that's broadly chromium and molybdenum. These elements, as well as d nine's, have this exception. Just remember these six elements; they either fall into the \( d^4 \) or \( d^9 \) exceptions for electron configurations. Now that we've seen these two examples, we'll attempt in the next video to see if we can give the correct configuration to molybdenum. Click on the next video and see how I approach this example question.

Electron Configuration Exceptions Exercise 1

Alright, so here it says with an atomic number \( z \) of 42, illustrate the exception to the electron configuration of molybdenum. So if we're looking at our periodic table, hopefully you have one out, the last noble gas we pass before we get to molybdenum is Krypton. And then we count to molybdenum. So \(5s^2 4d^4\). But, we know that we can't have this electron configuration because it's a neutral transition metal. Can't end with \(d^4\) or \(d^9\). We're going to have the promotion of an electron from the \(s\) orbital to the \(d\) orbital. So its correct electron configuration is actually Krypton \(5s^1 4d^5\). And in that way, we make the \(d\) orbitals half-filled which is more stable. If we were to write out its electron orbital diagram here, we'd have Krypton, \(5s\), and these will be \(4d\). So, \(5s^1\) and then \(4d^5\) means that it's half-filled and therefore more stable. Hopefully, you guys are able to attempt this even without my help. Remember, we have our 6 elements here that tend to fall into the \(d^4\) or \(d^9\) category and therefore represent exceptions next the next practice question. Remember when it comes to ions, first write the electron configuration of the neutral form then look at the ionic form. Come back and see if your answer matches up with mine.

Electron Configuration Exceptions Exercise 2

Alright, so here write the condensed electron configuration of the following ion. So we have to do it first for neutral silver. So if you're looking at your periodic table, initially you would see it as Krs2d9. It is one of the exceptions. So we have the promotion of an electron from the s orbital to the d orbital. So now its real electron configuration is Krs1d10. Here though, we're dealing with silver one ion so we've lost one electron. Remember, we lose electrons from the highest n value. Here, this electron's in the 5th shell and these 10 are in the 4th shell. We lose them from the highest shell number so we lose it from the s5 one orbital. So the electron configuration of silver one ion would just be Krd10. So we'd write it as krypton here. Our s5 has no electrons in it. So don't write any electrons within it. And then, d4 10. So up, up, up, up, up. Then we come back around and do down. So that would represent the electron configurations of neutral silver, silver ion and then its electron orbital diagram. Here we can see that it wants to do this because doing this promotion of an electron from the s orbital to the d orbital gives us a completely filled in d orbital set. So remember, d orbitals are most stable when they're either half-filled or completely filled. That explains this type of exception that arises. So hope you guys are able to follow along in terms of these exceptions. So keep in mind, you're just adding this to the list of things that you're refreshing in terms of electron configuration. We've gone over full electron configuration, condensed, electron configuration of ions, and, of course, transition metals and their exceptions. So keep this in mind because later on, it'll be used as justification for some of the organic reactions we'll see.

Write the condensed electron configuration of the following ion:

Cu⁺ (Z = 29)

Alright. So at this point, we're almost pros in terms of exceptions. Here we have the copper one ion. Remember, when it comes to the electron configuration of an ion, first look at its neutral form. So copper, its neutral form, if we're looking at the periodic table, initially we would see argon, 4s2, 3d9. But we know that copper is one of the 6 transition metals that has an exception. We know that they have d4 to d9. There are exceptions where we take an electron from the s orbital and promote it up to the d orbital. So here we take one electron from the s orbital and promote it up to 3d. So the correct configuration of copper is argon 4s1 3d10. Now copper 1 ion means we've lost one electron. Remember, you lose it from the highest n value. So here n equals 4 and here n equals 3. So we take it away from the 4s orbital. So copper 1 ion's electron configuration now is argon 3d10. Now that we've done this practice question, move on to the next one. We didn't talk about palladium before, but palladium also represents another transition metal with an exception to its electron configuration. From the question, look to see if you can figure out what its electron configuration would be. Attempt to do it for the neutral form and then for the plus 2 ion form. Once you do that, come back and see if your answer matches up with mine.

The palladium atom possesses a unique electron configuration, in which its d-orbital would prefer to be completely filled. Based on this information, provide the electron configuration of both Pd⁰ and Pd²⁺.

The question says, the palladium atom possesses a unique electron configuration in which its d orbital would prefer to be completely filled. Based on this information, provide the electron configuration of both palladium and palladium 2 plus. So for palladium 0, that just means it's neutral, it has no charge. Alright. So we have palladium s2, s2d8. Now, with electron orbital diagrams, you know, s5 has one electron orbital, and d has 5 electron orbitals. So it has 8. So we half fill first. Now they give us a clue in this question. They said that it has a unique electron configuration. So, obviously, this is not correct that in order for it to be completely filled, we need to add an electron here and here. And to do that, we would take both these electrons from the s5 orbital and donate each to one of these electron orbitals. So that would mean that, palladium's neutral form would be krypton d10. We take these electrons here and we would put 1 here and 1 here. So that's pretty unique. We're used to seeing the promotion of just one electron from an s orbital to a d orbital but palladium is very unique. It actually has both of its s orbital electrons being promoted to the d orbital electron. This is pretty stable because again remember d orbitals are most stable when they're half filled or in this case totally filled. Later on when we see, organic reactions involving palladium we're going to say that palladium is unique to some of these reactions because of this special situation with its electron configuration. So this would be the electron configuration of its neutral form. Palladium 2 ion means that we lose 2 electrons And the only place that can come from is here. Right? These electrons are in the 4th shell. Remember the noble gas portion think of it as a locked box which we don't go into to grab any electrons. The electrons that we remove are the electrons that are outside of this noble gas configuration lockbox. So here for palladium 2 ion, its electron configuration would just be Krypton d8 because we have to lose 2 electrons. So you grab the 2 electrons from here. Now, if we didn't know about this special unique circumstance, you should have still gotten this as the correct answer for the palladium 2 ion. Because even if you had thought this was the correct configuration initially, you know that you'd have to lose 2 electrons and they would come from here because this is the 5th shell. So for this example, remember palladium is pretty unique where we have this as the electron configuration of its neutral form and remember the exceptions that we learned with d4 and d9 transition metals. Those are all the different types of exceptions you should prepare yourself for.

Valence Electrons of Transition Metals

In this video, we're going to take a look at the valence electrons of elements. Now, remember, the valence electrons represent the outer shell electrons that are involved in the formation of chemical bonds. For main group elements, the number of valence electrons is equal to their group number. For instance, a group 1 element such as sodium would have one valence electron. A group 4 element like silicon would have 4 valence electrons, all the way up to group 8 like argon, which would have 8 valence electrons. This is only true for main group elements, your group A elements. Remember, transition metals are group B elements, and as a result, they have to be treated differently.

When it comes to transition metals, we're going to say the valence of a transition metal, and any transition metal, we're going to claim as being metal M, is equal to your s orbital electrons plus your d orbital electrons. If we take a look here at our periodic table, we have a more abbreviated ending for each one of these transition metals. For example, vanadium, which is technically, as we saw earlier, Ars2d3. We don't concern ourselves with the argon portion because we're going to say the valence electrons are equal to s plus d electrons. Here are our s and our d electrons.

So, if we quickly look at all the abbreviated endings of all these transition metals, we can figure out the number of valence electrons. Here we have scandium, which is s2d1. That's 2 s orbital electrons plus 1 d orbital electron for a total of 3 valence electrons. All these elements have that in common. All of them have 2 s orbital electrons with 1 d orbital electron. Titanium and its group below it, if you add them up, that's 2+2, so that's 4 valence electrons. Then this would be 5, 6. Now tungsten and seaborgium, remember, we said that they are not exceptions to the electron configuration. So they're just s2d4. But they still add up to 6 total valence electrons. This here would be 7, 8, 9, 10. Add all these up, you would see that they add up to 11. And this would just add up to 12.

So remember, when it comes to main group elements, we're going to say that the group number is equal to the valence electrons. But when it comes to transition metals, it's different. For transition metals, we look at the s and the d electrons and we add them together. Remember, we're only concerned with the d block transition metals. We're not going to worry about the f block transition metals because those elements, due to their high atomic masses, a lot of them are kind of weird and radioactive. So we don't really worry about them when it comes to organic reactions.

Again, when it comes to transition metals, just find the number of s and d electrons, add them together, and that equals your valence electron number. Now that we've basically seen how to figure out the valence electron number of transition metals, click on to the next video and see how we approach this question where we're asked to figure out the total number of valence electrons for a nickel plus 1 ion.

Valence Electrons of Transition Metals Example 1

Alright. So in this question it asks, how many valence electrons are present in the following ion? Here we're given the nickel one ion, its atomic number is 28. Now, you can approach this in two different ways. We can do it the easy way or we can do it the longer way. If we look up here, we see that nickel is right here. In its neutral form, nickel would have 10 valence electrons. Plus 1 means that we have lost 1 electron. As a result, we should expect it to have 9 valence electrons. So it's as simple as that, and that's our answer.

But let's say your professor wants you to show how it only has 9 valence electrons. Well, in that case, we'd have to write the electron configuration of neutral nickel, then find the electron configuration of the nickel ion. Nickel, as we can see up here, has an abbreviated electron configuration of Ars2d8. It starts off with argon because that's the last noble gas we pass before we get to it. s2d8. Now, the nickel plus one ion means we've lost one electron. Remember, the electron that we lose has to come from the highest n-value orbital. Here, because this number is 4, that means n=4. So we're dealing with the 4th shell. Here, 3 means n=3, we're dealing with the 3rd shell. We're going to lose the electron from the higher shell number, which means it's going to come from the 4s orbital. So, the electron configuration of the nickel ion would come out to be Ars1d8. If we add up the number of s-orbital electrons and d-orbital electrons, that will come out to 1+8=9.

We could do it the quick way, simply saying it should have 10 valence electrons when it's neutral. Plus 1 means we've lost one of those valence electrons, so now it has 9. Or you could do it the longer way where we have to give the full electron configuration of the nickel plus one ion. We give its condensed electron configuration here. Look at how many s and d electrons it has. Add them up to get our final answer.

Now that we've seen this example, attempt to do this practice one here. For this one, I want us to provide the condensed electron configuration and the number of valence electrons for the following ion. In other words, I want you to go through the longer route of doing all this work to figure out the number of valence electrons. So it's good practice, and it helps to justify why we have x number of valence electrons for ion 3 ion. Good luck, everyone.

Provide the condensed electron configuration and the number of valence electrons for the following ion:

Fe³⁺ (Z = 26)

Alright. So, this question asks you to provide the condensed electron configuration and the number of valence electrons for the following ion. Here we have iron 3 ion. If you looked up above, you would see that we could figure out the electron configuration of the neutral iron atom. It would come out to [Ar] 4s23d6. Now, iron 3 plus means that we've lost 3 electrons. Remember, the electrons that we're going to lose first come from the highest shell number, where n equals 4 and here n equals 3. So for the iron 3 ion, it comes out to [Ar]. All the 4s electrons are gone because we need to lose 3; they're both gone. Then we need to lose one more. That one more electron that we need to lose is going to come from the 3d orbitals. What we're left with is 3d5. Remember, valence electrons for d electrons and in total, we have 5 d electrons. So those 5 d electrons would mean that we have 5 valence electrons remaining. That would be our final answer. Hopefully, you were able to follow along and got the answer before I wrote it down. Just remember, when it comes to valence electrons for main group elements, it's just equal to their group number. But for transition metals, because they are group B elements, it's equal to s plus d orbital electrons. Remember this, and you'll always be able to figure out the number of valence electrons for any given transition metal atom and ion.