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.