Predict the number of unpaired electrons for each of the following. 

(a) Sc³⁺

(b) Co²⁺

Video transcript

Hello everyone. Today, we have the following problem how many unpaid electrons are present in each of the following transition metal ions? So first, we have iron two plus. So we want to first refer to the periodic table and note that iron has 26 electrons. And so we first, we want the next thing we want to do is we want to write out the noble gas configuration for that. So we look at the closest preceding noble gas, which is argon. We put that in brackets and then we proceed until we reach iron. So we go to the next row and we have for two and then we have 3d 6. And this is the neutral configuration. We see that our iron has a positive two charge. So we must remove two electrons from the highest energy level. And so that will be our 42. And so our configuration for iron two plus is argon 3d 6, which means we have six electrons in our 3d orbitals. So first, we follow the off ball principle which notes that the lowest energy orbitals are filled first and then we pair them up. So we have 12345, followed by six electrons. And so we know that we have a four unpaired electrons. And so because of that, we can get rid of entero A and C. Now, for our second one, we have zirconium and zirconium has atomic number of 40. And so, once again, we will do the noble gas configuration by looking at the closest preceding noble gas, which is krypton, we put that in brackets. And then as we go to the next row, we have five S two followed by four D two. And we looked at the charge, we have a positive four charge. So we must remove four electrons from the highest energy orbital. So what we can do is we can remove two from our five S orbital and we can remove two from our four D orbital, which actually just means that we have a configuration of krypton, meaning that we have zero unpaired electrons because all of them are paired up leaving us with an answer of D. And with that, we've answered the question overall, I hope this helped and it's on next time.