Identify a +2 cation that has the following ground state electron...

Question

Identify a +2 cation that has the following ground state electron
configurations: (b) [Ar]3d9

Accepted Answer

Hello. Everyone in this video. We're kind of trying to find out what our actual concern is with the element but we're only given the charge of the ion as well as the electron configuration of the ion. So actually our first step, so step one is going to find our element. So how do we find our elements while we're given this electric configuration and were given the charge this two plus charge in the case that we're losing two electrons. So we can simply add back our two electrons into this electron configuration to get the electron configuration of our element. So we're adding the two into the D orbital. So our actual electric configuration for the element is going to be a R. For us to and three D wall one plus two is three. No, let's see here. So I separated my puree table into four different colors representing each block. Our yellow block is going to be our s block. Our green block is going to be our D block. Our blue block is going to be our P block. And our pink block is going to be our F block. Let's also label our rose. So 23456 and seven. So here we are given a noble gas configuration. So our gun is going to be our starting point. So argon is located right over here in this black outline and this is going to be our starting point. So when, how I think of a electron configuration, it's kind of like we're reading a book. We're going from left to right. Are reading and reading and reading until we get to the element that we desire. So like I said, we're starting at a R. So let's go ahead and indicate that brief, read that part. Moving on to the for us to four represents which rover on S for which block which is S is yellow. So here, so we've read two elements, so every red one and two. Then we're moving on to the D block. So this number for the D block here, we know that whenever ruining the electron configuration, we're subtracting one from the actual real number. So we'll add back to find out which row are on. So three plus one is four. We're on the fourth row on the D block. Reading 12 and three. So we're stopping next at this right over here and once we've read all the way there, you can see that that's actually our stopping point R three D three. So this is actually going to be our original element. So that's going to be weak. Now step two is adding the charge. So now that we know it is the element V. We just need to charge for it to be the ion right? Or the Canton. Well we have our two plus hurts the problem gave us. So simply we're just bringing it back, we have a two plus charge and this is going to be the ion from the given information of a two plus charge. And this noble gas configuration here, and this is going to be our final answer for this problem.

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