Write the electron configuration for each ion.
d. Mo3 +

Question

Write the electron configuration for each ion. 
d. Mo3 +

Accepted Answer

Hi everyone in this video, we're going to find the ground state electron configuration of F. A two plus. So F is iron. It's located right over here on the P R. A table and Iron has atomic number of 26. So there's 26 electrons. And because we have a two plus charge in the problem, it means that there's two less electrons. So subtracting two, we have 24 electrons total. Alright, so we have four different sections here in this periodic table. We have what I have colored and yellow. That's going to be our S block. We have this blue color, which is going to be our P block. And then we have our green, which is our D block. So we also have different roads and that's just the first row, second row, 3rd and 4th, I'm stopping on the fourth because that is where our element is. So like reading a book, we're going from life to rights. So let's just take the first row. So we have this hydrogen and this helium. And basically what we're doing is just reading all the way or essentially kind of just reading up until we get to the element that the problem is asking for. So again, we have the hydrogen and the helium first and that is in yellow color and that is what we have as our s block. It's in the first row. And because we have to read past this first road to get to you our element, we have read both. So Put that as one S 2. Now. Now that we have done the first row, we move on to second row. And again we go from left to right and first we have this two and that's going to be our second row and there's two elements and it's in yellow which is R. S. Block. So that's going to be to us too. Moving on to this blue chunk here, that's again our second row, but in blue, that's our P block. And we have to breed all across that to get to iron. So that's going to be written us two P six. Now moving on to the third row again, we have these two in the S. Block in the third row. So that's three s. 2 Going across, we have the blue block in the 3rd row. So that's P. So we have three P. Six. Moving on, we have the fourth row and again we're reading these two and the yellow and the yellow color which is R. S. Block. So that is four S. Two. Now we enter into the green region which we have not before and it's a little bit weird here. So instead of reading off just the normal fourth row, we're actually going to subtract one. So this is our fourth row and it would be actually written as three. And because the color is green were in the D. Block, so local right three deep and let's see we're doing 12345 and six. So put a little six here. So that's just the regular electron configuration of just iron. But now we have to take into account that two plus charge in our problem. So two plus, like I said, that means two less electrons. So which electrons are we going to remove in this configuration? We're going to remove the one with the highest principle quantum number. So we have one, two, 3 4. Another three of course four is going to be our biggest quantum number. So subtracting two, we're going to just remove those two And the ground state electron configuration of FP two plus is going to be really honest. One S two to us too. Two P six, three us two, three P six. And again, we're canceling out the four S two. So now we have three D six. So this is going to be our final answer for the problem.

Write the electron configuration for each ion. d. Mo3 +

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