Given the subshells 1s, 2s, 2p, 3s, 3p and 3d, identify those tha...

Question

Given the subshells 1s, 2s, 2p, 3s, 3p and 3d, identify those that meet the following descriptions. 
(e) Contains the outermost electrons in a beryllium atom

Accepted Answer

Hello everyone today. We are being given the following problem among the following sub shells, three D four, S four, P four, D four, F five, S five P and five D. Which contained the outermost electrons in a selenium atom. So the first thing you wanna do is you want to determine what our atomic number of selenium is And that's going to be 34. And so this determines that we're also going to have 34 electrons since it is a neutral selenium atom. And it's important to note that this selenium is in the 4th period or the fourth or other periodic table. And so we also want to make note of some special characteristics of our orbital's. So we have S P and D. R s orbital hold a maximum of two electrons. Our p orbital holds a maximum of six electrons and our D orbital holds a maximum of 10 electrons. And so now we can construct our electron configuration or electronic and fake for short. So we're gonna start off with our first row our first period, which is one and we always start off with one S two. That means that we have the first two electrons filled in the s orbital. After that we move on to our two S two orbital, meaning that we have two electrons filled those first two elements on the periodic table in the second row are filled. We didn't move across All the way to the right hand side. It's still going to be two P six because we filled in those six electrons, we moved to the next row, which would be three S two. Because we filled those two electrons three P six four S 2 because we're in the 4th row and then we get into our D orbital. So then we're going to have three D orbital's and R D orbital's always start one less from our s orbital's. So we went from R four S, two, R three D. And so we're filling in all those electrons. And so once we move back to P, that is when we go back to our four. And so we have a four p and then we have four since selenium is the fourth element in our p orbital. Now we have to determine in these tough shells that we just listed below which continued outermost electrons. And so simply put that is going to be The last one that we have, which is going to be our four p Not four P 4, just our four p. And with that we have answered our question overall, I hope that this helped until next time

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