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The ground state electron configuration of Po (At: No. 84) is OA [Xe] 652 6d10 6p4 0 B [Xe] 652 4f14 5d10 6p4 0 c [Xe] 652 Sd10 6p4 0 D. [Xe] 6s2 5f14 '6a10 6p...

Question

The ground state electron configuration of Po (At: No. 84) is OA [Xe] 652 6d10 6p4 0 B [Xe] 652 4f14 5d10 6p4 0 c [Xe] 652 Sd10 6p4 0 D. [Xe] 6s2 5f14 '6a10 6p4 0 E [Xe] 6s2 6f14 6d10 Gp4

The ground state electron configuration of Po (At: No. 84) is OA [Xe] 652 6d10 6p4 0 B [Xe] 652 4f14 5d10 6p4 0 c [Xe] 652 Sd10 6p4 0 D. [Xe] 6s2 5f14 '6a10 6p4 0 E [Xe] 6s2 6f14 6d10 Gp4



Answers

Write the orbital diagram for the ground state of terbium. The electron configuration is $[\mathrm{Xe}] 4 f^{9} 6 s^{2}$

Hello there in question 86 we are writings, an electron configurations. Also, we want these configurations to be for the ground state, so that means we are filling in these electrons from lowest energy orbital's upward As Thea ALF bow principle would dictate eso. Make sure you have a periodic table handy and let's go ahead and get started. In letter. A were asked to do the electron configuration for Hillary um, which is t in looking at the periodic table. We see its atomic number 52 which means it has 52 protons. But since it's an atom, it also has 52 electrons. Um, tell Laurie, um, is found in the fifth period. Looking at your periodic table. It's in the fifth period. So as I write the electron configuration, we could do the shorthand, which means we go to the period before that and pick the noble gas noble gas. Of course, the last element in the period, and we put that in brackets, so that represents those core electrons, um, those electrons up until we get to the period that's DeLorean has found it. In other words, the 1st 36 electrons. Now we just need to go across the fifth period to complete the electron configuration. So the first thing we passed through are the S group elements. So we're gonna fill up the s s is filled with two. So this is the fifth period, so it's five s to then we come to the D Block, remember? The D Block is always one energy level behind. So I am next filling in the four DS and DS can hold 10, and then we come to the five p if we look at the location of Tillery. Um, um, it is in the fourth group of the P Group elements in that fourth column. So that means it ends with a P four. We could double check our work by taking the 36 electrons from KR adding the two from the S, the 10 from the D and the four from the P. And sure enough, we should come up with 52 which means this is our electron configuration moving on to let her be let her be asks us to do the electron configuration for caesium, which is number 55. That is going to bump us into the sixth period. So in the bracket, I will put the noble gas from the fifth period, which is in in in the sixth period. We see that caesium is in the very first group, which is an s block element. Specifically, the first group is s Juan. So that is it. That is all we need for this electron configuration again. If we take X e, which is number 54 add one electron, we get 55. So that is correct. Let her see. Let her see a selenium s E. This is number 34. Selenium is going to be found in that fourth period. Looking at the noble gas in the third period we see it is are gone. So are gone, goes in our brackets and then we work our way across the fourth period, so I'm going to fill in for us to Then we come to the D block one energy level behind. So it's gonna be three d 10 and then finally, selenium is in that same group as tolerant waas back in letter A. They're both in the P four group is just that selenium is in the fourth period, so it's going to end with four p four and that would be your completed configuration for that letter. D his platinum so platinums PT it's number 78 and it is in group six. So XY goes in the brackets, however platinum as we work across to fill in this electron configuration, you may recall that platinum is one of the exceptions in platinum one of the S electrons. So in other words, we come to success. Normally we would put to their but one of the s electrons actually bumps up to the five d success in the five D are very, very close and energy. And so platinum is an exception that one of those electrons bumps up to the five D right after the success. We would still fill in the EFS. So four f 14 remember that F is to energy levels behind. And then we get to five D If we look at platinums position, it is in the eighth D Group there in the transition metals. It's in the eighth group for the transition metals. But we picked up that extra electron from the success, so that actually makes it five d nine and again. Platinum is just one of those exceptions to our normal filling of the electrons into their orbital's. So that would be the correct configuration for platinum in letter E. We're doing Oz me, um, ask me in his number 76. So if we look at that, that is just two electrons short of where we were for platinum. So we're going to start out the same way XY ask me, um, is not an exception. So we're gonna fill in success tubes. We go across the sixth period. Next we come 24 f, and we fill that in, and then we're going to end with five d. And we look where Oz me, um, is there it is in the sixth B block or the sixth group in the D Block, so we're going to end it with five D six quick little circle around that before we go on to letter F letter F is chromium. Crimea is number 24. Chromium is going to be another exception. So we're going to start off we're going to put are gone in brackets. And then as we work across the fourth period, the first thing we come to is for us. However, I mentioned, it's an exception. It's an exception because normally if we followed the normal rules, we would have for us to three d four. But Orbital's that air half full are more stable than those that are empty. And since the four s and the three D are very close and energy, what you actually get is that electron that would normally be the second electron s orbital goes into the empty D to give us three d five, and that is the configuration for chromium. All right, thank you so much for watching.

Yeah, tonnes rules to find the ground state, L. S of C E, G D and P. T. And they give you the electronic configurations. Okay so for part A we have two six S. Electrons and they'll have opposite spins but the other two electrons will have positive in S values. So there's four electrons total. The two in the success state will it's zero. So you have S is equal to one half I just one half plus one half plus one half which is equal to one. So for L the Pauli exclusion principle will not restrict our values invincible. So for the two estates you have 00 The F state is three D. Heat. It's too so this is equal to five. Alright then part B. We have 26 S electrons, one for F electron and 15 D electron. So the two success electrons will have opposite spin values. The four F. Show can hold 14. So all seven of those electrons will have a positive one half spend. And this single five D electron will also have a positive so that is eight electrons with a positive one half spin. So as is equal to four. For L. You have the value for success is zero. All seven of those four F electrons have the same spin as we just said. So they have to have different types of L values. That's three, 210 negative one, negative two negative three. And this friends too zero. So only from the five D electron will we get him stable and stable is equal to two, so L is equal to two. Okay, for part C. The four F shell is full so it won't contribute to L. Or S. So we can exclude the four S. Shell. So again, the I've D shell it has five electrons. So it can it has 95 of them will have positive one half. The other four will have negative one half the success will have a positive one half. So this gives us a value of us equals one. So the six X electron will have in Aliquippa zero. The 55 D electrons will have em stable. So we have zero from six S. And then from the five D electrons, we have positive to positive one zero negative one negative two. So the four remaining five D electrons, we'll have positive to positive one zero negative one, which gives us an L equal to two.

So some barium is an F. Block element to draw. Its orbital diagram will keep the noble gas core of zen on, and then we have four F. So that's seven possible orbital's 4567 And then we have six s, which has one orbital, so for four F will have six electrons. They'll each occupy separate orbital's because there's um they don't pair up until they have to, And then for the six s, we have two electrons, one of each spin, and that's the orbital diagram.

Okay, so we're gonna write some electron configurations. Okay, so nitrogen Is gonna be one S 2 to us too, two P three. You can see that it has seven electrons and that it's kind of three over in the piece of level section of the periodic table. Mm. So then we've got sodium which has 11 electrons. So that's one s. 2 to us too, two P six. And then three S 1. You can see it's in that first group one Which is S one column and it's the 3rd row down. And the next one, we've got neon Neons got 10 electrons. So that's just one fewer than the sodium. So we're just going to remove that. We've got one s. 2 to us too. Two P six. And then we've got nickel. Okay. and Nickel has 28 electrons. So we've got 1 S2, two S 2, two p 6. When we get to the third energy level 32 three P six. And then for us to and after that comes three D. And we need eight, so three D eight there. And then finally, We've got silicon, which is 14 electrons, So one has to to us, too, two P six, three S two, three P two.


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