Question
OUTCOMES 2 The element that corresponds to the electron configuration 1s22s22p63s23p64s'3d5 A) titanium B) vanadium chromium D) manganese E) iron3, Give the ground state electron A) [Kr]Ss-4d"Sp6 configuration for L B) [Kr]Ss'4dW0Sps C) [Kr]4dWSp6 D) [Kr]Ss?Sp" E) [Kr]Ss"5d"Sp6
OUTCOMES 2 The element that corresponds to the electron configuration 1s22s22p63s23p64s'3d5 A) titanium B) vanadium chromium D) manganese E) iron 3, Give the ground state electron A) [Kr]Ss-4d"Sp6 configuration for L B) [Kr]Ss'4dW0Sps C) [Kr]4dWSp6 D) [Kr]Ss?Sp" E) [Kr]Ss"5d"Sp6


Answers
What is the expected ground-state electron configuration for each of the following elements? (a) tellurium; (b) cesium; (c) selenium; (d) platinum; (e) osmium; (f) chromium.
Here, we're essentially asked to determine the electron configuration of FE two plus, See you two plus and cR two plus. So if we look at the periodic table, essentially the nearest noble gas neighbor for all of these is the noble gas argon. Yeah. And first, with regards to chromium, essentially losing both as electrons is most likely. And we could have essentially 34 for a copper, essentially. Normally we have for a copper, we have uh typically it's four S 13 D 10 in order to have the stabilization resulting from fully filled Dior bubbles for a C U two plus, we're going to lose an S electron and we're also going to lose electron from Dior bubbles. And this would be three D 9 for FE two plus, where irons ground state is normally for us to three D 6 with the noble gas losing essentially the normal S electrons is likely going to be most probable, And this would result in 36. And this gives our final answers
Here we are continuing to look at electron configurations. So, for example, we've got Philip Krypton followed by 45 and we are populating our not degenerate orbital's. So here we have high spins, we've got 12345. So we jump into the hospital before returning back to the T2, 500 electrons. Next we have M3 plus to fill up to Krypton, followed by four D 3. So again, we have two e in the T two Orbital's and so we have three unpowered electron here, all in the T two. Finally, we have Ceo three plus when we fill up to argon, Followed by three D 6, draw our orbital's E and ET two, where we have 123456, where we have five, where we have four and paired electrons 1, 2, 3, 4.
The electronic configuration of titanium to be positive. It's equipped to configuration followed by four. The I am. The energy level diagram can be represented us. Does the state war with him and this is A. G or become Yeah, it has a total of for you on beard electrons. 2nd 1 is more libidinal. Mhm Sorry Mm. or three positive. The electronic configuration will be krypton 40 three. The energy level type from can be 12, three. There's a state to this e.g.. It has a total of three um Beard electrons and option C. We have cobalt three positive great clothes arkan Followed by three D six. Yeah. One fully beard two. Okay, so mhm Yeah, it has a total of four on beard electrons.
So here we're looking at an element and it has a condensed electron configuration off three D to for us to So this is known, asked Titanium. We have to unpowered electrons present on. We will just draw out thes orbital's below. We're gonna be drawing our Vaillant Orbital's so we can start off with our five d orbital's with are associated with n equals three and then we have our for us here. I'll just labeled in three deep for us and you can see we have two electrons in our for us two electrons in our three d. So for us is completely filled now moving on to our three d. We have to unpadded electrons, as I said before, so we singly occupy our over tools before doubly occupying them. And that is to avoid paying repairing energy penalty