can you tell me why the electron configuration for a lead atom is?

tc

[Xe] (4f^14 )(5d^10)(6s^2)(6p^2)
can u plz show me how to get this...

Answer
Okay try and follow me with this, if I had my scanner here I'd do this on paper and scan it in for you but this is the best we can do now:

-Pull out a periodic table & look up Lead (Pb) it has an atomic number of 82.
-Okay, your instructor should have given you the list of the oder that the electrons would fill in each sublevel, like the following
1s
2s 2p
3s 3p 3d
4s 4p 4d 4f
5s 5p 5d 5f
6s 6p 6d
7s 7p (and such...)

The way that the electrons will fill each sublevel goes diagonally down starting with like so: 1s 2s 2p 3s 3p 4s (notice here that I didn't continue on with 3d but INSTEAD i went diagonaly down-left to 4s because this is of lower energy than 3d-continue...go back up to level 3) 3d 4p 5s (notice diaginally down-left)

Now you know that you must know that each "sublevel"(i.e. s, p, d, f) has a certain number of orbital:
s = 1
p = 3
d = 5
f = 7
And in each orbital there can be two electrons occupying that spot therefore the max. number of electrons per sublevel is:
electrons in s = 1 x 2e- = 2e-
electrons in p = 3 x 2e- = 6e-
electrons in d = 5 x 2e- = 10e-
electrons in f = 7 x 2e- = 14e-

Now to your question:
Lets write the sublevels up to where you will have up to 82 electrons:
1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^10, 4p^6, 5s^2, 4d^10, 5p^6, 6s^2, 4f^14, 5d^10, 6p^2

Okay now gather all the exponents and add to see if you get 82 electrons (expoenents represent the number of electrons in each sublevel):
2 + 2 + 6 + 2 + 6 + 2 + 10 + 6 + 2 + 10 + 6 + 2 + 14 + 10 + 2 = 82

This is the long way of writing the elctron configuration of lead in order to write the short way you use a nobel gas to replace some of those earlier electrons. This method requires looking at the periodic table and looking for the near noble gas with a fewer number of electrons than lead. Therefore go up a row and the noble gas turns out to be [Xe]. You know that Xe's atomic number is 54 and therefore has 54 electrons already. Take 82 (Pb-lead) - 54 (Xe) = 28

So you can write [Xe] and then add on another 28 electrons.
Well, just go back up to the long-hand method: 1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^10, 4p^6, 5s^2, 4d^10, 5p^6, 6s^2, 4f^14, 5d^10, 6p^2

add up the electrons until you get 54 which is equal to [Xe]:
2 + 2 + 6 + 2 + 6 + 2 + 10 + 6 + 2 + 10 + 6 = 54 Therefore everything up to 5p^6 represents [Xe] so you've accrounted for most of those orbitals, you only have 28 left which is essentially the (6s^2, 4f^14, 5d^10, 6p^2) part:
2 + 14 + 10 + 2 = 28 (there's your remaning 28 electrons)

Therefore the short-hand way is: [Xe] 6s^2 4f^14 5d^10 6p^2

I hope this helps!!

Good luck!

Here's also a website for you to read and review there seems to be some examples at the bottom. http://www.fordhamprep.org/gcurran/sho/sho/lessons/lesson36.htm

Love,
Mary

Does anybody know where to download Monkey Jam Animation software??

T9X

I;m trying to make some stop motion stuff. Does anybody know a site where I can download the MonkeyJam software besides www.giantscreamingrobotmonkeys.com?
That site is not responding.

Any help would be GREATELY appreciated.
Thank You

Answer
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