Chemistry

How do you write the electron configuration of Curium (Cm) and copper (Cu)? I'm confused of writing electron configuration with d and f orbitals. Please help. Thanks.

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  1. Cm has 96 electrons. The previous noble gas is radon, Rn, with 86 electrons. It is a 7th row element and a member of the actinide series which starts with Ac, 3rd element on the 7th row and 7th element on the actinium series. The electron configuration is:
    [Rn] 7s^2, 6d^1, 5f^7
    Note that the
    [Rn] accounts for 86 electrons.
    7s^2 shows a full s-subshell in the 7th row (2 electrtons).
    6d^1 shows 1 d-electron 1 shell below the last one (7th) since it is the first element in the d-block of the 7th row.
    5f^7 shows 7 f-electrons, 2 shells below the 7th.

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  2. and Cu is 4s^2,3d^9 ?
    Just want to make sure:
    So if it is Holmium (Ho), is the electron configuration 6s^2, 5d^1, 4f^10 ?

    Thank You.

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  3. 5f^14 and 5f^7 is stable ( 5f^7 : all spin is up), so you cannot drive 5f^8 instead of 5f^7 6d^1

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  4. You can check out electron configurations of the elements by going to www.webelements.com , click on the elements, then scroll down to electron shell properties found on the right side of the page.

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  5. I don't get it.
    Why is Cm [Rn].5f7.6d1.7s2, but Ho [Xe].4f11.6s2 ? Why isn't Ho [Xe].6s^2, 5d^1, 4f^10 ?
    Help needed. Thanks

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  6. http://www.uwuses.uwaterloo.ca/~cchieh/cact/c120/eleconfg.html

    The above is a site with a lot of reading but it has some pictures and diagrams that may make it a little easier to understand than we can do on these boards. I think one important detail for you to remember is that no set of rules will allow one to fill the energy levels and make no mistakes. That is, there are exceptions that must be memorized. For example, the electron configuration for 28Ni is 1s2 2s2 2p6 3s2 3p6 3d8 4s2. Adding one more electron to make 29Cu we might expect to find 1s2 2s2 2p6 3s2 3p6 3d9 4s2 but that is not correct. The addition of the one electron changes the 3d and 4s energy levels just enough that it is 1s2 2s2 2p6 3s2 3p6 3d10 4s1. So this is an exception we must memorize. In the 3d series of elements (21 through 30), Cr and Cu are the two elements that don't follow the "correct" pattern. We have exceptions in the 4d, 4f, and 5f series, also.

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  7. Ho is the element of the Lanthanide series, the Lanthanides have the electrons are arranged into the orbital 4f.If you start from the element La (Z= 57), write the electron configuration of La,Ce(Z= 58),..., to Ho(Z=67) , you will understand why Ho=[Xe]4f11 6s2, it cannot 6s2, 5d1, 4f10 because the electrons are arranged into the lower energy level first, the energy level of 5d >4f

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