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  1. Jul 26, 2004 #1
    First, want to thank Gokul (one of the posters here) for so much help over the last little while. Really learned a lot so far. I have more questions...

    I believe I understand number 3, it tells you the number of electrons so basically, I just have to look up using the periodic table..right?

    so, for number 3, a) Beryllium, b) Nitrogen c) Magnesium d) Phosphorus e) Helium ?

    I dont get how exactly you write the electon configuration and what the s, p and so mean...so numbers 1 and 2, I dont get.

    1. use the periodic table to write the electron configuration of
    a) Sr
    b) the element with atomic number 52
    c) Ta
    d) Gd
    e) U

    2. Write the electron configurations for the following substances, using the orbital notation described previously:
    a) zinc atom
    b) vanadium atom
    c) chloride ion (Cl-)
    d) aluminum ion (Al3+)
    e) gallium (Ga)
    f) bromine (BR)

    3. Identify the elements whose neutral atoms have the following electron configurations:
    a) 1stothe22stothe2
    b) 1stothe22stothe23stothe2
    c) 1stothe22stothe22ptothe63stothe2
    d)1stothe22stothe22ptothe63stothe23ptothe3
    d)1stothe2

    Thanks
     
  2. jcsd
  3. Jul 26, 2004 #2
    anybody know?
     
  4. Jul 26, 2004 #3

    movies

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    The s and the p refer to atomic orbitals. Each orbital can hold two electrons. The number before the orbital refers to the energy level. Each energy level has orbitals which the electrons occupy, typically filling the lowest energy orbitals first. For each energy level there is one s orbital. Then for each energy level 2 or greater there are also 3 p orbitals. Then for each energy level 3 or greater there are 5 d orbitals. Then for each energy level 4 or greater there are 7 f orbitals.

    Anyway, to write the electron configuration, you first figure out how many electrons you have to work with, and then you start filling in the orbitals starting with the lowest energy orbital (1s). It gets a little bit trickier when you get to the d-orbitals though because in most cases the 4s orbitals get filled before the 3d orbitals.

    When I was taking gen. chem. I used a table like this to remember the filling order:

    1s
    2s 2p
    3s 3p 3d
    4s 4p 4d 4f
    5s 5p 5d 5f
    6s 6p 6d 6f

    If you write that out and then draw diagonal lines descending from right to left through the numbers, you can get the correct orbital filling order.

    E.g. Phosphorous: 1s2 2s2 2p6 3s2 3p3
     
  5. Jul 30, 2004 #4
    Exceptions in Electronic Configuration:

    Its just a few elements (copper, chromium, molybdenum, nickel, platinum, tungsten, Nb, etc) that you need to take care of. And that knowledge comes easily with experience. Copper and chromium are the most common examples.
     
  6. Aug 7, 2004 #5
    I dont get these questions either:

    1. How many NODAL surfaces are there for?

    a) a 2s orbital
    b) a 3p(x) orbital

    and

    2. Explain why the first ionization energy for sodium is much smaller than the second ionziation energy for sodium?
     
  7. Aug 12, 2004 #6
    anybody have any idea on my last 2 questions?

    Any help would be appreciated. I have no idea.
     
  8. Aug 20, 2004 #7

    chem_tr

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    my humble attempt

    Hello,

    I will try to do my best. I saw movies' message far after I finished my message; mine is a little more detailed, maybe annoying, but principally based on the same logical pattern.

    2. Write the electron configurations for the following substances, using the orbital notation described previously:
    a) zinc atom
    b) vanadium atom
    c) chloride ion (Cl-)
    d) aluminum ion (Al3+)
    e) gallium (Ga)
    f) bromine (Br)


    First, let me write the corresponding atomic numbers of these elements:

    Zn:30; V:23; Cl-:18; Al3+:10; Ga:31; Br: 35

    Now, I will show an "incredibly easy" method of writing electronic configurations of elements as complex as thorium. Just try to comprehend the following rules:

    a) The first shell (i.e., 1s2) takes only s orbital with a maximum of two electrons. In this notation, 1 is called "shell number" and shown with the letter n, while s is called "orbital shape provider", and shown with the letter l. The number equivalents are one less than their letters, thus making s=0; p=1; d=2; and f=3.

    b) The sum of n+l is very important in deciding the correct place in the configuration; the higher n+l numbers cause the orbital to be of higher energy, thus written after the lower ones.

    c) If two orbitals have the same sum of n+l, the one with greater n has a greater energy; thus written after the one with smaller n.

    With these in mind, I will write the required configurations without looking anywhere, just typing on my keyboard:

    30Zn: 1s2 2s2 2p6 3s2 3p6 4s2 3d10; n+l sums are as follows: 1 2 3 3 4 4 5

    23V: 1s2 2s2 2p6 3s2 3p6 4s2 3d3; n+l sums are as follows: 1 2 3 3 4 4 5

    17Cl-: 1s2 2s2 2p6 3s2 3p6; n+l sums are just like the ones above.

    13Al3+: 1s2 2s2 2p6

    31Ga: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p1

    35Br: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5

    3. Identify the elements whose neutral atoms have the following electron configurations:
    a) 1stothe22stothe2
    b) 1stothe22stothe23stothe2
    c) 1stothe22stothe22ptothe63stothe2
    d)1stothe22stothe22ptothe63stothe23ptothe3
    d)1stothe2


    1s²2s²: A total of four protons make this element Beryllium.

    1s²2s²2p²: Total=6 protons, so this is a carbon atom. Note that 2p has the n+l number of 3, just like 3s; so 2p orbital must be filled first. Therefore, 1s²2s²3s² configuration is impossible, unless irradiation-like measures are adopted (vacant 2p orbitals may not be shown here).

    1s²2s²2p63s²: This configuration is correct, though; making the total of 12 protons to indicate magnesium atom.

    1s²2s²2p63s²3p³: This configuration is also valid. Total proton number is 15, so this element is phosphorus

    1s²: This is helium's electronic configuration.

    Regards
    chem_tr
     
    Last edited: Aug 20, 2004
  9. Aug 20, 2004 #8
    Seems like you have it all covered so I will just read here.

    The Bob (2004 ©)
     
  10. Aug 21, 2004 #9

    chem_tr

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    There are questions unanswered already

    Well, I am not familiar with number one, so I will try to explain number two only.
    Let's write sodium's electronic configuration first:
    11Na: 1s2 2s2 2p6 3s1

    So, you can easily see that in the valence orbital (i.e., the outmost filled orbital) has a single electron which can be easily given to form a Na+ cation. However, if you try to take one more electron from this system with an energy similar to the first one, you fail; you have argon's electronic configuration in your hands, which is very stable. You will have to apply an energy 3 or 4 times greater than the first.

    Regards
    chem_tr
     
  11. Aug 21, 2004 #10

    GCT

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