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Help: Writing the 4 quantum numbers for each of the 8 electrons?

  1. May 21, 2012 #1
    1. The problem statement, all variables and given/known data

    An oxygen atom has a total of 8 electrons. Write the four quantum numbers for the each of the eight electrons in the ground state?

    I cant seem to figure this out. I know this...

    O = 1s^2 2s^2 2p^4

    But from that, I dont know how to finish the question???



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 21, 2012 #2
     
  4. May 21, 2012 #3
    slightly lol... :)
     
  5. May 21, 2012 #4
    In case your class notes or text aren;t clear, This looks like a reasonable explanation. The executive summary:
    n corresponds to the shell number. So the first shell has n=1, the second shell has n=2, etc.
    l corresponds to the subshell:
    0 for s
    1 for p
    2 for d
    etc.
    m corresponds to the orbitals (which do not show up directly in the electron configuration notation).
    s correspond to the electron spin.

    So the electrons of Li (1s22s1) have quantum number
    (n,l,m,s) = (1,0,0,+1/2) and (1,0,0,-1/2) and (2,0,0,+1/2)...or the third could be (2,0,0,-1/2).

    With the link above and the example, you ought to be able to figure it out. If I say any more I'll pretty much be doing the problem for you.
     
  6. May 21, 2012 #5
    Can you please look at this example:

    Write a complete set of quantum numbers for the electrons in Boron?

    So Boron is.... 1s2 2s2 2p1

    1: n=1, L=0, ml=0, ms=+1/2
    2: n=1, L=0, ml=0, ms=-1/2
    3: n=2, L=?, ml = ? ms = ?

    Having trouble with 3, because the answer says the L = 0....why? and why does ml = 0? and why does ms = +1/2

    I dont get it? :(
     
  7. May 21, 2012 #6
    L (switching to your notation) can have any value from 0 up to n-1. So if n=2, L is 0 or 1 (corresponding to the s and p subshells, respectively).

    ml is the integers from -L to L, so if you have n=2 and L=0 mL must be 0 (i.e. all integers between 0 and 0, inclusive). ms will always be ± 1/2. This gives us, for the valence electron of boron, either
    n=2, L=?, ml = 0 ms = +1/2
    or
    n=2, L=?, ml = 0 ms = -1/2
    For n=3, L can be 0, 1 or 2 (corresponding to the s,p and d subshells, respectively). Since the question asks about the ground state (i.e., the lowest energy) you should start with the s subshell (in other words, 3s1 ), which is L=1.

    I am guessing your text has the convention that you assign +1/2 before -1/2...I don't know if that is a universal convention.

    The cool thing about all this is that when n=1, there is one L value and there are two sets of the four quantum numbers.
    The first shell of the atom has one subshell and holds two electrons.
    The first period of the periodic table has two elements, and you can argue that He should be above Li based on electron configuration, putting both those elements in one part of the table.

    When n=2, there are two values for L there are 8 sets of the four quantum numbers.
    The second shell of the atom has two subshells and holds eight electrons.
    The second period of the periodic table has two sections and eight elements.

    See the pattern emerging? (I tried to use bold, italics and underlining to show the correspondence).
     
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