General Chemistry

posted by .

The four quantum numbers for the last electron placed in the orbitals of a certain element are: n=3 l=2 m of l=0 and m of s=-1/2. What is the element?

I need help with this part of chemistry I have NEVER understood this ever. I know how to do the configurations but not this. I need some explanation on how to do it.

  • General Chemistry -

    I'm sorry I can draw this thing out on the board but I'll try to explain it. First, however, you must learn the rules.
    n = principal quantum number and can have any whole number beginning with 1.

    l = any whole number beginning at 0 but can be no larger than n-1

    ml = any whole number (including zero) and it can be as small as -l and as large as +l

    ms can be +1/2 or -1/2.
    Get all of this in your mind and we go to the set up.
    ============================
    Here is what we do. We count by n until we get to n=3, l=2, ml = 0 and ms = 1/2
    n = 1
    l can be 0 (only)
    ml can be 0 (only)
    ms can be +1/2 and -1/2. So we put two electrons in for the +1/2 and -1/2 and that spells He.

    n=2
    l = 0................or 1
    ml = 0............or -1...0...+1
    ms = +/-1/2
    We put electrons into each of these holes or 8 electrons total. Together with the two we had above makes 10 and we are to Ne.

    n = 3
    l = 0........or 1..........or 2
    ml = 0.....-1, 0, +1.....-2,-1,0,+1,+2
    ms +/-1/2
    Add electrons in pairs (two for the ml = 0, 6 for the ml of -1,0,+1, and 2 for ml of -2, 2more for ml = of -1 and ONE more for ml = 0.
    Remembering we had 10 from n = 2, that plus what we have added here makes 13 more or 23 total. Must be V.
    It might help if you drew something like this on a sheet of paper; however, it get laborious to do this every time and you don't need to do that. You can take a shortcut. How?
    We take a little stock and look at the periodic table.
    n = 1 (period 1 takes you through He)=2e
    n = 2 (period 2 takes you through Ne)=10e
    n = 3(you wanted n = 3 so we know the element is somewhere after Na.
    For l=2, that's a d electron (s electrons are l = 0, p electrons are l =1, d electrons are l = 2, f electrons are l = 3). So a d electrons means it must be somewhere in the 3d transition series (n = 3, l =2). Remembering to fill the 4s next, we add those to get us through K and Ca, then we count what we have so far. Through Ca is 20, adding 1 more gives us Sc, another gives Ti, and the last one gives V. What makes Ti and Sc different.
    Sc is n = 3, l = 2, ml = -2
    Ti is n = 3, l = 2, ml = -1
    V is n = 3, l = 3, ml = 0
    etc.
    I hope this helps. If you draw something like I did above two or three times I can assure you it gets easier.

  • General Chemistry -

    Well I'm studying for my final and he gave us an answer key and with a spin down it says that the answer is nickel.. SOO I'm trying to make sense of it but I understand what n is what l is but how do you figure out what the elements are from ml? And all of that put together?

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Chem

    From the information below, identify element X. a. The wavelength of the radio waves sent by an FM station broadcasting at 97.1 MHz is 30.0 million (3.00 X lo7) times greater than the wavelength corresponding to the energy difference …
  2. Chemistry

    ELement A has a valence electronic structure of ns2np1. Element B has the valence electronic structure of ns2np5. If the value of n for element A is 3, what is the set of 4 quantum numbers that would describe the energy of the electron …
  3. science

    1s2 2x2 2p6 3s2 3p6 is the electron configuration ofr for which ion?
  4. chemistry

    using only the periodic table, write the expected ground-state electron configurations for: a. element number 116 b. an element with three unpaired 5d electrons c. the halogen with electrons in the 6p atomic orbitals
  5. chemistry

    Suppose you live in a different universe where different amount of quantum numbers is required to describe the atomic orbitals. These quantum numbers have the following rules: N principal 1,2,3,…. L orbital =N M magnetic -1. 0. +1 …
  6. Chemistry

    The four quantum numbers for the last electron placed in the orbitals of a certain element are: n=3 l=2 m of l=0 and m of s=-1/2. What is the element?
  7. chemistry

    If given the quantum numbers: 2,1,-1,-1/2 give the element. I understand the first two numbers and getting to the p orbitals of the second period, but I am not sure on how to designate -1 and -1/2 to Carbon (the answer) Also for: 4,1,0,-1/2 …
  8. Chemistry

    Element X has the highest first electron affinity in its period; the ground state electron configuration of it's most common ion is [Kr] 5s^2 4d^10 5p^6 a Element Y is the second largest element in it's period; it's valence electrons …
  9. chemistry

    Quantum numbers arise naturally from the mathematics used to describe the possible states of an electron in an atom. The four quantum numbers, the principal quantum number (n), the angular momentum quantum number (ℓ), the magnetic …
  10. Chemistry

    Element x has the highest first electron affinity in its period, the ground state electron configuration of its common is: [Kr] 5s2 4d10 5p6 Element Y is the second largest element in its period; its valence electron are in orbital(s) …

More Similar Questions