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Chemistry

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Here are the effective nuclear charges, Zeff, for the valence electrons of four different atoms. 1.30, 1.95, 2.60, 3.25 In order, these values could belong to what list of elements?

Sr, Ca, Mg, Be
Be, Mg, Ca, Sr
C, B, Be, Li
(or)
Li, Be, B, C

Here are the effective nuclear charges, Zeff, for the valence electrons of four different atoms. 2.20, 2.20, 2.20, 2.20 These values could belong to what list of elements?

Na, K, Rb, Cs
(or)
Na, Mg, Al, Si

  • Chemistry - ,

    To estimate Zeff, compare the number of protons to the number of inner electrons in an atom. Look for trends in how the estimates change from one atom to the next in each list.

  • Chemistry - ,

    no idea

  • Chemistry - ,

    kklklk

  • Chemistry - ,

    This seems complicated, but it's easier than you'd think. Your given a list of elements, and it states that the effective nuclear charge will be in the order given. The easiest way to solve this problem is to solve the first element, and see if it's Z eff is 1.30, and then check the following elements.
    It's easiest to do this by listing the electron configuration for each of the elements. I'll do this with Li as an example, and let you solve for the rest.
    Li: 1S2, 2S1

    Remember that we set up Slater's rule formula like this:
    (atomic #)-[(o.35(Valence shell -1)+0.85(Next level down)+1(remainder)]
    The atomic number of Li is 3.
    Li has 1 electron in its valence shell.(2S1)
    Li also has 2 electrons on the next level down. (1S2)
    After these shells are accounted for, zero are remaining.
    our formula is then set up:
    3-[0.35(1-1)+0.85(2)+1(0)] = 1.30

    Li fits the first charge of the charges given.

    Repeat this process in the next elements, and in the following equation to find the effective nuclear charge.

    To help you, the correct answer is:
    Part A: Li, Be, B, C
    Part B: Na, K, Rb, Cs
    But be sure to do the work yourself to understand how Slater's rule works.

    A more difficult element that isn't listed, for a reference look at Iodine (I).
    I: 1S2; 2S2,2P6; 3S2,3D10,3P6; 4S2,4D10,4P6; 5S2, 5P5.

    I's atomic number is 53.
    I has 7 electrons in it's valence shell (5S2, 5P5).
    I has 18 electrons on the next level down (4S2, 4D10, 4P6).
    I has 28 remainder electrons on its core shell's.
    our formula is then set up:
    53-[0.35(7-1)+0.85(18)+1(28)] = 7.6
    7.6 is Iodine's effective nuclear charge, if I did my math right, found by using Slater's rule.

  • Chemistry - ,

    How would you know how to do this problem without knowing Slater's rule? we've yet been taught that rule so I'm not sure how else I would figure it out.

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