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Atom's Radius

  1. Jun 11, 2005 #1
    Hi all. I'm a student trying to learn some chemistry concepts on my own. So, i'll asking for some help a bit here. Look forward to hairing from yall too.

    To the question..

    I don't understand why the radius of an atom gets larger as u move to the left of the periodic table. Since atoms have move electrons as you move to the right, i'm thinking, that the radius should also increase in that direction too. Move electrons, larger radius--but this isn't the case. How come?
     
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  3. Jun 12, 2005 #2
    Educated guess: As you move to the right, the amount of electrons increases, but likwise the number of protons in the nucleus is larger. The effective nuclear charge on the outer electrons is stronger and thus the atom is contracted a bit more.
     
  4. Jun 12, 2005 #3
    I would guess because those elements are in the same period. This means they have the same level of subshells avaliable, and while there in increase in electrons as well as protons as you go to the right, the effective nuclear charge is greater on the electrons, thus 'pulling' the valence electrons to reduce the atomic radii.

    To the left, its the opposite, *given the fact that it the elements are in the same period for fair comparison*. The overall nuclear charge on the elctrons is 'weaker' than that of the elements on the right (due to reduced proton number AND WITH THE SAME SUBSHELL), and there is a a more 'loose' hold on the electrons, thus increasing the atomic radii.
     
  5. Jun 12, 2005 #4
    As both whozum and Bladibla said, the increase in protons increase the attraction that are had on the electrons. This decreases the radius of the atom.

    Here is an explaination on a very good website and here is a periodic table that will show this.

    The Bob (2004 ©)
     
    Last edited: Jun 12, 2005
  6. Jun 12, 2005 #5

    Gokul43201

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    Specifically,in The Bob's link, click "Table of Contents"; scroll down to the Periodic Table section under Science Trek and click "Periodic Properties" about midway down that section.

    Whozum and Bladibla have said essentially the same thing.
     
  7. Jun 12, 2005 #6
    remember that atomic radii also increase as you go DOWN the periodic table.

    also, part of the reason for this is that as the electron orbitals are filled, they shield some of the charge of the protons in the nucleus from the outer orbitals. Thus, the + charge that electrons in the new orbital (i.e. each row, going down the periodic table) feel will effectively decrease as the orbitals become filled.

    Slater's rules are a good "rule of thumb" for finding this shielding effect:

    http://www.unb.ca/fredericton/science/chem/2201/Slater's_rule.html

    these can then be applied to the potential energy function:

    [tex]V(r_{ij}) = \frac{-Z^* e^2}{4\pi \epsilon r_{ij}}[/tex]

    and solved for [itex]r_{ij}[/itex], the radius, by setting the potential energy equal to the known values for the orbitals (where Z* is the effective charge of the nucleus felt by the electron, that is, Z minus the loss due to electron sheilding). A more theoretical way, that wouldn't depend upon knowing the energy values a priori, would be to solve the Schrodinger equation, but this can be difficult.

    If you solve this for varius elements on the period table, you will see the general trend of the radii increasing. ..you will also see that there are some exceptions!
     
  8. Jun 12, 2005 #7

    GCT

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    Adding to what other's have said, you'll need to consider in detail aspects of effective nuclear charge. The latter increases as you move to the right. Why is this so? It's due to the trend in electron shielding...try to read up on this subject and if you have any further specific questions I'll try answering them.
     
  9. Jun 12, 2005 #8
    makes sense....but the comment about the shielding effect brings up another question. If more electrons means more shielding for the outer shells causing an increase in radius. Then wouldn't it follow that the radius gets larger as you move to the right.The only explaination i can come up with is that the charge of the protons is so great that it overcomes this shielding effect
     
  10. Jun 12, 2005 #9

    Gokul43201

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    I'm completely stumped by this.

    A couple of questions, please :

    1. What do the subscripts i,j index ?

    2. How do I find these known values of the Potential Energies of the orbitals ?
     
  11. Jun 12, 2005 #10
    A guess here (so corrections please) but is mass affects the force of gravity then why not mass on the 'force' of charge. Protons have a greater mass so a greater pull.

    The Bob (2004 ©)
     
  12. Jun 12, 2005 #11
    [tex]r_{ij}[/tex] would be the same, in this case, as [tex]\sqrt{x^2 + y^2 + z^2}[/tex] where we take the origin to be the center of the nucleus. Since we are talking about one electron here with respect to a fixed origin, i guess the index notation really isn't needed.

    A chemist would look them up in a table. I remember doing this in inorganic chemistry. A physicist would probably turn up their nose at this.
     
  13. Jun 12, 2005 #12
    alright i think i understand so for. I have another question about atomic structure. My book states that there is greater energy in the orbitals as u move further away from the nucleus of an atom. It just states this and doesnt explain. anyone know why this is so?
     
  14. Jun 13, 2005 #13
    I'll try and answer this.
    It's because greater energy is required to 'maintain' the promoted electron at a greater distance from the effective nuclear charge.

    Note: 'orbitals' and 'Energy level' has many swapped uses. Energy level is a more accurate descritpion, however..
     
  15. Jun 13, 2005 #14

    Gokul43201

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    Could you please indulge me and point me towards such a table ? I really am curious because the potential energy does not commute with the Hamiltonian.
     
  16. Jun 13, 2005 #15

    Gokul43201

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    The total energy of an electron bound to a nucleus must be negative. When you separate the electron greatly from the nucleus (ionize the atom), the energy of interaction between the two goes to zero, and so, a stationary free electron has zero energy, which is greater than any negative number. This tells you why it's plausible that the energy increases with distance.

    A better explanation is slightly mathematical. The potential energy is always negative between an electron (negative charge) and nucleus (positive charge). The kinetic energy can be shown (in a few lines) to always be related to the potential energy through the relation, KE = -PE/2, which is a positive number. Hence, the total energy is given by, E = PE + KE = PE - PE/2 = PE/2, and is always a negative number proportional to the potential energy. The absolute value of the potential energy falls off as roughly 1/r with distance from the nucleus, so the total energy must increase (being the negative of this absolute value) with increasing r.
     
  17. Jun 13, 2005 #16

    ZapperZ

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    I'm curious about such a table also. Considering that the p, d, f, etc orbitals are highly eccentric with respect to the geometry of the electron distribution, I'm skeptical that one can actually look up a value for potential energy of a single orbital.

    Zz.
     
  18. Jun 13, 2005 #17

    GCT

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    The relative increase in shielding is less then the effect of the increasing proton number. Think about what happens as you successively add electrons starting from the 1s2 orbital for instance, how effectively do electrons in the 1s2 orbital shield each other relative to when you arrive at the p sub-orbitals?
     
  19. Jun 13, 2005 #18


    yes, i see now. thanks for your help guys
     
  20. Jun 13, 2005 #19
    I am glad I found this, I often wondered this same thing.
     
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