Ionization Energy

Discussion in 'Chemistry' started by chikis, Feb 8, 2012.

  1. chikis Registered Senior Member

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    Go back to the periodic table and study it carefully. You will see that ionization increases from left to right across a period. It decreases down a given group.
    As for atomic radi, you will see that it decreases across a period from left to right but as you reach towards the noble gases, you will see that their atomic radi are larger. In such circustance, the ionization energies of noble gases should be lower but it won't because they have already attained their octet structure and will hardly loose electron. Therefore, their ionization energy are high.
     
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  3. chikis Registered Senior Member

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    So are you saying because helium have only two electron filling it shell, is no longer octet?
     
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  5. Nasor Valued Senior Member

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    Sure. Because radius decreases left to right and bottom to top.
    You're just wrong about this. http://environmentalchemistry.com/yogi/periodic/atomicradius.html

    Note how Ne<F, Ar<Cl, etc.
     
    Last edited: Mar 16, 2012
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  7. chikis Registered Senior Member

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  8. hypervalent_iodine Registered Member

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    Except that you're wrong. I had wanted to post an image of the periodic table showing relative atomic radii for a clearer illustration, but apparently my low post count prohibits that. A strange way to counter spam bots, but no matter. Look it up and you'll see what I mean.

    It doesn't have an octet because it only has a single s orbital available to it. However, that orbital is still completely filled in He in exactly the same way as the 2s and 2p orbitals are filled in the later noble gases. As was said in another post, the fact that they have particularly high ionization energy is because these atoms simply don't want to have any of their electrons removed. They are already in a very stable, inert state as it is.

    If you have a look at the electron configurations of the atoms, the periodic trend becomes quite understandable and as much as I hate personifying chemistry, for the purposes of your question I'll make a few concessions. Atoms are happiest when their highest energy orbitals are filled (see the octet rule). Sodium, for instance, has a single valence electron. To have a complete highest energy orbital it can either pick up 7 electrons or lose 1 - I think it's pretty obvious which one it would rather do. On the other end of the periodic table we have the halogens, which have 7 valence electrons. To have a full set of orbitals, it can either lose 7 or gain 1; again, the answer is pretty obvious as to which it would rather.

    Going back to your original question, I think that the easiest way to think about it is by considering the definition of what ionization energy actually is. Simply, it describes the energy that is needed in order to remove a single electron from an atom. Given what I've just said and with the examples of the halogens and the alkali metals in mind, you should now be able to understand why sodium, which needs to lose 1 electron to gain a full outer orbital, has a much lower ionization energy (i.e. the energy required to remove a single electron) than say, chlorine, which would need to lose 7.
     
  9. Pete It's not rocket surgery Moderator

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  10. origin Trump is the best argument against a democracy. Valued Senior Member

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    Do you have any evidence for this or is this just your idea. The orbital shape is generally discussed in PChem where the overridding mecanism is the wave like aspects of the electron. The Schrodinger Equation as it is applied to electons orbitals is the heart of the discussion. No where, that I am aware of, are there references to magnetic fields shaping the orbitals. I am at a loss to understand how the 'left hand rule' could be applied to the non-particle aspects of an electron orbital.
     
  11. chikis Registered Senior Member

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    I want you to click on the link below and watch that pictorial representation of atomic radi of elements and tell me what you noticed.

    http://www.crystalmaker.com/support...er/atomicradii/resources/VFI_Atomic_Radii.jpg

    It is true that atomic radi decreases from left to right across a period. One would except that trend to extend to the noble gases. If that happenes, a more justification will be given for the high ionization energies exihibited by the noble gases. The otherwise is the case from the pictorial representation of atomic radi of elements which is in the link I pasted in. You will see that from that picture, that the atomic radi began decreasing from each period, (left to right) of the periodic table.

    If you watch that table well, you will notice the following:
    In the first period, hydrogen has an atomic radi which is smaller than that of helium and helium is a noble gas

    In the second period, Litium has large atomic radi but you will notice that after litium, the next elements after litium as you move from left to right, say from berylium, their atomic radi start getting smaller till Neon the noble gas which have atomic radi which is almost the same size to that of litium.

    When you get to period three, as usuall, the atomic radi of sodium is large, as you move from sodium through magnesium, their atomic radi start getting smaller up till chlorine, then to the argon whose atomic radius is large almost to the size of sodium.

    As you move through the proceeding periods, you will see how from the begining of each period how the akali metal has large atomic radi. But if you move from left to the right, the atomic radi start getting smaller till the extreme end. The noble at each end is always large and having it atomic radius size equal to the akali metal in that period.

    From the above ilustration, I can tell you that the noble gases exhibit high ionization energies for the simple fact that they have atained their octet structure and will hardly ionize.

    Is not wise to judge it high ionization energy using the properties of it atomic radius since the may be other sources having information about the atomic radi of element which may be opposite to what is in the link I pasted in thereby causing unecessary confusion.
     
  12. Nasor Valued Senior Member

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    That's a picture of "VFI" atomic radii, which is based on data related to the length of the covalent bonds that the elements form when they're part of a molecule. The VFI method of determining atomic radii works fairly well for elements that actually take part in covalent bonding, but is not a useful way to determine the atomic radius for noble gases, since they only form very very weak covalent bonds. The stronger a covalent bond, the shorter it will be. Strong, short covalent bonds normally go hand-in-hand with short atomic radii, which is why the VFI method of measuring radius works well for elements other than noble gasses. Any attempt to determine the atomic radius of a noble gas with the VFI method will be skewed by the fact that noble gasses only form very very weak (and therefore very long) covalent bonds.

    Note that this was partly explained in the text at the bottom of your picture: "These data are based on interatomic distances in the structures of the elements."

    Pete's chart, and probably whatever hypervalent_iodine wanted to link to, suffer from the same problem. Although Pete's chart doesn't actually say how the radius values were calculated, I would bet money that it is also VFI radii, or some similar method that depends on bond length.

    Here is another link from the same source as your graphic:http://www.crystalmaker.com/support/tutorials/crystalmaker/atomicradii/index.html

    Go down to the bottom and look at the actual numbers for atomic radii. You'll see that the radii decrees all the as way across the row from left to right, with no discrepancy for the noble gases.

    Edit: If anyone here believes that the noble gasses really have a smaller atomic radius (by which I mean the average distance from the nucleus to the valence electrons), I defy you to provide a coherent explanation of why that would be, especially considering Slater's rules...
     
    Last edited: Mar 27, 2012
  13. Nasor Valued Senior Member

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    Don't listen to wellwisher; he has a demonstrated history of saying things in the chemistry forum that are not just wrong, but borderline insane. He appears to have little to no actual education in chemistry, and bases his understanding of chemistry on a strange mixture of the Bohr model of the atom and a highschool-level understanding of electromagnetism. Notice how it apparently doesn't occur to him that the last two electrons are hard to remove because they are the two 1s electrons...
     
  14. Robittybob1 Banned Banned

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    When you say "last two" are you implying they are the two with the highest energy requirement or are you saying they are always the last to be removed. Is it like knocking the easy ones off first, and only then the "last two", or if you were to use the right energy frequencies you can knock the "last two" off the atom to begin with?
     
  15. Nasor Valued Senior Member

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    They are definitely the two that require the most energy to remove. I don't believe that they would necessarily be the last two to be removed if you were to ionize the atom; I think you might knock a "core" electron off before a valence electron, depending on your ionization scenario, although I'm not sure how the odds of removing a 1s compares to a 2s if you're doing something like, say, bombarding the atom with hard x-rays that have enough energy to ionize any of the inner electrons.
     
  16. Robittybob1 Banned Banned

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    In the process of refreshing on ionization came across this Shielding Effect.
    So it is always easier to knock the valence shell electrons off first.
     
  17. Nasor Valued Senior Member

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    It is "easier" in terms of taking less energy, but I don't know about the relative odds of knocking off a valence electron vs. a core electron if the atom is hit by a photon that has more than enough energy to knock off either.
     
  18. Robittybob1 Banned Banned

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    It would be like body guards, more likely to hit them than the inner circle.
     
  19. chikis Registered Senior Member

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    In other to make things more clearer, define that word "octet" in precise term as used in this concept.
     
  20. Trippy ALEA IACTA EST Staff Member

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    Duet = Two.
    Quartet = Four.
    Given that an 8 sided shape is an octagon, how many do you think 'Octet' is?

    The answer is eight.

    In highschool you get taught the Octet rule. That is. Among the first twenty elements, with the exception of Hydrogen and Helium, the most electrons an element can have in its valence shell is eight.

    From the octet rule, and the aufbau principle we derive lewis dot diagrams which provide a reasonably good model for the stoichometry, but not the shape of the first twenty elements.

    When I went through Highschool we also had to learn the first twenty elements.
    Harry He Likes Beer By Cups, Not Over Frothy, Never Nasty Mugs, All Sisters Please Stop Climbing Around Kinky Caves.

    Somewhere between the end of highschool and the beginning of university, should you choose to pursue chemistry, you learn the reason for this. There are these things called orbitals. Chemists label these S, P, D, & F (there's also a hypothetical G).

    The reason that the octet rule works is because there is one S orbital and three P orbitals, with each orbital holding two electrons of opposing spin. This gives us the total of eight electrons observed in the first twenty elements.
     
  21. chikis Registered Senior Member

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    Define that word "shielding" in a more simple term as it relates to the topic on discusion.
     
  22. Nasor Valued Senior Member

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    Canceling out some of the attractive force between another electron and the nucleus.
     
  23. chikis Registered Senior Member

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    Thanks man, I think I now have a better understanding on it.
    I can also picture it this way:
    the word shielding originated from the word shield which can be defined as a broad piece of metal or another suitable material held by straps or handle attached to one side which is used as protection against missiles or blows.
    Now that the word has being applied to this subject, it can be viewed as a protection against the attraction of the outer electron to towards the nucleus. This is because, the shell or shells before the outermost electron serve as a shield, shieding (protecting) the outermost electron from the attraction of the nucleuos.
    Therefore the more shells before an outer electron, the more the shielding effect. How about that?
     
    Last edited: Apr 4, 2012

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