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Why do noble gas electron configurations have large radii?

  1. Sep 24, 2015 #1
    In my lecture, we were told that a nitrogen with a negative 3 charge has the largest radius compared to most of the other atoms in the same period. How is that possible? It has more protons attracting the valence electrons closer to the center, but the prof said that because three electrons are added, an exception occurs. Why does that exception occur?
     
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  3. Sep 25, 2015 #2

    DrDu

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    First you should note that an isolated nitrogen ion with charge -3 does not exist. Rather, these ionic radii refer to the average distance of ions in compounds like nitrides and there the atoms resemble very little free ions in the gas phase.
     
  4. Sep 25, 2015 #3
    We were told that once a certain atom in a period reaches the electron configuration of a noble gas, its radius is increased in size. How is that possible?
     
  5. Sep 25, 2015 #4
    I am not sure what you mean with an exception happening, but nitrogen with a -3 charge (regardless of whether it exists or not) has the largest radius. With that said, it's difficult to look at ionic radii as they are difficult to measure and the ionic radii varies with the co-ordination of the ion and which ions are surrounding it. As such there are some inconsistencies. The general trend for atomic radii (increases down a group, decreases across a period) and ionic radii (increases down a group, varies) can be explained if you look at the electronic structure of the atom/ion.

    For trends in atomic/ionic radii down a group, the atoms get bigger because an extra layer of electrons is added. Compare the electric structure between nitrogen and phosphorus for example:
    Nitrogen: 1s2, 2s2, 2p3
    phosphorus: 1s2, 2s2, 2p6, 3s2, 3p3; phosphorus has an entire n=2 energy level over nitrogen providing electron shielding.

    For trends in atomic radii across a period, the size of the atoms decreases. The number of protons in the nucleus increase across the period; the increased in # of protons increases nuclear attraction for the electrons, so the electrons are pulled in more tightly. The electron shielding is the same for the atoms across a period, so the only factor affecting atom size is the number of protons.

    The trend for ionic radii across a period can be explained similarly. You just have to look at the positive and negative ions separately. The number of protons increases across the period. This tends to pull the electrons more towards the center of the atom, thus decreasing ionic radii. This is true for both negative and positive ions. However, the negative ions in the same period generally have an extra layer of electrons (so there is a big jump from, say, Al3+ to P3-, in ionic radii).

    As for noble gases, you generally want to ignore those. As DrDru said, atomic radii refers to the average distance of ions in compounds, and noble gases like neon and argon don't form bonds, so their atomic radii is more difficult to measure (it's based on their van der Waals radius).
     
  6. Sep 26, 2015 #5

    Borek

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    No, nitrogen with a -4 charge has a larger radius (and doesn't exist just like nitrogen with a -3 charge).
     
  7. Sep 27, 2015 #6

    James Pelezo

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    Generally, #p+ < #e- => electron-electron repulsion at the valence level => larger radius. In the isoelectric series, 7N(3-) > 8O(2-) > 9F(-) the #p+ the number of protons increase in series without changing the electron configurations. The increase in proton numbers functions to provide a stronger electrostatic attraction and reduces the ionic radius. The values given depend upon the analytical methods used which typically is related to measurement while the element is bonded in a structure. The exception is the Noble Gas radii which is based upon the Van der Waals radii. It is not sensible to compare radii of Noble Gas elements to radii of bonded elements. Neon radii are 0.154 - 0.160 nm depending on the source reference which is larger than the measured F(-) radii in a bonded system.

    Trends in ionic radius for some more isoelectronic ions
    upload_2015-9-27_12-40-45.png
     

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  8. Sep 27, 2015 #7

    Borek

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    Definitely, but it also means that the radius of 6B3- is even larger. So, why is the 7N3- listed as having

     
  9. Sep 27, 2015 #8

    James Pelezo

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    I'm confused with your notation 6B3-. Shouldn't this be 6C4-? Boron as I understand has only 5 protons and carries a valence of 3 electrons and would most likely assume a +3 oxidation state giving the notation 5B3+. As a metalloid, it may form compounds where 5B5- functions as an anion, but I've no knowledge of such a configuration. If you know of some, I would be delighted to know about them. As far as the ionic radii question goes, I would concede 6C4- would have a radius greater than the nitride ion as nitrogen has more positive charge than carbon. Both would tend to gain electrons by path of least resistance (octet rule) to achieve a noble gas configuration during bonding. As for why 7N3- is touted as the largest anion, I have no idea why. I can only speculate that when the subject of periodic trends in atomic and ionic radii are presented, most texts present Li, Be, B as cations with decreasing radii, completely skip over carbon and begin anion trends with N followed by O, F & Ne. You are right in noting that 6C4- (if that is what you are suggesting) does have a larger ionic radius than 7N3-.
     
  10. Sep 28, 2015 #9

    Borek

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    Sorry, my mistake. Call it a senior moment :wink:

    Yes, that's what I was aiming at. For me question as asked is unclear and based on a statement that is either wrong, or incomplete.
     
  11. Sep 28, 2015 #10

    James Pelezo

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    I would be guessing, but maybe the 'exception' is Nitrogen marks that element in the periodic trend when the tendency to gain electrons is more frequent, i.e., easier to gain 3 electrons) than lose 5 electrons to achieve a Noble Gas configuration. Again, this may be one of those cases when after covering Li, Be, B, as cations, Carbon is ignored and Nitrogen is presented as the 1st element with an exclusive tendency to gain electrons to achieve an octet. In doing so, the electron-electron repulsion without an increase in atomic number would result in a larger ionic radius than O,F and Ne. I would have bridged that trend with a note on CO2 versus CH4. Carbon in CO2 has a 4+ oxidation state and would have followed the decrease in cation radii trend, vs Methane (CH4 ) in which Carbon carries a 4- oxidation state and results in a larger radius than either 6C4+ and 7N3- . After this, molecules containing the remaining elements in that series would tend to gain electrons, present a large radius during bonding with the trend of decreasing radius following 7N3-, but as noted earlier, this is speculation as to what and how this topic is presented. Interesting note though.
     
  12. Sep 28, 2015 #11

    DrDu

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    The real point is that there isn't anything like N##^{3-}##. At best, there is something like nitrogen with an oxidation number -III, but this is a completely formal concept.
     
  13. Sep 28, 2015 #12

    James Pelezo

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    Exactly ... It is simple, it helps define reactivity of elements and leads well into bonding concepts.
     
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