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Electron Configuration

  1. Jun 4, 2013 #1
    In the "d" orbital of an atom, if it's almost half filled or filled it will get an electron from the nearest "s" orbital to gain a better stability.
    My question is why not "p" orbital doesn't do such?

    ns2 (n-1)d9 → ns1 (n-1)d10

    ns2 np5 -/→ ns1 np6 ???:cry:
     
  2. jcsd
  3. Jun 5, 2013 #2

    Borek

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    Staff: Mentor

    These s/d pairs are much closer in the energy level than possible p/d pairs.
     
  4. Jun 5, 2013 #3
    well my problem is why don't p orbitals drag s level orbitals? Is that because s-d distance is lesser than s-p distance?
     
  5. Jun 6, 2013 #4

    Borek

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    Honestly, I am not sure. It is either that the the energy gain of having the p filled up is smaller than having the d filled up, or that the energy difference between ns/(n-1)d is smaller than the energy difference between s/p.

    Note that these are very hand wavy explanations, and they lack rigor - which means you can bend them to fit whatever you want them to be. I hate putting too much meaning to such models, you would be just deluding yourself you know what is going on. There are definitely better answers, but they require much more rigorous models (QM).
     
  6. Jun 7, 2013 #5
    Are there no theories behind that abnormal d filling? :confused:

    Having said that, even my school teachers didn't say anything about how it happens. They just taught that Cr and Cu has a rather different electron configuration. :grumpy:
     
  7. Jun 7, 2013 #6

    Borek

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    They say there is an energy gain connected with half/full filled d, so 3d54s1 is preferred over 3d44s2 and 3d104s1 is preferred over 3d94s2. But this is hand wavy.
     
  8. Jun 8, 2013 #7
    Borek is correct, the d orbital is close in energy to the s orbital so when an electro jumps from s to d, the energy is much less than jumping from a p or f to a d. They are close enough to allow this.
     
  9. Jun 9, 2013 #8
    So there will be stability only when a s electron jumps to d, not p! And the inter distance is the reason. :D Thank you both of you!!! I think I can understand now.
     
  10. Jun 10, 2013 #9
    Yes, I found this guy's explanation very accurate. Check it out.

    'At the beginning of the first transition series (scandium), the 4s orbital is lower in energy than the 3d orbitals. Thus Sc has the configuration [Ar] 4s2 3d1. This continues as far as V which has the configuration [Ar] 4s2 3d3. We might expect Cr to have the configuration [Ar] 4s2 3d4 but, due to the shape of the 3d orbitals and the increasing nuclear charge as we go across the series, the 3d orbitals are now closer in energy to 4s than they were at Sc. The energy difference is so small, in fact, that it is smaller than the 'pairing' energy to put two electrons in the 4s orbital. Consequently the configuration [Ar] 4s1 3d5 is lower in energy than [Ar] 4s2 3d4. For Mn, the next electron goes into the 4s orbital as a paired electron in 4s is still lower in energy than a paired electron in 3d. As we go further to the right, the 3d orbitals continue to fill up but, by the time we get to Cu, the energy of 3d is now lower than 4s, so [Ar] 4s1 3d10 is preferred to [Ar] 4s2 3d9.

    So you see that the unusual electronic configurations of Cr and Cu are a consequence of the increasing stability of the 3d subshell with respect to the 4s subshall as you go across the series due to the increasing nuclear charge. It is interesting to note that in ALL transition metal cations, because of the increased effective nuclear charge in the ion, the 3d subshell is always lower in energy than 4s. This is why 4s electrons are always lost first on ionisation of transition metals'
     
  11. Jun 11, 2013 #10
    Yea and from this phrase I think when the configuration is around 4s2 3d4, the atmosphere there, around the atom behaves like the boiling point where the water molecules easily moves away. I mean, just like that, an equilibrium. :D
    And thanks again for this one. It's worthy!
     
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