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Configuration of Cr

  1. Aug 13, 2015 #1

    Suraj M

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    We all know that the configuration of Cr is [Ar]4s¹3d⁵
    now is this the Ground state or the excited state of Chromium ?
     
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  3. Aug 13, 2015 #2

    blue_leaf77

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    That electronic configuration is build on the basis of lower energy filled first. Therefore that must be the ground state.
     
  4. Aug 13, 2015 #3

    Suraj M

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    but 4s is with lower energy than 3d so it should be filled first so 4s²3d⁴ would be the ground state?
     
  5. Aug 13, 2015 #4

    blue_leaf77

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    The real thing for many electron system is not as simple as it is for one electron system where the energy levels is not dependent on the quantum numbers other than the principal quantum number. The other quantum numbers such as spin and orbital angular momentum begin to influence the energy level because of the presence of screening of the other electrons. For example, oribtal with angular momentum ##l=2## (oribtal d) has a wavefunction such that the peak occurs at relatively large distance from the nucleus, therefore the electron in this orbital will feel more screening of the nucleus charge by the other electrons and consequently higher in energy. On the contrary, electron sitting in s orbital has a wavefunction which peaks closer to the nucleus causing it to be bound more strongly. Furthermore, spin direction can also affect the energy due Pauli exclusion principle. Long story short, many electron system is so complicated that up to now only approximate solution have been proposed and put to practice.
     
  6. Aug 13, 2015 #5

    Suraj M

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    Ok another argument i can come up with is, the elements in the d block have electrons being added into the d orbital of the penultimate shell, but then if we say that 4s¹3d⁵ is the ground state config. of Cr then to get Mn there is an electron entering the 4s orbital, isn't this contradicting to the fact that Mn is on the d block. I'm sorry if I'm wrong, please point out any misconception i have, Thank you
     
  7. Aug 13, 2015 #6

    DrClaude

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  8. Aug 13, 2015 #7

    Suraj M

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    I'm not really aware of all the terms mentioned on that website.
     
  9. Aug 13, 2015 #8

    DrClaude

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    You don't need to understand everything... yet! :wink:
     
  10. Aug 13, 2015 #9

    Suraj M

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    Could you just tell me what that 'level' is?
     
  11. Aug 13, 2015 #10

    blue_leaf77

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    Those are the so-called "coupling term" business. This comes from the fact that individual orbital and spin angular momentum operator do not commute with the total Hamiltonian, and hence cannot be used as the specification for states. Instead it's the total orbital and total spin angular momenta which commute with the Hamiltonian, this gives rise to the so-called term (2nd column in that link). In this coupling term, each shell model configuration, e.g. 4s¹3d⁵ are split into further levels/terms with different energies, two of them are 4s¹3d⁵ (7S) and 4s¹3d⁵(5S) which correspond to the first and second row in that table. The term belonging to 4s²3d⁴ occurs for the first time in the 3rd row, which is 4s²3d⁴ (5D), whose energy is higher than the ground state. That's what DrClaude meant in comment #6.
     
  12. Aug 13, 2015 #11

    Suraj M

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    could you tell where I'm wrong in the post #5. about what?
     
  13. Aug 14, 2015 #12

    blue_leaf77

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    I am not sure if I get what you meant in #5. The configuration of Mn is 4s23d⁵, so it's still belonging to d block, what's wrong with it.
     
  14. Aug 14, 2015 #13

    Suraj M

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    What I'm trying to say is that, as Mn has a configuration 4s²3d⁵ and comparing this to the configuration of the previous element Cr-4s¹3d⁵, its like saying that the differentiating electron entered the 4s orbital. Could I interpret it like that?
     
  15. Aug 14, 2015 #14

    DrClaude

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    Not really, because you're not simply adding an electron when going from Cr to Mn, you are also changing the nuclear charge. See for instance my posts in this previous thread.
     
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