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Why is the electronic configuration of platinum $$\mathrm{[Xe] 4f^{14} 5d^9 6s^1}$$

and not $\mathrm{[Xe] 4f^{14} 5d^{10} 6s^0}$ or $\mathrm{[Xe] 4f^{14} 5d^8 6s^2}$?

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    $\begingroup$ Related question on Niobium… still unanswered! $\endgroup$ – F'x Dec 3 '12 at 9:31
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The Madelung energy ordering rule gives the energy of the orbitals approximately:

1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s < 5f < 6d < 7p

That would speak for [Xe] 4f14 5d8 6s2 if you follow the Aufbau principle (from German Aufbau = setup). But Pt is an exception (as there are some). A rule of thumb is that half-filled shells are stabilized. So that means in the case of Pt [Xe] 4f14 5d9 6s1.

The real answer is much more complicated. It comes from relativistic effects, electron correlation and shielding effects. There is an interplay between the attraction of nucleus and the electrons, and the electron repulsion between all electrons. The heavier atoms become, the more important relativistic effects become, since the inner electrons are moving much faster as they are in a stronger electric field from the higher charge of the nucleus. Often it is said that the outer orbitals are less compact than the inner ones, which is true if one calculates and analyses them, but a stronger effect is that outer electrons are shielded and therefore the nucleus attraction is weaker. So there is no simple way, learn the exceptions or solve the Schrödinger/Dirac equations.

Also be careful when speaking about orbitals. An orbital is a one-electron wavefunction and they are more a chemical concept than reality in multi-electron atoms.

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    $\begingroup$ If half-filled or full shells are more stable, why stop at [Xe] 6s1 4f14 5d9? Why not go all the way and have [Xe] 4f14 5d10? $\endgroup$ – user3932000 Oct 29 '16 at 20:01
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At this point in the periodic table, s (outer) and d almost have same energy. So, it is better having 4 fully filled and 2 half filled orbitals, than having 5 fully filled and 1 empty ( this empty 6s creates a problem as it is not energetically favored ), since in the first case all of them are stable whereas, in the second the empty 6s is not. The initial configuration is : [Xe] 4f14 5d8 6s2, then one electron is transfered from 6s to 5d, so that all orbitals become stable, either through full filling or half filling, which is better then having one empty and unstable. This makes it : [Xe] 4f14 5d9 6s1. It cannot be the other two because, in both of, one orbital is empty and unstable.

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    $\begingroup$ yes but empty orbitals are not unstable. Orbitals don't exist unless electrons are in them. $\endgroup$ – Ben Norris Dec 5 '12 at 11:42
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A quicker, less lengthy answer to your question:

Because it is in Group 10, Period 6 and all elements in that group are required to end with the 5d and 6s orbitals. And since it MUST use 6s, it is natural to put 1 electron in it (simply to show it exists, based on the element's placement on the periodic table). Also, because of the Aufbau principle, the other arrangements you mention simply cannot be.

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  • $\begingroup$ Why must is use the 6s? Why is is natural to put only one electron in it rather than 2 (or 0)? This doesn't really answer the question. $\endgroup$ – bon Nov 13 '15 at 14:54
  • $\begingroup$ @bon I think it must use 6s because that is the outermost electron orbital. $\endgroup$ – user3932000 Oct 29 '16 at 20:02
  • $\begingroup$ @user3932000 There are an infinite number of orbitals it's just that most of them are not occupied. There is no reason why the 6s has to be occupied like the answer is suggesting. $\endgroup$ – bon Nov 1 '16 at 12:25
  • $\begingroup$ @bon I misunderstood what the OP was saying; I thought they were saying the 5d orbital must be filled by using the 6s electrons. I agree with you that there's no reason why the atom must have the 6s orbital occupied, since it could just as easily be [Xe] 6s0 4f14 5d10. $\endgroup$ – user3932000 Nov 2 '16 at 6:33

protected by orthocresol Aug 8 '18 at 15:00

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