# Anomalous electronic configuration of platinum

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 $\mathrm{[Xe] 4f^{14} 5d^8 6s^2}$?

• Related question on Niobium… still unanswered! – F'x Dec 3 '12 at 9:31

The Madelung energy ordering rule gives the energy of the orbitals approximately:

$$\mathrm{1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s < 5f < 6d < 7p}$$

That would speak for $$\mathrm{[Xe] 4f^{14} 5d^{8} 6s^{2}}$$ if you follow the Aufbau principle (from German Aufbau = setup). But $$\ce{Pt}$$ 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 $$\ce{Pt}$$, $$\mathrm{[Xe] 4f^{14} 5d^9 6s^1}$$.

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.

• 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? – user3932000 Oct 29 '16 at 20:01

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.

• yes but empty orbitals are not unstable. Orbitals don't exist unless electrons are in them. – Ben Norris Dec 5 '12 at 11:42