Violations to Aufbau principle in ground states

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Violations to Aufbau principle in ground states

Postby Ender84 » Sat Feb 23, 2013 6:34 pm UTC

Hi! Sorry if this seems like a basic question, but it was extra credit on my chem honors test and I'm curious as to the answer. The question stated: Platinum, Gold, and Mercury all violate the Aufbau principle in their ground states. However, Thallium, Lead, and Bismuth do not. Try to explain this.

(I realize that this is because it makes the platinum, gold, and mercury most stable, but I have no idea as to the actual explanation.)
Thanks for the help! :D

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Re: Violations to Aufbau principle in ground states

Postby algorerhythms » Mon Feb 25, 2013 10:21 pm UTC

It's been a long time since I've studied the Aufbau Principle, so I may be wrong, but I seem to remember it having to do with states where the D orbital is completely filled being more stable, so in a few elements such as gold and mercury, one or two electrons that according to the Aufbau Principle would go to other orbitals end up in the D orbital.
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Re: Violations to Aufbau principle in ground states

Postby Jorpho » Tue Feb 26, 2013 5:20 am UTC

What we need for starters is ... ta_page%29 .
78 Pt platinum : [Xe] 4f14 5d9 6s1
79 Au gold : [Xe] 4f14 5d10 6s1
80 Hg mercury : [Xe] 4f14 5d10 6s2
81 Tl thallium : [Xe] 4f14 5d10 6s2 6p1
82 Pb lead : [Xe] 4f14 5d10 6s2 6p2
83 Bi bismuth : [Xe] 4f14 5d10 6s2 6p3
It doesn't make any kind of sense as far as I can see to say that mercury violates the Aufbau principle. Sounds just plain wrong to me: all the orbitals are filled. ... figuration says that the 5d orbitals and the 6s orbitals are roughly equivalent in energy, and that there is greater stability in gold to be gained from having the d orbitals filled than there is in having the 6s orbital filled. And apparently for platinum there is also a net gain even if only one of the d orbitals is half-filled.

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Re: Violations to Aufbau principle in ground states

Postby stianhat » Tue Feb 26, 2013 10:50 am UTC

The 6s subshell is more stable because of the 4f subshell being useless at screening the nuclear charge. If you take a look at the geometry of the f-shells this is quite apparent.

Thus, the 6s shell, when filled with electrons, is more stable, because the electrostatic force from the nucleus makes the 6s shell a more comfortable place for negatively charged electrons. You will also see this when ionizing the heavy and noble metals, the 6s electrons are usually not the first ones to go.

At higher atomic numbers the relativistic effects that arise from the size of the shells coupled with the speed of electron movement dominates contraction for all shells so the effect of couloumbic contraction dissipates.

I think.

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