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I understand how fcc structure enables plastic deformation in metals, but why is gold, in particular, the most malleable and ductile of fcc metals? Is there something about the electronic structure of gold (such as # of free electrons per atom, density of states in the conduction band, etc) that enable gold to be "soft" under compressive forces, yet strongly cohesive under tension, more so than any of the other fcc metals?

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  • $\begingroup$ Physical Ductility of the Elements $\endgroup$ – StephenG Sep 2 '18 at 5:51
  • $\begingroup$ It is the heaviest physically and chemically stable element with a single s-electron on the top orbit (and all lower energy orbits complete). $\endgroup$ – safesphere Sep 2 '18 at 7:02
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There are several reasons: The comments already mention the electronic structure described in https://www.failurecriteria.com/physicalductilit.html

But there is another one: the high purity of Au that is readily available. Since impurities hinder dislocation movement they cause embrittlement. With the Ellingham diagram a special property of Au can be understood: Notice the red circle wich highlights the position auf Au2O3 which is already above 0 kJ/mole at 0 K.

Since the dissociation of Au2O3 into Au + O2 leads to an increase of entropy the slope of the Au curve is positive in this diagram like for the other metals. This means that Au will never reach the negative regime and therefore not form oxides which sets Au apart from all other metals. This is the reason that gold is found in the form of nuggets and not as ore where it would mix with other metal oxides. Therefore gold is a very soft material (not just in theory) that can be hammered into thin sheets as thin as 100 nm thickness

Ellingham diagram

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