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?
[Physics] Reason for gold being the most malleable and ductile of metals
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The characteristic feature of a conducting medium is its presence of free electric charges on its surface, which are unbound. Unbound, free electrons do not have a restoring force and therefore, have no natural frequencies; however, they will always oscillate at the driving frequency. When an impinging EM wave oscillates these free electrons at optical frequencies, the free electrons will re-emit light with exactly the same driving frequencies. Since all of the visible frequencies are re-emitted, the metal surface looks “silvery."
The question is a lot more interesting when we address why gold or copper is colored, not silvery. Intuitively, if the free electrons are the cause of the silvery-look, then it must be the bound electrons in, say gold, which is responsible for its golden color. What else can it be? Bound electrons do have restoring forces and therefore, they have resonant frequencies. Because bound electrons in conductors behave the same as bound electrons in dielectrics, I can use this link to model the index of refraction as
$$n^2(ω) = 1 + \frac{Ne^2}{m ϵ_0}\left[\frac{-f_e}{ω^2-iβω} + \frac{f_j}{ω_{0j}^2-ω^2+iβ_jω}\right]$$
The first bracketed term is the contribution from the free electrons (note no natural frequencies $ω_0$, so no absorption) whereas the second term arises from the bound electrons (natural frequencies $ω_{0j}$ imply absorption). So for gold to appear a reddish-yellowish color (λ ≥ 500 nm), the blues and greens (λ ≤ 500 nm) must interact more with the bound electrons. I then expect gold to have larger skin depth values for λ ≤ 500 nm and smaller ones for λ ≥ 500 nm. On the other hand, I expect silver to have smaller skin depth values for all optical frequencies since there is apparently no absorption of any wavelengths. So the key is the skin depth values.
Skin Depth Values:
The skin depth $\delta$ is determined from the absorption coefficient where $\delta = 1/\alpha$. Using the website refractiveindex.info provided by Stefan Bischof’s post, I obtained $\alpha$-values and plotted $\delta$ vs. $\lambda$ for six different metals (Au, Ag, Cu, Al, Ni and Pt) using Excel. (Note that I am not showing the origin in the plot.)
Focusing on gold and silver only, the data shows exactly what I expected. Between 400 nm ≤ λ ≤ 500 nm, gold’s skin depth is increasing with longer wavelength to just about 490 nm, implying that the blues and greens are more likely to interact with the bound electrons and get absorbed. After the 500 nm peak, the skin depth steadily decreases implying that the red end of the visible spectrum is getting re-emitted by the free electrons. Silver on the other hand, has smaller δ-values (especially around 500 nm) suggesting that there is little interaction with the bound electrons and reflects all frequencies alike. Overall, the trend is that the silvery metals (Ag, Al, Ni and Pt) all have smaller skin depths while the colored metals (Au and Cu) have larger skin depths, as expected. I believe that this suggests that among other things, penetration depth will help to determine whether bound electrons are playing a role in the color of the metal.
Many thanks to Stefan Bischof for approaching the question in the manner he did.
Your confusion stems from a fundamental misunderstanding about drift velocity. Drift velocity is not the average speed of electron motion, but instead is the average velocity vector. The average speed of free electron motion in a metal can be approximated to be the Fermi speed
$$v_F = \sqrt{\frac{2E_F}{m_e}}$$
where $E_F$ is the Fermi energy. This is incredibly fast - inserting $E_F=10$ eV gives a result that is well over $1000$ km/s.
These electrons are traveling in a solid, though, which is rife with objects to collide with, including other electrons. Therefore, the mean free path of electrons in a metal (i.e. the distance an electron travels until it collides) is typically less than $1$ nm. Therefore, these electrons almost instantaneously collide with something else. A large number of these collisions would serve to essentially randomize the direction of travel of any given electron. When you add a bunch of uniformly-randomly-distributed vectors of roughly equal length together, the resultant is essentially zero, regardless of the actual length of the vectors you added. Therefore, the average velocity vector of an electron should be close to zero, and certainly should be much smaller than its average speed, since its velocity is pointed in an essentially random direction.
When an electric field is applied to a metal, it accelerates electrons in a certain direction, and therefore alters the probability distribution of electron velocity. Velocities in the direction of the field become less probable, and velocities against the direction of the field become more probable. The longer the electric field is allowed to act on a freely-moving electron, the more this probability distribution is distorted. But, as was previously discussed, the time between collisions is quite small due to the density of metals. This means that the electric field can only alter the velocity distribution slightly, which shifts the average velocity vector (i.e. the drift velocity) slightly away from zero.
Another misunderstanding arises from the false assumption that the speed of an electrical signal in a metal is equal to the either the Fermi speed or the drift velocity. In reality, it is unrelated to either of those things. Instead, for conductors the speed of an electrical signal is given by the group velocity of an electromagnetic wave:
$$v_g=\frac{d\omega}{dk}$$
where $\omega(k)$ is the dispersion relation, and is in general derived from the band structure of the material in question. For a good (i.e. close-to-ideal) conductor, the dispersion relation is
$$\omega(k)=\frac{2k^2}{\mu\sigma}$$
for a material with conductivity $\sigma$ and permeability $\mu$. Then the group velocity is
$$v_g=\sqrt{\frac{8\omega}{\mu\sigma}}$$
which, for copper, with $\sigma= 5.96\times 10^7$ S/m and $\mu\approx\mu_0=4\pi\times 10^{-7}$ H/m, and for a plane wave with frequency $1$ GHz, the group velocity is roughly $25$ km/s, and increases with increasing frequency.
EDIT:
The conductivity of a material $\sigma$ is defined by
$$\mathbf{J}=\sigma \mathbf{E}$$
for current density $\mathbf{J}$ and applied electric field $\mathbf{E}$. This essentially means that it's the average number of electrons passing through a unit area per unit time, per unit applied electric field. The higher the conductivity, the less electric field it takes to get electrons to flow. One simple model (specifically, the Drude model) based on similar arguments as above finds that for a material with electron density $n$ and mean time between collisions $\tau$, for DC currents one has
$$\sigma = \frac{ne^2\tau}{m_e}$$
Resistivity $(\rho)$ is defined as the inverse of conductivity. Therefore, again from the Drude model for DC currents, one has
$$\rho = \frac{m_e}{ne^2\tau}.$$
Best Answer
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