DRAFT Gold, Liquid, and Soft: Relativistic Chemistry, Band Structure, and Conduction

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elements

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science

Why is gold gold, mercury liquid and lead soft?

Author

Stephen J. Mildenhall

Published

2025-08-18

Modified

2025-08-20

A Quick Tour Before the Math

Gold is yellow, copper is reddish but most metals are silvery. Mercury is a liquid and lead is soft but most metals are hard. These facts are familiar, but the reasons sit at the intersection of quantum mechanics and relativity. Heavy nuclei drive relativistic changes to the radial shapes and energies of atomic orbitals; when those atoms form a solid, the altered orbital energies shape the bands near the Fermi level, which sets both the optical response (color) and how easily electrons carry charge and heat.

[FIGURE: Ionization energies for Cu, Ag, Au (from the Mendeleev code). Place near here.]

Valence electron

In an isolated atom, electrons outside the filled core (e.g., Cu: 4s¹; Au: 6s¹, with a filled d shell just below). In a solid, these orbitals hybridize into bands that participate in bonding and transport.

Valence band (electron sea)

In a metal, overlapping atomic valence orbitals form delocalized bands. The loosely bound, itinerant electrons are often described as an “electron sea”.

Delocalization

In a crystal, valence electrons are not tied to individual atoms. Their wavefunctions spread over many nuclei, forming extended Bloch states. This spreading — delocalization — is what allows electrons to move freely and carry current in metals.

Why Most Metals Are Silvery—and Why Copper and Gold Are Not

At $T=0$, the highest occupied single-particle energy in a solid is called the Fermi level. At finite $T$, it is the energy $E_F$ where the occupation probability is 50%.

Most metals look silvery because there are plenty of electronic states available at energies just above and just below the Fermi level, so the surface reflects essentially the entire visible spectrum with little wavelength preference.

The occupation of states at temperature $T$ follow the Fermi–Dirac distribution \[ f(E) = \frac{1}{1 + e^{(E - E_F)/(k_B T)}}. \]

Density of States (DOS)

$g(E)$ counts how many single-particle states per unit energy exist at energy $E$. The number of electrons near $E$ is $n(E) = g(E)\,f(E)$.

For silver (Ag) the filled $4d$ band lies well below the $5s$ conduction band. Interband transitions from $4d \to 5s$ require ultraviolet photons, so the visible range is reflected — silver appears colorless/shiny.

For copper (Cu) and gold (Au), the top of the filled $d$ band sits much closer to the $s$-derived conduction band. Visible photons can promote electrons across this smaller separation, leading to selective absorption:

copper absorbs blue–green ($\sim 2.1$ eV) and reflects orange–red;
gold absorbs blue ($\sim 2.3$ eV) and reflects yellow–red.

Interband Transition

An optical excitation moving an electron from a filled band (e.g., $d$) to an empty/partially filled band (e.g., $s$/conduction). If the required energy falls in the visible, the material appears colored.

[FIGURE: Place a schematic band diagram for Cu/Ag/Au. d-band near but below $E_F$ for Cu/Au; d-band deeper (UV transition) for Ag.]

Where Relativity Enters: s Contraction, d/f Expansion

The crucial difference for heavy elements is relativity. In atoms with large nuclear charge $Z$, inner electrons move at a significant fraction of $c$. The correct one-electron description uses the Dirac equation, which alters the radial part of the wavefunctions.

orbitals with nonzero probability density at the nucleus (s, and relativistic $p_{1/2}$) are stabilized and contract inward;
to remain orthogonal, other orbitals (d, f) are pushed outward and effectively raised in energy.

This s-contraction / d,f-expansion is not a separate force on d/f; it is the knock-on consequence of orthogonality once s has moved inward.

Relativistic Contraction

For heavy $Z$, Dirac solutions increase contact density for s (and $p_{1/2}$) orbitals, pulling their radial probability inward and lowering their energy. Orthogonality then pushes d/f outward.

Orthogonality

Quantum wavefunctions for different orbitals must be mathematically orthogonal: \[ \int \psi_i^*(\mathbf{r})\,\psi_j(\mathbf{r})\,d^3r = 0 \quad (i\ne j). \] If an s orbital contracts inward, it “occupies” more of the inner radial region. To remain orthogonal, d and f orbitals shift outward producing the compensating expansion effect.

[FIGURE: Radial functions for hydrogenic $R_{n0}$ and $u=rR$ (nonrelativistic), plus the Dirac near-origin schematic for a heavy $Z$ (from the provided code). Place here.]

Applied to gold (Z = 79): the $6s$ orbital is pulled down (stabilized), and the $5d$ orbitals are pushed up (destabilized) relative to a nonrelativistic picture. The result is a $5d \to 6s$ interband threshold in the visible (blue), which gold correspondingly absorbs.

By contrast, silver (Z = 47) is not heavy enough: its $4d$ remains sufficiently low that $4d \to 5s$ lies in the ultraviolet — hence silver’s broadly flat, silvery reflectance.

Copper (Z = 29) is lighter still, yet the $3d$–$4s$ separation is already small even nonrelativistically, so the absorption threshold falls in the blue–green, giving copper its color.

Mercury is a Liquid, Lead Is Soft: The Inert Pair Effect

Mercury (Z = 80) has configuration $[\text{Xe}]\,4f^{14} 5d^{10} 6s^{2}$. The relativistic contraction of $6s^2$ is extreme — those electrons sit tight to the nucleus and are reluctant to delocalize. Metallic bonding in elemental Hg is therefore weak, and the melting point plunges below room temperature: mercury is a liquid.

Lead (Z = 82) has $[\text{Xe}]\,4f^{14} 5d^{10} 6s^{2} 6p^{2}$. The $6s^2$ pair is likewise stabilized and relatively inert, leaving $6p^2$ to do most of the bonding. Fewer electrons participate, bonds are weaker, and the metal is soft with a low melting point relative to “typical” strong-bond metals.

Inert Pair Effect

In heavy p-block elements (Tl, Pb, Bi), the outer s² pair (e.g., 6s²) becomes chemically and metallurgically less available due to relativistic stabilization. Consequences: preferred lower oxidation states (e.g., Pb²⁺) and weaker metallic bonding (soft Pb, liquid Hg).

What “Conduction” Means: Electrical and Thermal

Electrical conduction in a metal is governed by the electrons within a few $k_B T$ of $E_F$. In a band picture, electrons occupy Bloch states with dispersion $E(\mathbf{k})$. The group velocity \[ \mathbf{v}(\mathbf{k}) = \frac{1}{\hbar}\,\nabla_{\mathbf{k}} E(\mathbf{k}) \] is the quantum analog of “speed.” An electric field shifts the distribution in $\mathbf{k}$-space via \[ \hbar\,\frac{d\mathbf{k}}{dt} = -e\,\mathbf{E}, \] changing the expectation value of velocity—and a current flows.

Bloch State

An electron state in a periodic crystal with wavefunction $\psi_{\mathbf{k}}(\mathbf{r})=u_{\mathbf{k}}(\mathbf{r})e^{i\mathbf{k}\cdot\mathbf{r}}$. Transport properties follow from $E(\mathbf{k})$.

Thermal conduction in metals is carried mostly by the same electrons (with a lattice/phonon contribution that is usually smaller in good metals). A temperature gradient broadens the energy distribution on the hot side; energetic electrons diffuse and carry heat. This shared role leads to the Wiedemann–Franz law: \[ \frac{\kappa}{\sigma T} \approx L, \] with the Lorenz number $L \approx 2.44\times 10^{-8}\,\text{W}\,\Omega\,\text{K}^{-2}$ in many simple metals.

Good Metal

A material where conduction is dominated by nearly free electrons with long mean free paths. Examples: copper, silver, gold, aluminum. Good metals obey the Wiedemann–Franz law closely because the same mobile electrons efficiently carry both charge and heat.

Phonon

A quantized lattice vibration. In insulators and semiconductors, phonons dominate thermal conductivity because there are no itinerant electrons.

Wiedemann–Franz Law

Empirical (and derivable in simple models): in metals, the ratio of thermal to electrical conductivity scales with temperature, indicating common carriers—electrons—transport both heat and charge.

What Changes in Compounds—and Why Most Compound Colors are Nonrelativistic

The metallic color story turns on d→s interband transitions near $E_F$. In compounds, electron clouds are reshaped by bonding and geometry, introducing new color mechanisms:

crystal-field splitting of d orbitals (geometry-dependent),
charge-transfer transitions (ligand ↔︎ metal),
band-gap-controlled absorption in extended solids (semiconductors, pigments),
spin–orbit effects growing important for heavy elements.

Most compound colors can be understood without leaning on relativity; Coulomb interactions and symmetry do the bulk of the work. Relativity resurfaces for heavy-element complexes (Au, Pt, Ir) via strong spin–orbit coupling and shifted orbital baselines.

Ligand (n)

Each of the atoms or groups attached to the central (usually the metal) atom of a co-ordination complex.

Crystal-Field Splitting

In a ligand field, metal d orbitals split into groups (e.g., $t_{2g}$, $e_g$) with separations often in the visible, producing vivid colors in transition-metal compounds.

Unifying Quantitative Thread

Atomic scale (Dirac picture): Relativity modifies the radial behavior: s (and $p_{1/2}$) contract; d/f expand by orthogonality. Net effect: energy of s decreases; energy of d/f increases.
Solid scale (band picture): Those atomic shifts feed into band energies: in Au, $5d$ rises and $6s$ falls relative to a nonrelativistic baseline—shrinking the interband threshold into the visible; in Hg, $6s^2$ is so stabilized that metallic bonding weakens drastically.
Transport and optics: Optical absorption probes interband separations; dc transport probes states and velocities near $E_F$. The same band structure that sets Au’s absorption edge also sets its conduction properties. In good metals, electrons carry both charge and heat, consistent with Wiedemann–Franz.

[FIGURE: Place a simple schematic showing d–s separation across Cu → Ag → Au → Hg, with the visible/UV threshold annotated.]

“Speeding Up” and Wavefunctions

It is natural to worry about talking of electrons “speeding up” when we model them with delocalized wavefunctions. The precise statement is: an external field shifts the occupation of Bloch states in $\mathbf{k}$-space, changing the expectation of group velocity. No classical trajectories are required; transport is an ensemble statement built from $E(\mathbf{k})$ and $f(E)$.

Explanation vs Prediction

We can now explain why gold is yellow, copper is red, mercury is liquid, and lead is soft. Could we have predicted these phenomena from first principles without data? In principle, yes: the Dirac equation plus electron–electron interactions plus crystal structure suffices. In practice, multi-electron relativistic problems require approximations (Hartree–Fock, DFT, beyond-DFT), and even modern computations lean on experiment for calibration and guidance. Our knowledge here is a bootstrapped partnership of observation and theory.

Figures to Insert

[FIGURE: Ionization energies Cu, Ag, Au] Use the Mendeleev script you ran; place the PNG here.
[FIGURE: Hydrogenic $R_{n0}$ vs $u=rR$ (nonrelativistic)] From the provided script R_u_nonrel.png.
[FIGURE: Dirac near-origin schematic for s-state, heavy Z] From the provided script dirac_schematic.png.
[FIGURE: Schematic band alignment (Cu/Ag/Au/Hg)] Simple diagram (you can sketch or generate) showing d-band position vs s-band and the visible vs UV threshold.

Minimal References

Bransden & Joachain, Physics of Atoms and Molecules — hydrogenic orbitals, fine structure, multi-electron atoms.
Ashcroft & Mermin, Solid State Physics — bands, DOS, transport, optical properties of metals.
Grant, Relativistic Quantum Theory of Atoms and Molecules — Dirac methods, relativistic orbital effects in heavy elements.

--- author: Stephen J. Mildenhall title: "DRAFT Gold, Liquid, and Soft: Relativistic Chemistry, Band Structure, and Conduction" description: 'Why is gold gold, mercury liquid and lead soft?' categories: - notes - elements - llm - science date: '2025-08-18' date-modified: last-modified draft: false toc: true toc-depth: 2 format: html: code-tools: true code-fold: false code-copy: true image: img/banner.png --- ![](img/banner.png){width=50%} # A Quick Tour Before the Math Gold is yellow, copper is reddish but most metals are silvery. Mercury is a liquid and lead is soft but most metals are hard. These facts are familiar, but the reasons sit at the intersection of quantum mechanics and relativity. Heavy nuclei drive relativistic changes to the radial shapes and energies of atomic orbitals; when those atoms form a solid, the altered orbital energies shape the bands near the Fermi level, which sets both the optical response (color) and how easily electrons carry charge and heat. [FIGURE: Ionization energies for Cu, Ag, Au (from the Mendeleev code). Place near here.] ::: {.callout-note} ## Valence electron In an isolated atom, electrons outside the filled core (e.g., Cu: 4s¹; Au: 6s¹, with a filled d shell just below). In a solid, these orbitals hybridize into bands that participate in bonding and transport. ::: ::: {.callout-note} ## Valence band (electron sea) In a metal, overlapping atomic valence orbitals form delocalized bands. The loosely bound, itinerant electrons are often described as an "electron sea". ::: ::: {.callout-note} ## Delocalization In a crystal, valence electrons are not tied to individual atoms. Their wavefunctions spread over many nuclei, forming extended Bloch states. This spreading — delocalization — is what allows electrons to move freely and carry current in metals. ::: # Why Most Metals Are Silvery—and Why Copper and Gold Are Not At $T=0$, the highest occupied single-particle energy in a solid is called the Fermi level. At finite $T$, it is the energy $E_F$ where the occupation probability is 50%. Most metals look silvery because there are plenty of electronic states available at energies just above and just below the Fermi level, so the surface reflects essentially the entire visible spectrum with little wavelength preference. The occupation of states at temperature $T$ follow the Fermi–Dirac distribution $$ f(E) = \frac{1}{1 + e^{(E - E_F)/(k_B T)}}. $$ ::: {.callout-note} ## Density of States (DOS) $g(E)$ counts how many single-particle states per unit energy exist at energy $E$. The number of electrons near $E$ is $n(E) = g(E)\,f(E)$. ::: For silver (Ag) the filled $4d$ band lies well below the $5s$ conduction band. Interband transitions from $4d \to 5s$ require ultraviolet photons, so the visible range is reflected — silver appears colorless/shiny. For copper (Cu) and gold (Au), the top of the filled $d$ band sits much closer to the $s$-derived conduction band. Visible photons can promote electrons across this smaller separation, leading to selective absorption: - copper absorbs blue–green ($\sim 2.1$ eV) and reflects orange–red; - gold absorbs blue ($\sim 2.3$ eV) and reflects yellow–red. ::: {.callout-note} ## Interband Transition An optical excitation moving an electron from a filled band (e.g., $d$) to an empty/partially filled band (e.g., $s$/conduction). If the required energy falls in the visible, the material appears colored. ::: [FIGURE: Place a schematic band diagram for Cu/Ag/Au. d-band near but below $E_F$ for Cu/Au; d-band deeper (UV transition) for Ag.] # Where Relativity Enters: s Contraction, d/f Expansion The crucial difference for heavy elements is relativity. In atoms with large nuclear charge $Z$, inner electrons move at a significant fraction of $c$. The correct one-electron description uses the Dirac equation, which alters the radial part of the wavefunctions. - orbitals with nonzero probability density at the nucleus (s, and relativistic $p_{1/2}$) are stabilized and contract inward; - to remain orthogonal, other orbitals (d, f) are pushed outward and effectively raised in energy. This s-contraction / d,f-expansion is not a separate force on d/f; it is the knock-on consequence of orthogonality once s has moved inward. ::: {.callout-note} ## Relativistic Contraction For heavy $Z$, Dirac solutions increase contact density for s (and $p_{1/2}$) orbitals, pulling their radial probability inward and lowering their energy. Orthogonality then pushes d/f outward. ::: ::: {.callout-note} ## Orthogonality Quantum wavefunctions for different orbitals must be mathematically orthogonal: $$ \int \psi_i^*(\mathbf{r})\,\psi_j(\mathbf{r})\,d^3r = 0 \quad (i\ne j). $$ If an s orbital contracts inward, it “occupies” more of the inner radial region. To remain orthogonal, d and f orbitals shift outward producing the compensating expansion effect. ::: [FIGURE: Radial functions for hydrogenic $R_{n0}$ and $u=rR$ (nonrelativistic), plus the Dirac near-origin schematic for a heavy $Z$ (from the provided code). Place here.] Applied to gold (Z = 79): the $6s$ orbital is pulled down (stabilized), and the $5d$ orbitals are pushed up (destabilized) relative to a nonrelativistic picture. The result is a $5d \to 6s$ interband threshold in the visible (blue), which gold correspondingly absorbs. By contrast, silver (Z = 47) is not heavy enough: its $4d$ remains sufficiently low that $4d \to 5s$ lies in the ultraviolet — hence silver’s broadly flat, silvery reflectance. Copper (Z = 29) is lighter still, yet the $3d$–$4s$ separation is already small even nonrelativistically, so the absorption threshold falls in the blue–green, giving copper its color. # Mercury is a Liquid, Lead Is Soft: The Inert Pair Effect Mercury (Z = 80) has configuration $[\text{Xe}]\,4f^{14} 5d^{10} 6s^{2}$. The relativistic contraction of $6s^2$ is extreme — those electrons sit tight to the nucleus and are reluctant to delocalize. Metallic bonding in elemental Hg is therefore weak, and the melting point plunges below room temperature: mercury is a liquid. Lead (Z = 82) has $[\text{Xe}]\,4f^{14} 5d^{10} 6s^{2} 6p^{2}$. The $6s^2$ pair is likewise stabilized and relatively inert, leaving $6p^2$ to do most of the bonding. Fewer electrons participate, bonds are weaker, and the metal is **soft** with a low melting point relative to "typical" strong-bond metals. ::: {.callout-note} ## Inert Pair Effect In heavy p-block elements (Tl, Pb, Bi), the outer s² pair (e.g., 6s²) becomes chemically and metallurgically less available due to relativistic stabilization. Consequences: preferred lower oxidation states (e.g., Pb²⁺) and weaker metallic bonding (soft Pb, liquid Hg). ::: # What “Conduction” Means: Electrical and Thermal Electrical conduction in a metal is governed by the electrons within a few $k_B T$ of $E_F$. In a band picture, electrons occupy Bloch states with dispersion $E(\mathbf{k})$. The **group velocity** $$ \mathbf{v}(\mathbf{k}) = \frac{1}{\hbar}\,\nabla_{\mathbf{k}} E(\mathbf{k}) $$ is the quantum analog of “speed.” An electric field shifts the distribution in $\mathbf{k}$-space via $$ \hbar\,\frac{d\mathbf{k}}{dt} = -e\,\mathbf{E}, $$ changing the expectation value of velocity—and a current flows. ::: {.callout-note} ## Bloch State An electron state in a periodic crystal with wavefunction $\psi_{\mathbf{k}}(\mathbf{r})=u_{\mathbf{k}}(\mathbf{r})e^{i\mathbf{k}\cdot\mathbf{r}}$. Transport properties follow from $E(\mathbf{k})$. ::: Thermal conduction in metals is carried mostly by the same electrons (with a lattice/phonon contribution that is usually smaller in good metals). A temperature gradient broadens the energy distribution on the hot side; energetic electrons diffuse and carry heat. This shared role leads to the Wiedemann–Franz law: $$ \frac{\kappa}{\sigma T} \approx L, $$ with the Lorenz number $L \approx 2.44\times 10^{-8}\,\text{W}\,\Omega\,\text{K}^{-2}$ in many simple metals. ::: {.callout-note} ## Good Metal A material where conduction is dominated by nearly free electrons with long mean free paths. Examples: copper, silver, gold, aluminum. Good metals obey the Wiedemann–Franz law closely because the same mobile electrons efficiently carry both charge and heat. ::: ::: {.callout-note} ## Phonon A quantized lattice vibration. In insulators and semiconductors, phonons dominate thermal conductivity because there are no itinerant electrons. ::: ::: {.callout-note} ## Wiedemann–Franz Law Empirical (and derivable in simple models): in metals, the ratio of thermal to electrical conductivity scales with temperature, indicating common carriers—electrons—transport both heat and charge. ::: # What Changes in Compounds---and Why Most Compound Colors are Nonrelativistic The metallic color story turns on d→s interband transitions near $E_F$. In compounds, electron clouds are reshaped by bonding and geometry, introducing new color mechanisms: - crystal-field splitting of d orbitals (geometry-dependent), - charge-transfer transitions (ligand ↔ metal), - band-gap-controlled absorption in extended solids (semiconductors, pigments), - spin–orbit effects growing important for heavy elements. Most compound colors can be understood without leaning on relativity; Coulomb interactions and symmetry do the bulk of the work. Relativity resurfaces for heavy-element complexes (Au, Pt, Ir) via strong spin–orbit coupling and shifted orbital baselines. ::: {.callout-note} ## Ligand (n) Each of the atoms or groups attached to the central (usually the metal) atom of a co-ordination complex. ::: ::: {.callout-note} ## Crystal-Field Splitting In a ligand field, metal d orbitals split into groups (e.g., $t_{2g}$, $e_g$) with separations often in the visible, producing vivid colors in transition-metal compounds. ::: # Unifying Quantitative Thread 1) Atomic scale (Dirac picture): Relativity modifies the radial behavior: s (and $p_{1/2}$) contract; d/f expand by orthogonality. Net effect: energy of s decreases; energy of d/f increases. 2) Solid scale (band picture): Those atomic shifts feed into band energies: in Au, $5d$ rises and $6s$ falls relative to a nonrelativistic baseline—shrinking the interband threshold into the visible; in Hg, $6s^2$ is so stabilized that metallic bonding weakens drastically. 3) Transport and optics: Optical absorption probes interband separations; dc transport probes states and velocities near $E_F$. The same band structure that sets Au’s absorption edge also sets its conduction properties. In good metals, electrons carry both charge and heat, consistent with Wiedemann–Franz. [FIGURE: Place a simple schematic showing d–s separation across Cu → Ag → Au → Hg, with the visible/UV threshold annotated.] # “Speeding Up” and Wavefunctions It is natural to worry about talking of electrons “speeding up” when we model them with delocalized wavefunctions. The precise statement is: an external field shifts the occupation of Bloch states in $\mathbf{k}$-space, changing the expectation of group velocity. No classical trajectories are required; transport is an ensemble statement built from $E(\mathbf{k})$ and $f(E)$. # Explanation vs Prediction We can now *explain* why gold is yellow, copper is red, mercury is liquid, and lead is soft. Could we have *predicted* these phenomena from first principles without data? In principle, yes: the Dirac equation plus electron–electron interactions plus crystal structure suffices. In practice, multi-electron relativistic problems require approximations (Hartree–Fock, DFT, beyond-DFT), and even modern computations lean on experiment for calibration and guidance. Our knowledge here is a bootstrapped partnership of observation and theory. # Figures to Insert - [FIGURE: Ionization energies Cu, Ag, Au] Use the Mendeleev script you ran; place the PNG here. - [FIGURE: Hydrogenic $R_{n0}$ vs $u=rR$ (nonrelativistic)] From the provided script `R_u_nonrel.png`. - [FIGURE: Dirac near-origin schematic for s-state, heavy Z] From the provided script `dirac_schematic.png`. - [FIGURE: Schematic band alignment (Cu/Ag/Au/Hg)] Simple diagram (you can sketch or generate) showing d-band position vs s-band and the visible vs UV threshold. # Minimal References - Bransden & Joachain, *Physics of Atoms and Molecules* — hydrogenic orbitals, fine structure, multi-electron atoms. - Ashcroft & Mermin, *Solid State Physics* — bands, DOS, transport, optical properties of metals. - Grant, *Relativistic Quantum Theory of Atoms and Molecules* — Dirac methods, relativistic orbital effects in heavy elements.