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Take a planar honeycomb lattice with two sites $A,B$ per primitive cell and one localized atomic orbital per site. In this explicitly idealized tight-binding model, all three nearest-neighbour hopping matrix elements are $-t$, with $t>0$; there is no on-site offset or further-neighbour hopping. Ignore electron interactions and treat physical spin as a separate degeneracy. Let the bond length be $b$, and choose the three vectors from an $A$ site to its neighbours asThe Bloch state amplitudes on $A,B$ form a two-component vector. A consistent phase convention gives the HamiltonianIts characteristic equation is $E^2=t^2|f{|}^2$, giving the two displayed energy bands. The two-component structure comes from the two sites, rather than two values of physical spin. This is a particularly small matrix model in which a crystal's geometry creates an unexpected low-energy law.

Historically, Philip Russell Wallace used a tight-binding model to investigate the electronic band structure of graphite in 1947. The material in that paper was layered graphite; the single-sheet model here isolates one simpler part of the problem.^[ref]

A molecular orbital with little or no net bonding interaction between the atoms in the stated orbital model. A symmetry-isolated atomic orbital can be exactly nonbonding in a minimal basis: the out-of-plane nitrogen $2p_z$ orbital in planar Ammonia has no matching hydrogen symmetry orbital. A mainly localized lone pair can acquire bonding admixture when a distortion permits additional mixing.

The orbital overlap of normalized real atomic orbitals is their inner product $S=\int {\varphi }_A{\varphi }_B$. It is a dimensionless measure of their common amplitude, not an electron count. For the Gaussian function toy, increasing $a$ from $1$ to $2$ reduces $S$ from $e^{-1}\approx 0.367879$ to $e^{-4}\approx 0.018316$. As $S$ tends to zero, both LCAO normalization denominators tend to $\sqrt{2}$.

Suppression of a ground state's orbital magnetic moment by the splitting and mixing of free-ion atomic orbitals in a crystal. A nondegenerate orbital state can generally be chosen real in zero field, giving zero expectation of the imaginary orbital angular momentum operator. Spin-orbit coupling to excited states still produces residual g-factor shifts and anisotropy, so quenching is not absence of every orbital response.

For a $2p$ atomic orbital, $l=1$. The radial factor obeysSubstitution into the radial Schrödinger equation for a central potential givesThus the energy eigenvalue isThe normalization constant cancels; $E_{\mathrm{h}}$ is the Hartree energy.

Use the classes $(E,2C_3,3{\sigma }_v)$ of the C3v point group. For the three hydrogen $1s$ atomic orbitals, the character of a representation counts orbitals fixed by each symmetry operation. Identity fixes three, a nontrivial threefold rotation fixes none, and a vertical mirror fixes one. ThereforeThe character inner product gives multiplicitiesThus ${\Gamma }_{\mathrm{H}}=A_1\oplus E$, with the $E$ contribution accounting for two independent symmetry orbitals.

In the C3v point group, the nitrogen valence atomic orbitals transform asThe $s$ orbital is invariant under every symmetry operation; $p_z$ is also invariant because each operation preserves the $z$ coordinate. The other two components mix as a two-dimensional irreducible representation.

For planar Ammonia, the nitrogen valence atomic orbitals have symmetries $2s:A_1^{\prime }$, $(2p_x,2p_y):E^{\prime }$ and $2p_z:A_2^{\prime \prime }$. The hydrogen symmetry orbitals span $A_1^{\prime }\oplus E^{\prime }$. Consequently the in-plane orbitals form a bonding $a_1^{\prime }$ orbital, a bonding $e^{\prime }$ pair and their antibonding partners. The out-of-plane $2p_z$ has no matching hydrogen combination and remains a nonbonding orbital. With the core omitted, the conventional valence occupation isThe preceding ammonia molecular orbital diagram shows the correlation. On pyramidalization, $A_1^{\prime }$ and $A_2^{\prime \prime }$ both correlate with $A_1$ of the C3v point group, permitting s–p orbital mixing. The pure planar lone pair can acquire bonding character, lowering its energy and helping stabilize the pyramid. Symmetry alone does not provide a numerical inversion barrier.^[ref]

For the tetrahedral symmetry classes $(E,8C_3,3C_2,6S_4,6{\sigma }_d)$, each set of four $s$ atomic orbitals hasThe counts are the fixed sites of each symmetry operation: a threefold rotation fixes one vertex of each tetrahedron, a twofold or improper fourfold operation fixes none, and a diagonal mirror fixes two. Character orthogonality reduces each set toThe dimensions $1+3=4$ provide a check. Thus the complete eight-orbital model contains two copies of each of $A_1$ and $T_2$, which can mix across the two species.

Let $P_{12}$ exchange the complete labels of the two electrons. The spatial factors obey $P_{12}{\Phi }_{\pm }=\pm {\Phi }_{\pm }$. The singlet state spin factor is antisymmetric, while each triplet state spin factor is symmetric. Thus in all four casesas required by the Pauli exclusion principle. In particular, the $M_S=0$ triplet state remains a triplet despite having opposite measured spin projections in each product term.

For a spin-independent Hamiltonian and the same fixed spatial atomic orbitals, direct evaluation giveswhere $J_{ab}$ is the direct Coulomb integral and $K_{ab}>0$ is the exchange integral for the helium $1s,2s$ pair. ThereforeThe antisymmetric spatial wavefunction vanishes when the two positions coincide and reduces their Coulomb repulsion. The lower energy of the maximum-spin member of this fixed electron configuration is the first of Hund's rules.

The Pauli exclusion principle requires the complete electron wavefunction to change sign on exchange of any two electrons, including both position and spin labels. Let $a=1s$ and $b=2s$ be normalized, orthogonal spatial atomic orbitals, and let $\alpha ,\beta$ denote spin up and down.

The symmetric and antisymmetric spatial functions areThe singlet state isThe three triplet states share the antisymmetric spatial factor:Every factor and complete product is normalized. Here the subscripts are $(S,M_S)$; the spatial electron configuration is $1s2s$ in each case.

Closed subshells have $L=S=0$, so only the open $p$ or $d$ subshell determines each atomic term. Apply Hund's rules in order: maximize $S$, maximize $L$ at that $S$, then choose the lowest $J$ below half filling or the highest $J$ above half filling.

For carbon, the two $2p$ electrons occupy different magnetic atomic orbitals with parallel spins, giving $S=1$. Their largest allowed $M_L$ at that spin is $1+0=1$, so $L=1$, a $P$ term. The $p^2$ subshell is less than half full, giving $J=|1-1|=0$. The ground level is ${}^3{P}_0$.

For vanadium, three parallel-spin $3d$ electrons give $S=3/2$. The largest $M_L$ compatible with the Pauli exclusion principle is $2+1+0=3$, hence $L=3$, an $F$ term. The $d^3$ subshell is less than half full, so $J=3-3/2=3/2$. The ground level is ${}^4{F}_{3/2}$.

For cobalt, the $d^7$ subshell is equivalent to three $d$ holes for the allowed $L,S$ terms, so its maximum-spin, maximum-$L$ term is again ${}^{4}F$. Explicitly, five spin-up electrons fill all $m_l$ values and two spin-down electrons can occupy $m_l=2,1$, giving $M_S=3/2$ and $M_L=3$. Since $d^7$ is more than half full, $J=3+3/2=9/2$. The ground level is ${}^4{F}_{9/2}$.

For the $p_z$ atomic orbital basis, only orbitals whose centres remain fixed contribute to a character. Each such orbital contributes $+1$ if its $z$ direction is preserved and $-1$ if it is reversed. In the standard D4h molecular point group class order,No nonidentity operation other than ${\sigma }_h$ leaves any carbon site fixed. Reflection in the molecular plane fixes all eight sites but reverses every $p_z$ function.

Molecular character reduction givesThe dimensions sum to $1+1+1+1+2\cdot 2=8$. The two copies of $E_g$ are separate equivalent irreducible representations which may mix with one another; they do not force all four functions to have a common energy.

Denote the two complex spatial atomic orbitals by ${\psi }_{+}$ and ${\psi }_{-}$, retaining the same spin state. With the phases specified by the supplied spherical harmonics, chooseThe two original atomic orbitals are orthonormal, so the squared coefficients add to one and these combinations are normalized and mutually orthogonal. Explicitly,which are real spatial wavefunctions. An overall sign on either orbital has no physical consequence.

These are the familiar $2p_x$ and $2p_y$ p orbitals. Each has two lobes of opposite wavefunction sign, with an atomic-orbital nodal plane through the atomic nucleus: $x=0$ for $2p_x$ and $y=0$ for $2p_y$. The sign is a phase convention, not a positive or negative electric charge; the probability density is nonnegative on both lobes.

Choose the phase of each inward-pointing ligand sigma bond donor lobe consistently. An operation taking one donor to another then acts as a permutation on the four donor functions. For the square, label these functions $L_1,L_2,L_3,L_4$ cyclically along $+x,+y,-x,-y$. Normalized symmetry-adapted linear combinations areThus the ligand group representation is ${\Gamma }_{\sigma }=A_{1g}+B_{1g}+E_u$. Its irreducible representations allow the first two combinations to mix with the matching nickel $d$ atomic orbitals; $E_u$ instead matches nickel $p_x,p_y$.

For the tetrahedron in the vertex order specified above, takeThe three latter functions transform like $x,y,z$. Hence ${\Gamma }_{\sigma }=A_1+T_2$. These are four ligand symmetry-adapted linear combinations, rather than the five metal $d$ functions of part (ii).

The angular factors of the nickel $3d$ atomic orbitals transform like quadratic Cartesian functions. In the D4h molecular point group, the assignments areAll five are even under inversion symmetry, hence the $g$ suffixes. The pair $(d_{xz},d_{yz})$ is mixed by a quarter-turn and spans a two-dimensional irreducible representation.

In the Td molecular point group, the corresponding result isThere is no inversion label because $T_d$ has no inversion centre. In both cases the dimensions add to five.

Use the printed numbering: site 1 is at the top on $+y$, sites 2 and 3 are on the right, site 4 is at the bottom, and sites 5 and 6 are on the left; $z$ is perpendicular to the ring. Let ${\chi }_j$ be consistently phased, orthonormal $p_z$ atomic orbitals at those sites.

In the Hückel method, $H_{jj}=\alpha$ and $H_{j,j\pm 1}=\beta$, including the 6–1 bond. Set $\lambda =(\epsilon -\alpha )/\beta$, with the usual $\beta <0$. The cyclic nearest-neighbour matrix has eigenvalues ${\lambda }_k=2\cos (2\pi k/6)$. One real, normalized orthonormal eigenbasis isEach row means $\psi ={\sum }_j{c}_j{\chi }_j$. The three lowest molecular orbitals hold the six $\pi$ electrons. Any real orthogonal mixing within either degenerate pair is an equally valid choice of molecular orbitals; the coefficient vectors specify exactly which sketches are used here.

The coloured circles show the signs and relative magnitudes of the lobes above the molecular plane. The lobes below it have the opposite signs; colours are quantum phases, not electric charges.

The nitrogen pair and the four-carbon set are separately invariant under the D2h point group. Construct symmetry-adapted linear combinations within these two sets; mixing equivalent representations across the sets is left for the Hückel method calculation.

For the nitrogen $p_z$ atomic orbitals, normalized combinations areFor the carbon $p_z$ atomic orbitals, takeThese six functions are orthonormal under the zero-overlap assumption. The suffixes $x,y,xy$ describe coefficient patterns across the carbon sites; the sign acquired by each underlying $p_z$ function must also be included when assigning an irreducible representation. In particular, reflection through the molecular plane changes the sign of every function, while inversion sends ${\chi }_1$ to $-{\chi }_4$. Ignoring that orbital sign would give incorrect $g/u$ labels.

The complete $\pi$ group representation is thereforeThe diagram shows all six ring symmetry-adapted linear combinations and the resulting energy ordering used in part (iii).

Let $a={\varphi }_{1s}$ and $b={\varphi }_{2s}$ be orthonormal atomic orbitals. The normalized spatial factors areFor the singlet state, use the antisymmetric spin factorFor the three components of the triplet state, useEach total wavefunction obeys the Pauli principle. At coincident spatial coordinates,Consequently the triplet pair probability density vanishes at coincidence, even for its $M=0$ component. This zero is stronger than the trivial statement that any exact point has zero integrated probability in a continuum.

The $3d$ and $4s$ atomic orbitals of chromium are close enough in energy that electron-electron terms can overturn the simple filling order. The configuration $[\mathrm{Ar}]3d^{5}4s^1$ permits five singly occupied $3d$ orbitals and a parallel-spin $4s$ electron. Relative to $3d^{4}4s^2$, it removes the $4s$ pair and permits more parallel-spin exchange interactions, lowering the Coulomb-and-exchange contribution enough to offset the orbital promotion cost.

As qualitative bookkeeping, six parallel electron spins give $\left(\genfrac{}{}{0ex}{}{6}{2}\right)=15$ same-spin pairs, compared with $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$ in high-spin $3d^{4}4s^2$. The individual exchange integrals are unequal, so this is an explanation of the tendency, rather than a quantitative atomic energy calculation. The observed ground state configuration is $\boxed{[\mathrm{Ar}]3d^{5}4s^1}$.^[ref]