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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 ${}^4{F}_{9/2}$.

Fill four spatial molecular orbitals with the eight $\pi$ electrons. For negative ${\beta }_S,{\beta }_L$ and $|{\beta }_S|>|{\beta }_L|$, the occupied energies are $\alpha +{\beta }_S+{\beta }_L$, $\alpha +{\beta }_S-{\beta }_L$, and the pair at $\alpha -\sqrt{{\beta }_S^2+{\beta }_L^2}$. ThereforeFor equal integrals, ${\beta }_S={\beta }_L=\beta <0$, this becomesAt equal bond lengths, the last two electrons occupy the frontier degeneracy at $E=\alpha$, so their distribution within that degenerate space does not change the bare Hückel method energy.

The uniform planar eight-electron ring has a partially filled frontier degeneracy. Bond-length alternation splits those levels, permitting the two frontier electrons to occupy the lowered orbital and opening a gap to the empty partner. This is the electronic mechanism for destabilizing the equal-bond planar geometry.

A controlled comparison keeps the mean resonance integral fixed. PutUsing part (i), the change from the uniform ring isThus a small alternation lowers the electronic energy, with a leading gain linear in $\delta$. An ordinary elastic cost starts quadratically, so the equal-bond planar state is unstable in this model. The equilibrium distortion requires those geometric and sigma-bond costs as well.

The eight $\pi$ electrons constitute a $4n$ cyclic count rather than the closed-shell $4n+2$ case associated with aromaticity. The result concerns the imposed planar geometry; it does not establish the unconstrained three-dimensional equilibrium shape of cyclooctatetraene.

The extra electron is bound to an approximately singly charged donor through a dielectric-screened Coulomb attraction. In the isotropic hydrogenic donor state model, replace the vacuum electron mass by $m^{\ast }$ and the vacuum permittivity by ${ϵ}_r{ϵ}_0$:At low temperature, the donor is mainly in its $1s$ state. Allowed electric-dipole transitions to $np$ states haveAssign the three printed lines to $n=2,3,4$. Their independently inferred binding energies areThey are mutually close, giving $R^{\ast }\simeq 10.63\mathrm{meV}$. The hydrogenic impurity absorption series accumulates at the donor-ionization limit. Above about $10.6\mathrm{meV}$, absorption can place the electron in the conduction-band continuum, explaining the continuous spectrum.

Using the supplied $m^{\ast }=0.2m_e$ and $R_H=13.61\mathrm{eV}$,The corresponding effective radius isThese are the parameters inferred from the printed lines within the stipulated scalar model, rather than substituted tabulated material constants. Real impurity spectra can depart from a single hydrogenic ladder through anisotropy and short-distance impurity effects.^[ref]

There are three useful temperature regimes.

At low temperature, dopant freeze-out keeps most donor electrons bound. Warming ionizes donors and increases conductivity through the rising mobile-electron concentration. For an uncompensated dilute donor system, negligible holes and spin degeneracy two, charge neutrality with partial donor ionization givesDeep in freeze-out, $n\ll N_D$, soThe activation scale is set by the shallow donor binding, rather than the whole band gap.

In the extrinsic regime, donors are largely ionized and $n\simeq N_D$ is nearly constant. The temperature dependence of conductivity then mainly follows the electron mobility. Increasing phonon scattering can make conductivity fall as temperature rises, even though donors remain ionized; impurity scattering can give a different mobility trend at lower temperature.

At sufficiently high temperature, thermally generated electron–hole pairs overwhelm the dopant population. The intrinsic, gap-activated increase in conductivity returns. The boundaries depend on donor concentration, compensation and scattering, so no one exponential describes the entire doped-sample curve.

Arsenic has five valence electrons, one more than silicon. As a substitutional donor impurity, it supplies a weakly bound extra electron which is largely thermally released into the conduction band at room temperature, leaving an ionized positive donor.

Consequently the mobile electron density increases far above the intrinsic value, so the conductivity increases substantially even at ppm dopant levels. The crystal remains macroscopically neutral: fixed donor charge compensates the added mobile charge.

Gallium has three valence electrons, one fewer than silicon, and acts as an acceptor impurity. A neutral acceptor can be described as a negatively charged centre binding an electron hole. Thermal release of these holes gives a p-type semiconductor.

At low temperature, the bound hole can absorb far-IR photons in transitions between acceptor levels. The simplest scalar hydrogenic acceptor state model predictsso one expects a discrete bound-hole absorption series followed by a photoionization continuum. In that idealization, the first lines would be at $3R_A^{\ast }/4$, $8R_A^{\ast }/9$ and $15R_A^{\ast }/16$, with threshold $R_A^{\ast }$.

The hole effective mass is not supplied, so the electron mass $0.2m_e$ cannot be reused to give numerical Ga line positions. In a real valence band, multiple hole bands, anisotropic masses and spin-orbit effects can split and complicate the series; acceptor spectra are generally less accurately described by one scalar hydrogenic ladder.^[ref]

For monatomic species, there are no molecular rotational or vibrational factors. Assume excited electronic states are negligible, so the internal partition sums are the ground-state degeneracies. A free electron has two spin states, givingwith the analogous expression for the ion. Since losing one electron changes the atomic mass negligibly, $(m_{{\mathrm{M}}^{+}}/m_{\mathrm{M}}{)}^{3/2}\simeq 1$. HenceThis is the ground-state form of the Saha ionization equation. Its assumptions include a dilute, nondegenerate ideal mixture, negligible interparticle interactions, thermal and chemical equilibrium, no additional ionization stages, and consistent treatment of internal degeneracies. Nuclear-spin factors may be omitted when they cancel between the atom and ion; they must not be counted on only one side.

Let $\alpha$ be the degree of ionization. Conservation of nuclei and charge neutrality, with no pre-existing electrons or other ions, giveSubstitution into the definition of $K_c$ givesThe physical root lies between zero and one:The second form avoids subtracting nearly equal terms at large $y$. The limits are $\alpha \simeq \sqrt{y}$ when $y\ll 1$ and $\alpha \simeq 1-1/y$ when $y\gg 1$. The dependence on initial number density matters: $K_c$ alone, without $c_{\mathrm{tot}}$ and the standard-state convention, does not specify the ionized fraction.

Each cyanide contributes charge $-1$, so the formal nickel oxidation state is $+2$ and the metal has $d^8$. The requested diagram retains only the five principally metal-$3d$ molecular orbitals; the mainly ligand bonding molecular orbitals lie below them and are omitted.

In the square, $d_{x^2-y^2}$ points directly towards all four ligand donors and gives the highest, strongly antibonding molecular orbital, $B_{1g}$. The $d_{z^2}$ function also has an allowed $A_{1g}$ interaction, but its in-plane overlap is smaller. Including the allowed nickel $4s$ mixing gives a principally $d$ $A_{1g}$ level with relatively weak antibonding character. The $B_{2g}$ and $E_g$ metal functions have no ligand partner in the sigma bond donor basis and are nonbonding in this minimal model. The useful qualitative ordering is thereforeThe three nonbonding functions are shown together because the sigma-only model does not determine their finer splitting. Additional electrostatic and pi backbonding interactions can change the ordering of the lower $d$ levels; molecular symmetry alone does not fix these energy gaps.^[ref]

In the tetrahedron, the ligand $T_2$ combinations interact with the metal $T_2$ functions, producing higher $d$-derived antibonding molecular orbitals. The metal $E$ pair has no sigma donor partner and remains nonbonding:The diagram fills both arrangements with eight metal electrons; the square is shown in its fully paired state.

For $n$ unpaired electrons, the spin-only magnetic moment iswhere ${\mu }_{\mathrm{B}}$ is the Bohr magneton and the second expression uses $g\simeq 2$.

For the square, a large separation of the highest $B_{1g}$ level favours pairing in the lower four molecular orbitals: $S=0$, $n=0$, ${\mu }_{\mathrm{eff}}=0$. If the cost of occupying $B_{1g}$ were smaller than the relevant electron pairing energy, a high-spin complex with two unpaired electrons and $S=1$ could instead occur, giving $\sqrt{8}{\mu }_{\mathrm{B}}\simeq 2.83{\mu }_{\mathrm{B}}$.

For ideal tetrahedral $d^8$, the occupation is $e^4{t}_2^4$. Hund's rules distribute four electrons over the three degenerate $t_2$ molecular orbitals as one pair and two single occupations. Thus $S=1$ and ${\mu }_{\mathrm{eff}}\simeq 2.83{\mu }_{\mathrm{B}}$. Merely increasing the $e$–$t_2$ splitting does not create a conventional paired tetrahedral $d^8$ alternative: four electrons must still occupy $t_2$.

Filling the three bonding molecular orbitals with two electrons each givesThree isolated ethene-like $\pi$ bonds, evaluated with the same Hückel method parameters, have $E_{\mathrm{loc}}=3\{2(\alpha +\beta )\}=6\alpha +6\beta$. Thus the signed delocalization energy isEquivalently, the positive stabilization is $-2\beta =2|\beta |$.

Benzene has a planar cyclic conjugated system with six $\pi$ electrons, satisfying the $4n+2$ count with $n=1$. All three bonding $\pi$ molecular orbitals are filled and the antibonding levels are empty, giving the closed-shell stabilization associated with aromaticity. The conclusion uses the continuous cyclic overlap as well as the electron count.

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 six $\pi$ electrons give the Hückel method energyThe appropriate localized reference contains two C=N $\pi$ bonds and one C=C $\pi$ bond. Three C=C bonds would incorrectly replace the nitrogen one-centre parameters. For one isolated C=N pair, the dimensionless matrix iswhose eigenvalues are $(1\pm \sqrt{5})/2$. Since $\beta <0$, the occupied bonding molecular orbital has energy $\alpha +(1+\sqrt{5})\beta /2$. Two doubly occupied C=N bonds and one doubly occupied C=C bond consequently giveTherefore the signed delocalization energy isThe positive stabilization is $1.82039|\beta |$, smaller than the benzene value $2|\beta |$ in the stated model but still stabilizing.

Pyrazine is aromatic: its planar cyclic conjugated system contains six $\pi$ electrons and has a filled set of three bonding $\pi$ molecular orbitals. Each ring nitrogen contributes one $p_z$ electron; its in-plane lone pair is outside this $\pi$ system. The heteroatoms change orbital energies and coefficients without interrupting cyclic conjugation or the $4n+2$ count. Aromaticity does not require the stabilization to equal benzene's.

If two fermions occupied the same complete spin orbital $u$, their antisymmetrized product would beThe zero vector cannot represent a normalized quantum state. This is the Pauli exclusion principle: at most one identical fermion occupies any complete one-particle state. Two electrons may occupy the same spatial orbital with opposite spins, because $a\alpha$ and $a\beta$ are different spin orbitals.

Using the same intrinsic-carrier approximation at $300\mathrm{K}$ givesSimilar mass density and coordination do not prove equal electron and hole effective masses; this result isolates the exponential band-gap effect.

A standard substitutional acceptor choice is Mg on a Ga site: a group-II atom replaces group III, leaving one fewer valence electron. Its size is sufficiently compatible with the cation sublattice to make substitution plausible. This identifies the intended acceptor chemistry; actual hole density also depends on activation, incomplete ionization and compensation. Size and valence rules alone do not guarantee a shallow, fully ionized acceptor.