Slater-Type Orbitals

STOs are the conceptual ancestor of essentially every basis set used in modern electronic-structure software, even though the actual numerical calculations mostly use Gaussian-type orbitals (GTOs) instead. The reason is entirely computational: products of two Gaussians centered at different nuclei collapse into a single Gaussian (the Gaussian product theorem), making multi-center integrals in molecular calculations vastly cheaper than the corresponding STO integrals. But GTOs have the wrong shape — no cusp at the nucleus and the wrong exponential tail — so practical basis sets (e.g., the STO-3G basis used in introductory computational chemistry courses) are built by literally fitting a sum of three Gaussians to reproduce a single Slater-type orbital's shape as closely as possible. Every time a chemist runs a Hartree–Fock or DFT calculation with an STO-nG or similarly-derived basis, Slater's 1930 exponential form is quietly present underneath the Gaussian machinery. Qualitatively, STOs (and the $Z_{\text{eff}}$/$\zeta$ language built on them) remain the everyday tool for explaining periodic trends in ionization energy, atomic and ionic radii, and electronegativity in inorganic and general chemistry — the same $Z_{\text{eff}}$ values that build STOs are exactly the numbers used to rationalize why atomic radius shrinks across a period, why electron affinities peak where they do, and why transition-metal ions preferentially lose $s$ electrons before $d$ electrons on ionization (directly foreshadowing the orbital-energy-ordering arguments used to explain Orbital Energy Ordering in Multielectron Atoms). In spectroscopy, Slater–Condon parameters — integrals over products of STOs — are still the standard language for parametrizing electron–electron repulsion in crystal-field and ligand-field theory, used routinely to interpret d–d and f–f transition energies in coordination complexes and lanthanide/actinide spectroscopy. In crystallography and charge-density analysis, multipole refinement of X-ray diffraction data (the Hansen–Coppens formalism) explicitly uses Slater-type radial functions to model the deformation density around atoms in a crystal, connecting a 1930 hand-calculation trick directly to state-of-the-art experimental electron-density mapping used to characterize bonding in modern materials.

Ask the tutor: How do Clementi–Raimondi effective nuclear charges differ numerically from plain Slater's-rule values, and why does that difference matter for predicting the 4s/3d energy crossover in the first-row transition metals?

Start from what a hydrogen-like orbital actually looks like. For an electron in an orbital with quantum numbers $n$, $\ell$, the exact solution of the one-electron Schrödinger equation is a product of an associated Laguerre polynomial (giving $n - \ell - 1$ radial nodes) and a decaying exponential $e^{-Zr/na_0}$. This is exact but has a problem the moment you leave hydrogen: in a many-electron atom, an electron in, say, a 3s orbital does not feel a bare nuclear charge $Z$. It feels $Z$ screened by every other electron, and the amount of screening depends on how much that electron's own charge distribution overlaps the core versus stays outside it. A 3s electron penetrates the core more than a 3p electron, which penetrates more than a 3d electron — so even electrons with the same $n$ see different effective charges. Real atoms need something more flexible than "plug in the true $Z$." Slater's fix, from the shielding rules, is to replace $Z$ with $Z_{\text{eff}} = Z - S$, where $S$ is built additively from tabulated contributions of other electrons depending on their $(n,\ell)$ group relative to the electron in question. That handles the screening. But there is a second, subtler problem: even with the correct $Z_{\text{eff}}$, a true hydrogenic radial function for, say, $n=3$ has two radial nodes, and reproducing those nodes accurately for a many-electron atom is not worth the algebraic cost — the chemically relevant part of the orbital is its outermost lobe, which is what overlaps with other atoms in a bond. Slater's second move is to throw away the nodes altogether and use the simplest possible nodeless radial function,

choosing the effective quantum number $n^*$ so that the large-$r$ decay rate and the location of the outer maximum still come out approximately right. The whole orbital, angular part included, is written

with $N$ a normalization constant and $Y_\ell^m$ the ordinary spherical harmonic — Slater orbitals keep the exact angular part, since that costs nothing and is already correct (angular nodes are set purely by $\ell$, $m$, not by screening). The single free parameter that does all the work is the orbital exponent

Large $\zeta$ means a tightly held, contracted orbital (large $Z_{\text{eff}}$, small $n^*$); small $\zeta$ means a diffuse, loosely held orbital. Because $\zeta$ packages together both the nuclear pull and the effective shell size, a single number tells you almost everything qualitative you need: relative orbital sizes, relative ionization energies, and relative degrees of orbital overlap in a bond.

Notice the key feature of the figure: the two curves agree reasonably well in the outer region — same rough peak location, same decay rate — and disagree badly near the core, where the true function must wiggle through a node to stay orthogonal to lower orbitals of the same $\ell$. Because bonding, ionization, and most spectroscopic properties are governed by the outer lobe, this is exactly the right place to spend your approximation budget. The rule for choosing $n^*$ (Slater's rule for effective principal quantum number) is not derived from first principles; it is an empirical fit chosen to make the STO's energy and size track real atomic behavior across the periodic table:

The break from integer values starting at $n=4$ reflects the fact that $d$ and $f$ orbitals of high shells penetrate the core so poorly, and are so strongly screened, that a literal $n^*=n$ would make them far too diffuse and far too weakly bound compared to experiment; damping $n^*$ below $n$ for $n\ge4$ corrects this without needing a separate, more complicated functional form.

The full unnormalized STO radial function, for a given atom, orbital, and its Slater-derived $Z_{\text{eff}}$ and $n^*$, is

where: $N$ is a normalization constant fixed by $\int_0^\infty |R|^2 r^2\,dr = 1$; $\zeta$ has units of inverse length once $a_0$ (the Bohr radius) is included explicitly, or is left in atomic units (bohr$^{-1}$) with $a_0=1$; $n^*$ is the effective principal quantum number from the table above, and $\ell$ enters only through the (unmodified) angular part $Y_\ell^m$.

Reading: the exponent $n^*-1$ controls how many powers of $r$ multiply the exponential before it takes over — larger $n^*$ pushes the peak of $R^2 r^2$ (the radial probability density) further out, exactly mimicking the outward shift of higher shells, without requiring any nodes to get there. The normalization constant follows from the standard Gamma-function integral $\int_0^\infty x^k e^{-ax}\,dx = k!/a^{k+1}$ for integer $k$; Slater orbitals are normally used with $n^*-1$ non-integer only rarely (mainly the $n=4$–6 damping), so the Gamma function generalizes this:

where: this is the standard closed form when $n^*$ is a positive integer; $(2n^*)!$ comes from evaluating $\int_0^\infty r^{2n^*} e^{-2\zeta r}\,dr = (2n^*)!/(2\zeta)^{2n^*+1}$ and taking the square root.

Reading: normalization is elementary calculus, not a special-function nightmare — this is the entire reason STOs were tractable by hand in 1930, decades before computers. The orbital energy predicted by this hydrogen-like ansatz, with $Z_{\text{eff}}$ and $n^*$ standing in for the true $Z$ and $n$, is

where: the formula is exactly the Bohr/hydrogenic energy formula with effective quantities substituted for the true ones — the entire predictive power of the Slater treatment rests on this one substitution being a good approximation.

Reading: because $\zeta$ already encodes both screening and shell-contraction effects, orbital energy trends across a period or down a group can be read directly off how $\zeta$ changes — no further calculation needed.

Example 1 — Foundational: constructing the STO for hydrogen's 1s orbital and checking against the exact solution. For H, $Z=1$, one electron, no shielding: $Z_{\text{eff}}=1$, and $n^*=n=1$ for the $n=1$ shell (no adjustment needed for the lowest shell). Then

The STO radial function is $R(r) = N r^{0}e^{-r} = Ne^{-r}$ (in bohr, $a_0=1$). Using the normalization formula with $n^*=1$:

So $R_{1s}^{\text{STO}}(r) = 2 e^{-r}$ — this is exactly the true hydrogenic 1s radial function, $R_{1s} = 2e^{-r/a_0}\cdot a_0^{-3/2}$ in the appropriate units. No approximation was made because hydrogen has no other electrons to screen and no radial nodes to discard at $n=1$ anyway — the STO is exact in this one special case. Checking $\langle r \rangle$: using the derived formula with $n^*=1$, $\zeta=1$:

which matches the well-known exact hydrogen 1s result $\langle r\rangle = 3a_0/2$ exactly. Limiting-case check: this is the correct behavior since for $n=1$, $\ell=0$ there is no orthogonality constraint to any lower orbital, so the "discard the node" approximation costs nothing — the STO and true hydrogenic function coincide identically.

Example 2 — Applied: comparing 3s and 3p orbital exponents in phosphorus, and predicting the $ns/np$ energy gap. Take phosphorus, $Z=15$, configuration $1s^2 2s^2 2p^6 3s^2 3p^3$. We need $Z_{\text{eff}}$ for a 3s electron and for a 3p electron using Slater's shielding rules (group the electrons as $(1s)(2s,2p)(3s,3p)$). For a 3s electron: shielding contributions are $0.35$ each from the other 4 electrons in the $(3s,3p)$ group (there are 4 other valence electrons: 1 other 3s + 3 in 3p), $0.85$ each from the 8 electrons in $(2s,2p)$, and $1.00$ from the 2 electrons in $1s$:

For a 3p electron, Slater's rules treat $s$ and $p$ of the same shell identically (same group), so the shielding calculation is identical:

With $n=3 \Rightarrow n^*=3$ for both, the simple Slater treatment gives $\zeta_{3s} = \zeta_{3p} = 4.80/3 = 1.60\ \text{bohr}^{-1}$ — identical orbital exponents and, naively, identical orbital energies $E = -13.6\,\text{eV}\times(1.60)^2 = -34.8$ eV for both. This is a known limitation: plain Slater's rules do not by themselves split $ns$ from $np$ within the same shell, because $Z_{\text{eff}}$ is computed identically for both. The real physical splitting (3s lower than 3p, a difference of several eV, reflecting greater 3s penetration of the core) has to be introduced either empirically (using spectroscopically fitted, separate $\zeta_{ns}$ and $\zeta_{np}$ values, as in more refined Slater–Zener orbital tables) or via a full self-consistent-field calculation.

Why this matters, and the correct qualitative fix: chemists handle this by remembering that Slater's simple rule is a first approximation — real 3s orbitals are more tightly bound than 3p because s-orbitals have nonzero amplitude at the nucleus and thus penetrate the shielding cloud of inner electrons more effectively. Refined parameter sets (Clementi–Raimondi effective charges, obtained by fitting STOs to numerical Hartree–Fock orbitals) give, for phosphorus, $Z_{\text{eff}}(3s) \approx 5.64$ and $Z_{\text{eff}}(3p) \approx 4.89$ — s indeed sees a larger effective charge than p, splitting the energies as physically required. This worked example is a genuine limiting case worth stating explicitly: as the shielding rule is applied without refinement, $ns$/$np$ splitting vanishes identically — a real gap in the crude Slater picture that more sophisticated parametrizations exist specifically to close.

Example 3 — Exam/competition level: using STOs to estimate the overlap-driven trend in bond covalency across isoelectronic species, and computing $\langle r\rangle$ for a 4s electron in Ca vs. a 3d electron in Ca (the $4s/3d$ crossover). This is the calculation behind one of the most cited "surprises" of transition-metal chemistry: why is the $4s$ orbital filled before $3d$ in period-4 atoms, even though $n=3$ is "lower" than $n=4$? Compute $Z_{\text{eff}}$ and $\langle r \rangle$ for both orbitals in neutral calcium ($Z=20$, configuration $[Ar]4s^2$).

4s electron. Group structure for Slater's rules: $(1s)(2s,2p)(3s,3p)(3d)(4s,4p)$. For a 4s electron, the shielding sum includes: the 1 other 4s electron at $0.35$; all 18 electrons in $n=1$–3 shells (i.e., everything with $n < 4$) contribute $1.00$ each (full shielding from lower shells for an $ns/np$ electron):

With $n=4 \Rightarrow n^*=3.7$ (the damped value):

3d electron (hypothetical occupation, as in Ca$^+$/Sc/early transition-metal comparisons). For a 3d electron, Slater's rules place all other electrons with $n<3$ or ($n=3$, lower $\ell$) at full shielding $1.00$ (d and f orbitals see essentially complete shielding from everything inside, including the same-shell $s,p$ electrons, because Slater groups d/f separately from s/p of the same shell) — take the reference configuration $[Ar]3d^1 4s^1$ relevant to comparing orbital energies directly:

With $n=3 \Rightarrow n^*=3$:

Interpretation. The 3d orbital ($\langle r\rangle \approx 5.25\,a_0$) is markedly more contracted than the 4s orbital ($\langle r\rangle \approx 9.42\,a_0$) despite having the smaller principal quantum number — 3d electrons are held in tight to a small effective charge $Z_{\text{eff}}=2.00$ but with a small $n^*=3$, while 4s electrons, though seeing a smaller $Z_{\text{eff}}=1.65$, spread out over the larger $n^*=3.7$ shell and, crucially, gain stabilization energy from the small residual probability density right at the nucleus (better core penetration, not fully captured by this simple radius estimate but corroborated by the energy formula). Using the energy formula directly:

This crude estimate actually predicts 3d below 4s for the $3d^14s^1$ configuration of Ca — consistent with the well-known experimental subtlety that for atomic calcium, once a 3d electron is present, $3d$ and $4s$ are close in energy and their exact ordering is configuration-dependent (a famous complication foreshadowed here and resolved properly only by full orbital-energy-ordering arguments). The qualitative lesson survives the crude numbers: d orbitals of a given $n$ are always more radially contracted than s/p orbitals of the next shell up, which is precisely why transition metals fill 4s before 3d in the neutral atom (4s is lower in energy for the empty-3d configuration relevant to building up Ca, Sc, Ti, …) yet lose 4s electrons preferentially on ionization (once 3d is populated, 4s becomes higher in energy) — the famous "4s before 3d fills, but 4s ionizes first" rule, whose mechanistic root is exactly this differential radial contraction visible directly in the STO $\langle r\rangle$ values above.