Radial Distribution and Orbital Penetration
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You should be comfortable with the definitions of the principal quantum number \(n\) and the orbital angular momentum quantum number \(l\), and you should understand that atomic orbitals represent probability densities rather than fixed paths.
Why this matters
Understanding where electrons actually spend their time explains why multi-electron atoms behave differently from hydrogen.
The shape of an orbital describes its angular distribution, but the radial distribution function tells us how far from the nucleus the electron is likely to be found. This distinction is critical because it reveals "penetration," where inner electrons shield outer ones imperfectly. This imperfect shielding determines effective nuclear charge, which in turn dictates ionisation energies, atomic radii, and the specific order in which orbitals are filled in the periodic table. Without this concept, the periodic table is just a grid; with it, the grid becomes a logical map of electron energy levels.
Radial Probability vs. Probability Density
It is a common source of confusion to conflate the probability density, \(\psi^2\), with the radial distribution function. The probability density tells you the likelihood of finding an electron at a specific point in space. However, because space is three-dimensional, the number of points available at a distance \(r\) from the nucleus increases as the surface area of a sphere, proportional to \(r^2\). Therefore, even if the probability density is highest at the nucleus (for an s-orbital), the actual probability of finding the electron in a thin shell at that radius is zero because the volume of that shell is zero.
The radial distribution function, often denoted as \(P(r)\) or \(4\pi r^2 R(r)^2\), accounts for this geometric factor. It represents the total probability of finding the electron anywhere on a spherical shell of radius \(r\) and infinitesimal thickness \(dr\). For a 1s orbital, the probability density is highest at \(r=0\), but the radial distribution function peaks at a finite distance. This peak corresponds to the most probable radius, which for hydrogen is the Bohr radius. For higher energy orbitals, this function exhibits multiple peaks, corresponding to "shells" of high probability, separated by nodes where the probability is zero.
Radial Nodes and Orbital Structure
As the principal quantum number \(n\) increases, the radial wavefunction \(R(r)\) develops more oscillations. These oscillations create radial nodes, which are spherical surfaces where the probability of finding the electron is exactly zero. The number of radial nodes is determined by the formula \(n - l - 1\). For example, a 2s orbital (\(n=2, l=0\)) has \(2 - 0 - 1 = 1\) radial node. A 3p orbital (\(n=3, l=1\)) has \(3 - 1 - 1 = 1\) radial node. A 3d orbital (\(n=3, l=2\)) has \(3 - 2 - 1 = 0\) radial nodes.
These nodes divide the orbital into distinct regions of high probability. The outermost peak of the radial distribution function generally defines the "size" of the orbital. However, the presence of inner peaks is the key to understanding penetration. An orbital with radial nodes has significant electron density closer to the nucleus than an orbital with the same \(n\) but higher \(l\). This is because the wavefunction does not vanish completely near the origin for lower \(l\) values.
Orbital Penetration and Shielding
Penetration refers to the ability of an electron in an orbital to spend a significant amount of time close to the nucleus, "penetrating" the cloud of inner electrons. The extent of penetration depends strongly on the value of \(l\). s-orbitals (\(l=0\)) have the highest penetration because their probability density is non-zero at the nucleus and they have the most radial nodes for a given \(n\). p-orbitals (\(l=1\)) have less penetration, and d-orbitals (\(l=2\)) have even less, with f-orbitals (\(l=3\)) having the least penetration for a given shell.
In multi-electron atoms, inner electrons shield outer electrons from the full nuclear charge. However, because s-electrons penetrate closer to the nucleus than p-electrons in the same shell, they experience a higher effective nuclear charge (\(Z_{\text{eff}}\)). Consequently, s-electrons are held more tightly and have lower energy than p-electrons in the same principal shell. This energy splitting, \(E_{ns} < E_{np} < E_{nd} < E_{nf}\), is a direct consequence of differential penetration and shielding. It explains why the 4s orbital fills before the 3d orbital in many transition metals, as the 4s electron penetrates the core more effectively than the 3d electron, lowering its energy relative to the 3d in the neutral atom context.
Effective Nuclear Charge and Periodic Trends
The effective nuclear charge is the net positive charge experienced by an electron, calculated as \(Z_{\text{eff}} = Z - S\), where \(Z\) is the atomic number and \(S\) is the shielding constant. Slater's rules provide a semi-empirical method to estimate \(S\). According to these rules, for an ns or np electron each other electron in the same (ns, np) group shields 0.35 (0.30 within 1s), each electron in shell \(n-1\) shields 0.85, and each electron in shells \(n-2\) and below shields 1.00; for an nd or nf electron, every electron in an inner group shields 1.00.
Because s-electrons penetrate more, they are shielded less effectively by other electrons in the same shell compared to p-electrons. This results in a higher \(Z_{\text{eff}}\) for s-electrons. For example, in Carbon (\(Z=6\)), the 2s electron experiences a higher \(Z_{\text{eff}}\) than the 2p electron. This difference in effective nuclear charge leads to the observed trend that atomic radii decrease across a period, as the increasing nuclear charge pulls the electron cloud closer.
The result
Quantum numbers and atomic orbitals
Solving the time-independent Schrödinger equation for an electron in a hydrogen-like atom, by separating variables in spherical coordinates, forces every stationary electron state to be labelled by three integers arising from the boundary conditions on the separated equation, plus a fourth from electron spin: the principal quantum number \(n=1,2,3,\dots\) (energy shell), the orbital angular momentum quantum number \(l=0,1,\dots,n-1\) (orbital shape), the magnetic quantum number \(m_l=-l,\dots,+l\) (orbital orientation), and the spin quantum number \(m_s=\pm\tfrac12\) (intrinsic electron spin). Each allowed \((n,l,m_l)\) triple defines one atomic orbital — a specific spatial probability-density pattern — and each orbital holds at most two electrons, one with each value of \(m_s\).
A common misconception is that the 4s orbital is always lower in energy than the 3d orbital. While this is true for neutral atoms like Potassium and Calcium, the energy levels shift upon ionisation. In transition metal cations, the 3d orbital often becomes lower in energy than the 4s orbital because the removal of 4s electrons reduces electron-electron repulsion and changes the shielding environment, making the 3d electrons more tightly bound. Therefore, when forming ions, electrons are removed from the 4s orbital before the 3d orbital, even though the 3d orbital was filled after the 4s in the neutral atom. C
Worked example
Problem. Calculate the effective nuclear charge (\(Z_{\text{eff}}\)) experienced by a 3p electron in a Phosphorus atom (\(Z=15\)) using Slater's rules. Assume the electron configuration is \(1s^2 2s^2 2p^6 3s^2 3p^3\).
- Identify the electron of interest: a 3p electron. According to Slater's rules, group the electrons: \((1s^2)(2s^2 2p^6)(3s^2 3p^3)\).
- Determine the shielding contribution from electrons in the same group (\(n=3\), ns/np). There are 4 other electrons in the \(3s^2 3p^3\) group (excluding the one we are calculating for). Each contributes 0.35. Contribution: \(4 \times 0.35 = 1.40\).
- Determine the shielding contribution from electrons in the \(n-1\) shell (\(n=2\)). There are 8 electrons in the \(2s^2 2p^6\) group. Each contributes 0.85. Contribution: \(8 \times 0.85 = 6.80\).
- Determine the shielding contribution from electrons in the \(n-2\) shell (\(n=1\)). There are 2 electrons in the \(1s^2\) group. Each contributes 1.00. Contribution: \(2 \times 1.00 = 2.00\).
- Sum the shielding constant \(S\): \(S = 1.40 + 6.80 + 2.00 = 10.20\).
- Calculate \(Z_{\text{eff}}\): \(Z_{\text{eff}} = Z - S = 15 - 10.20 = 4.80\).
Units check. Effective nuclear charge is a dimensionless quantity representing the net charge in units of elementary charge \(e\).
Three quick tests
- How many radial nodes does a 4d orbital have?
Answer
One: \(n - l - 1 = 4 - 2 - 1 = 1\). - Which orbital in the n=3 shell has the highest penetration?
Answer
3s. Lower l values have higher penetration. s (l=0) > p (l=1) > d (l=2). - Why does the 4s orbital fill before the 3d orbital in neutral atoms?
Answer
Because the 4s orbital has greater penetration towards the nucleus than the 3d orbital, allowing it to experience a higher effective nuclear charge and thus lower energy in the neutral atom context.
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Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.