Atomic and Ionic Radii
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You should be comfortable calculating effective nuclear charge (\(Z_{\text{eff}}\)) using Slater's rules and understand how shielding affects the attraction between the nucleus and valence electrons.
Why this matters
Atomic and ionic radii are the fundamental geometric parameters that determine how atoms pack together in solids and how they interact in chemical bonds.
Unlike macroscopic objects, atoms do not have a sharp, well-defined boundary. The electron cloud fades exponentially into space, meaning the "size" of an atom is a convention rather than a fixed physical edge. Consequently, atomic radius is defined operationally based on the context: covalent radius for bonded atoms, van der Waals radius for non-bonded contacts, and ionic radius for ions in crystal lattices. Understanding these trends allows chemists to predict lattice energies, bond lengths, and the structural stability of materials, bridging the gap between quantum mechanical probability distributions and observable macroscopic properties.
The Concept of Atomic Size
Because electrons are quantum particles described by wavefunctions, they do not orbit the nucleus in fixed paths like planets. Instead, they occupy orbitals where the probability of finding an electron is highest near the nucleus but extends indefinitely outward. This makes defining a precise "edge" for an atom impossible. To work around this, chemists define radii based on measurable distances between nuclei in specific environments. For example, in a covalent bond, the bond length is the distance between two nuclei. If we assume the two atoms are identical, the covalent radius is half that bond length. In a crystal of a noble gas, where atoms are held together by weak van der Waals forces, the distance between nearest neighbours defines the van der Waals radius, which is significantly larger than the covalent radius because the atoms are not pulled together by shared electrons. Similarly, ionic radii are derived from the distances between cations and anions in ionic crystals. These definitions are not absolute constants; they depend on the coordination number (the number of neighbouring ions) and the specific chemical environment. Therefore, when discussing atomic size, we must always specify the type of radius being used.
Periodic Trends in Atomic Radius
The size of an atom is governed by two competing factors: the principal quantum number \(n\) of the valence shell and the effective nuclear charge \(Z_{\text{eff}}\) experienced by those valence electrons. As established in the previous lecture, \(Z_{\text{eff}} = Z - S\), where \(S\) is the shielding constant.
Across a period from left to right, the principal quantum number \(n\) remains constant, but the nuclear charge \(Z\) increases. The added electrons enter the same shell and provide only partial shielding. Consequently, \(Z_{\text{eff}}\) increases, pulling the electron cloud closer to the nucleus. This results in a steady decrease in atomic radius. For instance, lithium has a larger radius than fluorine because the fluorine nucleus exerts a stronger pull on its valence electrons.
Down a group from top to bottom, the principal quantum number \(n\) increases. Each new period introduces a new, larger electron shell further from the nucleus. Although the nuclear charge \(Z\) increases significantly, the inner shells shield the valence electrons effectively. The increase in \(n\) dominates the trend, causing the valence electrons to be, on average, further from the nucleus. Thus, atomic radius increases down a group. For example, potassium is larger than sodium because its valence electron is in the \(n=4\) shell compared to sodium's \(n=3\) shell.
Atomic vs Ionic Radii
When atoms form ions, their sizes change dramatically due to the loss or gain of electrons. Cations (positive ions) are always smaller than their parent atoms. When a metal atom loses one or more valence electrons to form a cation, the number of electron shells may decrease, or the remaining electrons experience a higher \(Z_{\text{eff}}\) because there is less electron-electron repulsion and less shielding. For example, sodium (\(Z=11\)) has the electron configuration \(1s^2 2s^2 2p^6 3s^1\). The sodium ion (\(\mathrm{Na^+}\)) has the configuration \(1s^2 2s^2 2p^6\). The outer \(3s\) shell is removed, and the remaining electrons are pulled closer to the nucleus by the unchanged nuclear charge of +11. Thus, \(r(\mathrm{Na^+}) < r(\mathrm{Na})\).
Anions (negative ions) are always larger than their parent atoms. When a non-metal atom gains electrons to form an anion, the added electrons increase electron-electron repulsion in the valence shell. This repulsion pushes the electron cloud outward. Additionally, the added electrons do not increase the nuclear charge, so the \(Z_{\text{eff}}\) per electron decreases slightly. For example, chlorine (\(Z=17\)) gains an electron to form \(\mathrm{Cl^-}\). The extra electron increases repulsion among the \(3p\) electrons, expanding the radius. Thus, \(r(\mathrm{Cl^-}) > r(\mathrm{Cl})\).
Isoelectronic Series
An isoelectronic series is a set of ions that have the same number of electrons and the same electron configuration. Because the electron count is identical, the shielding \(S\) is roughly constant for all ions in the series. The only variable is the nuclear charge \(Z\). Therefore, the ion with the highest nuclear charge exerts the strongest pull on the electron cloud and has the smallest radius. Conversely, the ion with the lowest nuclear charge has the weakest pull and the largest radius.
Consider the isoelectronic series: \(\mathrm{O^{2-}}\), \(\mathrm{F^-}\), \(\mathrm{Na^+}\), \(\mathrm{Mg^{2+}}\), and \(\mathrm{Al^{3+}}\). All have 10 electrons (configuration \(1s^2 2s^2 2p^6\)). The nuclear charges are \(Z=8, 9, 11, 12, 13\) respectively. As \(Z\) increases, the radius decreases. \(\mathrm{Al^{3+}}\) is the smallest because the 13 protons pull the 10 electrons most tightly. \(\mathrm{O^{2-}}\) is the largest because the 8 protons pull the same 10 electrons least tightly. This trend is a direct consequence of Coulomb's law: the force of attraction is proportional to the product of the charges and inversely proportional to the square of the distance.
The result
Periodic trends
Define the effective nuclear charge felt by an electron as \(Z_{\text{eff}}=Z-S\), where \(S\) (the shielding constant) is computed from Slater's rules: for an electron in an \(ns\) or \(np\) orbital, sum \(0.35\) for every other electron in the same \(n s,np\) group, \(0.85\) for every electron in the \(n{-}1\) shell, and \(1.00\) for every electron in shells \(n{-}2\) and below. Then, moving left to right across a period, \(Z_{\text{eff}}\) for the valence electron increases (atomic radius decreases, first ionisation energy increases); moving down a group, \(Z_{\text{eff}}\) stays roughly constant while the valence principal quantum number \(n\) increases (atomic radius increases, first ionisation energy decreases).
A common misconception is that ionic radii are absolute, fixed values. In reality, ionic radii are relative parameters derived from crystal structures. The radius of an ion depends on its coordination number (the number of nearest neighbours) and the specific lattice energy of the crystal. For example, the radius of \(\mathrm{Na^+}\) is slightly different in \(\mathrm{NaCl}\) (coordination number 6) than in \(\mathrm{Na_2O}\) (coordination number 4). When comparing ionic sizes, one must ensure the data refers to the same coordination environment or use a consistent scale (such as the Shannon radii) to avoid errors in predicting lattice energies and structural preferences. C
Worked example
Problem. Consider the isoelectronic series \(\mathrm{P^{3-}}\), \(\mathrm{S^{2-}}\), \(\mathrm{Cl^-}\), \(\mathrm{K^+}\), and \(\mathrm{Ca^{2+}}\). Arrange these ions in order of increasing ionic radius. Given that the ionic radius of \(\mathrm{Cl^-}\) is approximately \(181\ \mathrm{pm}\) and the ionic radius of \(\mathrm{K^+}\) is approximately \(138\ \mathrm{pm}\), estimate the relative size difference between the largest and smallest ion in the series, assuming a linear trend with nuclear charge \(Z\).
- Identify the electron configuration. All ions have 18 electrons, corresponding to the configuration of Argon (\(1s^2 2s^2 2p^6 3s^2 3p^6\)).
- List the nuclear charges \(Z\) for each ion: \(\mathrm{P^{3-}}\) (\(Z=15\)), \(\mathrm{S^{2-}}\) (\(Z=16\)), \(\mathrm{Cl^-}\) (\(Z=17\)), \(\mathrm{K^+}\) (\(Z=19\)), \(\mathrm{Ca^{2+}}\) (\(Z=20\)).
- Apply the isoelectronic trend: as \(Z\) increases, the radius \(r\) decreases. Therefore, the order of increasing radius is the reverse of the order of increasing \(Z\).
- Order: \(\mathrm{Ca^{2+}} < \mathrm{K^+} < \mathrm{Cl^-} < \mathrm{S^{2-}} < \mathrm{P^{3-}}\).
- Estimate the size difference. The smallest ion is \(\mathrm{Ca^{2+}}\) and the largest is \(\mathrm{P^{3-}}\). We are given \(r(\mathrm{Cl^-}) = 181\ \mathrm{pm}\) and \(r(\mathrm{K^+}) = 138\ \mathrm{pm}\). The difference in \(Z\) between \(\mathrm{Cl^-}\) and \(\mathrm{K^+}\) is \(19 - 17 = 2\). The difference in radius is \(181 - 138 = 43\ \mathrm{pm}\). This suggests a change of approximately \(21.5\ \mathrm{pm}\) per unit change in \(Z\) in this region. From \(\mathrm{K^+}\) (\(Z=19\)) to \(\mathrm{Ca^{2+}}\) (\(Z=20\)), \(\Delta Z = 1\), so estimated \(r(\mathrm{Ca^{2+}}) \approx 138 - 21.5 = 116.5\ \mathrm{pm}\). From \(\mathrm{Cl^-}\) (\(Z=17\)) to \(\mathrm{S^{2-}}\) (\(Z=16\)), \(\Delta Z = -1\), so estimated \(r(\mathrm{S^{2-}}) \approx 181 + 21.5 = 202.5\ \mathrm{pm}\). From \(\mathrm{S^{2-}}\) (\(Z=16\)) to \(\mathrm{P^{3-}}\) (\(Z=15\)), \(\Delta Z = -1\), so estimated \(r(\mathrm{P^{3-}}) \approx 202.5 + 21.5 = 224\ \mathrm{pm}\). The difference between the largest (\(\mathrm{P^{3-}}\)) and smallest (\(\mathrm{Ca^{2+}}\)) is approximately \(224 - 116.5 = 107.5\ \mathrm{pm}\).
Units check. The input radii are in picometres (\(\mathrm{pm}\)), and the estimated difference is also in picometres, which is consistent with standard ionic radii magnitudes.
Reality check. Tabulated (Shannon, six-coordinate) radii are about 212 pm for P3-, 184 pm for S2- and 100 pm for Ca2+, so the true spread is about 112 pm. The straight-line estimate gets the order and the size of the spread right, but the radius is not really linear in \(Z\).
Three quick tests
- Why is the ionic radius of \(\mathrm{Na^+}\) smaller than the atomic radius of \(\mathrm{Na}\)?
Answer
\(\mathrm{Na^+}\) has lost its outer \(3s\) electron, reducing the principal quantum number \(n\) and increasing the effective nuclear charge felt by the remaining electrons, pulling them closer to the nucleus. - Arrange the following isoelectronic ions in order of decreasing radius: \(\mathrm{N^{3-}}\), \(\mathrm{O^{2-}}\), \(\mathrm{F^-}\), \(\mathrm{Ne}\), \(\mathrm{Na^+}\).
Answer
\(\mathrm{N^{3-}} > \mathrm{O^{2-}} > \mathrm{F^-} > \mathrm{Ne} > \mathrm{Na^+}\). As nuclear charge \(Z\) increases, the radius decreases for isoelectronic species. - Explain why the atomic radius decreases from left to right across Period 3.
Answer
Across Period 3, the principal quantum number \(n\) remains constant at 3, but the nuclear charge \(Z\) increases. The added electrons provide incomplete shielding, leading to an increase in \(Z_{\text{eff}}\), which pulls the valence electrons closer to the nucleus.
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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.