Ionisation Energy
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You should be comfortable with the Bohr model of the hydrogen atom, the concept of effective nuclear charge (\(Z_{\text{eff}}\)), and Slater’s rules for calculating shielding. Familiarity with Coulomb’s law and the inverse-square relationship between force and distance is assumed.
Why this matters
Ionisation energy is the fundamental metric for chemical reactivity and bond strength.
Understanding why it is difficult to remove an electron from an atom explains the periodic table’s structure, the stability of noble gases, and the driving force behind the formation of ionic and covalent bonds. It bridges the gap between quantum mechanical orbital occupancy and observable macroscopic chemical behaviour, serving as a primary predictor of elemental properties.
The Energy Cost of Removal
Ionisation energy is defined as the minimum energy required to remove the most loosely bound electron from a neutral gaseous atom to form a gaseous cation. Think of an electron as being held in place by the electrostatic attraction of the positively charged nucleus. To remove it, you must do work against this attraction, much like lifting a weight against gravity. The "cost" of this removal depends on two competing factors: how strongly the nucleus pulls (which increases with nuclear charge) and how far away the electron is (which increases with orbital size and shielding by inner electrons). If the nucleus pulls harder or the electron is closer, the energy required is higher. If inner electrons shield the outer electron from the nuclear pull, or if the electron is in a larger orbital further from the nucleus, the energy required is lower. This balance dictates the periodic trends we observe.
Quantitative Definition and Photoionisation
The first ionisation energy, \(I_1\), is usually quoted in \(\mathrm{kJ\,mol^{-1}}\) (per mole of atoms). It represents the enthalpy change for the process: \[ \mathrm{X(g) \to X^+(g) + e^-} \] While historically measured via spectroscopy, the concept is directly linked to the photoelectric effect. light behaves as a stream of photons with energy \(E=hf\). For an isolated atom, photoionisation occurs only if the photon energy exceeds the binding energy of the electron. This provides a direct experimental method to determine \(I_1\): the threshold frequency \(f_0\) corresponds exactly to the ionisation energy. \[ I_1 = h f_0 \] If the incident light frequency is below this threshold, no electrons are ejected, regardless of intensity. This confirms that ionisation is a quantum event dependent on individual photon energy, not cumulative thermal energy.
Periodic Trends via Effective Nuclear Charge
The variation of \(I_1\) across the periodic table is governed by \(Z_{\text{eff}}\). Moving left to right across a period, the nuclear charge \(Z\) increases by one for each element, while electrons are added to the same principal shell \(n\). According to Slater’s rules, the shielding constant \(S\) increases only slightly (by \(0.35\) per electron in the same \(ns, np\) group). Consequently, \(Z_{\text{eff}}\) increases significantly. The valence electrons are pulled closer to the nucleus, reducing the atomic radius \(r\). Since electrostatic potential energy is proportional to \(-Z_{\text{eff}}/r\), the binding energy increases, and \(I_1\) rises. \[ \text{Across a period: } Z_{\text{eff}}\uparrow \implies r\downarrow \implies I_1\uparrow \] Moving down a group, the principal quantum number \(n\) increases. Although \(Z\) increases, the number of inner shielding electrons also increases substantially. Slater’s rules assign a shielding contribution of \(1.00\) for all electrons in shells \(n-2\) and below. Thus, \(Z_{\text{eff}}\) remains relatively constant. However, the valence electron is in a higher energy orbital, further from the nucleus. The increased distance \(r\) dominates the interaction, weakening the nuclear attraction and lowering \(I_1\). \[ \text{Down a group: } n\uparrow, r\uparrow \implies I_1\downarrow \]
Anomalies and Subshell Stability
While the general trend is smooth, deviations occur due to electron configuration stability. Two key factors cause these anomalies: subshell half-filling and full-filling, and the difference in penetration between \(s\) and \(p\) orbitals. First, half-filled (\(p^3\)) and fully filled (\(p^6\)) subshells exhibit extra stability due to exchange energy and symmetrical charge distribution. This makes it harder to remove an electron than expected. For example, Nitrogen (\(1s^2 2s^2 2p^3\)) has a higher \(I_1\) than Oxygen (\(1s^2 2s^2 2p^4\)), despite Oxygen having a higher \(Z\). In Oxygen, one \(2p\) orbital contains a pair of electrons. The electron-electron repulsion within this orbital makes one electron easier to remove, lowering \(I_1\) relative to Nitrogen. Second, \(s\) orbitals penetrate closer to the nucleus than \(p\) orbitals of the same shell. Therefore, \(s\) electrons are more tightly bound. This explains why Beryllium (\(2s^2\)) has a higher \(I_1\) than Boron (\(2s^2 2p^1\)). The \(2p\) electron in Boron is shielded by the \(2s\) electrons and is further from the nucleus on average, making it easier to remove.
The results
Ionisation and the photoelectric effect
When light of frequency \(f\) strikes a material with work function \(\phi\) (the minimum energy needed to remove an electron), the maximum kinetic energy of any ejected electron is \(KE_{\max}=hf-\phi\), and no electrons are ejected at all if \(f\) is below the threshold frequency \(f_0=\phi/h\), no matter how intense the light. For an isolated gas-phase atom, the same relation holds with the work function replaced by the atom's first ionisation energy \(I_1\): photoionisation occurs only for \(hf\ge I_1\).
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 ionisation energy is a constant property of an element. In reality, the \(n\)-th ionisation energy is always significantly higher than the \((n-1)\)-th because you are removing an electron from an increasingly positive ion. For example, the first ionisation energy of Sodium is low, but the second is very high because you are removing an electron from the stable Neon-like core of \(\mathrm{Na^+}\). This jump in energy is used to determine the valency of an element. C
Worked example
Problem. Calculate the threshold wavelength \(\lambda_0\) (in nm) of light required to photoionise a gaseous Sodium atom, given that the first ionisation energy of Sodium is \(496\ \mathrm{kJ\,mol^{-1}}\). Use \(h = 6.626 \times 10^{-34}\ \mathrm{J\,s}\), \(c = 2.998 \times 10^8\ \mathrm{m\,s^{-1}}\), and \(N_A = 6.022 \times 10^{23}\ \mathrm{mol^{-1}}\).
- Convert the molar ionisation energy to energy per atom (Joules): \[ E_{\text{atom}} = \frac{496 \times 10^3\ \mathrm{J\,mol^{-1}}}{6.022 \times 10^{23}\ \mathrm{mol^{-1}}} = 8.236 \times 10^{-19}\ \mathrm{J} \]
- Use the photoionisation condition \(E_{\text{atom}} = h f_0 = \frac{hc}{\lambda_0}\) to solve for \(\lambda_0\): \[ \lambda_0 = \frac{hc}{E_{\text{atom}}} \]
- Substitute the constants: \[ \lambda_0 = \frac{(6.626 \times 10^{-34}\ \mathrm{J\,s})(2.998 \times 10^8\ \mathrm{m\,s^{-1}})}{8.236 \times 10^{-19}\ \mathrm{J}} \]
- Calculate the value: \[ \lambda_0 = \frac{1.986 \times 10^{-25}\ \mathrm{J\,m}}{8.236 \times 10^{-19}\ \mathrm{J}} = 2.411 \times 10^{-7}\ \mathrm{m} \]
- Convert to nanometres: \[ \lambda_0 = 241.1\ \mathrm{nm} \]
Units check. \(\frac{\mathrm{J\,s} \cdot \mathrm{m\,s^{-1}}}{\mathrm{J}} = \mathrm{m}\), which is consistent with wavelength units.
Three quick tests
- Which has a higher first ionisation energy: Magnesium or Aluminium?
Answer
Magnesium. Mg has a stable \(3s^2\) configuration, while Al has a \(3p^1\) electron that is further from the nucleus and shielded by the \(3s\) electrons. - Why is the second ionisation energy of Lithium much higher than its first?
Answer
The first electron is removed from the \(2s\) orbital. The second electron must be removed from the stable, inner \(1s\) core of the \(\mathrm{Li^+}\) ion, which is much closer to the nucleus and experiences a higher \(Z_{\text{eff}}\). - If the threshold frequency for photoionising potassium atoms is \(1.05 \times 10^{15}\ \mathrm{Hz}\), what is the ionisation energy in \(\mathrm{kJ\,mol^{-1}}\)?
Answer
\(E = h f N_A = (6.626 \times 10^{-34})(1.05 \times 10^{15})(6.022 \times 10^{23}) \approx 419\ \mathrm{kJ\,mol^{-1}}, close to the tabulated value\).
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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.