Ionisation Enthalpy: Successive Values and Periodic Trends
Ionisation enthalpy is one of the most useful ideas in the periodic table. It explains why sodium forms Na+ and not Na2+, why metals lose electrons easily, and why noble gases hardly react. In school exams you are asked for trends, for exceptions such as Be and B or N and O, and for short numericals on energy per atom. Class 11 NCERT and ICSE both cover it, so it is worth learning once and properly.
The definition and the formulas
X+(g) → X2+(g) + e− second ionisation enthalpy, IE2
For a one-electron species (H, He+, Li2+): IE = 1312 × Z2 / n2 kJ/mol
Energy per atom = IE (kJ/mol) × 1000 / NA (in joules)
Photon wavelength: λ = hc / E
The state is gaseous. The atom must be isolated, in the gas phase and in its ground state, so that no other effect (such as the lattice energy of a solid) mixes in. Units: kJ/mol for a mole of atoms, or electron volts (eV) per atom. Z is the nuclear charge and n the shell from which the electron is removed. NA = 6.022 × 1023 per mole, h = 6.626 × 10−34 J s and c = 3.00 × 108 m/s.
The expression 1312 Z2/n2 works only when the species has a single electron. For atoms with many electrons, repulsion between electrons and shielding change the value, so you cannot use it for Na or Mg.
What decides the value
- Atomic size. The larger the atom, the farther the outer electron is from the nucleus, and the easier it is to remove.
- Nuclear charge. More protons pull the electron harder.
- Shielding. Inner electrons cancel part of the nuclear pull on the outer electron.
- Subshell type. An s electron is held more tightly than a p electron of the same shell because it penetrates closer to the nucleus. Half-filled and fully filled subshells are extra stable.
Trends in the periodic table
| Direction | General trend of IE1 | Main reason |
|---|---|---|
| Across a period (left to right) | Increases, with small dips | Nuclear charge rises while shielding stays about the same, so size shrinks |
| Down a group | Decreases | New shells are added, size grows, shielding grows |
| Successive (IE1 < IE2 < IE3) | Always increases | Fewer electrons remain, so the same nuclear charge pulls harder |
Worked example 1: hydrogen in joules, eV and wavelength
The ionisation enthalpy of hydrogen is 1312 kJ/mol. Find the energy per atom, the value in eV, and the longest wavelength of light that can just ionise a hydrogen atom.
Energy per atom = 1312 × 1000 / 6.022 × 1023 = 2.179 × 10−18 J.
In eV: 2.179 × 10−18 / 1.602 × 10−19 = 13.6 eV.
λ = hc / E = (6.626 × 10−34 × 3.00 × 108) / 2.179 × 10−18 = 9.12 × 10−8 m = 91 nm. This is in the ultraviolet region, so visible light cannot ionise hydrogen.
Worked example 2: He+ and Li2+
Find the energy needed to remove the last electron from He+ (Z = 2) and from Li2+ (Z = 3), both in the ground state, n = 1.
He+: 1312 × 22 / 12 = 1312 × 4 = 5248 kJ/mol.
Li2+: 1312 × 32 / 12 = 1312 × 9 = 11 808 kJ/mol.
The energy grows with Z2, so tripling the nuclear charge makes the electron nine times harder to remove.
Worked example 3: finding the group from successive values
An element has approximate successive ionisation enthalpies of 738, 1451 and 7733 kJ/mol. How many valence electrons does it have?
IE2 / IE1 = 1451 / 738 = 1.97. IE3 / IE2 = 7733 / 1451 = 5.33.
The values rise gently for the first two electrons and then jump by more than five times. The third electron comes from a stable inner shell. So the element has 2 valence electrons and belongs to group 2. These values match magnesium (textbook tables differ slightly in the last digit).
Compare sodium: IE1 about 496 and IE2 about 4562 kJ/mol. The ratio is 4562 / 496 = 9.2, so the big jump comes after the first electron, which tells us sodium has 1 valence electron.
Worked example 4: energy to ionise 1 g of sodium
Use IE1 of sodium = 496 kJ/mol and molar mass 22.99 g/mol. Find the energy to ionise 1.00 g of gaseous Na atoms, and the wavelength of a photon that can ionise one atom.
Moles = 1.00 / 22.99 = 0.0435 mol. Energy = 0.0435 × 496 = 21.6 kJ.
Per atom: 496 000 / 6.022 × 1023 = 8.24 × 10−19 J (5.14 eV). λ = hc / E = (6.626 × 10−34 × 3.00 × 108) / 8.24 × 10−19 = 2.41 × 10−7 m = 241 nm, again in the ultraviolet.
The exceptions your teacher will ask about
- Be is higher than B. In Be the electron is removed from a filled 2s orbital. In B it comes from a 2p orbital, which is higher in energy and easier to lose.
- N is higher than O. N has a half-filled 2p subshell. In O one 2p orbital holds a pair of electrons, and the repulsion between the pair makes one of them easier to remove.
- Noble gases have the highest value in their period because their shells are full and the atoms are small.
Common mistakes
- Mixing up ionisation enthalpy and electron gain enthalpy. One removes an electron from an atom, the other adds one.
- Forgetting the gaseous state. Solid sodium needs extra energy to become gaseous atoms first.
- Using 1312 Z2/n2 for many-electron atoms. It is valid only for H, He+, Li2+ and similar one-electron species.
- Saying that IE2 is smaller than IE1. Successive values always increase.
- Mixing units. Keep kJ/mol for moles and J or eV for one atom.
Where this appears in school exams
| Question type | What to do |
|---|---|
| Arrange elements by IE1 | Use period and group trends, then check the Be/B and N/O exceptions |
| Identify the group from successive IE values | Find the first big jump in the ratio |
| Energy or wavelength numerical | Convert kJ/mol to J per atom with NA, then use hc/E |
| One-electron species | Use 1312 Z2/n2 |
The exact mark weightage changes between boards and years, so check your board’s current syllabus and sample paper.
Use the interactive periodic table to compare the elements discussed above.
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