1.3.6 (HL)—Ionization energy (IE)
- Syllabus
- First assessment 2025
- Objective
- 1.3.6
- Level
- HL
First ionization energy is the energy required to remove one mole of electrons from one mole of gaseous atoms. The convergence limit in an emission spectrum corresponds to ionization.
| Trend | Explanation |
|---|---|
| Across a period | Generally increases as effective nuclear charge increases |
| Down a group | Generally decreases because the outer electron occupies a higher shell |
| Be → B dip | The electron removed from B is in a higher-energy p subshell |
| N → O dip | Pairing in a p orbital makes one electron easier to remove |
E=hfandc=λf
Worked example — hydrogen convergence limit
The local course book gives λ=9.12×10−8m. First, f=c/λ=(3.00×108ms−1)/(9.12×10−8m)=3.29×1015s−1. Then Ephoton=hf=(6.63×10−34Js)(3.29×1015s−1)=2.18×10−18J. Convert one-photon energy to one mole and joules to kilojoules: IE=(2.18×10−18)(6.02×1023)/1000=1.31×103kJmol−1. This is the molar energy for the first ionization process H(g)→H+(g)+e−.
At the convergence limit, convert wavelength or frequency to energy per photon with E = hf, then multiply by the Avogadro constant and convert J mol⁻¹ to kJ mol⁻¹. Across-period trends are general patterns; subshell energy and electron pairing explain the named dips.
X(g)→X+(g)+e−firstionizationenergyinkJmol−1
Questions combine spectral calculations using the convergence limit with explanations of periodic first-IE trends and their sublevel-related discontinuities.
determine / explain
Convert molar ionization energy to energy per atom before using E=hf or E=hc/λ, and for trend explanations link nuclear charge, shielding, shell/sublevel energy and electron repulsion to the ease of removing the specified electron.
Using molar energy directly in E=hf without dividing by Avogadro's constant
Representative question
Determine the frequency of electromagnetic radiation, in s−1, equivalent to the first ionization energy of phosphorus. Use sections 1, 2 and 9 of the data booklet.
« 1.012×106 J mol−1/6.02×1023= » 1.68×10−18 «J»
≪1.68×10−18 J/6.63×10−34 J s−1=>2.54×1015μS−1 »
Marking guidance:
Award [2] for the correct final
answer.
Retrieve the chain: emission lines reveal discrete levels; capacities, sublevels, orbitals, and spin rules build configurations; first and successive ionization energies then reveal how electrons are held and arranged.
When checking an answer, ask: Did I link a line to a transition? Did I use 2n² and the filling rules? Did I explain an ionization trend or count electrons before a successive-IE jump?