General Chemistry
Ionization Energy
Energy required to remove an electron — predicts reactivity and bonding
Ionization energy (IE) is the energy required to remove the most loosely bound electron from a gaseous atom. First IE: removes first electron. Second IE: removes second (always higher than first). Trends: increases across periods (more nuclear charge), decreases down groups (electrons farther from nucleus). Anomalies (Be > B, N > O) reveal electron configuration effects. Low IE: easy to lose electrons → metallic, reactive (alkalis). High IE: hard to remove → nonmetallic, stable (noble gases). Predicts bond formation and oxidation states.
- First IE definitionX(g) → X⁺(g) + e⁻; energy required
- H first IE1312 kJ/mol
- He first IE2372 kJ/mol (highest among stable elements)
- Cs first IE376 kJ/mol (lowest among stable)
- Trend across periodIncreases (more effective nuclear charge)
- Trend down groupDecreases (electrons farther from nucleus)
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Why IE matters
- Bonding type. Ionic vs covalent.
- Reactivity. Metals with low IE are reactive.
- Oxidation states. Successive IEs predict states.
- Periodic trends. Confirms electron configuration model.
- Spectroscopy. Photoelectron spectroscopy.
- Material design. Solar cells, semiconductors.
- Plasma physics. Ionization at high T.
Common misconceptions
- IE is constant. Different for each successive electron.
- Fluorine has lowest IE. Highest IE (excluding noble gases).
- Lower IE means smaller atom. Opposite — larger atoms easier to ionize.
- Trends have no exceptions. Be > B, N > O are notable.
- IE is always positive. Yes — by definition; energy required.
- IE measures only outermost electron. Multiple IEs possible per atom.
Frequently asked questions
What is ionization energy?
Energy required to remove an electron from a gaseous atom in its ground state. X(g) + IE → X⁺(g) + e⁻. Always positive (energy must be added). Measured in kJ/mol or eV/atom (1 eV ≈ 96.5 kJ/mol). First IE = first electron; second IE = second; etc. Successive IEs always increase.
Why does IE increase across a period?
Effective nuclear charge increases as more protons added but electrons go to same shell. Outer electrons feel stronger pull → harder to remove. Li IE = 520 kJ/mol; Ne IE = 2081 kJ/mol. Most dramatic across short periods (rows 2-3) where shells aren't yet large.
Why does IE decrease down a group?
Outer electrons farther from nucleus, more shielded by inner shells. Despite more total protons, the effective charge experienced by outermost electron is similar. Distance dominates. Li (2nd row): 520 kJ/mol. K (4th): 419. Rb (5th): 403. Cs (6th): 376. Steady decrease.
Why is Be IE higher than B?
Anomaly. Be: [He] 2s² (full 2s subshell). B: [He] 2s² 2p¹. The lone 2p electron is higher in energy than the paired 2s — easier to remove. Despite B having more protons, removing 2p is easier than removing 2s. Similar: N (half-filled 2p³, stable) > O.
What about successive ionization energies?
Each successive IE is larger. Big jumps occur at noble-gas configuration. Sodium: IE1 = 496, IE2 = 4562 (huge jump — removing core electron). After IE1, you have Na⁺ = [Ne]; very stable. Same pattern for all metals — they lose few electrons before encountering huge jump.
How does IE predict bonding?
Low IE elements (alkalis): easily form cations → ionic bonds. High IE: nonmetals; rarely form cations; tend to gain electrons (form anions or covalent bonds). IE difference between two atoms predicts whether bond is more ionic or covalent. Combined with electron affinity gives even better picture.
How is IE measured?
Photoelectron spectroscopy (PES). Bombard sample with high-energy photons; measure kinetic energy of ejected electrons. KE = hv - IE. Different orbital electrons have different IEs — gives profile of orbital structure. Modern PES distinguishes 1s, 2s, 2p, etc. Foundation of orbital theory experimental verification.