Explain electron-affinity factors and group trends
Use nuclear attraction, distance, shielding and electron repulsion to explain first electron-affinity patterns in Groups 16 and 17.
Section 23 · Chemical energetics
A Level; Cambridge International Chemistry 9701, 2028–2030 syllabus, Topic 23. Use the syllabus sign convention for lattice enthalpy (gaseous ions forming the ionic solid); state phases, signs, units and temperature. Written practice is self-checked
Before you begin
- Use AS enthalpy changes and Hess's law, formulae for ions, balanced equations, and temperature in kelvins.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which factor helps explain why an added electron is less strongly attracted farther from the nucleus?
What you will learn
- Explain how effective nuclear attraction and distance affect electron gain
- Describe first electron-affinity trends in Groups 16 and 17
- Explain the small-atom exceptions using electron repulsion
Balance attraction against distance and shielding
An incoming electron is attracted to the nucleus. A stronger effective nuclear attraction and shorter distance tend to make electron gain more exothermic; greater shielding and a larger atomic radius weaken the attraction and tend to make it less exothermic.
The incoming electron also experiences repulsion from electrons already in the atom. In a very small orbital, this repulsion can reduce the energy released even though the nucleus is close.
Use the Group 16 and 17 exceptions accurately
Down each group the outer region is farther from the nucleus and more shielded, so first electron affinity generally becomes less exothermic. However, oxygen is less exothermic than sulfur and fluorine is less exothermic than chlorine: the compact 2p orbitals in O and F create greater electron–electron repulsion for the added electron.
Compare the first member with the second before stating the down-group trend. Chlorine's first electron affinity is more exothermic than fluorine's; sulfur's is more exothermic than oxygen's. Avoid describing every step as a perfectly smooth trend.
Pause and explain
Why is chlorine's first electron affinity more exothermic than fluorine's?
Worked example
Explain why chlorine's first electron affinity is more exothermic than fluorine's, even though fluorine is above chlorine in Group 17.
Show the worked solution
- The incoming electron enters a compact 2p orbital in fluorine.
- That small orbital has stronger electron–electron repulsion.
- The reduction in repulsion for chlorine makes its electron gain more exothermic overall.
Answer Fluorine's compact 2p orbital creates greater repulsion for the incoming electron. In chlorine, the electron enters a larger 3p orbital with less repulsion, so chlorine's first electron affinity is more exothermic than fluorine's.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Describe the general first-electron-affinity trend down Group 17 and give the important first-pair exception.
Give me a hint
Separate the overall distance/shielding trend from the compact-orbital exception.
Compare my reasoning
- Down the group, radius and shielding increase.
- The attraction for an added electron generally weakens.
- F is an exception because its 2p orbital is especially compact.
First electron affinity generally becomes less exothermic down Group 17. Fluorine is less exothermic than chlorine because the incoming electron experiences strong repulsion in fluorine's compact 2p orbital.
Look for these in your work
- General trend stated
- Distance/shielding linked
- F/Cl exception explained
Why is sulfur's first electron affinity more exothermic than oxygen's?
Give me a hint
Compare the 2p and 3p regions and account for the incoming electron.
Compare my reasoning
- O's added electron enters a compact 2p orbital.
- S's electron enters a larger 3p orbital.
- Reduced electron–electron repulsion makes S electron gain more exothermic.
Oxygen's compact 2p orbital has stronger repulsion for the new electron. Sulfur's larger 3p orbital reduces this repulsion, so sulfur's first electron affinity is more exothermic.
Look for these in your work
- Orbitals compared
- Repulsion direction correct
- Observed exception explained
An explanation says, ‘Down Group 16, electron affinity gets less exothermic only because nuclear charge falls.’ Identify two corrections.
Give me a hint
Check how proton number, shielding, radius and the O/S anomaly change down a group.
Compare my reasoning
- Nuclear charge does not fall down the group.
- Shielding and radius increase, reducing attraction at the outer region.
- O is less exothermic than S because of compact-orbital repulsion.
Nuclear charge increases down Group 16, while shielding and radius also increase; the added electron is farther from the nucleus and the trend is generally less exothermic. Oxygen is a first-pair exception because its compact 2p orbital increases repulsion.
Look for these in your work
- Nuclear-charge claim corrected
- Distance/shielding used
- Exception included
Common mistakes
- Claiming electron affinity becomes steadily more exothermic down both groups
- Using nuclear charge alone without distance or shielding
- Forgetting repulsion in the compact O and F orbitals
- Confusing a more negative value with a less exothermic process
What can you explain now?
Which factor helps explain why an added electron is less strongly attracted farther from the nucleus?
Compare with the explanation
Greater distance and shielding
Both reduce the effective attraction felt by the incoming electron.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
How do attraction, shielding and orbital size work together to explain why O and F are less exothermic than S and Cl?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.