🧪AP Chemistry Review
Unit 1 FRQ (Photoelectron Spectroscopy) with Feedback
Unit 1 FRQ (Photoelectron Spectroscopy) with Feedback
AP Chemistry Free Response Question for Electron Spectroscopy
AP Chemistry Section II (Free Response) contains 7 questions completed in 105 minutes: 3 long free-response questions worth 10 points each and 4 short free-response questions worth 4 points each. Photoelectron spectroscopy may appear in short-answer or parts of longer questions and typically asks students to interpret peaks, relate peak position to binding energy, and justify claims using nuclear charge and electron shielding ideas.
How to answer this AP FRQ
For PES FRQs, earn points by:
- Identifying the element from the number and relative size of peaks
- Writing the electron configuration requested
- Justifying peak shifts with chemical reasoning
On AP Chemistry FRQs, a justification should explicitly connect the claim to nuclear charge and the attraction between the nucleus and electrons. Avoid vague statements such as “smaller radius” unless you tie them to stronger Coulombic attraction and higher binding energy.
Consider the following photoelectron spectroscopy data:
a) Based on the graph, identify the element and write the full electron configuration.
b) A student claims that the 1s peak for sulfur (S) would be further to the right than the 1s peak of this element. Is this student correct? Justify your answer.
FRQ Writing Samples & Feedback
FRQ Practice Submission 1
a) 1s^2 2s^2 2p^6 3s^2 3p^2 Silicon
b)This student is incorrect because the distance between the 1s orbital and the nucleus for a sulfur atom is less than the distance for a silicon atom due to the increased number of protons in sulfur that allow for a strong attraction and hold on the inner electrons. This means that more energy would be required for an s orbital electron to be removed and the 1s peak would be further left than the 1s peak for Silicon.
Teacher FRQ Feedback
I think the only change would be referencing the attractive force to the electrons to the nucleus, not the distance (Though true). A better statement would sound "Since sulfur has more protons than silicon, sulfur has a greater nuclear charge, thus a greater attraction to the 1s electrons." Nice!
FRQ Practice Submission 2
a) Silicon is the element. The full electron configuration is: 1s^2 2s^2 2p^6 3s^2 3p^2.
b) The student is incorrect - the 1’st peak for sulfur would NOT be further to the right than the 1st peak for silicon. This is because sulfur has greater nuclear charge, so its nucleus attracts the 1s electrons more strongly. As a result, more energy is required to remove a 1s electron from sulfur, so the 1s peak would be further to the left.
Teacher FRQ Feedback
Your identification of silicon is correct, but the prompt asked for the full electron configuration, not shorthand notation. A complete response is: 1s^2 2s^2 2p^6 3s^2 3p^2.
Teacher's Solutions
a) The element is silicon. It has an electron configuration of 1s^2 2s^2 2p^6 3s^2 3p^2.
b) This student is incorrect. The 1s peak for sulfur would be further to the left on the photoelectron spectrum. This is due to the fact that sulfur has more protons in the nucleus and therefore has a greater nuclear charge. This greater positive charge attracts the electrons in sulfur more strongly, thus moving the peak further to the left.