AP Chemistry Quiz: Photoelectron Spectroscopy

What this quiz covers

This quiz focuses on Photoelectron Spectroscopy, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. 2s
2. 1s
3. 2p (correct answer)
4. 3s
5. 3p

Explanation: This question assesses understanding of photoelectron spectroscopy. In photoelectron spectroscopy, binding energy shows energy to remove electrons, higher for core levels. The plot increases energy left, with right peaks for valence. Peak heights are proportional to electrons in subshells. A tempting distractor is E 3p, assuming period 3, but three peaks with rightmost 3 match 2p³ for nitrogen-like atoms. A transferable strategy is that inner electrons have higher binding energy, and p subshell heights range from 1 to 6.

Question 2

1. K (correct answer)
2. Ca
3. Ar
4. Na
5. Cl

Explanation: This question assesses understanding of photoelectron spectroscopy. In photoelectron spectroscopy, binding energy quantifies attraction, greater for inner shells. Spectra show left-higher energy, from 1s outward. Heights indicate subshell electrons, for configuration matching. A tempting distractor is D Na, with four peaks, but six peaks of 2,2,6,2,6,1 match potassium's 1s²2s²2p⁶3s²3p⁶4s¹. A transferable strategy is that inner electrons have higher binding energy, and the rightmost peak identifies valence electrons.

Question 3

1. The conclusion is correct because a larger rightmost peak always indicates a higher principal energy level.
2. The conclusion is incorrect because peak height indicates number of electrons, not the period by itself. (correct answer)
3. The conclusion is correct because the largest peak must correspond to the 3p subshell.
4. The conclusion is incorrect because binding energy increases to the right in PES.
5. The conclusion is correct because the rightmost peak always corresponds to the 1s subshell.

Explanation: This question assesses understanding of photoelectron spectroscopy. In photoelectron spectroscopy, binding energy measures nuclear pull on electrons, higher for inner shells. Spectra plot increasing energy left, from core to valence. Peak heights correspond to electron counts per subshell. A tempting distractor is C, thinking largest peak is always 3p, but height indicates electrons, not period, so the conclusion errs without considering peak count. A transferable strategy is that inner electrons have higher binding energy, and analyze both height and position for accurate interpretation.

Question 4

1. Electrons in the 3s subshell experience less shielding than electrons in the 2p subshell.
2. Electrons in the 2p subshell are, on average, closer to the nucleus than electrons in the 3s subshell and are held more tightly. (correct answer)
3. The 2p peak is taller than the 3s peak, so it must have higher binding energy.
4. Electrons in the 3s subshell have the same binding energy as electrons in the 2p subshell because both are s and p orbitals.
5. Electrons in the 2p subshell have lower binding energy because p electrons are always less tightly bound than s electrons.

Explanation: This question tests photoelectron spectroscopy. Electrons in different principal energy levels (shells) have different binding energies, with inner shells having higher binding energy. The 2p electrons are in the n=2 shell while 3s electrons are in the n=3 shell, making 2p electrons closer to the nucleus on average and more tightly bound. This results in higher binding energy for 2p compared to 3s, regardless of the s vs p orbital type. Students might incorrectly think peak height determines binding energy (choice C) or that orbital type matters more than shell number (choice E). Remember that binding energy primarily depends on distance from nucleus: inner electrons have higher binding energy than outer electrons.

Question 5

1. Ne
2. Na
3. Mg (correct answer)
4. Al
5. Si

Explanation: This question tests photoelectron spectroscopy. The spectrum shows only four peaks (1s, 2s, 2p, 3s) with no 3p peak, meaning there are no electrons in the 3p subshell. The peak heights sum to 12 electrons total (2+2+6+2), corresponding to magnesium (Mg) with configuration 1s²2s²2p⁶3s². The absence of a 3p peak is crucial - it means all valence electrons are in the 3s subshell. Students might incorrectly choose aluminum (Al) thinking it could have an empty 3p, but Al would show a 3p peak with one electron. Remember that PES only shows peaks for occupied subshells; empty subshells produce no signal.

Question 6

1. Mg
2. Si
3. P
4. Al (correct answer)
5. Na

Explanation: This question tests photoelectron spectroscopy. The spectrum shows five peaks with the 3p peak having height 1, indicating just one electron in the 3p subshell. Combined with a 3s peak of height 2, this represents the electron configuration ending in 3s²3p¹, which is aluminum (Al) with 13 total electrons. The complete configuration is 1s²2s²2p⁶3s²3p¹. Students might incorrectly choose sodium (Na) or magnesium (Mg), but these wouldn't have any 3p electrons and would show only four peaks. Remember that the presence and height of peaks directly indicate which subshells contain electrons and how many.

Question 7

1. N (correct answer)
2. Be
3. Ne
4. Li
5. B

Explanation: This problem involves photoelectron spectroscopy analysis. In PES, the highest binding energy peak corresponds to 1s electrons (closest to nucleus), and peak height indicates electron count in each subshell. The spectrum shows peaks with heights 2, 2, and 3, representing 1s (2e⁻), 2s (2e⁻), and 2p (3e⁻) respectively. Adding these gives 7 total electrons, identifying the element as nitrogen (N) with electron configuration 1s² 2s² 2p³. Boron might seem plausible since it also has electrons in all three subshells, but it would show a 2p peak height of only 1, not 3. The strategy is to use peak heights to count total electrons and match to the atomic number of elements in the answer choices.

Question 8

1. 3p
2. 2s
3. 2p
4. 3s (correct answer)
5. 1s

Explanation: This problem tests photoelectron spectroscopy concepts. In PES, binding energy decreases as you move from inner to outer electron shells (1s > 2s > 2p > 3s), and peak height corresponds to the number of electrons in each subshell. The peak heights 2, 2, 6, and 1 (from highest to lowest binding energy) represent the electron counts in each subshell. Since s subshells hold maximum 2 electrons and p subshells hold maximum 6 electrons, the heights correspond to: 1s (2), 2s (2), 2p (6), and 3s (1). The peak with height 1 represents a partially filled subshell with only one electron, which must be the 3s subshell. Students might incorrectly think the smallest peak corresponds to 1s because it's the "first" subshell, but 1s is always fully filled with 2 electrons in neutral atoms beyond hydrogen. Remember: the outermost electrons have the lowest binding energy and can be partially filled.

Question 9

1. 2p electrons are more penetrating and therefore are held more tightly than 2s electrons.
2. 2s electrons experience less shielding and are held more tightly by the nucleus than 2p electrons. (correct answer)
3. 2s electrons have a higher principal energy level than 2p electrons.
4. 2p electrons have a greater nuclear charge than 2s electrons.
5. 2s electrons have lower binding energy because they are in an s subshell.

Explanation: This question tests photoelectron spectroscopy. Within the same principal energy level, s electrons have higher binding energy than p electrons because s orbitals penetrate closer to the nucleus and experience less shielding from inner electrons. The 2s electrons spend more time near the nucleus than 2p electrons, making them more tightly bound despite being in the same shell. This greater penetration means 2s electrons feel a stronger effective nuclear charge. Students might incorrectly think 2p electrons are held more tightly (choice A) or confuse principal energy levels (choice C). Remember that within the same shell, penetration follows the pattern s > p > d > f, leading to binding energies in the same order.

Question 10

1. Atom Y has more core electrons than atom X.
2. Atom Y has a greater nuclear charge than atom X. (correct answer)
3. Atom Y has fewer protons than atom X.
4. Atom Y must have a larger atomic radius than atom X.
5. Atom Y must have fewer valence electrons than atom X.

Explanation: This question tests photoelectron spectroscopy. When comparing atoms in the same period, higher binding energy for the same electron shell indicates stronger nuclear attraction. Since atom Y's 1s electrons have higher binding energy than atom X's 1s electrons, atom Y must have more protons (greater nuclear charge) pulling on those electrons. Being in the same period means they have the same number of electron shells, so the difference must be in nuclear charge. Students might incorrectly think higher binding energy means fewer protons (choice C) or relates to atomic radius (choice D). Remember that across a period, increasing nuclear charge leads to higher binding energies for all electron shells.

Question 11

1. Atom Y has a greater nuclear charge than atom X. (correct answer)
2. Atom Y has more 2p electrons than atom X.
3. Atom Y has a lower nuclear charge than atom X.
4. Atom Y has fewer total electrons, so electron-electron repulsion is smaller.
5. Atom Y must be an anion, which increases binding energy for all electrons.

Explanation: This question examines photoelectron spectroscopy principles. In PES, binding energy reflects how tightly electrons are held by the nucleus - stronger nuclear attraction results in higher binding energy for all electron shells. When atom Y shows higher binding energies than atom X for all corresponding peaks (1s, 2s, and 2p) while maintaining the same relative peak heights (same electron configuration), this indicates Y has a greater nuclear charge. The increased positive charge in Y's nucleus attracts all electrons more strongly, shifting all peaks to higher binding energy. Option C suggesting lower nuclear charge would produce the opposite effect (lower binding energies). The key insight is that when all peaks shift together to higher energy while maintaining the same pattern, it's due to increased nuclear charge pulling all electrons more tightly.

Question 12

1. Na
2. Mg (correct answer)
3. Ne
4. Al
5. Si

Explanation: This question tests photoelectron spectroscopy interpretation. In PES, binding energy decreases as you move to outer electron shells, and peak height corresponds to the number of electrons in each subshell. The four peaks with heights 2, 2, 6, and 2 (from highest to lowest binding energy) represent 1s (2e⁻), 2s (2e⁻), 2p (6e⁻), and 3s (2e⁻) respectively. This gives a total of 12 electrons, corresponding to magnesium (Mg) with electron configuration 1s² 2s² 2p⁶ 3s². Sodium might seem like a candidate since it has 3s electrons, but it would show a 3s peak height of only 1, not 2. The key is recognizing that each peak height directly tells you the electron count in that subshell, allowing you to build the complete electron configuration.

Question 13

1. 3p electrons are closer to the nucleus than 3s electrons.
2. The 3s electrons experience less shielding and are more penetrating than 3p electrons. (correct answer)
3. The 3p subshell is filled before the 3s subshell, so it must have lower binding energy.
4. Peak height determines binding energy, so the taller peak must be at higher binding energy.
5. 3s electrons have a larger number of electrons than 3p, increasing electron-electron repulsion and binding energy.

Explanation: This question explores photoelectron spectroscopy and orbital penetration effects. In PES, electrons in the same principal energy level (like 3s and 3p) can have different binding energies due to differences in orbital shape and penetration. The 3s electrons have higher binding energy than 3p electrons because s orbitals are more penetrating - they have greater probability density near the nucleus compared to p orbitals. This increased penetration means 3s electrons experience less shielding from inner electrons and feel stronger nuclear attraction, resulting in higher binding energy. Option C incorrectly suggests peak height determines binding energy, but height only indicates electron count, not energy. The fundamental principle is that within the same principal level, s electrons are more tightly bound than p electrons due to their greater penetrating ability.

Question 14

1. The peak with height 6 at intermediate binding energy
2. The peak with height 2 at the highest binding energy (correct answer)
3. The peak with height 2 at the lowest binding energy
4. The peak with height 2 at the second-lowest binding energy
5. None of the peaks could correspond to $1s$ electrons because $1s$ electrons are always lowest in binding energy

Explanation: This question tests understanding of photoelectron spectroscopy and electron shell ordering. In PES, binding energy increases from left to right, meaning electrons closest to the nucleus (highest binding energy) appear rightmost. The 1s electrons are always closest to the nucleus and thus have the highest binding energy of all electrons in an atom. Looking at the four peaks from lowest to highest binding energy (2, 2, 6, 2), the rightmost peak with height 2 at the highest binding energy represents the 1s² electrons. Students might incorrectly choose the leftmost peak thinking 1s comes first in electron configuration notation, but in PES, 1s appears rightmost due to highest binding energy. Remember: innermost electrons (1s) always have the highest binding energy in PES.

Question 15

1. A = 2s, B = 1s, C = 2p
2. A = 1s, B = 2s, C = 2p (correct answer)
3. A = 2p, B = 2s, C = 1s
4. A = 1s, B = 2p, C = 2s
5. A = 2s, B = 2p, C = 1s

Explanation: This question involves photoelectron spectroscopy. In photoelectron spectroscopy, binding energy is higher for electrons in lower energy levels or subshells closer to the nucleus. Peak height is proportional to the number of electrons in that subshell. The peaks are ordered from high binding energy (inner subshells) to low binding energy (outer subshells). A tempting distractor is C A=2p B=2s C=1s, but that would place the 1s at lowest binding energy, which is incorrect because 1s has the highest binding energy. The correct assignment is A=1s (height 2, highest BE), B=2s (height 2), C=2p (height 6, lowest BE). A transferable strategy is that inner electrons have higher binding energy.

Question 16

1. 1s$^2$ 2s$^2$ 2p$^6$ 3s$^2$ 3p$^3$
2. 1s$^2$ 2s$^2$ 2p$^6$ 3s$^2$ 3p$^5$
3. 1s$^2$ 2s$^2$ 2p$^6$ 3s$^1$ 3p$^2$
4. 1s$^2$ 2s$^2$ 2p$^6$ 3s$^2$ 3p$^1$ (correct answer)
5. 1s$^2$ 2s$^2$ 2p$^6$ 3s$^2$ 3p$^6$

Explanation: This question involves photoelectron spectroscopy. In photoelectron spectroscopy, higher binding energy peaks are for inner electrons. Peak height is proportional to electron occupancy in the subshell. The peaks are assigned to subshells from high to low binding energy: 1s,2s,2p,3s,3p. A tempting distractor is A 3p5, but that would make the 3p peak larger than the 3s peak, contrary to the description that 3p is noticeably smaller. The configuration with 3p1 and 3s2 makes the 3p peak smaller than 3s, consistent with the spectrum. A transferable strategy is that inner electrons have higher binding energy.

Question 17

1. The claim is correct because a 3p peak height of 6 uniquely identifies sulfur.
2. The claim is incorrect because peak height indicates binding energy, not the number of electrons.
3. The claim is correct because the highest-binding-energy peak height of 2 proves there are 2 valence electrons.
4. The claim is incorrect because the highest-binding-energy peak must correspond to valence electrons in sulfur.
5. The claim is incorrect because a last peak height of 6 indicates a filled p subshell, which is more consistent with argon than sulfur. (correct answer)

Explanation: This question involves photoelectron spectroscopy. In photoelectron spectroscopy, binding energy decreases from inner to outer subshells. Peak height is proportional to electrons in the subshell. The five peaks with heights 2,2,6,2,6 suggest configuration 1s2 2s2 2p6 3s2 3p6. A tempting distractor is A, but while the last peak is 6, that matches argon, not sulfur which has 3p4 with height 4. The student's claim is incorrect because the last peak of 6 indicates a filled 3p subshell, consistent with argon. A transferable strategy is that inner electrons have higher binding energy.

Question 18

1. Ca (correct answer)
2. K
3. Mg
4. Ar
5. Al

Explanation: This question assesses understanding of photoelectron spectroscopy. In photoelectron spectroscopy, binding energy reflects holding strength, greater for inner shells. The spectrum has left-higher energy, mapping subshells outward. Peak heights show electron numbers, helping identify elements. A tempting distractor is D Ar, with five peaks, but six peaks of 2,2,6,2,6,2 match calcium's 1s²2s²2p⁶3s²3p⁶4s². A transferable strategy is that inner electrons have higher binding energy, and more peaks indicate higher periods.

Question 19

1. 2s (correct answer)
2. 2p
3. 1s
4. 3s
5. 3p

Explanation: This question assesses understanding of photoelectron spectroscopy. In photoelectron spectroscopy, binding energy measures electron attraction, higher for core subshells like 1s. Peaks increase in binding energy left, with sequence from inner to outer. Heights are proportional to subshell electrons, distinguishing s from p. A tempting distractor is B 2p, confusing with next subshell, but for second-period elements, the second peak after 1s is 2s with height 2. A transferable strategy is that inner electrons have higher binding energy, and s subshells typically have height 1 or 2.

Question 20

1. Lithium (correct answer)
2. Neon
3. Helium
4. Hydrogen
5. Beryllium

Explanation: This question involves photoelectron spectroscopy. In photoelectron spectroscopy, binding energy is greater for electrons in inner shells. The peak height corresponds to the number of electrons in the shell or subshell. With two peaks, the spectrum likely corresponds to an element with electrons in two shells. A tempting distractor is B Helium, but Helium has only one peak for 1s2, not two peaks. The two peaks with heights 2 (high BE) and 1 (low BE) match lithium with 1s2 2s1. A transferable strategy is that inner electrons have higher binding energy.