AP Chemistry Quiz: Periodic Trends

What this quiz covers

This quiz focuses on Periodic Trends, 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. Li
2. Na
3. K (correct answer)
4. Li and Na are tied for lowest because they are both metals
5. Na and K are tied for lowest because they are in the same group

Explanation: This question tests understanding of periodic trends, specifically ionization energy down a group. As we move down Group 1 from Li to Na to K, nuclear charge increases, but more importantly, the number of electron shells increases from 2 to 3 to 4. The additional inner electron shells provide shielding that reduces the effective nuclear charge felt by the valence electron, and the valence electron is progressively farther from the nucleus. Both factors make it easier to remove the valence electron, so ionization energy decreases down the group. K has the lowest first ionization energy because its valence electron is farthest from the nucleus and most shielded. The distractor suggesting Li and Na are tied ignores the clear trend down groups. When comparing ionization energies, remember that it decreases down a group due to increased distance and shielding effects.

Question 2

1. Sr
2. Ba (correct answer)
3. They are the same because both are in Group 2
4. Neither; ionization energy increases down a group due to increased nuclear charge
5. Sr, because it has fewer occupied energy levels and therefore holds electrons less tightly

Explanation: This question tests understanding of periodic trends, specifically ionization energy down a group. As we move down Group 2 from Sr (Period 5) to Ba (Period 6), nuclear charge increases, but the number of electron shells also increases from 5 to 6. The additional inner electron shell in Ba provides more shielding that reduces the effective nuclear charge felt by the valence electrons, and the valence electrons are farther from the nucleus. Both factors make it easier to remove a valence electron from Ba compared to Sr, so Ba has the lower first ionization energy. The distractor claiming Sr has lower ionization energy because it has fewer shells contradicts the actual trend. When comparing ionization energies down a group, remember that ionization energy decreases due to increased distance and shielding effects outweighing increased nuclear charge.

Question 3

1. Si
2. S (correct answer)
3. P
4. P and S are tied for greatest because they are nonmetals
5. Si and P are tied for greatest because they are adjacent in the period

Explanation: This question tests understanding of periodic trends, specifically ionization energy across a period. As we move from Si to P to S across Period 3, nuclear charge increases from 14 to 15 to 16 protons, while all three elements have the same number of electron shells (3). The increased nuclear charge creates a stronger hold on valence electrons without additional shielding layers to counteract it, generally causing ionization energy to increase across the period. S, with the highest nuclear charge among these three, holds its valence electrons most tightly and requires the most energy to remove one. The distractor suggesting P and S are tied because they're nonmetals ignores the effect of nuclear charge differences. When comparing ionization energies across a period, consider that it generally increases left to right due to increasing effective nuclear charge.

Question 4

1. C
2. B
3. Be (correct answer)
4. B and C are tied for largest because they are both nonmetals/metalloids
5. All three are approximately equal because they are in the same period

Explanation: This question tests understanding of periodic trends, specifically atomic radius across a period. As we move from Be to B to C across Period 2, nuclear charge increases from 4 to 5 to 6 protons, while all three elements have the same number of electron shells (2). The increased nuclear charge pulls the electron cloud more tightly without additional shielding layers to counteract it, causing atomic radius to decrease across the period. Be, with the lowest nuclear charge among these three, has the weakest pull on its electrons and therefore the largest atomic radius. The distractor suggesting all three are equal ignores the significant effect of increasing nuclear charge on atomic size. When determining atomic radius trends, remember that radius decreases left to right across a period due to increasing effective nuclear charge.

Question 5

1. F
2. Cl because it is above Br in the group.
3. Cl
4. Br (correct answer)
5. F because it has the greatest effective nuclear charge.

Explanation: This question tests understanding of periodic trends, specifically atomic radius down a group. F, Cl, and Br are Group 17 elements with valence electrons in the n=2, n=3, and n=4 shells respectively. As we move down the group, each element adds a complete electron shell, significantly increasing the distance between the nucleus and valence electrons. While nuclear charge increases (9, 17, 35 protons), the additional electron shells provide substantial shielding that reduces the effective nuclear charge felt by outer electrons. Therefore, Br has the largest atomic radius, followed by Cl, then F with the smallest. The distractor suggesting F has the largest radius due to greatest effective nuclear charge confuses the concept—greater effective nuclear charge actually leads to smaller atomic radius. When comparing atomic radii in a group, remember that atomic radius increases down a group due to additional electron shells outweighing increased nuclear charge.

Question 6

1. Ne
2. Ar (correct answer)
3. They have the same atomic radius because both have full valence shells
4. Ne has the greater radius because it has fewer protons attracting the electrons
5. Cannot be predicted because noble gases do not follow periodic trends

Explanation: This question assesses understanding of periodic trends in atomic radius. Down a group, even for noble gases, nuclear charge increases, but valence electrons enter higher energy levels, increasing their average distance. Shielding from additional inner shells mitigates the nuclear attraction. Thus, effective nuclear charge does not rise much, and radius grows, with Ar having a greater radius than Ne. A tempting distractor is C, suggesting they have the same radius due to full valence shells, but full shells do not prevent the downward increase from shielding and distance. To compare radii in a group, including noble gases, always weigh the effects of shielding and electron distance against nuclear charge.

Question 7

1. Li
2. K (correct answer)
3. Na
4. All three have the same first ionization energy because they are in the same group
5. Na and K tie for the lowest because both are metals

Explanation: This question assesses understanding of periodic trends in first ionization energy. Moving down a group, the nuclear charge increases, but this is outweighed by the addition of new electron shells, increasing the distance of valence electrons from the nucleus. Shielding also increases significantly due to more inner electron shells, reducing the effective nuclear charge felt by the outermost electrons. As a result, it becomes easier to remove a valence electron, leading to lower first ionization energy down the group, with K having the lowest among Li, Na, and K. A tempting distractor is D, stating all have the same ionization energy because they are in the same group, but this ignores the downward trend driven by increased shielding and electron distance. When evaluating ionization energies down a group, focus on how added electron shells weaken the nuclear attraction on valence electrons.

Question 8

1. Cl
2. Br
3. All three have the same radius because they have the same number of valence electrons
4. Cannot be predicted from periodic position alone
5. F (correct answer)

Explanation: This question assesses understanding of periodic trends in atomic radius. Down a group, nuclear charge increases with more protons, but valence electrons are added to higher principal energy levels, increasing their distance from the nucleus. Shielding grows due to additional inner electron shells, which screen the valence electrons from the full nuclear charge. Thus, the effective nuclear charge remains similar or slightly decreases, allowing the atomic radius to increase down the group, making F the smallest among F, Cl, and Br. A tempting distractor is D, suggesting all have the same radius due to the same number of valence electrons, but this overlooks the expansion from added energy levels and shielding. To predict radii down a group, consider how increased electron distance and shielding counteract the rising nuclear charge.

Question 9

1. F (correct answer)
2. N
3. C
4. O
5. All have the same electronegativity because they are in the same period

Explanation: This question assesses understanding of periodic trends in electronegativity. Across a period from left to right, nuclear charge increases as protons are added to the nucleus. Shielding stays similar because electrons are filling the same energy level, resulting in a higher effective nuclear charge. This stronger pull makes atoms more effective at attracting electrons in bonds, increasing electronegativity, with F having the highest among C, N, O, and F. A tempting distractor is E, claiming all have the same electronegativity because they are in the same period, but this is wrong as the trend clearly rises across the period due to escalating effective nuclear charge. For electronegativity comparisons, always evaluate the effective nuclear charge, which intensifies from left to right in a period.

Question 10

1. O (correct answer)
2. N
3. C
4. C, N, and O have the same electronegativity because they are in the same period.
5. C because it has the lowest atomic number.

Explanation: This question tests understanding of periodic trends, specifically electronegativity across a period. C, N, and O are consecutive elements in Period 2, with nuclear charges of 6, 7, and 8 protons respectively, while all have valence electrons in the same n=2 shell. As nuclear charge increases across the period, the ability to attract electrons in a chemical bond increases because the nucleus exerts a stronger pull on bonding electrons without additional shielding layers. Oxygen, with the highest nuclear charge and smallest atomic radius of the three, has the greatest electronegativity. The distractor claiming all three have the same electronegativity ignores the fundamental principle that electronegativity increases across a period. When evaluating electronegativity trends, remember that across a period, electronegativity increases due to increasing effective nuclear charge and decreasing atomic radius.

Question 11

1. Al
2. Mg
3. Na (correct answer)
4. Na, Mg, and Al have approximately the same atomic radius because they are in the same period.
5. Al because it has the greatest atomic number.

Explanation: This question tests understanding of periodic trends, specifically atomic radius across a period. As we move from Na to Mg to Al across Period 3, the nuclear charge increases from 11 to 12 to 13 protons, while all three elements have the same number of electron shells (n=3). The increased nuclear charge pulls the electron cloud more tightly without additional shielding layers to counteract this effect, causing atomic radius to decrease from left to right. Therefore, Na has the largest atomic radius, followed by Mg, then Al with the smallest. The distractor suggesting all three have the same radius ignores the critical role of increasing nuclear charge across a period. When comparing atomic radii, always consider that within a period, atomic radius decreases from left to right due to increasing effective nuclear charge.

Question 12

1. Ne because it has the greatest effective nuclear charge.
2. Ar because it is in Period 3.
3. Kr (correct answer)
4. Ne
5. Ar

Explanation: This question tests understanding of periodic trends, specifically atomic radius down a group. Ne, Ar, and Kr are noble gases in Group 18 with complete electron shells at n=2, n=3, and n=4 respectively. As we move down the group, each element adds a complete electron shell, substantially increasing the distance between the nucleus and outermost electrons. While nuclear charge increases from 10 to 18 to 36 protons, the additional electron shells provide significant shielding that reduces the effective nuclear charge experienced by valence electrons. Therefore, Kr has the largest atomic radius, followed by Ar, then Ne with the smallest. The distractor suggesting Ne has the largest radius due to greatest effective nuclear charge misunderstands that higher effective nuclear charge creates smaller, not larger, atomic radius. When comparing atomic radii down a group, remember that size increases due to additional electron shells.

Question 13

1. Mg
2. Na
3. Al (correct answer)
4. All three have approximately the same atomic radius because they are in the same period
5. Cannot be determined without numerical radius data

Explanation: This question assesses understanding of periodic trends in atomic radius. As we move across a period from left to right, the nuclear charge increases because each subsequent element has one more proton in the nucleus. Shielding remains relatively constant since electrons are added to the same principal energy level, which does not significantly increase the shielding effect. Consequently, the effective nuclear charge experienced by the valence electrons increases, pulling them closer to the nucleus and decreasing the atomic radius, so Al has the smallest radius among Na, Mg, and Al. A tempting distractor is D, which claims all three have the same radius because they are in the same period, but this is incorrect as the trend shows a clear decrease across the period due to rising effective nuclear charge. To compare atomic radii across a period, always consider the effective nuclear charge, which strengthens from left to right.

Question 14

1. F (correct answer)
2. Br
3. Cl and Br are tied for smallest because they are in the same group
4. Cl
5. F and Cl are tied for smallest because they are both nonmetals

Explanation: This question tests understanding of periodic trends, specifically atomic radius down a group. As we move down Group 17 from F to Cl to Br, nuclear charge increases, but more significantly, the number of electron shells increases from 2 to 3 to 4. While nuclear charge would tend to pull electrons closer, the addition of entire new electron shells places valence electrons progressively farther from the nucleus, and inner shells provide shielding that reduces the effective nuclear charge. The effect of additional shells dominates, causing atomic radius to increase down the group. F has the smallest atomic radius because it has the fewest electron shells. The distractor suggesting Cl and Br are tied ignores the clear size difference between Period 3 and Period 4 elements. When comparing atomic radii, remember that radius increases down a group due to additional electron shells.

Question 15

1. Al
2. Ga
3. In (correct answer)
4. Al and Ga are tied for largest because they are in the same group
5. Ga and In are tied for largest because they are both heavier than Al

Explanation: This question tests understanding of periodic trends, specifically atomic radius down a group. As we move down Group 13 from Al (Period 3) to Ga (Period 4) to In (Period 5), nuclear charge increases significantly, but more importantly, the number of electron shells increases from 3 to 4 to 5. The addition of entire new electron shells places valence electrons progressively farther from the nucleus, and inner shells provide shielding that reduces the effective nuclear charge felt by outer electrons. The effect of additional shells dominates over increased nuclear charge, causing atomic radius to increase down the group. In has the largest atomic radius because it has the most electron shells. The distractor suggesting Al and Ga are tied ignores the clear difference in electron shell number. When comparing atomic radii down a group, focus on the dominant effect of additional electron shells.

Question 16

1. Ne
2. He
3. Ar (correct answer)
4. He and Ne are tied for lowest because they are both in Period 2 or above
5. All three are the same because noble gases have full valence shells

Explanation: This question tests understanding of periodic trends, specifically ionization energy down a group. As we move down Group 18 from He to Ne to Ar, nuclear charge increases, but the number of electron shells also increases from 1 to 2 to 3. The additional inner electron shells provide shielding that reduces the effective nuclear charge felt by the valence electrons, and the valence electrons are progressively farther from the nucleus. Both factors make it easier to remove an electron despite the full valence shell configuration. Ar has the lowest first ionization energy because its valence electrons are farthest from the nucleus and most shielded. The distractor claiming all three are the same ignores the distance and shielding effects. When analyzing ionization energy trends, remember that even for noble gases, ionization energy decreases down a group.

Question 17

1. Al
2. Mg
3. Na (correct answer)
4. Na, Mg, and Al have approximately the same atomic radius because they are in the same period
5. Mg and Al are tied for largest because they are adjacent in the period

Explanation: This question tests understanding of periodic trends, specifically atomic radius across a period. As we move from left to right across Period 3 (Na → Mg → Al), nuclear charge increases from 11 to 12 to 13 protons, while all three elements have the same number of electron shells (3). The increased nuclear charge pulls the electron cloud more tightly without additional shielding layers to counteract it, causing atomic radius to decrease across the period. Therefore, Na has the largest atomic radius, followed by Mg, then Al with the smallest. The distractor suggesting all three have the same radius ignores the significant effect of increasing nuclear charge. When comparing atomic radii, always consider that radius decreases left to right across a period due to increasing effective nuclear charge.

Question 18

1. P
2. Cl (correct answer)
3. They have the same electronegativity because they are in the same period
4. Neither; electronegativity cannot be compared across a period
5. P, because it is farther left in the period and therefore attracts electrons more strongly

Explanation: This question tests understanding of periodic trends, specifically electronegativity across a period. As we move from P to Cl across Period 3, nuclear charge increases from 15 to 17 protons, while both elements have the same number of electron shells (3). The increased nuclear charge in Cl creates a stronger attraction for bonding electrons without additional shielding to offset it, causing electronegativity to increase across the period. Cl, being further right in the period with higher nuclear charge, has greater electronegativity than P. The distractor claiming P has greater electronegativity reverses the actual trend, confusing leftward position with higher electronegativity. When determining electronegativity trends, remember it increases left to right across a period due to increasing effective nuclear charge's pull on bonding electrons.

Question 19

1. N
2. C
3. C, N, and O have approximately the same electronegativity because they are in the same period
4. N, because it is in Group 15 and therefore attracts electrons most strongly
5. O (correct answer)

Explanation: This question tests understanding of periodic trends, specifically electronegativity across a period. As we move from C to N to O across Period 2, nuclear charge increases from 6 to 7 to 8 protons, while all three elements have the same number of electron shells (2). The increased nuclear charge creates a stronger attraction for bonding electrons without additional shielding to offset it, causing electronegativity to increase across the period. Oxygen, with the highest nuclear charge among these three, has the greatest ability to attract electrons in a chemical bond. The distractor claiming N has the greatest electronegativity confuses group number with the actual trend. When determining electronegativity, remember it increases left to right across a period due to increasing effective nuclear charge.

Question 20

1. B
2. C (correct answer)
3. Be
4. B because it is in Group 13.
5. Be because it is a metal.

Explanation: This question tests understanding of periodic trends, specifically atomic radius across a period. Be, B, and C are consecutive elements in Period 2 with nuclear charges of 4, 5, and 6 protons respectively, all with valence electrons in the n=2 shell. As we move from Be to B to C, the increasing nuclear charge pulls the electron cloud more tightly without additional electron shells to provide shielding. This results in a progressive decrease in atomic radius across the period, making C the smallest of the three atoms. The distractor suggesting Be has the smallest radius because it's a metal incorrectly assumes metallic character determines size rather than position in the periodic table. When comparing atomic radii across a period, always remember that atomic radius decreases from left to right due to increasing effective nuclear charge.