Health Education Systems Inc (HESI) A2 Exam Quiz: Periodic Trends
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Periodic TrendsQuestion 1 of 20

Based on periodic trends in electronegativity, which of the following pairs of atoms would form the most polar covalent bond?

C-O
N-F
Si-S
H-F
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Health Education Systems Inc (HESI) A2 Exam Quiz

Health Education Systems Inc (HESI) A2 Exam Quiz: Periodic Trends

Practice Periodic Trends in Health Education Systems Inc (HESI) A2 Exam with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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 Health Education Systems Inc (HESI) A2 Exam.

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.

All questions

Question 1

Based on periodic trends in electronegativity, which of the following pairs of atoms would form the most polar covalent bond?

  1. C-O
  2. N-F
  3. Si-S
  4. H-F (correct answer)
Explanation: When you encounter questions about polar covalent bonds, focus on electronegativity differences between atoms. The greater the difference in electronegativity values, the more polar the bond becomes. To find the most polar bond, you need to calculate the electronegativity difference for each pair. Using the periodic table trends (electronegativity increases up and to the right), the approximate values are: H = 2.1, C = 2.5, N = 3.0, O = 3.5, F = 4.0, Si = 1.8, S = 2.5. Choice D (H-F) has the largest electronegativity difference: 4.0 - 2.1 = 1.9. This creates the most polar covalent bond because fluorine is the most electronegative element and strongly pulls electrons away from hydrogen. Choice A (C-O) has a difference of 3.5 - 2.5 = 1.0, making it moderately polar but not the most polar option. Choice B (N-F) shows a difference of 4.0 - 3.0 = 1.0, which is the same as C-O and therefore less polar than H-F. Choice C (Si-S) has the smallest difference of 2.5 - 1.8 = 0.7, making it the least polar bond among the options. Remember that fluorine forms the most polar covalent bonds because of its extreme electronegativity. When you see H-F as an option in bond polarity questions, it's often the answer since hydrogen and fluorine represent nearly opposite ends of the electronegativity scale among commonly bonded elements.

Question 2

Which of the following elements would require the most energy to remove its outermost electron?

  1. Lithium (Li)
  2. Carbon (C)
  3. Fluorine (F)
  4. Neon (Ne) (correct answer)
Explanation: When you encounter questions about removing electrons, you're dealing with ionization energy—the energy required to remove an electron from an atom. The key principle is that ionization energy increases across a period (left to right) and decreases down a group on the periodic table. Neon (Ne) requires the most energy to remove its outermost electron because it's a noble gas with a complete outer electron shell. This stable electron configuration (2s²2p⁶) means electrons are held very tightly by the nucleus, making them extremely difficult to remove. Noble gases have the highest ionization energies in their respective periods. Looking at why the other options are incorrect: Lithium (A) has the lowest ionization energy among these choices because it's an alkali metal with just one loosely held valence electron that it readily loses. Carbon (B) has moderate ionization energy—higher than lithium but significantly lower than the elements further right in its period. Fluorine (C) does have very high ionization energy since it strongly attracts electrons, but it's still lower than neon's because fluorine actually "wants" to gain electrons to complete its outer shell, not lose them. The pattern to remember: ionization energy generally increases as you move right across a period, with noble gases having the highest values. When you see ionization energy questions on the HESI, look for the element furthest right and highest up on the periodic table—but remember that noble gases are the peak of each period's ionization energy trend.

Question 3

Element A is in Group 1 and Element B is in Group 17 of the same period. Which statement accurately compares the two elements?

  1. Element A has a higher first ionization energy than Element B.
  2. Element B has a smaller atomic radius and a higher electronegativity than Element A. (correct answer)
  3. Element B has a larger atomic radius and a lower ionization energy than Element A.
  4. Both elements have a similar tendency to lose electrons because they are in the same period.
Explanation: When you encounter questions comparing elements across the periodic table, focus on the key trends: atomic radius decreases and electronegativity increases as you move left to right across a period, while ionization energy generally increases in the same direction. Element A (Group 1) is an alkali metal, while Element B (Group 17) is a halogen. Since they're in the same period, Element B is significantly further right on the periodic table. This positioning means Element B has more protons in its nucleus, creating a stronger pull on its electrons and resulting in a smaller atomic radius. Additionally, Element B's higher nuclear charge and smaller size give it much higher electronegativity - halogens are among the most electronegative elements, desperately wanting to gain electrons to complete their outer shell. Answer B correctly identifies both trends: Element B has a smaller atomic radius and higher electronegativity than Element A. Answer A is backwards - Element A (alkali metal) has a much lower first ionization energy than Element B because it readily loses its single valence electron. Answer C completely reverses the atomic radius trend and incorrectly states ionization energy relationships. Answer D misunderstands electron behavior - Element A wants to lose electrons (low ionization energy) while Element B wants to gain electrons (high electronegativity), making their tendencies opposite, not similar. Remember this pattern: as you move left to right across any period, atoms get smaller, hold electrons more tightly, and become more electronegative. Group 1 and Group 17 elements represent the extremes of this trend.

Question 4

An unknown element, X, has a higher first ionization energy than phosphorus (P) and a smaller atomic radius than silicon (Si). Based on periodic trends, which of the following is most likely element X?

  1. Aluminum (Al)
  2. Sulfur (S) (correct answer)
  3. Germanium (Ge)
  4. Potassium (K)
Explanation: When you encounter periodic trends questions, focus on how atomic properties change predictably across periods (left to right) and down groups (top to bottom). You need to identify an element with higher first ionization energy than phosphorus and smaller atomic radius than silicon. First ionization energy increases across a period and decreases down a group, while atomic radius decreases across a period and increases down a group. Phosphorus is in period 3, group 15, and silicon is in period 3, group 14. So you're looking for an element that's likely to the right of phosphorus on the periodic table (higher ionization energy) but positioned to have a smaller radius than silicon. Sulfur (S) fits perfectly. Located in period 3, group 16, sulfur is immediately to the right of phosphorus, giving it a higher first ionization energy due to increased nuclear charge. Since sulfur is also to the right of silicon in the same period, it has a smaller atomic radius because the electrons are pulled closer by the greater nuclear charge. Choice A (Aluminum) is to the left of both phosphorus and silicon, so it has lower ionization energy and larger atomic radius than both. Choice C (Germanium) is in period 4 below silicon, giving it a much larger atomic radius despite being in the same group. Choice D (Potassium) is in period 4, group 1, making it much larger and having much lower ionization energy than the period 3 elements. Remember: when comparing periodic trends, always locate elements on the periodic table first, then apply the left-to-right and top-to-bottom rules systematically.

Question 5

The concept of effective nuclear charge (ZeffZ_{eff}) helps explain periodic trends. Which trend is most directly explained by the fact that ZeffZ_{eff} increases steadily from left to right across a period?

  1. The decrease in atomic radius. (correct answer)
  2. The increase in atomic radius down a group.
  3. The decrease in electronegativity down a group.
  4. The similar chemical properties of elements within a group.
Explanation: When you encounter periodic trends questions, focus on how effective nuclear charge (ZeffZ_{eff}) drives most patterns across the periodic table. Effective nuclear charge represents the net positive charge experienced by valence electrons after accounting for shielding by inner electrons. As you move left to right across a period, the nuclear charge increases while electrons are added to the same energy level. Since inner electron shielding remains relatively constant, ZeffZ_{eff} increases steadily. This stronger pull draws valence electrons closer to the nucleus, directly causing atomic radius to decrease across a period. Choice A correctly identifies this fundamental relationship. Choice B describes a trend down a group, not across a period. Down a group, atomic radius increases because new electron shells are added, making atoms larger despite increasing nuclear charge. Choice C also refers to a down-group trend where electronegativity decreases due to increased distance between nucleus and valence electrons, not the across-period ZeffZ_{eff} increase mentioned in the question. Choice D relates to vertical similarities in the periodic table caused by identical valence electron configurations, which has nothing to do with the horizontal ZeffZ_{eff} trend. For HESI success, remember that ZeffZ_{eff} is the master concept behind most periodic trends. When you see questions about trends "across a period," immediately think about increasing ZeffZ_{eff} and its effects: smaller atoms, higher ionization energy, and greater electronegativity. Always match the direction of the trend (horizontal vs. vertical) with what the question asks.

Question 6

An unknown element is observed to have a relatively low first ionization energy, a large atomic radius, and it reacts vigorously with water. This element is most likely located in which area of the periodic table?

  1. The upper right section (p-block nonmetals)
  2. The lower left section (s-block) (correct answer)
  3. The middle section (d-block transition metals)
  4. The upper left section (s-block)
Explanation: When you encounter a question about predicting an element's location based on its properties, think about the periodic trends and how they correlate with position. The key properties here—low first ionization energy, large atomic radius, and vigorous reaction with water—all point to highly reactive metals. Elements with low ionization energies easily lose electrons, while large atomic radii indicate the outer electrons are far from the nucleus and weakly held. The vigorous water reaction suggests a metal that readily forms ionic compounds. These characteristics increase as you move down and left on the periodic table, reaching their peak in the lower left section where the alkali metals (Group 1) and alkaline earth metals (Group 2) reside. Option A (upper right p-block nonmetals) is incorrect because nonmetals typically have high ionization energies, small radii, and don't react with water to produce hydrogen gas like reactive metals do. Option C (d-block transition metals) is wrong because while some transition metals react with water, they generally have moderate ionization energies and smaller radii than s-block metals. Option D (upper left s-block) describes elements like lithium and beryllium, which are less reactive than their lower counterparts due to smaller size and higher ionization energies. The lower left s-block metals like cesium, rubidium, and barium perfectly match all three described properties, making B the correct answer. Study tip: Remember that reactivity in metals increases as you move down and left on the periodic table—this pattern appears frequently on chemistry assessments.

Question 7

A neutral chlorine atom (Cl) gains one electron to form a chloride ion (Cl⁻). Which statement correctly describes the change in radius?

  1. The radius increases because increased electron-electron repulsion expands the electron cloud. (correct answer)
  2. The radius decreases because the new electron is pulled tightly by the nucleus.
  3. The radius remains the same because no new electron shells are added.
  4. The radius increases because the added electron requires the formation of a new, larger shell.
Explanation: When you encounter questions about ionic radius changes, focus on the balance between nuclear attraction and electron-electron repulsion within the same electron shell. When a chlorine atom gains an electron to become Cl⁻, you're adding one more electron to the same outermost shell (the third shell). The nucleus still has 17 protons, but now there are 18 electrons instead of 17. This creates an imbalance: the nuclear charge remains constant, but there are more electrons in the same space repelling each other. The correct answer is A because the increased electron-electron repulsion in the third shell causes the electron cloud to expand outward, making the chloride ion larger than the neutral chlorine atom. Think of it like adding another person to an already crowded room - everyone spreads out more to reduce the discomfort. Answer B is incorrect because while the nucleus does attract the new electron, this attractive force is outweighed by the increased repulsion between the 18 electrons now sharing the same shells. Answer C misses the key point that even within the same shell, more electrons mean more repulsion and expansion. Answer D is wrong because the electron enters an existing shell (the third shell), not a new one - chlorine already has electrons in its outermost shell. Remember: when atoms gain electrons to form anions, they get larger due to increased electron-electron repulsion. When they lose electrons to form cations, they get smaller due to reduced repulsion and often loss of entire electron shells.

Question 8

Which of the following elements has the largest atomic radius?

  1. Sodium (Na)
  2. Silicon (Si)
  3. Sulfur (S)
  4. Cesium (Cs) (correct answer)
Explanation: When you encounter questions about atomic radius, you need to understand how atomic size changes across the periodic table. Atomic radius follows predictable trends based on an element's position. Atomic radius increases as you move down a group (column) because each row adds another electron shell, making atoms larger. Conversely, atomic radius decreases as you move across a period (row) from left to right because increasing nuclear charge pulls electrons closer to the nucleus. Looking at the periodic table positions: Cesium (Cs) is in Period 6, Group 1; Sodium (Na) is in Period 3, Group 1; Silicon (Si) is in Period 3, Group 14; and Sulfur (S) is in Period 3, Group 16. Since cesium is much further down the periodic table than the others, it has significantly more electron shells, making it the largest atom among these choices. Option A (Sodium) is incorrect because while it's in the same group as cesium, it's only in the third period, so it has fewer electron shells. Option B (Silicon) is wrong because it's in the same period as sodium but further to the right, meaning stronger nuclear charge pulls its electrons closer. Option C (Sulfur) is the smallest of these atoms since it's in the same period as sodium and silicon but even further right, experiencing the greatest nuclear attraction. Remember this pattern: atomic radius increases down groups and decreases across periods. When comparing atoms from different periods, the one lowest on the periodic table will typically be largest due to additional electron shells.

Question 9

Of the elements listed below, which one has the greatest tendency to attract a bonding pair of electrons?

  1. Oxygen (O) (correct answer)
  2. Sulfur (S)
  3. Selenium (Se)
  4. Tellurium (Te)
Explanation: This question tests your understanding of electronegativity, which measures an atom's ability to attract bonding electrons toward itself in a chemical bond. When you see elements from the same group (column) on the periodic table, remember that electronegativity follows predictable trends. Oxygen (A) has the highest electronegativity among these choices because electronegativity decreases as you move down a group in the periodic table. All four elements belong to Group 16 (the chalcogens), but they're arranged vertically with oxygen at the top. As atomic size increases going down the group, the nucleus becomes farther from the bonding electrons, reducing its pull on those electrons. Looking at why the other options are incorrect: Sulfur (B) is directly below oxygen, so it has lower electronegativity due to its larger atomic radius. The extra electron shell makes it less effective at attracting bonding electrons. Selenium (C) sits below sulfur, making it even larger and less electronegative. Tellurium (D) is the largest of these atoms, positioned at the bottom of this group, giving it the weakest pull on bonding electrons. The electronegativity values confirm this trend: oxygen (3.4) > sulfur (2.6) > selenium (2.6) > tellurium (2.1) on the Pauling scale. Study tip: For HESI chemistry questions about periodic trends, remember that electronegativity increases as you move up a group and across a period from left to right. Fluorine is the most electronegative element, followed by oxygen, nitrogen, and chlorine.

Question 10

Consider the following isoelectronic species: S²⁻, Cl⁻, K⁺, and Ca²⁺. Which of these has the smallest ionic radius?

  1. S²⁻
  2. Cl⁻
  3. K⁺
  4. Ca²⁺ (correct answer)
Explanation: When you encounter questions about ionic radius in isoelectronic species, you need to understand how nuclear charge affects electron attraction. Isoelectronic species have the same number of electrons but different numbers of protons in their nuclei. All four ions (S²⁻, Cl⁻, K⁺, and Ca²⁺) are isoelectronic with 18 electrons, giving them the same electron configuration as argon. However, they have different nuclear charges: sulfur has 16 protons, chlorine has 17, potassium has 19, and calcium has 20. The key principle is that greater nuclear charge pulls the electron cloud more tightly toward the nucleus, resulting in a smaller ionic radius. Since Ca²⁺ has the highest nuclear charge (20 protons) attracting the same 18 electrons, it exerts the strongest pull on the electron cloud, making it the smallest ion. This makes D the correct answer. Looking at the wrong choices: A) S²⁻ has the lowest nuclear charge (16 protons), so it has the weakest pull on electrons and actually the largest radius. B) Cl⁻ with 17 protons has a stronger pull than sulfur but weaker than the cations, giving it the second-largest radius. C) K⁺ with 19 protons pulls electrons more strongly than the anions but less than calcium, making it larger than Ca²⁺. Remember this pattern: among isoelectronic species, ionic radius decreases as nuclear charge increases. The more protons pulling on the same number of electrons, the smaller the ion becomes.

Question 11

An element with high metallic character would most likely be characterized by which of the following properties?

  1. A small atomic radius and a high electronegativity.
  2. A tendency to form anions and a high ionization energy.
  3. A large atomic radius and a low first ionization energy. (correct answer)
  4. A high electronegativity and a large atomic radius.
Explanation: When you encounter questions about metallic character, think about where metals are located on the periodic table and what properties make them behave as metals. Metallic character increases as you move down a group and decreases as you move across a period from left to right. Elements with high metallic character readily lose electrons to form positive ions (cations). This behavior is directly related to two key atomic properties: atomic radius and ionization energy. Large atoms hold their outermost electrons more loosely because these electrons are farther from the nucleus and experience less nuclear attraction due to shielding by inner electrons. Consequently, less energy is required to remove these outer electrons, resulting in low first ionization energy. Option C correctly identifies both characteristics: a large atomic radius and low first ionization energy. These properties work together to make electron loss favorable, which is the hallmark of metallic behavior. Option A describes nonmetals, which have small atomic radii and high electronegativity values. Option B also describes nonmetals - they tend to gain electrons to form anions and have high ionization energies because they hold onto their electrons tightly. Option D combines contradictory properties; high electronegativity is associated with nonmetals and small atomic radii, not large ones. Remember this pattern for the HESI: metallic character correlates with ease of electron loss. Look for large size and low ionization energy as indicators of high metallic character, while small size and high electronegativity point to nonmetallic behavior.

Question 12

Which of the following statements about periodic trends is generally NOT true?

  1. Atomic radius increases from top to bottom within a group.
  2. Electronegativity decreases from top to bottom within a group.
  3. First ionization energy decreases from left to right across a period. (correct answer)
  4. Anions are larger than their corresponding neutral atoms.
Explanation: Periodic trends questions test your understanding of how atomic properties change systematically across the periodic table. These patterns are driven by two key factors: nuclear charge (number of protons) and electron shielding. Let's examine each trend statement. Moving down a group, atomic radius does increase (A is true) because you're adding electron shells, making atoms larger despite increased nuclear charge. Electronegativity decreases down a group (B is true) because the outer electrons are farther from the nucleus and more shielded, reducing the atom's ability to attract electrons. When atoms gain electrons to form anions, they become larger than their neutral counterparts (D is true) because electron-electron repulsion increases while nuclear charge stays constant. However, first ionization energy actually increases from left to right across a period, not decreases (C is false). As you move across a period, nuclear charge increases while electron shielding remains roughly constant. This means outer electrons are held more tightly, requiring more energy to remove them. For example, it takes much more energy to remove an electron from fluorine than from lithium. Answer choice C represents the opposite of the actual trend, making it the statement that is NOT true. For HESI success, remember that periodic trends follow logical patterns based on atomic structure. Across periods: atomic radius and metallic character decrease while ionization energy and electronegativity increase. Down groups: atomic radius increases while ionization energy and electronegativity decrease. Understanding the "why" behind these trends will help you answer confidently even when trends are stated backwards.

Question 13

The reactivity of alkali metals (Group 1) increases down the group. This is best explained by the fact that as you move down the group, the...

  1. nuclear charge decreases, making it easier to lose an electron.
  2. electronegativity increases, making the atoms more likely to react.
  3. valence electron is further from the nucleus, resulting in a lower ionization energy. (correct answer)
  4. number of neutrons increases, which destabilizes the atom's valence shell.
Explanation: When you encounter questions about periodic trends, focus on how atomic structure changes as you move through the periodic table and how those changes affect chemical properties. Alkali metals become more reactive down Group 1 because their valence electrons become progressively easier to remove. As you move from lithium to cesium, each element has an additional electron shell. This means the outermost electron is located further from the positively charged nucleus. The increased distance weakens the electrostatic attraction between the nucleus and the valence electron, resulting in lower ionization energy. Since alkali metals react by losing their single valence electron, lower ionization energy translates directly to higher reactivity. Answer C correctly identifies this relationship. Answer A incorrectly states that nuclear charge decreases down the group. Actually, nuclear charge increases as you add more protons, but this effect is outweighed by the increased distance and electron shielding. Answer B wrongly suggests electronegativity increases down the group—it actually decreases, and electronegativity measures an atom's tendency to attract electrons, not lose them. Answer D incorrectly focuses on neutrons and valence shell destabilization. While neutron count does increase, neutrons don't directly affect chemical reactivity since they're electrically neutral. For HESI chemistry questions about periodic trends, remember that atomic size and distance effects often trump nuclear charge effects. When atoms get larger, their outer electrons become easier to remove, making metals more reactive and nonmetals less reactive.

Question 14

An element in Group 2 of the periodic table is most likely to form an ion with what charge, and how will the ion's radius compare to the neutral atom?

  1. A charge of +2, and a smaller radius. (correct answer)
  2. A charge of -6, and a larger radius.
  3. A charge of +2, and a larger radius.
  4. A charge of -2, and a smaller radius.
Explanation: When you encounter questions about ion formation, focus on electron configuration and the drive for atoms to achieve stable, noble gas configurations. Group 2 elements (alkaline earth metals like magnesium and calcium) have two valence electrons in their outermost shell. To achieve the stable electron configuration of the nearest noble gas, these atoms lose their two valence electrons, forming ions with a +2 charge. When electrons are removed, the remaining electrons experience a stronger pull from the nucleus since there are fewer electrons to shield each other from the nuclear charge. This increased effective nuclear charge pulls the electron cloud closer, making the ion smaller than the neutral atom. Looking at the options: Choice A correctly identifies both the +2 charge and smaller radius. Choice B suggests a -6 charge, which is impossible since Group 2 elements only have two valence electrons to lose or gain, and gaining six electrons would be energetically unfavorable. Choice C has the correct charge but incorrectly suggests a larger radius—this would only occur if electrons were added, not removed. Choice D proposes a -2 charge, which would require Group 2 elements to gain electrons rather than lose them, contradicting their tendency to achieve stability by losing valence electrons. For HESI chemistry questions, remember this pattern: metals lose electrons to form positive ions that are smaller than the parent atom, while nonmetals gain electrons to form negative ions that are larger. Group number often tells you how many electrons metals will lose.

Question 15

When a neutral atom of calcium (Ca) forms a calcium ion (Ca²⁺), what happens to its radius?

  1. The radius increases because the remaining electrons repel each other more strongly.
  2. The radius decreases because the entire valence electron shell is removed. (correct answer)
  3. The radius remains the same because the number of protons in the nucleus is unchanged.
  4. The radius decreases because the nucleus itself becomes smaller.
Explanation: When atoms form ions, you're witnessing a fundamental change in electron configuration that directly affects atomic size. Understanding how electron loss impacts atomic radius is crucial for predicting ionic properties. When calcium loses two electrons to form Ca²⁺, it doesn't just lose any two electrons—it loses its entire outermost electron shell (the 4s orbital). Neutral calcium has the electron configuration [Ar] 4s², meaning its valence electrons occupy the fourth energy level. After ionization, Ca²⁺ has the configuration [Ar], so its outermost electrons are now in the third energy level. Since atomic radius is determined by the distance of the outermost electrons from the nucleus, removing an entire electron shell dramatically decreases the ion's size. Answer A incorrectly suggests the radius increases due to electron-electron repulsion. While fewer electrons do mean less repulsion, this effect is overwhelmed by the loss of the entire outer shell. Answer C makes the common mistake of focusing on nuclear charge. Though the number of protons stays constant, what matters for size is the electron configuration relative to that nuclear charge. Answer D incorrectly states the nucleus shrinks—nuclear size doesn't change during ionization. The key pattern to remember: when atoms lose electrons to form cations, especially when they lose entire electron shells, the resulting ions are always significantly smaller than the original atoms. For HESI questions about ionic size, focus on electron shell changes rather than just electron count.

Question 16

The chemical reactivity of the halogens (Group 17) is highest at the top of the group (Fluorine) and decreases down the group. This trend is primarily due to the increase in...

  1. ionization energy, which makes it harder for the atom to gain an electron.
  2. metallic character, which enhances the ability to form anions.
  3. the number of valence electrons, which completes the octet more easily.
  4. atomic radius and electron shielding, which reduces the effective nuclear pull on an incoming electron. (correct answer)
Explanation: When you encounter questions about periodic trends in chemical reactivity, focus on how atomic structure changes affect an atom's ability to gain or lose electrons. For halogens, reactivity depends on how easily they can attract and capture an electron to complete their outer shell. Fluorine is the most reactive halogen because it's small and has high electronegativity. As you move down Group 17 from fluorine to iodine, the atoms get larger due to additional electron shells. This increased atomic radius creates more distance between the nucleus and any incoming electron. Additionally, the inner electron shells create a "shielding effect" that blocks some of the nuclear charge from reaching the outer regions where new electrons would be added. Together, these factors reduce the effective nuclear pull on an incoming electron, making it harder for larger halogens to attract electrons and thus reducing their reactivity. This makes answer D correct. Answer A is incorrect because ionization energy relates to removing electrons, not gaining them. While ionization energy does decrease down the group, this doesn't directly explain halogen reactivity. Answer B is wrong because halogens actually become more metallic (less nonmetallic) down the group, which decreases their tendency to form anions. Answer C misses the point entirely - all halogens have seven valence electrons and need one more to complete their octet; the number doesn't change down the group. Remember: For halogen reactivity questions, think about atomic size and electron shielding effects. Smaller atoms with less shielding are more reactive because they can better attract electrons.

Question 17

Which of the following elements exhibits the most pronounced metallic character?

  1. Germanium (Ge)
  2. Phosphorus (P)
  3. Barium (Ba) (correct answer)
  4. Aluminum (Al)
Explanation: When you encounter questions about metallic character, think about periodic trends. Metallic character refers to how readily an element loses electrons and exhibits typical metallic properties like conductivity, malleability, and luster. This property increases as you move down a group and decreases as you move across a period from left to right. To determine which element has the most pronounced metallic character, you need to consider their positions on the periodic table. Barium (Ba) is located in Group 2, Period 6 - it's an alkaline earth metal positioned far down and to the left on the periodic table. This placement gives it the strongest tendency to lose electrons and exhibit metallic behavior among these options. Looking at the incorrect choices: Aluminum (Al) is a metal, but it's in Period 3 and further right than barium, giving it less metallic character than Ba. Germanium (Ge) is a metalloid in Group 14, Period 4 - it has some metallic properties but significantly less than true metals. Phosphorus (P) is a nonmetal in Group 15, Period 3, with very little metallic character. The key pattern is that elements become more metallic as you move down and to the left on the periodic table. Barium's position in the lower left region makes it the most metallic among these choices. Study tip: Remember the diagonal rule - metallic character increases toward the bottom-left corner of the periodic table. When comparing metallic character, always check which element is furthest down and left.

Question 18

As one moves from left to right across a period in the periodic table, the atomic radius tends to decrease. What is the primary reason for this trend?

  1. The number of electron shells increases, causing the atom to expand.
  2. The increasing number of protons exerts a stronger electrostatic pull on the electrons in the same energy level. (correct answer)
  3. The mass of the nucleus increases, compressing the electron cloud through gravitational force.
  4. The electron-electron repulsion in the valence shell decreases, allowing the shell to contract.
Explanation: When you encounter questions about periodic trends, focus on the fundamental forces acting within atoms: the attraction between positively charged protons and negatively charged electrons. As you move left to right across a period, each element gains one proton in its nucleus and one electron in the same outer energy level. The key insight is that while electrons are added to the same shell, protons are added to the nucleus. This creates an increasingly positive nuclear charge that pulls more strongly on all electrons, including those in the outermost shell. Since the electrons aren't effectively shielding each other from this increased nuclear charge (they're in the same energy level), the entire electron cloud gets pulled closer to the nucleus, decreasing atomic radius. Option A incorrectly suggests electron shells increase across a period - this only happens when moving down a group (vertically). Option C mentions gravitational force, but electromagnetic forces are vastly stronger than gravity at the atomic scale and are responsible for atomic structure. Option D discusses decreasing electron-electron repulsion, but repulsion actually increases as more electrons are added to the same shell; however, this effect is overwhelmed by the stronger nuclear attraction. Remember this pattern: across a period, nuclear charge increases faster than electron shielding, so atoms contract. For HESI chemistry questions, always consider which fundamental force - nuclear attraction or electron repulsion - dominates in the scenario described.

Question 19

Which element is expected to show the largest increase between its first and second ionization energies?

  1. Magnesium (Mg)
  2. Aluminum (Al)
  3. Silicon (Si)
  4. Sodium (Na) (correct answer)
Explanation: When you encounter questions about ionization energy trends, focus on electron configuration and how much energy is required to remove electrons from different shells. Ionization energy is the energy needed to remove an electron from an atom. The first ionization energy removes the outermost electron, while the second removes the next electron. The key insight is that removing an electron from a lower energy level (closer to the nucleus) requires dramatically more energy than removing one from the outermost shell. Sodium (Na) has the electron configuration 1s²2s²2p⁶3s¹. Its first ionization energy removes the single 3s electron, leaving Na⁺ with a stable, filled second shell (1s²2s²2p⁶). The second ionization energy must remove an electron from this much lower, more tightly held 2p orbital. This represents a huge jump because you're breaking into a completed inner shell that's much closer to the nucleus. Choice A (Mg) goes from 3s² to 3s¹ to 2p⁶, so both electrons come from relatively similar energy levels. Choice B (Al) removes from 3p¹ then 3s², again similar energy levels. Choice C (Si) removes from 3p² then 3p¹, which are essentially the same energy level. Only sodium involves removing the first electron from an outer shell, then breaking into a completely different, much lower energy shell for the second ionization. Remember this pattern: elements with one electron in their outermost shell (Group 1 metals) show the most dramatic increase between first and second ionization energies because the second electron comes from a filled inner shell.

Question 20

The first ionization energy of beryllium (Be) is higher than that of boron (B), which is an exception to the general trend. This is best explained because beryllium has a...

  1. larger nuclear charge than boron, holding electrons more tightly.
  2. filled 2s subshell, which is a more stable electron configuration. (correct answer)
  3. smaller atomic radius than boron, increasing the nuclear pull.
  4. higher metallic character, making it harder to remove an electron.
Explanation: When you encounter ionization energy questions, focus on electron configuration stability and the energy required to remove electrons. Ionization energy generally increases across a period due to increasing nuclear charge, but there are notable exceptions. Beryllium's first ionization energy is indeed higher than boron's because beryllium has a completely filled 2s subshell (1s² 2s²), making it exceptionally stable. Removing an electron from this filled, stable configuration requires more energy. Boron, with configuration 1s² 2s² 2p¹, has that single 2p electron that's easier to remove since it's alone in the 2p subshell and slightly farther from the nucleus than the 2s electrons. Choice A is incorrect because while beryllium does have a smaller nuclear charge than boron (4 protons vs. 5 protons), this would predict beryllium should have lower ionization energy, not higher. Choice C misses the main point—though beryllium is smaller, the key factor is electron configuration stability, not just atomic size. Choice D is wrong because higher metallic character actually correlates with lower ionization energy (metals readily give up electrons), and beryllium being more metallic than boron would suggest easier electron removal. Remember this pattern: filled or half-filled subshells create exceptions to periodic trends due to their extra stability. On the HESI, when you see ionization energy exceptions, immediately think about electron configurations—particularly look for filled s subshells, half-filled p or d subshells, or completely filled subshells as explanations for unexpected values.