AP Chemistry Unit 1: Atomic Structure and Properties — Free Review Games.

Carbon has atomic number 6 (6 protons, 6 electrons when neutral). Carbon-14 has mass number 14, so it has 14 - 6 = 8 neutrons.

The number of protons (atomic number) uniquely identifies each element. Changing the proton count changes the element.

Isotopes have the same number of protons but different numbers of neutrons, giving them different mass numbers.

Atomic mass is a weighted average of naturally occurring isotopes. Chlorine is about 75% Cl-35 and 25% Cl-37, giving an average of approximately 35.5 amu.

Oxygen has 8 electrons. Following the aufbau principle: 1s² 2s² 2p⁴ fills orbitals in order of increasing energy.

This configuration has 15 electrons total (2+2+6+2+3), corresponding to phosphorus (Z = 15).

Coulomb's law states F = kq₁q₂/r². Mg²⁺ and O²⁻ have higher charges (2+ and 2-) and small ionic radii, producing the strongest attraction.

Ionization energy generally increases across a period, but Al has a slight dip compared to Mg because Al removes an electron from a 3p orbital (lower energy than 3s). The order is Na < Al < Mg < Si.

The peak heights represent electron counts per subshell: 1s² 2s² 2p⁶ 3s² 3p³. Total = 15 electrons, which is phosphorus.

Na's first ionization removes the 3s¹ valence electron. The second must remove from the stable, tightly held 2p⁶ core, requiring dramatically more energy.

The large jump between the 3rd and 4th ionization energies indicates that the first three electrons are valence electrons and the 4th comes from an inner core shell.

S and Cl are in the same period. Cl has one more proton but the additional electron enters the same shell, so increased nuclear charge pulls electrons inward, reducing radius.

1s electrons are in the innermost shell closest to the nucleus with essentially no shielding, so they experience the full nuclear charge and require the most energy to remove.

Fluorine's very small 2p orbitals create strong electron-electron repulsion when an additional electron is added, partially offsetting the strong nuclear attraction.

Average mass = (0.20 × 10) + (0.80 × 11) = 2.0 + 8.8 = 10.8 amu.

The d subshell contains 5 orbitals (ml = -2, -1, 0, +1, +2), and each orbital holds 2 electrons, so the 3d subshell holds a maximum of 10 electrons. The value 6 applies to a p subshell (3 orbitals), and 14 applies to an f subshell (7 orbitals).

The angular momentum quantum number (l) defines the shape of an orbital: l=0 is a spherical s orbital, l=1 is a dumbbell-shaped p orbital, l=2 is a d orbital, and so on. The principal quantum number (n) determines energy and size, ml determines orientation in space, and ms describes electron spin.

Sulfur (Z=16) is in Group 16 with the electron configuration [Ne]3s²3p⁴. The valence shell is n=3, containing 2 + 4 = 6 electrons total. Valence electrons are those in the outermost principal energy level only. The value 8 is not possible because the n=3 shell's s and p subshells hold at most 8 electrons, but sulfur has only 6 in that shell.

Neon has 10 electrons. Filling by the Aufbau principle: 1s² (2 electrons), 2s² (4 total), 2p⁶ (10 total) gives the complete configuration 1s²2s²2p⁶. Choice A accounts for only 8 electrons, choice C incorrectly places 4 electrons in 2s (which holds only 2), and choice D represents an excited state where one electron has jumped from 2p to 3s.

In the Bohr model, electrons occupy fixed energy levels. When a photon of exactly the right energy is absorbed, the electron is promoted to a higher (higher n) energy level, moving farther from the nucleus. A downward transition (to lower n) releases a photon rather than absorbing one. Ejection of the electron entirely requires energy equal to or greater than the ionization energy — more than a typical photon absorbed between bound levels.

The principal quantum number n (allowed values 1, 2, 3, ...) determines the energy level of an electron and correlates directly with the average distance between the electron and the nucleus. Higher n means higher energy and greater average distance. Shape is described by the angular momentum quantum number (l), spatial orientation by the magnetic quantum number (ml), and spin by ms.

[Ne] accounts for 10 electrons (1s²2s²2p⁶). Adding 3s²3p¹ contributes 3 more electrons for a total of 13, which corresponds to aluminum (Z=13). Magnesium (Z=12) would be [Ne]3s², silicon (Z=14) would be [Ne]3s²3p², and sodium (Z=11) would be [Ne]3s¹. Identifying the total electron count is the key step.

Neutral iron is [Ar]4s²3d⁶. When transition metals form cations, electrons are removed from the 4s subshell before the 3d subshell, because 4s electrons are higher in energy in the presence of a cation's reduced electron shielding. Removing both 4s electrons gives Fe²⁺ = [Ar]3d⁶. Choice B incorrectly removes 3d electrons rather than 4s electrons, which is a common error.

A 3p electron requires n=3 and l=1. For l=1, the allowed ml values are −1, 0, and +1 only. Choice B satisfies all rules: n=3, l=1 (p orbital), ml=0 (valid), ms=+½ (valid). Choice A fails because ml=2 exceeds the allowed range for l=1. Choice C has l=2, which designates a d orbital, not p. Choice D has n=2, so it describes a 2p electron rather than a 3p electron.

Across a period, each successive element has one more proton but the additional electrons enter the same valence shell. Core electrons provide the most shielding, and the core electron count stays constant across a period. The net result is a steady increase in effective nuclear charge (Zeff), which draws valence electrons progressively closer to the nucleus, decreasing atomic radius. Choice A describes the trend going down a group, where n increases — the opposite direction.

Copper's anomaly arises because a completely filled 3d subshell has extra stability from maximized exchange energy — the symmetric distribution of electrons among the five 3d orbitals lowers the total energy. Promoting one 4s electron to complete the 3d subshell is energetically favorable. The same logic applies to chromium ([Ar]4s¹3d⁵), where a half-filled 3d is similarly stabilized. Choice A is factually incorrect; the actual observed configuration is [Ar]4s¹3d¹⁰.

Because position and momentum cannot both be precisely known at the same moment, electrons cannot be assigned definite trajectories. Instead, the quantum mechanical model describes orbitals as probability distributions — three-dimensional regions where there is a high probability of locating the electron. This replaces the Bohr model's fixed circular orbits. Choice C confuses momentum-position uncertainty with energy quantization, and choice D describes the Pauli exclusion principle, not the uncertainty principle.

Shielding is most effective when electrons reside in inner shells (lower n) that are located between the nucleus and the outer electrons. Across Period 3, each additional electron enters the n=3 shell — the same shell as existing valence electrons. Same-shell electrons have very little shielding effect on one another. As a result, Zeff increases by nearly one unit for each additional proton, and valence electrons are drawn progressively closer to the nucleus across the period.

The configuration [Ar]3d⁵4s² has 7 electrons beyond argon (18), giving Z=25, which is manganese. By Hund's rule, the five 3d electrons each occupy a separate 3d orbital before any pairing occurs, so all five are unpaired. The 4s electrons are paired with each other, contributing 0 additional unpaired electrons. Iron (Z=26) is [Ar]3d⁶4s² with 4 unpaired 3d electrons, and chromium (Z=24) has the anomalous configuration [Ar]3d⁵4s¹.

Binding energy in PES is the energy required to eject an electron from a specific subshell. Core electrons (1s, 2s, etc.) are held closest to the nucleus and experience the strongest attraction, so they require the most energy to remove and appear at the highest binding energy in a PES spectrum. Valence electrons are farther from the nucleus and more loosely held, appearing at lower binding energy. Choice B describes low-binding-energy electrons, the opposite of what was asked.

Nitrogen ([He]2s²2p³) has one electron in each of the three 2p orbitals — a half-filled configuration with no paired 2p electrons and extra stability from exchange energy. Oxygen ([He]2s²2p⁴) must place a second electron into one 2p orbital; the repulsion between the two electrons sharing that orbital makes it easier to remove one, even though oxygen has a higher nuclear charge. Choice C is incorrect — oxygen's atomic radius is actually smaller than nitrogen's due to greater nuclear charge.

Al³⁺, Mg²⁺, and Na⁺ each contain 10 electrons but have nuclear charges of Z=13, 12, and 11 respectively. More protons attract the same electron cloud more strongly, resulting in a smaller ionic radius. Al³⁺ with the highest nuclear charge is the smallest, and Na⁺ with the lowest nuclear charge is the largest. This principle — greater nuclear charge means smaller size in an isoelectronic series — applies broadly to comparing ions with equal electron counts.

The dramatic jump in ionization energy occurs between the 3rd IE (2745 kJ/mol) and the 4th IE (11,578 kJ/mol) — roughly a fourfold increase. This large gap signals that the 4th electron must be removed from a core shell, meaning the first three electrons were valence electrons. The element therefore has 3 valence electrons and is aluminum (Al, Group 13, configuration [Ne]3s²3p¹). If there were only 2 valence electrons, the jump would appear between the 2nd and 3rd IE.

Magnesium (1s²2s²2p⁶3s²) has exactly four occupied subshells, producing four PES peaks. Listed from lowest to highest binding energy, the electron counts per subshell are: 3s²(2), 2p⁶(6), 2s²(2), 1s²(2), giving the ratio 2:6:2:2. Neon has three subshells (three peaks). Sodium also has four subshells but the outermost 3s¹ peak has area 1, giving 1:6:2:2. Silicon has five subshells (five peaks) with the outermost 3p²(2) giving 2:2:6:2:2.

Although fluorine has greater electronegativity, its 2p subshell is very small and already holds five electrons. Adding a sixth 2p electron forces it into a compact orbital where it experiences strong repulsion from existing electrons, partially offsetting the energy released. Chlorine's 3p orbitals are significantly larger and more diffuse, so the added electron experiences less repulsion, and the process releases more energy overall. Choice A is incorrect: fluorine (Z=9) has fewer protons than chlorine (Z=17), not more.

Iron's ground-state configuration is [Ar]4s²3d⁶. For transition metal cations, electrons are always removed from the 4s subshell before the 3d subshell. Removing the two 4s electrons and one 3d electron yields Fe³⁺ = [Ar]3d⁵. Manganese ([Ar]4s²3d⁵) losing three electrons — 4s² then one 3d — would give Mn³⁺ = [Ar]3d⁴, not [Ar]3d⁵. Choice D is wrong because losing one electron from [Ar]4s¹3d⁵ gives Cr⁺, not Cr³⁺.

Using the (n + l) rule: 4s has n+l = 4+0 = 4; 3d has n+l = 3+2 = 5; 4p has n+l = 4+1 = 5; 4d has n+l = 4+2 = 6. When n+l values tie (3d and 4p both equal 5), the subshell with lower n is lower in energy, so 3d < 4p. The resulting order from lowest to highest energy is 4s < 3d < 4p < 4d. Choice A incorrectly places 3d below 4s, which would mean d orbitals fill before s orbitals, contradicting the Aufbau filling sequence.

The second isotope has an abundance of 100% − 50.69% = 49.31%. The weighted average is: (78.918 × 0.5069) + (80.916 × 0.4931) = 40.003 + 39.900 = 79.903 amu, which rounds to 79.904 amu — the accepted atomic mass of bromine (Br). Choice B (79.917) results from an arithmetic error in the weighting. Choices C and D are the masses of the individual isotopes, not the weighted average, and would only be correct if the element were monoisotopic.

Both Ca²⁺ and Cl⁻ have the electron configuration [Ar] with 18 electrons, but Ca²⁺ has 20 protons and Cl⁻ has only 17 protons. The greater nuclear charge in Ca²⁺ exerts a stronger attractive force on the same 18 electrons, contracting the electron cloud and producing a smaller ionic radius. Cl⁻, with fewer protons holding the same number of electrons, is the larger ion. Having the same electron configuration does not mean identical size — nuclear charge is the deciding factor.

The dip at Al vs. Mg: magnesium ends with a filled 3s² subshell, while aluminum's additional electron enters the higher-energy 3p subshell, which is also partially shielded by the filled 3s. The 3p¹ electron is therefore easier to remove than either 3s electron from Mg. The dip at S vs. P: phosphorus ([Ne]3s²3p³) has each 3p orbital singly occupied — a stable half-filled configuration with no paired electrons. Sulfur ([Ne]3s²3p⁴) must pair one 3p orbital; the resulting electron-electron repulsion lowers the IE₁ below that of P despite sulfur's higher nuclear charge. Choice B is factually wrong on both counts: Al (Z=13) has more electrons than Mg (Z=12), and sulfur's atomic radius is smaller than phosphorus's.

The Aufbau principle requires filling subshells in order of increasing energy. Using the (n+l) rule, the 4s subshell has n+l = 4+0 = 4, while 3d has n+l = 3+2 = 5, so 4s is lower in energy and fills before 3d. Therefore, after 3p is filled, 4s is next. Choice A is incorrect because 3d, despite its lower principal quantum number, is higher in energy than 4s in neutral atoms.

A d subshell has angular momentum quantum number l=2, so the magnetic quantum number ml can take the values -2, -1, 0, +1, and +2 — five distinct values, each corresponding to one orbital. Since each orbital can hold up to 2 electrons, the d subshell has a maximum capacity of 10 electrons. Choice A (3) corresponds to a p subshell, and choice C (7) corresponds to an f subshell.

Atomic radius increases down a group because each successive element adds a new principal energy level, placing valence electrons farther from the nucleus. All four elements are in Group 1. Rubidium is in Period 5 with valence electrons in n=5, making it the largest. Lithium in Period 2 is the smallest of the four. This trend is driven by increasing shell size, which outweighs the effect of increasing nuclear charge within a group.

The n=3 level contains three subshells: 3s (1 orbital, 2 electrons), 3p (3 orbitals, 6 electrons), and 3d (5 orbitals, 10 electrons). The total capacity is 2+6+10 = 18 electrons. Choice A (8) accounts for only 3s and 3p, omitting 3d. Choice D (32) corresponds to the n=4 level, which also includes a 4f subshell with 14 electrons.

The angular momentum quantum number l defines orbital shape: l=0 gives a spherical s orbital, l=1 gives a dumbbell-shaped p orbital, and l=2 gives a d orbital. The principal quantum number n determines the energy level and overall size of the orbital. The magnetic quantum number ml specifies the orbital's orientation in space among degenerate orbitals of the same subshell, not its shape.

Isotopes are defined as atoms of the same element with the same number of protons (atomic number Z) but different numbers of neutrons. Because an element's identity is determined entirely by its proton count, isotopes always share the same number of protons and, in neutral atoms, the same number of electrons. Mass number (protons + neutrons), neutron count, and atomic mass all differ between isotopes of the same element.

First ionization energy generally increases across a period. Neon, as a noble gas with a completely filled valence shell (2s²2p⁶), has the highest first ionization energy in Period 2 at approximately 2081 kJ/mol. Removing any electron from neon requires breaking into a stable, filled octet configuration. Fluorine (1681 kJ/mol) is second highest. Although nitrogen has a slightly higher first ionization energy than oxygen due to the extra stability of its half-filled 2p subshell, neither exceeds neon.

The ground-state electron configuration of sulfur is [Ne]3s²3p⁴. The valence shell is n=3, containing 2 electrons in 3s and 4 electrons in 3p, for a total of 6 valence electrons. This places sulfur in Group 16 of the periodic table. The [Ne] core electrons are not valence electrons. Choice D (8) would represent a complete octet, which sulfur achieves only upon gaining two electrons to form S²⁻.

Across a period, protons are added to the nucleus while electrons enter the same principal energy level (n=3). Electrons in the same shell shield one another very ineffectively — the shielding constant for a same-shell electron is only about 0.35, compared to about 0.85 for an (n-1) shell electron. Because each new proton adds a full unit of nuclear charge while the corresponding valence electron provides only partial shielding, Zeff increases steadily from Na to Cl. Choice A is incorrect because same-shell electrons do not provide significant shielding.

Hund's rule states that electrons fill degenerate orbitals singly with parallel (same) spins before any pairing occurs. For carbon's two 2p electrons, each should occupy a separate 2p orbital with the same spin direction, minimizing electron-electron repulsion and maximizing exchange energy. Choice A violates Hund's rule by pairing in one orbital before the others are occupied. Choice B places electrons in separate orbitals but with antiparallel spins, which is also inconsistent with Hund's rule. Choice D violates the Pauli exclusion principle, which forbids two electrons in the same orbital from having identical spin quantum numbers.

The three quantum numbers directly identify the subshell. The principal quantum number n=4 specifies the fourth energy level. The angular momentum quantum number l=0 indicates an s subshell (l=0 is always s, l=1 is p, l=2 is d). The magnetic quantum number ml=0 is the only allowed value for l=0, confirming a single orbital. Therefore this electron is in the 4s subshell. Choice A (3s) incorrectly uses n=3, and choices C and D require l=1 and l=2, respectively.

All three species have 18 electrons but differ in nuclear charge: S²⁻ has Z=16, Cl⁻ has Z=17, and Ar has Z=18. With the same number of electrons, a greater nuclear charge pulls the electron cloud more tightly toward the nucleus, resulting in a smaller radius. Therefore the species with the fewest protons (S²⁻) has the largest, most diffuse electron cloud, and Ar with the most protons has the smallest. The correct order from largest to smallest is S²⁻ > Cl⁻ > Ar. Choice A incorrectly reverses this trend.

Photon energy is given by E = hν, where h is Planck's constant and ν is frequency. Since energy is directly proportional to frequency, the first source at 6.0 × 10¹⁴ Hz has exactly twice the photon energy of the second at 3.0 × 10¹⁴ Hz. Intensity (brightness) describes the number of photons delivered per unit time and area, not the energy per photon. Choice A incorrectly states that longer wavelength gives higher energy — the relationship is actually inverse (E = hc/λ). Choice D is incorrect because E = hν is linear, not quadratic, in frequency.

Neutral iron has the configuration [Ar]3d⁶4s². When iron forms a 2+ cation, it loses two electrons. In transition metal cations, 4s electrons are removed before 3d electrons because once the atom is ionized, the 3d subshell is lower in energy than 4s. Removing both 4s electrons yields [Ar]3d⁶ for Fe²⁺. Choice B incorrectly retains the 4s electrons while removing 3d electrons, which reverses the correct order of electron removal. Choice C represents an excited-state configuration for neutral iron.

In multielectron atoms, s orbitals have nonzero electron density at the nucleus (no angular nodes), allowing them to penetrate closer than p orbitals of the same n. This greater penetration means 2s electrons are less effectively shielded by inner electrons and experience a higher effective nuclear charge, lowering their energy relative to 2p. In hydrogen, which has only one electron and no shielding, 2s and 2p are degenerate. Choice D reverses the truth — the energy split between 2s and 2p arises specifically because of electron shielding in multielectron atoms.

The noble gas core [Xe] accounts for 54 electrons. Adding each subshell: 6s² contributes 2 electrons (total: 56), 4f¹⁴ contributes 14 electrons (total: 70), and 5d¹⁰ contributes 10 electrons (total: 80). Therefore Z=80, which is mercury (Hg). Choice A (Z=70, ytterbium) has the configuration [Xe]4f¹⁴6s², lacking the filled 5d subshell. Choice B (Z=78, platinum) has an anomalous configuration [Xe]4f¹⁴5d⁹6s¹. Choice D (Z=86, radon) is the next noble gas after Xe and includes 6p⁶.

Let x be the fractional abundance of ⁶³X, so (1−x) is the fractional abundance of ⁶⁵X. Setting up the weighted average: 62.93x + 64.93(1−x) = 63.55. Expanding: 62.93x + 64.93 − 64.93x = 63.55, which gives −2.00x = −1.38, so x = 0.69 or 69%. Verification: (0.69)(62.93) + (0.31)(64.93) = 43.42 + 20.13 = 63.55 amu. Choice D (31%) is the percent abundance of the heavier isotope ⁶⁵X, not ⁶³X.

The large jump in successive ionization energies reveals how many valence electrons an element has. The first three ionization energies increase gradually (577 → 1817 → 2745 kJ/mol), but IE₄ = 11,577 is roughly four times larger than IE₃ = 2745. This dramatic increase means the fourth electron must be removed from a stable, fully filled inner shell. Therefore the element has exactly 3 valence electrons, placing it in Group 13. For a Group 1 element, the large jump would appear between IE₁ and IE₂; for Group 2, between IE₂ and IE₃; for Group 14, between IE₄ and IE₅.

Each peak in a PES corresponds to electrons in a distinct subshell. The ratio 2:2:6:2 indicates four subshells containing 2, 2, 6, and 2 electrons respectively, totaling 12 electrons. This matches 1s²:2s²:2p⁶:3s², which is the configuration of magnesium (Z=12). Silicon (Z=14) would show five peaks with ratios 2:2:6:2:2. Sulfur (Z=16) would show 2:2:6:2:4. Neon (Z=10) has only three occupied subshells, producing three peaks with ratio 2:2:6. The highest-binding-energy peak corresponds to 1s (area ratio 2), and the tallest peak corresponds to 2p (area ratio 6).

Chromium's anomalous configuration results from the exceptional stability of a half-filled 3d subshell (3d⁵). When all five 3d orbitals each hold one electron with parallel spins, exchange energy — a quantum mechanical stabilization arising from electrons of the same spin interchanging positions — is maximized. This stabilization outweighs the small energy cost of promoting one electron from 4s to 3d. Choice A is incorrect: 3d is not universally lower than 4s across all Period 4 transition metals in neutral atoms; the energies are close and context-dependent. Choice D is factually wrong — [Ar]3d⁴4s² does not violate the Pauli exclusion principle.

In magnesium, the outermost electrons are in the 3s subshell. In aluminum, the thirteenth electron enters the 3p subshell. The 3p orbital is higher in energy than 3s and is partially shielded from the nucleus by the filled 3s subshell beneath it, reducing the effective nuclear charge experienced by the 3p electron. Despite Al having Z=13 versus Mg's Z=12, the reduced Zeff on the 3p electron makes it easier to remove. This is a recognized exception to the simple rule that first ionization energy increases uniformly across a period. Choice A is a minor contributing factor but not the primary explanation for this specific anomaly.

For any principal quantum number n, the angular momentum quantum number l must satisfy 0 ≤ l ≤ n−1. For n=2, l can only be 0 or 1. The set in choice C has l=2 for n=2, which violates this rule and is therefore impossible. Choice A is valid: n=3 allows l up to 2, and for l=2, ml ranges from −2 to +2. Choice B is valid: n=4 allows l up to 3, and for l=3, ml ranges from −3 to +3. Choice D is valid: n=1 requires l=0 and ml=0, and both spin values are allowed.

The de Broglie wavelength is given by λ = h/mv, where h is Planck's constant, m is the electron mass, and v is the electron's speed. Because all options involve electrons (constant mass), wavelength is inversely proportional to speed: higher speed gives shorter wavelength. The electron moving at 10% of the speed of light has the greatest speed of the four options and therefore the shortest de Broglie wavelength. This inverse relationship illustrates why high-energy electrons are used in electron microscopy — their short wavelengths allow resolution of very small features.

Nitrogen has the configuration [He]2s²2p³, in which each of the three 2p orbitals holds exactly one electron — a half-filled arrangement with maximum exchange energy stability. Adding an electron to form N⁻ requires forcing a second electron into one of the occupied 2p orbitals, introducing electron-electron repulsion and breaking the half-filled stability. In contrast, oxygen ([He]2s²2p⁴) already has one paired 2p orbital, so adding one more electron to form O⁻ does not disrupt a special arrangement, making the process more energetically favorable. Choice C is insufficient — a one-unit difference in nuclear charge alone cannot explain why nitrogen bucks the general trend of increasing electron affinity across a period.

The ion R³⁺ = [Ar]3d⁵ contains 18 (Ar core) + 5 = 23 electrons. The neutral atom therefore has Z = 23 + 3 = 26, identifying it as iron (Fe) with ground-state configuration [Ar]3d⁶4s². Forming Fe³⁺ requires removing both 4s electrons and one 3d electron, yielding [Ar]3d⁵. The resulting half-filled 3d subshell — five d electrons with parallel spins across all five d orbitals — has maximum exchange energy stabilization, which is why Fe³⁺ is one of iron's most common and thermodynamically stable oxidation states. Choice C is incorrect: Mn³⁺ (Z=25, minus 3 electrons = 22 electrons) gives [Ar]3d⁴, not 3d⁵. Choice D gives Cr³⁺ = [Ar]3d³, also not 3d⁵.

Moving down a group, each successive element has one additional principal energy level, so the outermost electrons occupy shells farther from the nucleus. Although nuclear charge also increases, the effect of additional inner-shell shielding outweighs the increased nuclear charge, resulting in a larger atomic radius. Choice A reaches the wrong conclusion: while nuclear charge does increase, shielding increases proportionally more, so radius grows rather than shrinks.

The Pauli exclusion principle states that no two electrons in the same atom can share all four quantum numbers (n, l, ml, ms). This limits each orbital to a maximum of two electrons with opposite spins. Choices A and D describe Hund's rule, which governs how electrons fill degenerate orbitals. Choice B describes the Aufbau principle.

Hund's rule states that electrons occupy degenerate orbitals singly before any pairing occurs, and all singly occupied orbitals have electrons with the same (parallel) spin. Nitrogen's 2p³ configuration places one electron in each 2p orbital with identical spin, giving three unpaired electrons. Choice D is wrong because Hund's rule requires parallel (same) spin in singly occupied degenerate orbitals, not alternating spins.

The quantum mechanical model treats electrons as matter waves and uses orbitals — probability distributions — to describe the likelihood of finding an electron in a given region of space. Unlike the earlier Bohr model, it does not assign electrons to fixed circular orbits or definite paths. The Heisenberg uncertainty principle makes it impossible to know both position and momentum precisely at the same time, ruling out choices C and D.

Electronegativity increases from left to right across a period because nuclear charge increases while atomic size decreases, strengthening the pull on bonding electrons. F (Z=9), O (Z=8), and N (Z=7) are all in Period 2. Fluorine is the most electronegative element on the periodic table, followed by oxygen, then nitrogen. The correct decreasing order is F > O > N.

Argon (Ar, Z=18) has the configuration [Ne] 3s² 3p⁶. Potassium (Z=19) adds one electron to the 4s subshell, which is lower in energy than the 3d for Period 4 elements, giving [Ar] 4s¹. Choice C ([Ar] 3d¹) incorrectly places the 19th electron in the 3d subshell; the 4s fills before 3d in the ground state of potassium. Choice A writes out the Ar core explicitly rather than using the noble gas shorthand.

A d subshell has angular momentum quantum number l = 2, so the magnetic quantum number ml can take the values −2, −1, 0, +1, and +2, giving 5 d orbitals. Each orbital holds a maximum of 2 electrons (with opposite spins), so the d subshell accommodates 5 × 2 = 10 electrons. Choice A (6) is the capacity of a p subshell (3 orbitals × 2), and choice C (14) is the capacity of an f subshell (7 orbitals × 2).

For a valid set: l must range from 0 to n−1; ml must range from −l to +l; ms must be ±1/2. Choice C: n=4 allows l=0,1,2,3, so l=3 is valid; ml ranges from −3 to +3, so ml=−2 is valid; ms=+1/2 is valid. Choice A fails because l=2 requires n ≥ 3, but n=2 only allows l=0 or 1. Choice B fails because l=1 restricts ml to −1, 0, or +1, not −2. Choice D fails because l=0 restricts ml to 0 only, not 1.

Phosphorus has a half-filled 3p³ subshell with all three electrons having parallel spin. This arrangement maximizes exchange energy — a quantum mechanical stabilization — making the configuration unusually stable. Sulfur (3p⁴) has one paired electron in a 3p orbital; that electron-electron repulsion within the same orbital makes it easier to remove than any of phosphorus's singly-occupied 3p electrons. This is a well-known exception to the general left-to-right increase in first ionization energy. Choice A is incorrect because sulfur actually has a smaller atomic radius than phosphorus.

Isotopes of the same element are defined by having the same atomic number — the same number of protons. For neutral atoms, the number of electrons always equals the number of protons to maintain electrical neutrality. Therefore, neutral atoms of the same element's isotopes share both their proton count and electron count. Isotopes differ by definition in their neutron count and therefore have different mass numbers and different atomic masses. Choice A is wrong because isotopes have different mass numbers and atomic masses.

Neutral copper has the anomalous configuration [Ar] 3d¹⁰ 4s¹ (choice D) because a completely filled 3d subshell is especially stable. When copper loses one electron to form Cu⁺, the 4s electron is removed first — in transition metal cations, 4s electrons are always removed before 3d electrons. Removing the single 4s¹ electron leaves [Ar] 3d¹⁰. Choice C ([Ar] 3d⁹) would result from removing a 3d electron instead, which is incorrect.

The kinetic energy of an ejected electron is given by KE = hf − Φ, where h is Planck's constant and f is the frequency of light. Increasing f increases hf, which directly increases KE. Increasing intensity (choice A) increases the number of ejected electrons per second but does not change the kinetic energy of each individual electron, since each photon still carries energy hf. Lower frequency (choice B) decreases KE and may not even eject electrons. A higher work function (choice D) increases Φ, which decreases KE.

The angular momentum quantum number l determines orbital shape: l=0 gives a spherical s orbital, l=1 gives a dumbbell-shaped p orbital, and l=2 gives the more complex d orbital shapes. The principal quantum number n determines the energy level and the overall size of the orbital. The magnetic quantum number ml specifies the orientation of the orbital in space among orbitals of the same l. The spin quantum number ms describes the intrinsic angular momentum of the electron, not the orbital shape.

Na⁺ has 10 electrons and 11 protons, so the nuclear charge per electron ratio is higher than in neutral Na (11 electrons, 11 protons), pulling electrons closer and shrinking the radius. Na⁻ has 12 electrons and 11 protons; the extra electron increases electron-electron repulsion, expanding the electron cloud. The correct order of increasing radius is Na⁺ < Na < Na⁻. Choice D reverses this ranking entirely.

Neutral Mn (Z=25) has the configuration [Ar] 3d⁵ 4s². When transition metals form cations, the 4s electrons are removed before the 3d electrons because, in cations, the 3d subshell is lower in energy than 4s. Removing both 4s² electrons from Mn yields [Ar] 3d⁵ for Mn²⁺. Choice B ([Ar] 3d⁵ 4s²) is the configuration of neutral Mn, not the ion. Choice A incorrectly removes 3d electrons instead of 4s electrons.

Energy levels in hydrogen are given by En = −13.6/n² eV. For n=4→2: ΔE = 13.6(1/4 − 1/16) ≈ 2.55 eV. For n=3→2: ΔE = 13.6(1/4 − 1/9) ≈ 1.89 eV. The n=4→2 transition releases more energy, so its photon has higher energy. Since E = hc/λ, higher photon energy corresponds to shorter wavelength. Choices A and D are wrong because they assign lower energy to the n=4→2 transition.

First ionization energy decreases as you move down a group in the periodic table. This occurs because valence electrons are in higher principal energy levels farther from the nucleus and are more effectively shielded by inner shells, reducing the effective nuclear charge they experience. Mg (Group 2, Period 3) has the highest IE₁ of the three; Ca (Period 4) is next; Ba (Period 6) has the lowest. In order of increasing IE₁: Ba < Ca < Mg.

Effective nuclear charge Zeff = Z − S, where S is the shielding constant. Core electrons reduce the nuclear charge felt by valence electrons, but the shielding is imperfect — inner electrons do not completely cancel out an equal number of protons. For sodium's 3s electron, Zeff is approximately 2.5 by Slater's rules, which lies between 1 and 11. Choice B is wrong because core electrons never provide perfect (100%) shielding; the outer electron still feels net nuclear attraction beyond just +1. Choice C is physically impossible — inner electrons cannot amplify the nuclear charge beyond Z.

Na ([Ne]3s¹) has only one valence electron. Its IE₁ removes the 3s electron relatively easily. IE₂ must pull an electron from the n=2 shell (the [Ne] core), which is much more tightly held due to its smaller principal quantum number, larger effective nuclear charge felt at n=2, and negligible shielding from only the 1s² electrons. This causes a dramatic jump — Na's IE₂ (~4562 kJ/mol) is roughly nine times its IE₁ (496 kJ/mol). Mg has two valence electrons, so both IE₁ and IE₂ remove 3s electrons; Mg's large jump appears between IE₂ and IE₃, not IE₁ and IE₂.

In the d⁵ configuration, all five d electrons have parallel spin. This maximizes exchange energy — a quantum mechanical stabilization that arises when electrons with the same spin can exchange positions. The total exchange energy gained by achieving d⁵ is large enough to offset the cost of promoting one s electron, making the d⁵s¹ arrangement energetically lower than d⁴s². Choice C misidentifies the driving force; the stabilization comes from the d⁵ arrangement, not from repulsion between s electrons. Choice D overgeneralizes; this anomaly only occurs in specific elements where the d⁵ (or d¹⁰) stabilization is large enough to tip the energy balance.

The total electron count is 2+2+6+2 = 12, matching magnesium (Z=12). Magnesium's ground-state configuration is 1s²2s²2p⁶3s², which has exactly four distinct subshells. The 1s subshell (2 electrons) appears at the highest binding energy, followed by 2s (2 electrons), 2p (6 electrons), and 3s (2 electrons, lowest binding energy). Neon (Z=10) has only 10 electrons and shows three peaks in a 2:2:6 pattern with no 3s electrons. Silicon (Z=14) would show 2:2:6:2 for the first four subshells but also has additional 3p electrons, giving a fifth peak.

From KE = p²/(2m), momentum p = √(2mKE). Since the proton mass is about 1836 times the electron mass, p_proton = √(2m_p KE) >> p_electron = √(2m_e KE) at the same KE. Because λ = h/p, larger momentum means shorter wavelength, so the proton has a shorter wavelength and the electron has a longer wavelength. Choice A reaches the wrong conclusion: smaller mass does mean smaller momentum, but smaller momentum gives longer wavelength, not shorter. Choice D incorrectly states that greater mass increases wavelength; greater mass increases momentum, which decreases wavelength.

The roughly nine-fold jump from IE₁ (496 kJ/mol) to IE₂ (4562 kJ/mol) reveals that the element has exactly one electron outside the noble gas core. Sodium (Na, [Ne]3s¹) matches this profile precisely — its actual IE₁ is 496 kJ/mol. After the single 3s valence electron is removed, the second ionization breaks into the n=2 Ne core, where electrons experience much higher effective nuclear charge and lower principal quantum number, requiring dramatically more energy. Magnesium (choice B) has two valence electrons, so its large jump falls between IE₂ and IE₃, not IE₁ and IE₂. Lithium (choice C) is Period 2, not Period 3.

With 8 electrons distributed among 5 d orbitals by Hund's rule: the first 5 electrons each singly occupy one orbital with parallel spin (↑ ↑ ↑ ↑ ↑), then the remaining 3 electrons pair up in three of those orbitals (↑↓ ↑↓ ↑↓ ↑ ↑). The result is two singly occupied orbitals with unpaired electrons. Because unpaired electrons possess net magnetic moments, Ni²⁺ is paramagnetic. Choice A is wrong because 8 electrons cannot fill 5 orbitals completely in pairs (that would require 10 electrons). Choice C is wrong because 8 orbitals would be needed for 8 unpaired electrons, but d has only 5 orbitals.

After IE₁, potassium becomes K⁺ with the electron configuration of argon ([Ne]3s²3p⁶) — a complete noble gas core. The second ionization energy of K requires pulling an electron from the n=3 shell, where electrons experience a much larger effective nuclear charge and lie much closer to the nucleus than n=4 electrons do. Calcium's IE₂, by contrast, removes a second 4s valence electron, which requires substantially less energy. This is the same principle behind Na's anomalously large IE₂. Choice C is partially true (pairing repulsion slightly lowers Ca's IE₂) but does not account for the large magnitude of the difference; the dominant factor is K⁺ having a noble gas core.

Each peak in a photoelectron spectrum corresponds to a distinct subshell, and peak intensity is proportional to the number of electrons in that subshell. The ratio 2:2:6:2:1 sums to 13 electrons total. Mapping from highest to lowest binding energy: 1s² : 2s² : 2p⁶ : 3s² : 3p¹ = 2:2:6:2:1. This matches aluminum (Z = 13, configuration 1s²2s²2p⁶ 3s²3p¹). Magnesium (Z = 12) would show only four peaks in ratio 2:2:6:2. Silicon (Z = 14) would show five peaks but with ratio 2:2:6:2:2, because its 3p subshell holds 2 electrons, not 1.

The Heisenberg uncertainty principle states that the product of the uncertainties in position and momentum has a fundamental minimum: Δx · Δp ≥ h/(4π). A circular Bohr orbit requires simultaneously knowing the exact position (a point on the circle) and exact momentum (direction and speed of travel) of the electron — a combination that is forbidden by the uncertainty principle. This is not an instrumental limitation but an inherent feature of quantum mechanics. Modern quantum mechanics therefore replaces orbits with orbitals (regions of probability density). Choice C is a common misconception; the principle is a property of nature, not of technology.

Beryllium's configuration is 1s²2s²: the 2s subshell is completely filled, and an added electron must enter the higher-energy 2p subshell. The extra stability of the filled 2s resists this, resulting in a near-zero electron affinity. Nitrogen's configuration is 1s²2s²2p³: the 2p subshell is exactly half-filled with one electron per orbital. Adding a fourth electron forces it to pair with an existing 2p electron, introducing electron-electron repulsion and reducing the energy released. Oxygen, by contrast, already has a doubly occupied 2p orbital and therefore does not lose this half-filled stability advantage when a fifth 2p electron is added.

The 2s orbital has a radial node but also has significant electron density very close to the nucleus — it penetrates the 1s core. This close approach means a 2s electron is not fully shielded by core electrons and experiences a higher effective nuclear charge (Zeff) than a 2p electron, which has a nodal plane through the nucleus and remains farther from it on average. Higher Zeff means stronger nuclear attraction and lower energy. This is why within any principal shell, subshell energy increases with l: s < p < d < f. Choice C is correct only for hydrogen (one-electron atom); in multi-electron atoms, shielding breaks the n-only degeneracy.

Using the formula: radial nodes = n − l − 1 = 4 − 1 − 1 = 2. A radial node is a spherical shell at a fixed distance r from the nucleus at which the radial wave function R(r) = 0, making the total probability density ψ² = 0 at that radius for all directions. This differs from an angular node, which is a plane or cone through the nucleus arising from the angular part of the wave function. A 4p orbital has 2 radial nodes and 1 angular node, for a total of 3 nodes, consistent with the general result that total nodes = n − 1 = 3. Choice A describes an angular node, not a radial node.

Check each element against all three criteria. Manganese: (1) [Ar]3d⁵ 4s² has 5 unpaired electrons, so it is paramagnetic [pass]; (2) Mn²⁺ removes both 4s electrons, giving [Ar]3d⁵, which is exactly half-filled [pass]; (3) IE₁(Mn) ≈ 717 kJ/mol < 800 kJ/mol [pass]. Iron fails criterion 2: Fe²⁺ = [Ar]3d⁶, which is not half-filled. Zinc fails criteria 1 and 2: Zn has a filled 3d¹⁰ 4s² configuration (diamagnetic), and Zn²⁺ = [Ar]3d¹⁰ (fully filled, not half-filled). Chromium fails criterion 2: Cr²⁺ loses the 4s¹ electron first then one 3d electron, giving [Ar]3d⁴, which is not half-filled. Only manganese satisfies all three conditions.