Atomic Structure and Periodicity: Quanta, Matter Waves, the Hydrogen Atom, Configurations and Periodic Trends

This chapter answers Section XL0.1 of XL0 Chemistry, the part of the Life Sciences (XL) paper that every XL candidate answers: “Planck’s quantum theory, wave particle duality, uncertainty principle, comparison between Bohr’s model and quantum mechanical model of hydrogen atom, electronic configuration of atoms and ions. Hund’s rule and Pauli’s exclusion principle. Periodic table and periodic properties: ionization energy, electron affinity, electronegativity and atomic size.” It builds the atom from the quantum of energy upwards: photons and the photoelectric effect, de Broglie waves and the uncertainty limit, the hydrogen atom as Bohr pictured it and as quantum mechanics solves it, the rules that fill orbitals in atoms and ions, and the periodic trends that follow from effective nuclear charge. The numericals are photon energies, photoelectron kinetic energies, de Broglie wavelengths, minimum uncertainties, Bohr energies and spectral wavelengths, and electron counts from quantum numbers.

1. Planck’s quantum theory and the photon

Classical physics treated radiant energy as continuous and could not explain the spectrum of a hot black body: it predicted intensity rising without limit at short wavelengths, which is not observed. In 1900 Max Planck proposed that energy is emitted or absorbed only in discrete packets, quanta, each of size E = hν, where h = 6.626 × 10⁻³⁴ J s is Planck’s constant and ν the frequency. Since c = νλ, E = hc/λ: the shorter the wavelength, the larger the quantum, and energy changes only in whole numbers of quanta (E = nhν). In 1905 Einstein applied the idea to light itself: light travels as photons, each carrying hν.

Worked example. A photon of 500 nm has E = (6.626 × 10⁻³⁴ × 3.00 × 10⁸)/(500 × 10⁻⁹) = 3.98 × 10⁻¹⁹ J, which is 2.48 eV (1 eV = 1.602 × 10⁻¹⁹ J). A mole of such photons carries 3.98 × 10⁻¹⁹ × 6.022 × 10²³ = 2.39 × 10⁵ J, that is 239 kJ mol⁻¹, the same order as bond energies. That is why visible and ultraviolet light can drive photochemistry: photosynthesis, vision and UV damage to DNA. A handy shortcut is E (eV) ≈ 1240/λ (nm).

The photoelectric effect settled the matter. Light ejects electrons from a metal only if its frequency exceeds a threshold frequency ν₀ fixed by the metal’s work function φ = hν₀, and then KE(max) = hν − φ. Raising the intensity raises the number of electrons, not their energy, and emission is instant. A wave model predicted the opposite on every count. Example: for a metal with φ = 2.00 eV, 500 nm light (2.48 eV) gives KE(max) = 0.48 eV, so a stopping potential of 0.48 V halts the current.

⚠️ Intensity is not frequency
Below the threshold, no amount of intensity ejects an electron. Above it, doubling the intensity doubles the photocurrent but leaves KE(max) unchanged; only a higher frequency raises KE(max). Always convert λ to joules or eV before subtracting φ, and keep both in the same unit.

2. Wave–particle duality and the uncertainty principle

Light shows wave behaviour (interference, diffraction) and particle behaviour (photoelectric and Compton effects). In 1924 de Broglie proposed the converse: every moving particle has a wave of wavelength λ = h/p = h/mv. For a particle of charge e accelerated from rest through a potential V, its kinetic energy is eV, so p = √(2meV) and λ = h/√(2meV). For an electron through 100 V: λ = 6.626 × 10⁻³⁴/√(2 × 9.109 × 10⁻³¹ × 1.602 × 10⁻¹⁹ × 100) = 1.23 × 10⁻¹⁰ m = 123 pm, comparable to the spacing of atoms in a crystal. Electron beams therefore diffract from crystals (Davisson and Germer, 1927), and the electron microscope resolves far finer detail than a light microscope because its wavelength is so short. A 0.15 kg ball at 40 m s⁻¹ has λ ≈ 1.1 × 10⁻³⁴ m, far too small to detect, which is why duality matters only for tiny masses.

A wave is spread out, so position and momentum cannot both be sharp. Heisenberg’s uncertainty principle (1927) states Δx · Δp ≥ h/4π, with Δp = mΔv. It is a property of nature, not of poor instruments. For an electron confined to Δx = 0.1 nm, about the size of an atom, Δv ≥ h/(4π m Δx) = 6.626 × 10⁻³⁴/(4π × 9.109 × 10⁻³¹ × 1.0 × 10⁻¹⁰) = 5.8 × 10⁵ m s⁻¹, as large as the speeds electrons have in atoms, so a sharply defined orbit is meaningless. For a 1 g bead located to 1 µm, Δv is about 5 × 10⁻²⁶ m s⁻¹, utterly negligible.

⚠️ Which form of the inequality?
Δx · Δp ≥ h/4π is the same as ≥ ħ/2; some older texts write h/2π or simply ≈ h, which changes the answer by a factor of 2 or 4π. Use the form the stem gives. Put the mass in kilograms and the length in metres; a mass in grams makes the answer a thousand times too large.

3. The hydrogen atom: Bohr’s model and the quantum-mechanical model

Bohr (1913) postulated that the electron moves in circular stationary orbits without radiating, that its angular momentum is quantised, mvr = nh/2π, and that light is emitted or absorbed only when it jumps between orbits, with ΔE = hν. For a one-electron species of nuclear charge Z this gives rₙ = 52.9 n²/Z pm and Eₙ = −13.6 Z²/n² eV (−2.18 × 10⁻¹⁸ Z²/n² J). Spectral lines follow 1/λ = R_H Z² (1/n₁² − 1/n₂²) with R_H = 1.097 × 10⁷ m⁻¹: the Lyman series ends on n₁ = 1 (ultraviolet), Balmer on n₁ = 2 (visible) and Paschen on n₁ = 3 (infrared). Worked example: for H, n = 3 → 2 gives 1/λ = 1.097 × 10⁷ × (1/4 − 1/9) = 1.524 × 10⁶ m⁻¹, so λ = 656 nm, the red Hα line. The ionisation energy is 13.6 eV for H, 54.4 eV for He⁺ and 122.4 eV for Li²⁺, and electrons falling from level n to the ground state can give up to n(n − 1)/2 distinct lines.

Schrödinger’s quantum-mechanical model replaces the orbit with a wavefunction ψ, an orbital, whose square ψ² gives the probability density of finding the electron. Solving the wave equation for hydrogen produces three quantum numbers (n, l, m_l), with spin (m_s) added as a fourth. For hydrogen the energy still depends only on n and equals −13.6/n² eV, so Bohr’s model gets the hydrogen spectrum right while getting the picture wrong.

Bohr’s model and the quantum-mechanical model of hydrogen compared
FeatureBohr’s modelQuantum-mechanical model
Where the electron isOn a fixed circular orbit of exact radiusSpread as a probability cloud (orbital); 52.9 pm is only the most probable 1s distance
Quantum numbersOne (n), imposed as a postulaten, l, m_l from the wave equation, plus spin m_s
Energy of hydrogen−13.6/n² eV−13.6/n² eV, the same; all orbitals of one n are degenerate
Ground-state angular momentumh/2πZero, since 1s has l = 0
Uncertainty principleViolated: exact radius and exact velocity togetherObeyed
Many-electron atoms, line intensities, splitting in a magnetic fieldFailsExplained (with approximations for many electrons)

4. Quantum numbers, Pauli, Hund and electronic configurations

  • Principal, n = 1, 2, 3 …: shell, size and (in hydrogen) energy. A shell holds n² orbitals and 2n² electrons.
  • Azimuthal, l = 0 to n − 1 (s, p, d, f): shape of the subshell; an orbital has l angular nodes.
  • Magnetic, m_l = −l to +l: orientation; a subshell has 2l + 1 orbitals.
  • Spin, m_s = +½ or −½. Radial nodes = n − l − 1 and total nodes = n − 1, so 2s has one radial node, 3p one radial and one angular, 3d none radial and two angular.

Three rules fill the orbitals. The Aufbau principle with the (n + l) rule fills lower n + l first, and for equal n + l the lower n first: 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p. Pauli’s exclusion principle: no two electrons in an atom share all four quantum numbers, so an orbital holds at most two electrons, with opposite spins. Hund’s rule of maximum multiplicity: in degenerate orbitals electrons enter singly with parallel spins before any pairing, so C (2p²) has two unpaired electrons, N (2p³) three and O (2p⁴) two. Parallel spins keep electrons apart, lowering repulsion, and gain exchange energy, which is also why half-filled and filled subshells are extra stable.

Two exceptions in the 3d row follow from that stability: Cr is [Ar] 3d⁵ 4s¹ and Cu is [Ar] 3d¹⁰ 4s¹. For ions, electrons leave the orbital of highest n first, so transition metals lose 4s before 3d: Fe [Ar] 3d⁶ 4s² gives Fe²⁺ [Ar] 3d⁶ and Fe³⁺ [Ar] 3d⁵; Cu²⁺ is [Ar] 3d⁹ and Zn²⁺ [Ar] 3d¹⁰. Anions add electrons to the next vacancy: Cl⁻ is [Ne] 3s² 3p⁶, isoelectronic with Ar. The count of unpaired electrons (five in Fe³⁺ and Mn²⁺, one in Cu²⁺) sets magnetic behaviour, which Section XL0.3 uses.

⚠️ 4s fills first and empties first
Fe²⁺ is [Ar] 3d⁶, never [Ar] 3d⁴ 4s². Once 3d holds electrons it lies below 4s in energy, so ionisation removes 4s electrons first. Check every ion by writing the neutral atom, including the Cr and Cu exceptions, before removing electrons.

5. The periodic table and periodic properties

The long form has 18 groups and 7 periods, arranged in s (groups 1–2), p (13–18), d (3–12) and f blocks. Properties recur because valence configurations recur. The driver is the effective nuclear charge, Z(eff) = Z − S, where S is the shielding by other electrons. Across a period Z rises by one each step but electrons in the same shell shield one another poorly, so Z(eff) rises and the outer electrons are held tighter; down a group a new, larger shell is added.

The four periodic properties at a glance
PropertyAcross a periodDown a groupExceptions worth knowing
Atomic sizeDecreasesIncreasesLanthanide contraction makes Zr and Hf almost equal in size
First ionisation energyIncreasesDecreasesBe > B, N > O, Mg > Al, P > S
Electron affinity (energy released)Increases, highest for halogensGenerally decreasesCl > F; noble gases, Be and N near zero or unfavourable
ElectronegativityIncreasesDecreasesF highest (about 4.0 on the Pauling scale)

Ionisation energy (IE). First IEs in kJ mol⁻¹: Li 520, Be 899, B 801, C 1086, N 1402, O 1314, F 1681, Ne 2081. B falls below Be because its electron is 2p, higher in energy and shielded by 2s²; O falls below N because its fourth 2p electron must pair, adding repulsion, while N has a stable half-filled 2p³. Successive IEs always rise, and a large jump marks the first electron taken from a noble-gas core: magnesium’s 738, 1451 and 7733 kJ mol⁻¹ reveal two valence electrons.

Electron affinity, electronegativity and size. Electron affinity is the energy released when a gaseous atom gains an electron; the electron gain enthalpy has the opposite sign. Cl (349 kJ mol⁻¹) exceeds F (328) because the incoming electron in fluorine’s small, crowded 2p shell meets strong repulsion, and adding a second electron to any anion (O⁻ → O²⁻) is endothermic. Electronegativity, the pull of a bonded atom on shared electrons, is not measured directly: Pauling derived it from bond energies (F ≈ 4.0, O ≈ 3.4, N ≈ 3.0, C ≈ 2.5, H ≈ 2.2), Mulliken took the average (IE + EA)/2, and Allred and Rochow used Z(eff)/r². A cation is smaller than its atom and an anion larger; in an isoelectronic series more protons pull the same electrons closer, so O²⁻ > F⁻ > Na⁺ > Mg²⁺ > Al³⁺.

🎯 Why a life scientist needs these trends
The electronegativity gap between O or N and H makes O–H and N–H bonds polar, and that polarity is what hydrogen bonds in water, proteins and DNA rely on. Ionic size decides which ion a binding site or channel accepts: a potassium channel passes K⁺ but excludes the smaller Na⁺.

Key takeaways

  • Energy comes in quanta E = hν = hc/λ. In the photoelectric effect KE(max) = hν − φ: frequency sets the electron energy, intensity only the number of electrons.
  • Matter waves have λ = h/mv, or h/√(2meV) after acceleration through V; the uncertainty principle Δx · Δp ≥ h/4π makes a sharp electron orbit meaningless but is negligible for everyday masses.
  • For one-electron species Eₙ = −13.6 Z²/n² eV and 1/λ = R_H Z²(1/n₁² − 1/n₂²). Bohr and wave mechanics agree on these energies but differ on orbits versus orbitals, the number of quantum numbers and the ground-state angular momentum (h/2π versus zero).
  • An orbital has n − l − 1 radial and l angular nodes; a shell has n² orbitals and 2n² electrons. Pauli allows two opposite-spin electrons per orbital; Hund fills degenerate orbitals singly with parallel spins first.
  • Cr is 3d⁵4s¹ and Cu 3d¹⁰4s¹; transition-metal ions lose 4s electrons before 3d, so Fe²⁺ is 3d⁶ and Fe³⁺ 3d⁵.
  • Rising Z(eff) across a period shrinks atoms and raises IE, electron affinity and electronegativity; the exceptions Be > B, N > O and Cl > F, the jump in successive IEs and isoelectronic radii are the favourite questions.

Practice questions (17)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. According to Planck’s quantum theory, the energy of a single quantum of radiation is directly proportional to which quantity?

    1. Its wavelength, since longer waves carry more energy in each quantum
    2. The intensity of the beam, since a brighter beam delivers more energy
    3. Its frequency, through E = hν with h the Planck constant
    4. The square of the wave amplitude, as the classical wave picture says
    Show answer

    Answer: C — Its frequency, through E = hν with h the Planck constant

    Planck’s quantum is E = hν = hc/λ, so the energy of one quantum rises with frequency and falls as wavelength grows. Intensity is the tempting answer, but it sets how many photons arrive each second, not the energy of each; the square of the amplitude is the classical measure of that same intensity.

  2. An electron (mass 9.109 × 10⁻³¹ kg, charge 1.602 × 10⁻¹⁹ C), initially at rest, is accelerated through a potential difference of 150 V. Taking h = 6.626 × 10⁻³⁴ J s and ignoring relativistic effects, its de Broglie wavelength, in pm to the nearest whole number, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 100

    Kinetic energy = eV = 1.602 × 10⁻¹⁹ × 150 = 2.403 × 10⁻¹⁷ J. Momentum p = √(2 × 9.109 × 10⁻³¹ × 2.403 × 10⁻¹⁷) = 6.617 × 10⁻²⁴ kg m s⁻¹. λ = h/p = 6.626 × 10⁻³⁴/6.617 × 10⁻²⁴ = 1.001 × 10⁻¹⁰ m = 100 pm. Forgetting the factor 2 under the root, or using 150 instead of 150 × 1.602 × 10⁻¹⁹ J, are the usual slips.

  3. Light of wavelength 400 nm falls on a metal surface whose work function is 2.25 eV. Taking h = 6.626 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹ and 1 eV = 1.602 × 10⁻¹⁹ J, the maximum kinetic energy of the emitted photoelectrons, in eV to two decimal places, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 0.85

    Photon energy E = hc/λ = (6.626 × 10⁻³⁴ × 3.00 × 10⁸)/(400 × 10⁻⁹) = 4.970 × 10⁻¹⁹ J = 4.970 × 10⁻¹⁹/1.602 × 10⁻¹⁹ = 3.102 eV. KE(max) = hν − φ = 3.102 − 2.25 = 0.852 eV, which is 0.85 to two decimal places. Subtracting the work function in eV from the photon energy in joules is the common error.

  4. Which of the following observations show either the particle nature of light or the wave nature of matter?

    1. White light is split into a spectrum of colours by a glass prism
    2. A beam of electrons passing through a thin crystal produces a diffraction pattern
    3. Below a threshold frequency no photoelectrons are emitted, however intense the light
    4. Light passing through two narrow slits forms alternate bright and dark fringes
    Show answer

    Answer: B — A beam of electrons passing through a thin crystal produces a diffraction pattern; C — Below a threshold frequency no photoelectrons are emitted, however intense the light

    Electron diffraction by a crystal shows that electrons, which are matter, behave as waves (de Broglie). The threshold frequency in the photoelectric effect can only be explained if light arrives as photons of energy hν, so it shows the particle nature of light. Dispersion by a prism and double-slit fringes are ordinary wave behaviour of light, which classical optics already explained; they show neither of the two properties asked about.

  5. The speed of a proton (mass 1.673 × 10⁻²⁷ kg) is known to within Δv = 500 m s⁻¹. Using Δx · Δp ≥ h/4π with h = 6.626 × 10⁻³⁴ J s, the minimum uncertainty in its position, in pm to the nearest whole number, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 63

    Δp = mΔv = 1.673 × 10⁻²⁷ × 500 = 8.365 × 10⁻²⁵ kg m s⁻¹. Δx = h/(4π Δp) = 6.626 × 10⁻³⁴/(12.566 × 8.365 × 10⁻²⁵) = 6.30 × 10⁻¹¹ m = 63 pm. Using h/2π in place of h/4π doubles the answer to 126 pm, the usual slip.

  6. For the ground state of the hydrogen atom, which quantity does the quantum-mechanical model predict differently from Bohr’s model?

    1. The orbital angular momentum, which is h/2π for Bohr but zero for a 1s electron
    2. The ionisation energy, which Bohr puts at 13.6 eV but wave mechanics well above it
    3. The level energy, which wave mechanics makes depend on l as well as on n for hydrogen
    4. The distance of 52.9 pm, which wave mechanics shows is never the most probable distance
    Show answer

    Answer: A — The orbital angular momentum, which is h/2π for Bohr but zero for a 1s electron

    Bohr quantises mvr = nh/2π, so the n = 1 electron has angular momentum h/2π, whereas a 1s orbital has l = 0 and orbital angular momentum √(l(l + 1)) h/2π = 0. Both models give Eₙ = −13.6/n² eV for hydrogen, hence the same 13.6 eV ionisation energy, and the radial probability of a 1s electron peaks at 52.9 pm, Bohr’s radius. Dependence of energy on l is real only in many-electron atoms, which makes it the tempting but wrong choice for hydrogen.

  7. Using 1/λ = R_H Z² (1/n₁² − 1/n₂²) with R_H = 1.097 × 10⁷ m⁻¹, the wavelength of the photon emitted when the electron of He⁺ (Z = 2) falls from n = 4 to n = 2, in nm to one decimal place, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 121.5

    1/λ = 1.097 × 10⁷ × 2² × (1/4 − 1/16) = 1.097 × 10⁷ × 4 × 3/16 = 8.228 × 10⁶ m⁻¹, so λ = 1.2154 × 10⁻⁷ m = 121.5 nm. It coincides with hydrogen’s Lyman-α line (n = 2 → 1) because Z²(1/4 − 1/16) = 3/4 = 1 − 1/4. Forgetting Z² gives 486.2 nm, hydrogen’s Hβ line.

  8. Which of the following statements about orbitals and quantum numbers are correct?

    1. A 3p orbital has one radial node and one angular node
    2. A 4d orbital has three angular nodes
    3. In the hydrogen atom the 3s, 3p and 3d orbitals all have the same energy
    4. The shell with n = 3 contains 18 orbitals
    Show answer

    Answer: A — A 3p orbital has one radial node and one angular node; C — In the hydrogen atom the 3s, 3p and 3d orbitals all have the same energy

    For 3p, l = 1 gives one angular node and n − l − 1 = 1 radial node. In hydrogen the energy depends only on n, so 3s, 3p and 3d are degenerate. A 4d orbital has l = 2 angular nodes (and one radial node), not three. The n = 3 shell has n² = 9 orbitals, which hold 18 electrons; 18 is the electron count, not the orbital count.

  9. What is the ground-state electronic configuration of a chromium atom (Z = 24)?

    1. [Ar] 3d⁴ 4s²
    2. [Ar] 3d⁶
    3. [Ar] 3d³ 4s² 4p¹
    4. [Ar] 3d⁵ 4s¹
    Show answer

    Answer: D — [Ar] 3d⁵ 4s¹

    Chromium is [Ar] 3d⁵ 4s¹: moving one electron from 4s to 3d gives six unpaired electrons with parallel spins and a half-filled 3d subshell, which gains exchange energy. [Ar] 3d⁴ 4s² is what the (n + l) filling order alone predicts, which is why it tempts; [Ar] 3d⁶ and a 4p electron are both far higher in energy.

  10. Which of the following ions has the ground-state configuration [Ar] 3d⁵?

    1. Fe²⁺ (iron, Z = 26)
    2. Fe³⁺ (iron, Z = 26)
    3. Co²⁺ (cobalt, Z = 27)
    4. Cr²⁺ (chromium, Z = 24)
    Show answer

    Answer: B — Fe³⁺ (iron, Z = 26)

    Iron is [Ar] 3d⁶ 4s². Ions lose the 4s electrons first, so Fe²⁺ is [Ar] 3d⁶ and Fe³⁺, which also loses one 3d electron, is [Ar] 3d⁵. Fe²⁺ is the usual wrong pick: removing only the two 4s electrons leaves 3d⁶, not 3d⁵. Cr²⁺ tempts those who remember chromium’s half-filled 3d⁵, but chromium is [Ar] 3d⁵ 4s¹, so Cr²⁺ loses the 4s electron and one 3d electron to become [Ar] 3d⁴. Co²⁺ is [Ar] 3d⁷.

  11. Four descriptions of the three 2p electrons of a nitrogen atom are given. Which one violates Hund’s rule but obeys Pauli’s exclusion principle?

    1. One electron in each of the three 2p orbitals, all three spins parallel
    2. Two electrons with the same spin in one 2p orbital, one in another orbital
    3. Two electrons of opposite spin in one 2p orbital, one in another orbital
    4. All three electrons crowded into one 2p orbital, the other two left empty
    Show answer

    Answer: C — Two electrons of opposite spin in one 2p orbital, one in another orbital

    Pairing two opposite-spin electrons while an empty degenerate 2p orbital is available breaks Hund’s rule, yet no two electrons share all four quantum numbers, so Pauli is obeyed. Two same-spin electrons in one orbital, and three electrons in one orbital, both break Pauli. One electron in each orbital with parallel spins is the true ground state and breaks neither rule.

  12. Using the allowed values of the quantum numbers and Pauli’s exclusion principle, the maximum number of electrons in an atom that can have n = 4 and m_l = −1 is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 6

    For n = 4, l can be 0, 1, 2 or 3. The value m_l = −1 needs l ≥ 1, so it occurs once each in 4p, 4d and 4f: three orbitals. Each orbital holds two electrons of opposite spin, giving 3 × 2 = 6. Forgetting spin gives 3; counting all n = 4 orbitals gives 32 electrons.

  13. Which sequence correctly orders the first ionisation energies of the second-period elements B, Be, C, N and O?

    1. Be < B < C < N < O
    2. B < Be < C < N < O
    3. Be < B < C < O < N
    4. B < Be < C < O < N
    Show answer

    Answer: D — B < Be < C < O < N

    The values are B 801, Be 899, C 1086, O 1314 and N 1402 kJ mol⁻¹. Boron’s electron is 2p, higher in energy and shielded by 2s², so it is easier to remove than beryllium’s 2s electron. Oxygen’s fourth 2p electron is paired and repelled, so it leaves more easily than an electron from nitrogen’s half-filled 2p³. The sequence that captures only the B < Be anomaly is the tempting near-miss.

  14. An element X has successive ionisation energies of 738, 1451, 7733 and 10540 kJ mol⁻¹. In which group of the periodic table does X most likely lie?

    1. Group 2, since the large jump follows removal of the second electron
    2. Group 1, since the first value is the smallest of all four values listed
    3. Group 13, since three electrons come out before the fourth value rises
    4. Group 14, since four ionisation energies are given for this element
    Show answer

    Answer: A — Group 2, since the large jump follows removal of the second electron

    IE₂/IE₁ is about 2, but IE₃/IE₂ is about 5.3. The third electron must come from a noble-gas core, so X has two valence electrons and lies in group 2 (these are magnesium’s values). The first IE is always the smallest of a series, so that fact says nothing about group 1, and the number of values quoted is not the number of valence electrons.

  15. A gaseous chlorine atom releases more energy than a gaseous fluorine atom when it gains an electron, although fluorine is the more electronegative element. Which explanation is correct?

    1. Fluorine’s outer electrons lie farther from the nucleus than chlorine’s, so it attracts weakly
    2. The added electron enters fluorine’s small, crowded 2p shell, where repulsion is strong
    3. The added electron in fluorine must occupy a new 3s shell lying far from its nucleus
    4. Chlorine places the extra electron in a vacant 3d orbital, which binds it more firmly
    Show answer

    Answer: B — The added electron enters fluorine’s small, crowded 2p shell, where repulsion is strong

    Fluorine is small, so the incoming electron is forced into a compact 2p shell already holding five electrons and meets strong electron–electron repulsion; in chlorine’s larger 3p shell that repulsion is diluted. Fluorine’s electrons are closer to the nucleus, not farther; its added electron completes 2p⁶ rather than entering 3s; and chlorine’s added electron goes into 3p, not into a vacant 3d orbital, so the d-orbital story is the tempting but wrong one.

  16. Which order of ionic radii is correct for the isoelectronic ions O²⁻, F⁻, Na⁺, Mg²⁺ and Al³⁺?

    1. Al³⁺ > Mg²⁺ > Na⁺ > F⁻ > O²⁻, since a larger charge makes a larger ion
    2. O²⁻ > F⁻ > Mg²⁺ > Na⁺ > Al³⁺, since Mg²⁺ holds more electrons than Na⁺
    3. O²⁻ > F⁻ > Na⁺ > Mg²⁺ > Al³⁺, since radius falls as nuclear charge rises
    4. F⁻ > O²⁻ > Na⁺ > Mg²⁺ > Al³⁺, since fluorine is more electronegative than oxygen
    Show answer

    Answer: C — O²⁻ > F⁻ > Na⁺ > Mg²⁺ > Al³⁺, since radius falls as nuclear charge rises

    All five ions have 10 electrons, the [Ne] configuration, while the nuclear charge rises from 8 (O) to 13 (Al). More protons pull the same electron cloud closer, so the radius falls steadily from O²⁻ to Al³⁺. The fully reversed order comes from assuming that a larger charge means a larger ion. Isoelectronic ions all hold the same number of electrons, so Mg²⁺ cannot have more than Na⁺, and electronegativity is a property of bonded atoms that does not decide the size of these ions; only the nuclear charge differs.

  17. Which of the following statements about periodic properties are correct?

    1. Atomic radius increases steadily from lithium to fluorine as electrons are added
    2. Zirconium and hafnium have almost the same atomic radius because of the lanthanide contraction
    3. Adding a second electron to a gaseous O⁻ ion releases energy
    4. On the Mulliken scale, electronegativity is the average of ionisation energy and electron affinity
    Show answer

    Answer: B — Zirconium and hafnium have almost the same atomic radius because of the lanthanide contraction; D — On the Mulliken scale, electronegativity is the average of ionisation energy and electron affinity

    The poor shielding by the 4f electrons that fill between them cancels the expected size increase from Zr (4d) to Hf (5d), so the two are almost equal in radius. Mulliken defined electronegativity as (IE + EA)/2. Atomic radius decreases, not increases, from Li to F because Z(eff) rises across the period. Adding an electron to the negative O⁻ ion is opposed by repulsion and absorbs energy, so the second step is endothermic.