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Quiz: Electron Configuration and Periodic Trends

Test your understanding of quantum mechanical models, electron configurations, electromagnetic radiation, photoelectron spectroscopy, and periodic trends with these questions.

1. What is the primary reason the Bohr model was eventually replaced by the quantum mechanical model?

  1. The Bohr model could not explain the spectra of atoms with more than one electron.
  2. The Bohr model incorrectly predicted the spectrum of hydrogen.
  3. The Bohr model required electrons to travel faster than the speed of light.
  4. The Bohr model placed the nucleus in the wrong location.
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The correct answer is A. The Bohr model successfully predicts the hydrogen emission spectrum but fails completely for multi-electron atoms. It cannot account for electron-electron repulsion or the complexity of multi-electron energy levels. The Bohr model did correctly predict the hydrogen spectrum (B) — that was its great success. It made no claim about electron speeds relative to light (C). Both models correctly place the nucleus at the center (D).

Concept Tested: Bohr Model vs. Quantum Mechanical Model

2. Which set of subshells is present in the third principal energy level (n = 3)?

  1. 3s only
  2. 3s and 3p
  3. 3s, 3p, and 3d
  4. 3s, 3p, 3d, and 3f
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The correct answer is C. For principal quantum number n, the available subshells are s, p, d, f … up to n types. For n = 3, the subshells are 3s, 3p, and 3d, for a maximum of 9 orbitals and 18 electrons. n = 1 has only s (A). n = 2 has s and p (B). The f subshell does not appear until n = 4 (D).

Concept Tested: Principal Quantum Number and Subshells

3. Which of the following correctly applies the Aufbau principle, Pauli exclusion principle, and Hund's rule to write the electron configuration of nitrogen (Z = 7)?

  1. 1s² 2s² 2p³ with all three 2p electrons in one orbital
  2. 1s² 2s² 2p³ with one electron in each of the three 2p orbitals
  3. 1s² 2s³ 2p²
  4. 1s² 2p⁵
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The correct answer is B. Nitrogen has 7 electrons. Aufbau gives 1s² 2s² 2p³. Hund's rule then requires that the three 2p electrons each occupy a separate 2p orbital with parallel spins before any pairing occurs. Placing all three 2p electrons in one orbital (A) violates the Pauli exclusion principle (maximum 2 per orbital) and Hund's rule. 1s² 2s³ 2p² (C) violates the Pauli exclusion principle because the s subshell holds only 2 electrons. 1s² 2p⁵ (D) violates the Aufbau principle by skipping the lower-energy 2s subshell.

Concept Tested: Electron Configuration, Aufbau Principle, Hund's Rule

4. What is the correct noble gas notation for iron (Fe, Z = 26)?

  1. [Ar] 3d⁸
  2. [Ne] 4s² 3d⁶
  3. [Kr] 4s² 3d⁶
  4. [Ar] 4s² 3d⁶
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The correct answer is D. Iron (Z = 26) has the full configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶. The nearest preceding noble gas is argon (Z = 18), so the noble gas notation is [Ar] 4s² 3d⁶. [Ar] 3d⁸ (A) incorrectly omits the 4s electrons and lumps all remaining electrons into 3d. Neon (B) accounts for only 10 electrons — the correct core for Period 3 elements, not Period 4. Krypton (Z = 36) comes after iron in the table; it cannot be a core for a lighter element (C).

Concept Tested: Noble Gas Notation and Electron Configuration

5. When iron (Fe) forms Fe³⁺, which electrons are removed and in what order?

  1. Three 3d electrons are removed first.
  2. Two 3d electrons and then one 4s electron are removed.
  3. Two 4s electrons are removed first, then one 3d electron.
  4. One 4s electron and two 3p electrons are removed.
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The correct answer is C. When transition metals form cations, the highest-n electrons are removed before lower-n d electrons. Iron's neutral configuration is [Ar] 4s² 3d⁶. Removing both 4s electrons gives Fe²⁺ ([Ar] 3d⁶); removing one 3d electron then gives Fe³⁺ ([Ar] 3d⁵). Removing 3d electrons first (A) is the most common misconception. Option (B) reverses the correct order. 3p electrons (D) are core electrons and are not removed under normal chemical conditions.

Concept Tested: Electron Configuration of Ions

6. A photon has a wavelength of 400 nm (visible violet light). A second photon has a wavelength of 800 nm (infrared). How do their energies compare?

  1. Both photons have the same energy because they travel at the same speed.
  2. The 400 nm photon has four times the energy of the 800 nm photon.
  3. The 800 nm photon has twice the energy of the 400 nm photon.
  4. The 400 nm photon has twice the energy of the 800 nm photon.
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The correct answer is D. From Planck's equation \(E = hc/\lambda\), energy is inversely proportional to wavelength. The 400 nm photon has half the wavelength of the 800 nm photon, so it carries twice the energy. Equal speed does not imply equal energy (A) — all photons travel at \(c\) in vacuum, but energy depends on frequency (and inversely on wavelength), not speed. Four times the energy (B) would require a 4:1 wavelength ratio, not 2:1. Option (C) reverses the inverse relationship — longer wavelength means lower energy.

Concept Tested: Planck's Equation and Photon Energy

7. A PES spectrum for sodium shows four peaks. Moving from highest to lowest binding energy, the peaks correspond to 1s, 2s, 2p, and 3s. The relative heights of these peaks are approximately 2:2:6:1. What does this height ratio reveal?

  1. The number of neutrons in the nucleus of each sodium isotope
  2. The relative energies of each subshell in electron volts
  3. The number of electrons in each subshell, consistent with the electron configuration [Ne] 3s¹
  4. The number of orbitals available in each subshell
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The correct answer is C. In PES, the relative area (height) of each peak is proportional to the number of electrons in that subshell. A 2:2:6:1 ratio matches 1s², 2s², 2p⁶, 3s¹ — exactly sodium's electron configuration ([Ne] 3s¹). Neutron counts (A) are not detected by PES. PES binding energy positions, not peak heights, indicate subshell energies in eV (B). The number of available orbitals in each subshell would give a ratio of 1:1:3:1 — the number of orbitals, not electrons (D).

Concept Tested: PES Data Interpretation

8. Moving across Period 3 from sodium (Na) to chlorine (Cl), effective nuclear charge (\(Z_{eff}\)) experienced by valence electrons increases steadily. Which periodic trend does this most directly explain?

  1. Atomic radius decreases from Na to Cl because higher \(Z_{eff}\) pulls valence electrons closer to the nucleus.
  2. Ionization energy decreases from Na to Cl because valence electrons are farther from the nucleus.
  3. Atomic radius increases from Na to Cl because more electrons are added.
  4. Electronegativity decreases from Na to Cl because larger atoms attract electrons less effectively.
Show Answer

The correct answer is A. As \(Z_{eff}\) increases across a period, the attractive force on valence electrons increases, pulling them closer to the nucleus and reducing atomic radius. This is why the smallest atoms in each period are on the right. Ionization energy increases (not decreases) from left to right — higher \(Z_{eff}\) holds electrons more tightly, making them harder to remove (B). Atomic radius increases going down a group, not across a period (C). Electronegativity increases (not decreases) from Na to Cl because stronger \(Z_{eff}\) pulls bonding electrons more strongly (D).

Concept Tested: Effective Nuclear Charge and Atomic Radius Trend

9. The first ionization energy (IE₁) of sodium is 496 kJ/mol, but the second ionization energy (IE₂) is 4562 kJ/mol — nearly nine times larger. What does this dramatic jump reveal?

  1. Sodium's second electron is a neutron that requires much more energy to remove.
  2. The second electron must be removed from an inner shell (core electron), confirming sodium has exactly one valence electron.
  3. The second ionization always requires twice the energy of the first ionization for all elements.
  4. Sodium's second electron is located in a 3p orbital, which is higher in energy than the 3s orbital.
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The correct answer is B. The large jump between IE₁ and IE₂ reveals that after removing sodium's single 3s valence electron, the second electron must come from the 2p core shell — much closer to the nucleus and held far more tightly. This directly confirms that sodium has one valence electron and supports its Group 1 classification. Neutrons are in the nucleus and cannot be removed by ionization energy (A). Successive ionization energies do not simply double — the pattern depends on shell structure (C). Sodium's second electron comes from the 2p subshell, not a 3p orbital (D).

Concept Tested: Successive Ionization Energy and Shell Structure

10. Fluorine has the highest electronegativity of all elements. Which combination of factors best explains this?

  1. Fluorine has a large atomic radius and a low ionization energy, making it easy to attract electrons.
  2. Fluorine has a high effective nuclear charge and a small atomic radius, creating a strong attraction for shared electrons in a bond.
  3. Fluorine is a noble gas with a completely filled valence shell, making it very stable.
  4. Fluorine has only one valence electron, giving it a strong drive to form bonds to complete its octet.
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The correct answer is B. Electronegativity measures the ability of a bonded atom to attract shared electrons. Fluorine (Period 2, Group 17) has a high \(Z_{eff}\) (9 protons, only 2 core electrons providing shielding) and a very small atomic radius, so bonding electrons are held close to the highly charged nucleus. Fluorine has a small atomic radius and very high ionization energy — not large and low (A). Fluorine is a halogen in Group 17 with 7 valence electrons, not a noble gas (C). Having 7 valence electrons, not 1, gives fluorine its strong attraction for an additional electron (D).

Concept Tested: Electronegativity and Effective Nuclear Charge