AP Chemistry Atomic Structure and Properties — Worked Answer Explanations

Unit 1 · 12 questions explained

1. Question 1 · Easy

   In a mass spectrometry experiment, a sample of neon gas produces peaks at $m/z=20,21,$ and $22$. What do the different $m/z$ values represent?

   • A
     Different ions of neon with varying charges

     Why not A: If charge varied while mass stayed 20, the ratio would shift differently.
   • B
     Neon atoms that lost different numbers of electrons

     Why not B: Electron loss changes charge, not mass; these peaks reflect different masses.
   • C
     Isotopes of neon with different numbers of neutronsCorrect
   • D
     Impurities in the neon sample

     Why not D: Peaks at 20, 21, and 22 match known neon isotopes exactly.
   Explanation

   ${}^{20}\mathrm{Ne}$, ${}^{21}\mathrm{Ne}$, and ${}^{22}\mathrm{Ne}$ are the three stable isotopes of neon. Each has 10 protons but 10, 11, or 12 neutrons respectively, giving different masses and thus different $m/z$ values at charge $+1$.

   Key takeaway

   Mass spectrometry separates ions by mass-to-charge ratio, revealing isotopic composition.

2. Question 2 · Easy

   An element has two naturally occurring isotopes: one with mass 10.013 u (19.9% abundance) and one with mass 11.009 u (80.1% abundance). What is the weighted average atomic mass?

   • A
     $10.51\text{u}$

     Why not A: Simple average of the two masses, ignoring abundances.
   • B
     $10.81\text{u}$Correct
   • C
     $11.01\text{u}$

     Why not C: Used only the heavier isotope's mass.
   • D
     $10.21\text{u}$

     Why not D: Multiplied by fractions in reverse (lighter isotope got 80.1%).
   Explanation

   Weighted average $=(10.013)(0.199)+(11.009)(0.801)=1.993+8.818=10.81\text{u}$. Each isotope's mass is weighted by its fractional abundance.

   Key takeaway

   Atomic mass is the sum of each isotope mass times its fractional abundance.

3. Question 3 · Easy

   Which electron configuration correctly represents a ground-state sulfur ($\mathrm{Z}=16$) atom?

   • A
     $[\mathrm{Ne}]3s^{2}3p^4$Correct
   • B
     $[\mathrm{Ne}]3s^{2}3p^6$

     Why not B: That is the configuration for argon (Z = 18).
   • C
     $[\mathrm{Ne}]3s^{2}3p^2$

     Why not C: That gives only 14 electrons — silicon, not sulfur.
   • D
     $[\mathrm{Ne}]3s^{1}3p^5$

     Why not D: Electron count is correct but this is an excited state.
   Explanation

   Sulfur has 16 electrons. Neon accounts for 10, leaving 6 for the third shell: $3s^2$ (2 electrons) and $3p^4$ (4 electrons), totaling 16.

   Key takeaway

   Fill orbitals in order of increasing energy: 1s, 2s, 2p, 3s, 3p, etc.

4. Question 4 · Easy

   Which of the following correctly orders the elements by increasing first ionization energy?

   $\mathrm{Li},\mathrm{Na},\mathrm{K}$

   • A
     $\mathrm{Li}<\mathrm{Na}<\mathrm{K}$

     Why not A: This is the order of increasing atomic radius, not ionization energy.
   • B
     $\mathrm{K}<\mathrm{Na}<\mathrm{Li}$Correct
   • C
     $\mathrm{Na}<\mathrm{K}<\mathrm{Li}$

     Why not C: Sodium and potassium are out of the correct periodic-trend order.
   • D
     $\mathrm{Li}<\mathrm{K}<\mathrm{Na}$

     Why not D: Potassium should have a lower ionization energy than sodium.
   Explanation

   First ionization energy decreases down a group as valence electrons are farther from the nucleus and more shielded. Going down Group 1: $\mathrm{Li}>\mathrm{Na}>\mathrm{K}$, so in increasing order: $\mathrm{K}<\mathrm{Na}<\mathrm{Li}$.

   Key takeaway

   First ionization energy decreases down a group due to increased atomic radius and shielding.

5. Question 5 · Medium

   A photoelectron spectrum of an unknown element shows peaks at binding energies of approximately 2470, 190, 99, and 11 eV. Based on the pattern of peaks and the known PES of the elements, which element is this most likely to be?

   • A
     Carbon

     Why not A: Carbon has only two distinct energy levels visible in its PES (1s and 2s/2p).
   • B
     Neon

     Why not B: Neon has three peaks (1s, 2s, 2p) and would show a different binding-energy pattern.
   • C
     Magnesium

     Why not C: Magnesium's 1s binding energy (~1311 eV) does not match the highest-energy peak given.
   • D
     SiliconCorrect
   Explanation

   Silicon (Z = 14, configuration $[\mathrm{Ne}]3s^{2}3p^2$) has four distinct subshell energies — 1s, 2s, 2p, and 3s/3p — producing four PES peaks. The high-energy peak at ~2470 eV corresponds to the core 1s electrons of a third-period element with 14 protons.

   Key takeaway

   The number of PES peaks equals the number of occupied subshells; peak position reflects nuclear charge and shielding.

6. Question 6 · Medium

   According to Coulomb's law, the energy of attraction between an electron and the nucleus is proportional to $\frac{Z_{\text{eff}}}{r}$. Which isoelectronic pair would have the electron held most tightly (most negative energy)?

   • A
     ${\mathrm{O}}^{2-}$

     Why not A: Lower nuclear charge (Z = 8) means weaker hold on 10 electrons.
   • B
     ${\mathrm{F}}^{-}$

     Why not B: Z = 9 gives intermediate attraction.
   • C
     $\mathrm{Ne}$

     Why not C: Z = 10 is higher than F and O, but Mg${}^{2+}$ has Z = 12.
   • D
     $\mathrm{M}{\mathrm{g}}^{2+}$Correct
   Explanation

   All four species are isoelectronic with 10 electrons. The one with the largest nuclear charge ($Z=12$ for $\mathrm{M}{\mathrm{g}}^{2+}$) exerts the strongest Coulombic attraction on those electrons, resulting in the smallest radius and most tightly held electrons.

   Key takeaway

   Among isoelectronic species, higher nuclear charge means smaller radius and stronger electron-nucleus attraction.

7. Question 7 · Medium

   An element's fifth ionization energy is dramatically larger than its fourth. What does this indicate about the element's electron configuration?

   • A
     The element is in Group 4 of the periodic table.Correct
   • B
     The element is in Group 5 of the periodic table.

     Why not B: A Group-5 element would show a large jump after the fifth ionization energy (between IE5 and IE6).
   • C
     The element has five valence electrons.

     Why not C: If there were five valence electrons, the large jump would occur after the fifth, not before.
   • D
     The fourth and fifth electrons occupy the same orbital.

     Why not D: Orbital pairing explains smaller differences, not the dramatic jumps seen here.
   Explanation

   A dramatic increase between the fourth and fifth ionization energies means the fifth electron must be removed from a full inner shell (much closer to nucleus). This indicates the element has exactly four valence electrons — placing it in Group 4.

   Key takeaway

   A large jump in successive ionization energies signals that the next electron is in a lower energy (inner) shell, revealing the number of valence electrons.

8. Question 8 · Medium

   A PES spectrum shows two peaks for an element: a very high binding-energy peak with relative area 1, and a low binding-energy peak with relative area 2. Which element is this?

   • A
     Helium

     Why not A: Helium shows one peak (1s) with area 2, not two peaks.
   • B
     LithiumCorrect
   • C
     Beryllium

     Why not C: Beryllium (1s${}^2$ 2s${}^2$) shows two peaks with equal relative areas of 2.
   • D
     Boron

     Why not D: Boron has three peaks (1s, 2s, 2p) not two.
   Explanation

   Lithium ($1s^{2}2s^1$) has two occupied subshells. The 1s subshell holds 2 electrons (high binding energy) and the 2s holds 1 electron (low binding energy), giving area ratio 2:1. The peak areas reflect the number of electrons in each subshell.

   Key takeaway

   PES peak area is proportional to the number of electrons in that subshell.

9. Question 9 · Medium

   The Bohr model correctly predicts the emission spectrum of hydrogen. However, it fails for multi-electron atoms primarily because it:

   • A
     Assumes circular orbits and ignores electron–electron repulsion.Correct
   • B
     Does not quantize electron energy levels.

     Why not B: The Bohr model does quantize energy levels; that is one of its central postulates.
   • C
     Ignores the wave nature of light entirely.

     Why not C: Bohr used photon energy ($E=h\nu$) and implicitly invokes the wave nature of light.
   • D
     Places electrons in the nucleus.

     Why not D: The Bohr model places electrons in defined orbits around the nucleus.
   Explanation

   The Bohr model treats electrons as particles in fixed circular orbits and has no mechanism for electron–electron repulsion, which becomes significant in multi-electron atoms. The quantum mechanical model addresses this through orbital shapes, shielding, and electron correlation.

   Key takeaway

   Bohr model works for one-electron atoms but fails for multi-electron systems because it ignores inter-electron repulsion.

10. Question 10 · Hard

    Atomic radii generally decrease across a period (left to right). What is the primary reason for this trend?

    • A
      Electrons are added to successively lower energy levels.

      Why not A: Across a period, electrons fill the same principal energy level.
    • B
      The principal quantum number $n$ decreases across a period.

      Why not B: $n$ remains constant across a period.
    • C
      Increasing nuclear charge pulls the electron cloud closer while the shielding from added electrons in the same shell is minimal.Correct
    • D
      Increased electron–electron repulsion compresses the electron cloud.

      Why not D: Repulsion actually tends to expand the cloud; it does not cause contraction.
    Explanation

    Across a period, $Z$ increases while electrons enter the same principal shell. Same-shell electrons shield each other poorly, so effective nuclear charge $Z_{\text{eff}}$ rises, drawing the electron cloud inward and shrinking atomic radius.

    Key takeaway

    Across a period, increasing $Z_{\text{eff}}$ with roughly constant shielding causes atomic radius to decrease.

11. Question 11 · Hard

    Which of the following correctly ranks the species by increasing atomic/ionic radius?

    $\mathrm{C}{\mathrm{l}}^{-},\mathrm{Cl},\mathrm{N}{\mathrm{a}}^{+},\mathrm{Na}$

    • A
      $\mathrm{N}{\mathrm{a}}^{+}<\mathrm{C}{\mathrm{l}}^{-}<\mathrm{Na}<\mathrm{Cl}$

      Why not A: Na is larger than Cl as a neutral atom in the same period.
    • B
      $\mathrm{N}{\mathrm{a}}^{+}<\mathrm{Na}<\mathrm{C}{\mathrm{l}}^{-}<\mathrm{Cl}$

      Why not B: The anion Cl${}^{-}$ is larger than neutral Cl; neutral Na is smaller than Cl${}^{-}$.
    • C
      $\mathrm{N}{\mathrm{a}}^{+}<\mathrm{Cl}<\mathrm{Na}<\mathrm{C}{\mathrm{l}}^{-}$Correct
    • D
      $\mathrm{Cl}<\mathrm{N}{\mathrm{a}}^{+}<\mathrm{Na}<\mathrm{C}{\mathrm{l}}^{-}$

      Why not D: Na${}^{+}$ has fewer electrons than Cl and is smaller, so Na${}^{+}$ should be smallest.
    Explanation

    $\mathrm{N}{\mathrm{a}}^{+}$ has 10 electrons and Z = 11 (smallest due to high $Z_{\text{eff}}$). Neutral Cl has 17 electrons/17 protons — smaller than Na (18 electrons/11 protons). Neutral Na is larger than Cl because of lower $Z_{\text{eff}}$ for its valence electrons. $\mathrm{C}{\mathrm{l}}^{-}$ has 18 electrons but only Z = 17, so electron repulsion expands it to the largest size.

    Key takeaway

    Anions are larger than their neutral parent atoms; cations are smaller. Compare isoelectronic species by Z.

12. Question 12 · Hard

    Spectral lines in the visible emission spectrum of hydrogen are produced when an electron transitions from a higher energy level to:

    Select the statement that is both correct and explains the underlying mechanism.

    • A
      $n=1$, releasing ultraviolet photons equal in energy to the transition.

      Why not A: Transitions to n = 1 (Lyman series) emit UV, not visible light.
    • B
      $n=2$, releasing a photon whose energy equals the difference in quantized energy levels.Correct
    • C
      $n=3$, releasing infrared radiation.

      Why not C: Transitions ending at n = 3 (Paschen series) produce infrared light, not visible.
    • D
      $n=2$, absorbing a photon to move to a lower energy state.

      Why not D: Emission involves releasing, not absorbing, a photon; electrons drop from higher to lower levels.
    Explanation

    The visible Balmer series corresponds to electron transitions from $n\ge 3$ down to $n=2$. The emitted photon carries exactly the energy difference $\Delta E=E_{\text{initial}}-E_{\text{final}}$, and for $n=3\to 2$, 4 $\to$ 2, etc., this energy falls in the visible range. $E=h\nu =hc/\lambda$.

    Key takeaway

    Visible hydrogen emission lines (Balmer series) result from electron transitions to n = 2; emitted photon energy equals the level energy difference.