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AP Chemistry – Units 1–3 Practice Set

Atomic structure, bonding, intermolecular forces.

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10 questions  ·  31 marks  ·  answers included  ·  AP Chemistry (College Board)
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This worksheet covers:

  • Units 1–3
  • Atomic structure
  • Periodicity
  • Bonding
  • Intermolecular forces

A calculator is permitted for Section C; explain your reasoning in full sentences as AP FRQs require.

Questions

Section A – Atomic Structure & Periodicity

Free response, no calculator needed
1

(a) Write the full electron configuration of a ground-state chlorine atom.(b) State the number of unpaired electrons in a ground-state chlorine atom and justify your answer.

Show answer
(a) $1s^2\,2s^2\,2p^6\,3s^2\,3p^5$(b) 1 unpaired electron: the five $3p$ electrons fill three orbitals such that two orbitals are doubly occupied and one orbital is singly occupied (Hund’s rule).
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2

Rank $\text{Na}$, $\text{Mg}$, and $\text{Al}$ in order of increasing first ionization energy, and explain the anomaly in the trend.

Show answer
Increasing order: $\text{Na} < \text{Al} < \text{Mg}$. Ionization energy generally increases across a period as nuclear charge increases, but Mg has a filled, extra-stable $3s^2$ subshell, while Al’s outer electron is in a higher-energy, less penetrating $3p$ orbital that is easier to remove, so Al has a lower first ionization energy than expected relative to Mg.
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3

Explain why atomic radius decreases across Period 3 from $\text{Na}$ to $\text{Cl}$, even though the number of electrons is increasing.

Show answer
Across the period, electrons are added to the same principal energy level (n = 3), while the number of protons in the nucleus also increases; the additional electrons do not shield each other effectively from the increasing nuclear charge, so the effective nuclear charge experienced by the outer electrons increases, pulling them closer to the nucleus and decreasing atomic radius.
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4

Identify the type(s) of intermolecular force(s) present in liquid $\text{NH}_3$ and explain how you can identify it/them from the molecule’s structure.

Show answer
Hydrogen bonding is present (as well as weaker dipole-dipole and dispersion forces): $\text{NH}_3$ has N-H bonds, where N is small and highly electronegative, and the N atom has a lone pair, satisfying the requirements for hydrogen bonding between $\text{NH}_3$ molecules.
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Section B – Bonding

Free response
5

For each molecule ($\text{CO}_2$, $\text{NH}_3$, $\text{CH}_4$): (a) state the molecular geometry predicted by VSEPR theory, and (b) state whether the molecule is polar or nonpolar overall, with a brief justification.

Show answer
$\text{CO}_2$: linear geometry; nonpolar overall (the two C=O bond dipoles point in opposite directions and cancel).$\text{NH}_3$: trigonal pyramidal geometry; polar overall (the lone pair on N means the bond dipoles do not cancel, giving a net dipole).$\text{CH}_4$: tetrahedral geometry; nonpolar overall (four identical C-H bond dipoles arranged symmetrically cancel).
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6

Compare the melting points of $\text{NaCl}$ (mp $\approx 801\,^\circ\text{C}$) and $\text{Cl}_2$ (mp $\approx -101\,^\circ\text{C}$), explaining the difference in terms of the type of bonding/forces present in each.

Show answer
$\text{NaCl}$ is an ionic solid held together by strong electrostatic attractions between oppositely charged $\text{Na}^+$ and $\text{Cl}^-$ ions throughout the lattice, requiring a large amount of energy to overcome. $\text{Cl}_2$ is a small nonpolar covalent molecule, and solid/liquid $\text{Cl}_2$ is held together only by weak London dispersion forces between molecules, which require much less energy to overcome; hence $\text{NaCl}$ has a far higher melting point than $\text{Cl}_2$.
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7

Predict which has stronger intermolecular forces, $\text{HF}$ or $\text{HCl}$, and justify your answer.

Show answer
$\text{HF}$ has stronger intermolecular forces. $\text{HF}$ can form hydrogen bonds, because F is very electronegative and has a small atomic radius, creating a highly polarized H-F bond and strong dipole interactions between molecules; $\text{HCl}$, although polar, cannot hydrogen bond (Cl is larger and less electronegative), so it relies only on weaker dipole-dipole and dispersion forces.
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Section C – Extended Response

Calculator allowed
8

A $1.110\,\text{g}$ sample of an unknown Group 2 metal chloride, $\text{MCl}_2$, is found to contain $0.709\,\text{g}$ of chlorine.(a) Calculate the mass of metal $M$ in the sample.(b) Calculate the moles of Cl atoms present, and hence the moles of $\text{MCl}_2$. ($A_r$ Cl $= 35.45$)(c) Calculate the molar mass of $M$ and identify it. ($A_r$: Mg 24.31, Ca 40.08, Sr 87.62, Ba 137.33)

Show answer
(a) mass of $M = 1.110 – 0.709 = 0.401\,\text{g}$(b) $n(\text{Cl}) = \dfrac{0.709}{35.45} = 0.0200\,\text{mol}$; $n(\text{MCl}_2) = \dfrac{0.0200}{2} = 0.0100\,\text{mol}$(c) molar mass of $M = \dfrac{0.401}{0.0100} = 40.1\,\text{g mol}^{-1}$, which matches calcium ($A_r = 40.08$), so $M$ is $\text{Ca}$.
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9

$\text{XeF}_4$ has 6 electron domains around the central Xe atom (4 bonding pairs and 2 lone pairs). (a) Describe the arrangement of the lone pairs and the resulting molecular geometry. (b) State whether the molecule is polar or nonpolar overall.

Show answer
(a) The two lone pairs occupy opposite (axial) positions to minimize repulsion, leaving the four bonding pairs in a plane around Xe; the molecular geometry is square planar.(b) Nonpolar overall — the four identical Xe-F bond dipoles are arranged symmetrically in the plane and cancel.
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10

The species $\text{O}^{2-}$, $\text{F}^-$, $\text{Na}^+$ and $\text{Mg}^{2+}$ are isoelectronic (each has 10 electrons).(a) Rank them in order of increasing ionic radius.(b) Explain this trend in terms of nuclear charge.

Show answer
(a) Increasing radius: $\text{Mg}^{2+} < \text{Na}^+ < \text{F}^- < \text{O}^{2-}$(b) All four species have the same number of electrons (10), but the nuclear charge increases from $\text{O}$ (8 protons) to $\text{F}$ (9) to $\text{Na}$ (11) to $\text{Mg}$ (12); a greater nuclear charge attracts the same electron cloud more strongly, pulling the electrons closer to the nucleus, so ionic radius decreases as nuclear charge increases.
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