AP Chemistry Molecular and Ionic Compound Structure and Properties — Worked Answer Explanations

Unit 2 · 12 questions explained

Below is a complete answer key for our AP Chemistry Molecular and Ionic Compound Structure and Properties practice questions. For each question you'll find the correct choice, a full written explanation of how to get there, and — for every wrong answer — a short note on exactly why it's tempting and where it goes wrong. Reading these straight through is one of the fastest ways to find the gaps in a unit before exam day.

Prefer to test yourself first? Take the timed Molecular and Ionic Compound Structure and Properties practice test and come back here to review, or head back to the Molecular and Ionic Compound Structure and Properties unit overview.

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Question 1 · Easy
How many lone pairs of electrons are on the central nitrogen atom in $N H_{3}$ ?
- A
  $0$
  Why not A: Nitrogen has 5 valence electrons; 3 are shared with H atoms, leaving a lone pair.
- B
  $1$ Correct
- C
  $2$
  Why not C: Two lone pairs would leave only 1 electron for bonding, not 3.
- D
  $3$
  Why not D: Three lone pairs would give nitrogen a negative formal charge and no N–H bonds.
Explanation
Nitrogen has 5 valence electrons. Three are used in N–H single bonds, leaving $5 - 3 = 2$ electrons = 1 lone pair on nitrogen in $N H_{3}$ .

Key takeaway
Count valence electrons minus bonding electrons to find lone pairs on the central atom.
Question 2 · Easy
Which of the following molecules is polar?
- A
  $B F_{3}$
  Why not A: BF $_{3}$ is trigonal planar; dipole moments cancel by symmetry.
- B
  $CC l_{4}$
  Why not B: Tetrahedral CCl $_{4}$ has four equal bond dipoles that cancel.
- C
  $C O_{2}$
  Why not C: Linear CO $_{2}$ has equal and opposite dipoles that cancel.
- D
  $H_{2} S$ Correct
Explanation
$H_{2} S$ is bent (like water) because the central S atom has 2 lone pairs. The two S–H bond dipoles do not cancel, giving a net dipole moment. The other molecules have symmetric geometries that cause their individual bond dipoles to sum to zero.

Key takeaway
A molecule is polar if it has polar bonds AND an asymmetric geometry so that the bond dipoles do not cancel.
Question 3 · Easy
What is the formal charge on the oxygen atom in $C O_{2}$ (a carbon–oxygen double bond)?
- A
  $- 2$
  Why not A: That would be the oxidation state-style charge, not formal charge.
- B
  $- 1$
  Why not B: Each O has 4 non-bonding and 4 bonding electrons in the double-bond structure.
- C
  $0$ Correct
- D
  $+ 1$
  Why not D: A positive formal charge on O would require it to have fewer electrons than its valence number.
Explanation
In $C O_{2}$ with two C=O double bonds, each O has 2 lone pairs (4 non-bonding $e^{-}$ ) and 4 bonding electrons shared (2 per bond). Formal charge $= 6 - 4 - \frac{4}{2} = 0$ .

Key takeaway
Formal charge = (valence electrons) $-$ (non-bonding electrons) $-$ $\frac{1}{2}$(bonding electrons).
Question 4 · Easy
Which hybridization is consistent with a carbon atom that forms two double bonds and no single bonds (as in $C O_{2}$ )?
- A
  $s p^{3}$
  Why not A: $s p^{3}$ corresponds to 4 sigma bond regions (tetrahedral), not 2.
- B
  $s p^{2}$
  Why not B: $s p^{2}$ gives 3 sigma bond regions (trigonal planar), not 2.
- C
  $s p$ Correct
- D
  $s p^{3} d$
  Why not D: $s p^{3} d$ is used by expanded-octet atoms like P or S, not carbon.
Explanation
The central C in $C O_{2}$ has 2 sigma bond regions, so it uses $s p$ hybridization. This leaves 2 unhybridized $p$ orbitals to form 2 $π$ bonds (one with each O), consistent with the two double bonds.

Key takeaway
Hybridization equals the number of sigma-bond regions (steric number): $sp$ = 2, $sp^2$ = 3, $sp^3$ = 4.
Question 5 · Medium
What is the electron-pair geometry and molecular geometry of $S F_{4}$ ?
- A
  Tetrahedral; tetrahedral
  Why not A: SF $_{4}$ has 5 electron domains (4 bonding + 1 lone pair), not 4.
- B
  Trigonal bipyramidal; trigonal bipyramidal
  Why not B: 5 electron domains give trigonal bipyramidal geometry only if all are bonding pairs.
- C
  Trigonal bipyramidal; see-sawCorrect
- D
  Octahedral; square planar
  Why not D: Octahedral electron-pair geometry requires 6 electron domains.
Explanation
S in SF $_{4}$ has 6 valence electrons. Four S–F bonds use 4 electrons; 1 lone pair remains, giving 5 electron domains. VSEPR: 5 domains → trigonal bipyramidal electron-pair geometry. With 1 lone pair in an equatorial position, the molecular geometry is see-saw.

Key takeaway
VSEPR step 1: count all electron domains (bonding + lone pairs) for electron-pair geometry. Step 2: remove lone-pair positions for molecular geometry.
Question 6 · Medium
In a Lewis structure, three resonance structures can be drawn for $S O_{3}$ . What is the best interpretation of these structures?
- A
  The molecule rapidly switches between the three structures.
  Why not A: Resonance structures are not interconverting species; they are representations of one structure.
- B
  The true structure is a hybrid in which the electron density is delocalized equally over all S–O bonds.Correct
- C
  The three structures represent three different $S O_{3}$ isomers.
  Why not C: Resonance structures have the same atom connectivity — they are not isomers.
- D
  Only the structure with the fewest formal charges is correct; the others are wrong.
  Why not D: All resonance contributors are used to build the hybrid; none is solely 'correct.'
Explanation
The three resonance structures of $S O_{3}$ differ only in where the double bond is placed. The actual molecule is a resonance hybrid: all three S–O bonds are equivalent (bond order $\approx 1.33$ ) and the $π$ electron density is delocalized equally over all three bonds.

Key takeaway
Resonance structures describe one real molecule whose electron density is a weighted average (hybrid) of all contributors.
Question 7 · Medium
Which of the following best explains why ionic compounds have high melting points compared with molecular compounds?
- A
  Ionic compounds are denser, so more energy is needed to expand the lattice.
  Why not A: Density is not the primary factor determining melting point.
- B
  Ionic bonds are covalent in nature and therefore very strong.
  Why not B: Ionic bonds are electrostatic, not covalent.
- C
  Electrostatic forces between oppositely charged ions in the crystal lattice are very strong.Correct
- D
  Ionic compounds cannot conduct electricity, which raises their melting points.
  Why not D: Electrical conductivity is unrelated to melting point.
Explanation
In an ionic crystal, every cation is surrounded by multiple anions (and vice versa). These omnidirectional electrostatic attractions, summed across the entire lattice (lattice energy), are very large. Breaking the lattice requires overcoming all of these interactions simultaneously, hence high melting points.

Key takeaway
High ionic melting points reflect strong, lattice-wide electrostatic forces; lattice energy increases with charge magnitude and decreasing ion size.
Question 8 · Medium
The bond angle in $H_{2} O$ ( $\approx 104.5°$ ) is less than the ideal tetrahedral angle of $109.5°$ because:
- A
  Hydrogen atoms are very small and provide less steric repulsion.
  Why not A: Smaller H atoms might allow a wider angle, not a narrower one.
- B
  Lone pairs occupy more space than bonding pairs and compress the H–O–H angle.Correct
- C
  The O–H bonds are polar, causing electrostatic repulsion between the hydrogen atoms.
  Why not C: Bond polarity and H–H repulsion are not the VSEPR explanation for the compressed angle.
- D
  Water uses $s p^{2}$ hybridization, which gives 120° angles reduced by lone pairs.
  Why not D: Water's O is $s p^{3}$ -hybridized, giving a tetrahedral base geometry, not trigonal planar.
Explanation
VSEPR: lone pairs have a larger effective volume than bonding pairs (they are held only by one nucleus). The two lone pairs on O exert greater repulsion on the two O–H bonds, compressing the H–O–H angle below the tetrahedral ideal of 109.5°.

Key takeaway
Lone pair–lone pair repulsion > lone pair–bonding pair repulsion > bonding pair–bonding pair repulsion, compressing bond angles.
Question 9 · Medium
Which of the following species has a trigonal planar molecular geometry?
- A
  $N H_{3}$
  Why not A: NH $_{3}$ is trigonal pyramidal, not planar, due to its lone pair.
- B
  $S O_{3}$ Correct
- C
  $PC l_{3}$
  Why not C: PCl $_{3}$ is trigonal pyramidal (like NH $_{3}$ ) because of a lone pair on P.
- D
  $Cl F_{3}$
  Why not D: ClF $_{3}$ has 2 lone pairs and a T-shaped molecular geometry.
Explanation
In $S O_{3}$ , sulfur forms 3 equivalent bonds (via resonance/delocalization) with no lone pairs on S. Three electron domains and zero lone pairs give a trigonal planar geometry with 120° bond angles.

Key takeaway
Trigonal planar geometry requires 3 bonding domains and 0 lone pairs on the central atom.
Question 10 · Hard
Rank the following in order of increasing bond length: C–C single bond, C=C double bond, $C \equiv C$ triple bond.
- A
  $C \equiv C < C = C < C - C$ Correct
- B
  $C - C < C = C < C \equiv C$
  Why not B: More bonds draw atoms closer; this order is backwards.
- C
  $C = C < C - C < C \equiv C$
  Why not C: Triple bond is shortest, not longest.
- D
  All are equal because the atoms are the same.
  Why not D: Bond order affects internuclear distance even between the same element pair.
Explanation
Higher bond order means the nuclei are held more closely together. Approximate lengths: C–C ≈ 154 pm, C=C ≈ 134 pm, C≡C ≈ 120 pm. So the triple bond is shortest.

Key takeaway
Higher bond order → shorter bond length and greater bond strength.
Question 11 · Hard
A student draws a Lewis structure for $N O_{2}^{-}$ (nitrite ion) in which nitrogen forms one single bond to one oxygen and one double bond to the other, with one lone pair on nitrogen. What is the formal charge on the nitrogen in this structure?
- A
  $+ 1$
  Why not A: With 5 valence electrons and this bonding, the formal charge is not +1.
- B
  $0$ Correct
- C
  $- 1$
  Why not C: Check the electron count: N has 5 valence electrons, 2 lone pair electrons, and 3 bonds (1 single + 1 double).
- D
  $- 2$
  Why not D: That would require 7 non-bonding electrons on N, far more than given.
Explanation
Nitrogen forms 1 single + 1 double bond (3 bonding pairs total) and has 1 lone pair (2 non-bonding electrons). Formal charge = $5 - 2 - \frac{6}{2} = 5 - 2 - 3 = 0$ .

Key takeaway
Formal charge = valence electrons $-$ lone pair electrons $-$ half the bonding electrons. Summing formal charges gives the ion's overall charge.
Question 12 · Hard
Using average bond enthalpies, estimate $Δ H$ for the reaction $H_{2} (g) + C l_{2} (g) \to 2 HCl (g)$ , given: $D (H - H) = 436 kJ/mol$ , $D (Cl - Cl) = 243 kJ/mol$ , $D (H - Cl) = 432 kJ/mol$ .
- A
  $+ 185 kJ/mol$
  Why not A: Forgot to subtract double the H–Cl bond energy formed.
- B
  $- 185 kJ/mol$ Correct
- C
  $- 679 kJ/mol$
  Why not C: Added all bond energies rather than (bonds broken) $-$ (bonds formed).
- D
  $+ 247 kJ/mol$
  Why not D: Used only one H–Cl bond formed instead of two.
Explanation
$Δ H \approx \sum D (bonds broken) - \sum D (bonds formed)$

Bonds broken: $1 \times 436 + 1 \times 243 = 679 kJ/mol$

Bonds formed: $2 \times 432 = 864 kJ/mol$

$Δ H = 679 - 864 = - 185 kJ/mol$

Key takeaway
$\Delta H \approx \sum D(\text{broken}) - \sum D(\text{formed})$; negative value means bond formation releases more energy than breaking required.