IB Chemistry Quiz: Understand The Covalent Model
20 questions · exam conditions
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Understand The Covalent ModelQuestion 1 of 20

What are the approximate bond angles and molecular geometry of phosgene, COCl₂, in which carbon is the central atom?

Trigonal planar, approximately 120°
Trigonal pyramidal, approximately 107°
Tetrahedral, approximately 109.5°
Bent, approximately 120°
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IB Chemistry Quiz

IB Chemistry Quiz: Understand The Covalent Model

Practice Understand The Covalent Model in IB Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Understand The Covalent Model, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

What are the approximate bond angles and molecular geometry of phosgene, COCl₂, in which carbon is the central atom?

  1. Trigonal planar, approximately 120° (correct answer)
  2. Trigonal pyramidal, approximately 107°
  3. Tetrahedral, approximately 109.5°
  4. Bent, approximately 120°
Explanation: The Lewis structure of COCl₂ has a central carbon atom with a double bond to the oxygen atom and single bonds to two chlorine atoms. This results in three electron domains around the central carbon atom. According to VSEPR theory, three electron domains arrange themselves in a trigonal planar geometry to minimize repulsion, leading to approximate bond angles of 120°. Since there are no lone pairs on the central carbon atom, the molecular geometry is the same as the electron domain geometry.

Question 2

Consider the molecules N₂, O₂, and F₂. Which sequence correctly lists these molecules in order of increasing bond length?

  1. N₂ < O₂ < F₂ (correct answer)
  2. F₂ < O₂ < N₂
  3. O₂ < F₂ < N₂
  4. N₂ < F₂ < O₂
Explanation: The bond length is inversely related to the bond order (the number of shared electron pairs). Nitrogen, N₂, has a triple bond (bond order 3). Oxygen, O₂, has a double bond (bond order 2). Fluorine, F₂, has a single bond (bond order 1). A higher bond order means a stronger attraction between the nuclei, pulling them closer together. Therefore, the triple bond in N₂ is the shortest, and the single bond in F₂ is the longest. The correct order of increasing bond length is N₂ < O₂ < F₂.

Question 3

In a paper chromatography experiment using a polar stationary phase (paper) and a non-polar solvent (mobile phase), a mixture is separated into three components with Rƒ values of 0.20, 0.55, and 0.85. What is the most likely deduction?

  1. The component with Rƒ = 0.85 is the most polar.
  2. The component with Rƒ = 0.20 has the lowest molar mass.
  3. The component with Rƒ = 0.20 is the most polar. (correct answer)
  4. The component with Rƒ = 0.55 is a non-polar substance.
Explanation: In chromatography with a polar stationary phase and a non-polar mobile phase, polar components will adsorb more strongly to the stationary phase and travel a shorter distance, resulting in a lower Rƒ value. Non-polar components will be more soluble in the mobile phase and travel further, resulting in a higher Rƒ value. Therefore, the component with the lowest Rƒ value (0.20) is the most polar.

Question 4

What is the correct description of the molecular geometry around each nitrogen atom and the overall polarity of the hydrazine molecule, N₂H₄?

  1. Geometry around each nitrogen: Trigonal planar; Overall polarity: Non-polar
  2. Geometry around each nitrogen: Trigonal pyramidal; Overall polarity: Non-polar
  3. Geometry around each nitrogen: Tetrahedral; Overall polarity: Polar
  4. Geometry around each nitrogen: Trigonal pyramidal; Overall polarity: Polar (correct answer)
Explanation: In hydrazine (H₂N-NH₂), each nitrogen atom is bonded to two hydrogen atoms and the other nitrogen atom, and also has one lone pair. This gives each nitrogen four electron domains, resulting in a trigonal pyramidal molecular geometry around each nitrogen. The N-H bonds are polar, and the presence of lone pairs and the trigonal pyramidal geometry at each end results in an asymmetrical distribution of charge. The bond dipoles do not cancel, making the molecule polar.

Question 5

A covalent compound has the molecular formula XY3\text{XY}_3 where X has 5 valence electrons and Y has 7 valence electrons. If the compound exhibits a trigonal pyramidal molecular geometry, what can be concluded about the bonding and electron distribution?

  1. The molecular geometry indicates that X violates the octet rule by forming coordinate covalent bonds with all Y atoms
  2. The compound must have resonance structures because the odd number of total valence electrons creates radical character
  3. The central atom X forms one double bond and two single bonds to achieve a stable electron configuration
  4. The central atom X forms three single bonds and has one lone pair, following the octet rule with sp³ hybridization (correct answer)
Explanation: When analyzing molecular geometry, you need to use VSEPR theory, which considers both bonding pairs and lone pairs of electrons around the central atom to predict shape. For XY3\text{XY}_3 with X having 5 valence electrons and Y having 7 valence electrons, let's determine the electron arrangement. The central atom X uses 3 of its 5 valence electrons to form single bonds with three Y atoms. This leaves X with 2 electrons (1 lone pair). Each Y atom contributes 1 electron to the bonding, achieving a stable configuration. The electron geometry around X is tetrahedral (4 electron domains: 3 bonding + 1 lone pair), but the molecular geometry is trigonal pyramidal because we only consider atom positions, not lone pairs. Option A is incorrect because X doesn't violate the octet rule - it has exactly 8 electrons (6 from bonding + 2 from the lone pair). Coordinate bonds aren't necessary here. Option B is wrong because the total valence electrons (5 + 3×7 = 26) is even, not odd, so there's no radical character or need for resonance structures. Option C is incorrect because forming double and single bonds would require different electron counts and wouldn't produce a trigonal pyramidal geometry. Option D correctly identifies that X forms three single bonds and retains one lone pair, following the octet rule with sp3sp^3 hybridization (mixing one s and three p orbitals to accommodate four electron domains). Remember: trigonal pyramidal geometry is the signature of AX3E\text{AX}_3\text{E} arrangement - three bonding pairs plus one lone pair around the central atom.

Question 6

Analyze the polar covalent bonds in PCl5\text{PCl}_5 and SF6\text{SF}_6. Despite both molecules containing polar bonds, they exhibit different overall molecular polarities. Which statement best explains this observation using covalent bonding principles?

  1. PCl₅ is nonpolar due to resonance structures, while SF₆ is polar because fluorine is more electronegative than chlorine
  2. SF₆ is polar because sulfur can form stronger covalent bonds than phosphorus, creating an unequal charge distribution
  3. Both molecules are nonpolar because the central atoms have equal sharing of electrons with all surrounding atoms
  4. PCl₅ has a net dipole moment due to its trigonal bipyramidal geometry, while SF₆ is nonpolar due to its octahedral symmetry (correct answer)
Explanation: When analyzing molecular polarity, you need to consider both individual bond polarities and the three-dimensional molecular geometry. Even when all bonds in a molecule are polar, the overall molecule can still be nonpolar if the bond dipoles cancel due to symmetrical arrangement. In PCl5\text{PCl}_5, phosphorus forms five polar P-Cl bonds arranged in a trigonal bipyramidal geometry. This geometry is asymmetrical - three chlorine atoms occupy equatorial positions while two occupy axial positions. The axial bonds are longer and at different angles than the equatorial bonds, preventing complete cancellation of the bond dipoles. This creates a net dipole moment, making PCl5\text{PCl}_5 polar overall. SF6\text{SF}_6 contains six polar S-F bonds arranged in a perfectly symmetrical octahedral geometry. All bond lengths are equal, and the bond dipoles point directly opposite each other, resulting in complete cancellation. Therefore, SF6\text{SF}_6 is nonpolar despite having polar bonds. Option A incorrectly invokes resonance structures for PCl5\text{PCl}_5 and misattributes SF6\text{SF}_6's polarity to electronegativity differences rather than geometry. Option B wrongly claims bond strength affects molecular polarity and incorrectly states SF6\text{SF}_6 is polar. Option C falsely suggests equal electron sharing makes bonds nonpolar, confusing this with geometric cancellation of dipoles. Remember this key principle: molecular polarity depends on both bond polarity and molecular geometry. Always visualize the 3D shape and consider whether bond dipoles cancel completely. Symmetrical geometries like octahedral typically yield nonpolar molecules, while asymmetrical ones often remain polar.

Question 7

Two isomeric compounds with the formula C2H2Cl2\text{C}_2\text{H}_2\text{Cl}_2 show different boiling points: one boils at 60°C and the other at 48°C. Using principles of covalent bonding and molecular structure, which factor most likely accounts for this difference?

  1. The isomers have different bond angles around the carbon atoms, affecting the strength of intramolecular covalent bonds
  2. The isomers exhibit different hybridization states of carbon, creating bonds of different strengths and affecting volatility
  3. One isomer has a permanent dipole moment while the other is nonpolar, leading to different intermolecular attractive forces (correct answer)
  4. One isomer can form hydrogen bonds while the other cannot, significantly affecting the intermolecular interactions
Explanation: When you encounter questions about isomers with different boiling points, focus on intermolecular forces rather than intramolecular bond properties. The molecular formula C2H2Cl2\text{C}_2\text{H}_2\text{Cl}_2 can form two structural isomers: 1,1-dichloroethene (both chlorines on the same carbon) and 1,2-dichloroethene (chlorines on different carbons). The key insight is that these isomers have dramatically different polarities. In 1,1-dichloroethene, both highly electronegative chlorine atoms pull electron density toward the same side of the molecule, creating a significant dipole moment. In contrast, 1,2-dichloroethene has chlorines on opposite sides, making their dipole effects cancel out and resulting in a nonpolar molecule. The polar isomer experiences stronger dipole-dipole intermolecular attractions, requiring more energy to overcome during boiling, hence the higher boiling point (60°C vs 48°C). This makes option C correct. Option A incorrectly suggests that bond angles affect intramolecular bond strength—bond angles don't significantly change covalent bond energies. Option B is wrong because both isomers have the same hybridization (sp2sp^2 carbons in alkenes), so bond strengths are essentially identical. Option D fails because neither isomer contains hydrogen bonding—the hydrogens are bonded to carbon, not highly electronegative atoms like oxygen or nitrogen. Remember: when comparing boiling points of isomers, always consider molecular polarity and the resulting intermolecular forces first. Intramolecular properties like bond strength rarely vary significantly between structural isomers.

Question 8

Consider a molecule where the central atom has the electron domain geometry of octahedral but the molecular geometry is square pyramidal. If this molecule undergoes a chemical reaction where one lone pair becomes a bonding pair, what changes would occur?

  1. The electron domain geometry changes to trigonal bipyramidal while the molecular geometry becomes tetrahedral
  2. The electron domain geometry remains octahedral while the molecular geometry becomes octahedral (correct answer)
  3. Both electron domain and molecular geometries change to trigonal bipyramidal due to the additional bonding pair
  4. The electron domain geometry becomes pentagonal while the molecular geometry becomes trigonal bipyramidal
Explanation: Square pyramidal molecular geometry with octahedral electron domain geometry means 5 bonding pairs and 1 lone pair around the central atom. When the lone pair becomes a bonding pair, there are now 6 bonding pairs and 0 lone pairs. The electron domain geometry remains octahedral (6 electron domains), but the molecular geometry becomes octahedral since all domains are now bonding. Options A and C incorrectly change the electron domain count. Option D uses non-standard geometry terms.

Question 9

Consider the bond dissociation energy data: C-C (347 kJ/mol), C=C (614 kJ/mol), C≡C (839 kJ/mol). A student predicts that the C≡C bond length should be exactly one-third the length of a C-C single bond based on the assumption that bond strength is inversely proportional to bond length. What is the primary flaw in this reasoning?

  1. Bond dissociation energy includes contributions from both σ and π bonds, while bond length is determined primarily by σ bond overlap (correct answer)
  2. The relationship between bond energy and bond length follows a logarithmic function rather than a simple inverse relationship
  3. Triple bonds involve sp hybridization which fundamentally changes the atomic radii compared to sp³ hybridized carbon atoms
  4. Bond dissociation energies are measured at different temperatures, making direct comparisons of bond strength invalid for length predictions
Explanation: The student's error is assuming that total bond energy directly correlates with bond length. While σ bonds determine internuclear distance, π bonds add energy without significantly affecting bond length. The C≡C bond has one σ and two π bonds, but the bond length is determined primarily by σ orbital overlap. Options B, C, and D contain incorrect assumptions about the energy-length relationship, hybridization effects, and measurement conditions.

Question 10

A student proposes that the compound AlF3\text{AlF}_3 should have a trigonal planar geometry based on VSEPR theory, treating it as a covalent molecule. However, experimental evidence shows AlF3\text{AlF}_3 has an ionic crystal structure. What is the most likely explanation for this discrepancy?

  1. The student correctly applied VSEPR theory, but the molecule exhibits resonance which changes the predicted geometry
  2. The student failed to account for the fact that aluminum typically forms only two covalent bonds due to its coordination preferences
  3. The high charge density of the Al³⁺ ion and small size of F⁻ lead to ionic bonding rather than covalent bonding, invalidating VSEPR predictions (correct answer)
  4. The student's prediction is correct, but experimental error or impure samples led to incorrect structural determination
Explanation: AlF₃ forms ionic bonds rather than covalent bonds due to the large electronegativity difference and the high charge density of Al³⁺. VSEPR theory applies to covalent molecules, so when bonding is primarily ionic, the structure is determined by ionic packing rather than electron pair repulsion. Option A incorrectly invokes resonance. Option B misrepresents aluminum's bonding capability. Option D dismisses experimental evidence without justification.

Question 11

In comparing the molecules BF3\text{BF}_3 and NF3\text{NF}_3, both have the same number of bonds between the central atom and fluorine. However, their molecular geometries and chemical properties differ significantly. Which factor most directly explains these differences?

  1. Boron has a smaller atomic radius than nitrogen, leading to stronger covalent bonds and different molecular geometry
  2. Nitrogen has a lone pair of electrons while boron has an incomplete octet, affecting both geometry and reactivity (correct answer)
  3. The electronegativity difference between B-F and N-F bonds creates different dipole moments and intermolecular forces
  4. Boron can utilize d-orbitals for bonding while nitrogen is limited to s and p orbitals in its valence shell
Explanation: BF₃ has trigonal planar geometry (incomplete octet, no lone pairs) while NF₃ has trigonal pyramidal geometry (complete octet with one lone pair). This lone pair difference affects both molecular shape via VSEPR theory and chemical reactivity (NF₃ can act as a Lewis base). Option A is incorrect because atomic radius doesn't explain geometry differences. Option C doesn't address the geometry difference. Option D is wrong because boron (period 2) has no accessible d-orbitals.

Question 12

Which row correctly identifies the electron domain geometry and the molecular geometry for ammonia, NH₃?

  1. Electron domain geometry: Tetrahedral, Molecular geometry: Trigonal pyramidal (correct answer)
  2. Electron domain geometry: Trigonal pyramidal, Molecular geometry: Trigonal pyramidal
  3. Electron domain geometry: Trigonal planar, Molecular geometry: Trigonal planar
  4. Electron domain geometry: Tetrahedral, Molecular geometry: Tetrahedral
Explanation: The Lewis structure of NH₃ shows the central nitrogen atom is bonded to three hydrogen atoms and has one lone pair of electrons. This gives a total of four electron domains. The arrangement of four electron domains is tetrahedral. The molecular geometry, which describes the arrangement of only the atoms, is trigonal pyramidal because the lone pair is not included in the shape's name.

Question 13

Which of the following places the substances in order of increasing boiling point?

  1. CH₄ < SiH₄ < NH₃ < H₂O (correct answer)
  2. CH₄ < NH₃ < SiH₄ < H₂O
  3. SiH₄ < CH₄ < NH₃ < H₂O
  4. CH₄ < SiH₄ < H₂O < NH₃
Explanation: Boiling points increase with the strength of intermolecular forces. CH₄ and SiH₄ are non-polar and have only London dispersion forces (LDFs); SiH₄ has more electrons and thus stronger LDFs and a higher boiling point than CH₄. NH₃ and H₂O both exhibit strong hydrogen bonding. Water's hydrogen bonds are stronger than ammonia's due to the greater polarity of the O-H bond and the ability of each water molecule to form more hydrogen bonds on average. Hydrogen bonds are much stronger than the LDFs in CH₄ and SiH₄. Thus, the correct order is CH₄ < SiH₄ < NH₃ < H₂O.

Question 14

Which statement correctly compares the polarity of tetrachloromethane (CCl₄) and dichloromethane (CH₂Cl₂)?

  1. Both are non-polar because they have a symmetrical tetrahedral arrangement of atoms.
  2. CCl₄ is polar because it contains polar C-Cl bonds, while CH₂Cl₂ is non-polar.
  3. CH₂Cl₂ is polar because its bond dipoles do not cancel, whereas CCl₄ is non-polar because its bond dipoles cancel. (correct answer)
  4. Both are polar because they contain polar covalent bonds between carbon and chlorine.
Explanation: Both molecules have a tetrahedral geometry. In CCl₄, the four C-Cl bonds are polar, but their dipoles are arranged symmetrically and cancel each other out, resulting in a non-polar molecule. In CH₂Cl₂, the C-Cl bonds are more polar than the C-H bonds. The arrangement of these different bonds is asymmetrical, so the bond dipoles do not cancel, resulting in a net dipole moment and a polar molecule.

Question 15

Which statement best describes the bonding and geometry of gaseous beryllium chloride, BeCl₂?

  1. The molecule is bent and polar, with the central beryllium atom obeying the octet rule.
  2. The molecule is linear and non-polar, with the central beryllium atom having an incomplete octet. (correct answer)
  3. The molecule forms double bonds between beryllium and chlorine to satisfy the octet rule for all atoms.
  4. The molecule is linear and polar because the individual Be-Cl bonds are polar covalent.
Explanation: The Lewis structure for BeCl₂ shows the central Be atom forming single bonds with two Cl atoms. This gives Be only four valence electrons, an exception to the octet rule known as an incomplete octet. With two electron domains and no lone pairs, VSEPR theory predicts a linear geometry and a 180° bond angle. Although the Be-Cl bonds are polar, the linear symmetry causes the bond dipoles to cancel, making the molecule non-polar.

Question 16

The ammonium ion, NH₄⁺, is formed when an ammonia molecule, NH₃, reacts with a hydrogen ion, H⁺. Which statement accurately describes a bond in the ammonium ion?

  1. It contains three covalent N-H bonds and one ionic bond to the fourth hydrogen.
  2. It contains three standard covalent bonds and one coordination bond, with all four bonds being identical. (correct answer)
  3. It contains four covalent N-H bonds formed by each atom contributing one electron to each bond.
  4. It contains three covalent bonds and one intermolecular hydrogen bond.
Explanation: Ammonia (NH₃) has a lone pair of electrons on the nitrogen atom. A hydrogen ion (H⁺) has an empty orbital. NH₃ donates its entire lone pair to form a new bond with H⁺. This type of covalent bond, where both electrons are contributed by one atom, is called a coordination or dative covalent bond. Once formed in the NH₄⁺ ion, this bond is indistinguishable from the other three N-H bonds.

Question 17

Silicon dioxide, SiO₂, has a melting point of approximately 1700 °C, while carbon dioxide, CO₂, sublimes at -78 °C. What is the best explanation for this large difference?

  1. SiO₂ is a covalent network solid requiring the breaking of strong covalent bonds, while CO₂ is a molecular substance with weak intermolecular forces. (correct answer)
  2. The Si=O double bonds in SiO₂ are significantly stronger than the C=O double bonds in CO₂, requiring more energy to break.
  3. SiO₂ has a much larger molar mass than CO₂, leading to exceptionally strong London dispersion forces.
  4. Silicon is a metalloid, which gives SiO₂ strong metallic properties, unlike the simple covalent molecule CO₂.
Explanation: The difference in physical properties arises from a fundamental difference in structure and bonding. SiO₂ is a giant covalent or covalent network solid, where each silicon atom is bonded to four oxygen atoms in a vast lattice. To melt it, strong covalent bonds must be broken, which requires a large amount of energy. CO₂ is a simple molecular substance composed of discrete O=C=O molecules held together by weak intermolecular London dispersion forces. Overcoming these weak forces requires very little energy.

Question 18

In which of the following substances can hydrogen bonding occur as the dominant intermolecular force between its molecules?

  1. Fluoromethane, CH₃F
  2. Methoxymethane, CH₃OCH₃
  3. Methanal, HCHO
  4. Methylamine, CH₃NH₂ (correct answer)
Explanation: For hydrogen bonding to exist between molecules of a substance, a hydrogen atom must be covalently bonded to a highly electronegative atom (N, O, or F). In CH₃F, CH₃OCH₃, and HCHO, all hydrogen atoms are bonded to carbon. Only in methylamine, CH₃NH₂, are there hydrogen atoms directly bonded to a nitrogen atom. This allows for hydrogen bonding to occur between CH₃NH₂ molecules.

Question 19

In the reaction between boron trifluoride (BF₃) and ammonia (NH₃), a single product is formed. Which of the following correctly describes the roles of the reactants and the bond formed?

  1. BF₃ acts as the electron pair donor, forming a coordination bond with NH₃.
  2. An electron is transferred from NH₃ to BF₃, and an ionic bond is formed.
  3. NH₃ acts as the electron pair donor, forming a coordination bond with BF₃. (correct answer)
  4. A standard covalent bond is formed by each reactant contributing one electron.
Explanation: The boron atom in BF₃ has an incomplete octet and is electron-deficient. The nitrogen atom in NH₃ has a lone pair of electrons. In the reaction, the electron-rich NH₃ donates its lone pair to the electron-deficient BF₃. A species that donates an electron pair is a Lewis base (NH₃), and one that accepts is a Lewis acid (BF₃). The resulting bond, where both electrons are from the same atom, is a coordination (or dative covalent) bond.

Question 20

Graphite is used as a lubricant and conducts electricity, whereas diamond is extremely hard and is an electrical insulator. Which statement best explains these differences?

  1. Diamond has stronger intramolecular covalent bonds than graphite, while graphite has weak ionic bonds between its layers.
  2. Graphite has layers of atoms that slide past one another and delocalized electrons, while diamond has a rigid 3D network with localized electrons. (correct answer)
  3. Diamond's tetrahedral structure contains mobile ions making it hard, while graphite's planar structure traps electrons.
  4. Graphite's softness is due to its non-polar nature, while diamond's hardness is due to its polar covalent bonds.
Explanation: Graphite consists of layers of covalently bonded carbon atoms. The layers are held together by weak London dispersion forces, allowing them to slide, which explains its use as a lubricant. Each carbon atom is bonded to three others, leaving one delocalized electron per atom that can move and conduct electricity. Diamond has a rigid, three-dimensional network structure where each carbon is bonded to four others. All electrons are localized in strong covalent bonds, making it very hard and an electrical insulator.