Understand The Covalent Model - IB Chemistry

1. A student proposes that the compound $\text{AlF}_3$ should have a trigonal planar geometry based on VSEPR theory, treating it as a covalent molecule. However, experimental evidence shows $\text{AlF}_3$ has an ionic crystal structure. What is the most likely explanation for this discrepancy?
   Answer: The high charge density of the Al³⁺ ion and small size of F⁻ lead to ionic bonding rather than covalent bonding, invalidating VSEPR predictions 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.
2. In comparing the molecules $\text{BF}_3$ and $\text{NF}_3$, 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?
   Answer: Nitrogen has a lone pair of electrons while boron has an incomplete octet, affecting both geometry and reactivity 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.
3. Using the electronegativity values in the data booklet, which of the following covalent bonds is the most polar?
   Answer: C–F in tetrafluoromethane Bond polarity is determined by the difference in electronegativity (ΔEN) between the two bonded atoms. Using data booklet values: ΔEN(N-H) = 3.0 - 2.2 = 0.8; ΔEN(C-F) = 4.0 - 2.6 = 1.4; ΔEN(O-H) = 3.4 - 2.2 = 1.2; ΔEN(Si-Cl) = 3.2 - 1.9 = 1.3. The greatest difference in electronegativity is for the C-F bond, making it the most polar bond among the choices.
4. In which of the following substances can hydrogen bonding occur as the dominant intermolecular force between its molecules?
   Answer: Methylamine, CH₃NH₂ 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.
5. Which of the following places the substances in order of increasing boiling point?
   Answer: CH₄ < SiH₄ < NH₃ < H₂O 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.
6. 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?
   Answer: NH₃ acts as the electron pair donor, forming a coordination bond with BF₃. 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.
7. Iodine, I₂, is only slightly soluble in water, while glucose, C₆H₁₂O₆, is very soluble. Which statement provides the best explanation for this observation?
   Answer: Glucose has many hydroxyl groups that form hydrogen bonds with water, while non-polar iodine cannot. The principle of 'like dissolves like' governs solubility. Water is a highly polar solvent that forms hydrogen bonds. Glucose has multiple polar hydroxyl (-OH) groups which can form strong hydrogen bonds with water molecules, making it very soluble. Iodine (I₂) is a non-polar molecule, and its interactions with polar water molecules are limited to weak London dispersion forces, resulting in low solubility.
8. Which statement correctly compares the polarity of tetrachloromethane (CCl₄) and dichloromethane (CH₂Cl₂)?
   Answer: CH₂Cl₂ is polar because its bond dipoles do not cancel, whereas CCl₄ is non-polar because its bond dipoles cancel. 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.
9. Ethanol (CH₃CH₂OH) and methoxymethane (CH₃OCH₃) are isomers with the same molar mass. Why is the boiling point of ethanol significantly higher than that of methoxymethane?
   Answer: Ethanol molecules can form hydrogen bonds, which are much stronger than the dipole-dipole forces in methoxymethane. Boiling involves overcoming intermolecular forces, not breaking intramolecular bonds. Ethanol has a hydrogen atom bonded to an oxygen atom, allowing it to form strong intermolecular hydrogen bonds. Methoxymethane is polar and has dipole-dipole forces, but it lacks an H atom bonded to an O, N, or F, so it cannot form hydrogen bonds with itself. Hydrogen bonds are significantly stronger than dipole-dipole forces, requiring more energy to overcome, thus giving ethanol a higher boiling point.
10. Which statement correctly compares the bond angles in methane (CH₄), ammonia (NH₃), and water (H₂O)?
    Answer: The bond angle in CH₄ is greater than in NH₃, which is greater than in H₂O, due to the increasing number of lone pairs. All three central atoms have four electron domains, so the ideal angle is 109.5°. However, lone pairs repel more strongly than bonding pairs. CH₄ has no lone pairs, so its angle is 109.5°. NH₃ has one lone pair, which compresses the H-N-H angle to about 107°. H₂O has two lone pairs, which cause even greater repulsion and compress the H-O-H angle to about 104.5°. Therefore, the bond angle decreases as the number of lone pairs increases: CH₄ > NH₃ > H₂O.
11. What are the approximate bond angles and molecular geometry of phosgene, COCl₂, in which carbon is the central atom?
    Answer: Trigonal planar, approximately 120° 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.
12. Analyze the polar covalent bonds in $\text{PCl}_5$ and $\text{SF}_6$. Despite both molecules containing polar bonds, they exhibit different overall molecular polarities. Which statement best explains this observation using covalent bonding principles?
    Answer: PCl₅ has a net dipole moment due to its trigonal bipyramidal geometry, while SF₆ is nonpolar due to its octahedral symmetry 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 $\text{PCl}_5$, 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 $\text{PCl}_5$ polar overall. $\text{SF}_6$ 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, $\text{SF}_6$ is nonpolar despite having polar bonds. Option A incorrectly invokes resonance structures for $\text{PCl}_5$ and misattributes $\text{SF}_6$'s polarity to electronegativity differences rather than geometry. Option B wrongly claims bond strength affects molecular polarity and incorrectly states $\text{SF}_6$ 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.
13. Graphene consists of a single layer of carbon atoms arranged in a hexagonal lattice. Which pair of properties is correctly explained by its structure?
    Answer: High tensile strength due to strong covalent bonds; good electrical conductivity due to delocalized electrons. Graphene is a giant covalent structure. The strong C-C covalent bonds within the single layer give it an extremely high tensile strength. Each carbon atom is bonded to three others, leaving one valence electron per atom that is delocalized across the entire sheet. These mobile, delocalized electrons allow graphene to be an excellent electrical conductor.
14. Which statement correctly describes the geometry and polarity of the sulfur difluoride molecule, SF₂?
    Answer: The molecule is bent and polar. The central sulfur atom in SF₂ has six valence electrons. It forms single bonds with two fluorine atoms and has two lone pairs of electrons. This results in four electron domains around the sulfur atom, leading to a bent (or V-shaped) molecular geometry. The S-F bonds are polar due to the large electronegativity difference. Because the molecule is bent, the bond dipoles do not cancel, resulting in a net molecular dipole, making SF₂ a polar molecule.
15. What is the total number of valence electrons and the number of non-bonding electrons on the chlorine atom in the Lewis structure of the hypochlorite ion, ClO⁻?
    Answer: Total valence electrons: 14; Non-bonding electrons on Cl: 6 To find the total valence electrons, sum the valence electrons of each atom and add one for the negative charge: 7 (from Cl) + 6 (from O) + 1 = 14. The Lewis structure consists of a single bond between Cl and O. To complete the octets, the oxygen atom has three lone pairs (6 non-bonding electrons) and the chlorine atom also has three lone pairs (6 non-bonding electrons).
16. 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?
    Answer: The component with Rƒ = 0.20 is the most polar. 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.
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?
    Answer: SiO₂ is a covalent network solid requiring the breaking of strong covalent bonds, while CO₂ is a molecular substance with weak intermolecular forces. 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.
18. 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?
    Answer: It contains three standard covalent bonds and one coordination bond, with all four bonds being identical. 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.
19. Which row correctly identifies the electron domain geometry and the molecular geometry for ammonia, NH₃?
    Answer: Electron domain geometry: Tetrahedral, Molecular geometry: Trigonal pyramidal 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.
20. 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?
    Answer: The electron domain geometry remains octahedral while the molecular geometry becomes octahedral 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.

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