Apply The Covalent Model - IB Chemistry

Consider the C–C bond in ethane (C₂H₆), the C=C bond in ethene (C₂H₄), and the C≡C bond in ethyne (C₂H₂). Which statement correctly compares the carbon-carbon bonds in these three molecules?
Answer: The bond in ethane is the longest and the bond in ethyne is the strongest. As the number of shared electron pairs (bond order) between two atoms increases, the electrostatic attraction between the nuclei and the electrons becomes stronger. This pulls the atoms closer together, decreasing bond length, and requires more energy to break, increasing bond strength. The order of bond order is ethane (single, 1) < ethene (double, 2) < ethyne (triple, 3). Therefore, the C-C single bond in ethane is the longest and weakest, while the C≡C triple bond in ethyne is the shortest and strongest. Statement A correctly identifies these two extremes.
Which statement best explains why the bond angle in H₂O (104.5°) is smaller than the bond angle in NH₃ (107°)?
Answer: H₂O has two lone pairs on the central atom while NH₃ has only one. Both molecules have central atoms with four electron domains. According to VSEPR theory, lone pairs of electrons repel more strongly than bonding pairs. Water's central oxygen atom has two lone pairs and two bonding pairs. Ammonia's central nitrogen has one lone pair and three bonding pairs. The greater repulsion from the two lone pairs in water pushes the bonding pairs closer together than the repulsion from the single lone pair in ammonia, resulting in a smaller H-O-H bond angle compared to the H-N-H bond angle.
A student draws two possible Lewis structures for $SO_2$ and needs to determine which better represents the actual bonding. Structure A shows single bonds with formal charges, while Structure B shows double bonds with different formal charges. Using formal charge analysis, which evaluation is correct?
Answer: Structure B is preferred because it places the smallest formal charges on the most electronegative atoms, with sulfur having a formal charge of +1 and each oxygen having a formal charge of 0 When evaluating Lewis structures, formal charge analysis helps determine which structure best represents actual molecular bonding. The key principle is that the most stable structure minimizes formal charges overall and places negative formal charges on the most electronegative atoms. For $SO_2$ , you need to calculate formal charges using the formula: Formal Charge = (valence electrons) - (nonbonding electrons) - ½(bonding electrons). Structure A uses single bonds, giving sulfur a formal charge of +2 and each oxygen -1. Structure B uses double bonds, resulting in sulfur having +1 and each oxygen having 0. Structure B is preferred because it achieves smaller formal charges overall and places the minimal formal charge on oxygen, the most electronegative atom. When oxygen atoms have formal charges of 0 rather than -1, this better reflects the actual electron distribution since oxygen, being highly electronegative, doesn't need additional negative charge to be stable. Option A incorrectly states that single bonds minimize formal charges on sulfur - actually, the double bonds in Structure B give sulfur a lower formal charge (+1 vs +2). Option B focuses on the octet rule violation but misses that formal charge distribution is more important for determining the best structure when both are reasonable. Option D incorrectly describes the formal charges in Structure B, stating sulfur has 0 and oxygen has -1, which matches Structure A's values, not B's. Remember: the best Lewis structure typically has the smallest formal charges on the most electronegative atoms, not necessarily the structure that strictly follows the octet rule.
Consider the molecules $BF_3$ , $NF_3$ , and $PF_3$ . A student predicts that all three should have similar bond angles since they all contain three fluorine atoms bonded to a central atom. Which analysis correctly evaluates this prediction?
Answer: The prediction is incorrect; $BF_3$ has 120° bond angles due to $sp^2$ hybridization, while $NF_3$ and $PF_3$ have smaller bond angles due to lone pair repulsion in $sp^3$ hybridization $BF_3$ has 3 electron domains (all bonding), giving $sp^2$ hybridization and 120° bond angles. Both $NF_3$ and $PF_3$ have 4 electron domains (3 bonding + 1 lone pair), giving $sp^3$ hybridization with bond angles less than 109.5° due to lone pair repulsion. Choice A ignores electron domain geometry. Choice C incorrectly suggests $PF_3$ has $sp^2$ hybridization. Choice D incorrectly assigns pyramidal geometry to $BF_3$ and assumes equal lone pair effects.
In comparing $BeCl_2$ and $H_2O$ , both molecules have two atoms bonded to a central atom, yet they have different molecular geometries. A student must explain this difference using valence bond theory. Which analysis correctly accounts for both molecular shapes?
Answer: $BeCl_2$ is linear due to $sp$ hybridization with 2 electron domains, while $H_2O$ is bent due to $sp^3$ hybridization with 4 electron domains including 2 lone pairs on oxygen When analyzing molecular geometry using valence bond theory, you need to consider both the hybridization of the central atom and the total number of electron domains (bonding pairs plus lone pairs) around it. For $BeCl_2$ , beryllium has 2 valence electrons and forms 2 bonds with chlorine atoms. This creates 2 electron domains around Be, requiring $sp$ hybridization. The two $sp$ hybrid orbitals arrange themselves 180° apart to minimize electron repulsion, creating a linear geometry. For $H_2O$ , oxygen has 6 valence electrons. It forms 2 bonds with hydrogen atoms and has 2 lone pairs, giving it 4 total electron domains. This requires $sp^3$ hybridization. While the four $sp^3$ orbitals arrange tetrahedrally, only two contain bonding pairs. The two lone pairs occupy more space than bonding pairs, compressing the H-O-H bond angle and creating a bent molecular geometry. Answer A incorrectly focuses on ionic versus covalent character rather than hybridization and electron domains. Answer B gets the hybridization wrong for both molecules - $BeCl_2$ uses $sp$ , not $sp^2$ , and $H_2O$ uses $sp^3$ , not $sp^2$ . Answer C completely reverses the geometries and incorrectly suggests beryllium has lone pairs. Answer D correctly identifies the hybridization and electron domain count for both molecules, explaining why $BeCl_2$ is linear ( $sp$ , 2 domains) and $H_2O$ is bent ( $sp^3$ , 4 domains with 2 lone pairs). Remember: molecular geometry depends on both hybridization and the presence of lone pairs - always count total electron domains, not just bonding pairs.
Which of the two liquids, OF₂ or SeF₂, would be predicted to have the higher boiling point and for what primary reason?
Answer: SeF₂, because it has a significantly greater number of electrons, leading to stronger London dispersion forces. Both OF₂ and SeF₂ have a bent molecular geometry and are polar. To compare their boiling points, we must compare the strength of their intermolecular forces (IMFs). SeF₂ has a much larger molar mass and many more electrons than OF₂ (Se has 34 electrons, O has 8). This leads to significantly stronger London dispersion forces (LDFs) in SeF₂. While dipole-dipole forces are also present, the large difference in the number of electrons makes the LDFs the dominant factor. The stronger LDFs in SeF₂ require more energy to overcome, resulting in a higher boiling point. D is incorrect as both have the same bent geometry.
The boiling points of the hydrogen halides are: HF (293 K), HCl (188 K), HBr (206 K), HI (238 K). Which statement best explains why HF has an anomalously high boiling point compared to the others?
Answer: The H-F bond is the most polar, allowing for the formation of strong hydrogen bonds. The boiling point trend for HCl, HBr, and HI increases with molar mass due to increasing London dispersion forces. HF deviates from this trend. Due to the high electronegativity of fluorine, the H-F bond is very polar, and this allows for the formation of strong intermolecular hydrogen bonds. These hydrogen bonds are much stronger than the dipole-dipole and London dispersion forces in the other hydrogen halides, requiring significantly more energy to overcome, resulting in a much higher boiling point. Intramolecular bond strength (C) does not determine boiling point. A and D state the opposite of the correct reasoning for intermolecular forces.
Which species has a different molecular geometry from its electron domain geometry?
Answer: H₂O Electron domain geometry describes the arrangement of all electron pairs (bonding and non-bonding) around the central atom, while molecular geometry describes only the arrangement of the atoms. A difference occurs when there are lone pairs on the central atom. In H₂O, the oxygen atom has four electron domains (2 bonding, 2 lone pairs), giving a tetrahedral electron domain geometry. However, the molecular geometry is described by the position of the atoms, which is bent. For CO₂, CH₄, and BF₃, the central atoms have no lone pairs, so their electron domain and molecular geometries are the same (linear, tetrahedral, and trigonal planar, respectively).
A student analyzes the bonding in $SF_4$ and $ClF_3$ . Both molecules have the same central atom hybridization but different molecular geometries. Which statement best explains this observation using VSEPR theory?
Answer: Both molecules have $sp^3d$ hybridization; $SF_4$ has seesaw geometry due to one lone pair, while $ClF_3$ has T-shaped geometry due to two lone pairs Both $SF_4$ and $ClF_3$ have 5 electron domains around the central atom ( $SF_4$ : 4 bonding + 1 lone pair; $ClF_3$ : 3 bonding + 2 lone pairs), giving $sp^3d$ hybridization. The different numbers of lone pairs result in different molecular geometries: seesaw for $SF_4$ and T-shaped for $ClF_3$ . Choice B incorrectly assigns geometries and lone pairs. Choice C uses wrong hybridization. Choice D uses incorrect hybridization and geometries.
Which substance is expected to have the highest solubility in hexane, C₆H₁₄?
Answer: Iodine (I₂) The principle of solubility is 'like dissolves like'. Hexane (C₆H₁₄) is a nonpolar solvent because it is a hydrocarbon with symmetrical C-C and C-H bonds. Therefore, it will best dissolve nonpolar solutes. Water and ammonia are both highly polar molecules capable of hydrogen bonding. Sodium chloride is an ionic compound. Iodine (I₂) is a nonpolar molecule because it is diatomic with a pure covalent bond. Thus, iodine will have the highest solubility in hexane due to the formation of London dispersion forces between the solute and solvent molecules.
The bond enthalpy of a N–N single bond is approximately 160 kJ mol⁻¹, and that of a N≡N triple bond is 945 kJ mol⁻¹. What can be deduced directly from this information?
Answer: The triple bond is more than five times as strong as the single bond. This question requires a direct mathematical comparison of the given values. Comparing the bond enthalpies: 945 kJ mol⁻¹ / 160 kJ mol⁻¹ ≈ 5.9. This value is greater than five, so statement A is a correct deduction from the data. Statement B incorrectly uses the single bond energy for the triple bond in N₂. Statement C is a common misconception; the data suggests the two π bonds contribute 945 - 160 = 785 kJ, so an average π bond (≈393 kJ) is stronger than this particular N-N σ bond. Statement D incorrectly relates bond strength and length; while the triple bond is shorter, the relationship is not a simple inverse ratio of three.
Which molecule is nonpolar despite containing polar covalent bonds?
Answer: SO₃ A molecule can be nonpolar if the polar bonds are arranged symmetrically, causing their bond dipoles to cancel out. SO₃ has a central sulfur atom with three double-bonded oxygen atoms and no lone pairs, resulting in a trigonal planar geometry. The S-O bonds are polar, but the symmetrical 120° arrangement means the vector sum of the dipoles is zero, making the molecule nonpolar. H₂S (bent), CH₃Cl (asymmetrical tetrahedral), and NCl₃ (trigonal pyramidal) all have asymmetrical shapes, resulting in a net molecular dipole moment, making them polar molecules.
What is the geometry of the carbonate ion, CO₃²⁻, and the approximate O-C-O bond angle?
Answer: Trigonal planar, 120° The Lewis structure for the carbonate ion, CO₃²⁻, shows a central carbon atom bonded to three oxygen atoms. Through resonance, the bonding is equivalent to three identical C-O bonds, each with a bond order of 4/3. There are three electron domains around the central carbon atom and no lone pairs. According to VSEPR theory, three electron domains arrange themselves in a trigonal planar geometry to minimize repulsion, resulting in bond angles of 120°.
Which molecule is polar and has a bond angle of approximately 107°?
Answer: PCl₃ To solve this, one must determine the Lewis structure, VSEPR geometry, and polarity for each molecule. PCl₃ has a central phosphorus atom with 3 bonding pairs and 1 lone pair. This gives it a trigonal pyramidal molecular geometry, which is polar due to its asymmetry. The electron domain geometry is tetrahedral, but the lone pair compresses the bond angles from 109.5° to approximately 107°. SO₂ is polar but bent with an angle closer to 120°. BF₃ is trigonal planar (120°) and nonpolar. CCl₄ is tetrahedral (109.5°) but nonpolar due to symmetry.
Both silicon dioxide (SiO₂) and carbon dioxide (CO₂) are oxides of Group 14 elements. Why does silicon dioxide have a much higher melting point than carbon dioxide?
Answer: Silicon dioxide exists as a covalent network structure, while carbon dioxide is a simple molecular substance with weak intermolecular forces. The key difference lies in their structure. Silicon dioxide (quartz) is a giant covalent network solid where each silicon atom is bonded to four oxygen atoms, and each oxygen to two silicons. Melting requires breaking these strong covalent bonds. Carbon dioxide is a simple molecular substance composed of discrete CO₂ molecules held together by weak London dispersion forces. Melting only requires overcoming these weak intermolecular forces. A is incorrect as SiO₂ has single bonds in a network, not double bonds. B is incorrect because SiO₂ is not a simple molecule. D is incorrect as larger atomic radius generally leads to weaker, longer bonds.
Which molecule contains a central atom that does not have a complete octet of valence electrons?
Answer: BeCl₂ The octet rule states that atoms tend to form bonds to achieve eight valence electrons. In BeCl₂, the central beryllium atom (Group 2) forms two single covalent bonds with two chlorine atoms. This results in the beryllium atom having only 4 valence electrons (2 bonds × 2 electrons/bond), which is an incomplete octet. The central atoms in H₂S (8 electrons), CCl₄ (8 electrons), and NF₃ (8 electrons) all satisfy the octet rule.
The boiling points of propan-1-ol (CH₃CH₂CH₂OH) and methoxyethane (CH₃OCH₂CH₃) are 370 K and 281 K respectively. Which statement best explains this significant difference?
Answer: Propan-1-ol can form hydrogen bonds between its molecules, which are much stronger than the forces in methoxyethane. Propan-1-ol and methoxyethane are structural isomers with the same molar mass (C₃H₈O), so London dispersion forces are comparable, making option A incorrect. Boiling involves overcoming intermolecular forces, not breaking intramolecular covalent bonds, making D incorrect. While methoxyethane is polar and has dipole-dipole forces (B), propan-1-ol contains a hydroxyl (-OH) group, allowing it to form strong intermolecular hydrogen bonds. Hydrogen bonds are significantly stronger than the dipole-dipole forces in the ether, accounting for the large difference in boiling points.
Methane (CH₄, Mᵣ = 16) is a gas at room temperature, while water (H₂O, Mᵣ = 18) is a liquid. What is the most important reason for this difference in physical state?
Answer: Water molecules are polar and form hydrogen bonds, while methane is nonpolar and only has weak London dispersion forces. The physical state at a given temperature depends on the strength of intermolecular forces (IMFs). Methane is a nonpolar molecule due to its symmetrical tetrahedral shape, so it only exhibits weak London dispersion forces. Water is a polar, bent molecule containing H atoms bonded to a highly electronegative O atom, which allows for strong hydrogen bonding between molecules. These hydrogen bonds are much stronger than the LDFs in methane, requiring more energy to overcome, thus water is a liquid while methane is a gas at room temperature, despite their similar molar masses.
Which statement correctly describes a property of graphite and the reason for it?
Answer: Graphite conducts electricity because it contains delocalized electrons within its layered structure. Graphite consists of layers of carbon atoms arranged in hexagonal rings. Within each layer, each carbon atom is covalently bonded to three others, leaving one delocalized valence electron per atom. These delocalized electrons are free to move along the layers, allowing graphite to conduct electricity. A describes diamond. C is incorrect; graphite has a very high melting point, and the forces between layers are weak intermolecular forces, not covalent bonds. D is incorrect; graphite is opaque and black due to the delocalized electrons absorbing light.
A student compares the bond lengths in $CO$ , $CO_2$ , and $CO_3^{2-}$ and notices that they follow the order: $CO < CO_2 < CO_3^{2-}$ . Which explanation best accounts for this trend using molecular orbital theory and resonance concepts?
Answer: $CO$ has the shortest bond due to triple bond character from molecular orbital overlap, $CO_2$ has intermediate length due to double bonds, and $CO_3^{2-}$ has the longest due to resonance delocalization reducing effective bond order $CO$ has a bond order of 3 (triple bond character) from molecular orbital analysis, giving the shortest bond. $CO_2$ has C=O double bonds with bond order 2. $CO_3^{2-}$ has resonance structures that delocalize electron density, giving a bond order of 4/3 (between single and double), resulting in the longest bonds. Choice B incorrectly focuses on hybridization rather than bond order. Choice C misapplies electronegativity concepts. Choice D oversimplifies based on molecular size rather than electronic structure.

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