4.7. Practice Problems

1. (a) Write the electron configuration of sodium (Na), magnesium (Mg), and chlorine (Cl). Identify the expected ions each forms and their electron configurations. (b) Determine the empirical formula of the compound formed between aluminum (Al) and sulfur (S). Explain your reasoning. (c) Rank the following ionic solids by expected melting point (lowest to highest), based on ionic charges and approximate ionic radii: NaCl, MgO, CaF₂. Briefly justify your ranking using lattice energy considerations.

This problem involves understanding ionic bonding and lattice energy. In part (a), students write electron configurations for sodium, magnesium, and chlorine. Na ([Ne] 3s¹) will lose one electron and form Na⁺, magnesium ([Ne] 3s²) will lose two electrons to form Mg²⁺, and chlorine ([Ne] 3s² 3p⁵) will gain one electron to form Cl⁻. In part (b), when combining aluminum, which typically forms Al³⁺, and sulfur that forms S²⁻, the empirical formula determined would be Al₂S₃ since two aluminum ions balance three sulfide ions. In part (c), students will rank ionic compounds based on ionic charge and size, where MgO (with 2+ and 2− ions) would have a higher melting point than NaCl (1+ and 1− ions) and CaF₂ (2+ and 1− ions) varies based on ionic radius effects. Lattice energy, which is the energy released upon formation, indicates stronger attractions result in higher melting points.

Think of ions forming as like characters at a dance. Sodium is a dancer who prefers to give up one dance partner (its electron) to become a perfect pair with chlorine, who loves collecting partners to feel complete (gaining an electron). The stronger the attraction between these couples (the ionic bond formed), the harder it is to separate them (higher melting points).

2. Draw the best Lewis structures (with formal charges) for each of the following, indicating resonance where applicable: (a) Nitrate ion, NO3− (b) Sulfate ion, SO42− (c) Ozone, O3 (d) Carbon dioxide, CO2.

In this problem, students are asked to construct Lewis structures, which represent the arrangement of electrons in molecules. For nitrate (NO₃⁻), students should note that it has three resonance forms where the charge is distributed. For sulfate (SO₄²⁻), similar resonance structures apply. In ozone (O₃), there are also two resonance forms, while carbon dioxide (CO₂) has a straightforward double bond between carbon and each oxygen. The goal is understanding how formal charges are balanced across atoms to get stable structures and how resonance allows for delocalization of electrons.

Imagine drawing different versions of a logo for a company. Each version (resonance structures) has slight changes while maintaining the essence (formal charge balance) of what makes the logo recognizable. Just like a company may have variations of its logo but still maintains its identity, molecules can have multiple Lewis structures that depict them correctly.

3. For each molecule below: (a) Predict the electron-domain geometry and molecular geometry using VSEPR theory. (b) Indicate approximate bond angles. (c) Determine whether the molecule is polar or nonpolar, and if polar, indicate the direction of the net dipole moment. (i) PCl5 (ii) ClF3 (iii) XeF4 (iv) NH4+ (v) HCN.

This problem tests the understanding of the VSEPR (Valence-Shell Electron Pair Repulsion) theory, which is used to predict molecular geometry based on electron domains. For PCl₅, with five bonding pairs and no lone pairs, the molecular shape is trigonal bipyramidal. ClF₃ has three bonds and two lone pairs, leading to a T-shaped geometry. XeF₄ has four bonds and two lone pairs which forms an octahedral geometry. NH₄⁺ is tetrahedral with four bonding pairs. HCN is linear with two atoms bonded directly. Polarity depends on the net dipole moment determined by geometry and individual bond polarities, where polar vectors do not cancel.

Imagine arranging furniture in a room. Each piece represents a bonding pair, and the space between them represents lone pairs. Depending on how you arrange the furniture (the electron pairs), you get different room shapes (molecular geometries). Some rooms can feel more open (nonpolar) while others may feel cramped (polar) due to how the furniture (bonds) directs the flow of movement.

4. (a) Rank the following substances in order of increasing boiling point, based on their predominant intermolecular forces. Provide brief justifications.
5. CH₃CH₂CH₂CH₃ (n-butane)
6. CH₃OH (methanol)
7. CH₃CH₂Cl (chloroethane)
8. CH₃C≡CH (1-propyne)
9. CCl₄ (carbon tetrachloride) (b) Explain why ice (solid H₂O) floats on liquid water (dipole interactions and hydrogen bonding). (c) Predict whether each of the following will dissolve more readily in water or in hexane (C₆H₁₄). Explain using “like dissolves like” and the types of intermolecular forces. (i) Benzene (C₆H₆) (ii) Ethylene glycol (HO–CH₂–CH₂–OH) (iii) Ammonium nitrate (NH₄NO₃) (iv) Iodine (I₂).

In this section, students explore the concept of intermolecular forces and their impact on boiling points and solubility. The ranking of substances will expose students to different types of intermolecular forces such as hydrogen bonding (methanol), dipole-dipole interactions (chloroethane), and London dispersion forces (n-butane and CCl₄). The order reflects the strength and nature of these forces. In part (b), students learn that ice has a lower density than liquid water due to the hydrogen bond arrangement, allowing ice to float. In part (c), 'like dissolves like' will help them determine solubility trends, with polar substances dissolving well in water (e.g., ethylene glycol and ammonium nitrate) while nonpolar ones like iodine and benzene dissolve better in hexane.

Consider cooking pasta in water. When you boil water, if you add salt (a polar substance), it dissolves easily because water can interact with those ions. However, oil (a nonpolar substance) doesn't mix with water. Just like in cooking, different materials need compatible environments (like polar with polar) to mix well, and this compatibility relates to their molecular interactions.

5. (a) Describe qualitatively the electron distribution in a metallic solid and explain how this distribution gives rise to electrical and thermal conductivity. (b) Compare the malleability and ductility of sodium metal and magnesium metal, given that sodium crystallizes in a body-centered cubic (BCC) lattice and magnesium in a hexagonal close-packed (HCP) lattice. Which is expected to be more malleable, and why?

Here, students discuss traits of metallic bonding, notably delocalized electrons that facilitate conductivity. These 'sea' of electrons allows electric current to flow easily through metals, as well as thermal energy transfer. In part (b), students compare sodium and magnesium based on their crystal structures, with sodium (BCC) generally being more malleable compared to magnesium (HCP) as the latter's denser packing can inhibit motion. The overall quality of these metals relates back to their ability to rearrange atoms under stress, showcasing properties of ductility.

Picture a large group of friends at a concert. Those who can move easily through the crowd (like delocalized electrons in a metallic solid) can swap places quickly, allowing the crowd to accommodate changes without disturbances (electrical and thermal conductivity). Meanwhile, in tightly packed dance circles (HCP), it's harder for friends to shift places, making it less flexible (lower malleability and ductility).

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