Question

The grid for Question 79 has nine lettered boxes, each of which contains an item that is used to answer the questions that follow. Items may be used more than once and there may be more than one correct item in response to a question. $$ \begin{aligned} &\text { Grid for Question } 79\\\ &\begin{array}{|l|l|l|} \hline \text { A } & \text { B } & \text { C } \\ \text { HCN } & \text { PO }_{4}^{3-} & \text { PH }_{3} \text { or } \mathrm{PF}_{3} \\ \hline \text { D } & \text { E } & \text { F } \\ \text { SiH }_{4} & \text { Cl }_{2} \mathrm{O} & \text { NH }_{2} \text { Cl } \\ \hline \text { G } & \text { H } & \text { I } \\ \text { HF or } \mathrm{F}_{2} & \text { CH }_{4} & \text { OF }_{2} \\ \hline \end{array} \end{aligned} $$ Place the letter(s) of the correct selection(s) on the appropriate line. (a) Electron-region geometry is the same as the molecular geometry_____ (b) Nonpolar molecule____ (c) Linear molecular geometry______ (d) Angular (bent) molecular geometry______ (e) Central atom is \(s p^{3}\) hybridized______ (f) Central atom is sp hybridized_____ (g) Which one in each pair of compounds has the lower boiling point?_____ (h) Which one in each pair of compounds has the higher vapor pressure?______ (i) Which one in each pair of compounds has the higher dipole moment?______ (j) Has dipole-dipole and hydrogen bonding intermolecular forces______

Short answer

(a) H; (b) H; (c) A; (d) I; (e) H; (f) A; (g) G; (h) G; (i) G; (j) G.

Step-by-step solution

Identify Electron-Region and Molecular Geometry (a)

From the grid, compare each compound's electron-region geometry and molecular geometry. For a molecule to have the same electron-region and molecular geometry, it should typically have no lone pairs on the central atom. **CH₄ (Option H)** fits this criterion as it has a tetrahedral shape with no lone pairs. Thus, the answer is (H).

Determine Nonpolar Molecule (b)

A nonpolar molecule has a symmetrical arrangement of polar bonds, causing the dipoles to cancel out. **CH₄ (Option H)** is a nonpolar molecule because it is a symmetrical tetrahedral molecule with equivalent C-H bonds. Thus, answer is (H).

Identify Linear Molecular Geometry (c)

Linear geometry typically involves molecules with two atoms or three atoms with no lone pairs on the central atom. **HCN (Option A)** is a linear molecule since it is composed of a straight line of atoms (H-C≡N). Thus, the answer is (A).

Identify Angular (Bent) Molecular Geometry (d)

Angular or bent geometry commonly results from molecules with a central atom with lone pairs. **OF₂ (Option I)** is a bent molecule due to the two lone pairs on the oxygen, causing a bent shape. Thus, the answer is (I).

Determine sp³ Hybridized Central Atom (e)

Central atoms with a tetrahedral arrangement have sp³ hybridization. These could include both hybridization and geometry indications. **CH₄ (Option H)** fits this criterion. Thus, the answer is (H).

Determine sp Hybridized Central Atom (f)

Molecules with linear geometry typically have sp hybridization. **HCN (Option A)**, being linear, implies sp hybridization of the central carbon atom. Thus, the answer is (A).

Lower Boiling Point in Pairs (g)

Boiling point is lowered by weaker intermolecular forces. Between pairs: In HF vs. F₂ (Option G), F₂, being nonpolar and having London forces only, has the lower boiling point. **Answer: F₂ (G)**.

Higher Vapor Pressure in Pairs (h)

Higher vapor pressure is usually linked to lower boiling points. From HF vs. F₂ (Option G), **F₂** has a higher vapor pressure due to weaker Van der Waals forces. **Answer: F₂ (G)**.

Higher Dipole Moment in Pairs (i)

Dipole moment relates to the difference in electronegativities. In HF vs. F₂ (Option G), **HF** has a dipole moment while F₂ does not. Thus, the answer is HF (G).

Dipole-Dipole and Hydrogen Bonding Intermolecular Forces (j)

Hydrogen bonding occurs when hydrogen is bonded to electronegative atoms like N, O, or F. **HF (Option G)** has hydrogen bonding due to the H-F bond. Thus, the answer is HF (G).

Key concepts

Hybridization

Understanding hybridization is essential when discussing the shapes and reactivities of molecules. Hybridization describes the mixing of atomic orbitals to form new, hybridized orbitals. These orbitals are then suited for the pairing of electrons to form chemical bonds in molecules.
In a molecule, hybridization occurs to minimize energy and make the molecule more stable. Common types of hybridization include:

• **sp Hybridization**: This involves the mixing of one s and one p orbital, resulting in two equivalent sp hybrid orbitals. Molecules with linear shapes, such as HCN (Option A in the grid), exhibit sp hybridization. Here, the central atom forms bonds at 180° angles.
• **sp³ Hybridization**: This type includes the mixing of one s and three p orbitals to form four equivalent sp³ hybrid orbitals. This results in a tetrahedral geometry, as seen in CH₄ (methane - Option H), where the central carbon atom has bonds at about 109.5° angles.

Recognizing the type of hybridization helps in predicting the geometry and reactivity of the molecule.

Intermolecular Forces

Intermolecular forces are forces of attraction or repulsion that act between neighboring particles. These forces are weaker compared to intramolecular forces like covalent bonds but significantly affect the physical properties of substances such as boiling points and vapor pressures.
Key types include:

• **Dipole-Dipole Forces**: These occur in polar molecules where partial charges on nearby molecules attract each other. This type of force is present in molecules with a permanent dipole moment, such as HCl.
• **Hydrogen Bonding**: A strong type of dipole-dipole attraction, hydrogen bonds form when hydrogen is covalently bonded to a highly electronegative atom like nitrogen, oxygen, or fluorine. For example, HF (Option G) exhibits hydrogen bonding.
• **London Dispersion Forces (Van der Waals Forces)**: These are present in all molecules, whether polar or nonpolar, and arise due to the movement of electrons creating temporary dipoles. Nonpolar molecules like F₂ (Option G) rely solely on these forces, leading to lower boiling points and higher vapor pressures compared to polar molecules like HF.

Understanding these forces is critical for predicting how substances will interact under different conditions.

Nonpolar Molecules

Nonpolar molecules are molecules in which the electrons are distributed more equally. They lack a positive or negative pole. Symmetry in the molecular shape often leads to nonpolarity, even if individual bonds are polar.
Take CH₄ (methane - Option H) as an example. In methane, the molecule is symmetrical, which means that any dipole moments from the C-H bonds cancel each other out. This makes CH₄ nonpolar despite each C-H bond having a slight polarity due to different electronegativities of carbon and hydrogen.
Key features of nonpolar molecules include:

• A symmetrical geometry, such as tetrahedral in the case of CH₄, which allows equal distribution of charge.
• Relying on London dispersion forces between molecules, as they do not have regions of positive or negative charge that can attract each other strongly.

These characteristics help in understanding why nonpolar molecules behave the way they do in different chemical and physical environments.