Problem 26 - Chemistry: The Central Science

Question

Draw the Lewis structure for each of the following molecules or ions, and predict their electron-domain and molecular geometries: (a) \(\mathrm{AsF}_{3}\), (b) \(\mathrm{CH}_{3}^{+}\), (c) \(\mathrm{BrF}_{3}\), (d) \(\mathrm{ClO}_{3}^{-}\), (e) \(\mathrm{XeF}_{2}\), (f) \(\mathrm{BrO}_{2}^{-}\).

Step-by-Step Solution

Answer

The Lewis structures and molecular geometries for the given molecules and ions are as follows: (a) AsF3: Lewis structure with As as central atom, connected to three F atoms. Electron-domain geometry is trigonal planar, with molecular geometry also being trigonal planar. (b) CH3+: Lewis structure with C as central atom, connected to three H atoms. Electron-domain geometry is trigonal planar, with molecular geometry also being trigonal planar. (c) BrF3: Lewis structure with Br as central atom, connected to three F atoms and two lone pairs. Electron-domain geometry is trigonal bipyramidal, with molecular geometry being T-shaped. (d) ClO3-: Lewis structure with Cl as central atom, connected to three O atoms and one lone pair. Electron-domain geometry is tetrahedral, with molecular geometry being trigonal pyramidal. (e) XeF2: Lewis structure with Xe as central atom, connected to two F atoms and three lone pairs. Electron-domain geometry is linear, with molecular geometry also being linear. (f) BrO2-: Lewis structure with Br as central atom, connected to two O atoms and two lone pairs. Electron-domain geometry is tetrahedral, with molecular geometry being bent or V-shaped.

Step 1: (a) AsF3 Lewis Structure and Geometry

1. Identify the central atom: As (Arsenic) is the central atom in AsF3. 2. Determine the total number of valence electrons: As has 5 valence electrons, and each F (Fluorine) has 7. So, 5 + (3 × 7) = 26 valence electrons are available. 3. Distribute the valence electrons: Connect As to the three F atoms with single bonds (2 electrons per bond). With 6 electrons used, 20 electrons are left. Add lone pairs to each F atom (6 electrons per F). No electrons are left, and the octet rule is satisfied for all atoms. 4. Identify the electron-domain geometry: As has 3 bond domains and no lone pairs, so it has a trigonal planar electron-domain geometry. 5. Identify the molecular geometry: As there are no lone pairs on the central atom, the molecular geometry is also trigonal planar.

Step 2: (b) CH3+ Lewis Structure and Geometry

1. Identify the central atom: C (Carbon) is the central atom in CH3+. 2. Determine the total number of valence electrons: C has 4 valence electrons, and each H (Hydrogen) has 1. So, 4 + (3 × 1) = 7 valence electrons are available. However, as it is a positively charged ion, we subtract one electron, leaving 6 valence electrons. 3. Distribute the valence electrons: Connect C to the three H atoms with single bonds (2 electrons per bond). With 6 electrons used, 0 electrons are left. The octet rule is satisfied for H, but not for C, which is normal for positively charged ions like this one. 4. Identify the electron-domain geometry: C has 3 bond domains and no lone pairs, so it has a trigonal planar electron-domain geometry. 5. Identify the molecular geometry: As there are no lone pairs on the central atom, the molecular geometry is also trigonal planar.

Step 3: (c) BrF3 Lewis Structure and Geometry

1. Identify the central atom: Br (Bromine) is the central atom in BrF3. 2. Determine the total number of valence electrons: Br has 7 valence electrons, and each F (Fluorine) has 7. So, 7 + (3 × 7) = 28 valence electrons are available. 3. Distribute the valence electrons: Connect Br to the three F atoms with single bonds (2 electrons per bond). With 6 electrons used, 22 electrons are left. Add lone pairs to each F atom (6 electrons per F) and 2 lone pairs on Br. No electrons are left, and the octet rule is satisfied for all atoms. 4. Identify the electron-domain geometry: Br has 3 bond domains and 2 lone pairs, so it has a trigonal bipyramidal electron-domain geometry. 5. Identify the molecular geometry: As there are 2 lone pairs on the central atom, the molecular geometry is T-shaped.

Step 4: (d) ClO3- Lewis Structure and Geometry

1. Identify the central atom: Cl (Chlorine) is the central atom in ClO3-. 2. Determine the total number of valence electrons: Cl has 7 valence electrons, and each O (Oxygen) has 6. So, 7 + (3 × 6) = 25 valence electrons are available. However, as it is a negatively charged ion, we add one electron, leaving 26 valence electrons. 3. Distribute the valence electrons: Connect Cl to the three O atoms with single bonds (2 electrons per bond). With 6 electrons used, 20 electrons are left. Add lone pairs to each O atom (6 electrons per O) and 1 lone pair on Cl. No electrons are left, and the octet rule is satisfied for all atoms. 4. Identify the electron-domain geometry: Cl has 3 bond domains and 1 lone pair, so it has a tetrahedral electron-domain geometry. 5. Identify the molecular geometry: As there is 1 lone pair on the central atom, the molecular geometry is trigonal pyramidal.

Step 5: (e) XeF2 Lewis Structure and Geometry

1. Identify the central atom: Xe (Xenon) is the central atom in XeF2. 2. Determine the total number of valence electrons: Xe has 8 valence electrons, and each F (Fluorine) has 7. So, 8 + (2 × 7) = 22 valence electrons are available. 3. Distribute the valence electrons: Connect Xe to the two F atoms with single bonds (2 electrons per bond). With 4 electrons used, 18 electrons are left. Add lone pairs to each F atom (6 electrons per F) and 3 lone pairs on Xe. No electrons are left, and the octet rule is satisfied for all atoms. 4. Identify the electron-domain geometry: Xe has 2 bond domains and 3 lone pairs, so it has a linear electron-domain geometry. 5. Identify the molecular geometry: As there are 3 lone pairs on the central atom, the molecular geometry is also linear.

Step 6: (f) BrO2- Lewis Structure and Geometry

1. Identify the central atom: Br (Bromine) is the central atom in BrO2-. 2. Determine the total number of valence electrons: Br has 7 valence electrons, and each O (Oxygen) has 6. So, 7 + (2 × 6) = 19 valence electrons are available. However, as it is a negatively charged ion, we add one electron, leaving 20 valence electrons. 3. Distribute the valence electrons: Connect Br to the two O atoms with single bonds (2 electrons per bond). With 4 electrons used, 16 electrons are left. Add lone pairs to each O atom (6 electrons per O) and 2 lone pairs on Br. No electrons are left, and the octet rule is satisfied for all atoms. 4. Identify the electron-domain geometry: Br has 2 bond domains and 2 lone pairs, so it has a tetrahedral electron-domain geometry. 5. Identify the molecular geometry: As there are 2 lone pairs on the central atom, the molecular geometry is bent or V-shaped.

Key Concepts

Electron-Domain Geometry

In the world of chemistry, understanding electron-domain geometry is essential in predicting molecular shapes. This concept refers to the arrangement of electron domains around a central atom. Electron domains include both bonding and non-bonding (lone) pairs of electrons.

For instance, in the molecule \(\text{AsF}_3\), arsenic (As) is the central atom surrounded by three single bonds with fluorine (F), creating three bonding domains. With no lone pairs, the geometry is trigonal planar. In contrast, for \(\text{BrF}_3\), bromine (Br) has three bond domains and two lone pairs, resulting in a trigonal bipyramidal arrangement.

The geometry is crucial as it gives insight into molecular interactions and properties. Here's how it usually works:

• Linear: 2 domains
• Trigonal Planar: 3 domains
• Tetrahedral: 4 domains
• Trigonal Bipyramidal: 5 domains
• Octahedral: 6 domains

Molecular Geometry

Molecular geometry is all about the shape of a molecule based on its atom arrangements. While electron-domain geometry considers both bonds and lone pairs, molecular geometry looks at the position of atoms only.

Take \(\text{BrF}_3\) as an example. Its electron-domain geometry is trigonal bipyramidal, but the placement of atoms makes it T-shaped due to the presence of two lone pairs. Similarly, \(\text{CH}_3^+\) has a trigonal planar molecular geometry as there are no lone pairs altering the shape from the electron-domain arrangement.

Understanding molecular geometry helps in predicting how a molecule will interact with others, impacting properties like polarity and reactivity.

• Bent or V-Shaped: typically due to lone pairs
• Trigonal Planar: flat triangular shape
• T-Shaped: influenced by lone pairs on a trigonal bipyramidal base
• Tetrahedral: a pyramid with triangular faces
• Linear: a straight line of atoms

Valence Electrons

Valence electrons are the outermost electrons of an atom and are key players in chemical bonding. They determine the chemical properties of an element. Recognizing valence electrons is crucial in building Lewis structures and predicting the behavior of molecules.

For example, arsenic in \(\text{AsF}_3\) has 5 valence electrons, while each fluorine provides 7. Thus, the total valence electrons count for \(\text{AsF}_3\) is 26. Each hydrogen in \(\text{CH}_3^+\) offers just 1 valence electron, totaling to 6 after removing one for its positive charge.

When you draw Lewis structures, you need to:

• Count the total valence electrons
• Use pairs of electrons to form bonds
• Satisfy the octet rule (or duet for hydrogen) for each atom

Understanding valence electrons helps design molecules and anticipate their stability and interactions.

Octet Rule

In chemistry, the octet rule is a guideline stating that atoms strive to have eight electrons in their valence shell, mirroring the stable configuration of noble gases. This rule is pivotal in understanding molecule formation and stability.

For example, in \(\text{ClO}_3^-\), chlorine (Cl) follows the octet rule by sharing electrons with three oxygen (O) atoms, utilizing additional electrons from lone pairs. Similarly, in \(\text{BrO}_2^-\), bromine (Br) achieves an octet rule compliance by forming bonds with oxygen and having lone pairs.

Keep in mind:

• Many atoms strive to fulfill the octet rule.
• Hydrogen is an exception, as it completes its shell with two electrons (duet rule).
• Molecules with more or fewer electrons than the octet may be more reactive.

This rule aids in predicting and understanding each atom's involvement in a molecule's hypothetical structure and behavior.