Problem 11 - Chemistry and Chemical Reactivity

Question

Draw a Lewis structure for each of the following molecules or ions: (a) \(\mathrm{BrF}_{3}\) (c) \(\mathrm{XeO}_{2} \mathrm{F}_{2}\) (b) \(\mathrm{I}_{3}^{-}\) (d) \(\mathrm{XeF}_{3}^{+}\)

Step-by-Step Solution

Answer

BrF3: T-shaped, XeO2F2: square planar, I3-: linear, XeF3+: trigonal planar.

Step 1: Determine Valence Electrons for BrF3

Br has 7 valence electrons and each F has 7 valence electrons. Therefore, \( ext{BrF}_3 \) has \( 1 \times 7 + 3 \times 7 = 28 \) valence electrons.

Step 2: Arrange Atoms for BrF3

Place Br in the center and three F atoms around it. Connect each F atom to Br with a single bond.

Step 3: Distribute Remaining Electrons on BrF3

Use the remaining electrons to complete the octets of the F atoms first, each requiring 6 more electrons. Place any remaining electrons on the Br atom.

Step 4: Drawing the BrF3 Lewis Structure

After distributing electrons, Br has 10 electrons around it in an expanded octet. The final structure has Br with three single bonds to F and two lone pairs on Br.

Step 5: Determine Valence Electrons for XeO2F2

Xe has 8 valence electrons, each O has 6, and each F has 7. Therefore, \( ext{XeO}_2 ext{F}_2 \) has \( 8 + 2 \times 6 + 2 \times 7 = 34 \) valence electrons.

Step 6: Arrange Atoms for XeO2F2

Place Xe in the center with 2 O atoms and 2 F atoms around it. Connect each with single bonds initially.

Step 7: Form Double Bonds with Oxygen in XeO2F2

Given the need to complete the octet for O, form double bonds with the Xe if necessary. Then, check for the completion of all octets.

Step 8: Distribute Remaining Electrons on XeO2F2

Complete octets of surrounding atoms and place any remaining electrons as lone pairs on Xe to form expanded octet if necessary.

Step 9: Finalize Lewis Structure for XeO2F2

Xe has an expanded octet, bonded to two O atoms with double bonds and two F atoms with single bonds, throughout using lone pairs to complete octets.

Step 10: Determine Valence Electrons for I3^-

Iodine has 7 valence electrons, so \( ext{I}_3^- \) has \( 3 \times 7 + 1 = 22 \) valence electrons (extra for the negative charge).

Step 11: Arrange Atoms for I3^-

Place the central I atom and two terminal I atoms around it with single bonds.

Step 12: Distribute Remaining Electrons on I3^-

Use lone pairs to complete the octets of the outer I atoms, and place the remaining electrons as lone pairs on the central I atom.

Step 13: Finalize Lewis Structure for I3^-

The central I atom has 3 lone pairs, and each terminal I atom holds 3 lone pairs, indicating a linear structure.

Step 14: Determine Valence Electrons for XeF3+

Xe has 8 valence electrons, each F has 7, but note the positive charge minus 1 electron, so \( ext{XeF}_3^+ \) has \( 8 + 3 \times 7 - 1 = 26 \) valence electrons.

Step 15: Arrange Atoms for XeF3+

Place Xe in the center with three F atoms around it, connecting them with single bonds.

Step 16: Distribute Remaining Electrons for XeF3+

Assign lone pairs to F atoms to satisfy their octets, and place any remaining electrons as lone pairs on Xe.

Step 17: Finalize Lewis Structure for XeF3+

In the final structure, Xe might need to hold as many as three lone pairs, implying a trigonal planar geometry.

Key Concepts

Valence Electrons

Valence electrons are the outermost electrons in an atom and are crucial for forming chemical bonds. Knowing the number of valence electrons allows you to predict how atoms will interact in a molecule. For example, Bromine (Br) and Fluorine (F) each have 7 valence electrons, which is why in the molecule BrF\(_3\), a total of 28 valence electrons need to be distributed.

When drawing a Lewis structure, start by counting the total valence electrons for all atoms involved. For ions, adjust the count by adding extra electrons for negative charges or removing electrons for positive charges.

• For \( ext{BrF}_3\), you would count \(1 \times 7 + 3 \times 7 = 28\) electrons.
• For \( ext{I}_3^-\), add an extra electron for the charge, resulting in \(22\) electrons.

This count helps you determine how atoms connect and share electrons to satisfy their octets.

Expanded Octet

Some elements, like Xenon (Xe) and Bromine (Br), can have more than eight electrons around them, forming an expanded octet. This occurs when elements use d orbitals in their outer shell to hold extra electrons.

In molecules like \( ext{BrF}_3\) and \( ext{XeO}_2 ext{F}_2\), the central atom can accommodate more than eight electrons:

• In \( ext{BrF}_3\), Bromine has 10 electrons: three single bonds with F and two lone pairs.
• In \( ext{XeO}_2 ext{F}_2\), Xenon holds 12 electrons, forming bonds with O and F atoms while utilizing lone pairs to complete its structure.

This ability allows these elements to form stable compounds even with non-standard electron configurations.

Molecular Geometry

The 3D arrangement of atoms in a molecule dictates its molecular geometry, which is significant for understanding reactions and properties. With Lewis structures, you determine geometry by counting bonds and lone pairs around the central atom.

The AXE notation is used where A is the central atom, X is the number of surrounding atoms, and E is the number of lone pairs:

• For \( ext{I}_3^-\), with three lone pairs around the central I, the result is a linear shape.
• For \( ext{XeF}_3^+\), the central Xe with lone pairs and bond pairs may adopt a trigonal planar geometry.

The unique 3D arrangement impacts the molecule's interaction with others, such as polarity and molecular attraction.

Lone Pairs

In a Lewis structure, lone pairs or non-bonding pairs are pairs of valence electrons not involved in bonding. These lone pairs affect the shape and reactivity of the molecule.

Many molecules have atoms with lone pairs influencing the geometry:

• In \( ext{BrF}_3\), Bromine has two lone pairs, altering its electronic geometry to a T-shaped form.
• For \( ext{I}_3^-\), each iodine atom holds three lone pairs, resulting in a linear molecular shape.

Lone pairs tend to take up more space than bonding pairs, affecting the bond angles and overall shape. They are essential in determining the final structure and predicting molecular interactions.