Question

Write the Lewis structure for (a) \(\mathrm{BrF}_{3}\) (b) \(\mathrm{I}_{3}^{-}\) (c) \(\mathrm{XeF}_{4}\)

Short answer

BrF3, I3-, and XeF4 have 28, 22, and 36 valence electrons, respectively, and their central atoms hold lone pairs.

Step-by-step solution

Count Valence Electrons for BrF3

Determine the total number of valence electrons for the molecule \( \mathrm{BrF}_3 \). Bromine (\( \mathrm{Br} \)) has 7 valence electrons, and each fluorine (\( \mathrm{F} \)) also has 7. Therefore, for \( \mathrm{BrF}_3 \), the total number of valence electrons is \( 7 + 3\times7 = 28 \) electrons.

Place Atoms and Bonds for BrF3

Place bromine as the central atom and surround it with three fluorine atoms. Draw a single bond between the bromine and each fluorine, using 2 electrons per bond. Therefore, \( 3 \times 2 = 6 \) electrons are used for bonds, leaving \( 28 - 6 = 22 \) electrons.

Distribute Remaining Electrons for BrF3

Distribute the remaining 22 electrons to satisfy the octet rule for the outer atoms (fluorine). Each \( \mathrm{F} \) atom gets 6 more electrons to complete its octet \( 3 \times 6 = 18 \), leaving 4 electrons. Place these 4 electrons (as lone pairs) on the central bromine.

Count Valence Electrons for I3-

Determine the total number of valence electrons for \( \mathrm{I}_3^- \). Each iodine (\( \mathrm{I} \)) has 7 valence electrons, plus 1 for the additional negative charge, resulting in \( 3\times7 + 1 = 22 \) valence electrons.

Place Atoms and Bonds for I3-

Place one iodine atom in the center with single bonds to the other two iodine atoms. This uses 4 electrons (2 per bond), leaving \( 22 - 4 = 18 \) electrons.

Distribute Remaining Electrons for I3-

Distribute the remaining electrons to complete the octets around the outer iodine atoms \( \mathrm{I} \). Provide 6 electrons per outer iodine \( 2\times6 = 12 \), leaving 6 electrons. Place these 6 electrons on the central iodine as lone pairs.

Count Valence Electrons for XeF4

Determine the total number of valence electrons for \( \mathrm{XeF}_4 \). Xenon (\( \mathrm{Xe} \)) has 8 valence electrons, and each fluorine has 7. Thus, \( 8 + 4\times7 = 36 \) valence electrons.

Place Atoms and Bonds for XeF4

Place xenon as the central atom, surrounded by the four fluorine atoms, using single bonds. This uses \( 4\times2 = 8 \) electrons, leaving \( 36 - 8 = 28 \) electrons.

Distribute Remaining Electrons for XeF4

Distribute the remaining electrons to the outer fluorine atoms, providing each \( \mathrm{F} \) with 6 electrons \( 4\times6 = 24 \), leaving 4 electrons. Place these 4 electrons as two lone pairs on the central xenon.

Key concepts

Valence Electrons

Valence electrons are the outermost electrons in an atom and are crucial in determining how an atom will bond with others. They are responsible for the chemical properties of an element. Knowing the number of valence electrons helps us predict the type of bonds an atom will make. For instance:

• Bromine ( \( \mathrm{Br} \)) has 7 valence electrons.
• Fluorine ( \( \mathrm{F} \)) also has 7 valence electrons.
• Iodine ( \( \mathrm{I} \)) contains 7 valence electrons as well.
• Xenon ( \( \mathrm{Xe} \)) possesses 8 valence electrons.

Understanding these numbers allows us to accurately distribute them around the atoms in molecules like \( \mathrm{BrF}_3 \), \( \mathrm{I}_3^- \), and \( \mathrm{XeF}_4 \).
This is the first step in drawing a Lewis structure as it indicates how many total electrons need to be placed among the atoms.

Octet Rule

The octet rule posits that atoms strive to have eight valence electrons in their outer shell. This is known as a complete octet, resembling the electron configuration of noble gases, which are stable. In practice, this rule guides the electron distribution around atoms in a molecular formula.
For example, in the molecule \( \mathrm{BrF}_3 \), each fluorine needs 6 additional electrons to complete its octet beyond the shared bonding pair. Similarly, in \( \mathrm{I}_3^- \), each iodine at the sides requires 6 extra electrons aside from their bond with the central iodine.

• Outer atoms generally follow the octet rule strictly.
• Central atoms, especially larger ones like bromine or iodine, can accommodate more than eight electrons if needed, following the rules of expanded octet.
• Exceptions can occur, particularly in molecules involving transition metals or compounds with odd numbers of electrons.

Using the octet rule, we ensure that these molecules are stable by fulfilling the electron requirements for each atom involved.

Molecular Geometry

Molecular geometry refers to the three-dimensional arrangement of atoms in a molecule. It greatly influences the molecule's properties, including its reactivity, polarity, and biological activity. Understanding molecular geometry requires considering:

• How electron pairs, both bonding and non-bonding (lone pairs), arrange themselves to minimize repulsion.
• Shapes that common molecular geometries include linear, trigonal planar, tetrahedral, trigonal bipyramidal, and octahedral.

For example, \( \mathrm{BrF}_3 \) has a T-shaped geometry due to three bonded pairs and two lone pairs on bromine, which lead to electron pair repulsions that shape the molecule. \( \mathrm{XeF}_4 \) is square planar with two lone pairs opposing each other and balancing the shape symmetrically around xenon.
These geometries arise because molecules adjust to positions that create the least electron pair repulsion, following the VSEPR (Valence Shell Electron Pair Repulsion) theory, making them crucial for predicting spatial arrangements.