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:ד: sp 32 O sp d 3 O sp d 3 sp 2 : F: sp B. F:

:ד: sp 32 O sp d 3 O sp d 3 sp 2 : F: sp B. F:

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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### Lewis Structure and Hybridization of BF₃ (Boron Trifluoride) #### Lewis Structure of BF₃ The Lewis structure of boron trifluoride (BF₃) is depicted as follows: - Boron (B) is at the center with three single bonds connecting to three fluorine (F) atoms. - Each fluorine atom has three pairs of lone electrons. - Boron has no lone pairs. ``` .. .. :F: :F: | | B--------F .. ``` Each fluorine atom conforms to the octet rule as it has eight electrons around it (six from lone pairs and two from the bond with boron). #### Hybridization The structure implies the central boron atom uses hybridized orbitals to form bonds with fluorine atoms. ##### Multiple Choice Question: What is the hybridization of the boron atom in BF₃? - ⭕ sp - ⭕ sp² - ⭕ sp³d - ⭕ sp³ - ⭕ sp²d The correct answer is **sp² hybridization**. In BF₃, boron forms three equivalently spaced sp² hybrid orbitals, which coordinate with the 2p orbitals of fluorine atoms. This results in a trigonal planar molecular shape with 120° bond angles.
Transcribed Image Text:### Lewis Structure and Hybridization of BF₃ (Boron Trifluoride) #### Lewis Structure of BF₃ The Lewis structure of boron trifluoride (BF₃) is depicted as follows: - Boron (B) is at the center with three single bonds connecting to three fluorine (F) atoms. - Each fluorine atom has three pairs of lone electrons. - Boron has no lone pairs. ``` .. .. :F: :F: | | B--------F .. ``` Each fluorine atom conforms to the octet rule as it has eight electrons around it (six from lone pairs and two from the bond with boron). #### Hybridization The structure implies the central boron atom uses hybridized orbitals to form bonds with fluorine atoms. ##### Multiple Choice Question: What is the hybridization of the boron atom in BF₃? - ⭕ sp - ⭕ sp² - ⭕ sp³d - ⭕ sp³ - ⭕ sp²d The correct answer is **sp² hybridization**. In BF₃, boron forms three equivalently spaced sp² hybrid orbitals, which coordinate with the 2p orbitals of fluorine atoms. This results in a trigonal planar molecular shape with 120° bond angles.
**Title: Understanding Hybridization in Simple Molecules** **Introduction** Hybridization is a concept used in chemistry to describe the mixing of atomic orbitals in atoms to form new hybrid orbitals. These hybrid orbitals are used to form chemical bonds in molecules. Identifying the hybridization of the central atom in a molecule provides insights into the molecule’s shape, bond angles, and bonding properties. **Problem Statement** For each molecule shown here, indicate the hybridization of the central atom. **Molecular Structures** 1. **Methane (CH₄) Structure:** - Central Atom: Carbon (C) - Attached Atoms: Four Hydrogen (H) **Options for Hybridization:** - \( sp^2 \) - \( sp \) - \( sp^3 \) - \( sp^3d \) - \( sp^3d^2 \) 2. **Ammonia (NH₃) Structure:** - Central Atom: Nitrogen (N) - Attached Atoms: Three Hydrogen (H) with one lone pair on Nitrogen **Options for Hybridization:** - \( sp \) - \( sp^3 \) - \( sp^3d \) - \( sp^2 \) - \( sp^3d^2 \) **Graphs/Diagrams Explanation** - The molecules are depicted as Lewis structures. - In the methane structure, the carbon atom (C) is at the center with single bonds to four hydrogen atoms (H), making it symmetrical and forming a tetrahedral shape. - In the ammonia structure, the nitrogen atom (N) is at the center with single bonds to three hydrogen atoms (H) and a pair of lone electrons above the nitrogen, forming a trigonal pyramidal shape due to the lone pair. **Objective** To identify the correct hybridization state of the central atom in each molecule, you need to consider the number of sigma bonds and lone pairs around the central atom. **Learning Outcome** Upon completion, you should be able to determine the hybridization of the central atom in various molecules, enhancing your understanding of molecular geometry and bonding. **Exercise** For each molecule, select the correct hybridization option for the central atom from the provided choices. This exercise will not only help in reviewing hybridization concepts but also in visualizing the molecular shapes based on their hybrid states. --- This section is
Transcribed Image Text:**Title: Understanding Hybridization in Simple Molecules** **Introduction** Hybridization is a concept used in chemistry to describe the mixing of atomic orbitals in atoms to form new hybrid orbitals. These hybrid orbitals are used to form chemical bonds in molecules. Identifying the hybridization of the central atom in a molecule provides insights into the molecule’s shape, bond angles, and bonding properties. **Problem Statement** For each molecule shown here, indicate the hybridization of the central atom. **Molecular Structures** 1. **Methane (CH₄) Structure:** - Central Atom: Carbon (C) - Attached Atoms: Four Hydrogen (H) **Options for Hybridization:** - \( sp^2 \) - \( sp \) - \( sp^3 \) - \( sp^3d \) - \( sp^3d^2 \) 2. **Ammonia (NH₃) Structure:** - Central Atom: Nitrogen (N) - Attached Atoms: Three Hydrogen (H) with one lone pair on Nitrogen **Options for Hybridization:** - \( sp \) - \( sp^3 \) - \( sp^3d \) - \( sp^2 \) - \( sp^3d^2 \) **Graphs/Diagrams Explanation** - The molecules are depicted as Lewis structures. - In the methane structure, the carbon atom (C) is at the center with single bonds to four hydrogen atoms (H), making it symmetrical and forming a tetrahedral shape. - In the ammonia structure, the nitrogen atom (N) is at the center with single bonds to three hydrogen atoms (H) and a pair of lone electrons above the nitrogen, forming a trigonal pyramidal shape due to the lone pair. **Objective** To identify the correct hybridization state of the central atom in each molecule, you need to consider the number of sigma bonds and lone pairs around the central atom. **Learning Outcome** Upon completion, you should be able to determine the hybridization of the central atom in various molecules, enhancing your understanding of molecular geometry and bonding. **Exercise** For each molecule, select the correct hybridization option for the central atom from the provided choices. This exercise will not only help in reviewing hybridization concepts but also in visualizing the molecular shapes based on their hybrid states. --- This section is
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