/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 54 What hybrid orbitals would be ex... [FREE SOLUTION] | 91影视

91影视

What hybrid orbitals would be expected for the central atom in each of the following? a. \(\mathrm{BrF}_{3}\) b. \(\mathrm{TeF}_{4}\) c. \(\mathrm{XeF}_{4}\) d. \(\mathrm{IF}_{4}^{-}\)

Short Answer

Expert verified
BrF鈧 has sp鲁d, TeF鈧 has sp鲁d, XeF鈧 has sp鲁d虏, and IF鈧勨伝 has sp鲁d虏 hybridization.

Step by step solution

01

Understanding Hybridization Concept

Hybridization is a concept in chemistry where atomic orbitals mix to form new hybrid orbitals. These hybrid orbitals influence molecular geometry and bonding properties. The type of hybridization can be determined using the steric number, which is the number of atoms bonded to the central atom plus the number of lone pairs on the central atom.
02

Determining Steric Number for BrF鈧

For BrF鈧, bromine is the central atom. It is surrounded by 3 fluorine atoms and has 2 lone pairs. The steric number is 3 (bonded atoms) + 2 (lone pairs) = 5. A steric number of 5 corresponds to sp鲁d hybridization.
03

Determining Steric Number for TeF鈧

In TeF鈧, tellurium (Te) is the central atom. It forms bonds with 4 fluorine atoms and has 1 lone pair. The steric number is 4 (bonded atoms) + 1 (lone pair) = 5. Hence, the hybridization is sp鲁d.
04

Determining Steric Number for XeF鈧

For XeF鈧, xenon (Xe) is the central atom. It is bonded to 4 fluorine atoms and has 2 lone pairs. The steric number is 4 (bonded atoms) + 2 (lone pairs) = 6, which suggests an sp鲁d虏 hybridization.
05

Determining Steric Number for IF鈧勨伝

In IF鈧勨伝, iodine (I) is the central atom. It bonds with 4 fluorine atoms and has 2 lone pairs. The charge does not affect the hybridization calculation. The steric number is 4 (bonded atoms) + 2 (lone pairs) = 6, thus indicating an sp鲁d虏 hybridization.
06

Summary of Results

Based on the steric numbers calculated, the hybridizations for the central atoms are as follows: BrF鈧 (sp鲁d), TeF鈧 (sp鲁d), XeF鈧 (sp鲁d虏), and IF鈧勨伝 (sp鲁d虏).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Steric Number
The concept of the steric number is a foundational idea in understanding hybridization. It helps predict the type of hybrid orbitals formed around a central atom. To determine the steric number, count all atoms bonded to the central atom and add the number of lone pairs present on the central atom. For example, in the molecule \(\mathrm{BrF}_{3}\), bromine is the central atom. It has three atoms attached (three fluorine atoms) and two lone pairs of electrons. Therefore, its steric number is \(3 + 2 = 5\). A steric number of five suggests an \(\mathrm{sp}^{3}\mathrm{d}\) hybridization. This approach helps in determining the hybridization across different molecules, essential for predicting molecular geometry and bonding properties.
Molecular Geometry
Molecular geometry is the three-dimensional arrangement of atoms in a molecule. It plays a crucial role in determining the molecule's physical and chemical properties. By using the steric number, you can predict the molecular geometry of a compound. For instance, a steric number of five typically leads to a trigonal bipyramidal geometry if there are no lone pairs. However, with lone pairs present, such as in \(\mathrm{BrF}_{3}\), the geometry slightly changes to a T-shaped structure. Other common geometries include tetrahedral, square planar, and octahedral, often determined by different combinations of bonded atoms and lone pairs. Understanding these shapes helps in visualizing how molecules interact with each other.
Bonding Properties
Bonding properties of a molecule are significantly influenced by its hybridization. The type of hybrid orbitals formed determines how atoms bond and the strength of these bonds. For example, \(\mathrm{sp}^{3}\mathrm{d}\) hybridization involves one \(s\), three \(p\), and one \(d\) orbital mixing to form five hybrid orbitals of equal energy. This configuration is present in molecules such as \(\mathrm{BrF}_{3}\) and \(\mathrm{TeF}_{4}\). When the central atom undergoes \(\mathrm{sp}^{3}\mathrm{d}^{2}\) hybridization, as in \(\mathrm{XeF}_{4}\) or \(\mathrm{IF}_{4}^{-}\), two \(d\) orbitals are involved, allowing for an octahedral configuration that influences bond angles and the polarity of the molecule. Recognizing these bonding characteristics is crucial for explaining chemical reactivity and interactions between different substances.
Central Atom
The central atom of a molecule is usually the atom bonded to the greatest number of other atoms. It plays a pivotal role in determining the hybridization and subsequently the molecular geometry and bonding properties. For example, in \(\mathrm{BrF}_{3}\), bromine serves as the central atom. It is the site of hybridization because it connects to three fluorine atoms and contains electron pairs that affect the shape and interactions of the molecule. Choosing the correct central atom is essential in examining the hybridization because it directly influences the calculation of the steric number, which then impacts the entire analysis of the molecular properties. The central atom, typically the least electronegative and most willing to share or acquire electrons, orchestrates the spatial configuration of the molecule.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.