/*! 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 29 (a) Explain why \(\mathrm{BrF}_{... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 29

(a) Explain why \(\mathrm{BrF}_{4}^{-}\) is square planar, whereas \(\mathrm{BF}_{4}^{-}\) is tetrahedral. (b) Water, \(\mathrm{H}_{2} \mathrm{O}\), is a bent molecule. Predict the shape of the molecular ion formed from the water molecule if you were able to remove four electrons to make \(\left(\mathrm{H}_{2} \mathrm{O}\right)^{4+}\).

Short Answer

Expert verified
(a) \(\mathrm{BrF}_{4}^{-}\) is square planar because it has 6 electron pairs with 2 lone pairs, resulting in the remaining 4 electron pairs forming a square plane to minimize repulsion. \(\mathrm{BF}_{4}^{-}\) is tetrahedral because it has 4 electron pairs with no lone pairs, resulting in a tetrahedral geometry that minimizes repulsions between electron pairs. (b) The molecular ion \(\left(\mathrm{H}_{2} \mathrm{O}\right)^{4+}\) will have a linear shape because when 4 electrons are removed, it results in no lone pairs on the central Oxygen atom and retains the 2 covalent bonds with Hydrogen atoms, giving a linear geometry that minimizes repulsions between the 2 electron pairs.

Step by step solution

01

Understand VSEPR Theory

VSEPR theory stands for Valence Shell Electron Pair Repulsion theory, which predicts the shapes of molecules based on the fact that electron pairs in the valence shell of an atom repel each other and will arrange themselves in such a way to minimize these repulsive forces. Electrons in chemical bonds, like single, double, and triple bonds as well as lone pairs, are counted as electron pairs.
02

Identify the Central Atom and Valence Electron Pairs for BrF4- and BF4-

First, we need to identify the central atom and the surrounding atoms for both \(\mathrm{BrF}_{4}^{-}\) and \(\mathrm{BF}_{4}^{-}\). For \(\mathrm{BrF}_{4}^{-}\), Br (Bromine) is the central atom, and it is surrounded by 4 F (Fluorine) atoms. For \(\mathrm{BF}_{4}^{-}\), B (Boron) is the central atom, and it is surrounded by 4 F (Fluorine) atoms. Next, we need to count the valence electron pairs for both molecules. For \(\mathrm{BrF}_{4}^{-}\), Br has a total of 7 valence electrons. It forms four single bonds with F atoms, which consist of 4 electron pairs. Since it has a negative charge, this means there is an additional electron, which is an additional electron pair. Therefore, Br has a total of 6 electron pairs, and 2 of them are lone pairs. For \(\mathrm{BF}_{4}^{-}\), B has a total of 3 valence electrons. It forms four single bonds with F atoms, which consist of 4 electron pairs. Since it has a negative charge, this means there is also an additional electron. However, this additional electron is not an additional lone pair. It is rather a contribution from an F atom to have an octet electron configuration. Therefore, B has a total of 4 electron pairs, and none of them are lone pairs.
03

Apply VSEPR Theory to BrF4- and BF4-

Applying VSEPR theory to \(\mathrm{BrF}_{4}^{-}\), with 6 electron pairs (with 2 lone pairs), it will have a square planar shape. This is because, for 6 electron pairs, an octahedral geometry is expected, but since there are two lone pairs, the remaining 4 electron pairs will form a square plane to minimize the repulsion between the two lone pairs. For \(\mathrm{BF}_{4}^{-}\), with 4 electron pairs (with no lone pairs), it will have a tetrahedral shape. Tetrahedral geometry is expected for 4 electron pairs, as it minimizes the repulsions between them.
04

Determine the Shape of (H2O)4+ using VSEPR Theory

First, consider the \(\mathrm{H}_{2} \mathrm{O}\) molecule. Oxygen is the central atom, and it forms two single bonds with Hydrogen atoms. Oxygen also has 2 lone pairs of electrons. When 4 electrons are removed to form \(\left(\mathrm{H}_{2} \mathrm{O}\right)^{4+}\), the resulting molecule has no lone pairs on the central Oxygen atom but retains the 2 covalent bonds with the Hydrogen atoms. Applying VSEPR theory to \(\left(\mathrm{H}_{2} \mathrm{O}\right)^{4+}\), with 2 electron pairs (with no lone pairs), it will have a linear shape. This is because, for 2 electron pairs, a linear geometry is expected, as it minimizes the repulsions between them.

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.

Molecular Geometry
Understanding molecular geometry is crucial for grasping how molecules form and behave. It refers to the three-dimensional arrangement of atoms within a molecule.

The molecular geometry is determined by the bonding and non-bonding electron pairs located in the outermost, or valence, shell of the central atom in a molecule. Electrons have negative charges and, like magnets with the same poles, they repel each other. This repulsion causes electron pairs to arrange themselves as far apart as possible within a molecule, creating specific geometric shapes.

For instance, in the case of \(\mathrm{BF}_{4}^{-}\), the central boron atom forms a tetrahedral shape as it is surrounded by four bonding pairs of electrons with no lone pairs, leading to an arrangement where the repulsion is minimized and the molecules form an equal distance from each other. The VSEPR theory, Valence Shell Electron Pair Repulsion theory, guides us in predicting these molecular structures.
Electron Pair Repulsion
Electron pair repulsion is a fundamental concept in chemistry that helps explain the shapes of molecules. According to this principle, electron pairs around a central atom position themselves as far apart from each other as possible to minimize the repulsive forces between them. Lone pairs, or non-bonding electrons, also contribute to this phenomenon.

Electron pair repulsion impacts molecular geometry significantly. For example, in \(\mathrm{BrF}_{4}^{-}\), bromine is surrounded by six pairs of electrons 鈥� four bonding pairs with fluorine and two lone pairs. The VSEPR theory predicts that to minimize repulsion, the molecule adopts a square planar geometry since lone pairs exert a greater repulsive force and need more space. This differs from \(\mathrm{BF}_{4}^{-}\), which lacks lone pairs and thus is tetrahedral, illustrating the importance of considering both bonding and non-bonding electron pairs in determining molecular shapes.
Chemical Bonding
Chemical bonding is the force that holds atoms together within molecules, forming as a result of interactions between electrons, specifically valence electrons. The three primary types of chemical bonds are ionic, covalent, and metallic bonds.

Covalent bonding is where atoms share pairs of electrons, and it highly influences molecular geometry. In our exercise, the covalently bonded molecules like \(\mathrm{BrF}_{4}^{-}\) and \(\mathrm{BF}_{4}^{-}\), as well as \(\mathrm{H}_{2}\mathrm{O}\), demonstrate different geometries based on the number of shared electron pairs (bonding pairs) and lone pairs.

As electron pairs are removed or added, like in \(\mathrm{H}_{2}\mathrm{O}^{4+}\), the molecule's geometry can change drastically. Upon removal of four electrons, the \(\mathrm{H}_{2}\mathrm{O}\) molecule would no longer have lone pairs and would result in a linear shape molecule, showing the direct relation between electron configuration and molecular structure.

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

The molecule shown here is difluoromethane \(\left(\mathrm{CH}_{2} \mathrm{~F}_{2}\right)\), which is used as a refrigerant called \(\mathrm{R}\) -32. (a) Based on the structure, how many electron domains surround the \(C\) atom in this molecule? (b) Would the molecule have a nonzero dipole moment? (c) If the molecule is polar, in what direction will the overall dipole moment vector point in the molecule? [Sections \(9.2\) and 9.3]

You can think of the bonding in the \(\mathrm{Cl}_{2}\) molecule in several ways. For example, you can picturethe Cl- -Cl bond containing two electrons that each come from the \(3 p\) orbitals of a \(\mathrm{Cl}\) atom that are pointing in the appropriate direction. However, you can also think about hybrid orbitals. (a) Draw the Lewis structure of the \(\mathrm{Cl}_{2}\) molecule. (b) What is the hybridization of each Cl atom? (c) What kind of orbital overlap, in this view, makes the \(\mathrm{Cl}-\mathrm{Cl}\) bond? (d) Imagine if you could measure the positions of the lone pairs of electrons in \(\mathrm{Cl}_{2}\). How would you distinguish between the atomic orbital and hybrid orbital models of bonding using that knowledge? (e) You can also treat \(\mathrm{Cl}_{2}\) using molecular orbital theory to obtain an energy level diagram similar to that for \(\mathrm{F}_{2}\). Design an experiment that could tell you if the MO picture of \(\mathrm{Cl}_{2}\) is the best one, assuming you could easily measure bond lengths, bond energies, and the light absorption properties for any ionized species.

The cyclopentadienide ion has the formula \(\mathrm{C}_{5} \mathrm{H}_{5}^{-}\). The ion consists of a regular pentagon of \(\mathrm{C}\) atoms, each bonded to two \(\mathrm{C}\) neighbors, with a hydrogen atom bonded to each \(\mathrm{C}\) atom. All the atoms lie in the same plane. (a) Draw a Lewis structure for the ion. According to your structure, do all five \(\mathrm{C}\) atoms have the same hybridization? Explain. (b) Chemists generally view this ion as having \(s p^{2}\) hybridization at each \(C\) atom. Is that view consistent with your answer to part (a)? (c) Your Lewis structure should show one nonbonding pair of electrons. Under the assumption of part (b), in what type of orbital must this nonbonding pair reside? (d) Are there resonance structures equivalent to the Lewis structure you drew in part (a)? If so, how many? (e) The ion is often drawn as a pentagon enclosing a circle. Is this representation consistent with your answer to part (d)? Explain. (f) Both benzene and the cyclopentadienide ion are often described as systems containing six \(\pi\) electrons. What do you think is meant by this description?

(a) What is the probability of finding an electron on the internuclear axis if the electron occupies a \(\pi\) molecular orbital? (b) For a homonuclear diatomic molecule, what similarities and differences are there between the \(\pi_{2 p}\) MO made from the \(2 p_{x}\) atomic orbitals and the \(\pi_{2 p}\) MO made from the \(2 p_{y}\) atomic orbitals? (c) Why are the \(\pi_{2 p}\) MOs lower in energy than the \(\pi_{2 p}^{*}\) MOs?

(a) Starting with the orbital diagram of a boron atom, describe the steps needed to construct hybrid orbitals appropriate to describe the bonding in \(\mathrm{BF}_{3}\) (b) What is the name given to the hybrid orbitals constructed in (a)? (c) Sketch the large lobes of the hybrid orbitals constructed in part (a). (d) Are there any valence atomic orbitals of B that are left unhybridized? If so, how are they oriented relative to the hybrid orbitals?
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Making Measurements

Read Explanation

Chemistry Branches

Read Explanation

Kinetics

Read Explanation

Chemical Reactions

Read Explanation

Chemical Analysis

Read Explanation
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.