The species \(\mathrm{PBr}_{4}^{-}\) has been synthesized and has been described as a tetrahedral anion. Comment on this description.

Short Answer

Expert verified
The description of the \(\mathrm{PBr}_{4}^{-}\) ion as a tetrahedral anion is accurate, considering the VSEPR theory. This theory predicts that the compound has a tetrahedral geometry due to its distribution of electrons.

Step by step solution

01

Identify the number of valence electrons

Identify the total number of valence electrons. Phosphorus (P) is in Group 5A, so it has 5 valence electrons, bromine (Br) is in Group 7A, so it has 7 valence electrons. As there are 4 bromine atoms, so we have a total of \(5+4*7=33\) valence electrons. However, the ion \(\mathrm{PBr}_{4}^{-}\) carries an extra negative charge which indicates an additional electron. Therefore, the total number of valence electrons is 34.
02

Setup the Lewis Structure

First, we place the least electronegative atom (P) in the center. Each of the four Bromine (Br) atoms is connected to the central Phosphorus (P) atom. We then distribute the remaining electrons as lone pairs on the surrounding atoms.
03

Comment on the description

The compound has one central atom (Phosphorus, P), surrounded by four other atoms (Bromine, Br). According to the VSEPR theory, such a configuration corresponds to a tetrahedral geometry. Therefore, the description of the \(\mathrm{PBr}_{4}^{-}\) ion as a tetrahedral anion is accurate.

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 Vaia!

Key Concepts

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

Valence Electrons
Valence electrons are the outermost electrons of an atom and are crucial in determining its chemical bonding behavior. These electrons can be shared, gained, or lost in the formation of chemical bonds. The number of valence electrons is the primary factor in determining how elements react with one another.

In our exercise, the focus is on the tetrahedral anion \(\mathrm{PBr}_{4}^{-}\), which is formed by phosphorus and bromine atoms. The phosphorus atom, located in Group 5A of the periodic table, has five valence electrons, while each bromine atom, found in Group 7A, has seven valence electrons. When combined, the \(\mathrm{PBr}_{4}^{-}\) anion overall has 34 valence electrons, taking into account the additional electron contributed by the negative charge of the anion.

Understanding valence electrons enables students to predict and draw the Lewis structure of molecules and polyatomic ions, a significant step in exploring chemical bonding and molecule geometry.
Lewis Structure
The Lewis structure, also known as a Lewis dot diagram, represents the valence electrons of an atom or a molecule using dots. It is a simplistic graphical representation that shows the arrangement of valence electrons around atoms.

In the exercise regarding the \(\mathrm{PBr}_{4}^{-}\) anion, the Lewis structure is constructed by first placing the central atom, which is usually the least electronegative, and connecting it with single lines to represent bonds with surrounding atoms. In this case, phosphorus is the least electronegative and forms single bonds with four bromine atoms. After forming the bonds, the remaining valence electrons are distributed as lone pairs to fulfill the octet rule, where each atom attempts to have eight electrons in its valence shell.

The accurate depiction of the Lewis structure is foundational for understanding molecule geometry, as it sets the stage for the application of VSEPR theory to determine the three-dimensional shape of the molecule or ion.
VSEPR Theory
VSEPR theory, which stands for Valence Shell Electron Pair Repulsion theory, is a model used to predict the shape of individual molecules based on the repulsion between the electron pairs in the valence shell of an atom. The theory posits that because electron pairs repel each other, molecules will adjust their shape so that these pairs are as far apart as possible, leading to a specific molecular geometry.

For the \(\mathrm{PBr}_{4}^{-}\) ion described in the exercise, VSEPR theory helps us understand why it is referred to as a tetrahedral anion. After arranging the valence electrons according to the Lewis structure, we apply VSEPR theory to predict the ion's shape. With one central atom bonded to four other atoms, no lone pairs on the central atom, and electron pairs spread out to minimize repulsion, the geometry that satisfies these conditions is tetrahedral. This spatial arrangement is consistent with the description provided, affirming that the \(\mathrm{PBr}_{4}^{-}\) ion indeed forms a tetrahedral geometry.

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

Use VSEPR theory to predict the geometric shapes of the following molecules and ions: (a) \(\mathrm{N}_{2}\); (b) HCN; (c) \(\mathrm{NH}_{4}^{+} ;\) (d) \(\mathrm{NO}_{3}^{-} ;\) (e) NSF.

Sketch the probable geometric shape of a molecule of (a) \(\mathrm{N}_{2} \mathrm{O}_{4}\left(\mathrm{O}_{2} \mathrm{NNO}_{2}\right) ;\) (b) \(\mathrm{C}_{2} \mathrm{N}_{2}(\mathrm{NCCN}) ;\) (c) \(\mathrm{C}_{2} \mathrm{H}_{6}\) \(\left(\mathrm{H}_{3} \mathrm{CCH}_{3}\right) ;(\mathrm{d}) \mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}\left(\mathrm{H}_{3} \mathrm{COCH}_{3}\right)\).

Of the following species, the one with a triple covalent bond is (a) \(\mathrm{NO}_{3}^{-} ;\) (b) \(\mathrm{CN}^{-} ;\) (c) \(\mathrm{CO}_{2} ;\) (d) \(\mathrm{AlCl}_{3}\).

Alternative strategies to the one used in this chapter have been proposed for applying the VSEPR theory to molecules or ions with a single central atom. In general, these strategies do not require writing Lewis structures. In one strategy, we write (1) the total number of electron pairs \(=[\) (number of valence electrons) \(\pm\) (electrons required for ionic charge) \(] / 2\) (2) the number of bonding electron pairs \(=\) (number of atoms) -1 (3) the number of electron pairs around central atom \(=\) total number of electron pairs \(-3 \times[\) number of terminal atoms (excluding \(\mathrm{H}\) )] (4) the number of lone-pair electrons = number of central atom pairs - number of bonding pairs After evaluating items \(2,3,\) and \(4,\) establish the VSEPR notation and determine the molecular shape. Use this method to predict the geometrical shapes of the following: (a) \(\mathrm{PCl}_{5} ;\) (b) \(\mathrm{NH}_{3} ;\) (c) \(\mathrm{ClF}_{3} ;\) (d) \(\mathrm{SO}_{2} ;\) (e) \(\mathrm{ClF}_{4}^{-}\); (f) \(\mathrm{PCl}_{4}^{+}\). Justify each of the steps in the strategy, and explain why it yields the same results as the VSEPR method based on Lewis structures. How does the strategy deal with multiple bonds?

Draw Lewis structures for the following species, indicating formal charges and resonance where applicable: (a) \(\mathrm{HCO}_{2}=\) (b) \(\mathrm{HCO}_{3}^{-}\) (c) \(\mathrm{FSO}_{3}^{-}\) (d) \(\mathrm{N}_{2} \mathrm{O}_{3}^{2-}\) (the nitrogen atoms are joined centrally with one oxygen atom on one \(\mathrm{N}\) and two on the other)

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.

Sign-up for free