/*! 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 115 The phosphorus trihalides \(\lef... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 115

The phosphorus trihalides \(\left(\mathrm{PX}_{3}\right)\) show the following variation in the bond angle \(\mathrm{X}-\mathrm{P}-\mathrm{X} : \mathrm{PF}_{3}, 96.3^{\circ} ; \mathrm{PCl}_{3}, 100.3^{\circ}\) ; \(\mathrm{PBr}_{3}, 101.0^{\circ} ; \mathrm{PI}_{3}, 102.0^{\circ} .\) The trend is generally attributed to the change in the electronegativity of the halogen. (a) Assuming that all electron domains are the same size, what value of the \(X-P-X\) angle is predicted by the VSEPR model? (b) What is the general trend in the \(X-P-X\) angle as the halide electronegativity increases? (c) Using the VSEPR model, explain the observed trend in \(X-P-X\) angle as the electronegativity of \(X\) changes. (d) Based on your answer to part (c), predict the structure of \(\mathrm{PBrCl}_{4}\)

Short Answer

Expert verified
In summary, the expected X-P-X bond angle according to VSEPR theory is 120掳. The observed trend in bond angles as the electronegativity of the halogen increases is attributed to increased electron density and repulsion, leading to smaller bond angles than predicted. The predicted structure of \(\mathrm{PBrCl}_4\) is a see-saw molecular geometry with the bromide ion and lone pair in equatorial positions and the three chloride ions in one equatorial and two axial positions.

Step by step solution

01

Calculate the expected X-P-X angle using VSEPR model

Using the VSEPR model, we can determine the electron domain geometry of a molecule. In the case of \(\mathrm{PX}_3\), there are three bonding pairs around the central P atom, and no lone pairs. In this trigonal planar geometry, the electron domains will be in the plane of the molecule and will be 120掳 apart from each other. So, according to VSEPR theory, the expected X-P-X bond angle is 120掳.
02

Identify the electronegativity trend for halogens

Now, let's determine the trend in electronegativity for the given halogens: F, Cl, Br, and I. As we move down a group in the periodic table, the electronegativity of the atom decreases. Therefore, the order of electronegativity is: F > Cl > Br > I
03

Explain the observed trend in X-P-X angle

As the electronegativity of X increases, the electron density around the X atoms increases, and they tend to repel each other more. This repulsion forces the X-P-X bond angle to compress slightly, resulting in a smaller bond angle than the predicted 120掳. This explains the observed trend in X-P-X angles for the phosphorus trihalides as the electronegativity of X changes.
04

Predict the structure of PBrCl4

Based on the VSEPR theory and the electronegativity trend of halogens, we can predict the structure of \(\mathrm{PBrCl}_4\). The phosphorus atom has a total of five valence electrons. In this ion, the central phosphorus atom is surrounded by four different ligands - one bromide ion and three chloride ions. In addition, phosphorus has one lone pair of electrons. In a molecule with five electron domains (four bonding pairs and one lone pair), the electron domain geometry is trigonal bipyramidal, and the molecular geometry is called see-saw. The lone pair will occupy an equatorial position since it is more repulsive when it is in the axial position. The bromide ion, which has lower electronegativity than the chloride ions, will also occupy an equatorial position since there is more repulsion between bonding electron pairs in equatorial positions. The three chloride ions will occupy the remaining equatorial and two axial positions. So, the predicted structure of \(\mathrm{PBrCl}_4\) is a see-saw molecular geometry with the bromide ion and lone pair in equatorial positions, and the three chloride ions in one equatorial and two axial positions.

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.

Phosphorus Trihalides
Phosphorus trihalides, represented as \( \text{PX}_3 \), are a group of compounds where phosphorus (P) is bonded to three halogen atoms (X). The halogens in these compounds can include fluorine (F), chlorine (Cl), bromine (Br), and iodine (I). These compounds are essential in understanding molecular geometry and bond angles as described by VSEPR theory.

In phosphorus trihalides:
  • The central phosphorus atom forms three single covalent bonds with halogen atoms.
  • The structure of these compounds is crucial in various chemical reactions, particularly in organic chemistry where they serve as reagents.
  • Their variability in bond angles highlights key principles about molecular geometry and electronegativity.
Understanding the properties and behaviors of phosphorus trihalides helps in predicting the structural outcomes of different chemical combinations.
Electronegativity
Electronegativity is a measure of an atom's ability to attract and hold onto electrons in a chemical bond. Halogens, such as fluorine, chlorine, bromine, and iodine, vary in electronegativity. The trend generally decreases as you move down the periodic table group of halogens.

Here is the order of electronegativity for the relevant halogens:
  • Fluorine (F) has the highest electronegativity.
  • Chlorine (Cl) follows fluorine in electronegativity.
  • Bromine (Br) is less electronegative than chlorine.
  • Iodine (I) has the lowest electronegativity among these halogens.
Electronegativity affects molecular geometry by influencing the electron density around the atoms. As atoms with higher electronegativity pull electrons closer, this increases electron repulsion, ultimately affecting bond angles. Consequently, the more electronegative the halogen, the smaller the bond angle due to compression from electron repulsion.
Molecular Geometry
Molecular geometry refers to the three-dimensional arrangement of atoms within a molecule. It dictates the molecule's shape, determining reactivity, polarity, phase of matter, color, magnetism, biological activity, and more. The VSEPR (Valence Shell Electron Pair Repulsion) theory is pivotal in predicting and understanding molecular geometry.

For \( \text{PX}_3 \) molecules:
  • Three bonded pairs and no lone pairs on phosphorus lead to a trigonal planar arrangement of electron domains.
  • In a perfect trigonal planar geometry, bond angles are predicted to be 120掳.
  • However, differences in electronegativity can lead to variations in these angles, as halogens with higher electronegativity create greater electron repulsion, compressing the angle.
By understanding these principles, the geometry and resulting properties of a wide variety of compounds can be predicted, enabling insights into their chemical behaviors.

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

(a) Using only the valence atomic orbitals of a hydrogen atom and a fluorine atom, and following the model of Figure 9.46, how many MOs would you expect for the HF molecule? (b) How many of the MOs from part (a) would be occupied by electrons? (c) It turns out that the difference in energies between the valence atomic orbitals of H and F are sufficiently different that we can neglect the interaction of the 1 s orbital of hydrogen with the 2\(s\) orbital of fluorine. The 1 s orbital of hydrogen will mix only with one 2\(p\) orbital of fluorine. Draw pictures showing the proper orientation of all three 2\(p\) orbitals on Finteracting with a 15 orbital on \(\mathrm{H} .\) Which of the 2\(p\) orbitals can actually make a bond with a 1\(s\) orbital, assuming that the atoms lie on the z-axis? (d) In the most accepted picture of HF, all the other atomic orbitals on fluorine move over at the same energy into the molecular orbital energy-level diagram for HF. These are called "nonbonding orbitals." Sketch the energy-level diagram for HF using this information and calculate the bond order. (Nonbonding electrons do not contribute to bond order.) (e) Look at the Lewis structure for HF. Where are the nonbonding electrons?

Draw sketches illustrating the overlap between the following orbitals on two atoms: (a) the 2 s orbital on each atom, (b) the 2\(p_{z}\) orbital on each atom (assume both atoms are on the \(z\) -axis), (c) the 2 s orbital on one atom and the 2\(p_{z}\) orbital on the other atom.

Consider the molecule \(\mathrm{BF}_{3}\). (a) What is the electron configuration of an isolated B atom? (b) What is the electron configuration of an isolated F atom? (c) What hybrid orbitals should be constructed on the B atom to make the B鈥揊 bonds in \(\mathrm{B} \mathrm{F}_{3}\)?(d) What valence orbitals, if any, remain unhybridized on the B atom in \(\mathrm{BF}_{3} ?\)

In the formate ion, \(\mathrm{HCO}_{2}^{-}\) , the carbon atom is the central atom with the other three atoms attached to it. (a) Draw a Lewis structure for the formate ion. (b) What hybridization is exhibited by the C atom? (c) Are there multiple equivalent resonance structures for the ion? (d) How many electrons are in the \(\pi\) system of the ion?

Sodium azide is a shock-sensitive compound that releases \(\mathrm{N}_{2}\) upon physical impact. The compound is used in automobile airbags. The azide ion is \(\mathrm{N}_{3}^{-} .\) (a) Draw the Lewis structure of the azide ion that minizes formal charge (it does not form a triangle). Is it linear or bent? (b) State the hybridization of the central Natom in the azide ion. (c) How many \(\sigma\) bonds and how many \(\pi\) bonds does the central nitrogen atom make in the azide ion?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Inorganic Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Ionic and Molecular Compounds

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.