/*! 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 22 Predict the molecular structure ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 22

Predict the molecular structure (including bond angles) for each of the following. a. \(\mathrm{PCl}_{3}\) b. \(\mathrm{SCl}_{2}\) c. \(\mathrm{SiF}_{4}\)

Short Answer

Expert verified
The molecular structures and bond angles for the given compounds are: a. \(\mathrm{PCl}_{3}\): Trigonal pyramidal geometry with bond angles of around \(109.5^{\circ}\). b. \(\mathrm{SCl}_{2}\): Bent or v-shaped geometry with bond angles of around \(104.5^{\circ}\). c. \(\mathrm{SiF}_{4}\): Tetrahedral geometry with bond angles of around \(109.5^{\circ}\).

Step by step solution

01

a. Predicting the molecular structure and bond angles for \(\mathrm{PCl}_{3}\)

1. Central atom: Phosphorus (P) is the central atom, as it is less electronegative than Chlorine (Cl). 2. Bonding electron pairs: \(3\) P-Cl bonds, making it \(3\) bonding electron pairs. 3. Lone electron pairs: To find the number of lone pairs, first determine the number of valence electrons for P and Cl. P has \(5\) valence electrons, and each Cl has \(7\). Since there are \(3\) Cl atoms, the total number of electrons is \((5 + 3\times 7) = 26\). Subtract \(3\times2\) electrons from the bonds, which leaves \(20\) electrons in \(10\) lone pairs. Two lone pairs are on the central atom P, and the others are on the Cl atoms. 4. With \(3\) bonding pairs and \(1\) lone pair on the central atom, we have a tetrahedral arrangement of electron pairs, a "trigonal pyramidal" molecular geometry, and bond angles of around \(109.5^{\circ}\).
02

b. Predicting the molecular structure and bond angles for \(\mathrm{SCl}_{2}\)

1. Central atom: Sulfur (S) is the central atom, as it is less electronegative than Chlorine (Cl). 2. Bonding electron pairs: \(2\) S-Cl bonds, making it \(2\) bonding electron pairs. 3. Lone electron pairs: To find the number of lone pairs, first determine the number of valence electrons for S and Cl. S has \(6\) valence electrons, and each Cl has \(7\). Since there are \(2\) Cl atoms, the total number of electrons is \((6 + 2\times 7) = 20\). Subtract \(2\times2\) electrons from the bonds, which leaves \(16\) electrons in \(8\) lone pairs. Two lone pairs are on the central atom S, and the others are on the Cl atoms. 4. With \(2\) bonding pairs and \(2\) lone pairs on the central atom, we have a tetrahedral arrangement of electron pairs, a "bent" or "v-shaped" molecular geometry, and bond angles of around \(104.5^{\circ}\).
03

c. Predicting the molecular structure and bond angles for \(\mathrm{SiF}_{4}\)

1. Central atom: Silicon (Si) is the central atom, as it is less electronegative than Fluorine (F). 2. Bonding electron pairs: \(4\) Si-F bonds, making it \(4\) bonding electron pairs. 3. Lone electron pairs: To find the number of lone pairs, first determine the number of valence electrons for Si and F. Si has \(4\) valence electrons, and each F has \(7\). Since there are \(4\) F atoms, the total number of electrons is \((4 + 4\times 7) = 28\). Subtract \(4\times 2\) electrons from the bonds, which leaves \(20\) electrons in \(10\) lone pairs. There are no lone pairs on the central atom Si, the others are on the F atoms. 4. With \(4\) bonding pairs and no lone pairs on the central atom, we have a tetrahedral arrangement of electron pairs, a "tetrahedral" molecular geometry, and bond angles of around \(109.5^{\circ}\).

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!

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 the MO model to determine which of the following has the smallest ionization energy: \(\mathrm{N}_{2}, \mathrm{O}_{2}, \mathrm{N}_{2}^{2-}, \mathrm{N}_{2}^{-}, \mathrm{O}_{2}^{+} .\) Explain your answer.

Which of the following are predicted by the molecular orbital model to be stable diatomic species? a. \(\mathrm{H}_{2}^{+}, \mathrm{H}_{2}, \mathrm{H}_{2}^{-}, \mathrm{H}_{2}^{2-}\) b. \(\mathrm{He}_{2}^{2+}, \mathrm{He}_{2}^{+}, \mathrm{He}_{2}\)

Place the species \(\mathrm{B}_{2}^{+}, \mathrm{B}_{2},\) and \(\mathrm{B}_{2}^{-}\) in order of increasing bond length and increasing bond energy.

What do each of the following sets of compounds/ions have in common with each other? a. \(\mathrm{SO}_{3}, \mathrm{NO}_{3}^{-}, \mathrm{CO}_{3}^{2-}\) b. \(\mathrm{O}_{3}, \mathrm{SO}_{2}, \mathrm{NO}_{2}^{-}\)

Consider the following molecular orbitals formed from the combination of two hydrogen \(1 s\) orbitals: a. Which is the bonding molecular orbital and which is the antibonding molecular orbital? Explain how you can tell by looking at their shapes. b. Which of the two molecular orbitals is lower in energy? Why is this true?
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Kinetics

Read Explanation

Chemistry Branches

Read Explanation

Chemical Analysis

Read Explanation

The Earths Atmosphere

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.