Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 10: Problem 63

Predict the geometry of sulfur dichloride \(\left(\mathrm{SCl}_{2}\right)\) and the hybridization of the sulfur atom.

Short Answer

Expert verified
The geometry of sulfur dichloride \(\mathrm{SCl}_{2}\) is bent shape or V-shaped and the hybridization of the sulfur atom is sp3.

Step by step solution

01

Determine Valence Electrons and draw the Electron Dot Structure

Sulfur has 6 valence electrons, and each Chlorine atom has 7 valence electrons. So, for \(\mathrm{SCl}_{2}\), the total number of valence electrons is \(6 + 2*7 = 20\). The Lewis dot structure can be drawn with Sulfur at the center (as it is less electronegative) and a Chlorine atom on each side, each bonded to the sulfur by a single bond, and the remaining 4 electrons placed as lone pairs on the sulfur atom.
02

Determine the Electron Pair Geometry and Molecular Geometry

The sulfur atom is surrounded by 4 regions of electron density, including 2 single bond pairs and 2 lone pairs. These 4 pairs try to position themselves as far apart as possible to minimize repulsion. Hence, the electron-pair geometry is tetrahedral. However, the molecular geometry is determined by the positions of the atoms only, which forms a V-shape or bent shape.
03

Determine the Hybridization

The hybridization of the atom can be determined from the electron pair geometry. In this case, with 4 regions of electron density around the sulfur atom, the sulfur atom is sp3 hybridized.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Draw a molecular orbital energy level diagram for each of these species: \(\mathrm{He}_{2}, \mathrm{HHe}, \mathrm{He}_{2}^{+} .\) Compare their relative stabilities in terms of bond orders. (Treat HHe as a diatomic molecule with three electrons.)

What is valence bond theory? How does it differ from the Lewis concept of chemical bonding?

Consider a \(\mathrm{N}_{2}\) molecule in its first excited electronic state; that is, when an electron in the highest occupied molecular orbital is promoted to the lowest empty molecular obital. (a) Identify the molecular orbitals involved and sketch a diagram to show the transition. (b) Compare the bond order and bond length of \(\mathrm{N}_{2}{ }^{*}\) with \(\mathrm{N}_{2}\), where the asterisk denotes the excited molecule. (c) Is \(\mathrm{N}_{2}{ }^{*}\) diamagnetic or paramagnetic? (d) When \(\mathrm{N}_{2} *\) loses its excess energy and converts to the ground state \(\mathrm{N}_{2}\), it emits a photon of wavelength \(470 \mathrm{nm}\), which makes up part of the auroras lights. Calculate the energy difference between these levels.

What is molecular orbital theory? How does it differ from valence bond theory?

Write the ground-state electron configuration for \(\mathrm{B}_{2}\). Is the molecule diamagnetic or paramagnetic?
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Physical Chemistry

Read Explanation

Making Measurements

Read Explanation

Chemical Reactions

Read Explanation

Inorganic Chemistry

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.

Sign-up for free