/*! 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 52 Which of these species has a lon... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 52

Which of these species has a longer bond, \(\mathrm{B}_{2}\) or \(\mathrm{B}_{2}^{+} ?\) Explain in terms of molecular orbital theory.

Short Answer

Expert verified
\(\mathrm{B}_{2}\) has a bond order of 3, whereas \(\mathrm{B}_{2^{+}}\) has a bond order of 2.5. Therefore, \(\mathrm{B}_{2^{+}}\) has a longer bond length compared to \(\mathrm{B}_{2}\).

Step by step solution

01

Write Electron Configurations

In order to perform our comparison, we must understand the electron configuration of each molecule. \(\mathrm{B}_{2}\) has a total of 10 electrons. The electron configuration, according to the MO theory, is \(\sigma_{1s}^{2}, \sigma_{1s}^{*2}, \sigma_{2s}^{2}, \sigma_{2s}^{*2}, \pi_{2p}^{2}\). \(\mathrm{B}_{2}^{+}\) has 9 electrons. Its electron configuration is therefore \(\sigma_{1s}^{2}, \sigma_{1s}^{*2}, \sigma_{2s}^{2}, \sigma_{2s}^{*2}, \pi_{2p}^{1}\).
02

Calculate Bond Order

Now, after obtaining the electron configuration for \(\mathrm{B}_{2}\) and \(\mathrm{B}_{2}^{+}\), calculate the bond order using the formula 'half the difference between number of bonding and antibonding electrons'. For \(\mathrm{B}_{2}\), the bonding electrons = 8 and the anti-bonding = 2, thus giving a bond order of 3. For \(\mathrm{B}_{2}^{+}\), the bonding electrons = 7 and the anti-bonding = 2, giving a bond order of 2.5.
03

Compare Bond Lengths

The bond length is inversely proportional to the bond order. Therefore, the molecule with the smaller bond order will have the longer bond length. \(\mathrm{B}_{2}^{+}\) has a lower bond order (2.5) than \(\mathrm{B}_{2}\) (3), and so it has a longer bond length.

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

Antimony pentafluoride, \(\mathrm{SbF}_{5}\), reacts with \(\mathrm{XeF}_{4}\) and \(\mathrm{XeF}_{6}\) to form ionic compounds, \(\mathrm{XeF}_{3}^{+} \mathrm{SbF}_{6}^{-}\) and \(\mathrm{XeF}_{5}^{+} \mathrm{SbF}_{6}^{-}\). Describe the geometries of the cations and anion in these two compounds.

How many atoms are directly bonded to the central atom in a tetrahedral molecule, a trigonal bipyramidal molecule, and an octahedral molecule?

Explain why an atom cannot have a permanent dipole moment.

Which of these molecules are linear? \(\mathrm{ICl}_{2}^{-}, \mathrm{IF}_{2}^{+}, \mathrm{OF}_{2}\), \(\mathrm{SnI}_{2}, \mathrm{CdBr}_{2}\).

The compounds carbon tetrachloride \(\left(\mathrm{CCl}_{4}\right)\) and silicon tetrachloride \(\left(\mathrm{SiCl}_{4}\right)\) are similar both in geometry and hybridization. However, \(\mathrm{CCl}_{4}\) does not react with water but \(\mathrm{SiCl}_{4}\) does. Explain the difference in their chemical reactivities. (Hint: The first step of the reaction is believed to be the addition of a water molecule to the Si atom in \(\mathrm{SiCl}_{4}\).)
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Inorganic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Organic 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.