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 shorter bond length than \(\mathrm{B}_{2}^{+}\) due to more filled bonding \(\pi_{2p}\) orbitals.

Step by step solution

01

Identify Electron Configurations

The electron configuration of \(\mathrm{B}_{2}\) is \(\sigma_{1s}^2 \sigma_{1s*}^2 \sigma_{2s}^2 \sigma_{2s*}^2 \pi_{2p}^4\). The electron configuration of \(\mathrm{B}_{2}^{+}\) is \(\sigma_{1s}^2 \sigma_{1s*}^2 \sigma_{2s}^2 \sigma_{2s*}^2 \pi_{2p}^3\) because it has one fewer electron.
02

Compare the Filled Orbitals

Both \(\mathrm{B}_{2}\) and \(\mathrm{B}_{2}^{+}\) have the same filled \(\sigma_{2s}\) and \(\sigma_{2s*}\) orbitals, but \(\mathrm{B}_{2}\) has both \(\pi_{2p}\) orbitals filled whereas \(\mathrm{B}_{2}^{+}\) has only three electrons in those orbitals.
03

Determine Bond Length

Since \(\mathrm{B}_{2}\) has more electrons in its bonding \(\pi_{2p}\) orbitals, its bonds are stronger and therefore shorter. Hence, \(\mathrm{B}_{2}\) has a shorter bond length than \(\mathrm{B}_{2}^{+}\).

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!

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

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