The bonding molecular orbital in \({{\rm{H}}_2}\) is lower in energy than an electron in \({\rm{1s}}\) orbital because the bonding molecular is more stable.

Short Answer

Expert verified

The bonding molecular orbital in \({{\rm{H}}_2}\) is lower in energy than an electron in \(1s\) orbital because the bonding molecular is more stable

Step by step solution

01

Definition of bonding molecular orbital

Bonding orbitals are used in molecular orbital theory (MO) to describe the attractive interactions between the atomic orbitals of two or more atoms in a molecule.

02

Explanation for lower energy

Bonded molecular orbital electrons in the \({{\rm{H}}_{\rm{2}}}\) molecule will be in between the two H nuclei and this electron is more stable than a stabilized electron because it interacts with two nuclei \(1s\) one atomic orbital of a separated hydrogen atom. Bonded molecular orbital \({{\rm{H}}_2}\) is less energy than electrons in \({{\rm{H}}_2}\)orbital because the bonding molecular is more stable

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

For many years after they were discovered, it was believed that the noble gases could not form compounds. Now we know that belief to be incorrect. A mixture of xenon and fluorine gases, confined in a quartz bulb and placed on a windowsill, is found to slowly produce a white solid. Analysis of the compound indicates that it contains \({\rm{77}}{\rm{.55\% }}\)Xe and \({\rm{22}}{\rm{.45\% \;F}}\)by mass.

(a) What is the formula of the compound?

(b) Write a Lewis structure for the compound.

(c) Predict the shape of the molecules of the compound.

(d) What hybridization is consistent with the shape you predicted?

Label the molecular orbital shown as σ or Π, bonding or antibonding, and indicate where the node occurs.


Consider nitrous acid, \({\rm{HN}}{{\rm{O}}_{\rm{2}}}{\rm{(HONO)}}\). (a) Write a Lewis structure. (b) What are the electron pair and molecular geometries of the internal oxygen and nitrogen atoms in the \({\rm{HN}}{{\rm{O}}_{\rm{2}}}\) molecule? (c) What is the hybridization on the internal oxygen and nitrogen atoms in \({\rm{HN}}{{\rm{O}}_{\rm{2}}}\)?

Another acid in acid rain is nitric acid, HNO3, which is produced by the reaction of nitrogen dioxide, NO2, with atmospheric water vapor. What is the hybridization of the nitrogen atom in NO2? (Note: the lone electron on nitrogen occupies a hybridized orbital just as a lone pair would.)

The bond energy of a C–C single bond averages \({\rm{347 kJ mo}}{{\rm{l}}^{{\rm{ - 1}}}}\); that of a

C ≡ C triple bond averages \({\rm{839 kJ mo}}{{\rm{l}}^{{\rm{ - 1}}}}\). Explain why the triple bond is not three times as strong as a single bond.

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