Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

In the molecular orbital model, compare and contrast \(\sigma\) bonds with \(\pi\) bonds. What orbitals form the \(\sigma\) bonds and what orbitals form the \(\pi\) bonds? Assume the \(z\) -axis is the internuclear axis.

Short Answer

Expert verified
In the molecular orbital model, σ bonds are formed by end-on overlapping of bonding atomic orbitals (such as s, p_z, and hybrids) along the internuclear axis (z-axis) and have cylindrical symmetry, resulting in stronger bonds. On the other hand, π bonds are formed by side-on overlapping of atomic orbitals (mainly p_x and p_y) perpendicular to the internuclear axis, with a nodal plane along the axis and electron density distributed above and below this plane, leading to weaker bonds due to less effective overlapping.

Step by step solution

01

Define and understand σ and π bonds

σ bonds (sigma bonds) are formed by end-on overlapping of bonding orbitals along the internuclear axis (usually considered along the z-axis). These bonds have cylindrical symmetry about the internuclear axis and tend to be stronger due to greater overlap compared to π bonds. π bonds (pi bonds) are formed by the side-on overlapping of bonding orbitals along the internuclear axis. These bonds have a nodal plane along the internuclear axis, and the electron density is distributed above and below this plane. Due to less effective overlapping, π bonds are generally weaker than σ bonds. It's important to note that a double bond is composed of one σ bond and one π bond, while a triple bond has one σ bond and two π bonds.
02

Identify contributing atomic orbitals for σ bonds

Atomic orbitals such as s, p_z, and hybridized orbitals (sp, sp^2, sp^3, etc.) can contribute to form σ bonds. For example, an s-s overlap, s-p_z overlap, and a p_z - p_z overlap can all result in σ bonds, as long as the overlapping occurs along the internuclear axis.
03

Identify contributing atomic orbitals for π bonds

Atomic orbitals like p_x, p_y, and hybridized orbitals (if necessary) can form π bonds. It is essential to remember that p_x and p_y orbitals lie perpendicular to the internuclear axis (z-axis), which allows for side-on overlapping and π bond formation. In summary, σ bonds are characterized by end-on overlapping of bonding atomic orbitals (such as s, p_z, and hybrids) along the internuclear axis and tend to be stronger due to more significant overlap. In contrast, π bonds result from side-on overlapping of atomic orbitals (mainly p_x and p_y) perpendicular to the internuclear axis and are generally weaker due to less effective overlapping.

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

Why does the molecular orbital model do a better job in explaining the bonding in \(\mathrm{NO}^{-}\) and \(\mathrm{NO}\) the hybrid orbital model?

Using the molecular orbital model, write electron configurations for the following diatomic species and calculate the bond orders. Which ones are paramagnetic? $$ \text {a} \mathrm{Li}_{2} \quad \text { b. } \mathrm{C}_{2} \quad \text { c. } \mathrm{S}_{2} $$

In Exercise 95 in Chapter 8 , the Lewis structures for benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) were drawn. Using one of the Lewis structures, estimate \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{C}_{6} \mathrm{H}_{6}(g)\) using bond energies and given that the standard enthalpy of formation of \(\mathrm{C}(g)\) is 717 \(\mathrm{kJ} / \mathrm{mol}\) . The experimental \(\Delta H_{\mathrm{f}}^{\circ}\) value of \(\mathrm{C}_{6} \mathrm{H}_{6}(g)\) is 83 \(\mathrm{kJ} / \mathrm{mol} .\) Explain the discrepancy between the experimental value and the calculated \(\Delta H_{\mathrm{f}}^{\circ}\) value for \(\mathrm{C}_{6} \mathrm{H}_{6}(g)\)

Consider the following molecular orbitals formed from the combination of two hydrogen 1s 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?

Why are \(d\) orbitals sometimes used to form hybrid orbitals? Which period of elements does not use \(d\) orbitals for hybridization? If necessary, which \(d\) orbitals \((3 d, 4 d, 5 d, \text { or } 6 d)\) would sulfur use to form hybrid orbitals requiring \(d\) atomic orbitals? Answer the same question for arsenic and for iodine.

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