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In a buckyball, three of the bonds around each hexagon are so-called double bonds. They result from adjacent atoms sharing a state that does not participate in the \(s p^{2}\) bonding. Which state is it, and is this extra bond a \(\sigma\) -bond or a \(\pi\) -bond? Explain.

Short Answer

Expert verified
The shared state that does not participate in \(s p^{2}\) bonding is the p electron state of the carbon atom. The extra bond is a \(\pi\) bond formed by the parallel overlap of p orbitals of adjacent carbon atoms.

Step by step solution

01

Understanding Bonding in Buckyball

A buckyball or fullerene is made up of 60 carbon atoms arranged in a structure of 20 hexagons and 12 pentagons, having a soccer ball shape. Each carbon atom is \(s p^{2}\) hybridized and forms 3 sigma bonds with 3 neighboring carbon atoms, one in each of three, and leaves one p orbital for forming \(\pi\) bond
02

Identify double bonds

When hexagonal structures are formed, there are three bonds that can be classified as double bonds. This is because in addition to \(\sigma\) bonds formed by \(s p^{2}\) hybridization, the carbon atoms are also forming \(\pi\) bonds by sharing their unpaired p electrons.
03

Determine bond type

The extra bond according to the problem, is not participating in \(s p^{2}\) hybridization. This means that it is not a \(\sigma\) bond. The unparticipated bond making the bond 'double' is a \(\pi\) bond, formed by the parallel overlap of p orbitals, and this contributes to the structure of hexagon in buckyballs.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding sp2 Hybridization
When we dive into the intricate world of chemistry, particularly the structure of molecules like the fascinating 'buckyball,' we encounter the concept of sp2 hybridization. This term refers to the mixing of one s orbital and two p orbitals from the same atom to form three new, equivalent orbitals known as sp2 hybrid orbitals.

Each carbon atom has an electron configuration that, when energized, leads to the promotion of an electron into a higher energy state, thus resulting in the hybridization process. In buckyballs, every carbon atom undergoes sp2 hybridization, creating three hybrid orbitals that form sigma bonds with neighboring carbon atoms. The remaining unhybridized p orbital is essential for pi bonding, which is crucial for the formation of double bonds in the buckyball structure.

Understanding sp2 hybridization is fundamental because it contributes to the unique shape and properties of molecules. The geometric consequence of this hybridization is a trigonal planar arrangement, promoting the formation of strong and stable chemical bonds.
Delving into the Nature of Pi Bonds
Proceeding to pi bonds, these result from the side-to-side overlap of adjacent unhybridized p orbitals. In a buckyball, once the sigma bonds are formed through sp2 hybridization, each carbon atom still has one unhybridized p orbital. These p orbitals extend above and below the plane of the atoms that are sigma bonded.

The pi bonds form when these lobes overlap with those from neighboring carbon atoms. This creates a region of electron density that does not lie along the line of the internuclear axis, which is the characteristic of the pi bond compared to the sigma bond. It's also important to note that pi bonds are generally weaker than sigma bonds and provide less overlap, but they are pivotal in conferring stability and unique chemical properties to the structure of the buckyball.
Sigma Bonds: The Backbone of Molecular Structure
Switching our focus to sigma bonds, they are the most common type of covalent bond formed between two atoms. Characterized by head-to-head orbital overlap, they allow the sharing of electrons along the axis connecting two atomic nuclei.

In the context of a buckyball, every carbon atom uses its three sp2 hybrid orbitals to form sigma bonds with three different neighboring carbon atoms. This ensures a firm and robust connectivity within the molecule, contributing to its overall stability and strength. Unlike pi bonds, these are free to rotate, which makes them pivotal in contributing to the flexibility and multiple conformations that a molecule like buckyball can exhibit.
The Fullerene Structure: A Geodesic Sphere of Carbon
Lastly, exploring the fullerene structure opens up a fascinating realm of molecular geometry. Fullerenes are a class of carbon allotropes, among which the buckyball, or C60, is a quintessential example. It resembles a geodesic dome with a spheroidal shape resemblant of a soccer ball consisting of pentagons and hexagons.

The unique arrangement of 20 hexagons and 12 pentagons, each vertex being a carbon atom, is a testament to carbon's versatile bonding capabilities. The curvature of the buckyball's surface is made possible because of the seamless integration of pentagonal and hexagonal rings, with sp2 hybridized carbon atoms at the vertices. The pi bonds, although they don't contribute to the spherical shape directly, play a key role in the electronic properties, making fullerenes potential candidates in various applications ranging from organic photovoltaics to drug delivery systems.

The understanding of fullerene structures not only epitomizes the beauty and intricacy of molecular architecture but also paves the way for innovative materials and technologies based on this extraordinary carbon assemblage.

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Most popular questions from this chapter

The magnetic field at the surface of a long wire of radius \(R\) and cairying a current \(I\) is \(\mu_{0} I / 2 \pi R\). How large a curtent could a 0.1 mm diameter niobium wire carry without exceeding its \(0.2 \mathrm{~T}\) critical field?

In Section 10.2. we discussed two-lobed \(p_{\text {re }} p_{\text {s }}\) and \(p_{z}\) states and four-lohed hybrid \(s p^{3}\) states. Another kind of hybrid state that sticks out in just one direction is the sp. formed from a single \(p\) state and an \(s\) state. Consider an urbitrary combination of the \(2 s\) state with the \(2 p_{x}\) state. Let us represent this by \(\cos \tau \psi_{2.0 .0}+\sin T \psi_{2.1 .0}\) (The trig factors ensure norinalization. In carrying out the integral, crow terms integrate to \(0 .\) leaving \(\cos ^{2}+\int\left|\psi_{2,0,0}\right|^{2} d b^{\prime}+\sin ^{2} \tau \int\left|\psi_{21,0}\right|^{2} d V\). which is \(\left.1 .\right)\) (a) Calculate the probability that an electron in such a state would be in the \(+z\) -hemisphere. (Nore: Here. the cross tems do nor integrate to \(0 .\) ) (b) What value of \(r\) leads to the maximum probability. what is the value of the maximum probability, and what is the cotresponding ratio of \(\psi_{2,0.0}\) to \(\phi_{2,1,0} ?\) (c) Using a computer, make a density (shading) plot of the probability density - density versus \(r\) and \(\theta\) for the \(r\) -value found in part (b).

In the boron atom, the single \(2 p\) electron does not completely fill any \(2 p\) spatial state, yet solid boron is not a conductor. What might explain this? (It may be helpful to consider again why beryllium is not an insulator.)

The photons emitted by an LED arise from the energy given up in electron-hole recombinations across the energy gap. How large should the energy gap be to give photons at the red end of the visible spectrum \((700 \mathrm{~nm}) ?\)

As a crude approximation, an impurity pentavalent atom in a (tetravalent) silicon lattice can be treated as a one-electron atom, in which the extra electron orbits a net positive charge of 1 . Because this "atom" is not in free space, however, the permitivity of free space, \(\varepsilon_{0}\) must be replaced by \(\kappa \varepsilon_{0}\), where \(\kappa\) is the dielectric constant of the surrounding material. The hydrogen atom ground-state energies would thus become $$ E=-\frac{m e^{4}}{2\left(4 \pi \kappa \varepsilon_{0}\right)^{2} h^{2}} \frac{1}{n^{2}}=\frac{-13.6 \mathrm{eV}}{\kappa^{2} n^{2}} $$ Given \(\kappa=12\) for silicon, how much ener gy is needed to frec a donor electron in its ground state? (Actually. the effective mass of the donor electron is less than \(m_{e}\), so this prediction is somewhat high.)

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