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(a) If the valence atomic orbitals of an atom are sp hybridized, how many unhybridized \(p\) orbitals remain in the valence shell? How many \(\pi\) bonds can the atom form? (b) Imagine that you could hold two atoms that are bonded together, twist them, and not change the bond length. Would it be easier to twist (rotate) around a single \(\sigma\) bond or around a double \((\sigma\) plus \(\pi)\) bond, or would they be the same?

Short Answer

Expert verified
(a) In an sp hybridized atom, there are 2 unhybridized p orbitals and it can form 2 π bonds. (b) It is easier to twist around a single σ bond than around a double bond (σ + π) due to the rigid nature of the π bond.

Step by step solution

01

(a) Determine the number of unhybridized p orbitals

sp hybridization occurs when one s-orbital combines with one p-orbital to form two hybrid orbitals. Since p orbitals are three in number (px, py, and pz), the remaining unhybridized p orbitals will be 2 (px and py if we assume the pz gets hybridized).
02

Determine the number of π bonds an atom with sp hybridization can form

An unhybridized p orbital can form one π bond. As there are two unhybridized p orbitals in an sp hybridized atom (from the previous step), the atom can form two π bonds.
03

(b) Comparing the ease of twisting around a single σ bond and a double bond (σ + π)

A single σ bond allows free rotation around the bond axis without affecting the bond length. In contrast, a π bond does not allow free rotation due to the parallel arrangement of electron cloud above and below the plane, holding the atoms together more rigidly. Thus, rotation around a double bond (σ+π) would require energy to break the π bond, making it harder to twist compared to a single σ bond. So, it would be easier to twist (rotate) around a single σ bond than around a double bond (σ + π).

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Key Concepts

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

Hybridization
Hybridization is a fascinating concept in chemistry that helps us understand how atoms form bonds and create molecules. When atoms bond, their orbitals, which are regions where electrons are likely to be found, can actually mix together. This mixing of orbitals is what we call hybridization. The new hybrid orbitals that form help atoms bind together in a stable manner. In the case of sp hybridization, a single s orbital combines with one p orbital. This combination results in two sp hybrid orbitals. These hybrid orbitals are used to form sigma bonds, which provide strong and stable single bonds between atoms. However, not all of the atom's p orbitals are used in this hybridization. When one p orbital is used, the other two p orbitals (like px and py or py and pz) stay unhybridized and can be used to form pi bonds. This allows the atom to participate in multiple bond formations by using its unhybridized orbitals.
Sigma Bonds
Sigma bonds are the foundation of molecular structures. They represent a type of covalent bond where the overlap of atomic orbitals occurs primarily along the axis connecting two nuclei. This head-on overlap results in a very strong and resilient bond. It's like the backbone of molecular bonding, holding the atoms firmly in place. An important feature of sigma bonds is that they allow for free rotation around the bond axis. Whenever you picture two atoms bonded like a door hinge, consider that hinge as a sigma bond.
- Sigma bonds result from either two p orbitals overlapping or a hybrid orbital overlapping with another orbital. - These bonds are the first bonds formed during molecular bonding, providing the structural integrity for the entire molecule. - Because rotation is possible around these bonds, they contribute to the flexibility and dynamic nature of molecules.
Pi Bonds
Pi bonds add complexity to the molecular picture by contributing to the formation of double or triple bonds. They arise from the side-on overlap of unhybridized p orbitals. Pi bonds sit above and below the atomic axis, unlike sigma bonds. When a molecule has double or triple bonds, pi bonds complement sigma bonds, adding strength but also rigidity.
- Pi bonds restrict rotation around the bond axis because the side-on overlap creates an electron cloud that holds the atoms together more tightly. - While they do not form directly between nuclei like sigma bonds, they enhance the overall bonding energy of a molecule. - The number of pi bonds an atom can form depends on its available unhybridized p orbitals, as seen in sp hybridization where two pi bonds can be possible.
Molecular Geometry
Molecular geometry focuses on the three-dimensional arrangement of atoms in molecules, which is determined largely by the type of bonding that occurs. By understanding hybridization and the presence of sigma and pi bonds, we can predict the shape and geometry of molecules.
- The geometry of a molecule dictates its physical and chemical properties. - For example, sp hybridized molecules typically exhibit linear geometry, resulting in straightforward shapes like those seen in molecules like carbon dioxide. - Other types of hybridizations, such as sp² and sp³, can create different geometries, like trigonal planar and tetrahedral, which affect molecular function and reactivity. - Understanding the principles of hybridization, sigma bonds, and pi bonds can greatly assist in predicting how a molecule will interact with others, its reactivity, and its role in larger chemical systems.

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Most popular questions from this chapter

The \(\mathrm{O}-\mathrm{H}\) bond lengths in the water molecule \(\left(\mathrm{H}_{2} \mathrm{O}\right)\) are \(96 \mathrm{pm}\), and the \(\mathrm{H}-\mathrm{O}-\mathrm{H}\) angle is \(104.5^{\circ} .\) The dipole moment of the water molecule is \(1.85 \mathrm{D}\). (a) In what directions do the bond dipoles of the \(\mathrm{O}-\mathrm{H}\) bonds point? \(\mathrm{In}\) what direction does the dipole moment vector of the water molecule point? (b) Calculate the magnitude of the bond dipole of the \(\mathrm{O}-\mathrm{H}\) bonds. (Note: You will need to use vector addition to do this.) (c) Compare your answer from part (b) to the dipole moments of the hydrogen halides (Table 8.3). Is your answer in accord with the relative electronegativity of oxvgen?

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