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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

(a) What is the difference between hybrid orbitals and molecular orbitals? (b) How many electrons can be placed into each MO of a molecule? (c) Can antibonding molecular orbitals have electrons in them?

Draw the Lewis structure for each of the following molecules or ions, and predict their electron-domain and molecular geometries: (a) \(\mathrm{AsF}_{3}\) (b) \(\mathrm{CH}_{3}^{+},(\mathbf{c}) \mathrm{Br} \mathrm{F}_{3}\) (d) \(\mathrm{ClO}_{3}^{-}\) (e) \(\mathrm{XeF}_{2}\) (f) \(\mathrm{BrO}_{2}^{-}\).

Methyl isocyanate, \(\mathrm{CH}_{3} \mathrm{NCO},\) was made infamous in 1984 when an accidental leakage of this compound from a storage tank in Bhopal, India, resulted in the deaths of about 3800 people and severe and lasting injury to many thousands more. (a) Draw a Lewis structure for methyl isocyanate. (b) Draw a ball-and-stick model of the structure, including estimates of all the bond angles in the compound. (c) Predict all the bond distances in the molecule. (d) Do you predict that the molecule will have a dipole moment? Explain.

Consider the \(\mathrm{H}_{2}^{+}\) ion. (a) Sketch the molecular orbitals of the ion and draw its energy-level diagram. (b) How many electrons are there in the \(\mathrm{H}_{2}^{+}\) ion? (c) Write the electron configuration of the ion in terms of its MOs. (d) What is the bond order in \(\mathrm{H}_{2}^{+}\) ? (e) Suppose that the ion is excited by light so that an electron moves from a lower-energy to a higher- energy MO. Would you expect the excited-state \(\mathrm{H}_{2}^{+}\) ion to be stable or to fall apart? (f) Which of the following statements about part (e) is correct: (i) The light excites an electron from a bonding orbital to an antibonding orbital, (ii) The light excites an electron from an antibonding orbital to a bonding orbital, or (iii) In the excited state there are more bonding electrons than antibonding electrons?

(a) Using only the valence atomic orbitals of a hydrogen atom and a fluorine atom, and following the model of Figure 9.46 , how many MOs would you expect for the HF molecule? (b) How many of the MOs from part (a) would be occupied by electrons? (c) It turns out that the difference in energies between the valence atomic orbitals of \(\mathrm{H}\) and \(\mathrm{F}\) are sufficiently different that we can neglect the interaction of the 1 s orbital of hydrogen with the 2 s orbital of fluorine. The 1 s orbital of hydrogen will mix only with one \(2 p\) orbital of fluorine. Draw pictures showing the proper orientation of all three \(2 p\) orbitals on F interacting with a 1 sorbital on \(\mathrm{H}\). Which of the \(2 p\) orbitals can actually make a bond with a 1 s orbital, assuming that the atoms lie on the \(z\) -axis? (d) In the most accepted picture of HF, all the other atomic orbitals on fluorine move over at the same energy into the molecular orbital energy-level diagram for HE. These are called "nonbonding orbitals." Sketch the energy- level diagram for HF using this information and calculate the bond order. (Nonbonding electrons do not contribute to bond order.) \((\mathbf{e})\) Look at the Lewis structure for HE. Where are the nonbonding electrons?

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