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\((\mathrm{a})\) If the valence atomic orbitals of an atom are \(s p\) 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? Explain.

Short Answer

Expert verified
In the case of \(sp\) hybridization, there are two unhybridized \(p\) orbitals remaining in the valence shell and the atom can form a maximum of two \(\pi\) bonds. Rotating around a single \(\sigma\) bond is easier because the electron density is concentrated in one bond, allowing rotation without much energy. In contrast, rotating around a double bond (\(\sigma\) + \(\pi\)) is more challenging as the side-by-side p orbitals forming the \(\pi\) bond must break to rotate, requiring more energy.

Step by step solution

01

Understanding Hybridization of Atomic Orbitals

Hybridization is the mixing of valence atomic orbitals to form new, equivalent orbitals that can be used for bonding with other atoms. In this case, we have \(s p\) hybridized orbitals, meaning one s orbital and one p orbital combine to form two new \(sp\) hybrid orbitals.
02

Calculating unhybridized p orbitals

In the valence shell of an atom, there are one s orbital and three p orbitals. Since one s and one p orbital are hybridized, we still have two unhybridized p orbitals left in the valence shell.
03

Calculating π bonds an atom can form

π bonds are formed when parallel unhybridized p orbitals overlap side by side. Since we have two unhybridized p orbitals remaining, the atom can form a maximum of two π bonds.
04

Understanding rotation around single and double bonds

A single bond consists of only one \(\sigma\) bond, while a double bond is composed of a \(\sigma\) bond and a \(\pi\) bond. The rotation around a bond involves the movement of atoms and electron density.
05

Rotation around single and double bonds

Rotating around a single \(\sigma\) bond is easier because the electron density between the two atoms is concentrated in one sigma bond, permitting rotation without much energy. In contrast, rotating around a double bond (which includes a \(\sigma\) and a \(\pi\) bond) is more challenging because the side-by-side p orbitals that form the \(\pi\) bond must break in order to rotate, requiring more energy. Therefore, it is easier to twist around a single \(\sigma\) bond than around a double (\(\sigma\) plus \(\pi\)) bond.

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Key Concepts

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

SP Hybridized Orbitals
The magic of chemistry often lies in the understanding of how atoms bond together to form molecules. One crucial concept within this arena is the hybridization of atomic orbitals. When we discuss sp hybridized orbitals, we're referring to a scenario where an atom's single s orbital mixes with a single p orbital. This process transforms the atom's valence orbitals, producing two new orbitals of equal energy—each being part sp and part unsp. These hybrid orbitals are oriented in a straight line, 180 degrees apart, which is crucial for forming certain types of molecular geometries, like the linear arrangement in a carbon dioxide molecule (CO2).

This type of hybridization is essential for understanding the structure and reactivity of molecules. It forms the basis for establishing strong sigma bonds, which hold atoms firmly together along the axis connecting them.
Unhybridized P Orbitals
While hybridization does wonders in creating equalized bonding orbitals, not all p orbitals may participate in the process. Unhybridized p orbitals are the p orbitals that remain untouched after an s orbital and one or two p orbitals have hybridized, depending on the type of hybridization involved. For instance, in sp hybridized atoms, we have one s and one p orbital hybridizing, leaving two unhybridized p orbitals per atom.

These remaining p orbitals are significant because they retain their perpendicular orientation to the hybridized orbitals and to each other, poised for further molecular interactions. They play a pivotal role in forming pi bonds, which adds to the complexity and richness of chemical bonding behavior.
Pi Bond Formation
Consider the role of those remaining, unhybridized p orbitals we just talked about. They're not just idly sitting there; they are ready to engage in pi bond formation. A pi bond is a type of covalent bond that's formed by the side-to-side overlap of two p orbitals from different atoms. It is this overlap that allows for the sharing of electrons in a space above and below the axis of the molecular framework.

Imagine two parallel lines with a ladder rung between them; that rung represents the pi bond. In the molecular universe, this translates to a pi bond being less strong than a sigma bond, because the electron cloud overlap is not directly between the nuclei of the bonding atoms. However, pi bonds are crucial for the double and triple bonds seen in organic molecules like ethene (C2H4).
Sigma and Pi Bonds
To fully appreciate molecular architecture, one must differentiate between sigma and pi bonds. A sigma bond is a bond formed directly between two atoms through the head-to-head or end-to-end overlap of orbitals, where the electron cloud is symmetrically distributed around the bond axis. In contrast, a pi bond, as we have learned, involves a side-to-side overlap and delivers a cloud of electron density above and below the bond axis.

Covalent molecules rely on sigma bonds for their basic structure. These bonds permit rotation since they don't lock the spatial orientation of the atoms they connect. Pi bonds, built on sigma bonds, restrict molecular rotation due to their parallel orbital setup. This distinction is pivotal when predicting molecular properties like the presence of geometric isomers or the rigidity of molecular structures.
Rotation around Chemical Bonds
The ability of molecules to rotate around bonds is not just theoretical trivia; it has practical implications in everything from polymer flexibility to enzyme function. Rotation around chemical bonds varies depending on whether those bonds are sigma or involve pi bonding. A pure sigma bond, such as the one in ethane (C2H6), allows free rotation. There's no pi bond to maintain specific parallel orientations, so the molecule can twirl without much ado.

However, introduce a pi bond, like in ethene (with a double bond), and everything changes. The pi bond locks the relative positions of the atoms into place, barring rotation without a significant energy input to break that bond. Understanding this rotational capacity is integral for grasping concepts such as stereochemistry and the physical properties of molecules like the difference between saturated and unsaturated fats.

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Most popular questions from this chapter

(a) Methane \(\left(\mathrm{CH}_{4}\right)\) and the perchlorate ion \(\left(\mathrm{ClO}_{4}-\right)\) are both described as tetrahedral. What does this indicate about their bond angles? (b) The \(\mathrm{NH}_{3}\) molecule is trigonal pyramidal, while \(\mathrm{BF}_{3}\) is trigonal planar. Which of these molecules is flat?

(a) Starting with the orbital diagram of a sulfur atom, describe the steps needed to construct hybrid orbitals appropriate to describe the bonding in \(\mathrm{SF}_{2}\). (b) What is the name given to the hybrid orbitals constructed in (a)? (c) Sketch the large lobes of the hybrid orbitals constructed in part (a). (d) Would the hybridization scheme in part (a) be appropriate for \(\mathrm{SF}_{4} ?\) Explain.

The molecules \(\mathrm{SiF}_{4}, \mathrm{SF}_{4}\), and \(\mathrm{XeF}_{4}\) have molecular formulas of the type \(\mathrm{AF}_{4}\), but the molecules have different molecular geometries. Predict the shape of each molecule, and explain why the shapes differ.

(a) Explain why \(\mathrm{BrF}_{4}^{-}\) is square planar, whereas \(\mathrm{BF}_{4}^{-}\) is tetrahedral. (b) Water, \(\mathrm{H}_{2} \mathrm{O}\), is a bent molecule. Predict the shape of the molecular ion formed from the water molecule if you were able to remove four electrons to make \(\left(\mathrm{H}_{2} \mathrm{O}\right)^{4+}\).

(a) What is the probability of finding an electron on the internuclear axis if the electron occupies a \(\pi\) molecular orbital? (b) For a homonuclear diatomic molecule, what similarities and differences are there between the \(\pi_{2 p}\) MO made from the \(2 p_{x}\) atomic orbitals and the \(\pi_{2 p}\) MO made from the \(2 p_{y}\) atomic orbitals? (c) Why are the \(\pi_{2 p}\) MOs lower in energy than the \(\pi_{2 p}^{*}\) MOs?

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