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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? Explain.

Short Answer

Expert verified
(a) There are two unhybridized p orbitals remaining in the valence shell, and the atom can form two π bonds. (b) It would be easier to twist around a single σ bond than around a double bond (σ plus π). This is because the π bond restricts rotation, which creates a higher energy barrier compared to a single σ bond.

Step by step solution

01

Identify hybrid orbitals and number of valence electrons

In sp hybridization, the s and p orbitals are combined to form two new hybrid orbitals, sp. Since there are three p orbitals in the valence shell (px, py, and pz), one of them is used for hybridization with the s orbital, leaving two p orbitals unhybridized.
02

Determine the number of remaining unhybridized p orbitals

Since one p orbital is used for hybridization, there are two unhybridized p orbitals remaining in the valence shell (lets say, py and pz).
03

Determine the number of π bonds possible

Each unhybridized p orbital can form one π bond. Therefore, the atom can form two π bonds. #Answer_Part_A# (a) There are two unhybridized p orbitals remaining in the valence shell, and the atom can form two π bonds. (b)
04

Understand the nature of σ and π bonds

A σ bond is formed by the head-on overlap of orbitals between two atoms, while a π bond is formed by the sideways overlap of orbitals. Because of this, σ bonds allow free rotation, while π bonds restrict rotation, making it difficult to twist the atoms around the bond without breaking the π bond.
05

Compare ease of rotation around a single σ bond and a double bond (σ plus π)

Since a single bond has only a σ bond, rotation around the bond is relatively easy. However, double bonds have both a σ and a π bond, and rotation around the bond requires breaking the π bond causing a higher energy barrier, making it more difficult and energetically unfavorable to twist around a double bond. #Answer_Part_B# (b) It would be easier to twist around a single σ bond than around a double bond (σ plus π). This is because the π bond restricts rotation, which creates a higher energy barrier compared to a single σ bond.

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Key Concepts

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

sp hybridization
Hybridization is a fundamental concept in chemistry used to describe how atomic orbitals mix to form new, hybridized orbitals. In the case of sp hybridization, one s orbital and one p orbital blend together. This mixing results in two identical sp hybrid orbitals.
  • The total available valence orbitals start with one s and three p orbitals.
  • In sp hybridization, one s and one p orbital (often the px) combine, leaving two p orbitals unhybridized (py and pz).
  • The sp hybrid orbitals are linear and often used in molecules requiring 180-degree bond angles.
Sp hybridization is common in carbon atoms found in molecules like acetylene (C₂H₂), where it accounts for the linear shape of the molecule.
unhybridized p orbitals
When some p orbitals remain unhybridized, they retain their original shape and orientation. In sp hybridization, while one p orbital mixes with an s orbital, the other two p orbitals remain unhybridized.
  • These unhybridized orbitals (often py and pz) remain perpendicular to each other and to the sp hybrid orbitals.
  • They play a crucial role in forming x o n vobr bonds.
  • Each unhybridized p orbital can participate in forming one x o n vobr bond by overlapping side-by-side with another p orbital from a neighboring atom.
In molecules like ethyne (C₂H₂), the two carbon atoms each have two unhybridized p orbitals. These p orbitals allow the formation of x o n vobr bonds, creating a triple bond between the carbon atoms with one σ and two x o n vobr bonds.
sigma and pi bonds
Sigma ( σ ) and pi ( π ) bonds are types of covalent bonds differing in their orbital overlap.

σ Bonds

Sigma bonds are the first bonds formed between two atoms and arise from the head-on overlap of orbitals.
  • They allow free rotation of atoms around the bond axis due to their symmetrical orbital overlap.

π Bonds

Pi bonds form when unhybridized p orbitals overlap sideways.
  • They are always formed after a σ bond — in double or triple bonds following the first σ bond.
  • The side-by-side overlap restricts rotation, increasing rigidity in the molecular structure.
Due to these differing properties, rotation around a σ bond is easier than around a σ plus π bond. In essence, the presence of a x o n vobr bond in double bonds creates an energy barrier that must be overcome for rotation, making it less flexible compared to single σ bonds.

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Most popular questions from this chapter

(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 these hybrid orbitals. (d) Would the hybridization scheme in part (a) be appropriate for \(\mathrm{SF}_{4} ?\) Explain.

(a) Explain why \(\mathrm{BrF}_{4}^{-}\) is square planar, whereas \(\mathrm{BF}_{4}^{-}\) is tetrahedral. (b) How would you expect the \(\mathrm{H}-\mathrm{X}-\mathrm{H}\) bond angle to vary in the series \(\mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2} \mathrm{~S}, \mathrm{H}_{2} \mathrm{Se}\) ? Explain. (Hint: The size of an electron pair domain depends in part on the electronegativity of the central atom.)

Butadiene, \(\mathrm{C}_{4} \mathrm{H}_{6}\) is a planar molecule that has the following carbocarbon bond lengths: (a) Predict the bond angles around each of the carbon atoms and sketch the molecule. (b) From left to right, what is the hybridization of each carbon atom in butadiene? (c) The middle \(C-\) bond length in butadiene \((1.48\) A) is a little shorter than the average \(\mathrm{C}-\mathrm{C}\) single bond length \((1.54 \hat{\mathrm{A}}) .\) Does this imply that the middle \(\mathrm{C}-\mathrm{Cbond}\) in butadiene is weaker or stronger than the average \(\mathrm{C}-\mathrm{C}\)? (\mathbf{d} ) Based on your answer for part ( c ),discuss what additional aspects of bonding in butadiene might support the shorter middle \(\mathrm{C}-\) C bond.

(a) The \(\mathrm{PH}_{3}\) molecule is polar. Does this offer experimental proof that the molecule cannot be planar? Explain. (b) It turns out that ozone, \(\mathrm{O}_{3}\), has a small dipole moment. How is this possible, given that all the atoms are the same?

(a) Draw a picture showing how two \(p\) orbitals on two different atoms can be combined to make a sigma bond. (b) Sketch a \(\pi\) bond that is constructed from \(p\) orbitals. (c) Which is generally stronger, a \(\sigma\) bond or a \(\pi\) bond? Explain. (d) Can two \(s\) orbitals combine to form a \(\pi\) bond? Explain.

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