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Chapter 11: Problem 15

Write Lewis structures for the following molecules, and then label each \(\sigma\) and \(\pi\) bond. (a) \(\mathrm{HCN} ;\) (b) \(\mathrm{C}_{2} \mathrm{N}_{2}\) (c) \(\mathrm{CH}_{3} \mathrm{CHCHCCl}_{3} ;\) (d) HONO.

Short Answer

Expert verified
The Lewis structures of the molecules are: (a) H-C鈮 (b) \(-C鈮-)-N=N- (c) H_3C-CHCHCCl_3 (d) H-O-N=O. In (a), there are two \(\sigma\) bonds and two \(\pi\) bonds; in (b), there are three \(\sigma\) bonds and four \(\pi\) bonds; in (c), there are nine \(\sigma\) bonds and no \(\pi\) bonds; and in (d), there are three \(\sigma\) bonds and one \(\pi\) bond.

Step by step solution

01

Draw Lewis Structures

A Lewis structure is a visual representation of the molecule that shows how the valence electrons are distributed among the atoms in the molecule. Lewis structures can be drawn for each of the molecules as follows: (a) \( \mathrm{HCN} \) (b) \(\mathrm{C}_{2} \mathrm{N}_{2}\) (c) \(\mathrm{CH}_{3} \mathrm{CHCHCCl}_{3}\) (d) HONO.
02

Identify Sigma and Pi Bonds

Each molecule can now be examined for sigma (\(\sigma\)) and pi (\(\pi\)) bonds. A sigma bond (\(\sigma\)) bond is formed by the end-to-end overlapping and Pi bond (\(\pi\)) is formed by the lateral or side by side overlapping of atomic orbitals. In each molecule, single bonds represent \(\sigma\) bonds. Each double bond has one \(\sigma\) and one \(\pi\) bond while each triple bond has one \(\sigma\) and two \(\pi\) bonds.
03

Label Sigma and Pi Bonds

Now each bond in the molecule is labelled as either a sigma (\(\sigma\)) or pi (\(\pi\)) bond based on the type of overlap forming the bond. Label each single bond as a \(\sigma\) bond. Label one of the bonds in a double bond as a \(\sigma\) bond and the other as a \(\pi\) bond. For a triple bond, label one of the bonds as a \(\sigma\) bond and the other two as \(\pi\) bonds.

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Key Concepts

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

Understanding Sigma Bonds
A sigma bond, often denoted as \( \sigma \), is the strongest type of covalent bond, formed by the head-on overlap of orbitals from two atoms. This overlap occurs along the axis connecting the two nuclei, making the \( \sigma \) bond symmetrical around this axis, which provides stability, allowing for free rotation of the bonded atoms.
  • Single bonds are almost always sigma bonds due to the strong overlap.
  • In multiple bonds, such as double or triple bonds, there is always exactly one sigma bond, with the others being pi bonds.
To visualize a sigma bond, think of two spheres meeting directly at their surfaces. Besides providing basic linkage between atoms, \( \sigma \) bonds also determine the molecular geometry. This is why understanding them is crucial for predicting how molecules behave and interact.
The Role of Pi Bonds
Pi bonds, depicted as \( \pi \) bonds, are formed from the side-to-side overlap of orbitals. These bonds are weaker than sigma bonds, as the overlap is less extensive compared to the end-to-end overlap.
  • A single pi bond is present in double bonds.
  • Triple bonds contain two pi bonds along with a sigma bond.
When visualizing pi bonds, imagine two lobes of overlapping atomic orbitals placed above and below the axis that runs between two bonded atoms. This lateral overlapping prevents the rotation around the bond \( \pi \) bonds are what give multiple bonds (like double and triple bonds) their rigidity. This constraint helps determine the overall shape and properties of the molecule.
Valence Electrons Distribution Demystified
When drawing Lewis structures, it's crucial to understand how valence electrons are distributed among atoms in a molecule. Valence electrons are responsible for the bonding and chemical properties of an element.
  • Lewis structures aim to show how valence electrons pair and form bonds.
  • They help in visualizing the distribution of electrons and the kinds of bonds鈥� like sigma or pi鈥� present in the molecule.
Each atom seeks to complete its outer shell with a noble gas configuration, typically an octet for most elements. In doing so, atoms share electrons, forming bonds distinguished by their electron sharing configuration鈥攕igma for sharing one pair of electrons head-on, or pi, for sharing above and below the bond axis. Understanding this electron distribution is key in predicting molecular geometry and reactivity.

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Most popular questions from this chapter

Methyl nitrate, \(\mathrm{CH}_{3} \mathrm{NO}_{3}\), is used as a rocket propellant. The skeletal structure of the molecule is \(\mathrm{CH}_{3} \mathrm{ONO}_{2}\). The N and three O atoms all lie in the same plane, but the \(\mathrm{CH}_{3}\) group is not in the same plane as the \(\mathrm{NO}_{3}\) group. The bond angle \(\mathrm{C}-\mathrm{O}-\mathrm{N}\) is \(105^{\circ},\) and the bond angle \(\mathrm{O}-\mathrm{N}-\mathrm{O}\) is \(125^{\circ} .\) One nitrogen-to-oxygen bond length is \(136 \mathrm{pm},\) and the other two are \(126 \mathrm{pm}\) (a) Draw a sketch of the molecule showing its geometric shape. (b) Label all the bonds in the molecule as \(\sigma\) or \(\pi\), and indicate the probable orbital overlaps involved. (c) Explain why all three nitrogen-to-oxygen bond lengths are not the same.

In which of the following molecules would you expect to find delocalized molecular orbitals: (a) \(\mathrm{C}_{2} \mathrm{H}_{4}\) (b) \(\mathrm{SO}_{2} ;\) (c) \(\mathrm{H}_{2} \mathrm{CO}\) ? Explain.

For each of the species \(\mathrm{C}_{2}^{+}, \mathrm{O}_{2}^{-}, \mathrm{F}_{2}^{+},\) and \(\mathrm{NO}^{+}\) (a) Write the molecular orbital diagram (as in Example \(11-6)\) (b) Determine the bond order, and state whether you expect the species to be stable or unstable. (c) Determine if the species is diamagnetic or paramagnetic; and if paramagnetic, indicate the number of unpaired electrons.

Furan, \(\mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O},\) is a substance derivable from oat hulls, corn cobs, and other cellulosic waste. It is a starting material for the synthesis of other chemicals used as pharmaceuticals and herbicides. The furan molecule is planar and the \(\mathrm{C}\) and \(\mathrm{O}\) atoms are bonded into a fivemembered pentagonal ring. The H atoms are attached to the C atoms. The chemical behavior of the molecule suggests that it is a resonance hybrid of several contributing structures. These structures show that the double bond character is associated with the entire ring in the form of a \(\pi\) electron cloud. (a) Draw Lewis structures for the several contributing structures to the resonance hybrid mentioned above. (b) Draw orbital diagrams to show the orbitals that are involved in the \(\sigma\) and \(\pi\) bonding in furan. [Hint: You need use only one of the contributing structures, such as the one with no formal charges.] (c) How many \(\pi\) electrons are there in the furan molecule? Show that this number of \(\pi\) electrons is the same, regardless of the contributing structure you use for this assessment.

Consider the molecules \(\mathrm{CO}^{+}\) and \(\mathrm{CN}^{-}\) and use molecular orbital theory to answer the following: (a) Write the molecular orbital configuration of each ion (ignore the 1 s electrons). (b) Predict the bond order of each ion. (c) Which of these ions is paramagnetic? Which is diamagnetic? (d) Which of these ions do you think has the greater bond length? Explain.
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