Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Dichlorobenzene, \(\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{Cl}_{2}\), exists in three forms (isomers). called ortho, meta, and para: Which of these would have a nonzero dipole moment? Explain.

Short Answer

Expert verified
Both ortho-dichlorobenzene and meta-dichlorobenzene have nonzero dipole moments due to their molecular geometry, which leads to a net polarization. Para-dichlorobenzene has a zero net dipole moment because the carbon-chlorine bonds are opposite each other on the benzene ring, causing their dipole moments to cancel each other out.

Step by step solution

01

Determine the molecular structure of each isomer

For ortho-dichlorobenzene (o-dichlorobenzene), the two chlorine atoms are attached to adjacent carbon atoms on the benzene ring. In meta-dichlorobenzene (m-dichlorobenzene), the two chlorine atoms are attached to carbon atoms with one carbon atom between them. Finally, in para-dichlorobenzene (p-dichlorobenzene), the chlorine atoms are attached to carbon atoms opposite each other on the benzene ring.
02

Analyze the polarity of individual bonds in each isomer

Since chlorine is more electronegative than carbon, the carbon-chlorine bond has a dipole moment, with the chlorine end being more negative. Therefore, each individual carbon-chlorine bond in the isomers of dichlorobenzene will be polar.
03

Analyze the molecular geometry of each isomer

The benzene ring is planar and has a hexagonal structure. In o-dichlorobenzene, the two carbon-chlorine bonds are adjacent to each other, forming a V-shape that creates a net dipole moment. In m-dichlorobenzene, the two carbon-chlorine bonds are separated by a carbon atom and are oriented at an angle to each other, and due to their position, their dipole moments do not cancel out, also resulting in a net dipole moment. In p-dichlorobenzene, the carbon-chlorine bonds are opposite each other on the benzene ring and, therefore, their dipole moments cancel each other, resulting in a zero net dipole moment.
04

Identify isomers with nonzero dipole moments

Based on our analysis, both ortho-dichlorobenzene and meta-dichlorobenzene have nonzero dipole moments, as their molecular geometry leads to a net polarization. Para-dichlorobenzene, on the other hand, has a zero net dipole moment due to the cancellation of the individual bond dipoles.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Molecular Structure
Understanding the molecular structure is crucial in predicting the physical properties of a compound, including its dipole moment. The molecular structure dictates how atoms are arranged in space and how they are bonded to each other. In the case of the three isomers of dichlorobenzene, their molecular structures differ in terms of the position of the chlorine atoms on the benzene ring.

For ortho-dichlorobenzene, the chlorine atoms are next to each other, while in meta-dichlorobenzene, there is one carbon atom between them. In para-dichlorobenzene, the chlorines are situated opposite each other. This arrangement impacts the resulting dipole moment due to the distribution of electrical charges and how they affect each other within the molecule.
Polarity of Bonds
The polarity of bonds is a measure of how electrons are distributed between two atoms in a bond. Electronegativity differences between bonded atoms cause electrons to be pulled more towards one atom, creating a polar bond. In the benzene derivatives we are examining, the bonds between carbon and the more electronegative chlorine are indeed polarized, with a partial negative charge on the chlorine and a partial positive charge on the carbon.

This polarity is consistent among all isomers of dichlorobenzene. Each carbon-chlorine bond contributes to the overall dipole moment of the molecule. Understanding the nature of these polar bonds is vital for analyzing the symmetrical aspects of the molecule, which in turn informs us about the presence or absence of an overall dipole moment.
Molecular Geometry
The molecular geometry or shape of a molecule also greatly influences its dipole moment. While the individual bonds may be polar, the three-dimensional arrangement of these polar bonds can lead to the cancellation of dipole moments. In ortho- and meta-dichlorobenzene, the asymmetrical arrangement of polar bonds results in a net dipole moment. Conversely, para-dichlorobenzene has a symmetrical geometry with the two polar bonds directly opposite each other, leading to a cancellation of dipole moments.

Such geometrical considerations not only hold the key to understanding the dipole moments but also help predict the collective effect of bond dipoles in complex molecular structures. Comprehending the spatial orientation of molecules allows us to explain and predict their physical and chemical behaviors, including reactivity and interaction with other molecules.
Isomerism in Benzene Derivatives
Isomerism is the phenomenon where compounds with the same molecular formula have different structural arrangements of atoms. In benzene derivatives, this often plays a significant role in determining properties such as dipole moment. The isomers ortho, meta, and para refer to the different positions the substituents can take on the benzene ring, leading to distinct molecular geometries and physical properties.

Ortho isomers, having adjacent substituents, usually have higher dipole moments than their meta and para counterparts. Meta isomers, with a single carbon space between substituents, exhibit a moderate level of dipole moments. In the case of para isomers, substituents are opposite to each other, an arrangement that often leads to a net dipole moment of zero as seen in para-dichlorobenzene. The study of isomerism is essential in the field of organic chemistry as it dictates how the same atoms can combine in different ways to produce substances with unique characteristics.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) Methane \(\left(\mathrm{CH}_{4}\right)\) and the perchlorate ion \(\left(\mathrm{ClO}_{6}\right)\) 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?

In ozone, \(\mathrm{O}_{3}\), the two oxygen atoms on the ends of the molecule are equivalent to one another. (a) What is the best choice of hybridization scheme for the atoms of ozone? (b) For one of the resonance forms of ozone, which of the orbitals are used to make bonds and which are used to hold nonbonding pairs of electrons? (c) Which of the orbitals can be used to delocalize the \(\pi\) electrons? (d) How many electrons are delocalized in the \(\pi\) system of ozone?

(a) Draw Lewis structures for ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\), ethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\), and acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\). (b) What is the hybridization of the carbon atoms in each molecule? (c) Predict which molecules, if any, are planar. (d) How many \(\sigma\) and \(\pi\) bonds are there in each molecule? (e) Suppose that silicon could form molecules that are precisely the analogs of ethane, ethylene, and acetylene. How would you describe the bonding about \(S i\) in terms of hydrid orbitals? Does it make a difference that Si lies in the row below \(\mathrm{C}\) in the periodic table? Explain.

(a) The nitric oxide molecule, NO, readily loses one electron to form the \(\mathrm{NO}^{+}\) ion. Why is this consistent with the electronic structure of NO? (b) Predict the order of the \(\mathrm{N}-\mathrm{O}\) bond strengths in \(\mathrm{NO}, \mathrm{NO}^{+}\), and \(\mathrm{NO}^{-}\), and describe the magnetic properties of each (c) With what neutral homonuclear diatomic molecules are the \(\mathrm{NO}^{+}\) and \(\mathrm{NO}^{-}\) ions isoelectronic (same number of electrons)?

The \(\mathrm{O}-\mathrm{H}\) bond lengths in the water molecule \(\left(\mathrm{H}_{2} \mathrm{O}\right)\) are \(0.96 \mathrm{~A}\), 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? 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 oxygen?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free