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

Short Answer

Expert verified
Ortho-dichlorobenzene and meta-dichlorobenzene have nonzero dipole moments as their bond dipoles do not cancel each other out. In contrast, para-dichlorobenzene has a zero dipole moment because the bond dipoles are equal in magnitude and opposite in direction, leading to their cancellation.

Step by step solution

01

Draw the Lewis structures of the three isomers

We will begin by drawing Lewis structures for ortho-dichlorobenzene (1,2-dichlorobenzene), meta-dichlorobenzene (1,3-dichlorobenzene), and para-dichlorobenzene (1,4-dichlorobenzene). Ortho-dichlorobenzene: ``` Cl | / \ --C---C-- |\ / \ | C Cl |/ \_/ --C---C-- \ H ``` Meta-dichlorobenzene: ``` Cl | / \ --C---C-- |\_ /_\ | C C-Cl | \_/_| |--C---C- ``` Para-dichlorobenzene: ``` Cl | / \ --C---C-- |/_\_/\ | C-Cl |:_:_:| --C---C-- ```
02

Analyze the molecular geometry

Now that we have the Lewis structures, we need to analyze the molecular geometry of each isomer and determine if they have a nonzero dipole moment or not. Ortho-dichlorobenzene: The two Cl atoms are adjacent to each other, leading to two bond dipoles that don't cancel each other out. These bond dipoles add up in the plane of the molecule, resulting in a nonzero dipole moment. Meta-dichlorobenzene: The two Cl atoms are opposite to each other on the benzene ring, resulting in bond dipoles that are not equal in magnitude but opposite in direction. This means that the bond dipoles partially cancel each other out, leading to a nonzero dipole moment. Para-dichlorobenzene: The two Cl atoms are positioned diametrically opposite to one another on the benzene ring. As a result, the bond dipoles are equal in magnitude and opposite in direction, meaning they completely cancel out each other. Thus, this isomer has a zero dipole moment.
03

Determine which isomers have nonzero dipole moments

Based on our analysis, we can conclude that ortho-dichlorobenzene (1,2-dichlorobenzene) and meta-dichlorobenzene (1,3-dichlorobenzene) have nonzero dipole moments, whereas para-dichlorobenzene (1,4-dichlorobenzene) has a zero dipole moment.

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.

Lewis Structures
When embarked on the journey of understanding molecules, Lewis structures are the fundamental diagrams we create to visually map out the valence electrons and their bonds within a molecule. For instance, when we observe the different isomers of dichlorobenzene, Lewis structures aide us in showcasing the specific placements of chlorine (Cl) atoms attached to the benzene ring. This representation is critical as it allows us to visualize the distinct structures of ortho-, meta-, and para-dichlorobenzene.

Despite the similarities in their molecular formulas, it's the Lewis structures that unveil the unique arrangement of atoms. In ortho-dichlorobenzene, the two Cl atoms sit next to each other, while in meta-dichlorobenzene, they are separated by a single benzene carbon. Para-dichlorobenzene showcases the Cl atoms at opposite ends of the benzene ring. These varying placements are not mere aesthetics – they have a profound impact on the molecule's characteristics and are pivotal when determining properties like the dipole moment.
Molecular Geometry
Molecular geometry goes beyond the simple connections between atoms; it deals with the three-dimensional arrangement of atoms in a molecule. By understanding the molecular geometry of different isomers of dichlorobenzene, we unlock the reasons behind their varying dipole moments.

The genuine shape of the molecule matters. For ortho-dichlorobenzene, the Cl atoms' proximity leads to a non-symmetrical arrangement creating a notable dipole moment. Conversely, para-dichlorobenzene's symmetrical geometry allows the opposing dipoles to cancel each other out, resulting in a molecule with no overall dipole moment. Meta-dichlorobenzene, sitting between these two extremes, has a less symmetrical shape than para but is more balanced than ortho, resulting in a nonzero yet smaller dipole moment.
Chemical Isomerism
Chemical isomerism is the phenomenon where compounds with the same molecular formula exhibit different chemical structures. This concept helps us to understand why molecules like dichlorobenzene, which share the same number of each type of atom, can have wildly differing physical and chemical properties.

Isomers like ortho-, meta-, and para-dichlorobenzene are prime examples of structural isomerism, specifically positional isomers, where the position of the substituent groups (in this case, chlorine atoms) varies on the benzene ring. Each position results in a distinctive distribution of electrons and thus contributes to different reactivity, boiling points, and as we've explored, dipole moments. Understanding isomerism is crucial for chemistry students not only to ace their assignments but also to later grasp the complexities of organic chemistry and the behavior of molecules in various environments.

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) Which geometry and central atom hybridization would you expect in the series \(\mathrm{BH}_{4}^{-}, \mathrm{CH}_{4}, \mathrm{NH}_{4}^{+} ?(\mathbf{b})\) What would you expect for the magnitude and direction of the bond dipoles in this series? (c) Write the formulas for the analogous species of the elements of period 3 ; would you expect them to have the same hybridization at the central atom?

Give the electron-domain and molecular geometries of a molecule that has the following electron domains on its central atom: (a) four bonding domains and no nonbonding domains, (b) three bonding domains and two nonbonding domains, (c) five bonding domains and one nonbonding domain, (d) four bonding domains and two nonbonding domains.

What is the hybridization of the central atom in (a) \(\mathrm{SiCl}_{4}\), (b) HCN, (d) \(\mathrm{TeCl}_{2}\) (c) \(\mathrm{SO}_{3}\)

Consider the Lewis structure for glycine, the simplest amino acid: (a) What are the approximate bond angles about each of the two carbon atoms, and what are the hybridizations of the orbitals on each of them? (b) What are the hybridizations of the orbitals on the two oxygens and the nitrogen atom, and what are the approximate bond angles at the nitrogen? (c) What is the total number of \(\sigma\) bonds in the entire molecule, and what is the total number of \(\pi\) bonds?

Sulfur tetrafluoride \(\left(\mathrm{SF}_{4}\right)\) reacts slowly with \(\mathrm{O}_{2}\) to form sulfur tetrafluoride monoxide \(\left(\mathrm{OSF}_{4}\right)\) according to the following unbalanced reaction: $$\mathrm{SF}_{4}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{OSF}_{4}(g)$$ The \(\mathrm{O}\) atom and the four \(\mathrm{F}\) atoms in \(\mathrm{OSF}_{4}\) are bonded to a central S atom. (a) Balance the equation. (b) Write a Lewis structure of \(\mathrm{OSF}_{4}\) in which the formal charges of all atoms are zero. (c) Use average bond enthalpies (Table 8.4 ) to estimate the enthalpy of the reaction. Is it endothermic or exothermic? (d) Determine the electron-domain geometry of \(\mathrm{OSF}_{4},\) and write two possible molecular geometries for the molecule based on this electron- domain geometry. (e) Which of the molecular geometries in part (d) is more likely to be observed for the molecule? Explain.

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