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

Short Answer

Expert verified
Ortho and meta dihydroxybenzene isomers have a nonzero dipole moment due to their asymmetrical molecular geometry, which leads to an uneven distribution of electron density. In contrast, para dihydroxybenzene has a symmetrical molecular geometry with equal and opposite dipole moments, resulting in a zero dipole moment.

Step by step solution

01

Draw the Lewis structures of the isomers.

First, we need to draw the Lewis structures for each of the isomers. Here are the Lewis structures for ortho, meta, and para dihydroxybenzene: Ortho: ``` O || H₁O -- C₁ C₂ -- O₂H // \\ \ HC₄──C₃──C₆──C₅ \\ / -------- H₆ ``` Meta: ``` O₁H \ C₁ // \ HC₄ C₂ -- O₂H \ // C₃──C₆ // \ HC₅ C₁ \ O₃H ``` Para: ``` O₁H \ C₁ // \ HC₄ C₂ \\ C₃ \ // C₆ ---O₂H // HC₅ ```
02

Analyze the molecular geometry.

We need to look at the molecular geometry of the structures to understand their shape and symmetry to help determine their polarity. Ortho: The molecular geometry of ortho dihydroxybenzene is not symmetrical. The hydroxyl groups are on adjacent carbon atoms. Meta: The molecular geometry of meta dihydroxybenzene is more symmetrical than ortho, but not completely symmetrical. The hydroxyl groups are separated by a single carbon atom. Para: The molecular geometry of para dihydroxybenzene is symmetrical. The hydroxyl groups are on opposite sides of the benzene ring.
03

Determine the polarity of the isomers.

To determine the polarity of each isomer, we need to consider the electronegativity of the atoms involved and the molecular geometry we determined in Step 2. Ortho: Due to the asymmetrical molecular geometry, there will be an uneven distribution of electron density, resulting in a nonzero dipole moment. Meta: The molecular geometry is not completely symmetrical, but the hydroxyl groups are separated by a single carbon atom. Due to this separation, there is a stronger combined dipole moment, resulting in a nonzero dipole moment. Para: The molecular geometry is symmetrical, meaning the dipole moments for the hydroxyl groups are equal and opposite, ultimately canceling each other out. This results in a zero dipole moment. So, ortho and meta dihydroxybenzene have nonzero dipole moments, while para dihydroxybenzene has a zero dipole moment.

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Key Concepts

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

Molecular Polarity
Molecular polarity refers to the distribution of electric charge across a molecule. It's crucial because it determines many physical properties of compounds, like solubility, boiling point, and intermolecular interactions.
To understand the polarity of a molecule, consider both the shape of the molecule and the electronegativity differences between atoms. Electronegativity is the tendency of an atom to attract electrons. When there’s a significant difference in electronegativity between bonded atoms, a dipole moment is created.
  • A nonpolar molecule has an equal sharing of electrons, resulting in no net dipole moment. This is often the case in symmetric molecules, where dipole moments cancel each other out.
  • A polar molecule results from an uneven distribution of electrons, leading to a net dipole moment.
In terms of dihydroxybenzene isomers, ortho and meta isomers are polar due to their asymmetric electron distribution. Their molecular geometries lead to a nonzero dipole moment. However, the para isomer, thanks to its symmetric geometry, balances the dipole moments, leading to zero dipole moment.
Lewis Structures
Lewis structures are diagrams showing the bonds between atoms in a molecule and the lone pairs of electrons that may exist in the molecule. They are a useful way to visualize the arrangement of electrons and how they participate in bonding.
The Lewis structures for ortho, meta, and para dihydroxybenzene help us understand the location of the hydroxyl groups relative to each other.
  • Ortho: The hydroxyl groups are adjacent on the benzene ring.
  • Meta: The hydroxyl groups are separated by one carbon atom.
  • Para: The hydroxyl groups are on opposite sides of the benzene ring.
These structures are valuable not only for identifying the location of atoms but also for predicting molecular properties such as polarity and reactivity. Drawing these structures helps see how molecular geometry will influence molecular polarity.
Molecular Geometry
Molecular geometry is about the three-dimensional arrangement of atoms in a molecule. It plays a key role in determining the physical and chemical properties of a molecule, especially its polarity.
For the isomers of dihydroxybenzene:
  • Ortho: It has an asymmetrical geometry since the hydroxyl groups are close, which tilts the electron cloud distribution.
  • Meta: Although more symmetrical than ortho, the uneven separation of hydroxyl groups gives it some level of polarity.
  • Para: Its symmetrical structure places hydroxyl groups opposite each other, leading to an even distribution of electron density, neutralizing any dipole moments.
Understanding the molecular geometry assists in predicting the behavior of a molecule in different environments, such as in solvents or during chemical reactions. These factors are essential for determining the suitability of a molecule in various applications, including pharmaceuticals and materials science.

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Most popular questions from this chapter

Indicate whether each statement is true or false. (a) \(p\) orbitals can only make \(\sigma\) or \(\sigma^{*}\) molecular orbitals. (b) The probability is always \(0 \%\) for finding an electron in an antibonding orbital. (c) Molecules containing electrons that occupy antibonding orbitals must be unstable. (d) Electrons cannot occupy a nonbonding orbital.

(a) Predict the electron-domain geometry around the central \(\mathrm{S}\) atom in \(\mathrm{SF}_{2}, \mathrm{SF}_{4}\), and \(\mathrm{SF}_{6}\). ( \(\mathbf{b}\) ) The anion \(\mathrm{IO}_{4}^{-}\) has a tetrahedral structure: three oxygen atoms form double bonds with the central iodine atom and one oxygen atom which carries a negative charge forms a single bond. Predict the molecular geometry of \(\mathrm{IO}_{6}{ }^{5-}\).

Describe the bond angles to be found in each of the follow- ing molecular structures: (a) trigonal planar, \((\mathbf{b})\) tetrahedral, (c) octahedral, (d) linear.

The following is part of a molecular orbital energy-level diagram for MOs constructed from 1 s atomic orbitals. (a) What labels do we use for the two MOs shown? (b) For which of the following molecules or ions could this be the energy-level diagram: $$ \mathrm{H}_{2} \mathrm{He}_{2}, \mathrm{H}_{2}^{+}, \mathrm{He}_{2}^{+}, \mathrm{or} \mathrm{H}_{2}^{-} ? $$ (c) What is the bond order of the molecule or ion? (d) If an electron is added to the system, into which of the MOs will it be added? [Section 9.7\(]\)

(a) Draw Lewis structures for chloromethane \(\left(\mathrm{CH}_{3} \mathrm{Cl}\right),\) chloroethene \(\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}\right)\), and chloroethyne \(\left(\mathrm{C}_{2} \mathrm{HCl}\right) .(\mathbf{b})\) 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?

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