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Two different compounds have the formula \(\mathrm{XeF}_{2} \mathrm{Cl}_{2}\) . Write Lewis structures for these two compounds, and describe how measurement of dipole moments might be used to distinguish between them.

Short Answer

Expert verified
The Lewis structures for the two compounds with the formula XeF2Cl2 are: Compound 1: :F: | :Cl-Xe-Cl: | :F: Compound 2: :F: | :Cl:Xe:Cl: | :F: Measurement of dipole moments can be used to distinguish between these two compounds. Compound 1 has a net dipole moment of zero, making it a nonpolar molecule, whereas Compound 2 has a non-zero net dipole moment, making it a polar molecule. Techniques such as infrared or Raman spectroscopy can be used to differentiate between these compounds based on their interactions with electromagnetic radiation.

Step by step solution

01

Determine the Possible Lewis Structure for Compound 1

Firstly, we will create a possible Lewis structure of the first compound. Xe has a total of 8 valence electrons, F has 7, and Cl has 7. We can start by putting the Xe atom in the center and arranging the F and Cl atoms around it. In compound 1, let's consider that the two F atoms are positioned opposite to each other and the Cl atoms are also on opposite sides. The structure would look like: ``` F | Xe - Cl | F ``` Now, we will add lone pairs. Each F atom will have 3 lone pairs, and each Cl atom will have 2 lone pairs. The central Xe atom has no lone pairs left because it has 8 electrons and each bond contributes 2 electrons. The Lewis structure for compound 1 is: ``` :F: | :Cl-Xe-Cl: | :F: ```
02

Determine the Possible Lewis Structure for Compound 2

For the second compound, let's consider that the F and Cl atoms are positioned adjacently. The structure would look like: ``` F | Cl - Xe - Cl | F ``` Adding the lone pairs like before, we have: ``` :F: | :Cl:Xe:Cl: | :F: ```
03

Understand the Concept of Dipole Moments

A dipole moment is a measure of the charge distribution in a molecule. It is described as the product of the charge and the distance between the atoms. The dipole moment (\(\mu\)) can be calculated using the formula: \[\mu = Q \times d\] Where: - \(Q\) denotes the charge difference between the bonded atoms. - \(d\) denotes the distance between the bonded atoms. Dipole moments are vector quantities. In a molecule, the overall dipole moment is determined by the vector sum of the individual bond dipole moments. A molecule with a net dipole moment is polar, whereas a molecule with no net dipole moment (the bond dipoles cancel each other out) is nonpolar.
04

Determine Dipole Moments for both Compounds

In compound 1, the dipole moments of the opposite Xe-F and Xe-Cl bonds will cancel each other out, leading to an overall net dipole moment of zero. This compound is nonpolar. In compound 2, the bond dipole moments of the Xe-F and Xe-Cl bonds do not cancel each other out, leading to an overall non-zero net dipole moment. This compound is polar.
05

Explain how Measurement of Dipole Moments can be used to Distinguish Between Them

By measuring the dipole moments of the two compounds, we can distinguish between them. Since compound 1 has a net dipole moment of zero, it is nonpolar. On the other hand, compound 2 has a non-zero net dipole moment, making it a polar molecule. Using methods such as infrared or Raman spectroscopy, polar and nonpolar molecules can be distinguished based on their interactions with electromagnetic radiation. Therefore, by measuring the dipole moments and observing the behavior of the molecules in these techniques, it is possible to differentiate between compound 1 and compound 2.

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Key Concepts

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

Dipole Moments
Dipole moments are essential for understanding the distribution of electric charges in a molecule. They arise when there is a separation of positive and negative charges, which typically occurs in polar covalent bonds. The dipole moment is a vector quantity consisting of both magnitude and direction.
It can be calculated using the formula: \[\mu = Q \times d\]Here, \(\mu\) represents the dipole moment, \(Q\) is the difference in charge between atoms, and \(d\) is the distance between the bonded atoms.
In simpler terms, imagine a tug of war where atoms pull on a shared pair of electrons. The stronger the pull (difference in electronegativity), the larger the dipole moment.
For example, in a molecule like water (H₂O), there is a significant dipole moment because the oxygen atom is much more electronegative than the hydrogen atoms, creating a particular orientation of charges with distinct polar regions.
Measuring and analyzing dipole moments can provide insights into the molecular structure and help determine if a molecule is polar or nonpolar.
Molecular Polarity
Understanding molecular polarity involves looking at the shape and charge distribution of a molecule. It directly depends on the arrangement of bonds and the symmetry of the molecule.
A molecule is polar if it has an uneven distribution of electron density, resulting in distinct positive and negative ends, or poles. Nonpolar molecules, on the other hand, generally have their dipoles cancel out due to a symmetric arrangement.
When considering molecular polarity, it's important to consider both the shape and the constituent atoms' electronegativity. Even if bond dipoles are present due to electronegativity differences, the overall molecule can be nonpolar if the molecular geometry allows these dipoles to cancel each other.
For instance, carbon dioxide (CO₂) is a linear molecule where oxygen atoms on each side pull equally in opposite directions, leading to cancellation of the dipoles and a nonpolar molecule overall. In contrast, ammonia (NH₃) is polar because its pyramidal shape leads to an unequal distribution of charge.
Being aware of whether a compound is polar or nonpolar helps predict its interactions in a chemical system and its solubility in various solvents.
XeF2Cl2 Compounds
Compounds with the formula \(\mathrm{XeF_2Cl_2}\) offer a great example to study molecular polarity and dipole moments. Xenon (Xe), despite being a noble gas atom, can form compounds with fluorine (F) and chlorine (Cl) due to the availability of vacant p and d orbitals. This gives rise to diverse molecular geometries that influence molecular properties.
In the first possible compound structure, when \(\mathrm{F}\) and \(\mathrm{Cl}\) atoms are arranged opposite each other around \(\mathrm{Xe}\), the molecular shape allows for the cancellation of dipoles due to the symmetrical arrangement. As a result, the molecule has an overall net dipole moment of zero, making it nonpolar.
In the second structure, if \(\mathrm{F}\) and \(\mathrm{Cl}\) atoms are positioned adjacently, the asymmetrical shape leads to a net dipole moment as the dipoles do not completely cancel out. This makes the molecule polar.
These two structural isomers can actually be distinguished by measuring their dipole moments using techniques like spectroscopy. A nonpolar compound would not interact with an electric field in the same way as a polar one would. Learning about compounds like \(\mathrm{XeF_2Cl_2}\) provides an in-depth understanding of how atomic arrangement and molecular geometry affect chemical behavior.

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Most popular questions from this chapter

Carbon and sulfur form compounds with each other with the formulas \(\mathrm{CS}_{2}\) and \(\mathrm{C}_{3} \mathrm{S}_{2}\) . Draw a Lewis structure for each compound that has a formal charge of zero for all atoms in the structure.

Identify the five compounds of \(\mathrm{H}, \mathrm{N},\) and \(\mathrm{O}\) described as follows. For each compound, write a Lewis structure that is consistent with the information given. a. All the compounds are electrolytes, although not all of them are strong electrolytes. Compounds \(\mathrm{C}\) and \(\mathrm{D}\) are ionic and compound \(\mathrm{B}\) is covalent. b. Nitrogen occurs in its highest possible oxidation state in compounds \(\mathrm{A}\) and \(\mathrm{C}\); nitrogen occurs in its lowest possible oxidation state in compounds \(\mathrm{C}, \mathrm{D}\), and \(\mathrm{E}\). The formal charge on both nitrogens in compound \(\mathrm{C}\) is \(+1\); the formal charge on the only nitrogen in compound \(\mathrm{B}\) is 0. c. Compounds \(\mathrm{A}\) and \(\mathrm{E}\) exist in solution. Both solutions give off gases. Commercially available concentrated solutions of compound A are normally 16\(M .\) The commercial, concentrated solution of compound \(E\) is 15\(M .\) d. Commercial solutions of compound E are labeled with a misnomer that implies that a binary, gaseous compound of nitrogen and hydrogen has reacted with water to produce ammonium ions and hydroxide ions. Actually, this reaction occurs to only a slight extent. e. Compound \(D\) is 43.7\(\% \mathrm{N}\) and 50.0\(\%\) O by mass. If compound D were a gas at STP, it would have a density of 2.86 \(\mathrm{g} / \mathrm{L} .\) f. A formula unit of compound \(\mathrm{C}\) has one more oxygen than a formula unit of compound \(\mathrm{D}\). Compounds \(\mathrm{C}\) and \(\mathrm{A}\) have one ion in common when compound \(\mathrm{A}\) is acting as a strong electrolyte. g. Solutions of compound \(\mathrm{C}\) are weakly acidic; solutions of compound \(\mathrm{A}\) are strongly acidic; solutions of compounds \(\mathrm{B}\) and \(\mathrm{E}\) are basic. The titration of 0.726 g compound \(\mathrm{B}\) requires 21.98 \(\mathrm{mL}\) of 1.000 \(\mathrm{M} \mathrm{HCl}\) for complete neutralization.

Write Lewis structures and predict whether each of the following is polar or nonpolar a. \(\mathrm{HOCN}\) (exists as \(\mathrm{HO}-\mathrm{CN} )\) b. \(\mathrm{COS}\) c. \(\mathrm{XeF}_{2}\) d. \(\mathrm{CF}_{2} \mathrm{Cl}_{2}\) e. \(\mathrm{SeF}_{6}\) f. \(\mathrm{H}_{2} \mathrm{CO}(\mathrm{C} \text { is the central atom })\)

Write electron configurations for the most stable ion formed by each of the elements Al, Ba, Se, and I (when in stable ionic compounds).

Write Lewis structures that obey the octet rule (duet rule for H) for each of the following molecules. Carbon is the central atom in \(\mathrm{CH}_{4}\) , nitrogen is the central atom in \(\mathrm{NH}_{3},\) and oxygen is the central atom in \(\mathrm{H}_{2} \mathrm{O}\) . a. \(\mathrm{F}_{2} \quad\) e. \(\mathrm{NH}_{3}\) b. \(\mathrm{O}_{2} \quad\) f. \(\mathrm{H}_{2} \mathrm{O}\) c. \(\mathrm{CO} \quad\) g. \(\mathrm{HF}\) d. \(\mathrm{CH}_{4}\)

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