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Predict whether each of the following molecules is polar or nonpolar: \((\mathbf{a}){CCl}_{4},(\mathbf{b}) \mathrm{NH}_{3},(\mathbf{c}) \mathrm{SF}_{4},(\mathbf{d}) \mathrm{XeF}_{4},(\mathbf{e}) \mathrm{CH}_{3} \mathrm{Br}\)

Short Answer

Expert verified
In summary: - \(CCl_{4}\) is nonpolar - \(NH_{3}\) is polar - \(SF_{4}\) is polar - \(XeF_{4}\) is nonpolar - \(CH_{3}Br\) is polar

Step by step solution

01

Determine the Shape and Geometry of Each Molecule

To determine whether the molecules are polar or nonpolar, we first need to know their shape and geometry. We can find the shape and geometry by considering the central atom's valence electrons and creating a Lewis structure for each molecule.
02

Analyzing \((\mathbf{a}){CCl}_{4}\)

For CCl4, the central atom is Carbon (C) and is surrounded by four Chlorine atoms. The CCl4 molecule has four C-Cl bonds with identical bond dipoles, and it has a tetrahedral shape. Since the bond dipoles cancel out in this symmetrical shape, it has no net molecular dipole moment, and CCl4 is a nonpolar molecule.
03

Analyzing \(\mathrm{NH}_{3}\)

For NH3, the central atom is Nitrogen (N) and is surrounded by three Hydrogen atoms and has one lone pair of electrons. The NH3 molecule has three N-H bonds with non-zero bond dipoles, and it has a trigonal pyramidal shape. In this case, the bond dipoles do not cancel, leading to a net molecular dipole moment. Therefore, NH3 is a polar molecule.
04

Analyzing \(\mathrm{SF}_{4}\)

For SF4, the central atom is Sulfur (S), surrounded by four Fluorine atoms and has one lone pair of electrons. The SF4 molecule has four S-F bonds with non-zero bond dipoles, and it has a seesaw shape. In this case, the bond dipoles do not cancel, resulting in a net molecular dipole moment. Therefore, SF4 is a polar molecule.
05

Analyzing \(\mathrm{XeF}_{4}\)

For XeF4, the central atom is Xenon (Xe), surrounded by four Fluorine atoms and has two lone pairs of electrons. The XeF4 molecule has four Xe-F bonds with non-zero bond dipoles, and it has a square planar shape. In this case, bond dipoles cancel out due to the symmetrical shape, and there is no net molecular dipole moment, making XeF4 a nonpolar molecule.
06

Analyzing \(\mathrm{CH}_{3}\mathrm{Br}\)

For CH3Br, the central atom is Carbon (C), surrounded by three Hydrogen atoms and one Bromine atom. The CH3Br molecule has three C-H bonds and one C-Br bond, with different bond dipoles. Its geometry is tetrahedral. However, due to the presence of the C-Br bond, the bond dipoles do not cancel, resulting in a net molecular dipole moment, making CH3Br a polar molecule.
07

Final Results

Based on our analysis: - CCl4 is nonpolar - NH3 is polar - SF4 is polar - XeF4 is nonpolar - CH3Br is polar

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Key Concepts

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

Molecular Geometry
Molecular geometry refers to the three-dimensional arrangement of atoms around a central atom in a molecule. Understanding this concept is crucial when predicting whether a molecule is polar or nonpolar. The arrangement and the angles formed by the atoms play a significant role in determining the molecular geometry.
To predict the molecular geometry, one must consider the number of bonding pairs and lone pairs of electrons around the central atom. This arrangement can influence the overall shape and symmetry.
  • Shapes like linear, trigonal planar, and tetrahedral typically arise when there are no lone pairs affecting the geometry.
  • In contrast, geometries like trigonal pyramidal and seesaw may arise due to lone pairs affecting the molecular shape and bond angles.
Recognizing these geometrical shapes is essential for understanding the behavior of molecules under different conditions.
Lewis Structures
Lewis structures are a simple way to represent atoms and their valence electrons in a molecule. They help visualize the electron arrangement and bonding in the molecule, which is fundamental in determining molecular geometry.
To draw a Lewis structure, start by identifying the central atom and arrange the surrounding atoms around it. Distribute the valence electrons to satisfy the octet rule for the outer atom first, then the central atom.
  • For example, in \( ext{NH}_3\), Nitrogen is the central atom surrounded by three hydrogen atoms. Nitrogen has a lone pair, which can affect its geometry.
  • Similarly, in \( ext{CCl}_4\), Carbon is the central atom and Chlorine atoms are arranged symmetrically, making a perfect tetrahedral shape.
Lewis structures provide a foundation for understanding the molecular geometry and predicting the behavior of the molecule.
Dipole Moment
The dipole moment is a measure of the separation of positive and negative charges in a molecule. It quantifies the polarity of a molecule, helping to determine whether a molecule is polar or nonpolar.
If the bond dipoles do not cancel each other out, the molecule will have a net dipole moment and be polar. Conversely, if they cancel out, the molecule is nonpolar.
  • In \( ext{NH}_3\), the shape is trigonal pyramidal, which causes bond dipoles not to cancel, resulting in a polar molecule.
  • Conversely, in \( ext{CCl}_4\), the symmetrical tetrahedral shape allows the bond dipoles to cancel, making it nonpolar.
Understanding dipole moments helps in predicting the interaction of molecules with other molecules and predicting properties like solubility.
Tetrahedral Shape
The tetrahedral shape is a common and important molecular geometry where a central atom is surrounded by four atoms at the corners of a tetrahedron. The bond angle in a perfect tetrahedron is approximately \(109.5^\circ\).
The symmetry of the tetrahedral shape often leads to nonpolar molecules, as the bond dipoles tend to cancel each other out.
  • For instance, \( ext{CCl}_4\) is a classic example of a tetrahedral molecule, featuring symmetry that allows dipole cancellation.
  • However, substituting one of the atoms with another less electronegative atom, such as in \( ext{CH}_3 ext{Br}\), can result in a molecule with a net dipole moment, thus being polar.
The tetrahedral shape is crucial in chemistry for understanding compound characteristics and real-world applications.
Trigonal Pyramidal Shape
The trigonal pyramidal shape is a molecular geometry that occurs when a central atom is bonded to three other atoms and has one lone pair. The lone pair affects the angles, leading to an asymmetrical shape and a bond angle of slightly less than \(109.5^\circ\).
This shape is inherently polar because the lone pair pushes the bonding pairs, creating an asymmetry that means bond dipoles do not cancel.
  • An example of a trigonal pyramidal molecule is \(\text{NH}_3\), where the lone pair on Nitrogen creates an imbalance, leading to a net dipole moment and thus a polar molecule.
This shape is profoundly important in chemical bonding and affects the physical properties of molecules, like boiling and melting points.

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Most popular questions from this chapter

Ethyl acetate, \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{2},\) is a fragrant substance used both as a solvent and as an aroma enhancer. Its Lewis structure is (a) What is the hybridization at each of the carbon atoms of the molecule? (b) What is the total number of valence electrons in ethyl acetate? (c) How many of the valence electrons are used to make \(\sigma\) bonds in the molecule? (d) How many valence electrons are used to make \(\pi\) bonds? (e) How many valence electrons remain in nonbonding pairs in the molecule?

In which of the following molecules can you confidently predict the bond angles about the central atom, and for which would you be a bit uncertain? Explain in each case. \((\mathbf{a}) \mathrm{H}_{2} \mathrm{S},(\mathbf{b}) \mathrm{BCl}_{3},(\mathbf{c}) \mathrm{CH}_{3} \mathrm{I},(\mathbf{d}) \mathrm{CBr}_{4,}(\mathbf{e}) \mathrm{TeBr}_{4}\)

What are the electron-domain and molecular geometries of a molecule that has the following electron domains on its central atom? (a) Three bonding domains and no nonbonding domains, (b) three bonding domains and one nonbonding domain, (c) two bonding domains and two nonbonding domains.

In the formate ion, \(\mathrm{HCO}_{2}^{-}\) , the carbon atom is the central atom with the other three atoms attached to it. (a) Draw a Lewis structure for the formate ion. (b) What hybridization is exhibited by the C atom? (c) Are there multiple equivalent resonance structures for the ion? (d) How many electrons are in the \(\pi\) system of the ion?

Indicate whether each statement is true or false. (a) \(s\) orbitals can only make \(\sigma\) or \(\sigma^{*}\) molecular orbitals. (b) The probability is 100\(\%\) for finding an electron at the nucleus in a \(\pi^{*}\) orbital. (c) Antibonding orbitals are higher in energy than bonding orbitals (if all orbitals are created from the same atomic orbitals). (d) Electrons cannot occupy an antibonding orbital.

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