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

Short Answer

Expert verified
(a) CCl4: Nonpolar (tetrahedral geometry, bond dipoles cancel out) (b) NH3: Polar (trigonal pyramidal geometry, bond dipoles do not cancel out) (c) SF4: Polar (seesaw geometry, bond dipoles do not cancel out) (d) XeF4: Nonpolar (square planar geometry, bond dipoles cancel out) (e) CH3Br: Polar (tetrahedral geometry, asymmetrical charge distribution) (f) GaH3: Nonpolar (trigonal planar geometry, bond dipoles cancel out)

Step by step solution

01

(a) CCl4

CCl4 has a central carbon atom surrounded by four chlorine atoms, with each chlorine atom forming a single bond with the carbon atom. The molecular geometry of CCl4 is tetrahedral due to the repulsion among the four bond pairs. This leads to a symmetrical shape, and the individual bond dipoles cancel out, resulting in a nonpolar molecule.
02

(b) NH3

NH3 has a central nitrogen atom surrounded by three hydrogen atoms, with each hydrogen atom forming a single bond with the nitrogen atom. There is also a lone pair of electrons on the nitrogen atom. The molecular geometry of NH3 is trigonal pyramidal due to the repulsion among the three bond pairs and the lone pair. This leads to an asymmetrical shape, and the bond dipoles do not cancel out, resulting in a polar molecule.
03

(c) SF4

SF4 has a central sulfur atom surrounded by four fluorine atoms, with each fluorine atom forming a single bond with the sulfur atom. There are also two lone pairs of electrons on the sulfur atom. The molecular geometry of SF4 is seesaw due to the repulsion among the four bond pairs and the lone pairs. This leads to an asymmetrical shape, and the bond dipoles do not cancel out, resulting in a polar molecule.
04

(d) XeF4

XeF4 has a central xenon atom surrounded by four fluorine atoms, with each fluorine atom forming a single bond with the xenon atom. There are also two lone pairs of electrons on the xenon atom. The molecular geometry of XeF4 is square planar due to the repulsion among the four bond pairs and the lone pairs. Despite the polar Xe-F bonds, the symmetry of the molecule causes the bond dipoles to cancel out, resulting in a nonpolar molecule.
05

(e) CH3Br

CH3Br has a central carbon atom surrounded by three hydrogen atoms and one bromine atom, with each atom forming a single bond with the carbon atom. The molecular geometry of CH3Br is tetrahedral due to the repulsion among the four bond pairs. However, the CH3Br molecule has an asymmetrical distribution of charge due to the polar C-Br bond, which does not cancel out with the nonpolar C-H bonds. This results in a polar molecule.
06

(f) GaH3

GaH3 has a central gallium atom surrounded by three hydrogen atoms, with each hydrogen atom forming a single bond with the gallium atom. The molecular geometry of GaH3 is trigonal planar due to the repulsion among the three bond pairs. This leads to a symmetrical shape, and the individual bond dipoles cancel out, resulting in a nonpolar molecule. In summary, the molecules CCl4, XeF4, and GaH3 are nonpolar, while NH3, SF4, and CH3Br are 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 plays a pivotal role in determining the properties of molecules, including their polarity. It refers to the three-dimensional arrangement of atoms within a molecule, which arises from the chemical bonding and the electron lone pairs present. Each molecular shape is influenced by the VSEPR (Valence Shell Electron Pair Repulsion) theory, which predicts that electron pairs around a central atom will arrange themselves to minimize repulsion.

For instance, when a molecule like carbon tetrachloride (CCl4) is formed, the four chlorine atoms are positioned symmetrically around the carbon atom in a tetrahedral shape due to the repulsion between the electron pairs in the four covalent bonds. This symmetrical arrangement leads to an overall nonpolar molecule because the bond dipoles cancel each other out. Conversely, ammonia (NH3) has a trigonal pyramidal geometry due to the lone pair of electrons exerting repulsion, which disrupts the symmetry and creates a polar molecule.
Bond Dipoles
Bond dipoles emerge from differences in electronegativity between two bonded atoms. Electronegativity is a measure of an atom's ability to attract shared electrons in a chemical bond. When two atoms with differing electronegativities form a bond, the shared electrons are drawn closer to the more electronegative atom, leading to a negative partial charge on that atom and a positive partial charge on the other.

In molecules with multiple bonds, like sulfur tetrafluoride (SF4), the dipoles associated with each bond may not cancel out if the geometry is asymmetrical. The seesaw shape of SF4 causes an uneven distribution of charge, making the molecule polar. Understanding the bond dipoles and how they influence molecular polarity is essential for predicting the behavior of molecules in different chemical reactions and interactions.
Symmetrical Shape
The symmetrical shape of a molecule is a determinant factor in its polarity. Symmetry in molecular geometry occurs when the shape of the molecule allows for an even distribution of charge, leading to the cancellation of the effects of bond dipoles. For example, xenon tetrafluoride (XeF4) has a square planar geometry. Despite having polar bonds, the symmetrical arrangement results in the cancellation of bond dipoles across the molecule, rendering it nonpolar.

Similarly, trihedral molecules like gallium trihydride (GaH3) with equal bond angles and an equal distribution of charge create a condition where the bond dipoles cancel out, resulting in a nonpolar molecule. In essence, when assessing molecular polarity, one must consider not just the individual bond polarities but also the overall symmetry of the molecular structure.

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Most popular questions from this chapter

(a) Explain why the following ions have different bond angles: \(\mathrm{C} 1 \mathrm{O}_{2}^{-}\) and \(\mathrm{NO}_{2}^{-}\). Predict the bond angle in each case. (b) Explain why the \(\mathrm{XeF}_{2}\) molecule is linear and not bent.

How many nonbonding electron pairs are there in each of the following molecules: (a) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{~S} ;\) (b) \(\mathrm{HCN}\); (c) \(\mathrm{H}_{2} \mathrm{C}_{2}\) (d) \(\mathrm{CH}_{3} \mathrm{~F}\) ?

(a) Draw a picture showing how two \(p\) orbitals on two different atoms can be combined to make a sigma bond. (b) Sketch a \(\pi\) bond that is constructed from \(p\) orbitals. (c) Which is generally stronger, a \(\sigma\) bond or a \(\pi\) bond? Explain. (d) Can two s orbitals make a \(\pi\) bond? Explain.

Propylene, \(\mathrm{C}_{3} \mathrm{H}_{4}\) is a gas that is used to form the important polymer called polypropylene. Its Lewis structure is (a) What is the total number of valence electrons in the propylene molecule? (b) How many valence electrons are used to make \(\sigma\) bonds in the molecule? (c) How many valence electrons are used to make \(\pi\) bonds in the molecule? (d) How many valence electrons remain in nonbonding pairs in the molecule? (e) What is the hybridization at each carbon atom in the molecule?

The Lewis structure for allene is Make a sketch of the structure of this molecule that is analogous to Figure 9.27. In addition, answer the following three questions: (a) Is the molecule planar? (b) Does it have a nonzero dipole moment? (c) Would the bonding in allene be described as delocalized? Explain.

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