Consider all the hydrogen halide molecules \(\mathrm{HX}\), where \(X\) is a group VIIA atom. (a) Which is the most polar? Why? (b) Which is the least polar? Why? (c) Draw all these molecules, showing their relative bond dipole moments.

Short Answer

Expert verified
The most polar hydrogen halide molecule is HF, due to the largest difference in electronegativity values between hydrogen (2.1) and fluorine (3.98). The least polar hydrogen halide molecule is HAt, due to the smallest difference in electronegativity values between hydrogen (2.1) and astatine (2.2). The bond dipole moments point from the hydrogen atom to the halogen atom in each molecule, with the arrow length representing the relative strength of the bond dipole moment.

Step by step solution

01

Identify the Group VIIA halogens

First, we need to identify the Group VIIA halogens. The Group VIIA elements, also known as halogens, are: - Fluorine (F) - Chlorine (Cl) - Bromine (Br) - Iodine (I) - Astatine (At) Now we can build the hydrogen halide molecules as HF, HCl, HBr, HI, and HAt.
02

Compare electronegativity values of halogens

To determine the polarity of each hydrogen halide molecule, we must compare the electronegativity values of the halogen atoms. Electronegativity refers to the ability of an atom to attract bonding electrons toward itself in a molecule. The greater the difference in electronegativity between two atoms, the more polar their bond is. The general electronegativity trend across the periodic table is that it increases as you go from left to right and decreases as you go down a group. Here are the approximate electronegativity values (in Pauling scale) for each halogen atom: - Fluorine (F): 3.98 - Chlorine (Cl): 3.16 - Bromine (Br): 2.96 - Iodine (I): 2.66 - Astatine (At): 2.2 (largely based on estimates, as it is less common in nature) Hydrogen (H) has an electronegativity value of 2.1.
03

Determine the most and least polar hydrogen halide molecules

Using the electronegativity values, we can determine the most and least polar hydrogen halide molecules. a) Most polar: Since electronegativity increases as you go up the group, the most polar molecule will have the largest electronegativity difference between hydrogen and the halogen. In this case, HF will have the greatest difference in electronegativity values (3.98 - 2.1 = 1.88) and is therefore the most polar molecule. b) Least polar: Conversely, the least polar molecule will have the smallest electronegativity difference between hydrogen and the halogen. As you go down the group, electronegativity decreases, so in this case, it will be HAt (2.2 - 2.1 = 0.1) with the least difference in electronegativity values and hence the least polar molecule.
04

Draw the hydrogen halide molecules and their bond dipole moments

To draw the molecules and their bond dipole moments, represent the direction of the dipole moment with an arrow pointing from the lower electronegative (hydrogen) atom towards the more electronegative (halogen) atom. Here are the hydrogen halide molecules and their bond dipole moments: HF: ``` H - F → ``` HCl: ``` H - Cl → ``` HBr: ``` H - Br → ``` HI: ``` H - I → ``` HAt: ``` H - At → ``` In each molecule, the arrow represents the bond dipole moment, pointing from the less electronegative hydrogen atom towards the more electronegative halogen atom. The length of the arrow can also be used to show the relative strength of the bond dipole moment, with a longer arrow representing a more polar bond (in the case of HF) and a shorter arrow representing a less polar bond (in the case of HAt).

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Key Concepts

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

Electronegativity
Electronegativity is a fundamental concept in chemistry that describes an atom's tendency to attract and hold onto electrons within a bond. Imagine it as a measure of an atom's greed for electrons. In a bond, if one atom is more electronegative than the other, it will pull the shared electrons closer to itself, creating a polarity in the bond. This difference in electronegativity between hydrogen and a halogen is what dictates the polarity of hydrogen halides.

On the Pauling scale, which is commonly used to quantify electronegativity, atoms increase in electronegativity from left to right across a period and decrease from top to bottom down a group. Therefore, in the context of hydrogen halides, the most electronegative halide, fluorine, forms the most polar bond with hydrogen, while astatine, which is the least electronegative, forms the least polar bond.
Bond Dipole Moment
The bond dipole moment is a vector quantity—it has both magnitude and direction—and it gives us a way to visualize the polarity of a chemical bond. Think of it as an arrow that points from the positively charged end of the bond (the less electronegative atom) to the negatively charged end (the more electronegative atom).

The size of the bond dipole moment is dependent on two factors: the difference in electronegativity between the two bonded atoms and the distance between the nuclei, also known as bond length. The greater the difference in electronegativity, the larger the dipole moment. Consequently, in hydrogen halides, HF has the largest bond dipole moment due to the substantial difference in electronegativity between hydrogen and fluorine, while HAt has the smallest bond dipole moment.
Periodic Table Trends
Understanding the trends in the periodic table is crucial for predicting the behavior of atoms, particularly their electronegativity, and, by extension, the properties of the molecules they form. As previously noted, in general, electronegativity increases across a period due to the increasing number of protons in the nucleus, which in turn pulls the electrons closer. Conversely, as you move down a group, the increase in the number of electron shells outweighs the effect of additional protons, resulting in a decrease in electronegativity.

These trends explain why hydrogen bonded to a halogen higher on the periodic table forms a more polar bond—there is a more significant attraction for the shared electrons. In our hydrogen halide example, we observed this trend firsthand, with the polarity declining from HF to HAt.
Chemical Bonding
Chemical bonding is the process that holds atoms together in molecules. Atoms bond to achieve a more stable electron configuration, often resembling the full outer shells of noble gases. There are several types of bonds, including ionic, covalent, and metallic bonds. Hydrogen halides are formed through covalent bonding, where hydrogen shares its lone electron with the halogen's unpaired electron, leading to a stable molecule.

It is important, however, to recognize that not all covalent bonds are equal. If the atoms involved in the bond have different electronegativities, the shared electrons will not be distributed evenly, giving rise to a polar covalent bond as seen in hydrogen halides. The outcome is a molecule that has a partial positive charge on one end (hydrogen) and a partial negative charge on the other (halogen), enabling it to engage in various intermolecular interactions and lending to its unique chemical properties.

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Most popular questions from this chapter

Which, if any, of these molecules do you expect to be polar: \(\mathrm{CO}_{2}, \mathrm{CS}_{2}\), or \(\mathrm{CSO}\) (carbon is the central atom in all three molecules)? Explain your answer.

Explain the difference between electron-group geometry and molecular shape. How do you use electron-group geometry when deciding what shape a molecule has?

The dot diagram \(\ddot{\mathrm{Q}}=\mathrm{S}=\ddot{\mathrm{O}}\) for \(\mathrm{SO}_{2}\) is incorrect. (a) Draw the correct dot diagram. (b) How do the correct and incorrect dot diagrams differ in their prediction of molecular shape and polarity?

There are exceptions to the predictions of VSEPR. Consider \(\mathrm{CH}_{3}\), known as a methyl radical. (a) Create a dot diagram for the methyl radical. How is it fundamentally different from other dot diagrams you have done? (b) Use VSEPR to predict the shape of the methyl radical and draw it with that shape (treat the odd electron as a single electron group). (c) The methyl radical is known to be planar with \(120^{\circ} \mathrm{H}-\mathrm{C}-\mathrm{H}\) angles. What steric number is being employed here, and what is the carbon atom doing with respect to the odd electron in determining molecular shape? (d) The \(\mathrm{CF}_{3}\) radical does obey VSEPR. Draw it according to its VSEPR-predicted shape. What steric number is being employed here? (e) The \(\mathrm{C}-\mathrm{H}\) bond is shorter than the \(\mathrm{C}-\mathrm{F}\) bond. When bonds are short, the atoms at the ends of the bonds can bang into each other (this is called steric congestion) unless a geometry is adopted to get around this. Use this knowledge to explain why \(\mathrm{CH}_{3}\) violates VSEPR, but \(C F_{3}\) does not.

Some molecules have central atoms with steric numbers greater than 4 . For example, the central sulfur atom in \(\mathrm{SF}_{6}\) has a steric number of 6 . Its actual geometry (called octahedral), is shown below (all the bond angles are \(90^{\circ}\) or \(180^{\circ}\) ). (a) Are the bonds in \(\mathrm{SF}_{6}\) polar, polar covalent, or ionic? Explain. (b) Would you predict the \(\mathrm{SF}_{6}\) molecule to be polar or nonpolar? Justify your answer.

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