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Indicate the bond polarity (show the partial positive and partial negative ends) in the following bonds. a. \(\mathrm{C}-\mathrm{O}\) d. \(\mathrm{Br}-\mathrm{Te}\) b. \(\mathrm{P}-\mathrm{H}\) e. \(\mathrm{Se}-\mathrm{S}\) c. \(\mathrm{H}-\mathrm{Cl}\)

Short Answer

Expert verified
a. C-O bond: \(\mathrm{C}^{δ+}-\mathrm{O}^{δ-}\) d. Br-Te bond: \(\mathrm{Te}^{δ+}-\mathrm{Br}^{δ-}\) b. P-H bond: Relatively non-polar e. Se-S bond: Relatively non-polar c. H-Cl bond: \(\mathrm{H}^{δ+}-\mathrm{Cl}^{δ-}\)

Step by step solution

01

Determining Electronegativity

Consult a periodic table or table of electronegativity values to find the electronegativity of each atom in each bond. a. C-O: Electronegativity of C: 2.55 Electronegativity of O: 3.44 d. Br-Te: Electronegativity of Br: 2.96 Electronegativity of Te: 2.1 b. P-H: Electronegativity of P: 2.19 Electronegativity of H: 2.20 e. Se-S: Electronegativity of Se: 2.55 Electronegativity of S: 2.58 c. H-Cl: Electronegativity of H: 2.20 Electronegativity of Cl: 3.16
02

Calculate Electronegativity Differences and Assign Polarity

Calculate the difference in electronegativity for each bond. The atom with the greater electronegativity will have a partial negative charge, while the atom with lower electronegativity will have a partial positive charge. a. C-O: Electronegativity difference: 3.44 - 2.55 = 0.89 Bond Polarity: \(\mathrm{C}^{δ+}-\mathrm{O}^{δ-}\) d. Br-Te: Electronegativity difference: 2.96 - 2.1 = 0.86 Bond Polarity: \(\mathrm{Te}^{δ+}-\mathrm{Br}^{δ-}\) b. P-H: Electronegativity difference: 2.20 - 2.19 = 0.01 Since the difference is very small, the bond between P and H is relatively non-polar. e. Se-S: Electronegativity difference: 2.58 - 2.55 = 0.03 Since the difference is very small, the bond between Se and S is relatively non-polar. c. H-Cl: Electronegativity difference: 3.16 - 2.20 = 0.96 Bond Polarity: \(\mathrm{H}^{δ+}-\mathrm{Cl}^{δ-}\) In summary: - C-O bond: \(\mathrm{C}^{δ+}-\mathrm{O}^{δ-}\) - Br-Te bond: \(\mathrm{Te}^{δ+}-\mathrm{Br}^{δ-}\) - P-H bond: Relatively non-polar - Se-S bond: Relatively non-polar - H-Cl bond: \(\mathrm{H}^{δ+}-\mathrm{Cl}^{δ-}\)

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Key Concepts

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

Electronegativity
Electronegativity is a measure of how strongly an atom attracts electrons in a chemical bond. It's a fundamental concept to understand bond polarity. Every element has its own electronegativity value.
These values can be found on periodic tables or specialized tables listing them. The higher the value, the more strongly the atom attracts shared electrons. For example, oxygen (O) has a higher electronegativity than carbon (C), which is why in a C-O bond, oxygen pulls the electrons closer to itself.
This difference in electronegativity is crucial because it helps predict whether a bond will be polar or non-polar. In general, the larger the electronegativity difference between two atoms, the more polar the bond will be, leading to a bond with one atom having a partial negative charge and the other a partial positive charge.
Partial Charges
Partial charges occur in polar bonds where there is an unequal sharing of electrons between atoms. When electrons are not shared equally, the atom with more electronegativity pulls the electrons closer, acquiring a partial negative charge (denoted as ext{δ-}), while the other atom becomes partially positive (denoted as ext{δ+}).
Take the example of the H-Cl bond: chlorine (Cl), with its higher electronegativity, pulls the electrons closer. This results in Cl being ext{δ-} and hydrogen (H) being ext{δ+}.
These partial charges are crucial for understanding molecular interactions such as hydrogen bonding, solubility, and reactivity. Partial charges in molecules influence how they interact with other molecules and are fundamental to many chemical processes in life and industry.
Molecular Structure
Molecular structure refers to the arrangement of atoms within a molecule. Understanding the structure is essential to predict how a molecule behaves and reacts.
Atoms in a molecule are arranged in a specific geometry that results from the types of bonds and the number of surrounding bonds each atom forms. For example, linear, trigonal planar, and tetrahedral are common geometries.
The bond polarity can influence molecular structure because polar bonds can lead to polar molecules, which have an overall charge difference across the molecule. This can affect the molecule’s properties such as boiling point, solubility, and biological activity. Hence, the shape and type of bonds a molecule has directly influence its role in chemical reactions.
Non-polar Bonds
Non-polar bonds occur when there is an equal sharing of electrons between the two atoms involved in the bond. This results in no partial charge across the bond, as the electron density is evenly distributed.
The P-H bond and Se-S bond are great examples of non-polar bonds in this context. Their electronegativity differences are so minimal that neither atom pulls the shared electrons more than the other.
Non-polar bonds lead to molecules that do not have charged areas, making them hydrophobic or water-repelling. They do not easily interact with polar substances and often play different roles in biological and chemical processes compared to polar molecules. Understanding non-polar bonds helps explain the behavior and interaction of substances at the molecular level.

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Most popular questions from this chapter

Which of the following statements is(are) true? Correct the false statements. a. The molecules \(\mathrm{SeS}_{3}, \mathrm{SeS}_{2}, \mathrm{PCl}_{5}, \mathrm{TeCl}_{4}, \mathrm{ICl}_{3}\), and \(\mathrm{XeCl}_{2}\) all exhibit at least one bond angle, which is approximately \(120^{\circ}\). b. The bond angle in \(\mathrm{SO}_{2}\) should be similar to the bond angle in \(\mathrm{CS}_{2}\) or \(\mathrm{SCl}_{2}\). c. Of the compounds \(\mathrm{CF}_{4}, \mathrm{KrF}_{4}\), and \(\mathrm{SeF}_{4}\), only \(\mathrm{SeF}_{4}\) exhibits an overall dipole moment (is polar). d. Central atoms in a molecule adopt a geometry of the bonded atoms and lone pairs about the central atom in order to maximize electron repulsions.

Write Lewis structures that obey the octet rule (duet rule for H) for each of the following molecules. a. \(\mathrm{H}_{2} \mathrm{CO}\) b. \(\mathrm{CO}_{2}\) c. \(\mathrm{HCN}\) Carbon is the central atom in all of these molecules.

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. \(\mathrm{F}_{2}\) e. \(\mathrm{NH}_{3}\) b. \(\mathrm{O}_{2}\) f. \(\mathrm{H}_{2} \mathrm{O}\) c. \(\mathrm{CO}\) g. HF d. \(\mathrm{CH}_{4}\)

The most common type of exception to the octet rule are compounds or ions with central atoms having more than eight electrons around them. \(\mathrm{PF}_{5}, \mathrm{SF}_{4}, \mathrm{ClF}_{3}\) and \(\mathrm{Br}_{3}^{-}\) are examples of this type of exception. Draw the Lewis structure for these compounds or ions. Which elements, when they have to, can have more than eight electrons around them? How is this rationalized?

Arrange the atoms and/or ions in the following groups in order of decreasing size. a. \(\mathrm{O}, \mathrm{O}^{-}, \mathrm{O}^{2-}\) b. \(\mathrm{Fe}^{2+}, \mathrm{Ni}^{2+}, \mathrm{Zn}^{2+}\) c. \(\mathrm{Ca}^{2+}, \mathrm{K}^{+}, \mathrm{Cl}^{-}\)

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