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Chapter 11: Problem 15

Which of the following molecules contain polar covalent bonds? a. phosphorus, \(P_{4}\) b. oxygen, \(\mathrm{O}_{2}\) c. ozone, \(\mathrm{O}_{3}\) d. hydrogen fluoride, HF

Short Answer

Expert verified
The molecule that contains a polar covalent bond is d. hydrogen fluoride, HF, due to the significant difference in electronegativity between hydrogen and fluorine (\(1.78\)).

Step by step solution

01

Review electronegativity values

The first step is to find the electronegativity values of the involved elements using a periodic table or reference such as the one provided by Pauling's scale. We need these values to determine the type of bond formed between atoms in the given molecules.
02

Determine bond types

We will now consider each molecule, calculate the difference in electronegativity between the atoms, and determine whether the bonds are polar covalent or not. a. Phosphorus, \(P_{4}\) Phosphorus has an electronegativity of 2.19. In this molecule, all phosphorus atoms are the same, so the difference in electronegativity is 0. Since there is no difference in electronegativity, the bonds are nonpolar covalent. b. Oxygen, \(\mathrm{O}_{2}\) Oxygen has an electronegativity of 3.44. In this molecule, both atoms are the same, so the difference in electronegativity is 0. The bonds in \(\mathrm{O}_{2}\) are nonpolar covalent. c. Ozone, \(\mathrm{O}_{3}\) Ozone is a molecule formed by three oxygen atoms. Since all the atoms are the same, the difference in electronegativity is 0. The bonds in ozone are nonpolar covalent. However, the molecule as a whole has a bent structure resulting in an overall molecular dipole, making the molecule polar. d. Hydrogen fluoride, HF Hydrogen has an electronegativity of 2.20, and fluorine has an electronegativity of 3.98. The difference in electronegativity is \(3.98 - 2.20 = 1.78\). Since there is a significant difference in electronegativity, the bond formed is polar covalent.
03

Identify molecules with polar covalent bonds

After analyzing the electronegativity values and bond types for each molecule, we can now determine which of them contain polar covalent bonds. From our analysis, only hydrogen fluoride (HF) contains a polar covalent bond. So, the answer is: d. hydrogen fluoride, HF

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Key Concepts

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

Electronegativity
Electronegativity is a key concept in chemistry that measures the ability of an atom to attract electrons in a chemical bond. Each element has a unique electronegativity value, which can be found in the periodic table, most commonly referenced on Pauling's scale.
  • Higher electronegativity means the atom attracts electrons more strongly.
  • Typically, nonmetals have higher electronegativity values than metals.
  • The difference in electronegativity between two bonded atoms is essential to determining the bond type.
For molecules, the difference in electronegativity between the atoms decides if the bond is polar covalent, nonpolar covalent, or ionic.
  • Small or zero difference implies nonpolar covalent bonds.
  • A moderate difference results in a polar covalent bond.
  • A large difference often leads to ionic bonds, where electrons are fully transferred.
Understanding electronegativity allows us to classify and predict the nature of bonds in a molecule which in turn affects its chemical properties.
Molecular Dipole
A molecular dipole occurs in a molecule where there is a distribution of electrical charge leading to positive and negative ends. This property is not solely determined by individual polar bonds but also by the molecule's shape.
Molecules can contain polar bonds but still be nonpolar overall if the molecular geometry allows the dipole moments to cancel each other out.
  • For example, carbon dioxide ( ext{CO}_2) has two polar C=O bonds, but the linear structure causes these forces to balance, resulting in a nonpolar molecule.
  • In contrast, water ( ext{H}_2 ext{O}) with its bent shape and polar O-H bonds, cannot neutralize its dipoles resulting in a molecular dipole.
In the case of ozone ( ext{O}_3), the molecule has a bent structure which does not allow the polar bonds' dipoles to cancel out, leading to an overall dipolar molecule despite each bond being nonpolar individually.
Nonpolar Covalent Bonds
Nonpolar covalent bonds occur when two atoms share a pair of electrons with equal or near-equal attraction. This usually happens between atoms of the same element or atoms with very similar electronegativity.
  • Examples include the hydrogen ( ext{H}_2), oxygen ( ext{O}_2), and nitrogen ( ext{N}_2) molecules where each bond involves the sharing of electrons equally.
  • Such uniform distribution of electron density results in no permanent dipole moment in the molecule.
It's essential to understand that while individual bonds might be nonpolar, the molecule's structure could still result in a polar molecule due to its shape and arrangement, like in the case of ozone ( ext{O}_3). Nonpolar covalent bonds are majorly important in molecular interactions, as they affect physical properties like solubility and melting/boiling points.

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Most popular questions from this chapter

On the basis of the electronegativity values given in Figure \(11.3,\) indicate whether each of the following bonds would be expected to be ionic, covalent, or polar covalent. a. \(\mathrm{H}-\mathrm{O}\) b. \(\mathrm{O}-\mathrm{O}\) c. \(\mathrm{H}-\mathrm{H}\) d. \(\mathrm{H}-\mathrm{Cl}\)

In each of the following molecules, which end of the molecule is negative relative to the other end? a. carbon monoxide, \(\mathrm{CO}\) b. iodine monobromide, IBr c. hydrogen iodide, HI

A(n) ______ chemical bond represents the equal sharing of a pair of electrons between two nuclei.

On the basis of their electron configurations, predict the formula of the simple binary ionic compounds likely to form when the following pairs of elements react with each other. a. aluminum, Al, and sulfur, S b. radium, \(\mathrm{Ra}\), and oxygen, \(\mathrm{O}\) c. calcium, \(\mathrm{Ca},\) and fluorine, \(\mathrm{F}\) d. cesium, \(\mathrm{Cs}\), and nitrogen, \(\mathrm{N}\) e. rubidium, Rb, and phosphorus, \(\mathrm{P}\)

For each of the following sets of elements arrange the elements in order of increasing electronegativity. a. \(\mathrm{K}, \mathrm{Cr}, \mathrm{Br}\) b. \(\mathrm{Mg}, \mathrm{Ra}, \mathrm{Ca}\) c. \(\mathrm{Cl}, \mathrm{Na}, \mathrm{Al}\)
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