Chapter 9: Problem 40

Using the periodic table only, arrange the members of each of the following sets in order of increasing bond length: (a) \(\mathrm{H}-\mathrm{F}, \mathrm{H}-\mathrm{I}, \mathrm{H}-\mathrm{Cl}\) (b) \(\mathrm{C}-\mathrm{S}, \mathrm{C}=\mathrm{O}, \mathrm{C}-\mathrm{O}\) (c) \(\mathrm{N}-\mathrm{H}, \mathrm{N}-\mathrm{S}, \mathrm{N}-\mathrm{O}\)

Short Answer

Expert verified

(a) \(\text{H}-\text{F} < \text{H}-\text{Cl} < \text{H}-\text{I}\) (b) \(\text{C}=\text{O} < \text{C}-\text{O} < \text{C}-\text{S}\) (c) \(\text{N}-\text{H} < \text{N}-\text{O} < \text{N}-\text{S}\)

Step by step solution

Understand the concept of bond length

Bond length is the average distance between the nuclei of two bonded atoms. Generally, bond length increases with the size of the atom and decreases with the bond order (single, double, triple bond).

Gathering periodic table information

Determine the atomic radii of the relevant atoms using the periodic table. This will help in comparing the bond lengths.

Analyzing set (a)

For \(\text{H}-\text{F}, \text{H}-\text{I}, \text{H}-\text{Cl}\), the atoms bonded with hydrogen are F (Fluorine), I (Iodine), and Cl (Chlorine). Since bond length increases with the size of the atom, and I > Cl > F in size, the order is: \(\text{H}-\text{F} < \text{H}-\text{Cl} < \text{H}-\text{I}\).

Analyzing set (b)

For \(\text{C}-\text{S}, \text{C}=\text{O}, \text{C}-\text{O}\), note that double bonds (like \(\text{C}=\text{O}\)) are shorter than single bonds. Comparing single bonds, Sulfur (S) is larger than Oxygen (O). So the order is: \(\text{C}=\text{O} < \text{C}-\text{O} < \text{C}-\text{S}\).

Analyzing set (c)

For \(\text{N}-\text{H}, \text{N}-\text{S}, \text{N}-\text{O}\), compare the atomic sizes of H (Hydrogen), S (Sulfur), and O (Oxygen). Sulfur is larger than Oxygen and Hydrogen is the smallest. Thus, the order is: \(\text{N}-\text{H} < \text{N}-\text{O} < \text{N}-\text{S}\).

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Key Concepts

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

Periodic Table

The periodic table is a fundamental tool in chemistry. It organizes elements based on their atomic number, electron configuration, and recurring properties. The layout is such that elements with similar characteristics are grouped together in columns, known as groups or families, and rows, known as periods. This organization helps us predict the behavior and properties of elements. For example:

Elements in the same group usually have similar chemical properties.
As you move from left to right across a period, the atomic number increases, and elements become less metallic.
As you move down a group, elements have more electron shells, increasing their size and changing their reactivity.

Understanding the periodic table allows us to compare the characteristics of different elements, such as their atomic radii and bonding tendencies, which are crucial for analyzing bond lengths.

Atomic Radii

Atomic radii refer to the size of an atom's constituent particles, primarily the distance from the nucleus to the outermost stable electron orbit. In the periodic table:

Atomic radii decrease across a period (left to right) due to the increasing nuclear charge, pulling electrons closer to the nucleus.
Atomic radii increase down a group as new electron shells are added, expanding the atomic size.

When discussing bond lengths, we see that the size of the atoms involved critically influences the distance between their nuclei when bonded. For example, in the set \(\text{H}-\text{F}, \text{H}-\text{I}, \text{H}-\text{Cl}\), Iodine (\text{I}) has the largest atomic radius, followed by Chlorine (\text{Cl}), and then Fluorine (\text{F}). Consequently, the corresponding bond lengths increase in the order \(\text{H}-\text{F} < \text{H}-\text{Cl} < \text{H}-\text{I}\).

Bond Order

Bond order is the number of chemical bonds between a pair of atoms. It indicates the stability of a bond: higher bond orders correspond to stronger and shorter bonds.

A single bond (bond order of 1) is longer and weaker than a double bond (bond order of 2).
A double bond is longer than a triple bond, which has a bond order of 3.
Higher bond orders result in shorter bond lengths due to the increased sharing of electrons, which pulls the atoms closer together.

For example, in the set \(\text{C}-\text{S}, \text{C}=\text{O}, \text{C}-\text{O}\), the bond order of \(\text{C}=\text{O}\) is 2 (a double bond) and thus shorter than both \(\text{C}-\text{S}\) and \(\text{C}-\text{O}\), which have bond orders of 1. Therefore, the order of increasing bond length is \(\text{C}=\text{O} < \text{C}-\text{O} < \text{C}-\text{S}\).

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Short Answer

Step by step solution

Understand the concept of bond length

Gathering periodic table information

Analyzing set (a)

Analyzing set (b)

Analyzing set (c)

Key Concepts

Periodic Table

Atomic Radii

Bond Order

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