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Chapter 22: Problem 38

Use Lewis structures and other information to explain the observation that (a) the oxygen-to-oxygen bond lengths in \(\mathrm{O}_{2}, \mathrm{O}_{3}\) and \(\mathrm{H}_{2} \mathrm{O}_{2}\) are \(121,128,\) and \(148 \mathrm{pm},\) respectively. (b) the oxygen-to-oxygen bond length of \(\mathrm{O}_{2}\) is \(121 \mathrm{pm}\) and for \(\mathrm{O}_{2}^{+}\) is \(112 \mathrm{pm}\). Why is the bond length for \(\mathrm{O}_{2}^{+}\) so much shorter than for \(\mathrm{O}_{2} ?\)

Short Answer

Expert verified
The bond lengths in \(\mathrm{O}_{2}, \(\mathrm{O}_{3}\) and \(\mathrm{H}_{2} \mathrm{O}_{2}\) are defined by the bond multiplicity, with double bonds being shorter than single bonds. The ionization of \(\mathrm{O}_{2}\) to \(\mathrm{O}_{2}^{+}\) results in an increase in the bond order and hence a decrease in bond length.

Step by step solution

01

Understanding Molecular Structures

Let's begin by understanding the molecular structures. \(\mathrm{O}_{2}\) is a diatomic molecule and its Lewis structure shows that it has a double bond. \(\mathrm{O}_{3}\) is a triatomic molecule and its Lewis structure shows that it has one double bond and one single bond, therefore has an average bond order of 1.5. \(\mathrm{H}_{2} \mathrm{O}_{2}\) on the other hand is a four atom molecule, and in its most stable conformation, it has a single bond between the two Oxygen atoms.
02

Comparing Bond Lengths

Bond length is inversely proportional to bond strength and directly proportional to bond multiplicity. Double bonds are generally shorter and stronger than single bonds. Hence, in \(\mathrm{O}_{2}\) the bond length is shorter (121 pm) than in \(\mathrm{H}_{2} \mathrm{O}_{2}\) (148 pm) where there is a single bond between the Oxygen atoms. In \(\mathrm{O}_{3}\), with an average bond order of 1.5, the bond length (128 pm) is intermediate between the two extreme cases.
03

Understanding Ionization

Now, let's analyze the effect of ionization on bond length. When a molecule is ionized, it loses an electron, often resulting in a higher bond order and thus a shorter bond length. This is the case for \(\mathrm{O}_{2}^{+}\). After losing an electron, \(\mathrm{O}_{2}^{+}\) will have a bond order slightly more than double bond, hence resulting in a shorter bond length (112 pm) as compared to \(\mathrm{O}_{2}\) (121 pm).

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