(a) Construct a Lewis structure for \(\mathrm{O}_{2}\) in which each atom achieves an octet of electrons. (b) Explain why it is necessary to form a double bond in the Lewis structure. (c) The bond in \(\mathrm{O}_{2}\) is shorter than the \(\mathrm{O}-\mathrm{O}\) bond in compounds that contain an \(\mathrm{O}-\mathrm{O}\) single bond. Explain this observation.

First, let's draw the Lewis structure for an O2 molecule. Oxygen has 6 valence electrons, so in a molecule of O2, there are a total of 12 valence electrons. To achieve an octet, each atom requires 2 more electrons. \(2 x 2 = 4\) electrons will be involved in the formation of bonds. Therefore, we have 8 remaining electrons to distribute around the structure as lone pairs. We can now draw the Lewis structure as follows: 1. Write the symbols for each element: O - O 2. Place a bonding pair of electrons (a line) between the symbols: O = O 3. Distribute the remaining 8 electrons as lone pairs, making sure each atom achieves an octet: :O = O: So the Lewis structure of O2 molecule is :O=O: with double bond and the remaining electrons are shown as lone pairs on each oxygen atom.

In order to achieve an octet of electrons, each oxygen atom in the O2 molecule requires 2 more electrons. A single bond, which consists of two shared electrons, is not enough to complete the octet for both oxygen atoms. Therefore, a double bond (two pairs of shared electrons) is formed between two oxygen atoms, allowing both of them to have a complete octet.

A double bond involves the sharing of two electron pairs, which leads to increased electron density between the two oxygen atoms compared to a single bond. These additional electrons in the double bond create a stronger attraction between the oxygen nuclei in O2 than those in compounds with an O-O single bond. As a result, the bond in O2 is shorter (and stronger) than the O-O single bond in other compounds.