(a) Construct a Lewis structure for \(\mathrm{O}_{2}\) in which each atom achieves an octet of electrons. (b) How many bonding electrons are in the structure? (c) Would you expect the \(\mathrm{O}-\mathrm{O}\) bond in \(\mathrm{O}_{2}\) to be shorter or longer than the \(\mathrm{O}-\mathrm{O}\) bond in compounds that contain an \(\mathrm{O}-\mathrm{O}\) single bond? Explain.

To construct the Lewis structure, we will follow these steps: 1. Determine the total number of valence electrons. Oxygen has 4 electron pairs (8 valence electrons). Since there are two oxygen atoms, the total number of valence electrons is 16. 2. Place the two oxygen atoms next to each other and join them with a single bond, using 2 of the 16 valence electrons. 3. Distribute the remaining 14 valence electrons around the oxygen atoms as lone pairs in order to achieve an octet of electrons for each atom: - Atom 1: (already has 2 shared electrons) + 6 lone electrons = 8 - Atom 2: (already has 2 shared electrons) + 6 lone electrons = 8 4. If an octet is not achieved, connect the atoms with more bonds. The Lewis structure for \(\mathrm{O}_{2}\) is: O = O / \ (o) (o)

In the given Lewis structure, there are two bonding electron pairs (equivalent to a double bond) between the oxygen atoms. Therefore, there are 4 bonding electrons in the structure.

In general, multiple bonds between atoms (double and triple bonds) are shorter than single bonds because the increased bonding forces pull the atoms closer together. Since \(\mathrm{O}_{2}\) has a double bond between the oxygen atoms, we would expect the \(\mathrm{O}-\mathrm{O}\) bond in \(\mathrm{O}_{2}\) to be shorter than the \(\mathrm{O}-\mathrm{O}\) bond in compounds containing an \(\mathrm{O}-\mathrm{O}\) single bond.