To find the bond order of OH- and OH+, you need to know the Lewis structures of these molecules or ions. Here's how you can determine the bond order step by step:
1. Draw the Lewis structure: Start by determining the arrangement of atoms and the number of valence electrons for each molecule or ion. For OH-, the Lewis structure can be represented as H-O- with three lone pairs on the oxygen atom, while OH+ can be represented as H-O with two lone pairs on the oxygen atom.
2. Count the total number of valence electrons: Add up the valence electrons for all the atoms. For OH-, you have 1 (from hydrogen) + 6 (from oxygen) + 1 (from the negative charge) = 8 valence electrons. For OH+, you have 1 (from hydrogen) + 6 (from oxygen) - 1 (from the positive charge) = 6 valence electrons.
3. Determine the number of bonding and antibonding electrons: Look at the Lewis structure and identify the number of shared electron pairs (bonding electrons) and the number of lone pairs on the central atom (antibonding electrons). In both OH- and OH+, there is only one bond between the oxygen atom and the hydrogen atom.
4. Calculate the bond order: The bond order is the difference between the number of bonding electrons and the number of antibonding electrons, divided by 2. So, for OH-, the bond order would be (2 - 6) / 2 = -2 / 2 = -1. For OH+, the bond order would be (2 - 4) / 2 = -2 / 2 = -1.
In this case, both OH- and OH+ have bond orders of -1. Note that negative bond orders represent unstable molecules, indicating that they are more likely to dissociate rather than exist as stable entities.