According to Molecular Orbital Theory (MOT), which species has the highest bond order?

To determine which species has the highest bond order according to Molecular Orbital Theory, it's essential to understand how bond order is calculated. The bond order can be calculated using the formula:

[ \text{Bond Order} = \frac{(\text{Number of bonding electrons} - \text{Number of antibonding electrons})}{2} ]

Among the given options, let's analyze the relevant molecular orbitals for each species:

1. O2+: This ion has one less electron than O2. The bond order for O2 can be calculated as follows. O2 has 10 bonding electrons and 6 antibonding electrons, giving a bond order of 2. For O2+, the removal of one electron (from a bonding orbital since it's the highest energy orbital) results in a bond order of 1.5.

2. N2: In N2, both electrons fill the bonding molecular orbitals effectively. It has 10 bonding electrons and no antibonding electrons, resulting in a bond order of 3.

3. NO: The nitrogen monoxide molecule has 11 valence electrons. The bond order can be calculated similarly, leading to a bond order of about 2.