a. B₂ is paramagnetic because it has two unpaired electrons, one in each of its p orbitals.
b. C₂ is diamagnetic because all of its electrons are paired.
c. O₂ is paramagnetic because it has two unpaired electrons, one in each of its p* orbitals.
d. NO has an odd number of electrons and, therefore, must be paramagnetic.
e. CO is diamagnetic because all of its electrons are paired.

a. B₂ has 2 bonding pairs and 1 antibonding pair so it has a bond order of 1.
b. C₂ has 3 bonding pairs and 1 antibonding pair so it has a bond order of 2.
c. O₂ has 4 bonding pairs and 2 antibonding pairs so it has a bond order of 2.
d. NO has 4 bonding pairs and 1.5 antibonding pairs so it has a bond order of 2.5.
e. CO has 4 bonding pairs and 1 antibonding pair so it has a bond order of 3.

a. Tetrahedral--sp³
b. Linear--sp
c. Octahedral--d²sp³
d. Trigonal Bipyramidal--dsp³
e. Trigonal Planar--sp²

Because each oxygen is equivalent, their hybridizations must also be equivalent. Therefore, we must choose between sp² and sp³ hybridization. For best overlap, it makes sense to assume that the resonating lone pair is in a p-orbital so that it is properly positioned to make a p bond to the central carbon. Because we must have that p-orbital free, i.e. not hybridized, the hybridization must be sp² for both oxygens.