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Response Details:
The first statement is true. Decreasing bond length is a sign of stronger bond energy/order. Think of it as more of a tighter bond.
Second statement is true. But note that this suggests a triplet ground state for molecular oxygen. (Two unpaired electrons to fill the two degenerate pi-star orbitals. This suggests one electron in each orbital, half-filling them. And by Hund's Rule, both would point up. Also recall the Pauli's Exclusion principle forbids those two electrons from occupying the same orbital; the half-fill the two degenerate pi-star)
So for the third statement, you really have to build the MO digagram. Fill the number of electrons. You know that O2 has two electrons in the pi-star. Take this as your reference. O2+ has only one electron in the pi-star (an antibonding orbital). So the bond order of O2+ is: 0.5*(e- in bonding orbital- electrons in antibonding orbitals) = 0.5*(8-3) = 5/2. Note that the bond order of O2 is 0.5*(8-4) = 2. That suggests that O2+ has a stronger bond energy. In fact, since O2- and O2 2- have more electrons in antibonding orbitals than even O2, you can predict the bond energies that O2 2- has the lowest, then O2-, O2, and O2+. (PM me if you are having trouble with this)
Whoops, I solved the next statement, which is true.
The number of unpaired electrons in O2 2- is 0. It has an additional two electrons to fill the antibonding pi orbtials, relative to O2. But O2 already has two electrons there. The result is that four electrons are used to populate the two orbitals. At the ground state, these electrons exactly go into each orbital as pairs (each pair has two electrons with opposite spin to satisfy Pauli; recall earlier that if only two electrons were used to fill the two orbitals, the electrons could singly occupy each orbital and would likely have the same spin. This may confuse you so PM if you have more questions)
The last statement is true.
Hope that helps.
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