Draw the MO diagrams of #"O"_2#, #"O"_2^(2-)#, and #"O"_2^(2+)#?

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Answer 1 · 2017-07-28T21:04:46

#"O"_2^(2+)# is more thermodynamically stable, as it has the stronger bond... triple bonds are clearly stronger than #1.5#-bonds.

You can read here if you want to know how to construct an MO diagram.

Given the MO diagram of #"O"_2#, you should be able to deduce the MO diagram of #"O"_2^(-)# and #"O"_2^(2+)#.

One trick I like to consider for bond strength is:

Each added electron in a bonding molecular orbital adds #color(red)ul(0.5)# to the bond order. That is, it strengthens the overall bond by half a bond of that type of interaction (#sigma, pi, delta, phi, . . . #).
Each added electron in an antibonding molecular orbital adds #color(red)ul(-0.5)# to the bond order. That is, it weakens the overall bond by half a bond of that type of interaction (#sigma, pi, delta, phi, . . . #).

#"O"_2#, of course, has a bond order of #2#... it is #:stackrel(..)"O"=stackrel(..)"O":#, with a double bond, so it only makes sense...

#"O"_2^(-)#, or #stackrel((-1//2))( :stackrel(...)"O")stackrel(--)("—")stackrel((-1//2))(stackrel(...)"O": )#, simply has one more electron in one of the #pi_(2px//y)^"*"# antibonding molecular orbitals. So, #"O"_2^(-)# has a bond order of #2 + (1 xx -0.5) = bb(1.5)#.
#"O"_2^(2+)#, or #stackrel((+))( :"O")-=stackrel((+))("O": )#, simply has two less electrons in the #pi_(2px//y)^"*"# antibonding molecular orbitals. Thus, it has a bond order of #2 - (2 xx -0.5) = bb(3)#, isoelectronic with #"N"_2#.

#"O"_2^(2+)# is thus more thermodynamically stable, as it has the stronger bond... triple bonds are clearly stronger than #1.5#-bonds.