Determine the bond order of the #"C"-"O"# bond in the following molecules, and then arrange the #"C"-"O"# bond lengths in increasing order? #"CO"#, #"CO"^(+)#, #"CO"^(2+)#, #"CO"_2#, #"CO"_3^(2-)#

Bond length increases from left to right on your list, i.e.

#r_(CO) < r_(CO^(+)) < [r_(CO^(2+)) = r_(CO_2)] < r_(CO_3^(2-))#

• #"CO"#: #3#
• #"CO"^(+)#: #2.5#
• #"CO"^(2+)#: #2#
• #"CO"_2#: #2#
• #"CO"_3^(2-)#: #1.bar(33)#

In order to determine this, we should reference an MO diagram.

We can see that the highest occupied molecular orbital (HOMO) is fully occupied, but the next-highest MOs are the #pi^"*"# antibonding lowest unoccupied molecular orbitals (LUMOs). Keep this in mind.

CO MOLECULE

#"CO"# by default has a triple bond, so by chemical intuition, its bond order is #3#. This can be verified by the usual equation:

#"BO" = 1/2("Bonding - Antibonding")#

#= 1/2([stackrel(2sigma)overbrace(2)+stackrel(1pi)overbrace(2(2)) + stackrel(3sigma)overbrace(2)] - [stackrel(2sigma^"*")overbrace(2)]) = 3#

CO+ CATION

#"CO"^(+)# has one electron removed from a bonding MO, so its bond order decreases by #1/2# to become #2.5#. Use the example above to verify via the equation.

CO#""^(bb(2+))# DICATION

#"CO"^(2+)# has an additional bonding electron removed relative to #"CO"^(+)#, so it has a bond order of #2#.

CO2 MOLECULE

#"CO"_2# looks like this:

#:stackrel(..)("O")="C"=stackrel(..)"O":#

So, its double bonds suggest a bond order of #2#.

CARBONATE ANION

By considering its resonance delocalization, the bond order on this molecular ion can be determined.

We should notice that for the #pi# bond order, there will be #2# #pi# electrons delocalized onto three bonds, making each bond on average having #2/3# of a #pi# electron, thereby giving a #pi# bond order of #1/3#.

Since perfect single bonds have a bond order of #1#, the total bond order on #"CO"_3^(2-)# is #bb(1.bar(33))#.

OVERALL

Overall, we have the bond orders:

• #"CO"#: #3#
• #"CO"^(+)#: #2.5#
• #"CO"^(2+)#: #2#
• #"CO"_2#: #2#
• #"CO"_3^(2-)#: #1.bar(33)#

You can choose how to order #"CO"^(2+)# and #"CO"_2# however you like, but the rest is already in order. The shortest bond corresponds to the strongest bond.

Hence, bond length increases down this list.