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-)#

1 Answer
Truong-Son N.
Apr 30, 2017

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.

Inorganic Chemistry, Miessler et al., Ch. 5.3, Fig. 5.14

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.

![http://upload.wikimedia.org/](useruploads.socratic.org)

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.

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