Order by bond length? #"NO"#, #"NO"^(+)#, #"NO"^(-)#

1 Answer
Sep 17, 2017

The bond strength increases going from #"NO"^(-)# to #"NO"^(+)#, and the bond length consequently shortens going from #"NO"^(-)# to #"NO"^(+)#.

#r_("N"-"O")^("NO"^(-)) > r_("N"-"O")^("NO") > r_("N"-"O")^("NO"^(+))#


To start with, #"O"_2# is (roughly) isoelectronic with #"NO"^(-)##""^(color(red)("[*]"))#, and #"O"_2# has two #pi^"*"# antibonding electrons, as you should know from its MO diagram from your textbook. Thus, so does #"NO"^(-)#.

#color(red)(["*"])# - except for the orbital ordering below the #pi^"*"# (the #2b_1,2b_2#) of these species

To prove this, here is the MO diagram of #"NO"# (Miessler et al., Answer Key):

(The original was this; I added the orbital depictions and symmetry labels.)

Quick overview of what the labels correspond to what MOs:

  • #1a_1# is the #sigma_(2s)# bonding MO.
  • #2a_1# is the #sigma_(2s)^"*"# antibonding MO.
  • #1b_1# is the #pi_(2p_x)# bonding MO.
  • #1b_2# is the #pi_(2p_y)# bonding MO.
  • #3a_1# is the #sigma_(2p_z)# bonding MO, but it's relatively nonbonding with respect to oxygen.
  • #2b_1# is the #pi_(2p_x)^"*"# antibonding MO.
  • #2b_2# is the #pi_(2p_y)^"*"# antibonding MO.
  • #4a_1# is the #sigma_(2p_z)^"*"# antibonding MO.

Note that for #"O"_2#, the #1b_1,1b_2# and #3a_1# orbitals are switched in energy. From this MO diagram, we can see that:

  • #"NO"^(+)# has #bb0# #pi^"*"# antibonding electrons.
  • #"NO"# has #bb(1)# #pi^"*"# antibonding electron.
  • #"NO"^(-)# has #bb(2)# #pi^"*"# antibonding electrons.

As the number of antibonding electrons increases, the #"N"-"O"# bond weakens, having acquired antibonding character (which as the name suggests, goes against making a bond).

Thus, the bond strength increases going from #"NO"^(-)# to #"NO"^(+)#, and the bond length consequently shortens going from #"NO"^(-)# to #"NO"^(+)#.