Draw the diagrams for #"NO"_2^-#, #"NO"_2^+#, and #"NO"_2#. The HOMO of #"NO"_2^-# shows it is somewhat anti-bonding. Would you expect the nonbonding electron pairs on nitrogen or oxygen to be more reactive? Discuss

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Feb 9, 2018

You've seen the molecular orbital (MO) diagram of #"CO"_2#:

Inorganic Chemistry, Miessler, Fischer, and Tarr, pg. 148

#"CO"_2# and #"NO"_2^+# are isoelectronic and thus have the same electron configuration. Thus, simply add one or two electrons into the #2b_(3u)# and #2b_(2u)# to get #"NO"_2# and #"NO"_2^-#, respectively.

Nitrogen atom has #2p# atomic orbitals lower by #"2.52 eV"#, and #2s# atomic orbitals lower by #"6.13 eV"# than with carbon atom.

That makes the HOMO (#1b_(2g), 1b_(3g)#) belongs more so to nitrogen in #"NO"_2^(+)# than to carbon in #"CO"_2#. It also means the LUMO (#2b_(3u), 2b_(2u)#) is less antibonding in #"NO"_2^(+)# than in #"CO"_2#.

The nonbonding electron pairs on oxygen in #"NO"_2^(-)# should be more reactive...

  • The HOMO on #"NO"_2^-# are the (#2b_(3u), 2b_(2u)#) MOs, which are half-filled. Those are the lone pair on nitrogen.

The #4a_g# is the LUMO then, and it belongs more to nitrogen than oxygen, as it is above nitrogen atom in energy, and nitrogen's atomic orbitals are above oxygen's in energy.

  • The #2a_g# and #2b_(1u)# nonbonding #sigma# orbitals, as well as the (#1b_(2g), 1b_(3g)#) #pi# orbitals, belong more to oxygen than to nitrogen. Those are the lone pairs on the oxygens.

The lone pair on nitrogen is in #pi# orbitals, meant for making #pi# bonds. But the lone pairs on oxygen are in #sigma# AND #pi# orbitals, meant for making #sigma# OR #pi# bonds.

Hence, upon protonation, the #"H"^(+)# goes onto the OXYGEN via the #sigma# orbitals #2a_g# and #2b_(1u)#, making #sigma# bonds.

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