Question

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

Answer 1

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

`"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.