Question

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

Answer 1

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"^(+))`

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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"^(+)`.