Electronic valence molecular orbital configuration of "O"_2^(+)?

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Would be very grateful for an explanation, i've never seen this before. cheers

1 Answer
Apr 25, 2018

(sigma_(2s))^2(sigma_(2s)^"*")^2(sigma_(2p_z))^2(pi_(2p_x))^2(pi_(2p_y))^2(pi_(2p_x)^"*")^1

Your first answer option is for "N"_2^(-).
Your second answer option is for "O"_2^(+).
Your third answer option is for "N"_2.

You should be able to draw the MO diagram for "N"_2^(-) given the information below.


You'll need the molecular orbital (MO) diagram of "O"_2. Begin with the atomic orbitals. Oxygen atom has 2s and 2p valence orbitals and 6 valence electrons:

Each oxygen contributes 6, so we distribute 12 valence electrons into the molecule to get "O"_2.

  • Two 2s orbitals combine to give a sigma_(2s) bonding and sigma_(2s)^"*" antibonding MO.
  • Two 2p_x orbitals combine to give a pi_(2p_x) bonding and pi_(2p_x)^"*" antibonding orbital. These are the same energy as the pi_(2p_y) counterparts.
  • Two 2p_y orbitals combine to give a pi_(2p_y) bonding and pi_(2p_y)^"*" antibonding orbital. These are the same energy as the pi_(2p_x) counterparts.
  • Two 2p_z orbitals combine to give a sigma_(2p_z) bonding and sigma_(2p_z)^"*" antibonding MO.

Lastly, "N"_2 would have the sigma_(2p_z) above the pi_(2p_x) and pi_(2p_y), whereas "O"_2 would have it below.

For "O"_2, you should get:

(sigma_(2s))^2(sigma_(2s)^"*")^2(sigma_(2p_z))^2(pi_(2p_x))^2(pi_(2p_y))^2(pi_(2p_x)^"*")^1(pi_(2p_y)^"*")^1

In your notation,

(s_(s))^2(s_(s)^"*")^2(s_(p))^2(p_(x))^2(p_(y))^2(p_(x)^"*")^1(p_(y)^"*")^1

If you want "O"_2^(+), take one electron out of the pi_(2p_y)^"*" orbital. For "O"_2^(+), you should therefore get:

color(blue)((sigma_(2s))^2(sigma_(2s)^"*")^2(sigma_(2p_z))^2(pi_(2p_x))^2(pi_(2p_y))^2(pi_(2p_x)^"*")^1)

In your notation,

color(blue)((s_(s))^2(s_(s)^"*")^2(s_(p))^2(p_(x))^2(p_(y))^2(p_(x)^"*")^1)