Question

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

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

Answer 1

`(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.

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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)`