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

#(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)#