What is the molecular orbital diagram for #C_2^-#?

The problem provides you with the MO diagram for the #"C"_2# molecule, so all you really have to do here is add an electron to that diagram.

You need to add an electron and not remove one because of the overall negative charge that exists on the molecule. As you know, a neutral carbon atom has a total of #6# electrons.

This, of course, implies that a #"C"_2# molecule has a total of

#2 xx "6 e"^(-) = "12 e"^(-)#

It thus follows that the #"C"_2^(-)# species will have

#"12 e"^(-) + "1 e"^(-) = "13 e"^(-)#

So, where would this extra electron go in terms of molecular orbitals?

It will be added to the lowest energy unoccupied molecular orbital, or lowest unoccupied molecular orbital, #"LUMO"#, that follows that highest energy occupied molecular orbital, or highest occupied molecular orbital, #"HOMO"#.

As you can see in the diagram, the two #2p_(pi)# orbitals, let's say #2p_(pix)# and #2p_(piy)#, are the highest energy occupied molecular orbitals.

The lowest energy unoccupied molecular orbital is #2p_(sigma)#, so that is where the extra electron will be added.

The electron configuration of the neutral #"C"_2# molecule is -- I'll use the notation given to you in the diagram

#"C"_ 2: (1s_ (sigma))^2 (1s_ (sigma)^"*")^2 (2s_ (sigma))^2 (2s_ (sigma)^"*")^2 (2p_ (pi))^4#

The electron configuration of the #"C"_2^(-)# ion will be

#"C"_ 2^(-): (1s_ (sigma))^2 (1s_ (sigma)^"*")^2 (2s_ (sigma))^2 (2s_ (sigma)^"*")^2 (2p_ (pi))^4 color(red)((2p_ (sigma))^1#

Notice that because the extra electron is added to a bonding MO, the bond order of the #"C"_2^(-)# will actually be higher than the bond order of the neutral #"C"_2# molecule.

#"B.O"_ ("C"_ 2^(-)) > "B.O"_ ("C"_ 2)#

I won't show the calculation here because I'm not sure you're familiar with bond orders yet

Also, an unpaired electron will make the #"C"_2^(-)# ion paramagnetic, i.e. it is attracted by an externally applied magnetic field.

On the other hand, the neutral #"C"_2# molecule has no unpaired electrons, so it is diamagnetic, i.e. it is not attracted by an externally applied magnetic field.