Is `C_2^+` paramagnetic or diamagnetic? Is `C_2` paramagnetic or diamagnetic?

`"C"_2` has a similar MO ordering to `"N"_2`.

From bottom to top, we have the following valence MOs:

1. `\mathbf(sigma_(2s))`, generated from the linear combination `2s_A + 2s_B` (head-on, in-phase)
2. `\mathbf(sigma_(2s)^"*")`, generated from the linear combination `2s_A - 2s_B` (head-on, out-of-phase)
3. `\mathbf(pi_(2p_x))`, generated from the linear combination `2p_(x,A) + 2p_(x,B)` (sidelong, in-phase)---degenerate with (4)
4. `\mathbf(pi_(2p_y))`, generated from the linear combination `2p_(y,A) + 2p_(y,B)` (sidelong, in-phase)---degenerate with (3)
5. `\mathbf(sigma_(2p_z))`, generated from the linear combination `2p_(z,A) + 2p_(z,B)` (head-on, in-phase)
6. `\mathbf(pi_(2p_x)^"*")`, generated from the linear combination `2p_(x,A) - 2p_(x,B)` (sidelong, out-of-phase)---degenerate with (7)
7. `\mathbf(pi_(2p_y)^"*")`, generated from the linear combination `2p_(y,A) - 2p_(y,B)` (sidelong, out-of-phase)---degenerate with (6)
8. `\mathbf(sigma_(2p_z)^"*")`, generated from the linear combination `2p_(z,A) - 2p_(z,B)` (head-on, out-of-phase)

You can see the compatible bonding orbital combinations here (reverse one of the orbitals' signs to achieve the corresponding antibonding orbital combination):

As a result, we have the following MO diagram (with atomic orbital energies from Inorganic Chemistry, Miessler et al., pg. 134):

Thus, with no unpaired electrons, neutral `"C"_2` is diamagnetic.

If `"C"_2 -> "C"_2^(+) + e^(-)`, then is `"C"_2^(+)` paramagnetic or diamagnetic? Paramagnetism requires unpaired electrons.